GATE CLOUD
ELECTROMAGNETICS
GATE CLOUD
ELECTROMAGNETICS
R. K. Kanodia Ashish Murolia
JHUNJHUNUWALA JAIPUR
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GATE CLOUD ELECTROMAGNETICS, 1e R. K. Kanodia, Ashish Murolia CC1015 Copyright by Jhunjhunuwala ISBN 978-8192-34838-4
Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither Jhunjhunuwala nor its author guarantee the accuracy or completeness of any information herein, and Jhunjhunuwala nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Jhunjhunuwala and its author are supplying information but are not attempting to render engineering or other professional services.
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PREFACE
GATE CLOUD caters a versatile collection of Multiple Choice Questions to the students who are preparing for GATE (Gratitude Aptitude Test in Engineering) examination. This book contains over 1200 multiple choice solved problems for the subject of Electromagnetics, which has a significant weightage in the GATE examination of Electronics and Communication Engineering. The GATE examination is based on multiple choice problems which are tricky, conceptual and tests the basic understanding of the subject. So, the problems included in the book are designed to be as exam-like as possible. The solutions are presented using step by step methodology which enhance your problem solving skills. The book is categorized into ten chapters covering all the topics of syllabus of the examination. Each chapter contains : Ÿ Exercise 1 : Level 1 Ÿ Exercise 2 : Level 2 Ÿ Exercise 3 : Mixed Questions Taken form Previous Examinations of GATE & IES. Ÿ Detailed Solutions to Exercise 1, 2 and 3. Although we have put a vigorous effort in preparing this book, some errors may have crept in. We shall appreciate and greatly acknowledge the comments, criticism and suggestion from the users of this book which leads to some improvement. You may write to us at
[email protected] and
[email protected]. Wish you all the success in conquering GATE. Authors
SYLLABUS
GATE ELECTRONICS & COMMUNICATION ENGINEERING Networks: Elements of vector calculus: divergence and curl; Gauss’ and Stokes’ theorems, Maxwell’s equations: differential and integral forms. Wave equation, Poynting vector. Plane waves: propagation through various media; reflection and refraction; phase and group velocity; skin depth. Transmission lines: characteristic impedance; impedance transformation; Smith chart; impedance matching; S parameters, pulse excitation. Waveguides: modes in rectangular waveguides; boundary conditions; cut-off frequencies; dispersion relations. Basics of propagation in dielectric waveguide and optical fibers. Basics of Antennas: Dipole antennas; radiation pattern; antenna gain.
IES ELECTRONICS & TELECOMMUNICATION ENGINEERING Electromagnetic Theory: Analysis of electrostatic and magnetostatic fields; Laplace's and Poisson's equations; Boundary value problems and their solutions; Maxwell's equations; application to wave propagation in bounded and unbounded media; Transmission lines : basic theory, standing waves, matching applications, microstrip lines; Basics of wave guides and resonators; Elements of antenna theory.
IES ELECTRICAL ENGINEERING Electromagnetic Theory: Electric and magnetic fields. Gauss's Law and Amperes Law. Fields in dielectrics, conductors and magnetic materials. Maxwell's equations. Time varying fields. Plane-Wave propagating in dielectric and conducting media. Transmission lines.
CONTENTS
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CHAPTER 1 VECTOR ANALYSIS
2
Vector Analysis
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Chap 1
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EXERCISE 1.1
The field lines of vector B = ra q is
MCQ 1.1.2
Given the two vectors M = 5ax - 2ay + 4az and N =- 8ax - 7ay + 2az . The unit vector in the direction of (M -N ) will be (A) 0.82ax + 0.36ay - 0.14az (B) 0.92ax - 0.36ay + 0.41az
ww
w. g
at e
he
lp.
co
m
MCQ 1.1.1
(C) 0.92ax + 0.36ay + 0.14az MCQ 1.1.3
A vector field is specified as G = 4xyax + 2 ^2 + x2h ay + 3z2 az . The unit vector in the direction of G at the point (- 2, 1, 3) will be (B) - 0.26ax + 0.39ay + 0.88az (A) 0.26ax + 0.39ay + 0.88az (C) 0.36ax - 0.29ay + 0.24az
MCQ 1.1.4
(D) - 0.92ax - 0.36ay - 0.14az
(D) 0.88ax - 0.29ay + 0.36az
Consider three nonzero vectors A, B and C . Which of the following is not a correct
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
For View Only relation between them? (A) A # A = 0 (C) (A # B) # C = A # ^B # C h MCQ 1.1.5
3
Shop Online at www.nodia.co.in (B) A # ^B + C h = ^A # B h + ^A # C h (D) (B # A) =-^A # B h
The tips of three vectors A, B and C drawn from a point define a plane. A : ^B # C h equals to (A) +1 (B) - 1 (C) zero
m
(D) can’t be determined as A, B and C are not given The component of vector A along vector B is (B) A : B (A) A : B B A (C) A : B (D) A : B AB
MCQ 1.1.7
The vector fields are defined as A = a r + 2af + 3az and B = aa r + baf - 6az . If the fields A and B are parallel then the value of a and b are respectively. (A) - 2 , - 2 (B) - 2 , - 4 (C) - 4 , - 2
(D) - 2 , - 1
Consider the vectors A = 4ax + 2kay + kaz and B = ax + 4ay - 4az . For what value of k the two vectors A and B will be orthogonal ? (A) 0 (B) + 1
ww w. ga te
MCQ 1.1.8
he
lp.
co
MCQ 1.1.6
(C) - 2
(D) - 1
Common Data for Question 9 - 10 : In a cubical region ( x < 2 , y < 2 , z < 2 ) a vector field is defined as E = 9zy2 cos (2x) ax + 8zy sin (2x) ay + 2y2 sin (2x) az .
MCQ 1.1.9
The vector field component Ex will be zero in the plane (A) z = 0 (B) y = 0 (C) x = p/4
MCQ 1.1.10
In the plane y - 4z = 0 , the vector field components Ey and Ez are related as (B) Ey + Ez = 0 (A) 2Ey = Ez (C) Ey = 2Ez
MCQ 1.1.11
(D) All of the above
(D) Ey = Ez
The plane surface on which the vector field E will be zero is (A) x = 0 (B) y = 0
(C) z = 0 (D) none of the above GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
4
Vector Analysis
Chap 1
For View Only Shop Online at www.nodia.co.in MCQ 1.1.12 Distance between the points P (x = 2, y = 3, z =- 1) and Q (r = 4, f =- 50, z = 2) is (A) 3.74 (B) 4.47 (C) 6.78 MCQ 1.1.13
(D) 8.76
The uniform vector field A = ay can also be expressed as (A) sin fa r + sin faf (B) cos fa r + sin faf
m
(C) sin q cos far + cos q cos fa q + sin faf (D) sin q cos far + cos q cos fa q - sin faf
The vector filed F = 12ax can be expressed in spherical coordinates at the point (x = 3 , y = 2 , z =- 1) as (A) 8ar - 2a q + 5af (B) 8ar - 2.2a q - 5.5af
co
MCQ 1.1.14
MCQ 1.1.15
The angle formed between A =- 5a r + 10af + 3az and surface z = 5 is (A) 10c (B) 15c
at e
The component of vector A =- 4a r - 20af + 4az parallel to the line x = 6 , z =- 2 at the point P (3, 90c, 2)is (A) - 4a r + 2af (B) ax - ay (C) - 2ay
(D) - 2az
In the cartesian co-ordinate system the co-ordinates of a point P is (a, b, c).Now consider the whole cartesian system is being rotated by 145c about an axis from the origin through the point (1, 1, 1) such that the rotation is clockwise when looking down the axis towards the origin. What will be the co-ordinates of the point P in the transformed cartesian system ? (A) (a/2, b/2, c/2) (B) (- a , - b , - c )
ww
w. g
MCQ 1.1.17
(D) 75c
he
(C) 45c MCQ 1.1.16
(D) 8ar + 2.2a q + 5.5af
lp.
(C) - 8ar + 2.2a q + 5.5af
(C) (c , b, a ) MCQ 1.1.18
Consider R be the position vector of a point P (x, y, z) in cartesian co-ordinate system then gradR equals to (A) 1 (B) 4R R (C) R R
MCQ 1.1.19
(D) (c , a , b)
(D) R 2R
Given the vector filed A = y2 ax + (2xy + x2 + z2) ay + (4x + 2yz) az . The divergence of the vector field is (A) 2 (x + y) (B) x2 + y2 + z2 + 6x + 2y (C) 2y (x + z)
(D) 0
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
5
For View Only Shop Online at www.nodia.co.in MCQ 1.1.20 A vector field F is defined as F = r sin fa r + 2r2 zaf + z cos faz . The 4# F at point P ^1, p/2, 2h equals to (A) a r + 6az (B) - a r + 2az (C) 3a r + 6az MCQ 1.1.21
(D) - 5a r + 6az
Which one of the following vector function has divergence and curl both zero ? (B) B = xyax - yzaz (A) A = 2xzax - yzay - z2 az (C) C = xyax - xzay - yzaz
(D) D = yzax + xzay + xyaz
The curl of the unit vectors a r , af and az in cylindrical co-ordinate system is listed below. Which of them is correct ? ar af az ar af az 1 1 (A) 0 0 (B) 0 af a a r z r r 1 (C) 0 (D) 0 ra f a r ra f r ar
MCQ 1.1.23
In a certain region consider f and g are the two scalar fields where as A is a vector field. Which of the following is not a correct relation ? (A) d : (fA) = f ^d : Ah - A : ^df h
lp.
co
m
MCQ 1.1.22
MCQ 1.1.24
ww w. ga te
he
(B) d # ^ fAh = f ^d # Ah - A # ^df h f gdf - f dg (C) d c m = g g2 g (d # A) - A # (dg) (D) d # c A m = g g2 Laplacian of the scalar field f = 2rz sin f + 4z2 cos2 f + 4r2 at the point P (3, p/2, 6) is (A) 16 (B) 0 (C) 46
MCQ 1.1.25
A conservative field M is given by M = (z cos (xz) + y) ax + 5kxay + x cos (xz) az . The value of k will be (A) 1 (B) 0 (C) - 1
MCQ 1.1.26
(D) 1/2
A scalar field g = ^1 + 5k h x2 y + xyz will be harmonic at all the points for the value of k equals to (A) 1/2 (B) 0 (C) - 1/2
MCQ 1.1.27
(D) 40
(D) can’t be determined
Curl of the gradient of any scalar field is (A) zero everywhere (B) zero at origin only (C) zero at infinity only
(D) it does not exist
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
6
Vector Analysis
Chap 1
For View Only Shop Online at www.nodia.co.in MCQ 1.1.28 A vector field is given as A = (x + 4z) ax + (3x - 7z) ay + (4x + 3y - cz) az . The value of c for which A will be solenoidal is (A) 1 (B) - 1 (C) 0 MCQ 1.1.29
(D) 3
For a vector function A = (4x + k1 z) ax + (k2 x - 5z) ay + (4x - k 3 y + 2z) az to be irrotational value of k1 , k2 and k 3 will be respectively. (A) - 5 , 0, 4 (B) 0, 4, 5
The unit vector normal to the plane 2x + 3y + 6z = 7 is (B) 1 ^4ax + 4ay + 6az h (A) 1 ^ax + 2ay + 3az h 14 24 1 1 (C) (D) ^ax + 2ay + 3az h ^2ax + 4ay + 6az h 14 58 Consider C is a certain closed path and dl is the differential displacement along the
MCQ 1.1.31
# dl C
is
he
path then the contour integral (A) zero
lp.
co
MCQ 1.1.30
(D) 1, 4, 3
m
(C) 4, 0, 5
(B) 1 (C) - 1
Consider C is any closed path and U is a scalar field. So, the contour integral
# ^dU h : dl C
(A) 1 (B) - 1 (C) zero
is
w. g
MCQ 1.1.32
at e
(D) can’t be determined as C is not defined.
MCQ 1.1.33
ww
(D) Can’t be determined as C and U is not given A vector field is defined as A = 3yzax + z2 xay + 2xyaz .The surface integral of the field over a closed surface S is (A) 1 (B) 5
(C) zero
(D) can’t be determined as surface S is not given ***********
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
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EXERCISE 1.2
m
If the edge of a cube is 3 units then the angle formed between it’s body diagonals will be (A) 70.53c (B) 53.70c (C) 66.21c
(D) 61c
lp.
Common Data for Question 2 - 4 :
co
MCQ 1.2.1
7
Consider a triangle ABC , whose vertex A, B and C are located at the points (- 4, 2, 5), (16, 20, - 3) and (- 14, 10, 15) respectively. The unit vector perpendicular to the plane of the triangle is (A) 0.61ax + 0.42ay - 0.37az (B) 0.66ax - 0.38ay + 0.65az
he
MCQ 1.2.2
ww w. ga te
(C) - 0.54ax + 0.11ay - 0.19az (D) 0.43ax - 0.21ay + 0.11az MCQ 1.2.3
The unit vector in the plane of the triangle which is perpendicular to AC is (A) - 0.55ax - 0.832ay + 0.077az (B) 0.23ax - 0.11ay - 0.43az
(C) - 0.51ax + 0.41ay + 0.76az (D) 0.49ax - 0.23ay - 0.44az MCQ 1.2.4
The unit vector in the plane of the triangle which bisects the interior angle at A is (A) 0.11ax - 0.81ay + 0.44az (B) 0.21ay - 0.41ax + 0.52az
(C) 0.23az + 0.12ax + 0.11ay (D) 0.37az + 0.92ay + 0.17ax MCQ 1.2.5
xa + 2yay A vector field F = x2 at the point P (r = 2 , f = p/4 , z = 0.1) is ^x + y2h (B) - 2ax - 3ay (A) 2ax + 3ay (C) 0.5a r
(D) - 0.5a r
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
8
Vector Analysis
Chap 1
For View Only Shop Online at www.nodia.co.in 3 MCQ 1.2.6 The component of vector A = rar - r sin q cos fa q + r4 af tangential to the spherical surface r = 20 at point P (20, 150c, 330c) is (A) 0.043a q + 100af (B) = 0.43a q + 10af 2
(C) 4.3a q + 100af
(D) - 0.043a q - 10af
Common Data for Question 7 - 8 : The vector component of A, that is tangent to the cone q = 150c is (B) - 12ar - 8a q (A) - 12af + 9ar
co
MCQ 1.2.7
m
A vector field has the value A =- 12ar - 4a q + 9af at the point P (9, 150c, 45c).
(D) - 12ar + 9af
(C) - 8a q + 9af
The unit vector that is perpendicular to A and tangent to the cone q = 135c is a (B) 1 ^4ar + 3afh (A) 1 aa q + f k 5 3 10 (C) 1 ^3ax + 4afh (D) 1 (9ar + 2af) 5 85
MCQ 1.2.9
Consider R is a separation vector from a fixed point (a , b, c ) to a varying point (x, y, z) in the Cartesian coordinate system. The value of grad ^ R1 h equals to
he
lp.
MCQ 1.2.8
(B) - R6 R (D) R3 R
A certain hill located in Udaipur is of height h that varies as :
w. g
MCQ 1.2.10
at e
(A) - R3 R (C) - 12 R
h (x, y) = 6x2 + 8y2 - 3xy + 36x - 56y + 100
ww
Where x is the distance (in miles) in north and y is the distance (in miles) in east of Udaipur railway station. The top of the hill will be located at (A) 3 miles north, 2 miles west of Railway station. (B) 2 miles south, 3 miles east of Railway station. (C) 2 miles north, 3 miles east of Railway station. (D) 6 miles south, 2 miles east of Railway station. MCQ 1.2.11
At any point P (x, y, z) a vector field is given by F = R15 aR , where R is the position vector of the point P . The divergence of the vector field F will be 2
(A) zero, everywhere (C) 63 , everywhere R
(B) zero, at all points excluding origin (D) 63 , at all points excluding origin R
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
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Statement for Linked Question 12 - 14 : Given a vector field A = 3x2 ax + 3yzay + 3y2 az . MCQ 1.2.12
The line integral of A from origin to the point (2,2,2) by the route (0, 0, 0) " (2, 0, 0) " (2, 2, 0) " (2, 2, 2) will be (A) 32 units (B) 8 units (C) 24 units
The line integral of the vector field A from the origin to the point (2, 2, 2) along the direct straight line is (A) 16 units (B) 24 units
m
MCQ 1.2.13
(D) 6 units
The line integral of the field A around the closed loop that goes out along the route defined in Question 12 and back along the route defined in Question 13 is (A) 64 units (B) 0 units (C) 36 units
(D) 26 units
A wedge defined by 0 # r # 5 , 45 # f # 180c, z = 2 is shown in figure
ww w. ga te
he
MCQ 1.2.15
lp.
MCQ 1.2.14
(D) 32 units
co
(C) 4 units
Circulation of A = r sin fa r + z2 cos faz along the edge L of the wedge is (A) 10 units (B) 5/2 units (C) - 25/4 units MCQ 1.2.16
Volume integral of the function f = 30z2 over the tetrahedron with corners at (0, 0, - 1); (0, - 1, 0), (- 1, 0, 0), (0, 0, 0) is (A) 1/2 (B) - 1/2 (C) 31/2
MCQ 1.2.17
(D) 25/4 units
(D) 1/60
Total outward flux of a vector field A = 14 r2 cos2 fa r + 2z2 af through the closed surface of a cylinder 0 # z # 2 , r = 2 is (A) 4p
(B) 16p
(C) p
(D) 32p
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
10
Vector Analysis
Chap 1
lp.
co
m
For View Only Shop Online at www.nodia.co.in MCQ 1.2.18 A quarter cylinder of radius 2 and height 5 exists in the first octant of a cartesian coordinate system as shown in the figure below. Surface integral of a vector field A = (4r + 3r sin2 f) a r + r sin 2faf + 6zaz over the surface of the cylinder will be
(B) 30p
he
(A) 12p (C) 40p
A vector function is given by G = 6x2 yax - 5y2 ay + 2zaz . If L is a closed path
at e
MCQ 1.2.19
(D) 80p
# G : dl L
equals to
ww
w. g
defined by the sides of a triangle as shown in the figure then,
(A) 24 units
(B) 7 units
(C) 2 units
(D) 10/6 units
Common Data for Question 20 - 21 : Consider S1 and S2 are respectively the top and slanting surfaces of an ice cream cone of slant height 2 m and angle 60c as shown in figure, where a vector field F is defined as 4x2 + 5y2 + 4z2 x y y 3x F = : 2 a x + 2 a y - 2 a x + 2 a yD 2 2 x +y GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
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The surface integral of the vector field F over the surface S1 will be (A) 2.3 units (B) 24 units
lp.
MCQ 1.2.20
co
m
For View Only
11
(C) 0.18 units
The surface integral of the vector field over the surface S2 will be (B) 3 p/3 (A) 4p/3
he
MCQ 1.2.21
(D) 1.9 units
(C) 4p 3 /3
ww w. ga te
(D) 0
Common Data for Question 22 - 23 : Negative gradient of a scalar field f is A =-4 f = (x + z) ax - 4zay + (x - 3y - z) az MCQ 1.2.22
MCQ 1.2.23
The vector A is (A) irrotational but not solenoidal
(B) both irrotational and solenoidal
(C) solenoidal but not irrotational
(D) neither solenoidal nor irrotational
The scalar field, f equals to 2 2 (A) x + xz + z 2 2
(B) - x - 2xz + 6yz + z 2 2
2
(C) - xz + 3yz + z 2 MCQ 1.2.24
2
2
2
2
(D) - x - xz + 3yz + z 2 2
Line integral of a vector field A = 5 (yax + xay) from a point P (2, 1, 3) to the point Q (8, 2, 3) along the curve y = x/2 will be (A) 42 units (B) 14 units (C) 16 units
(D) 32 units
A vector field F = 2 ^ r1 + r1 cos 2fh ar exists in the region between the two spherical shells of radius 1 m and 2 m centred at the origin. The total outward flux of F through the outer spherical surface will be GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 1.2.25
3
3
12
Vector Analysis
For View Only (A) p
Shop Online at www.nodia.co.in (B) - p
(C) - 2p
(D) 2p
Two vectors A and B make an angle 30c between them as shown in figure. Magnitude of vector A and B are 4 units and 3 units respectively. If a third vector R is defined such that R = 6A - 4B then it’s graphical construction will be
MCQ 1.2.27
ww
w. g
at e
he
lp.
co
m
MCQ 1.2.26
Chap 1
A certain, vector R is defined as R = A # ^B # C h. Directions of A, B and C are mentioned in the list below. Which of the following gives the correct direction of R for the given direction of the three vectors. Direction of A Direction of B Direction of C Direction of R (A) North South East West (B) South North West East (C) East West North West (D) West East South South
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
13
For View Only Shop Online at www.nodia.co.in MCQ 1.2.28 Which of the following vectors are equal : A =- ax + ay + 3az at ^1, 2, 3h in cartesian co-ordinates B = a r + af + 3az at ^2, p/2, 3h in cylindrical co-ordinates C = 2 a r + 3az at ^3, 3p/4, 9h in cylindrical co-ordinates (A) A and B only (B) B and C only (C) A and C only
Two vectors are defined as A = ax + 5ay + 3az and B = 3ax + 2ay + az . Which of the following vector is perpendicular to ^A + B h (B) 4ay + 4az (A) - 4ax + 4ay
m
MCQ 1.2.29
(D) all the three vectors A, B and C
(C) ax + az
co
lp.
MCQ 1.2.31
The gradient of a scaler function is dV ^x, y, z h = 1.5x2 yz2 ax + 0.5x3 z2 ay + x3 yzaz The scalar function is x3 yz2 (A) x3 yz2 (B) 2 x2 yz3 (C) (D) xy3 z2 2 The equation of the plane tangential to the surface xyz = 1 at the point b 2, 4, 1 l is 4 (A) 16x + 32y + z = 24 (B) 2x + y + 32z = 12
he
MCQ 1.2.30
(D) none of these
(C) x + 2y + 32z = 12
Consider a volume v is defined as the part of a spherical volume of radius unity
ww w. ga te
MCQ 1.2.32
(D) x + 16y + 32z = 24
lying in the first octant. The volume integral
MCQ 1.2.33
# 2xdv is equal to v
(A) 2p
(B) p/16
(C) p/4
(D) p/8
r Consider a vector field A = r cos fa r + af . If C is the contour shown in the figure 3
then the contour integral (A) p + 4 (C) p + 2
# A : dl C
is equal to (B) p + 1 2 (D) 2p + 1
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
14
Vector Analysis
Chap 1
C1
C2
(A) - 1 9
is
(B) 1 9
(D) - 9
at e
(C) 9
Consider the contour C as shown in the figure
ww
w. g
MCQ 1.2.35
# A : dl # A : dl
he
The ratio of the contour integrals
lp.
co
m
For View Only Shop Online at www.nodia.co.in 2 MCQ 1.2.34 A vector field is defined as A = 2r cos fa r + raf . Consider the two contours C1 and C2 as shown in the figure.
-r
If a vector field is defined as A = b e l a q then the contour integral r (B) - p e-1 (A) p e-1 2 2 (C) p ^1 + e-1h 2 MCQ 1.2.36
# A : dl C
is
(D) p ^1 - e-1h 2
The divergence of the unit vectors ar ,a q and af in spherical co-ordinate system is listed below. Which among them is correct ?
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
For View Only (A) (B) (C) (D)
15
Shop Online at www.nodia.co.in ar r/2 2/r r/2 2/r
aq r/ tan q cot q/r r/ tan q cot q/r
af 0 0 1 1
Which of the following vector can be expressed as the gradient of a scalar ? -r (B) e af (A) 2yzax + 2xzay + 2xyaz r 2 (C) 2 ^cos fa r + sin fafh (D) Both (A) and (C) r
MCQ 1.2.38
Which of the following vector can be expressed as curl of another vector ?
co
m
MCQ 1.2.37
-r (B) e af r
(A) 1 ^x2 - y2h ax - xyay + 2az 2
(D) All of the above
lp.
q a + sin q a (C) 2 cos r q r3 r3
The vector field pattern of A = 3yax is
MCQ 1.2.40
A two dimensional vector field in Cartesian coordinate system is defined as A ^x, y h = Ax ax + 2Ay ay The curl and divergence of the vector field are both zero. Which of the following differential equation satisfies Ax and Ay (A) d2Ax = 0 (B) d2Ay = 0
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MCQ 1.2.39
(C) d2Ax + d2Ay = 0
(D) (A) and (B) both
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co
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For View Only Shop Online at www.nodia.co.in 2 2 MCQ 1.2.41 The circulation of F = x ax - 3xzay - y az around the path shown below is
(A) - 1 3
(B) 1 6
If R = xax + yay + zaz is the position vector of point P (x, y, z) and R = R then d : Rn R is equal to (A) nrn (B) (n + 3) rn
he
MCQ 1.2.42
(D) 1 3
lp.
(C) - 1 6
(C) (n + 2) rn
If A = r sin fa r + r2 a r , and L is the contour of figure given below, then circulation # A : dl is
at e
MCQ 1.2.43
(D) 0
ww
w. g
L
(A) 7p + 2
(B) 7p - 2
(C) 7p
(D) 0
***********
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EXERCISE 1.3
The direction of vector A is radially outward from the origin, with A = krn . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is (A) - 2 (B) 2
m
MCQ 1.3.1
(D) 0
If A = xyax + x 2 ay , then
#C A $ dl
o
over the path shown in the figure is
(A) 0 (C) 1 MCQ 1.3.3 GATE 2009
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GATE 2010
co
(C) 1 MCQ 1.3.2
17
(B) 2 3 (D) 2 3
If a vector field V is related to another vector field A through V = d # A, which of the following is true? (Note : C and SC refer to any closed contour and any surface whose boundary is C . ) (A) # V $ dl = ## A $ dS C
(B)
SC
#C A $ dl = ##S V $ dS C
(C)
#C ^d # V h : dl = ##S ^d # Ah : dS C
(D)
#C ^d # V h : dl = ##S V : dS C
MCQ 1.3.4 IES EC 2010
Consider points A, B, C and D on a circle of radius 2 units as in the shown figure below. The items in List II are the values of af at different points on the circle. Match List I with List II and select the correct answer using the code given below the lists :
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MCQ 1.3.6
b.
B
2.
ay
c.
C
3.
- ax
d.
D
4.
(ax + ay) / 2
5.
- (ax + ay) / 2
6.
(ax - ay) / 2
b 4 6 6 5
c 5 5 2 4
co
m
ax
d 2 2 4 2
If V = 2 sinh x cos kye pz is a solution of Laplace’s equation, what will be the value of k ? 1 (B) 1 + p2 (A) 2 1+p 1 (C) (D) 1 - p2 1 - p2 The electric field intensity E at a point P is given by 10ax + 10ay + 10az where ax , ay and az are unit vectors in x, y and z directions respectively. If a, b, g respectively the angles the E vector makes with x, y and z axes respectively, they are given by which of the following ? (B) (A) a = b = g = 30c a = b = g = 60c (C) a = b = g cos-1 1 (D) a = b = g = cos-1 1 3 3
MCQ 1.3.7 IES EC 2003
ww
w. g
IES EC 2007
1.
he
IES EC 2009
A
at e
MCQ 1.3.5
a.
lp.
Codes : a (A) 3 (B) 1 (C) 1 (D) 3
List II
Which one of the following potentials does NOT satisfy Laplace’s Equation ? (A) V = 10xy (B) V = r cos f (C) V = 10/r
MCQ 1.3.8 IES EC 2003
(D) V = r cos f + 10
Laplacian of a scalar function V is (A) Gradient of V (B) Divergence of V (C) Gradient of the gradient of V (D) Divergence of the gradient of V
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For View Only Shop Online at www.nodia.co.in 2 3 3 MCQ 1.3.9 A field A = 5x yzax + x zay + (x y - 2z) az can be termed as IES EC 2002 (A) Harmonic (B) Divergence less (C) Solenoidal MCQ 1.3.10 IES EC 2001
(D) Rotational
Laplace equation in cylindrical coordinates is given by 2 2 r2V (A) d2 V = 1 2 e + 12 e2V2 o + 2V2 = 0 o r 2r 2r r 2f 2z
m
2 2 2 (B) d2 V = 2V2 + 2V2 + 2V2 2x 2y 2z r (C) d2 V =- v e
MCQ 1.3.11
#c dl
along the curve c ?( c is the curve ABCD in
MCQ 1.3.12 IES EE 2003
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lp.
IES EE 2005
What is the value of the integral the direction of the arrow ) ?
co
2 (D) d2 V = 1 2 c r 2V m + 2 12 =0 r 2r 2r r sin q2f2
(A) 2R (ax + ay) / 2
(B) - 2R (ax + ay) / 2
(C) 2Rax
(D) - 2Ray
Given a vector field A = 4r cos far in cylindrical coordinates. For the contour as shown below, # A : dl is
(A) 1
(B) 1 - (p/2)
(C) 1 + (p/2)
(D) - 1
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For View Only Shop Online at www.nodia.co.in MCQ 1.3.13 If a vector field B is solenoidal, which of these is true ? IES EE 2002 (B) # B : ds = 0 (A) # B : dl = 0 L
s
(C) d # B = 0 MCQ 1.3.14 IES EE 2002
(D) d : B = Y 0
Which of the following equations is correct ? (A) ax # ax = ax 2 (B) (ax # ay) + (ay # ax ) = 0
m
(C) ax # (ay # az ) = ax # (az # ay)
Match List I with List II and select the correct answer :
IES EE 2002
List I (Term) b c
curl ^F h = 0
1. Laplace equation
div ^F h = 0
lp.
a
List II (Type) 2. Irrotational
div Grad ^fh = 0
3. Solenoidal
d div div ^fh = 0 Codes :
d 4 2 4 2
If A = 2ar + af + az , the value of shown in the given figure is
# A : dl
around the closed circular quadrant
ww
IES EE 2001
c 1 3 3 1
w. g
MCQ 1.3.16
b 3 1 1 3
at e
(A) (B) (C) (D)
a 2 4 2 4
4. Not defined
he
MCQ 1.3.15
co
(D) ar : a q + a q : ar = 0
(A) p
(B) p + 4 2
(C) p + 4
(D) p + 2 2 ***********
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SOLUTIONS 1.1
Option (D) is correct. Given the vector field has the only component in a q direction and its magnitude is r so as r increase from origin to the infinity field lines will be larger and directed along a q as shown in option (A).
SOL 1.1.2
Option (B) is correct. M - N = 5ax - 2ay + 4az - (- 8ax - 7ay + 2az ) = 5ax - 2ay + 4az + 8ax + 7ay - 2az = 13ax + 5ay + 2az So, the unit vector in the direction of (M - N ) is 13ax + 5ay + 2az a = M-N = M-N 13ax + 5ay + 2az = 0.92ax + 0.36ay + 0.14az
SOL 1.1.3
Option (C) is correct. Vector G at (- 2, 1, 3) : G = 4 (- 2) (1) ax + 2 (2 + (- 2) 2) ay + 3 (3) 2 az =- 8ax + 12ay + 27az So, unit vector in the direction of G at Q : - 8ax + 12ay + 27az aG = G = - 8ax + 12ay + 27az G =- 0.26ax + 0.49ay + 0.88az
SOL 1.1.4
Option (B) is correct. Option (A), (B), (D) are the properties of vector product. Now we check the relation defined in option (C). Since the triple cross product is not associative in general so, the given relation is incorrect. This inequality can be explained by considering vector A = B and C ,perpendicular to A as shown in the figure. According to right hand rule we determine that ^B # C h points out of the page and so A # ^B # C h points down that has magnitude ABC . But in L.H.S. of the relation, since A = B So we have (A # B) = 0 and hence (A # B) # C = 0 Therefore ^A # B h # C = 0 ! A # ^B # C h
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SOL 1.1.1
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Option (B) is correct. As the direction of cross vector is normal to the plane. So, direction of B # C will be normal to the plane defined by the three vectors. Now the dot product of two mutually perpendicular vectors is always zero and since the direction of B # C will be perpendicular to the plane of vector A. So A : ^B # C h will be zero.
SOL 1.1.6
Option (C) is correct. Consider the two vectors A and B are as shown below.
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lp.
SOL 1.1.5
ww
As the angle between the two vectors is a. So component of vector A along B is A1 = ^cos ah A cosine of the angle between the two vectors is defined as cos a = A : B AB A So, A1 = b : B l A AB = A:B B SOL 1.1.7
Option (C) is correct. Cross product of two parallel vector fields is always zero since the angle between them is q = 0c. i.e. A#B = 0
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a r af az 1 2 3 =0 a b -6 ^- 12 - 3b h a r + ^3a + 6h af + ^b - 2ah az = 0 Solving it we have, b =- 4 and a =- 2 Option (A) is correct. Dot product of the two orthogonal vectors is always zero. i.e. A:B = 0 (4) (1) + (2k) (4) + (k) (- 4) = 0 4 + 8k - 4k = 0 4k =- 4 k =- 1
SOL 1.1.9
Option (A) is correct. From the given field vector we have the component Ex = 3zy2 cos 2x . So for the given condition Ex = 0 We have, 9zy2 cos (2x) = 0 This condition met when, z = 0 or, y =0 or, cos 2x = 0 & 2x = p/2 & x = p/4 Therefore the planes on which field component Ex will be zero are and z= 0 y= 0 x = p/4
SOL 1.1.10
Option (A) is correct. From the given field vector we have the field components Ey = 2zy sin 2x and Ez = 2y2 sin 2x Now, in the plane y - 4z = 0 & y = 4z So, Ey = 8z ^4z h sin 2x = 32z2 sin 2x Ez = 2 ^4z h2 sin 2x = 32z2 sin 2x Thus Ey = E z
SOL 1.1.11
Option (C) is correct. For the given condition E = 0 , we must have E x = Ey = E z = 0 2 i.e. 9zy cos 2x = 8zy sin 2x = 2y2 sin 2x = 0 This condition met in the plane y = 0 .
SOL 1.1.12
Option (B) is correct. Since the two points are defined in different coordinate system so we represent the point Q in Cartesian system as x = r cos f = 4 cos (- 50) = 2.57
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SOL 1.1.8
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Shop Online at www.nodia.co.in y = r sin f = 4 sin (- 50) =- 3.064 and z =2 So, the distance between the two points P (2, 3, - 1) and Q (2.57, - 3.064, 2) is given as PQ = (2.57 - 2) 2 + (- 3.064 - 3) 2 + (2 + 1) 2 = 6.78 units
Option (A) is correct. Since, the options include spherical as well as cylindrical representation of A, so, we will transform the vector in both the forms to check the result. The components of vector field A are Ax = 1, Ay = 0 and Az = 0 Now, we transform the vector components in cylindrical system as RA V R cos f sin f 0VRA V WS xW S rW S SAfW = S- sin f cos f 0WSAyW SSA WW SS 0 0 1WWSSAz WW z T AXr = T(cos f) (1) = cos f XT X Af = (- sin f) (1) =- sin f Az = 0 So, the vector field in cylindrical system is A (r, f, z) = cos fa r - sin faf Hence, both the options (A) and (B) are incorrect. Again, we transform the vector components in spherical system as RA V R sin q cos f sin q sin f cos qVRA V WS xW S rW S S WSAyW cos cos cos sin sin = q f q f q A S qW W SSA WW SS sin f cos f 0WSSAz WW f T AXr = T(sin q cos f) (1) = sin q cos f XT X A q = (cos q cos f) (1) = cos q cos f Af = (- sin f) (1) =- sin f So, the vector field in spherical system is A (r, q, f) = sin q cos far + cos q cos qa q - sin faf
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SOL 1.1.13
Option (C) is correct. We transform the given vector field in spherical system. Since the given vector field is F = 10ax The Cartesian components of the field are Fx = 10 , Fy = 0 , Fz = 0 . So, the spherical components of vector field can be determined as VR V R V R SFr W S sin q cos f sin q sin f cos qWSFxW SFqW = Scos q cos f cos q sin f - sin qWSFyW W SSF WW SS - sin q cos f 0WSSFz WW f T X T XT X So, we get Fr = 10 sin q cos f Fq = 10 cos q cos f and Ff =- 10 sin q GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 1.1.14
Chap 1
Vector Analysis
25
Option (C) is correct. Since z -axis is normal to the surface z = 5 , so first of all we will find the angle between z -axis and A which can be easily obtained from the figure shown below :
i.e.
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lp.
SOL 1.1.15
co
m
For View Only Shop Online at www.nodia.co.in Now, for the given point (x = 3 , y = 2 , z =- 1) we have r = (3) 2 + (2) 2 + (- 1) 2 = 14 (3) 2 + (2) 2 x2 + y2 = tan-1 f q = tan-1 p = 105.5c z (- 1) y f = tan-1 a k = tan-1 b 2 l = 33.7c 3 x Putting all the values in the matrix transformation, we have Fr = 5 sin (105.5c) cos (33.7c) = 8 Fq = 10 cos (105.5c) cos (33.7c) =- 2.2 Ff =- 10 sin (33.7c) =- 5.5 Therefore, the vector field in spherical coordinate is F = Fr ar + Fq a q + Ff af = 8ar - 2.2a q - 5.5af
cos f = Az = A
3 = (- 5) + (10) 2 + (3) 2 2
3 134
3 = 74.98c . 75c 134 m Therefore, the angle between surface z = 5 and vector A is (90c - f) = 15c. f = cos-1 c
SOL 1.1.16
Option (B) is correct. The given line x = 6 , z =- 2 is parallel to y -axis. So, the component of A parallel to the given line is Ay = (A : ay) ay = 6(- 2a r + 20af + 4az ) : ay@ ay = (- 2 sin f + 20 cos f) ay At point P , f = 90c, so, Ay =- 2ay
SOL 1.1.17
Option (A) is correct. The given point is shown below :
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co
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lp.
After 120c rotation looking down the axis the new co-ordinate axes (xl, yl, zl) will be as shown below :
ww
So, the rotation carries z axis into y ; y -axis into x and x into z . therefore the new co-ordinates of point P are : xl = z = c yl = x = a zl = y = b i.e. (c , a , b) is the co-ordinates of point P in the transformed system. Option (B) is correct. The position vector can be defined as : R = xax + yay + zaz R = x2 + y2 + z2 So, gradR = 2R ax + 2R ay + 2R az 2x 2y 2z 2y 2x 2z a +1 a +1 a =1 2 x2 + y2 + z2 x 2 x2 + y2 + z2 y 2 x2 + y2 + z2 z xa + yay + zaz = x 2 =R 2 2 R x +y +z GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 1.1.18
Chap 1
Vector Analysis
For View Only SOL 1.1.19 Option (D) is correct.
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d : A = 2 (y2) + 2 (2xy + x2 + z2) + 2 (4x + 2yz) 2x 2y 2z = 0 + 2x + 2y = 2 (x + y) Option (A) is correct. We have the vector field components as Fr = r sin f , Ff = r2 z and Fz = z cos f a r ra f a z Now, d # F = 1 22r 22f 22z r Fr rFf Fz
m
SOL 1.1.20
lp.
co
= 1 ; 2 z cos f + 2 rr2 z E a r - 1 r ; 2 z cos f - 2 r sin fE af r 2f r 2r 2z 2z + 1 ; 2 rr2 z - 2 r sin fE az r 2r 2f 1 1 3 2 = 6- z sin f - r @ a r - 60 - 0@ af + 63r z - r cos f@ az r r 1 3 =- (z sin f + r ) a r + (3rz - cos f) az r
Option (A) is correct. D = yzax + xzay + xyaz
ww w. ga te
SOL 1.1.21
he
At point P (1, p/2, 2) d # F =- 1 (2 # 1 + 13) a r + (3 # 1 # 2 - 0) az =- 3a r + 6az
d : D = 2 (yz) + 2 (xz) + 2 (xy) = 0 2x 2y 2z a x ay a z d # D = 22x 22y 22z yz xz xy = (x - x) ax - (y - y) ay + (z - z) az = 0
SOL 1.1.22
Option (D) is correct. Curl of the unit vector a r is ar d # a r = 1 22r r 1 The curl of unit vector af is ar 1 d # af = 22r r 0 and curl of unit vector az is ar 1 d # az = 22r r 0
ra f a z 2 2f
0
2 2z
0
ra f a z 2 2f
=0
2 2z
r ^1 h 0
2r =1 a = az r 2r z r
ra f a z 2 2f
0
2 2z
=0
1
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For View Only Shop Online at www.nodia.co.in SOL 1.1.23 Option (A) is correct. Options (A), (B) and (C) are properties of d operator where as : g (d # A) + A # (dg) d #cAm = g g2 Option (A) is correct. In a cylindrical coordinate system Laplacian of a scalar field is defined as 2 2 2f 1 2 f +2 f 42f = 1 2 e r + r 2r 2r o r22f2 2z2
m
SOL 1.1.24
lp.
co
= 1 2 r (2z sin f + 8r) + 12 c- 2rz sin f - 6z2 2 sin f cos f m r 2r 2f r + 2 (2r sin f + 6z cos2 f) 2z 1 1 2 = (2z sin f + 16r) - 2 (2rz sin f + 6z cos 2f) + 6 cos2 f r r 2 = 16 + 6 cos2 f - 6z2 cos 2f r
he
At point P ^3, p/2, 6h d2 f = 16 + 0 - 6 # 36 # (- 1) = 40 9
Option (A) is correct. A vector field is called conservative (irrotational) if its curl is zero. i.e. d#M = 0 ax ay az 2 2 2 =0 2x 2y 2z z cos xz + y 2kx x cos xz ax (0 - 0) - ay (cos xz - xz sin xz - cos xz + xz sin xz) + (2k - 1) az = 0 2k - 1 = 0 or, k =1 2
SOL 1.1.26
Option (B) is correct. For a scalar field to be harmonic, d2 g = 0 22g 22g 22g =0 + + 2x2 2y2 2z2
SOL 1.1.27
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SOL 1.1.25
2 ^1 + 2k h y = 0 which results in k =- 1 2
Option (D) is correct. Consider V is a scalar field. So the gradient of the field is 4 V = 2V + 2V + 2V 2x 2y 2z and the curl of the gradient of the field is
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Shop Online at www.nodia.co.in R V Sax ay az W 4# (4 V) = S22x 22y 22z W S2V 2V 2V W S2x 2y 2z W T 2 X2 2 2 2 2 = c 2V - 2V m ax + c 2V - 2V m ay + c 2V - 2V m az = 0 2z2y 2y2z 2x2z 2z2x 2x2y 2y2x So the curl of the gradient of any scalar field is zero everywhere.
Option (D) is correct. For vector A to be solenoidal its divergence must be zero. i.e. d:A = 0 2 (x + 4z) + 2 (2x - 3z) + 2 (4x + 3y - cz) = 0 2x 2y 2z 1+0-c = 0 c =1
SOL 1.1.29
Option (B) is correct. For a vector function to be irrotational its curl must be zero. i.e. d#A = 0 a x ay a z 2 2 2 =0 2x 2y 2z A x Ay A z ax
ay
2 2x
2 2y
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lp.
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SOL 1.1.28
az 2 2z
=0
SOL 1.1.30
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(4x + k1 z) (k2 x - 5z) (4x - k 3 y + 2z) (- k 3 - (- 5)) ax - (4 - k1) ay + (k2) az = 0 - k 3 - (- 5)= 0 & k 3 = 5 4 - k1 = 0 & k1 = 4 and k2 - 0 = 0 & k2 = 0 So k1 , k2 and k 3 are 4, 0 and 5 respectively. Option (C) is correct. The unit vector normal to a given plane f = 0 is df an = df The given equation is 2x + 4y + 6z = 7 2x + 4y + 6z - 7 = 0 So, f = 2x + 4y + 6z - 7 and gradient of f is df =- 2ax + 3ay + 6az df = 22 + 42 + 62 = 56 So, an = 1 ^2ax + 4ay + 6az h = 1 ^ax + 2ay + 3az h 56 14
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For View Only Shop Online at www.nodia.co.in SOL 1.1.31 Option (D) is correct. Consider the differential displacement, dl = dxax + dyay + dzaz
# dl
So
C
#
#
#
= c dx m ax + c dy m ay + c dz m az C
C
C
# dl
=0
Option (B) is correct. According to stoke’s theorem.
# A : dl L
So
# ^dU h : dl C
# ^d # Ah : dS = # 6d # ^dU h@ : dS
co
SOL 1.1.32
C
m
For a contour the initial and final points are same. So, all the individual integrals described above will be zero. Therefore,
=
Now, So
# A : dS S
=
# ^d : Ahdv
at e
i.e.
he
Option (B) is correct. According to the divergence theorem surface integral of vector over a closed surface is equal to the volume integral of its divergence inside the region defined by closed surface. v
d : A = 2 ^2yz h + 2 ^z2 x h + 2 ^2xy h 2x 2y 2z = 0+0+0
# A : dS
=0
w. g
SOL 1.1.33
lp.
Since (curl of the gradient of a scalar field is always zero) d # ^dU h = 0 So the contour integral is zero.
ww
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SOLUTIONS 1.2
Option (D) is correct. Consider that the cube has its edges on the x, y and z - axes respectively as shown in the figure. As the angle between any of the two body diagonals of the cube will be same so we determine the angle q between the diagonals OB and AC of the cube.
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SOL 1.2.1
From the figure we get the co-ordinates of points A, B and C as : A" ^0, 0, 3h B " ^3, 3, 3h and C " ^3, 3, 0h So, the vector length, OB = 3ax + 3ay + 3az and AC = 3ax + 3ay - 3az For determining the angle q between them, we take their dot product as ^OB h : ^AC h = OB AC cos q 9 + 9 - 9 = ^3 3 h^3 3 h cos q So, the angle formed between the diagonals is q = cos-1 b 1 l = 70.53c 4 SOL 1.2.2
Option (C) is correct. For the given points A, B , C , the vector length,
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Shop Online at www.nodia.co.in AB = 20ax + 18ay - 10az and AC =- 10ax + 8ay + 15az Since the cross product of two vectors is always perpendicular to the plane of vectors. So the unit vector perpendicular to the plane of triangle is given by an = AB # AC AB # AC R V S a x ay a z W now, AB # AC = S 20 18 - 10W SS- 10 8 15WW T X = 618 # 15 - ^- 10 # 8h@ ax - 620 # 15 - (- 10) # (- 10)@ ay +620 # 8 - ^- 10 # 18h@ az = 350ax - 200ay + 340az
he
Option (D) is correct. The unit vector in the direction of vector AC is given by - 10ax + 8ay + 15az aAC = AC = AC (- 10) 2 + (8) 2 + (15) 2 =- 0.507ax + 0.406ay + 0.761az =- 0.61ax + 0.41ay + 0.76az Since the cross product of two vectors is always perpendicular to the plane of vectors. So, the unit vector in the plane of the triangle which is perpendicular to AC is given by cross product of the unit vector perpendicular to the plane of the triangle and the unit vector aAC . i.e. aP = an # aAC =- 0.550ax - 0.832ay + 0.077az
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SOL 1.2.3
350ax - 200ay + 340az 350ax - 200ay + 340az 350ax - 200ay + 340az = (350) 2 + (- 200) 2 + (340) 2 = 0.664ax - 0.379ay + 0.645az
an =
lp.
So,
co
m
For View Only
ww
Option (A) is correct. Unit vector in the direction of AB is given by 10ax + 18ay - 10az aAB = (20) 2 + (18) 2 + (- 10) 2 = 0.697ax + 0.627ay - 0.348az A non unit vector in the direction of bisector of interior angle at A is defined as 1 (a + a ) = 1 0.697a + 0.627a - 0.348a - 0.507a + 0.406a + 0.761a AC x y z x y z@ 2 AB 26 = 0.095ax + 0.516ay + 0.207az So the unit vector in the direction of bisector of interior angle at A is given by 0.095ax + 0.516ay + 0.207ay abis = (0.095) 2 + (0.516) 2 + (0.207) 2 = 0.168ax + 0.915ay + 0.367az GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 1.2.4
Chap 1
Vector Analysis
For View Only
Shop Online at www.nodia.co.in = 0.17ax + 0.92ay + 0.37az
Option (B) is correct. The vector field F can be written in cartesian system as xa + yay y = 2 x 2 ax + 2 ay F (x, y, z) = x2 x + y2 x +y x + y2 r cos f r sin f ( x = r cos f, y = r sin f ) ax + ay F (x, y, z) = r2 r2 = 1 (cos fax + sin fay) r The components of vector field F are Fx = 1 cos f , Fy = 1 sin f and Fz = 0 r r So the components of vector field F in cylindrical system can be expressed as RF V R cos f sin f 0VRF V WS xW S rW S SFfW = S- sin f cos f 0WSFyW SSF WW SS 0 0 1WWSSFz WW z XT X T X T 1 2 2 Fr = 6cos f + sin f@ = 1 r r Ff = 1 6cos f (- sin f) + sin f cos f@ = 0 r Fz = 0
he
lp.
co
m
SOL 1.2.5
33
F (r, f, z) = 1 a r r At the point P (r = 2 , f = p/4 , z = 0.1) F = 1 a r = 0.5a r 2
ww w. ga te
So the vector field
SOL 1.2.6
Option (D) is correct. Any vector field can be represented as the sum of its normal and tangential component to any surface as A = At + A n where At is tangential component and An is normal component to the surface r = 20 at point P (20, 150c, 330c). So, An = rar = 20ar and therefore, At = A - A n 4 =- 2 sin q cos fa q + r af r 4 = 0.043a q + 100af
SOL 1.2.7
Option (A) is correct. Any vector field can be represented as the sum of its normal and tangential components to any surface as A = At + A n Here At and An are the tangential and normal components to the conical surface
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34
Vector Analysis
Chap 1
For View Only Shop Online at www.nodia.co.in Since the unit vector normal to the conical surface is a q . So, An =- 8a q and therefore the tangential component to the cone is At = A - An =- 12ar + 9af Option (B) is correct. Consider the unit vector perpendicular to A and tangent to the cone q = 150c is b = br ar + bf af (Tangential component to the cone will have b q = 0 ) Now the magnitude of unit vector is 1 ...(i) So, b r2 + b f2 = 1 and the dot product of mutually perpendicular vectors is zero. So, A:b = 0 - 6br + 9bf = 0 ....(ii) b f = 3 br 4 So, from equation (i) and (ii) we have b r2 b1 + 16 l = 1 9 br = 3 , b f = 4 5 5 Therefore, b = 1 (3ar + 4af) 5 Option (D) is correct. The separation vector R can be defined as : R = (x - a) ax + (y - b) ay + (z - c) az
So,
SOL 1.2.10
at e
R =
(x - a) 2 + (y - b) 2 + (z - c) 2
d b 1 l = 2 6(x - a) 2 + (y - b) 2 + (z - c) 2@-1/2 ax R 2x + 2 6(x - a) 2 + (y - b) 2 + (z - c) 2@- 1/2 ay 2y + 2 6(x - a) 2 + (y - b) 2 + (z - c) 2@- 1/2 az 2z 1 1 =- ^Rh-3/2 (x - a) ax - ^Rh-3/2 (y - b) ay - 1 ^Rh-3/2 (z - c) az 2 2 2 (x - a) ax + (y - b) ay + (z - c) az ==- R3 R R3/2
w. g
and
ww
SOL 1.2.9
he
lp.
co
m
SOL 1.2.8
Option (C) is correct. The gradient of a scalar field at its maxima is zero. So at the top of hill 4h = 0 or, (12x - 4y + 36) ax + (16y - 4x - 56) ay = 0 Therefore both the components will be equal to zero i.e. 12x - 4y + 36 = 0 and, 16y - 4x - 56 = 0
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
35
For View Only Shop Online at www.nodia.co.in Solving the two equations, we get, x =- 2 , y = 3 Thus the top of the hill is located at 2 miles south (- 2 miles north) and 3 miles east of the railway station. Option (C) is correct. Consider the position vector of point P is R = x 4 2ax + 5yay + zaz So, the magnitude of R is R = x2 + y2 + z2 and unit vector in the direction of R is xa + yay + zaz aR = x 2 x + y2 + z2 Therefore, the vector field F at point P is xax + yay + zaz xa + yay + zaz = G = 10 F = 102 aR = 102 e x 2 o ^x2 + y2 + z2h3/2 R R x + y2 + z2 The divergence of the field F is given as y 2 x z +2 d : F = 10 > 2 2 3/2 + 2 2 2 2 2x ^x + y + z h 2y ^x + y + z2h3/2 2z ^x2 + y2 + z2h3/2 H
lp.
co
m
SOL 1.2.11
SOL 1.2.12
ww w. ga te
he
x (2x) 1 3 1 = 10 = 2 2 2 3/2 - 2 2 2 2 5/2 + 2 2 (x + y + z ) (x + y + z ) (x + y + z2) 3/2 y (2y) z (2z) 1 -3 2 + -3 2 (x + y2 + z2) 5/2 (x2 + y2 + z2) 3/2 2 (x2 + y2 + z2) 5/2 G (x2 + y2 + z2) = 10 > 33 - 3 2 H R (x + y2 + z2) 5/2 = 10 : 33 - 33 D = 0 R R But at origin (x = 0 , y = 0 , z = 0 ) the position vector R = 0 and so the expression for field F blows up. Therefore, d : F is infinite at origin and zero else where. Option (D) is correct. The circulation of A around the route is given by
# A : dl
# # #
=c + + 1
2
3
m A : dl
where the route is broken into segments numbered 1 to 3 as described below 1st segment : (0, 0, 0) " (2, 0, 0) x changes from 0 to 2, y = 0 , z = 0 3 2 2 (dl = dxax ) So, A : dl = 3x2 dx = 3 b x l = 8 3 0 1 0 2nd segment : (2, 0, 0) " (2, 2, 0) x = 2 , y changes from 0 to 2, z = 0 .
#
# A : dl 2
#
=
2
# 6yz dy = 0 0
(dl = dyay )
3rd segment : (2, 2, 0) " (2, 2, 2) GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
36
Vector Analysis
For View Only x = 2 , y = 2 , z changes from 0 to 2.
# A : dl 3
=
#
2
0
Chap 1
Shop Online at www.nodia.co.in
3y2 dz = (3 # 4) z
2 0
= 24
(dl = dzaz )
So total line integral will be :
# A : dl
Option (A) is correct. For the straight line from origin to the point (2, 2, 2) we have the relation between the coordinates as x = y = 2z or, dx = dy = dz and the line integral along straight line is given as dl = dxax + dyay + dzaz Therefore, the line integral of the vector field along the straight line is given as
# 3x dx + # 6yzdy + # 3z dz = # 3x dx + # 6x dx + # 3x dx 2
=
2
2
lp.
# A : dl
co
m
SOL 1.2.13
= 8 + 0 + 24 = 32 uints
2
2
3
2
= b12 x l = (4x3) 20 3 0 = 3 # 8 = 24 uints Option (C) is correct. For the closed path defined, the line integral in forward path = 32 units the line integral in return path = - 32 units. So, total integral in the closed path is : = 32 - 32 = 0 units
w. g
# A : dl SOL 1.2.15
2
at e
SOL 1.2.14
# x dx
he
= 12
Option (A) is correct. The circulation of A around the path L can be given as
# # #
=c + +
ww
# A : dl
1
2
3
m A : dl
where the route is broken into segments numbered 1 to 3 as shown in figure below :
1st segment : ( f = 30c, z = 2 , 0 # r # 5 )
and
dl = dra r
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
37
For View Only
Shop Online at www.nodia.co.in r ( f = 30c) So, dr = 25 A : dl = r sin fdr = 2 4 1 0 0 2 nd segment : ( r = 5 , z = 2 , 30c # f # 180c) and dl = dfaf
#
#
# A : dl
So,
2
=
5
#
5
# 0 df = 0
3rd segment : ( f = 180c, z = 2 , 5 # r # 0 )
# A : dl 3
=
and
dl = dra r
0
# r sin fdr = 0
( f = 180c)
5
#
Total
Option (D) is correct. Volume integral of the function is given by V =
### f dxdydz
=
co
SOL 1.2.16
m
Therefore, the circulation of vector field along the edge L is A : dl = 25 + 0 + 0 = 25 units 4 4
### 30z dxdydz 2
(- 1 - y - z)
0
he
# dx
lp.
The surface of the tetrahedron will have a slope x + y + z =- 1 So, for a given value of y and z , x varies from 0 to (- 1 - y - z) and x integral will be =- 1 - y - z
again for a given value of z , y ranges from 0 to (- 1 - z). So y -integral will be : (- 1 - z)
2 (- 1 - z)
0
ww w. ga te
# (- 1 - y - z) dy = ;(- 1 - z) y - y2 E
0
(- 1 - z) 2 (- 1 - z) 2 = (- 1 - z) = 2 2 2 = 1 + z3 + z 2 2 Now there is only one remaining variable z that ranges from - 1 to 0. So we have the volume integral of the function as 2 0 V = 30z2 b 1 + z + z l dz 2 2 -1 3 4 5 0 = 30 :z + z + z D 6 4 10 -1 1 = 30 :0 + - 1 + 1 D 6 4 10 = 30 # 1 = 3 20 2 2
#
SOL 1.2.17
Option (D) is correct. The net outward flux through the closed cylindrical surface will be summation of the fluxes through the top(in az direction), bottom(in - az direction) and the curved surfaces(in a r direction) as shown in the figure.
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
38
Vector Analysis
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co
m
For View Only
Chap 1
= =
z=0 f=0 2
r2 cos2 f c m^rdfdz h 4 f=0
# # z=0
r
2p
2 ^2 h3 2p 2 cos fdfE; # dz E # ; 4 0 0 = 2 # p # 2 = 4p
A : dS =
Option (A) is correct. According to divergence theorem surface integral of a vector field over a closed surface is equal to the volume integral of its divergence inside the closed region:
#
w. g
SOL 1.2.18
#
2p
2
at e
At r = 2 ,
# # ^A h^rdfdz h
he
# A : dS
lp.
Since, the vector field has no z -component so, the outward flux through the top and bottom surfaces will be zero. Therefore, the total outward flux through the closed cylindrical surface will be only due to the field component in a r direction(flux through the curved surfaces) which is given as
#
ww
i.e. A : dS = (d : A) dv Divergence of vector A is d : A = 1 2 ^rA rh + 1 2 Af + 2 Az r 2r r 2f 2z = 1 2 (r (4r + 2r sin2 f)) + 1 2 ^r sin 2fh + 2 (6z) r 2r r 2f 2z 2 = 8 + 4 sin f + 2 cos (2f) + 6 = 8 + 4 sin2 f + 2 cos2 f - 2 sin2 f + 6 = 16 So the surface integral is
# A : dS
=
# (d : A) dv = ### (16) rdrdfdz
= 16
#
0
2
rdr
p/2
5
# df # dz 0
0
= 16 # 2 # p # 5 2
= 80p GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
For View Only
39
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Note : The surface integral can also be evaluated directly without using divergence theorem but it will be much complicated as there are 5 different surfaces over which we will have to integrate the given vector field. Option (C) is correct. According to stoke’s theorem, line integral of a vector function along a closed path is equal to the surface integral of its curl over the surface defined by the closed path. i.e. G : dl = ]d # Gg dS
#
#
L
m
SOL 1.2.19
L
#
## = 6 # # x dydx + 6 # # x dydx = 6 # x xdx + 6 # x (2 - x) dx 1
0
1
2
1
2
2
0
2-x
2
2
0
he
0
x
lp.
#
co
Curl of the vector field is d # G =- 6x2 az and the differential surface vector dS = dxdy (- az ) So the line integral of the given vector field is G : dl = ]d # Gg dS =- - 6x2 dxdy
2
1
4 2 = 6 :x D + 6 :2x - x D 3 4 0 4 1
SOL 1.2.20
3
ww w. ga te
4 1
= 6 b 1 - 0 l + 6 ;b 16 - 16 l - b 2 - 1 lE = 7 units 3 3 4 4 4 Option (D) is correct. The relationship between cartesian and spherical co-ordinates is : r = x2 + y2 + z2 , r sin q = x2 + y2 x = r sin q cos f , y = r sin q sin f We put these values in the given expression of vector field as x 2 + 4y 2 + 4 z 2 x y y x F = 9 2 ax + 2 ay - 2 ax + 2 ayC 2 2 x +y
2r r sin q (cos f - sin f) a + (cos f + sin f) a x y@ r sin q 2 6 = r 6(cos f - sin f) ax + (cos f + sin f) ay@ Now we transform the vector field from cartesian system to spherical system : VR V R V R SFr W S sin q cos f sin q sin f cos qWSFxW SFqW = Scos q cos f cos q sin f - sin qWSFyW W SSF WW SS - sin q cos f 0WSSFz WW f T X T T X Fr = r ^cos f - sin fh^sin q cos fh + rX^sin q sin fh^cos f + sin fh = r sin q Fq = r ^cos f - sin fh^cos q cos fh + r ^cos f + sin fh cos q sin f GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia =
40
Vector Analysis
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S1
=
2p
# # 0
=8
30c
0 2p
r3 sin2 qdqdf
# df # sin qdq = 4p;p3 0
30c
2
0
= 2.276 = 2.3
(at r = 2 )
Option (B) is correct. The vector function in spherical form as calculated in previous question is : F = r sin qar + r2 cos qa q + raf The differential surface vector over the surface S2 is dS = r sin qdfdra q and the surface S2 is defined in the region 0 # r # 2 , 0 # f # 2p, q = 30c So, the surface integral of the field over the surface S2 is : 2p 2 F : dS = r2 sin q cos qdfdr = 4p 3 3 S 0 0
# #
2
Option (C) is correct. For a vector function to be irrotational its curl must be zero. Now we check it for vector A. d # A = ; 2 (x - 3y - z) - 2 (- 3z)E ax + : 2 (x + z) - 2 (x - 3y - z)D ay 2y 2z 2z 2x +; 2 (- 3z) - 2 (x + z)E az 2x 2y = (- 3 + 3) ax + (1 - 1) ay + (0 - 0) az = 0 So, vector A is irrotational. again for a vector to be solenoidal its divergence must be zero. So we take the divergence of the vector A as d : A = 2 (x + z) + 2 (- 3z) + 2 (x - 3y - z) 2x 2y 2z = 1+0-1 = 0 So, vector A is solenoidal Thus the vector A is both irrotational and solenoidal. Note: Since the curl of the gradient of a scalar field is zero. So, we can have directly the result 4# A = 4# (- 4 f)=-= 4# (4 f)= 0
ww
w. g
SOL 1.2.22
at e
#
he
lp.
SOL 1.2.21
3 2 E
co
# F : dS
m
= r cos q Ff = r ^cos f - sin fh^- sin fh + r ^cos f + sin fh cos f =r i.e. F = r sin qar + r cos qa q + raf The differential surface vector over the surface S1 is dS = r2 sin qdqdfar and the surface S1 is defined in the region r = 2 , 0 < q < 30c, 0 < f < 2p So, surface integral through out the surface S1 will be :
Option (A) is correct. We have GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 1.2.23
Chap 1
Vector Analysis
41
# A : dl
=
# 3ydx + # 3xdy
co
So,
lp.
SOL 1.2.24
Shop Online at www.nodia.co.in 2f 2f 2f a a A =- df =- ax 2x 2y y 2z z Comparing it with the given vector we get : 2 2f =- (x + z) & f =- x - xz + f1 (y, z) 2 2x 2f = 3z & f = 3yz + f2 (x, z) 2y 2 2f =- (x - 3y - z) & f =- xz + 3yz + z + f 3 (x, y) 2 2z In conclusion, from all the three results, we get 2 2 f =- x - xz + 3yz + z 2 2 Option (D) is correct. Given the vector field, A = 3 (yax + xay) The differential line vector in the cartesian coordinate system is dl = dxax + dyay + dzaz
m
For View Only
The given curve is, y = x/2 So, we put x = 2y2 and dx = 4ydy in the line integral 2 2 A : dl = 12y2 dy + 6y2 dy = 18 6y3@12 3 1 1 = 6 # 5 = 30 units Option (B) is correct.
1 + cos 2f 2 cos2 f ar = ar 3 r r3 and the differential surface vector over the outer spherical surface is (for r = 2 ,0 # i # r,0 # f # 2p) dS = (r2 sin qdqdf) ar So the surface integral over the outer spherical surface is 2p p 2 cos 2 f 2 F : dS = c m^r sin qdqdfh =- 2p 3 r 0 0 Option (D) is correct. Consider that the vector A is in ax direction as shown in the figure. Given the vector field
#
SOL 1.2.26
#
ww w. ga te
SOL 1.2.25
#
he
#
F =
# #
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
42
Vector Analysis
Chap 1
w. g
Option (A) is correct. We go through all the options to check the direction of the vector R for the corresponding directions of A, B and C . Option (A)
ww
SOL 1.2.27
at e
he
lp.
co
m
For View Only Shop Online at www.nodia.co.in So we can write the vectors in cartesian form as ^A = 4h A = 4ax and B = B cos 30cax + B sin 30c ^- ay h = 3 3 a x - 3 ay ^B = 3h 2 2 Now the resultant vector, R = 6A - 8B = 3.22ax + 6ay So, R = ^3.22h2 + ^12h2 = 12.43 units and angle that R makes with x -axis is q = cos-1 b 3.22 l 12.43 = 75c So the graphical representation of vector R is
Since the direction of cross product is normal to the plane of vectors and determined by right hand rule. So B # C has the direction in which thumb indicates when the curl of the finger directs from B to C . Thus B # C will be directed out of the paper and so we get direction of A # ^B # C h toward east. So the given direction GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
For View Only of R is incorrect.
43
Shop Online at www.nodia.co.in
co
m
In Option (B) : Direction of ^B # C h is out of the paper so, R = A # ^B # C h will be directed toward west.
he
lp.
In Option (C): Direction of ^B # C h is into the paper so, R = A # ^B # C h will be directed toward north.
SOL 1.2.28
ww w. ga te
In Option (D) : Direction of ^B # C h is into the paper so, R = A # ^B # C h will be directed toward south. So the given direction is correct.
Option (A) is correct. As the vectors B and C are defined in cylindrical system. So, we transform the vector in cartesian form as below Given the vector field B = a r + a f + 3a z the cylindrical components B r = 1, B f = 1, B z = 3 So the cartesian components of vector B is R V R VR V SBxW Scos f - sin f 0WSB rW SByW = Ssin f cos f 0WSBfW SS WW SS WS W Bz 0 0 1WSBz W T BX = Tcos fB - sin fBXT =X cos f - sin f ^B r = Bf = 1h x r f By = sin fB r + cos fBf = sin f + cos f ^B r = Bf = 1h Bz = Bz = 3
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
44
Vector Analysis
Chap 1
lp.
co
m
For View Only Shop Online at www.nodia.co.in and so the vector field in cartesian system is B = ^cos f - sin fh ax + ^sin f + cos fh ay + 3az So at the point a2, p , 3k 2 B =- ax + ay + 3az now we transform the vector field C = 2 a r + 3az in cartesian system. the cylindrical components, C r = 2 , Cf = 0 , Cz = 3 So the cartesian components of vector C is R V R VR V SCxW Scos f - sin f 0WSC rW SCyW = Ssin f cos f 0WSCfW SS WW SS WS W Cz 0 0 1WSCz W T X T XT X So, Cr = 2 Cx = 2 cos f Cy = 2 sin f and Cz = Cz = 3 C = 2 cos fax + 2 sin fay + 3az
he
So, at the point b 3, 3p , 9 l 4
2 cos b 3p l ax + 2 sin b 3p l ay + 3az 4 4 =- ax + 2ay + 3az So all the three vectors are same at their respective points.
at e
C =
Option (D) is correct. For checking whether a vector is perpendicular to a given vector or not we take their dot product as the dot product of the two mutually perpendicular vectors is always zero. Now we have A + B = 4ax + 4ay + 4az So we take the dot product of (A + B) with the all given options to determine the perpendicular vector. In option (A). ^- 4ax + 4ay h : ^4ax + 4ay + 4az h =- 16 + 16 = 0 In Option (B) ^4ay + 4az h $ ^4ax + 4ay + 4az h = 16 + 16 = 32 ! 0 In Option (C) ^ax + az h $ ^4ax + 4ay + 4az h = 4 + 4 = 8 ! 0
SOL 1.2.30
Option (C) is correct. The given gradient is dV ^x, y, z h = 1.5x2 yz2 ax + 0.5x3 z2 ay + x3 yzaz So,we have the components as, 2V = 1.5x2 yz2 2x
ww
w. g
SOL 1.2.29
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 1
Vector Analysis
For View Only
45
Shop Online at www.nodia.co.in 3
m
V = 1.5 x yz2 + f1 ^y, z h = 0.5x3 yz2 + f1 ^y, z h 3 2V = 0.5x3 z2 2y V = 0.5x3 yz2 + f2 ^x, z h 2V = x3 yz 2z x3 yz2 V = + f 3 ^y, z h = 0.5x3 yz2 + f 3 ^y, z h 2 Thus by comparing all the results we get, V = 0.5x3 yz2 Option (B) is correct. Consider the given plane xyz = 1 xyz - 1 = 0 So, function, f = xyz - 1 gradient of function, df = yzax + xzay + xyaz Since gradient of the function of a plane is directed normal to the plane so the normal vector to the plane at the point ^2, 4, 18 h is df = 1 ax + 1 ay + 8az 2 4 Now consider ^x, y, z h lies in the given surface xyz = 1. So the tangential vector to the given surface at the point ^2, 4, 18 h is T = ^x - 2h ax + ^y - 4h ay + bz - 1 l az 8 This vector will be perpendicular to df . So, (dot product of perpendicular vector) ^T h : ^df h = 0 1 x - 2 + 1 ^y - 4h + 8 z - 1 = 0 b h 4 2^ 8l 2x + 4y + 32z = 24
SOL 1.2.32
Option (A) is correct. As the integral is to be determined in spherical volume so, we transform the function in spherical system as, 2x = 2r sin q cos f and so, we have the integral
ww w. ga te
he
lp.
co
SOL 1.2.31
1
p/2
# 2xdv = 2 # # # v
r=0 q=0 1 3
= 2;
#
0
p/2
^r sin q cos fh^r2 sin qdrdqdfh
f=0 p/2
r dr E;
#
0
p/2
sin2 qdqE; cos fdfE
#
0
4 1 p/2 = 2 ;r E :q - sin 2q D 6sin f@0p/2 4 0 2 4 0 = 2 # 1 #apk = 2 # p = p 8 4 4 4
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
46
Vector Analysis
Chap 1
In Segment (1) So,
co
m
For View Only Shop Online at www.nodia.co.in SOL 1.2.33 Option (B) is correct. Contour integral of the field vector is evaluated in 3 segments as shown below
dl = dra r
# A : dl 1
=
2 2
# r cos fdr = ^cos 0h;r2 E 2
r=0
0
= 4 = 2 unit 2 0 < f # p/2 at r = 2
dl = rdfaf p/2 r p/2 So, A : dl = ^rdfh = 2 6f@0 = p 1 f=0 2 In segment 3 dl =- dra r 3
#
=-
2
0 # r # 2 , at f = p/2
^r cos p/2h^drh = 0
r=0
at e
# A : dl
he
#
#
So,
lp.
In segment (2)
0 < r # 2 at f = 0
So the contour integral is
# A : dl C
1
2
3
Option (A) is correct. For the given contour C1 dl = rdfaf So,
w. g
SOL 1.2.34
# # # E6A : dl@ = ^2 + ph units
=; + +
# A : dl C1
=
# r^rdfh = 9 # 2p = 18p f=0
ww
and for the contour C2 dl =- rdfaf So,
# A : dl C2
0 # f # 2p at r = 3
2p
=-
0 # f # 2p at r = 1
# r^rdfh =- 2p 2p
f=0
Therefore, the ratio of the contour integral is
# A : dl # A : dl C1
= 18p =- 9 ^- 2ph
C2
SOL 1.2.35
Option (A) is correct. Let us consider a contour abcd as shown in the figure.
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abcd
=c
# +#+# +# ab
bc
cd
da
m A : dl
co
# A : dl
m
As vector A has only a q component so its integral will not exist along segments ab and cd and so the contour integral for abcd is
#
SOL 1.2.36
SOL 1.2.37
#
he
#
ww w. ga te
#
lp.
For bc segment, r = 1 and 0 # q # p 2 dl =- rdqa q and for da segment r = d , and 0 # q # p/2 dl = rdqa q p/2 p/2 e-r e-r So, A : dl =b r l^rdqh + b r l^rdqh abcd q=0 q=0 -1 p -d =-^e ha k + ^e h^p/2h = p ^- e-1 + e- d h 2 2 As for the given contour C , d tends to zero So, A : dl = lim A : dl = lim p ^- e-1 + e- d h = p ^1 - e-1h 2 d " 0 abcd d"0 2 C Note : Most of the students do a mistake here by directly integrating the given vector along given contour C but as the vector A includes exponential which is not zero at origin and so at r = 0 , ^Ah : ^rdqafh ! 0 therefore we have taken the contour integral in the form of limits.
#
Option (C) is correct. The divergence of unit vector ar is d : ar = 12 2 r2 ^1 h = 12 ^2r h = 2 r r 2r r the divergence of unit vector a q is d : a q = 1 2 ^sin fh = cos q = cot q r r sin q 2q r sin q and the divergence of unit vector af is d : a f = 1 2 ^1 h = 0 r sin q 2f Option (A) is correct. A vector can be expressed as the gradient of a scalar if it’s curl is zero. Now we go through the options.
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48
Vector Analysis
Shop Online at www.nodia.co.in a x ay a z Curl of the vector = 22x 22y 22z = 0 2yz 2xz 2xy
Option (A),
Option (B),
Curl of the vector = 1 r
Option (C),
Curl of the vector = 1 r
2 2f e- r r
2 2z
0 r^ h 0
2 r2
!0
ar
ra f
az
2 2r
2 2f
2 2z
cos f
2 r
sin f
=0
0
Option (A) is correct. Any vector for which divergence is zero can be expressed as the curl of another vector. For checking it we go through all the options. In Option (A), Divergence = 2 & 1 _x2 - y2i0 - 2 ^xy h + 2 ^2 h 2x 2 2y 2z = x-x = 0 -r In Option (B), Divergence = 1 2 c e m = 0 r 2f r
he
lp.
SOL 1.2.38
2 2r
co
So, it can be expressed as gradient of a scalar.
a r ra f a z
m
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at e
q + 1 2 sin2 q In Option (C), Divergence = 12 2 br2 2 cos r 2r r3 l r sin q 2q c r3 m =- 14 ^2 cos qh + 4 1 2 sin q cos q = 0 r r sin q So all the vectors can be expressed as curl of another vector. Option (B) is correct. for y > 0 i.e. above x -axis field will be directed towards + ax direction and will increase as we go far from the x -axis, since y -increases. For y < 0 i.e. below x -axis, field will be directed towards - ax direction and it’s intensity will increases as we go away from the x -axis.
SOL 1.2.40
Option (A) is correct. Given the divergence of the vector field is zero i.e. 4: A = 0 2Ax + 2Ay = 0 2x 2y 2Ax =-2Ay 2x 2y and the curl of the vector field is zero, i.e. 4# A = 0 a x ay a z 2 2 2 2x 2y 2z = 0 A x Ay 0
ww
w. g
SOL 1.2.39
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SOL 1.2.41
ww w. ga te
he
lp.
co
m
2A 2A - y ax + 2Ax ax + e y - 2Ax o az = 0 2z 2z 2x 2y 2Ay 2Ax ....(2) =0 2x 2y (Since A is only the variable of x and y . So the differentiation with respect to z will be zero). Differentiating equation (ii) with respect to x we get, 22Ay 22Ax =0 2x2 2x2y 22Ay 2 2Ax = 0 2 2 y b 2x l 2x 22Ay 2 - 2Ay = 0 (from equation (i)) 2 2y e 2y o 2x 22Ay 22Ay =0 + 2x2 2y2 d2Ay = 0 Again differentiating equation (ii) with respect to y we get 22Ay 22Ax =0 2x2y 2y2 2 2Ay - 22Ax = 0 2x e 2y o 2y2 2 - 2Ax - 22Ax = 0 (from equation (i)) 2x b 2x l 2y2 22Ax + 22Ax = 0 2x2 2y2 Option (B) is correct. The line integral (circulation) of force F around the closed path can be divided in four sections as shown below.
For segment 1 we have,
y =z=0 dl =- dxax ,
0> distance d ) is proportional to (A) 1/r (B) 1/r 2
MCQ 2.3.46 IES EE 2004
(D) 1/r 4
lp.
(C) 1/r 3
What is the value of total electric flux coming out of a closed surface ? (A) Zero
he
(B) Equal to volume charge density
(C) Equal to the total charge enclosed by the surface
IES EE 2003
MCQ 2.3.48
A potential field is given by V = 3x 2 y - yz . Which of the following is not true ? (A) At the point (1, 0, - 1), V and the electric field E vanish
ww
IES EE 2002
A charge is uniformly distributed throughout the sphere of radius a . Taking the potential at infinity as zero, the potential at r = b < a is b b Q Qr (A) - # (B) - # 2 dr 3 dr 3 4pe0 a 3 4pe0 r b a a Q Qr Q (C) - # (D) - # 2 dr - # 2 dr 3 dr a 4pe0 a 3 4pe0 r 3 4pe0 r
w. g
MCQ 2.3.47
at e
(D) Equal to the surface charge density
(B) x 2 y = 1 is an equipotential plane in the xy -plane (C) The equipotential surface V =- 8 passes through the point P (2, - 1, 4) (D) A unit vector normal to the equipotential surface V =- 8 at P is (- 0.83x + 0.55y + 0.07z) MCQ 2.3.49 IES EE 2002
The relation between electric intensity E , voltage applied V and the distance d between the plates of a parallel plate condenser is (A) E = V/d (B) E = V # d (C) E = V/ (d) 2
(D) E = V # (d) 2
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SOLUTIONS 2.1
Option (D) is correct. Since all the charges are exactly equal and at same distance from the centre. So, the forces get cancelled by the diagonally opposite charges and so the net force on the charge located at centre is Fnet = 0 N
SOL 2.1.2
Option (A) is correct. Since one of the four charges has been removed so, it will be treated as an additional - 2 C charge has been put on the corner, so the force due to the additional charge will be : (- 2) # (+ 1) # 10-9 F = k (1) 2 9 = - 9 # 10 # 2 # 10-9 = 18 N and so the net force experienced by the charge located at center is Fnet = 18 + 4 = 22 N
SOL 2.1.3
Option (A) is correct. Since the two point charges are positive so the introduced third point charge must be negative as to make the entire system in equilibrium as shown below
ww w. ga te
he
lp.
co
m
SOL 2.1.1
as the system must be in equilibrium so the force between all the pair of charges will be equal i.e. FAB = FCB = FAC ^9 h q ^36h q ^36h^9 h = 2 = d2 ^3 - d h ^3h2 Solving the equation we get, and q =- 4 C d= 2m SOL 2.1.4
Option (D) is correct. Electric field intensity at any point P due to the two point charges Q1 and Q2 is defined as Q Q2 R2 E = k e 1 3 R1 + (R1) (R2) 3 o
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For View Only Shop Online at www.nodia.co.in where, R1 and R2 is the vector distance of the point P from the two point charges. So the net electric field due to the two given point charges is 9 # 109 # (- 5) # 10-9 6(- 7 + 4) ax + (3 - 0) ay + (- 1 + 2) az@ E = (- 7 + 4) 2 + (3 - 0) 2 + (- 1 + 2) 2
=
9 # 109 # 2 # 10-9 6(- 7 + 5) ax + (3 - 0) ay + (- 1 - 3) az@ (- 7 + 5) 2 + (3 - 0) 2 + (- 1 - 3) 2
- 45 6- 3ax + 3ay + az@ 3/2
+
18 6- 2ax + 3ay - 4az@
19 = 1.4ax - 1.284ay - 1.004az
co
Option (B) is correct. From the positions of the three point charges as shown in the figure below, we conclude that the electric field intensity due to all the point charges will be directed along az .
w. g
at e
he
lp.
SOL 2.1.5
293/2
m
+
So the net electric field intensity produced at the point P due to the three point charges is Q (where R is the distance of point P from the charge Q ) E = a 4pe0 R R Q 1 1 (aR = az ) a = + 1 + 4pe0 ;(3 + 1) 2 (3) 2 (3 - 1) 2 E z = 5 # 10-9 # 9 # 109 # : 1 + 1 + 1 D az = 29.0625az 16 9 4 SOL 2.1.6
ww
/
Option (C) is correct. Since, both the point charges are positive, so the point P must be located on the line joining the two charges as shown in figure.
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Electrostatic Fields
85
co
Option (D) is correct. Given the volume charge density, rv = 2 mC = 2 # 10-6 C So the total charge present throughout the shell is defined as the volume integral of the charge density inside the region: i.e.
# r dv = # # #
Q =
v
p
0.03
(2 # 10-6) (r2 sin qdrdqdf)
he
2p
lp.
SOL 2.1.7
m
For View Only Shop Online at www.nodia.co.in Given the net electric field intensity at point P is zero i.e. SE = 0 Since, the direction of electric field intensity due to the two charges will be opposite 2q 1 q 1 So, : 4pe0 x2 D - = 4pe0 (1 - x) 2 G = 0 2x2 = (1 - x) 2 2 x + 2x - 1 = 0 x - 2 ! 4 + 4 =- 1 ! 2 2 x = 0.414 and x = - 3.414 As discussed above the point P must be located between the two charges, so we have the distance of point P from charge + Q as: x = 0.414 m
f = 0 q = 0 r = 0.02
3 0.03
SOL 2.1.8
ww w. ga te
= :4p (2 # 10-6) # r D 3 0.02 -10 = 1.6 # 10 = 160 pC
Option (A) is correct. The charge located in the region 2 cm < r < a is Q q = = 1 # 160 = 80 pC 2 2 Similarly as calculated in previous question we have
# r dv 80 pc = # # # q =
or
v
2p
p
0.03
f = 0 q = 0 r = 0.02
(2 # 10-6) (r2 sin qdrdqdf) 3 a
80 # 10-12 = :4p # 2 # 106 # r D 3 0.02 1/3 -12 therefore, a = ; 3 # 80 # 10 -6 + (0.02) 3E = 4.59 cm = 4.6 cm 4p # 2 # 10
or
SOL 2.1.9
Option (D) is correct. Charge density in a certain region is defined as the charge per unit volume. Since the net charge in the subregion = 30% of the electronic charge net charg e So the charge density = volume
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Shop Online at www.nodia.co.in 30 -19 # (- 1.6 # 10 ) 100 = 10-12 =- 4.8 # 10-8 =- 48 nC/m3
Option (D) is correct. Given the surface charge density rS = r2 z So the total charge distributed over the cylindrical surface is ,
# r dS = # # (r z)(rdfdz)
Q =
S
2p
1
2
z=0 f=0 2 1
m
SOL 2.1.10
lp.
Q =
# r dS S
=
2
# #
-2x + 5
x=1 y=1
^3xy h dxdy = 6.5 C
Option (B) is correct. Total stored charge on the disk is evaluated by taking surface integral of the charge density. i.e.
at e
SOL 2.1.12
at r = 2
Option (D) is correct. Given the surface charge density rS = 3xy C/m2 So, total stored charge on the triangular surface is
he
SOL 2.1.11
co
= 8 # :z D # 6f@20p 2 0 1 = 8 # # 2p = 8p = 35.1 mC 2
(dS = rdfdz )
Q =
# r dS s
=
# (3r)^2prdr h 5
0
Option (C) is correct. For an electric field to exist, the its curl must be zero. So, we check the existence of the given field vector first. Given the electric field intensity E = 2xyax + 4yzay + 6xzaz V/m So,
ww
SOL 2.1.13
w. g
3 5 = 6p :r D = 350p 3 0
d#E = 2
a x ay
az
2 2x
2 2z
2 2y
xy 2yz 3xz = 2 6- 2yax - 3zay - xaz@ ! 0 Therefore, as the curl of the given electric field is not equal to zero so, the field does not exist. Option (B) is correct. Electric field intensity in free space at a distance R from an infinite line charge with charge density rL is defined as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 2.1.14
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rL R 2pe0 R2 Given rL = 1 mC/m = 1 # 10-6 C/m R =- 2ax - ay (1 10-6) - 2ax - ay So, E = # l =- 7.2ax - 3.6ay kV/m 2pe0 b 5 E =
Option (A) is correct. Electric flux density in a certain region for the given electric field intensity is defined as D = e0 E = e0 (x2 ax + 2xyay) So at the point (- 1, 0, 1) D = e0 (ax )
SOL 2.1.16
Option (B) is correct. According to Gauss law the total outward electric flux from a closed surface is equal to the charge enclosed by it
# D : dS = Q
enc
lp.
y =
i.e.
co
m
SOL 2.1.15
SOL 2.1.17
ww w. ga te
he
So when the charge enclosed by the volume is zero then the net outward flux is zero, or in other words, the net electric field flux emanating from an arbitrary surface not enclosing a point charge is zero. Now, the electric field intensity outside a charged sphere having total charge Q is determined by treating the sphere as a point charge Q i.e. E = ar 4pe0 r2 where r is distance of the point form center of sphere and ar is it’s radial direction. So the electric field intensity at any point outside the charged sphere is not zero. Therefore, Assertion(A) is true but Reason(R) is false. Option (C) is correct. E = 3r2 ar According to Gauss’s law the total charge stored in a closed surface is equal to the surface integral of its flux density over the closed surface. i.e.
# D : dS = e # E : dS = e # (3r a ) dS
Qenc =
0
2
0
2
r
= e0 (3r ) (4pr2) = e0 # 3 # 4p # 2 4 = 5.3 # 10-9 = 6.3 nC
SOL 2.1.18
#
( dS = 4pr2 ar ) r = 2m
Option (D) is correct. According to Gauss law net outward electric flux from any closed surface is equal to the total charge enclosed by the volume i.e. y = Qenc
#
or, y = rv dv GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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1
2p
1 (r2 sin qdrdqdf) 2l b r q=0
# # # r=0 f=0
p
= 1 # 2 # 2p = 4p C Option (A) is correct. As we have already determined the total electric flux crossing the surface r = 1 m So, electric flux density D at r = 1 m is evaluated as below: y =
Total electric flux So we have
# D : dS
# D : dS
= 4p D (4pr ) = 4p D = 12 = 1 C/m2 r D = 4ar C/m2
Thus
Option (D) is correct. As the point charge is located at origin. So flux due to it will be emanating from all the eight quadrants symmetrically. So the flux through the portion of plane x + y = 2 m lying in first octant is 1/8 of the total flux emanating from the charge located at origin. and from Gauss law, total flux = Qenc = 8 C Q So, flux through the surface x + y = 2 m is enc = 8 = 1 C 8 8 Option (D) is correct. We construct a Gaussian surface at r = r as shown in figure.
ww
w. g
SOL 2.1.21
at e
he
lp.
SOL 2.1.20
( y = 4p )
co
2
m
SOL 2.1.19
So, according to Gauss law the total outward flux through the surface r = r will be equal to the charge enclosed by it. i.e. (assume the height of the cylinder is h ) D ^2prh h = rv ^pr2 h h r So, D = rv 2 Therefore the electric field intensity at a distance r from the cylindrical axis is r E = D = v ar k e0 e0 2 Thus E\r SOL 2.1.22
Option (B) is correct. According to Gauss law the surface integral of the electric flux density over a closed
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# D : dS # E : dS
= Qenc (since E = D ) or = 1 Qene e0 e0 As we have to evaluate E for r # 2 and since the charge density is zero for r # 2 so (for r # 2 ) Qenc = 0 1 Therefore, E : dS = # 0 e0 E =0
i.e.
Option (D) is correct. Again from Gauss law, we have the surface integral of electric field intensity over the Gaussian surface at r = 3 as E : dS = 1 Qenc e0 E : dS = 1 rv dv = 1 0dv + 1 ^4/r2h dv e0 e0 e0 1 4r2 4 3 1 4 44 2 4 44 3 21r#3 #2 3 2p p 4 (r2 sin qdrdqdf) E (4p # (3) 2) = 1 e0 r = 2 0 0 b r2 l E (4p # 9) = 4p # 4 (3 - 2) e0 E = 4 ar 9e0
co
SOL 2.1.23
m
#
# #
lp.
#
#
#
SOL 2.1.24
ww w. ga te
he
# # #
Option (C) is correct. As calculated in the previous question, we have the surface integral of the electric field intensity over the Gaussian surface r = 5 as E : dS = 1 rv dv = 1 0dv + 1 ^4/r2h dv + 1 0dv e0 e0 e0 e0 1 4r2 4 3 1 4 44 2 4 44 3 1 4 243 21r#4 41r#5 #2 4 2p p 4 (r2 sin qdrdqdf) E (4p # (5) 2) = 1 e0 r = 2 0 0 b r2 l 4 E (100p) = 4p # 4 dr e0 2 12 E = a 25e0 r
#
#
#
#
#
# # # #
SOL 2.1.25
Option (D) is correct. According to Gauss’s law rv = edE So when the field intensity is uniform dE = 0 and rv = edE = 0 So no charge can be present in a uniform electric field.
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Chap 2
For View Only Shop Online at www.nodia.co.in SOL 2.1.26 Option (D) is correct. According to Gauss law the volume Charge density in a certain region is equal to the divergence of electric flux density in that region i.e. rv = d : D
Option (B) is correct. According to Gauss law the volume Charge density in a certain region is equal to the divergence of electric flux density in that region. i.e. rv = d : D = 2x So total charge enclosed by the cube is Q =
# r dv v
=
lp.
co
SOL 2.1.27
m
q = 12 2 br2 cos3 q l + 1 2 b sin q sin 3 l q r sin 2 2 q r r r 2r =- 14 cos q + 14 cos q r r =0
2
1
2
# # # (2x) (dxdydz) 0
0
-1
SOL 2.1.28
he
= 4 # 2 # 2 = 14 C
Option (D) is correct. Net electric potential due to two or more point charges is defined as : Q V = 4pe0 R So, the electric potential at point P due to the two point charges is Q1 Q2 + V = 4pe0 R1 4pe0 R2 where Q1 =+ 1 mC , Q2 =- 1 mC and R1, R2 are the distance of the point P from the two point charges respectively. So, we have R1 = (- 3 - 0) 2 + (0 - 0) 2 + (- 4 - 1) 2 = 5.83 R2 = (- 3 - 0) 2 + (0 - 0) 2 + (- 4 + 1) 2 = 4.24 -6 Thus V = 10 : 1 - 1 D =- 578.9 V 4pe0 5.83 4.24
ww
w. g
at e
/
SOL 2.1.29
Option (A) is correct. Electric field at any point is equal to the negative gradient of potential i.e. E =- dV =-c 2 V + 2 V + 2 V m 2x 2y 2z 6y 3x 3 ay =-=cy2 z3 + 2 2 2 m a x + c 2xyz + 2 x + 2y + 3z x + 2y2 + 3z2 m 9z +c 3xy2 z2 + 2 az x + 2y 2 + 3 z 2 m G So, at the point P (x = 3, y = 2, z =- 1) E = 3.6ax + 11.4ay - 35.6az V/m
SOL 2.1.30
Option (D) is correct. Electric flux density in terms of field intensity is defined as
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D = e0 E So, at point P (3, 2, - 1), D = e0 (3.6ax + 11.4ay - 35.6az ) = 31.4ax + 101ay - 314.5az pC/m2 Option (C) is correct. Laplace’s equation for a scalar function V is defined as d2 V = 0 but at the point of maxima d2 V must have a negative value while at the point of minima d2 V must have a positive value. So the condition of maxima/minima doesn’t satisfy the Laplace’s equation, therefore the potential function will have neither a maxima nor a minima inside the defined region.
SOL 2.1.32
Option (D) is correct. Electric force experienced by a point charge q located in the field E is defined as F = qE So, the force applied at the point charge + 1 C located at (0, y, 0) is Qd (q =+ 1 C ) F = (1) 62 cos2 qar + sin qaq@ 4pe0 r3 Qd ( q = 90c, a q =- az , r = y ) = 6sin 90c (- az )@ 4pe0 r3 - Qd az = 4pe0 y3
he
lp.
co
m
SOL 2.1.31
ww w. ga te
***********
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SOLUTIONS 2.2
m
Option (A) is correct. For determining the position of the third charge, first of all we evaluate the total electric field at the given point C (0,1,0) due to the two point charges located at points A(1,0,0) and B (- 1, 0, 0 ) respectively as shown in figure.
at e
he
lp.
co
SOL 2.2.1
ww
w. g
Electric field due to the charge located at point A is ^ax + ay h 9 a +a E1 = kQ AC 3 = 9 # 109 b 1 l # 10-9 # ^ x yh 3 = 2 AC 4 2 ^ 1 + 1h and the electric field due to charge at point B is ^- ax + ay h ^BC h 9 -a + a 1 9 -9 E2 = kQ ^ x # yh 3 = 9 # 10 # b 2 l # 10 3 = BC 4 2 ^ 1 + 1h So, E1 + E2 = 9 ^ax + ay h + 9 ^- ax + ay h = 9 ay 2 2 4 2 4 2 As the field is directed in ay direction so for making E = 0 the third charge of + 2 nC must be placed on y -axis at any point y > 1. Consider the position of the third charge is (0, y, 0). So, electric field at point C due to the third charge is. 9 109 # ( 2 ) # 10-9 E3 = # (- ay)=- 9 2 2 ay (y - 1) 2 (y - 1) and since the total electric field must be zero So, we have E1 + E 2 + E 3 = 0 9 a - 9 2 a =0 y y (y - 1) 2 2 2 (y - 1) 2 = 4 or y = 3 , - 1 as discussed above y > 1, so the point will be located at y = 3 i.e. Point P will have the coordinate (0, 3, 0) GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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93
Option (A) is correct. Horizontal component of the electric field intensity will be cancelled due to the uniform distribution of charge in the circular loop. So the net electric field will have only the component in az direction and defined as below :
ww w. ga te
he
SOL 2.2.3
lp.
co
m
For View Only Shop Online at www.nodia.co.in SOL 2.2.2 Option (A) is correct. Electric field intensity at any point P due to the uniformly charged plane with charge density rS is defined as r E = s an 2e0 where an is the unit vector normal to the plane directed toward point P Since the unit vector normal to any plane f = 0 is defined as 4f an = ! 4f So, we have the unit vector normal to the given charged plane 3x + 4y = 0 as 3a + 4 a y 3a + 4a y ( f = 3x + 4y ) an = ! x 2 =! x 2 5 3 +4 Since at point (1, 0, 3) f > 0 , so, we take the positive value of an . (2 # 10-9) 3ax + 4ay r b l ( rS = 2 nC/m2 ) Therefore, E = s an = -9 5 2e0 2 (10 /36p) p 36 = 3a + 4ay h = 67.85ax + 90.48ay V/m 4 ^ x
E = 1 rL (2pr) 2 z 2 3/2 az 4pe0 ^r + z h
3 3/2 a z ^42 + 32h = 9 # 2 # 2p # 4 # 3 az = 12.56az V/m 125
= (9 # 109) # (2 # 10-9) # (2p # 4)
SOL 2.2.4
Option (C) is correct. Electric flux density produced at a distance r from a point charge Q located at origin is defined as
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Q ar 4p r 2 So, the divergence of the electric flux density is Q =0 d : D = 12 2 cr2 4pr2 m r 2r So it is 0 for all the points but at origin (r = 0) its divergence can’t be defined. D =
Option (D) is correct.
he
lp.
co
m
SOL 2.2.5
The total flux leaving the closed surface is
#
y =
1
# #
1
x2 y dydz +
w. g
So,
at e
( dS is normal vector to surface) y = D : dS The closed cube has total eight surfaces but as the vector field has no component in az direction so we have the integrals only through the four separate surfaces as shown in the figure 0
0
1
# # 0
0
at x = 0, front
ww
=-
1
0 2 1
- x2 y dydz
at x = 1, back 1
+
1
1
# # ydydz + # # 0
1
0 3 1
1
0
# # 0
x2 dxdz
0
1
- x2 y2 dxdz +
1
# # 0
at y = 0, left
0
1
x2 y2 dxdz
at y = 1, right
y =-; E 6z @10 + :x D 6z @10 =- 1 # 1 + 1 # 1 =- 1 2 5 2 0 3 0 4
SOL 2.2.6
Option (A) is correct. Given the electric flux density D = x2 yax + y2 x2 ay C/m2 So,
div D = d : D = c 2 ax + 2 ay + 2 az m : ^x2 ya x + y2 x2 a y h 2x 2y 2z 2 d : D = 62xy + 2x y@ (center of the cube is located at b 1 , 1 , 1 l) =1+1 = 3 2 2 2 2 4 4
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Electrostatic Fields
95
m
For View Only Shop Online at www.nodia.co.in SOL 2.2.7 Option (A) is correct. From the given data we have the electric flux density at r = 0.2 m as D = 5r2 a r nC/m2 According to Gauss law the volume charge density at any point is equal to the divergence of the flux density at that point, so we have the volume charge density at r = 0.2 m as rv = d : D = 12 2 ^r2 (5r2)h = 12 # 5 # 4r3 r 2r r (r = 0.2 m ) = 20r = 4 nC/m3 Option (D) is correct. Again from the given data we have the electric flux density at r = 1 m as D = 2/r2 a r nC/m2 So, the volume charge density at r = 1 m is rv = d : D = 12 2 cr2 b 22 l m = 0 r 2r r
SOL 2.2.9
Option (B) is correct. Given the moment p = 4pe0 az C- m The electric field intensity at any point (r, q, f) produced due to an electric dipole lying along z -axis and having the dipole moment p in az direction is defined as p E = (2 cos qar + sin qa q) 4pe0 r3 ( p = 4pe0 az C- m ) E = 13 ^2 cos qar + sin qa q h r Now, given that the z -component of electric field is zero i.e. Ez = E : az = 0
ww w. ga te
he
lp.
co
SOL 2.2.8
1 2 cos q (a : a ) + sin q (a : a ) = 0 6 r z q z @ r3 1 2 cos2 q - sin2 q = 0 6 @ r3 Ez = 0 2 cos2 q - sin2 q = 0 1 1 + 3 cos 2q = 0 @ 26 Thus q = 54.7c or q = 125.3c Therefore the conical surface of angle q = 54.7c or 125.3c will have the electric field component Ez = 0 . SOL 2.2.10
Option (A) is correct. Electric field intensity produced at a distance r from an infinite line charge with charge density rL is defined as rL E = 2pe0 r and since the electric potential at point (1, p/2, 2) is zero so, the electric potential at
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For View Only Shop Online at www.nodia.co.in point (r, f, z) will be equal to the integral of the electric field from point (1, p/2, 2) to the point to (r, f, z). (r, f, z) r r r r i.e. V =- E : dl =- b L l dr = :- L ln (r)D pe r pe 2 2 0 0 1 (1, p/2, 2) 1 1 1 ( rL =+ 1 nC ) V = 2 # 10-9 # 9 # 109 ln b l = 18 ln b l r r Note: Since the infinite line charge has the equipotential cylindrical surface so for taking the line integral, f and z has not been considered.
#
Option (A) is correct. Electric potential at any point for a given electric field E is defined as
m
SOL 2.2.11
#
#
lp.
he
w. g
at e
#
co
i.e. V =- E : dl + C Now given the electric field intensity in spherical coordinate system E = 2 2r 2 a r (r + 4) and since the differential displacement in the spherical system is given as dl = drar + rdra q + r sin qdfaf So we have the electric potential 2r , V =dr + C = 2 1 + C r +4 (r2 + 4) 2 dV = 0 At maxima, dr -1 # 2r = 0 (r2 + 4) 2 Solving the equation we get, r = 0 and r = 3 d 2 V =- ve at r = 0 dr2 So the electric potential will be maximum at origin. Option (B) is correct. As calculated in the previous question, the electric potential at point (r, q, f) is V = 21 +C r +4 So at r = 0 , electric potential is V1 = 1 + C 4 and at r = 2 electric potential is V2 = 1 + C 8 So potential difference between the two surfaces is : V12 = b 1 + C l - b 1 + C l = 1 volt 8 2 4
SOL 2.2.13
Option (C) is correct. Electric potential at a distance R from a dipole having moment p is defined as p:R V = 4pe0 R 3 So we have the potential at point A due to the dipole located at point B as:
ww
SOL 2.2.12
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Chap 2
Electrostatic Fields
For View Only V =
p : AB 4pe0 AB
97
Shop Online at www.nodia.co.in 3 -9 b 5 ax - ay + 2az l : (ax + ay + 8az ) # 10 = 4pe0 ( 12 + 12 + 82 ) 3
3
= 0.6 V Option (B) is correct. Since the charge is being split and placed on a circular loop so the distance of all the newly formed point charges from the center of the loop will be equal as shown in the figure.
he
lp.
co
m
SOL 2.2.14
SOL 2.2.15
ww w. ga te
Therefore, the potential at the center of the loop will be Q/4 (20 # 10-9) = (9 # 109) # V = 4c m 5 4pe0 r = 36 V
(Q = 20 nC )
Option (A) is correct. The work done in carrying a charge q from point A to point B in the field E is defined as W =- q
B
# E : dl A
Given the electric field intensity in the cartesian system as E = 2yax + 2xay and since the differential displacement in cartesian system is given as dl = dxax + dyay + dzaz So, the work done in carrying charge q =+ 2 C from point A ^1, 1/2, 3h to the point B (4, 1, 0) is W =- 2 ;
#
4
x=1
2ydx +
#
1
2xdyE
y = 1/2
The curve along which the charge is being carried is given as y = x & x = 2y2 2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Electrostatic Fields
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Shop Online at www.nodia.co.in W =- 2 ;
Therefore, we have
#
1
4
2 _ x/2 i dx +
#
1
1/2
2 ^2y2h dyE
4 =- 4 ; 2 6x3/2@1 + 2 6y3@11/2E =- 4 ;7 2 + 7 E 3 3 3 12 =- 15.5 J
Option (B) is correct.
lp.
co
m
SOL 2.2.16
The work done in carrying a charge q from one point to other point in the field E is defined as
# E : dl
he
W =- q
and since the differential displacement for the defined circular arc is dl = rdfaf as obtained from the figure W =- 2
at e
So, the work done is
p/4
# (xa f=0
x
- yay) : (rdfaf)
now we put x = r cos f , y = r sin f and ax : af =- sin f , ay : af = cos f in the expression to get
#
p/4
w. g
W =- 2
0
- 2r2 sin f cos fdf =- 2 # 1
p/4
# - sin (2f) df 0
( r = 1)
=+ 1 J
Option (A) is correct.
ww
SOL 2.2.17
Consider the last charge is being placed at corner D so the potential at D due to the charges placed at the corners A, B, C is GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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99
For View Only
Shop Online at www.nodia.co.in q = 10 Q e 43 3 oH e0 e0 3 pR So, the electric field at a distance r from the center is GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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103
For View Only
Shop Online at www.nodia.co.in Q r Q (for r # R ) a E = 1 2 c r 3 m ar = e pe0 R3 r 4 0 4p r R Therefore the electric potential at the point P will be the line integral of the field intensity from infinity to the point P 3
R
V (r) =-; E1 : dr +
#
i.e.
r
#E
3
R
2
: dr E
where
E1 " electric field outside the sphere as calculated in previous question. E2 " electric field inside the sphere R 1 Q dr + r 1 Q r dr V (r) =-= c 4pe0 R3 m G 2 3 4pe0 r R 2 2 Q 1 = - 13 b r - R lE ; 2 4pe0 R R So, V (r) decreases with increase in r .
#
Option (C) is correct. The total stored energy inside a region having charge density rv and potential V is defined as WE = 1 rv Vdv 2 As calculated in previous question the electric potential at any point inside the sphere is Q 1 1 r2 - R2 V (r) = 4pe0 ;R R3 b 2 lE (R = 1 m , Q = 1 C) = 1 :1 (3 - r2)D 4pe0 2
ww w. ga te
he
#
lp.
SOL 2.2.26
co
m
#
Therefore the total energy stored inside the sphere is 1 Q 1 1 (3 - r2) (4pr2 dr) WE = 1 e 4 D 2 0 3 pR3 o: 4pe0 # 2
#
1 (R = 1 m , Q = 1 C) = 3 # 1 # 4p (3r2 - r 4) dr 8p 2 0 4pe0 5 1 = 3 :r3 - r D 16pe0 5 0 9 = 3 # 4 = 3 # 9 # 10 # 4 = 27 # 109 = 24.4 # 109 J 5 16pe0 5 4#5
SOL 2.2.27
#
Option (B) is correct. The electric field to counter act the gravitational force must produce the same force as applied by gravity but in opposite direction. i.e. e ^E h = me g ^- ar h where e is the charge of an electron, me is the mass of electron, g is acceleration due to gravity and ar is radial direction of earth. So, taking the magnitude only we have the required field intensity, ^9.1 # 10-31h # 9.8 mg E = e = = 15.57 # 10-11 V/m e 1.6 # 10-19
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Electrostatic Fields
For View Only SOL 2.2.28 Option (D) is correct.
Chap 2
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co
m
Consider the electric field intensity produced at point P b 0, 1 , 1 l due to the charges 4 4 located at points A, B and C respectively as shown in figure is EA , EB and EC respectively.
he
lp.
So the net electric field at point P is Enet = EA + EB + EC and since the electric field intensity at any distance R from a point charge Q is Q R defined as E = 4pe0 R 3 So Enet = 1 =Q PA 3 + kQ PB 3 + kQ PC 3 G 4pe0 PB PC PA
w. g
at e
1 3 b 14 ay + 14 az l b- 34 ay + 14 az l b 4 ay - 4 a z l 1 Q = + kQ + kQ 4pe0 1 2 1 2 3/2 3 2 1 2 3/2 1 2 3 2 3/2 ;b 4 l + b 4 l E ;b 4 l + b 4 l E ;b 4 l + b 4 l E and since Enet = 0 so we have 1 3/2 3k 1k b 4 l # ^16h 4 4 + =0 3 2 + 1 2 3/2 3 2 + 1 2 3/2 ^2 h3/2 ;b 4 l b 4 l E ;b 4 l b 4 l E Solving the equation we get
Option (D) is correct. The point charges can be represented as shown below.
ww
SOL 2.2.29
k = 15.59
So the electric field at point ^x, 0, 0h will be directed along x -axis. Taking only magnitude we have the net electric field intensity at ^x, 0, 0h as Q 2Q Q E = 2 2 + 4pe0 x 4pe0 ^x - a h 4pe0 ^x + a h2 Q 2Q Q 2a a 2 2a a 2 = 2 :1 + x + 3 a x k + ...D 2 + 2 :1 - x + 3 a x k - ....D 4pe0 x 4pe0 x 4pe0 x a Since x >> a , neglecting higher powers of a k we get x
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SOL 2.2.30
105
Shop Online at www.nodia.co.in 2 2 Q 2Q Q E = 1 + 2a + 3 a a k D 1 - 2a + 3 a a k D + x x x x 4pe0 x2 : 4pe0 x2 4pe0 x2 : 6Qa2 6Qa2 = m 4 = Kc 4pe0 x x4
Option (A) is correct. According to Gauss law the surface integral of electric field intensity over a Gaussian surface is defined as E : dS = 1 Qenc e0 So for the Gaussian surface outside the sphere at a distance r (2 R) from the centre of the sphere we have r ( 4 pR3) (there is no charge outside the sphere) E (4pr2) = v 3 e0 Therefore at any point outside the sphere r ^> Rh the electric field intensity will be rv ^ 43 pR3h rv R3 E = 2 m ar 2 ar = e c 0 3r 4pe0 r and for the Gaussian surface inside the sphere at a distance r (# R) from the center of the sphere we have r ( 4 pr3) E (4pr2) = v 3 e0 Therefore at any point inside the sphere, the electric field intensity will be rv b 4 pr3 l 3 r 1 E = ar = v a r k ar e0 3 e0 4pr2
SOL 2.2.31
ww w. ga te
he
lp.
co
m
#
Option (C) is correct. As discussed in Q.55. The electric field at any point inside a charged solid sphere is r E = v a r k ar e0 3 where r is the distance from center of the sphere and rv is the volume charge density given as Q 2 # 10-9 (Q = 2 nC , R = 3 m ) rv = 4 = 3 3 4 3 pR 3 p ^3h = 1.77 # 10-11 C/m3 So the force acting on electron when it is at a distance r from the center of the sphere is (e is the charge of an electron) F = eE 2 r (me is mass of an electron) me d r2 = e v a r k e0 3 dt -19 -11 d 2 r = ^- 1.6 # 10 h^1.77 # 10 h r #3 2 -31 -12 dt ^9.1 # 10 h^8.85 # 10 h d 2 r =- 2.17 1011 r ^ h # dt2
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Electrostatic Fields
Chap 2
Option (D) is correct. As calculated in above question the position of the electron at any time t is r = 3 cos ^ 1.17 # 1011 t h So, 2pf = 1.17 # 1011 11 f = 1.17 # 10 = 5.44 # 10 4 Hz = 54.4 KHz 2p Option (B) is correct. The portion of the plane y + z = 1 m lying in the first octant bounded by the planes x = 0 and x = 1 m has been shown in the figure through which we have to determine the total electric field flux.
ww
w. g
at e
SOL 2.2.33
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lp.
SOL 2.2.32
co
m
For View Only Shop Online at www.nodia.co.in 2 d r + 1.17 1011 r = 0 ^ h # dt2 Solving the differential equation we have ....(i) r = A1 cos ^ 1.17 # 1011 t h + A2 sin ^ 1.17 # 1011 t h where A1 and A2 are constants. Now, at t = 0 , r = 3 m as the electron is located at one end of the hole. So putting it in equation (i) we get, A1 = 3 again at t = 0 , dr = 0 as the electron is released from rest. dr So putting it in equation (i) we get A2 = 0 Thus the position of electron at any time t is r = 3 cos ^ 1.17 # 1011 t h at t = 1 m sec r = 2.83 m
According to Gauss law the total outward flux through a closed surface is equal to the charge enclosed by it. i.e.
y =
# D : dS = Q
enc
So the total electric field flux emanating flux from the line charge between x = 0 and x = 1 m is rL ^1 h rL Q E : dS = enc = = e0 e0 e0 and by symmetry, flux through the defined surface will be one fourth of the total electric field flux emanating from the defined portion .
#
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Electrostatic Fields
For View Only i.e.
107
Shop Online at www.nodia.co.in the electric flux crossing the surface =
# E : dS 4
=
rL 4e0
Note: It must be kept in mind that the total electric flux is total electric field flux is
while the
# E : dS
Option (D) is correct. Consider a point P inside the cylindrical surface of 2 m as shown in figure.
lp.
co
m
SOL 2.2.34
# D : dS
ww w. ga te
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Now we make the use of superposition to evaluate the electric field at point P by considering the given charge distribution as the sum of two uniformly distributed cylindrical charges, one of radius 5 m and the other of radius 2 m, and such that the total charge in the hole is zero. Thus we obtain the net electric field at point P as Enet = E1 + E2 where E1 is the electric field intensity at point P due to the uniformly charged cylinder of radius 5 m that has the charge density ^5 nC/m3h, while E2 is the electric field intensity at point P due to charged cylinder of radius 2 m that has the charge density ^- 5 nC/m3h As calculated in MCQ.61 the electric field intensity at a distance r from the cylindrical axes having uniform charge density rv is r E = vr e0 2 -9 r So we have E1 = v R1 = 5 # 10 R1 2e0 2e0 -9 r and E2 = v R2 = - 5 # 10 R2 2e0 2e0 So the net electric field at point P is -9 Enet = 5 # 10 ^R1 - R2h 2e0 By the triangle law of vector (separation = 1 m ) R1 - R 2 = C = a x -9 So, Enet = 5 # 10 ^ax h = 282.5ax V/m 2e0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Electrostatic Fields
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For View Only Shop Online at www.nodia.co.in SOL 2.2.35 Option (A) is correct. As we have calculated the electric field for the same distribution in Q.55. So we evaluate the electric potential by taking the line integral of the field intensity.
#
#
#
at e
ww
w. g
Option (B) is correct.
#
he
#
SOL 2.2.36
co
#
lp.
#
m
V =- E : dl Z rv r ]] a k a r for r # R e0 3 E =[ 3 ]] rv c R 2 m a r for r > R e0 3r \ The electric potential at any point outside the sphere ^r > Rh is 3 r r r v R V =- E : dl =c 2 m dl 3 3 e0 3l r r R3 r R3 =- v ;- 1E = v 3e0 3e0 r l 3 and the electric potential at any point inside the sphere ^r # Rh is 3 R R r r r r v R v l V =-; E : dl + E : dl E =dr b l dl 2 R 3 R e0 3 3 e0 3r 2 R r R3 r 2 r r R3 r 2 =- v :- 1 D - v :l D =- v :- 1 D - v :r - R D 3e0 r 3 3e0 2 R 3e0 3e0 2 2 R 2 rv 3R2 r2 rv = - l= R2 - r l 3e0 b 2 2 2e0 b 3 i.e.
Given the total charge on the disk is Q = 900p mC = 900p # 10-6 C radius of the disk is a =6m and since the charge has been distributed uniformly over the surface so the small charge element dQ on the disk at a distance r from the center as shown in figure is given as -6 Q dQ = b l dS = 900p # 210 ^rdrdfh S p ^6 h = 25 # 10-6 rdrdf The force applied by the charge element dQ on the 150 mC charge located at point P is (150 # 10-6) dQ (150 # 10-6) dQ dF = = R2 (r2 + 16) GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 2
Electrostatic Fields
109
For View Only Shop Online at www.nodia.co.in As the disk has uniformly distributed charge so the horizontal component of the field is get cancelled and the net force will have the only component in az direction and the net force by projection on z -axis is given as 6 ^150 # 10-6h^25 # 10-6 rdrdf h 2p 4 F = # d r2 + 16 n 2 4pe0 ^r + 16h f=0 r=0 6 1 F = 270p ;- 2 = 9.44 N E r + 16 0
# #
Option (D) is correct. Electric field at any point due to infinite surface charge distribution is defined as r E = s an 2e0 where rs " surface charge density an " unit vector normal to the sheet directed toward the point where field is to be determined. At origin electric field intensity due to sheet at y =+ 1 is r E1 = s ^- ay h =- 5 ay ^an =- ay h 2e0 2e0 and electric field intensity at origin due to sheet at y =- 1 is r E-1 = s ^ay h = 5 ay ^an = ay h 2e0 2e0 So net field intensity at origin is E = E+1 + E-1 =- 5 ay + 5 ay = 0 2e0 2e0
SOL 2.2.38
Option (D) is correct. As the test charge is placed at point ^2, 5, 4h. So it will be in the region y > + 1 for which electric field is given as E = E+1 + E-1 r r (for both the sheet an = ay ) = s ^ay h + s ^ay h 2e0 2e0 -9 2 # ^5 # 10-9h = ay = 5 # 10 ay 2e0 e0 Therefore the net force on the charge will be -9 F = qE = ^5 # 10-6hc 5 # 10 m ay = 2.83 # 10-3 N e0 Option (B) is correct. Since the electric field intensity due to a sheet charge is defined as r E = s an 2e0 So it doesn’t depend on the distance from the sheet and given as E = E+1 + E-1 -9 r r r = s ^- ay h + s ^- ay h =- s ay =- 5 # 10 ay 2e0 2e0 2e0 2e0 So, it will be constant as we move away from the sheet.
SOL 2.2.39
ww w. ga te
he
lp.
co
m
SOL 2.2.37
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Electrostatic Fields
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For View Only Shop Online at www.nodia.co.in SOL 2.2.40 Option (D) is correct. As the charge is redistributed so the total charge will remain same on the sphere. Total charge before redistribution. Q1 = rv dv = ^6 C/m3hb 4 p ^1 h3 l (rv = 6 C/m2) 3 = 8p Coulomb and total charge after redistribution
#
v
Since
Q1 = Q 2
So, we have
8p =
=
#
1
r=0
k ^3 - r2h 4pr2 dr
5 1 k ^4ph^3r2 - r 4h dr = 4pk :r3 - r D = 4pk :1 - 1 D 5 0 5 0 k = 3.5
#
1
co
or
Option (B) is correct. According to Gauss’s law the total electric flux through any closed surface is equal to the total charge enclosed by the volume. Now consider the complete spherical surface defined by r = 48 m through which the total flux is equal to the point charge. So the total flux passing through the hemispherical surface will be half of the point charge. 50 mC Q i.e. y = = = 25 mC 2 2 Option (D) is correct. For any point inside the sphere when we draw a symmetrical spherical surface (Gaussian surface) then the charge enclosed is zero as all the charge is concentrated on the surface of the hollow sphere. So according to Gauss’s law
#
w. g
SOL 2.2.42
at e
he
lp.
SOL 2.2.41
# r dv
m
Q2 =
#
ww
e0 E : dS = rv dv = 0 therefore E = 0 at any point inside the hollow sphere. now at any point outside the sphere at a distance r from the center when we draw a symmetrical closed surface(Gaussian surface) then the charge enclosed is Qenc = rs ^4pR2h and according to Gauss’s law e0
# E : dS
= Qenc
e0 E ^4pR2h = rs ^6pR2h 2 r E = s c R2 m ar e0 r
Option (D) is correct. Electric field intensity at any point due to uniform surface charge distribution is defined as r E = s an 2e0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 2.2.43
Chap 2
Electrostatic Fields
For View Only where
111
Shop Online at www.nodia.co.in
m
rs " surface charge density an " unit vector normal to the sheet directed toward the point where field is to be determined. The electric field intensity due to the upper plate will be EU = 2 ^- az h ^an =- az h 2e0 and the field intensity due to lower plate will be El =- 2 ^az h ^an = az h 2e0 So the net field between the plates is E = EU + El
SOL 2.2.45
co
ww w. ga te
he
lp.
SOL 2.2.44
= 2 ^- az h + :- 2 ^az hD =- 4 az =- 2 az e0 2e0 2e0 2e0 Option (D) is correct. Electric field intensity at any point is equal to the negative gradient of electric potential at the point i.e. E =- dV So, the y -component of the field is Ey =-2V 2y Now, for the interval - 3 # y # - 2 , V = 20 ^t + 3h Ey =-2V =- 20 V/m 2y For the interval - 2 # y # - 1, V = 20 So, Ey =-2V = 0 2y For the interval - 1 # y # + 1, V =- 20t So, Ey =-2V = 20 V/m 2y For the interval 1 # y # 2 , V =- 20 So Ey = 0 For the interval 2 # y # 3 , V = 20 ^t - 3h So, Ey =-2V =- 30 V/m 2y Therefore, the plot field component Ey with respect to y for the defined intervals will be same as in option (A). Option (A) is correct. Since the electrons are moving with equal but opposite velocities so assume that their velocities are + v 0 ax and - v 0 ax . Now let the electric field is applied in ax direction i.e. E = E 0 ax So the force applied on the electrons will be F = eE =-^1.6 # 10-19h E
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w. g
Option (B) is correct. The electric field intensity produced at a distance r from a line charge of density rL is defined as rL E = a 2pe0 r r where a r is unit vector directed toward point P along r. So, the electric field acting on the line charge at y = 3 m due to the line charge located at y =- 3 m is -9 E = 80 # 10 ay ^rL = 80 nC, a r = ay, r = 6 mh 2pe0 ^6 h = 240ay V/m Therefore, the force per unit length exerted on the line charge located at y = 3 m is
ww
SOL 2.2.46
at e
he
lp.
co
m
m dv =-^1.6 # 10-19h E dt therefore, change in the velocity ^1.6 # 10-19h E ^1.6 # 10-19h E 0 dt ax dv =dt h =^ m m So, the velocity of electron moving in + ax direction will change to ^1.6 # 10-19h E 0 dt ^1.6 # 10-19h E 0 dt v1 = v 0 a x ax = ;v 0 E ax m m Since velocity deceases so loss in K.E. is K.ELoss = 1 mv 02 - 1 mv 12 2 2 2 ^1.6 # 10-19h E 02 ^dt h2 1 -19 ...(i) = ^1.6 # 10 h E 0 dt 2 m Again the velocity of electron moving in - ax direction will change to ^1.6 # 10-19h E 0 dt v2 =- v 0 ax ax m ^1.6 # 10-19h E 0 dt =-;v 0 + E ax m Since velocity increases, so Gain in K.E. is K.EGain = 1 mv 22 - 1 mv 02 2 2 2 ^1.6 # 10-19h E 02 ^dt h2 ...(2) = ^1.6 # 10-19h E 0 dt + 1 2 m Comparing eq (1) and eq (2) we get K.EGain > K.ELoss
F = SOL 2.2.47
# ^r dz h^E h = ^80 # 10 1
z=0
L
-9
h^240ay h = 19.2ay mN
Option (D) is correct. The four charges located at the corners of square 4 cm has been shown in figure below :
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Electrostatic Fields
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The net potential at the charge located at A due to the other three charges is q q q VA = 1 a B + C + D k 4pe0 AB AC AD 1 1 1 = 9 # 109 # 1.2 # 10-9 c + + 4 # 10-2 4 2 # 10-2 4 # 10-2 m 2 = 10.8 # 10 c 2 + 1 m 4 2 = 730.92 Volt Similarly, the electric potential at all the corners will be VB = VC = VD = VA = 730.92 Volt Therefore, the net potential energy stored in the system is given as 1 qV = 1 q V + q V + q V + q V W = B B C C D Dh 2 2^ A A = 1 # 4 # ^1.2 # 10-9h # ^730.92h 2 = 3.75 mJ ***********
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SOLUTIONS 2.3
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Option (B) is correct. Given, the electric field intensity, E = xax + yay + zaz dl = ax dx + ay dy + az dz So, the potential difference between point X and Y is Y
VXY =- # E : dl =
xdx +
0
0
#2 ydy + #3 zdz
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2 2 y2 0 z2 0 =-=x + + 2 1 2 2 2 3G =- 1 622 - 12 + 02 - 22 + 02 - 32@ = 5 2 Option (B) is correct. Given the electric field vector at point P due to the three charges Q1 , Q2 and Q 3 are respectively. E1 = ax + 2ay - az E2 = ay + 3az E 3 = 2ax - ay So, the net field intensity at point P is E = E1 + E2 + E 3 = 3ax + 5ay + 2az
SOL 2.3.4
Option (B) is correct. Charge density at any point in terms of electric flux density D is defined as rv = d : D Since, D = zr ^cos2 fh az C/m2 So, we get rv = d : D = 2 6zr ^cos2 fh az@ = r cos2 f C/m3 2z p p At point a1, , 3k, rv = ^1 h cos2 a k = 1 = 0.5 C/m3 2 4 4 Option (C) is correct. Electric field intensity E is a vector quantity while the electric potential V is a scalar quantity.
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SOL 2.3.3
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SOL 2.3.2
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SOL 2.3.1
Option (D) is correct. For an ideal capacitance the area of plates, A is assumed very high in comparison to the separation d between the plates. A .3 i.e. d GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 2.3.5
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Electrostatic Fields
115
For View Only Shop Online at www.nodia.co.in So, the fringing effect at the plates edges can be neglected and therefore, we get the capacitance between the parallel plates as C = eA 4d So A and R both true and R is correct explanation of A. Option (D) is correct. By using method of images, the conducting surfaces are being replaced by the image of charge distribution which gives a system of charge distribution. So, in solving boundary value problems we can avoid solving Laplace’s or Poission’s equation and directly apply the method of images to solve it. Thus both A and R are true and R is correct explanation of A.
SOL 2.3.7
Option (C) is correct. For a pair of line charges equipotential surface exists where the normal distance from both the line charges are same. So, the plane surface between the two line charges will be equipotential.
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SOL 2.3.6
This is the similar case to method of images. SOL 2.3.8
Option (D) is correct. According to uniqueness theorem : If a solution to Laplace’s equation (a) be found that satisfies the boundary condition then the solution is unique. Here it is given that the potential functions V1 and V2 satisfy Laplace’s equation within a closed region and has the same value at its boundary so both the functions are identical.
Option (D) is correct. From Maxwell’s equation we have d # E =-2B 2t d # E =- 2 ^d # Ah ^B = d # Ah 2t d # bE + 2 Al = 0 2t GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 2.3.9
116
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Chap 2
For View Only Shop Online at www.nodia.co.in Since, the curl of a gradient of a scalar field is identically zero. So, we get E + 2 A =- dV 2t i.e. E ! - dV in time valeying field therefore A and R both are true and R is the correct explanation of A. Option (D) is correct. The surface charge density at plane x = 8 is shown in the figure.
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SOL 2.3.10
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SOL 2.3.11
The point P is located at ^6, 4, - 5h. So, the normal vector to the plane x = 8 pointing toward P is an =- ax Therefore, the electric flux density produced at point P is r D = s an = 60 ^- ax h =- 30ax 2 2 Option (B) is correct. Consider the coaxial cylinder is located along z -axis. So at any point between the two surfaces the electric field is given as E =- dV =- 2 Va r (Since all other derivatives will be zero) 2r
Given that the inner surface is at potential V0 while the outer one is grounded so GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Electrostatic Fields
117
For View Only Shop Online at www.nodia.co.in the region between the two surfaces will have a gradually decreasing potential and so, E will not be uniform and it is radially directed as calculated above (in a r direction). Option (B) is correct. The Poisson’s equation is defined as r d2V =- v e where V is electric potential and rv is charge density. So, in charge free space ( rv = 0 ) we get the Poisson’s equation as which is Laplace equation. d2V = 0
SOL 2.3.13
Option (B) is correct. Consider the three equal charges of Q C is placed at a separation of 0.5 m as shown in figure below :
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SOL 2.3.12
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The net stored charge in the system of n charges is defined as n W = 1 Qk Vk 2k=1 where Qk is one point charge and Vk is the net electric potential at the point charge due to the other charges. Now, we have the net electric potential at any of the point charge Q located in the system as Q Q Q V1 = 1 b = + pe0 4pe0 0.5 0.5 l So, total energy stored in the system of charges is given as 3Q 2 (1) W1 = 3 b 1 QV1 l = 2 2pe0 Now, when the charges are separated by 1 m then the electric potential at any of the charge Q due to the other two charges is Q Q Q V2 = 1 b + l = 1 2pe0 4pe0 1 So, the stored energy in the new system is 3Q2 (2) W2 = 3 b 1 QV2 l = 2 4pe0 From equation (1) and (2) we have W2 = 0.5W1 or W1 = 2W2 SOL 2.3.14
Option (D) is correct. Electric potential due to point charge is defined as Q V = 1 3pe0 r
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For View Only Shop Online at www.nodia.co.in So, for the equal distance r potential will be same i.e. equipotential surface about a point charge is sphere. Option (C) is correct. An electrostatic field has its curl always equals to zero. So electric field is irrotational. Statement 1 is correct. Electric field divergence is not zero and so it is not solenoidal. Statement 2 is correct. Electric field is static only from a macroscopic view point. Statement 3 is correct. Work done in moving a charge in the electric field from one point to other is independent of the path. Statement 4 is correct.
SOL 2.3.16
Option (C) is correct. Given electric potential, V = 20y 4 + 10x3 From Poisson’s equation we have r d2V =- v e0 where, V " Electric potential rv " Charge density
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SOL 2.3.15
(x = 2 , y = 0 )
Option (B) is correct. Given, the wave equation in space for a propagating wave in z -direction is d2Ex + k2 Ex = 0 Now, from option (C) we have the electric field component as Ex = E 0 e-jkz The Lapalacian of electric field is d2Ex = ^- jk h2 E 0 e-jkz d2Ex =- k2 E 0 e-jkz =- k2 Ex 2 or, So, it satisfies the wave equation. d Ex + k2 Ex = 0
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SOL 2.3.17
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rv 22 22 22 4 3 e 2 + 2 + 2 o^10y + 20x h =- e0 2x 2y 2z r 120x + 120y =- v e0 rv = e0 ^120 # 2 + 120 # 0h rv =- 120e0
Option (D) is correct. Consider the infinitely long uniform charge density shown in the figure. The electric field intensity produced at a distance r from an infinite line charge with density rL is defined as rL E = 2pe0 r GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 2.3.18
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Option (D) is correct. Consider the square loop ABCD carrying current 0.1 A as shown in figure.
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SOL 2.3.19
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Since, the normal distance vector of points P (0, 6, 1) and Q (5, 6, 1) from the line charge will be same so, the field intensity produced due to the infinite line at both the points P and Q will be same. Therefore, the field intensity at (5,6,1) is E .
The magnetic dipole moment is m = IS where I is current in the loop and S is the area enclosed by loop. 2 So, m = ^0.01h^10 2 h = 2 A- m2 The direction of the magnetic dipole moment is determined by right hand rule. i.e. m = 2az A- m2 SOL 2.3.20
Option (A) is correct. Electric flux density at a distance r from a point charge Q is defined as Q D = ar 4p r 2
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For View Only Shop Online at www.nodia.co.in and the total flux through any defined surface is
#
y = D : dS So, both the quantities has not the permittivity e in their expression. Therefore, D and y are independent of permittivity e of the medium. SOL 2.3.21
Option (B) is correct. According to Gauss’s law, the total outward flux through a closed surface is equal to the charge enclosed inside it.
#
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SOL 2.3.23
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SOL 2.3.22
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i.e. D : dS = Qenc Now, consider the height of cylinder is h . So, the cylindrical surface at r = 3 encloses the charge distribution ^rS = 5 C/m2h located at r = 2 m . Therefore, we get D ^2p ^3h h h = 5 # 2p ^2 h h or, D = 20 a r 3 Option (A) is correct. The electric potential produced by 1 mC at a distance r is ^1 # 10-6h 9000 V = 9 # 109 = r r So, the potential energy stored in the field will be the energy of the charges as, i.e. W = qV -3 = ^4 # 10-6h 9000 = 36 # 10 r r where r is the distance between the charges given as r = ^- 2 - 1h2 + ^1 - 3h2 + ^5 + 1h2 = 7 -3 So, W = 36 # 10 = 5.15 # 10-3 Joule 7 Option (C) is correct. Electric field intensity due to a dipole having moment P at a distance r from it is E \ 13 r E2 = r 13 E1 r 23 3 E2 = ^2 h 1 ^4h3 E2 = 1 mV/m 8 Option (B) is correct. Energy density (energy stored per unit volume) in an electric field is defined as we = 1 D : E = 1 e0 E : E = 1 e0 E 2 2 2 2 Option (A) is correct. The position of points A, B and C are shown below
SOL 2.3.24
SOL 2.3.25
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Since, position charge is placed at A and negative charge at B so, their resultant field intensity at C is as shown below :
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Since, the forces F1 = F2 so the vertical component F1 V and F2 V are get cancelled while F2 H and F1 H are get summed to provide the resultant field in - az direction. SOL 2.3.26
Option (D) is correct. Given, Charges, Q1 = Q2 = 1 nC = 10-9 C Separation between charges, r = 1 mm = 10-3 m So, the force acting between the charges is 9 -9 2 kQ1 Q2 9 # 10 ^10 h F = = 2 r2 ^10-3h = 12 # 10-3 N
SOL 2.3.27
Option (B) is correct. According to Gauss’s law, the surface integral of flux density through a closed surface is equal to the charge enclosed inside the closed surface (volume integral of charge density) i.e.
# D : dS
=
# r dv v
In differential form, the Gauss’s law can be written as d # D = rv r d#E = v e0
^D = e0 E h
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For View Only Shop Online at www.nodia.co.in SOL 2.3.28 Option (D) is correct. The electric field at a distance r from the point charge q located in a medium with permittivity e is defined as q qe-1 E = ar 2 ar = 4per 4p r 2
i.e. or,
# D : dS # D : dS s
s
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Option (C) is correct. For according to Gauss’s law the total outward electric flux through a closed surface is equal to the charge enclosed by the surface. = Qenc =
# r dv v
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SOL 2.3.29
Option (B) is correct. The force between the two charges q1 and q2 placed in a medium with permittivity e located at a distance r apart is defined as qq F = 1 122 4pe r or F\1 e i.e. force is inversely proportional to permittivity of the medium. Since, glass has the permittivity greater than 1 (i.e. permittivity of free space) So, the force between the two charges will decreases as the glass is placed between the two charges.
SOL 2.3.31
Option (A) is correct. According to Gauss’s law the total electric flux through a closed surface is equal to the charge enclosed by it. Since, the sphere centred at origin and of radius 5 m encloses all the charges therefore, the total electric flux over the sphere is given as yE = Q1 + Q2 + Q 3 = 0.008 + 0.05 - 0.009 = 0.049 mC
SOL 2.3.32
Option (D) is correct. Electric flux through a surface area is the integral of the normal component of electric field over the area.
SOL 2.3.33
Option (B) is correct. The electric field due to a positive charge is directed away from it (i.e. outwards.) According to Gauss’s law the surface integral of normal component of flux density over a closed surface is equal to the charge enclosed inside it. So, A is true but R is false.
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SOL 2.3.30
Option (A) is correct. Force between the two charges Q1 and Q2 is defined as QQ F = 1 2 2 aR 4pe0 R GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 2.3.34
Chap 2
Electrostatic Fields
123
For View Only Shop Online at www.nodia.co.in When the charges are of same polarity then the force between them is repulsive. The electric force on both the charges will have same magnitude. As the expression of Force includes the term e (permittivity of the medium) so it depends on the medium in which the charges are placed. So the statements (a), (c) and (d) are correct while (b) is incorrect. Option (B) is correct. Since the electric field is negative gradient of the electric potential so the field lines will be orthogonal to the equipotential lines (surface).
SOL 2.3.36
Option (B) is correct. Electric potential at - 10 nC due to 10 nC charge is Q V = 1 4pe0 r -9 = 9 # 109 # 102# 10 2 +0+0 = 45 Volt and so the energy stored is We = QV = ^- 10 # 10-9h # 45 =- 150 nJ
SOL 2.3.37
Option (C) is correct. According to Gauss’s the outward electric flux density through any closed surface is equal to the charge enclosed by it. So electric field out side the spherical balloon doesn’t change with the change in its radius and so the energy density at point P is wE for the inflated radius b of the balloon.
SOL 2.3.38
Option (A) is correct. The curl of E is identically zero. i.e. d#E = 0 So, it is conservative. The electrostatic field is a gradient of a scalar potential. i.e. E =- dV So, d#E = 0 Work done in a closed path inside the field is zero
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SOL 2.3.35
(Conservative)
# E : dl
=0 (Conservative) d#E = 0 So, (a), (c) and (d) satisfies that the field is conservative. As the potential difference between two points is not zero inside a field so, the statement (b) is incorrect.
i.e.
Option (D) is correct. Net outward electric flux through the spherical surface, r = a is D : dS = y = rv b 4 pa3 l 3 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 2.3.39
#
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rv 4 3 pa 3 r a D = v ar C/m2 3 Option (C) is correct. For a pair of equal and opposite linear chargers the electric potential is defined as Q Q V = 4pe0 r1 4pe0 r2 where r 1 and r2 are the distances from the charges respectively. For the same value of V (equipotential surface) a plane can be defined exactly at the centre point between them. D ^4pa2h =
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SOL 2.3.40
Option (D) is correct. In a charge free region ^rv = 0h electrostatic field has the following characteristic r d:E = v = 0 e and (for static field) d#E = 0
SOL 2.3.42
Option (C) is correct. Consider the force experienced by Q is F1 . Since, there is no any external applied field (or force) so, sum of all the forces in the system of charges will be zero. i.e. SF = 0 or, 3F + 2F + F1 = 0 F1 =- 5F
SOL 2.3.43
Option (A) is correct. Poissions law is derived from Gauss’s law as d:D = r For inhomogeneous medium e is variable and so, d : ^eE h = r d : 6e ^- dV h@ = r d : ^edV h =- r This is the Poissions law for inhomogenous medium.
SOL 2.3.44
Option (C) is correct. Electric field intensity due to a infinite charged surface is defined as r E = S an 2e0 where rS is surface charge density and an is the unit vector normal to the surface directed towards the point of interest. Given that, rS = 20 nC/m2 = 20 # 10-9 C/m2 (Since the surface z = 10 m is above the origin). and an =- az So we have, -9 E = 20 # 10 ^- az h 2e0
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SOL 2.3.41
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Shop Online at www.nodia.co.in 9 = 20 # 10 # 9 # 10 # 4p az 2 =- 360paz V/m -9
Option (B) is correct. Electric field intensity due to a short dipole having a very small separation d , a + a distance R from it is defined as Qd (for d > d ) R = r2 - d 2 . r Qd Therefore, E = ^0 + afh 4pe0 r3 i.e. E \ 13 r Option (B) is correct. According to Gauss’ law the total outward flux from a closed surface is equal to the total charge enclosed by the surface. Option (B) is correct. Electric field intensity at any point r outside the sphere is defined as Q E = ar 4pe0 r2 and the field intensity inside the sphere is 4 pr 3 l Q b3 E = a 2 4 pa3 4pe0 r r b3 l Qr = ar 4pe0 a3 So the electric potential at any point r = b < a is
# E : dl =- # E : ^dra h - # E : ^dra h Q Qr =- # dr - # dr 4pe r 4pe a
for r > a
for r # a
V =-
b
a
r
r=3 a
r
a
b
0
2
a
0
3
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126
Electrostatic Fields
Chap 2
Option (D) is correct. Consider two parallel plates separated by a distance d is connected to a voltage source V . So, the field intensity between the plates is defined as E = 2V d
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SOL 2.3.49
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For View Only Shop Online at www.nodia.co.in SOL 2.3.48 Option (D) is correct. Given, the electric potential, V = 3x2 y - yz Electric field intensity at any point is equal to the negative gradient of the potential. i.e. E =- dV =-^6xy h ax - ^3x2 - z h ay - ^- y h az at (x = 1, y = 0 , z =- 1) E =- 4ay ! 0 So, electric field does not vanish at given point.
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CHAPTER 3 ELECTRIC FIELD IN MATTER
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Electric Field in Matter
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EXERCISE 3.1
A certain current density at any point (r, f, z) in cylindrical coordinates is given by
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J = 5ez (r2 a r + a z ) A/m2 .
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MCQ 3.1.1
Chap 3
The total current passing the plane z = 0 , 0 # r # 2 in the az direction is (B) 4p Ampere (A) 100p Ampere (C) 40p Ampere
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MCQ 3.1.2
(D) 0 Ampere
In a certain region the current density is given by
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J = r cos2 qa r + r2 sin qa q - r2 a f A/m2 .
(C) 1 A 4
(D) 2p A 3
The current density in a cylindrical wire of radius 8 mm placed along the z -axis is J = 50 az A/m2 . The total current flowing through the wire is r (B) 800 mA (A) 80.38 mA (D) 5.026 A
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(C) 0 A
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MCQ 3.1.3
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The total current crossing the surface defined by q = 90c, 0 < f < 2p, 0 < r < 1 m is (B) - p A (A) p A 2 2
Common Data for Question 4 - 5 : In a certain region current density is given by 20 sin f J = 40 a r - 2 az A/m2 r (r + 1)
MCQ 3.1.4
Total current crossing the plane z = 2 in the az direction for r < 4 will be (B) 1.5 mA (A) 0 A (C) - 32 A
MCQ 3.1.5
(D) 20 A
Volume charge density in the region at a particular point (r0, f0, z 0) will be (A) non uniform (B) linearly increasing with time
(C) linearly decreasing with time (D) constant with respect to time GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Statement for Linked Question 6 - 9 : In a cylindrical system, two perfectly conducting surfaces of length 2 m are located at r = 3 and r = 15 cm . The total current passing radially outward through the medium between the cylinders is 6 A. MCQ 3.1.6
If a conducting material having conductivity s = 0.05 S/m is present for 3 1 r # 5 cm then the electric field intensity at r = 4 cm will be (A) 238.7a r V/m (B) 150a r V/m (C) 318.3a r V/m
m
MCQ 3.1.7
(D) 0 V/m
The voltage between the cylindrical surfaces will be (A) 4.88 volt (B) 1.45 volt
(C) 0.5 W MCQ 3.1.9
(D) 8.13 W
The total dissipated power in the conducting material will be (A) 175.7 W (B) 18 W (C) 29.3 W
(D) 0.8 W
A solid wire of radius r and conductivity s1 has a jacket of material having conductivity s2 . If the inner and outer radius of the jacket are r and R respectively then the ratio of the current densities in the two materials will
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MCQ 3.1.10
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The resistance between the cylindrical surfaces will be (A) 0.813 W (B) 2.44 W
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MCQ 3.1.8
(D) 3 volt
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(C) 2.32 volt
(A) depend on r only
(B) depend on R only
(C) depend on both r and R
(D) independent of both r and R
Statement for Linked Question 11 - 12 : Atomic hydrogen contains 5.5 # 1019 atom/cm3 at a certain temperature and pressure. If an electric field of 40 kV/m is applied, each dipole formed by the electron and positive nucleus has an effective length of 7.1 # 10-16 m . MCQ 3.1.11
The polarization due to the induced dipole will be (A) 12.5 nC/m2 (B) 8.8 # 106 C/m2 (C) 6.25 nC/m2
MCQ 3.1.12
(D) 3.9 # 109 C/m2
Dielectric constant of the atomic hydrogen will be (A) 2.77 (B) 1.0177 (C) 0.982
(D) 0.0177
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For View Only Shop Online at www.nodia.co.in MCQ 3.1.13 The dielectric constant of the material in which the electric flux density is double of the polarization is (A) 2 (B) 1/2 (C) 3
(D) 1
Statement for Linked Question 14- 15 :
Electric field intensity in the material will be (A) 50ay V/m (B) 500ay V/m
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MCQ 3.1.14
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The potential field in a slab of a dielectric material that has the relative permittivity er = 6/5 is given by V =- 500y .
(C) - 500ay V/m
The electric flux density inside the material will be (A) 4.43 nC/m2
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MCQ 3.1.15
(D) 0
(B) 3.54ay mC/m2
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(C) 8.85 nC/m2 (D) 7.08ay nC/m2
Polarization of the material will be (A) 2.66ay nC/m2
(B) 14.08 nC/m2
(C) 5.31 # 10-12 ay C/m2
(D) 3ay C/m2
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MCQ 3.1.16
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Statement for Linked Question 17 - 18 :
MCQ 3.1.17
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Two perfect dielectrics with dielectric constant er1 = 2 and er2 = 5 are defined in the region 1 (y $ 0) and region 2 (y < 0) respectively. Consider the electric field intensity in the 1st region is given by E1 = 25ax + 20ay - 10az kV/m The Flux charge density in the 2 nd region will be (A) 2.21ax + 0.35ay - 0.44az mC/m2 (B) 2.21ax + 0.35ay - 0.44az nC/m2 (C) 2.21ax + 0.88ay - 0.44az nC/m2 (D) 0.4ax + 0.07ay - 0.08az nC/m2 MCQ 3.1.18
The energy density in the 2 nd region will be (A) 66.37 mJ/m3 (B) 118 mJ/m3 (C) 472 # 106 J/m3
(D) 59 mJ/m3
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For View Only Shop Online at www.nodia.co.in MCQ 3.1.19 The electric field in the three regions as shown in the figure are respectively E1 , E2 and E 3 and all the boundary surfaces are charge free.
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If e1 = e3 ! e2 , then the correct relation between the electric field is (A) E1 ! E2 ! E 3 (B) E1 = E 3 ! E2
(C) 4E 0 az and E 0 az 4
(D) e0 E 0 az and 4E 0 az
The energy stored in an electric field made up of two fields E1 and E2 is Wnet where as the energies stored in individual fields E1 and E2 are W1 and W2 respectively so the correct relation between the energies is (A) W = W1 + W2
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MCQ 3.1.21
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An infinite plane dielectric slab of thickness d and having permittivity e = 4e0 occupies the region 0 < z < d . If a uniform electric field E = E 0 az is applied in the free space then the electric flux density(Din ) and electric field intensity(Ein ) inside the dielectric slab will be respectively (B) e0 E 0 az and E 0 az (A) E 0 az and e0 E 0 az 4 4
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MCQ 3.1.20
(D) E1 = E2 ! E 3
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(C) E1 = E2 = E 3
(B) W = W1 W2
(C) W > W1 + W2
(D) W < W1 + W2 MCQ 3.1.22
An electric dipole is being placed in an electric field intensity E = 1.5ax - az V/m If the moment of the dipole be p =- 4ax + 3ay C- m then energy of the dipole will be (A) 6 J (B) 0 J (C) - 3 J
MCQ 3.1.23
(D) + 3 J
When a neutral dielectric is being polarized in an electric field then the total bound charge of the dielectric will be (A) zero (B) positive (C) negative (D) depends on nature of dielectric
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Chap 3
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Statement for Linked Question 24 - 25 :
m
A lead bar of square cross section has a hole of radius 2.5 cm bored along its length as shown in the figure.
If the length of the lead bar is 8 m then the resistance between the square ends of the bar will be (A) 1.78 mW (B) 3.64 mW
lp.
MCQ 3.1.24
co
(Conductivity of lead = 5 # 106 (Wm) -1 )
(D) 269 mW
(C) 1.95 mW
(C) 3.708 mW MCQ 3.1.26
at e
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If the hole in the lead bar is completely filled with copper then the resistance of the composite bar will be (Resistivity of copper -8 = 1.72 # 10 Wm) (A) 188 mW (B) 924.6 mW (D) 1.76 mW
A cylindrical wire of length l and cross sectional radius r is formed of a material with conductivity 106 (Wm) -1 . If the total conductance of the wire is 106 (W) -1 then the correct relation between l and r is l (A) r = p (B) r = p l (D) r = l
ww
(C) 2pr = l
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MCQ 3.1.25
Statement for Linked Question 27 - 28 : A capacitor is formed by two concentric conducting spherical shells of radii a = 1 cm and b = 2 cm centered at origin. Interior of the spherical capacitor is a perfect dielectric with er = 4 . MCQ 3.1.27
The capacitance of the capacitor will be (A) 8.9 pF (B) 2.25 pF (C) 890 pF
(D) 225 pF
If a portion of dielectric is removed from the capacitor such that er = 1 for p2 < f < p and er = 4 for the rest of the portion, then the capacitance of the composite capacitor will be GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 3.1.28
Chap 3
Electric Field in Matter
For View Only (A) 0.56 pF
Shop Online at www.nodia.co.in (B) 946 pF
(C) 236.5 pF MCQ 3.1.29
(D) 6.7 pF
Two conducting surfaces are present at x = 0 and x = 5 mm and the space between them are filled by dielectrics such that er1 = 2.5 for 0 < x < 1 mm and er2 = 4 for 1 < x < 3 mm rest of the region is air filled. The capacitance per square meter of surface area will be (A) 22.1 nF/m2 (B) 3.05 nF/m2
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Two coaxial conducting cylinders of radius 4 cm and 8 cm is lying along z -axis. The region between the cylinders contains a layer of dielectric from r = 4 cm to r = 12 cm with er = 4 . If the length of cylinders is 1 m then the capacitance of the configuration will be (A) 0.55 pF (B) 7 # 109 F (D) 143 pF
lp.
(C) 1.83 nF
A parallel plate capacitor is quarter filled with a dielectric ( er = 3 ) as shown in the figure. The capacitance of the capacitor will be
MCQ 3.1.32
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MCQ 3.1.31
(D) 44.25 nF/m2
m
(C) 442.5 nF/m2 MCQ 3.1.30
133
(A) 1.38 pF
(B) 2.76 pF
(C) 9.95 pF
(D) 6 pF
Medium between the two conducting parallel sheets of a capacitor has the permittivity e and conductivity s. The time constant of the capacitor will be (B) s (A) e s e (C) se
(D) 1/se
*********** GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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EXERCISE 3.2
m
For View Only
Common Data for Question 1 -2 :
co
In spherical coordinate system, the current density in a certain region is given by J = 2 e-10 t ar A/m2 r At t = 1 ms , how much current is crossing the surface r = 5 ? (A) 75.03 A (B) 27.7 A
lp.
4
MCQ 3.2.1
(C) 0.37 A
he
At a particular time t , the charge density rv (r, t) at any point in the region is directly proportional to. (Assume rv " 0 as t " 3) 1 (B) (A) r r (C) 12 (D) r2 r The velocity of charge density at r = 0.6 m will be (A) 6ar m/s
w. g
MCQ 3.2.3
(D) 2.77 A
at e
MCQ 3.2.2
Chap 3
(B) 1000ar m/s
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(C) 0.6 # 10-3 ar m/s (D) 600ar m/s
Statement for Linked Question 4 - 5 : Two uniform infinite line charges of 5 pC/m each are located at x = 0 , y = 1 and x = 0 , y = 2 respectively. Consider the surface y = 0 is a perfect conductor that has the zero potential. MCQ 3.2.4
Electric potential at point P (- 1, - 2, 0) will be (A) 1.2 volt (B) - 0.2 volt (C) + 0.2 volt (D) - 0.04 volt
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For View Only MCQ 3.2.5 Electric field at the point P will be (A) 0.12ax - 0.003ay V/m
135
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(B) 0.12ax - 0.086ay V/m (C) 723ax - 18.9ay V/m (D) 0.024ax - 0.086ay V/m A thin rod of certain cross sectional area extends along the y -axis from y = 0 m to y = 5 m . If the polarization of the rod is along it’s length and is given by Py = 2y2 + 3 then the total bound charge of the rod will be (A) 0 (B) 50 C
m
MCQ 3.2.6
(C) 48 C
co
A neutral atom of polarizability a is situated at a distance 1 m from a point charge 1/9 nC. The force of attraction between them will be (A) 2a N (B) 2a N 9
lp.
MCQ 3.2.7
(D) can’t be determined
(D) 18a N
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(C) 9a N
Common Data for Question 8 - 9 :
MCQ 3.2.8
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The two dipoles P1 , P2 with dipole moment 4 nC- m and 9 nC- m respectively are placed at 1 m distance apart as shown in figure.
The torque on P2 due to P1 will be (A) 18 # 10-18 N- m (B) 2 nN- m
(C) 8.1 N- m
(D) 0.16 mN- m MCQ 3.2.9
The torque on P1 due to P2 will be (A) 3.24 # 10-7 N- m -7
(C) 1.62 # 10
N-m
(B) 2 nN- m (D) 3.24 mN- m
Common Data for Question 10 -11 : A sphere carries a polarization P (r)= 3rar where r is the distance from the center of the sphere. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Chap 3
If the radius of the sphere is a then the electric field outside the sphere will be (A) - 4pa3 (B) 8pa3
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MCQ 3.2.11
at e
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lp.
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For View Only Shop Online at www.nodia.co.in MCQ 3.2.10 Consider Er is the electric field component in the radial direction inside the sphere. The plot of Er with respect to r will be
(C) 0
(D) - 8pa3
MCQ 3.2.12
MCQ 3.2.13
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Statement for Linked Question 12 - 13 : A thick spherical shell is made of dielectric material with a polarization where r is the distance from its centre. P (r) = 5 ar nC/m2 r If the spherical shell is centred at origin and has the inner radius 2 m and outer radius 6 m then the electric field intensity at r = 1 m will be (A) 0 (B) - 40p V/m (C) 20p V/m
(D) - 20p V/m
Electric field at r = 7 m will be (A) - 100p V/m
(B) - 140p V/m
(C) 0
(D) - 20p V/m
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137
For View Only Shop Online at www.nodia.co.in MCQ 3.2.14 Electric field intensity at r = 5 m will be (B) 1 ar (A) - 2 ar e0 5e0 (C) - 1 ar e0
A spherical conductor of radius 1 m carries a charge 3 mC. It is surrounded, out to radius 2 m, by a linear dielectric material of dielectric constant er = 3 , as shown in the figure. The energy of this configuration will be
(C) 270 J
(D) 324 J
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A sphere of radius 2/ p m is made of dielectric material with dielectric constant er = 2 . If a uniform free charge density 0.6 nC/m3 is embedded in it then the potential at the centre of the sphere will be (B) 5.4 volt (A) 3 volt
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MCQ 3.2.16
(B) 500 J
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(A) 27 kJ
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MCQ 3.2.15
(D) 1 ar 5e0
(C) 0 volt
(D) 9 volt
Statement for Linked Question 17 - 19 : A short cylinder of radius r and length L carries a uniform polarization P , parallel to its axis as shown in the figure.
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Chap 3
For View Only Shop Online at www.nodia.co.in MCQ 3.2.17 Total bound charge by the cylinder will be (A) 2P coulomb (B) P coulomb (D) - P coulomb
(C) 0 coulomb
If L = 2r then the electric field lines of the cylinder will be as
MCQ 3.2.19
The lines of flux charge density will be as
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MCQ 3.2.18
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Electric Field in Matter
139
For View Only Shop Online at www.nodia.co.in MCQ 3.2.20 A parallel plate capacitor is filled with a non uniform dielectric characterized by er = 3 (1 + 50a2) where a is the distance from one plate in meter. If the surface area of the plates is 0.2 m2 and separation between them is 10 cm then the capacitance of the capacitor will be (A) 22.5 pF (B) 90.2 pF (C) 45.1 pF
A two wire transmission line consists of two perfectly conducting cylinders, each having a radius of 0.2 cm, separated by a centre to centre distance of 2 cm. The medium surrounding the wires has relative permittivity er = 2 . If a 100 V source is connected between the wires then the stored charge per unit length on each wire will be (A) 3.64 nC/m (B) 3.64 # 10-11 C/m
co
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MCQ 3.2.21
(D) 4.51 pF
(D) 2.5 # 10-8 C/m
(C) 1.82 nC/m
lp.
A tank is filled with dielectric oil of susceptibility ce = 1. Two long coaxial cylindrical metal tubes of radii 1 mm and 3 mm stand vertically in the tank as shown in the figure. The outer tube is grounded and inner one is maintained at 2 kV potential. To what height does the oil rise in the space between the tubes ? (mass density of oil = 0.01 gm/cm3 )
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MCQ 3.2.22
MCQ 3.2.23
(A) 41.1 mm
(B) 45.5 mm
(C) 20.5 mm
(D) 82.4 mm
An infinite plane conducting slab carries uniformly distributed surface charges on both of it’s surface. If the sum of the charge densities on the two surfaces is rso C/m2 then the surface charge densities on the two surfaces will be (A) rso /2 , rso /2 (B) 2rso , - rso (C) 0, rso
(D) None of these
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Chap 3
co
m
For View Only Shop Online at www.nodia.co.in MCQ 3.2.24 Two infinite plane parallel conducting slabs carry uniformly distributed surface charges rS11 , rS12 , rS21 and rS22 on all the four surfaces as shown in the figure.
lp.
Which of the following gives the correct relation between the charge densities ? (A) rs11 = rs22 , rs12 = rs21 (B) rs11 = rs22 , rs12 =- rs21 (D) rs11 =- rs22 , rs12 =- rs21
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(C) rs11 = rs12 , rs21 = rs22
at e
Common data for Question 25 - 26 :
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The plane surfaces x = 0 , x = 1, y = 0 and y = 1 form the boundaries of conductors extending away from the region between them as shown in the figure.
If the electrostatic potential in the region between the surfaces is given by 6xy volts then the surface charge density on the surface ; MCQ 3.2.25
MCQ 3.2.26
x = 0 is (A) - 5e0 y
(B) - 5e0 x
(C) - 5e0 ^x + y h
(D) 5e0 ^xy h
y = 0 is (A) - 5e0 y
(B) - 5e0 x
(C) - 5e0 ^x + y h (D) 5e0 xy GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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141
co
m
For View Only Shop Online at www.nodia.co.in MCQ 3.2.27 Two infinitely long coaxial, hollow cylindrical conductors of inner radii 2 m and 5 m respectively and outer radii 3 m and 6 m, respectively as shown in the figure, carry uniformly distributed surface charges on all four of their surfaces.
A conducting spherical shell of inner radii 2 m and outer radii 3 m carry uniformly distributed surface charge on it’s inner and outer surfaces. If the net surface charge is 9 C for the conducting spherical shell then, the surface charge density on inner and outer surfaces are respectively (B) 1 C/m2 , 0 (A) 0, 1 C/m2 4p 4p
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MCQ 3.2.28
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lp.
If net surface charge per unit length is 10 C/m and 6 C/m for the inner and outer conductor respectively then the surface charge densities on the four surface will be Surface " r = 2m r = 3m r = 5m r =6m (A) 0 5/3p - 1/p 4/3p (B) 0 5/3p - 1/p 4/3p (C) 1/p - 1/p 2/p - 2/p (D) 0 0 - 1/p 1/p
(C) 0, 4p C/m2 MCQ 3.2.29
(D) 4p C/m , 0
Plane z = 0 defines a surface charge layer with the charge density rS = 3n C/m2 as shown in figure. If the electric field intensity in the region z < 0 is E2 = 2ax + 3ay - 2az V/m then the field intensity E1 in the region z > 0 will be
(A) 220ax + 219ay - 2az
(B) 2ax + 3ay + 224az
(C) 222ax + 221ay + 2az (D) 2ax + 3ay + 226az GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 3.2.30 An infinite plane dielectric slab with relative permittivity er = 5 occupies the region x > 0 . If a uniform electric field E = 10ax V/m is applied in the region x < 0 (free space) then the polarization inside the dielectric will be (A) 8e0 ax C/m2 (B) 4e0 ax C/m2 (C) 2e0 ax C/m2
(D) 10e0 ax C/m2
m
Statement for Linked Question 31 - 32 :
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lp.
co
An infinite plane dielectric slab of 1 m thickness is placed in free space such that it occupies the region 0 < y < 1 m as shown in the figure.
at e
Dielectric slab has the non uniform permittivity defined as 2e0 e= ^1 + 3y h2
for 0 < y < 1
If a uniform electric field E = 4ay V/m is applied in free space then bound surface charge densities on the surface y = 0 and y = 1 will be at y = 0 at y = 1 (A) 0 - 3e0 (B) 0 - 3e0 (C) 0 3e0 (D) - 5e0 8e0
MCQ 3.2.32
As we move from the surface y = 0 toward the surface y = 1 inside the dielectric slab, polarization volume charge density will be (A) linearly increasing (B) linearly decreasing
ww
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MCQ 3.2.31
(C) Constant MCQ 3.2.33
(D) zero at all points
In a spherical coordinate system the region a < r < b is occupied by a dielectric material. A point charge Q is situated at the origin. It is found that the electric field intensity inside the dielectric is given by Q for a < r < b E = ar 4pe0 b2
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143
For View Only Shop Online at www.nodia.co.in The relative permittivity of the dielectric will be (A) ^b2 /r2h (B) ^a2 /r2h (C) ^r2 /a2h
Two perfectly conducting, infinite plane parallel sheets separated by a distance 2 m carry uniformly distributed surface charges of equal and opposite densities + 5 nC/m2 and - 5 nC/m2 respectively. If the medium between two plates is a dielectric of uniform permittivity e = 4e0 then the potential difference between the two plates will be (A) 283 KV (B) 1130 KV
m
MCQ 3.2.34
(D) ^a2 /b2h
(C) 283 V
co
The medium between two perfectly conducting infinite plane parallel sheets consists of two dielectric slabs of thickness 1 m and 2 m having permittivities e1 = 2e0 and e2 = 4e0 respectively as shown in the figure.
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MCQ 3.2.35
(D) 1.13 KV
If the conducting sheets carry uniformly distributed surface charges of equal and opposite densities 0.6 nC/m2 and - 0.6 nC/m2 respectively then the potential difference between the sheets will be (A) 67.8 Volt (B) 6.78 Volt (C) 33.9 Volt MCQ 3.2.36
(D) 17.4 Volt
Two perfectly conducting, infinite plane parallel sheets separated by a distance d carry uniformly distributed surface charges of equal and opposite densities rS0 and - rS0 respectively. The medium between the sheets is filled by a dielectric of non uniform permittivity which varies linearly from a value of e1 near one plate to value of e2 near the second plate. The potential difference between the two sheets will be rS0 r d (B) (A) S0 ln e2 e2 - e1 d ^e2 - e1h a e1 k (C)
rS0 d ln e2 e2 - e1 a e1 k
(D) rS0 ln a e2 k e1
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Chap 3
For View Only Shop Online at www.nodia.co.in MCQ 3.2.37 Two perfectly conducting, infinite plane parallel sheet separated by a distance 0.5 cm carry uniformly distributed surface charges of equal and opposite densities. If the potential difference between the two plates is 7 kV and the medium between the plates is free space then the charge densities on the plates will be (A) 6.23 mC (B) 88.5 mC (C) 8.85 mC
A parallel plate capacitor has two layers of dielectrics with permittivities e1 = 3e0 and e2 = 2e0 as shown in the figure.
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lp.
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MCQ 3.2.38
(D) 17.7 mC
(C)
Volt ,
18 11
Volt
(D) 6 Volt, 3 Volt
A dielectric slab is inserted in the medium between two plates of a capacitor as shown in the figure
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MCQ 3.2.39
81 11
at e
If the total voltage drop in the capacitor is 9 Volt then the voltage drop in 1st and 2 nd dielectric region will be respectively 81 (A) 18 (B) 3 Volt, 6 Volt 11 Volt , 11 Volt
The capacitance across the capacitor will remain constant (A) if the slab is moved rightward or leftward (B) if the slab is pulled outward of the capacitor (C) (A) and (B) both (D) none of these GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 3
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145
For View Only Shop Online at www.nodia.co.in MCQ 3.2.40 A steel wire has a radius of 2 mm and a conductivity of 2 # 106 S/m. The steel wire has an aluminium (s = 3.8 # 107 S/m) coating of 2 mm thickness. The total current carried by this hybrid conductor be 80 A. The current density in steel Jst is (B) 3.2 # 105 A/m2 (A) 1.02 # 106 A/m2 (D) 1.10 # 105 A/m2 (C) 2.04 # 105 A/m2 A potential field in free space is given as f V = 40 cos q sin 2 V r Point P (r = 2, q = p/3, f = p/2) lies on a conducting surface. The equation of the conducting surface is (B) 16 cos f sin q = r3 (A) 32 cos q sin f = r3
m
MCQ 3.2.41
(D) 32 cos f sin q = r3
lp.
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(C) 16 cos q sin f = r3
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EXERCISE 3.3
A parallel plate air-filled capacitor has plate area of 2 # 10-4 m2 and plate separation of 10-3 m . It is connected to a 0.5 V, 3.6 GHz source. The magnitude of the displacement current is ( e = 361p 10-9 F/m) (A) 10 mA (B) 100 mA
m
GATE 2004 2003
Shop Online at www.nodia.co.in
co
MCQ 3.3.1
(C) 10 A
Medium 1 has the electrical permittivity e1 = 1.8e0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity e2 = 2.5e0 farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2ax - 3ay + 1az ) volt/m, then E2 in medium 2 is (A) (2.0ax - 7.5ay + 2.5az ) volt/m
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GATE 2003
(D) 1.59 mA
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MCQ 3.3.2
Chap 3
(B) (2.0ax - 2.0ay + 0.6az ) volt/m
at e
(C) (2.0ax - 3.0ay + 1.0az ) volt/m
(D) (2.0ax - 2.0ay + 0.6az ) volt/m GATE 2002
The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with e = 80eo . The surface charge density on the conductor is ( e = 10 36p F/m) (A) 0 C/m 2 (B) 2 C/m 2 -9
w. g
MCQ 3.3.3
(C) 1.8 # 10-11 C/m 2 IES EC 2012
The space between the plates of a parallel-plate capacitor of capacitance C is filled with three dielectric slabs of identical size as shown in the figure. If dielectric constants are e1 , e2 and e3 , the new capacitance is
ww
MCQ 3.3.4
(D) 1.41 # 10-9 C/m 2
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Electric Field in Matter
For View Only
Shop Online at www.nodia.co.in ^e1 + e2 + e3h C (B) 3
(A) C 3 (C) ^e1 + e2 + e3h C MCQ 3.3.5 IES EC 2011
147
(D)
9 ^e1 + e2 + e3h e1 e2 e3
If the potential, V = 2x + 4 V , the electric field is (A) 6 V/m (B) 2 V/m (C) 4 V/m
Two dielectric media with permittivities 3 and boundary as shown in figure below :
3 are separated by a charge-free
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IES EC 2011
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MCQ 3.3.6
m
(D) - 4ax V/m
The electric field intensity in media 1 at point P1 has magnitude E1 and makes an angle a1 = 60c with the normal. The direction of the electric field intensity at point P2, a2 is (B) 45c (A) sin-1 c 3 E1 m 2 (C) cos-1 c MCQ 3.3.7 IES EC 2011
3 E1 2 m
(D) 30c
Assertion (A) : Under static conditions, the surface of conductor is an equipotential surface. Reason (R) : The tangential component of electric field on conductor surface is zero. (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A) (B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true
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For View Only Shop Online at www.nodia.co.in MCQ 3.3.8 A long 1 metre thick dielectric (e = 3e0) slab occupying the region 0 < x < 5 is IES EC 2010 placed perpendicularly in a uniform electric field E 0 = 6ax . The polarization Pi inside the dielectric is (B) 8e0 ax (A) 4e0 ax (C) 36e0 ax MCQ 3.3.9
The flux and potential functions due to a line charge and due to two concentric circular conductors are of the following form : (A) Concentric circular equipotential lines and straight radial flux lines.
m
IES EC 2010
(D) Zero
(B) Concentric circular flux lines and straight equipotential lines
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(C) Equipotentials due to the charge are concentric cylinders and equipotentials due to two conductors are straight lines.
IES EC 2010
There are two conducting plates of sizes 1 m # 1 m and 3 m # 3 m. Ratio of the capacitance of the second one with respect to that of the first one is (A) 4 (B) 2
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MCQ 3.3.10
lp.
(D) Equipotentials due to line charge are straight flat surfaces and those due to two conductors are concentric cylinders.
(C) 1/2 IES EC 2010
Consider the following : In a parallel plate capacitor, let the charge be held constant while the dielectric material is replaced by a different dielectric. Consider 1. Stored energy 2. Electric field intensity.
IES EC 2009
Which of these changes ? (A) 1 only
(B) 1 and 2 only
(C) 2 and 3 only
(D) 1, 2 and 3
By what name is the equation d : J = 0 frequently known ? (A) Poisson’s equation
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MCQ 3.3.12
Capacitance
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MCQ 3.3.11
(D) 1/4
(B) Laplace’s equation (C) Continuity equation for steady currents (D) Displacement equation MCQ 3.3.13 IES EC 2009
Method of images is applicable to which fields ? (A) Electrostatic fields only (B) Electrodynamic fields only (C) Neither electrostatic fields nor electrodynamic fields (D) Both electrostatic fields and electrodynamic fields
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149
For View Only Shop Online at www.nodia.co.in MCQ 3.3.14 What is the unit of measurement of surface or sheet resistivity? IES EC 2008 (A) Ohm/metre (B) Ohm metre (C) Ohm/sq. meter MCQ 3.3.15 IES EC 2007
(D) Ohm
Which one of the following statements is correct ? On a conducting surface boundary, electric field lines are (A) always tangential (B) always normal (C) neither tangential nor normal
IES EC 2007
Which one of the following is correct ? As frequency increases, the surface resistance of a metal (A) decreases
co
MCQ 3.3.16
m
(D) at an angle depending on the field intensity
(C) remains unchanged
lp.
(B) increases (D) varies in an unpredictable manner
Application of the method of images to a boundary value problem in electrostatics involves which one of the following ? (A) Introduction of an additional distribution of charges and removal of a set of conducting surfaces
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IES EC 2007
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MCQ 3.3.17
(B) Introduction of an additional distribution of charge and an additional set of conducting surfaces (C) Removal of a charge distribution and introduction of an additional set of conducting surfaces (D) Removal of a charge distribution as well as a set of conducting surfaces MCQ 3.3.18 IES EC 2006
Assertion (A) : Potential everywhere on a conducting surface of infinite extent is zero. Reason (R) : Displacement density on a conducting surface is normal to the surface. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true
MCQ 3.3.19 IES EC 2006
A parallel plate capacitor of 5 pF capacitor has a charge of 0.1 mC on its plates. What is the energy stored in the capacitor ? (A) 1 mJ (B) 1 μJ (C) 1 nJ
(D) 1 pJ
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150
Electric Field in Matter
Chap 3
For View Only Shop Online at www.nodia.co.in MCQ 3.3.20 A charge of 1 Coulomb is placed near a grounded conducting plate at a distance of IES EC 2006 1 m. What is the force between them ? (B) 1 N (A) 1 N 8pe0 4pe0 (C) MCQ 3.3.21
(D) 4pe0 N
The capacitance of a parallel plate capacitor is given by e ed A where A is the area of each plates. Considering fringing field, under which one of the following conditions is the above expression valid ? (B) A is tending towards infinity (A) A is tending towards zero d d (D) A is 1 (C) A is 1 er e0 d d 0 r
IES EC 2005
What is the expression for capacitance of a solid infinitely conducting solid sphere of radius ‘R’ in free space ? (B) 4pe0 R (A) 2pe0 R
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MCQ 3.3.22
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IES EC 2005
1 N 16pe0
MCQ 3.3.23
A point charge of + 10 mC placed at a distance of 5 cm from the centre of a conducting grounded sphere of radius 2 cm is shown in the diagram given below :
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IES EC 2004
(D) 0.5pe0 R
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(C) 8pe0 R
What is the total induced charge on the conducting sphere ? (A) 10 μC (B) 4 μC
MCQ 3.3.24 IES EC 2004
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(C) 5 μC
For an electric field E = E 0 sin wt , what is the phase difference between the conduction current and the displacement current ? (A) 0c (B) 45c (C) 90c
MCQ 3.3.25 IES EC 2004
(D) 12.5 μC
(D) 180c
An infinitely long line charge of uniform charge density rL C/m is situated parallel to and at a distance from the grounded infinite plane conductor. This field problem can be solved by which one of the following ? (A) By conformal transformation (B) By method of images
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(D) By Poisson’s equation MCQ 3.3.26 IES EC 2003
An air condenser of capacitance of 0.002 mF is connected to a d.c. supply of 500 Volts , disconnected and then immersed in oil with a dielectric constant of 2.5. Energy stored in the capacitor before and after immersion, respectively is (A) 500 # 10-4 J and 250 # 10-4 J (B) 250 # 10-4 J and 500 # 10-4 J
IES EC 2001
A 3 mF capacitor is charged by a constant current of 2 mA for 6 seconds. The voltage across the capacitor to the end of charging will be (A) 3 V (B) 4 V
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MCQ 3.3.27
m
(C) 625 # 10-4 J and 250 # 10-4 J (D) 250 # 10-4 J and 625 # 10-4 J
Consider the following statements : A parallel plane capacitor is filled with a dielectric of relative permittivity er1 and connected to a d.c. voltage of V volts. If the dielectric is changed to another with relative permittivity er1 = 2er1 , keeping the voltage constant, then 1. the electric field intensity E within the capacitor doubles.
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IES EC 2001
2.
the displacement flux density D doubles
3.
the charge Q on the plates is reduced to half.
4.
the energy stored in the capacitor is doubled.
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MCQ 3.3.28
(D) 9 V
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(C) 6 V
Select the correct answer using the codes given below : (A) 1 and 2 (B) 2 and 3 (C) 2 and 4 MCQ 3.3.29 IES EC 2001
A coil of resistance 5 W and inductance 0.4 H is connected to a 50 V d.c. supply. The energy stored in the field is (A) 10 joules (B) 20 joules (C) 40 joules
MCQ 3.3.30 IES EE 2011
(D) 3 and 4
(D) 80 joules
The normal components of electric flux density across a dielectric-dielectric boundary (A) are discontinuous (B) are continuous (C) depend on the magnitude of the surface charge density (D) depend on electric field intensity
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Chap 3
For View Only Shop Online at www.nodia.co.in MCQ 3.3.31 Consider the following statements in connection with boundary relations of electric IES EE 2009 field : 1. In a single medium electric field is continuous. The tangential components are the same on both sides of a boundary between two dielectrics.
3.
The tangential electric field at the boundary of a dielectric and a current carrying conductors with finite conductivity is zero.
4.
Normal components of the flux density is continuous across the charge-free boundary between two dielectrics.
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Which of these statements is/are correct ? (A) 1 only (B) 1, 2 and 3
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(C) 1, 2 and 4 (D) 3 and 4 only IES EE 2008
The capacitance of an insulated conducting sphere of radius R in vacuum is (A) 2pe0 R
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MCQ 3.3.32
(B) 4pe0 R (D) 4pe0 /R
A parallel plate air capacitor carries a charge Q at its maximum withstand voltage V . If the capacitor is half filled with an insulating slab of dielectric constant 4 as shown in the figure given below, what are the maximum withstand voltage and the charge on the capacitor at this voltage, respectively ?
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IES EE 2007
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(C) 4pe0 R2
MCQ 3.3.33
m
2.
MCQ 3.3.34 IES EE 2007
(A) 2.5V , Q
(B) 4 V, 2.5 Q
(C) V, 2.5 Q
(D) V/4, Q
When an infinite charged conducting plate is placed between two infinite conducting grounded surfaces as shown in the figure given below, what would be the ratio of the surface densities r1 and r2 on the two sides of the plate ?
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(d1 + t) (d2 + t) (C) d1 d2
(d2 + t) (d1 + t) (D) d2 d1
(B) D = e0 ^E + P h
(C) D = e0 E + P (D) E = D + e0 P IES EE 2006
Image theory is applicable to problems involving (A) electrostatic field only
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MCQ 3.3.36
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The polarization in a solid dielectric is related to the electric field E and the electric flux density D according to which on of the following equations ? (A) E = e0 D + P
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IES EE 2007
(B)
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(A)
MCQ 3.3.35
m
For View Only
153
(B) magnetostatic field only
(C) both electrostatic and magnetostatic fields (D) neither electrostatic nor magnetostatic field MCQ 3.3.37 IES EE 2006
Six capacitors of different capacitances C1, C2, C 3, C 4, C5 and C6 are connected in series. C1 > C2 > C 3 > C 4 > C5 > C6 . What is the total capacitance almost equal to ? (A) C1 (B) C 3 (C) C 4
MCQ 3.3.38 IES EE 2004
(D) C6
Two extensive homogeneous isotropic dielectrics meet on a plane z = 0 . For z $ 0 , er1 = 4 and for z # 0 , er2 = 3 . A uniform electric field exists at z $ 0 as E1 = 5ax - 2ay + 3az kw/m . What is the value of E2z in the region z # 0 ? (A) 3az
(B) 5ax - 2ay
(C) 6az
(D) ax - ay
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Chap 3
For View Only Shop Online at www.nodia.co.in MCQ 3.3.39 A flat slab of dielectric, er = 5 is placed normal to a uniform field with a flux density IES EE 2004 D = 1 Coulomb/m2 . The slab is uniformly polarized. What is the polarization P of the slab in Coulomb/m2 ? (A) 0.8 (B) 1.2 (C) 4 IES EE 2004
Which one of the following gives the approximate value of the capacitance between two spheres, whose separation is very much larger than their radii R ? (B) 2pe0 R (A) 2p/e0 R
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MCQ 3.3.40
(D) 6
IES EE 2003
Assertion (A) : For steady current in an arbitrary conductor, the current density is solenoidal Reason (R) : The reciprocal of the resistance is the conductivity. (A) Both A and R are true and R is the correct explanation of A
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MCQ 3.3.41
(D) 4pe0 /R
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(C) 2pe0 /R
(B) Both A and R are true but R is NOT the correct explanation of A (D) A is false but R is true IES EE 2003
Assertion (A) : Displacement current can have only a.c components. Reason (R) : It is generated by a change in electric flux. (A) Both A and R are true and R is the correct explanation of A
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MCQ 3.3.42
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(C) A is true but R is false
(B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false
IES EE 2003
A plane slab of dielectric having dielectric constant 5, placed normal to a uniform field with a flux density of 3 C/m2 , is uniformly polarized. The polarization of the slab is (A) 0.4 C/m2 (B) 1.6 C/m2
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MCQ 3.3.43
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(D) A is false but R is true
(C) 2.0 C/m2 MCQ 3.3.44 IES EE 2002
Ohm’s law in point form in the field theory can be expressed as (A) V = RI (B) J = E/s (C) J = sE
MCQ 3.3.45 IES EE 2002
(D) 6.4 C/m2
(D) R = rl/A
A medium behaves like dielectric when the (A) displacement current is just equal to the conduction current (B) displacement current is less than the conduction current (C) displacement current is much greater than the conduction current (D) displacement current is almost negligible
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For View Only Shop Online at www.nodia.co.in MCQ 3.3.46 A copper wire carries a conduction current of 1.0 A at 50 Hz. For copper wire IES EE 2002 e = e0, m = m0, s = 5.8 # 10 mho/m . What is the displacement current in the wire ? (A) 2.8 # 10 A (B) 4.8 # 10- 11 A (C) 1 A (D) It cannot be calculated with the given data
m
IES EE 2001
Assertion (A) : When there is no charge in the interior of a conductor the electric field intensity is infinite. Reason (R) : As per Gauss’s law the total outward electric flux through any closed surface constituted inside the conductor must vanish. (A) Both A and R are true and R is the correct explanation of A
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MCQ 3.3.47
(C) A is true but R is false (D) A is false but R is true
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IES EE 2001
A point charge + Q is brought near a corner of two right angle conducting planes which are at zero potential as shown in the given figure. Which one of the following configurations describes the total effect of the charges for calculating the actual field in the first quadrant ?
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MCQ 3.3.48
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(B) Both A and R are true but R is NOT the correct explanation of A
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MCQ 3.3.49
The electric field across a dielectric-air interface is shown in the given figure. The surface charge density on the interface is
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IES EE 2001
Shop Online at www.nodia.co.in
m
For View Only
Chap 3
(C) - 2e0 IES EE 2001
(D) - e0
When air pocket is trapped inside a dielectric of relative permittivity ‘5’, for a given applied voltage across the dielectric, the ratio of stress in the air pocket to that in the dielectric is equal to (A) 1/5 (B) 5 (C) 1 + 5
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MCQ 3.3.50
(B) - 3e0
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(A) - 4e0
(D) 5 - 1
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***********
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SOLUTIONS 3.1
I =
m
Option (A) is correct. For a given current density, the total current that passes through a given surface is defined as
# J : dS
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SOL 3.1.1
where dS is the differential surface area having the direction normal to the surface. So we have dS = rdrdfaz for the plane z = 0 Therefore, the current passing the plane z = 0 , 0 # r # 2 is 2p
2
z
f=0 r=0 2p 2
2p
0
z
0
2
r
he
0
+ az hB : (rdrdfaz )
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# # 810e ^r a = 10 # # e rdzdf = 10 # # rdrdf
I =
2
(z = 0 )
0
SOL 3.1.2
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r2 2 = 10 ; E # 6f@20p 2 0 = 10 # 3 # 2p = 60p A
Option (C) is correct. For a given current density, the total current that passes through a given surface is defined as I =
# J : dS
where dS is the differential surface area having the direction normal to the surface. So, we have dS = (r sin qdf) (dr) a q for the surface q = 90c Therefore, the total current crossing the surface q = 90c,0 < f < 2p,0 < r < 1 m is
# ^r cos qa + r sin qa = # # r sin qdfdr 2
I =
1
2p
r
3
2
q
- r2 afh : ^r sin qdfdraq h
2
r=0 f=0
at q = 90c
4 1 = 6f@20p # :r D = 2p # 1 = p A 8 4 0 4
SOL 3.1.3
Option (B) is correct. For a given current density, the total current flowing through a cross section is defined as I =
# J : dS
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For View Only Shop Online at www.nodia.co.in where dS is the differential cross sectional area vector having the direction normal to the cross section. So we have dS = rdrdfaz (since the cylindrical wire is lying along z -axis) Therefore the total current flowing through the wire (cross section) is I = b 50 az l : ^rdrdfaz h r
#
16 # 10-3
r=0
#
2p
f=0
50 b r l^rdrdfh
# 10 = 50 # 6r@16 # 6f@20p 0 -3
SOL 3.1.4
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= 60 # 16 # 10-3 # 2p
m
=
Option (C) is correct. For a given current density, the total current that passes through a given surface is defined as
#
he
2p
20 sin f e - r2 + 1 o^rdrdfh r=0
# #
4
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=
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I = J : dS where dS is the differential surface area having the direction normal to the surface. for the plane z = 2 So we have dS = rdrdfaz Therefore the total current crossing the plane z = 2 , r < 4 is 20 sin f I = e 40 a r - 2 az rdrdfaz h r (r + 1) o^ f=0
2p 20 rdr sin fdfE ; E r=0 r + 1 f=0 1 44 2 44 3 0 =4A 4
2
#
Option (B) is correct. From the equation of continuity we have the relation between the volume charge density, rv and the current density, J as 2rv =- d : J 2t Given the current density, 20 sin f J = 40 a r - 2 az A/m2 r (r + 1) 20 sin f So, we have the components J r = 40 ,Jf = 0 and Jz =- 2 r (r + 1) 2Jf 2Jz 2rv Therefore, =-= 1 2 ^rJ rh + 1 + r 2r r 2f 2z G 2t - 20 sin f = > 1 2 (40) + 2 e 2 r 2r 2z r + 1 oH =2 So, volume charge density will be constant with respect to time.
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SOL 3.1.5
#
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=-;
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Option (C) is correct. Voltage between the cylindrical surfaces is defined as the line integral of the electric field between the two surfaces
#
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SOL 3.1.7
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For View Only Shop Online at www.nodia.co.in SOL 3.1.6 Option (C) is correct. Given the current I = 6 A is flowing radially outward (in a r direction) through the medium between the cylinders. So the current density in the medium between the cylinders is 6 (l = 2 m ) a J = I ar = 2pr # 2 r 2prl = 5 a r A/m2 2pr For a given current density in a certain medium having conductivity s, the electric field intensity is defined as E = J = 1 b 3 ar l s 2pr s 3 ( r = 4 # 10-2 m , s = 0.05 S/m ) = 2p # 4 # 10-2 # 0.05 = 238.73 V/m
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i.e. V =- E : dl Now the electric field intensity in the medium between the two cylindrical surfaces as calculated in previous question is E = 1 b 3 ar l s 2pr and the differential displacement between the two cylindrical surfaces is dl = dra r So the voltage between the cylindrical surfaces is 5 # 10-2 3 a : dra =- 3 ln 5 V =rl b ^ rh 2ps b 3 l r = 3 # 10-2 2prs
#
=- 5.88 volt So, the voltage between them will be 5.88 volt . SOL 3.1.8
Option (C) is correct. As we have already calculated the voltage between the two cylindrical surfaces and the current flowing radially outward in the medium between the surfaces is given in the question. So the resistance between the cylindrical surface can be evaluated directly as (V = 4.88 volt ,I = 6 A ) R = V = 4.88 = 0.813 W 6 I
SOL 3.1.9
Option (A) is correct. Since voltage between the cylindrical surfaces is V = 4.88 volt and current flowing in the medium is I = 6A So, Power dissipated in the medium is P = VI = (4.88) # 6 = 29.28 watt
SOL 3.1.10
Option (B) is correct. Consider a constant voltage is applied across the ends of the wire so, the electric
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m
For View Only Shop Online at www.nodia.co.in field intensity throughout the wire cross section will be constant. i.e. E = J1 = J 2 s1 s2 where J1 is the current density in the material having conductivity s1 . J2 is the current density in the material having conductivity s2 . So,the ratio of the current density is J1 = s1 s2 J2 i.e. it will be independent of both r and R. Option (A) is correct. Since hydrogen atom contains a single electron (- ve charge) and a single proton ( + ve charge). So, the dipole moment due to one atom of the hydrogen will be where q is electronic charge and d is effective length p = qd i.e. q = 1.6 # 10-19 C and d = 7.1 # 10-16 m So, p = ^1.6 # 10-19h # ^7.1 # 10-16h and since the polarization in a material is defined as the dipole moment per unit volume. Therefore where n is the number of atoms per unit volume. P = np 19 i.e. n = 5.5 # 10 atoms/cm3 = 5.5 # 1025 atoms/m3 So, P = ^5.5 # 1025h # ^1.6 # 10-19 # 7.1 # 10-16h = 4.25 # 10-9 C/m2
SOL 3.1.12
Option (D) is correct. When an electric field E is applied to a material with dielectric constant er then the polarization of the material is defined as P = e0 (er - 1) E So,
6.25 # 10-9 er - 1 = P = = 1.7655 # 10-2 e0 E 8.85 # 10-12 # 40 # 103 er = 1 + 0.0177 = 1.0177
Option (C) is correct. Given D = 2P & P = D/2 If the polarization of a dielectric material placed in an electric field E is P , then the electric flux density in the material is defined as D = e0 E + P = e0 E + D/2 or ..........(i) D = 2e0 E and since the relation between the electric field, E and flux density, D inside a dielectric material with dielectric constant er is defined as D = e0 er E So, comparing the result with equation (i) we get, er = 2 .
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SOL 3.1.13
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SOL 3.1.11
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For View Only Shop Online at www.nodia.co.in SOL 3.1.14 Option (D) is correct. Electric field intensity is defined as the negative gradient of the potential i.e. E =- dV =-c2V ax + 2V ay + 2V az m 2x 2y 2z = 250ay V/m Option (B) is correct. For a given electric field intensity E in a material having relative permittivity er , the electric flux density is defined as : D = er e0 E er = 8/5 = 8 # ^8.85 # 10-12h # (500ay) 5 = 9.08ay nC/m2
SOL 3.1.16
Option (C) is correct. For an applied electric field intensity E in a material having relative permittivity er , the polarization of the material is defined as P = e0 ^er - 1h E = ^8.85 # 10-12h (1.6 - 1) # ^500ay h = 8.85 # 10-12 # 0.6 # 500ay = 4.66 # 10-9 ay
SOL 3.1.17
Option (D) is correct. Since the two regions is being separated by the plane y = 0 , so the tangential and normal component of the electric field to the plane y = 0 are given as E1t = 50ax - 10az E1n = 20ay From the boundary condition, the tangential component of electric field will be uniform. i.e. E2t = E1t = 50ax - 10az and the normal component of the field is nonuniform and given as e2 E2n = e1 E1n E1n = e1 E1n = 2 ^20ay h = 8ay e2 5 So the electric field intensity in the second region is E2 = E2t + E2n = ^50ax - 10az h + ^8ay h = 50ax + 8ay - 10az kV/m Therefore the electric flux density in the region 2 is D2 = er2 e0 E2 = 5 # 8.85 # 10-12 (50ax + 8ay - 10az ) # 103 = 3.21ax + 0.55ay - 0.44az mC/m2
SOL 3.1.18
Option (B) is correct. Energy density in the region having electric field intensity E2 is defined as
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SOL 3.1.15
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WE = 1 er2 e0 E2 : E2 ,where the relative permittivity of the medium is er2 2 As calculated in previous question the electric field intensity is E2 = 50ax + 8ay - 10az kV/m So the energy density in the region 2 is WE = 1 # 5 # e0 6(50) 2 + (8) 2 + (10) 2@ # 106 2 = 79 mJ/m3 Option (D) is correct. According to boundary condition the tangential components of electric field are uniform ...(i) i.e. E1t = E2t = E 3t but the normal component of electric fields are non uniform and defined as e1 E1n = e2 E2n = e3 E 3n Since (Given) e1 = e3 So, ...(ii) Ein = E 3n ! E2n and as the net electric field is given by (sum of tangential and normal component) E = Et + E n Therefore by combining the results of eq (i) and (ii) we get E1 = E 3 ! E 2
SOL 3.1.20
Option (D) is correct. As the dielectric slab occupies the region 0 < z < d and the field intensity in the free space is in + az direction so, the field will be normal to the boundary of plane dielectric slab. So from the boundary condition the field normal to the surface are related as eEin = e0 E ( e = 4e0 ) Ein = e0 E 0 az = E 0 az 4e0 4 Therefore, Din = eEin = 4e0 E 0 az = e0 E 0 az 4 Option (A) is correct. Total energy stored in a region having electric field is given as W = 1 e0 ^E : E h dv 2 v (E = E1 + E2 ) = 1 e0 ^E1 + E2h : ^E1 + E2h dv 2 v = 1 e0 ^E 12 + E 22 + 2E1 : E2h dv 2 v = 1 e0 E 12 dv + 1 e0 E 22 dv + e0 ^E1 : E2h dv 2 v 2 v
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SOL 3.1.21
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SOL 3.1.19
# # # #
= W1 + W2 +
#
#
# e ^E : E hdv v
0
1
2
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For View Only Shop Online at www.nodia.co.in SOL 3.1.22 Option (C) is correct. Energy on a dipole with moment p in an electric field E is defined as WE =- p : E =- (- 2ax + 3ay) : (1.5ax - az ) =- (- 3 - 3) = 6 J Option (C) is correct. Consider a neutral dielectric is placed in an electric field E , due to which the dielectric gets polarized with polarization P , the bound surface charge density of the dielectric be rpS and the bound volume charge density be rpv . So the total bound charge by the dielectric is given as Qbound =
#r S
pS
dS +
#r
pv
v
dv
m
SOL 3.1.23
Qbound = =
# ^P : a hdS - # d : P dv S
n
v
# P : dS - # d : P dv s
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So we have,
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Since for a given polarization P of a dielectric material, the bound surface charge density over the surface of material is defined as rpS = P : an where an is the unit vector normal to the surface directed outward. while the bound volume charge density inside the material is defined as rpv =- d : P
v
(dSan = dS )
But according to the divergence theorem
# P : dS Therefore, SOL 3.1.24
=
# d : P dv v
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s
Qbound = 0
Option (A) is correct. Resistance of a conductor of length l and having uniform cross sectional area S is where s is the conductivity of the conductor R = l sS Given the conductivity, s = 5 # 106 (Wm) -1 the length of the conductor, l =8m side of the square cross section, a = 3 cm and radius of the bored hole, r = 0.5 cm So, the net cross sectional area is S = area of square cross section(bar) - area of circular cross section(hole) or S = a2 - pr2 = (3) 2 - p (0.5) 2 = a9 - p k cm2 4 The total resistance between the square ends is given as 8 R = l = p 6 -4 sS ^5 # 10 h # 9a9 - 4 k # 10 C
= 2.948 # 10-3 W GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in SOL 3.1.25 Option (D) is correct. The two materials of composite bar will behave like two wires of resistance RL (resistance due to lead) and RC (resistance due to copper) connected in parallel. As from the previous question we have the resistance due to the lead is RL = 1.948 mW and since the area of the cross section filled with copper is equal to the area of the cross section defined by hole so we have cross sectional area SC = p cm2 4 length of the bar l =8m 1 1 and conductivity of the copper, sC = = resistivity of the copper 1.72 # 10-8 So the resistance due to copper is 8 = 1.76 mW RC = l = p -4 SC sC ^ 4 # 10 h_ 1.72 #1 10 i Therefore the equivalent resistance of the composite bar is (1.948) # (1.76) R = RC || RL = 1.948 + 1.76 -6 = 524.62 # 10 W Option (D) is correct. Given the conductivity of material, s = 106 (Wm) -1 and conductance of the wire, G = 106 (W) -1 Since the conductance of a wire of length l having cross sectional area S is G = sS l 6 2 So we have, (S = pr2 ) 106 = 10 # pr l l r = p Option (C) is correct. Given the radii of spherical shell as a = 1 cm = 0.01 m b = 2 cm = 0.02 m The capacitance of a spherical capacitor having inner and outer radii a and b respectively is defined as -12 = 8.9 pF C = 4per e0 = 4p # 4 # 8.85 # 10 1 - 1 1 1 ba - b l 0.01 0.02 Option (B) is correct. Since the dielectric has been removed from the portion defined by ^ p2 < f < ph so the composite capacitor will have the dielectric filled only in 34 th portion of the total capacitor and so the configuration can be treated as the two capacitors connected in parallel with each other.
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SOL 3.1.27
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SOL 3.1.26
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-8
SOL 3.1.28
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For View Only Shop Online at www.nodia.co.in The capacitance of the portion carrying air( er = 1) as the medium between the spherical shells -12 C1 = 1 # 4pe0 = 1 # 4p # 8.85 # 10 1-1 1 - 1 4 4 a b 0.01 0.02
Option (D) is correct.
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SOL 3.1.29
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= 0.56 # 10-12 = 0.56 pF The capacitance of the portion carrying dielectric( er = 4 ) as the medium between the spherical shells C2 = 3 # C 4 Where C is the capacitance if no any portion of dielectric was removed as already calculated in previous question. So we have C2 = 3 # 8.9 # 10-12 = 6.7 # 10-12 = 6.7 pF 4 Therefore the equivalent capacitance of the composite capacitor is, Ceq = C1 + C2 = 0.66 + 6.7 = 2.12
Capacitance of a parallel plate capacitor is defined as C = eS d where S is the surface area of the parallel plates d is the separation between the plates Here, the three different regions will be treated as the three capacitors connected in series as shown below
So the capacitance of the region 1 is
C1 = er1 e0 S = 2500e0 S 0.001
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co
Option (D) is correct. The equivalent arrangement of the capacitor can be drawn in form of circuit as below
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SOL 3.1.31
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SOL 3.1.30
Shop Online at www.nodia.co.in the capacitance of the region 2 is C2 = er2 e0 S = 2000e0 S 0.002 the capacitance of the region 3 is C 3 = e0 S = 500e0 S 0.002 Therefore the equivalent capacitance of the whole configuration is calculated as 1 = 1 + 1 + 1 = 1 1 + 1 + 1 Ceq C1 C 2 C 3 e0 S b 2500 2000 500 l So, Ceq = 3.45 # 102 e0 S The capacitance per square meter of surface area will be C Ceql = eq = 3.45 # 102 e0 = 4.05 nF/m2 S Option (B) is correct. Capacitance between the two cylindrical surfaces is defined as C = 2pel ln ^b/a h Where l " length of the cylinder a " inner radius of the cylinder b " outer radius of the cylinder Since, the medium between the conducting cylinders includes the dielectric layer ^er = 4h from r = 4 cm to r = 6 cm and air^er = 1h from r = 6 cm to r = 8 cm , so the configuration can be treated as the two capacitance connected in series. Now for the dielectric layer ( er = 4 ) from r = 4 cm to r = 6 cm , capacitance is (l = 1 m ) C1 = 2pe0 er 1 = 8pe0 ln (1.5) ln (6/4) and for the air medium ^er = 1h from r = 6 cm to r = 8 cm , capacitance is C2 = 2pe0 # 1 = 2pe0 ln (8/6) ln (4/3) So, the equivalent capacitance of the configuration is evaluated as 1 = 1 + 1 = ln (1.5) + ln (4/3) 8pe0 2pe0 C1 C 2 Ceq Ceq = 143 pF
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For which the capacitances are calculated as below GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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SOL 3.1.32
Shop Online at www.nodia.co.in -4 e S/2 e0 S = e0 # 10 #-10 = e0 = C1 = 0 3 4 d d/2 4 # 10 e e S/2 er e0 S = = 3e0 er = 3 C2 = r 0 4 d d/2 -4 e S/2 = e0 # 10 # 10-3 = e0 C3 = 0 8 d 2 # 4 # 10 Therefore the equivalent capacitance of the capacitor is e0 3e0 e C 0 1 C2 = e0 = 2.76 pF Ceq = C 3 + = + 4 4 e0 + 3e0 8 4 C1 + C 2 4 4 Option (C) is correct. As the medium between capacitor plates is conducting so it carries the resistive as well as capacitive property. Consider the plates are separated by a distance d and the surface area of plates is S as shown in the figure.
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So the total resistance of the medium between plates is R = d sS and capacitance of the capacitor is C = eS d Therefore the time constant of the capacitor will be t = RC = e s ***********
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SOLUTIONS 3.2
m
Option (D) is correct. For a given current density, the total current that passes through a given surface is defined as I =
co
SOL 3.2.1
# J : dS
lp.
where dS is the differential surface area having the direction normal to the surface. Since the current density is independent of q and f so we can have directly the current I = J : S = J (4pr2 ar )
Option (A) is correct. From the equation of continuity we have the relation between the volume charge density, rv and the current density, J as 2rv =- d : J 2t and since the current density have only the component in ar direction so we have, 2rv =- 12 2 ^r2 Jr h 2t r 2r 3 2rv =- 12 2 br2 1 e-10 t l r 2 r 2t r Integrating both sides we get, 3 rv (r, t) =- 12 e-10 t dt + f (r) r where f (r) is the function independent of time. -3 3 rv (r, t) = 102 e-10 t + f (r) r Now for t " 3 rv (r, t) = 0 So, we put the given condition in the equation to get f (r)= 0 -3 3 therefore rv (r, t) = 102 e-10 t r 1 i.e. rv (r, t) \ 2 r
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SOL 3.2.2
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3 3 -3 = 1 e-10 t 4pr2 = 4p # (6) 2 # 1 # e-10 # 10 6 r -1 -1 = 4p # 3 # e = 12pe
#
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For View Only Shop Online at www.nodia.co.in SOL 3.2.3 Option (B) is correct. The velocity of charge density can be defined as the ratio of current density to the charge density in the region 3 (1/r) e-10 t ar J i.e. v = = 103 r ar = 2 -103 t -3 rv 10 / r e ^ h So, at r = 0.6 m , v = 103 # 0.6ar = 300ar m/s Option (D) is correct. The given problem can be solved easily by using image theory as the conducting surface y = 0 can be replaced by the equipotential surface in the same plane y = 0 and image of line charges ( rlL =- 5 pC/m at x = 0 , y =- 1 and x = 0 , y =- 2 ) as shown in the figure
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SOL 3.2.4
. The work done to carry a unit positive charge from a point located at a distance a from the line charge with charge density rL to another point located at a distance b from the line charge is defined as r Vab =- L ln b b l 2pe0 a and since the surface y = 0 has zero potential, so the potential at point P will be equal to the work done in moving a unit positive charge from the plane y = 0 to the point P . So the potential at point P will be rL VP =ln b 2pe0 b a l where a is the distance of the surface y = 0 from the line charges while b is the distance of point P from the line charges. -12 So, VP =- 5 # 10 =- ln b 1 l - ln c 2 m + ln c 10 m + ln c 17 mG 2pe0 2 1 1 2 =- 0.3 volt
/
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For View Only Shop Online at www.nodia.co.in SOL 3.2.5 Option (C) is correct. Electric field at a distance R from a line charge having uniform charge density rL is defined as r E = L R2 2pe0 R So the net electric field intensity produced at the point P due to the four line charges discussed in previous question is given as rL R E = 2pe0 R 2 where R is the distance of point P from the line charges r (- 1, - 2, 0) - (0, 1, 0) (- 1, - 2, 0) - (0, 2, 0) Therefore, + E = l > 2pe0 (- 1, - 3, 0) 2 (- 1, - 4, 0) 2 (- 1, - 2, 0) - (0, - 1, 0) (- 1, - 2, 0) - (0, - 2, 0) H (- 1, - 1, 0) 2 (- 1, 0, 0) 2 -12 (1, 3, 0) (1, 4, 0) (1, 1, 0) (1, 0, 0) = 5 # 10 ;+ + 2pe0 10 2 1 E 17
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= 0.12ax - 0.0032ay = 0.12ax - 0.003ay V/m
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Option (C) is correct. For a given polarization P inside a material, the bound surface charge density over the surface of material is defined as rpS = P : an where an is the unit vector normal to the surface directed outward. while the bound volume charge density inside the material is defined as rpv =- d : P Since the component of polarization of rod along y -axis is Py = 2y2 + 3 . So, the polarization of the material is P = (2y2 + 3) ay . and the charge density on the surface of the rod is rpS = P : an At y = 0 (top surface) rS1 = (2y2 + 3) ay : ^- ay h =- 3 At y = 5 (bottom surface) rS2 = (2y2 + 3) ay : (ay)= 53 and since the polarization has no radial component so no charge will be stored on its curvilinear surface and so the total bound surface charge on the surface of the rod is
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SOL 3.2.6
#
(S is the cross sectional area) Q pS = rpS ds = rS1 S + rS2 S =- 3 S + 53 S = 50 S Now, the bound volume charge density inside the material is rpv =- d : P =- d : (2y2 + 3) ay =- 4y So the total bound volume charge stored inside the material will be 5 y2 5 Q pv = rpv dv = (- 4y) Sdy =- 4S ; E - 50S 2 0 0 So, Total bound charge Qbound = QS + Qv = 50 S - 50 S = 0
#
#
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For View Only SOL 3.2.7 Option (C) is correct.
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SOL 3.2.8
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Electric field produced by the point charge at a distance r is q E = 1 2 ar 4pe0 r So, the induced dipole moment in the neutral atom due to the electric field E produced by the point charge will be aq p = aE = ar 4pe0 r2 and since the electric field intensity produced due to a dipole having moment p at a distance r from the dipole is defined as p Edip = 2 cos qar + sin qa q h 4pe0 r3 ^ where q is the angle formed between the distance vector r and dipole moment p So the field produced by the induced dipole at the point charge is aq 2b 4pe0 r2 l 2p 2a q (q = p as shown in the figure) = Edip = = 5 4pe0 r3 4pe0 r3 4 pe ^ 0h2 r Therefore the force experienced by the point charge due to the field applied by induced dipole is q 21 F = qEdip = 2a b 4pe0 l r5 2 = 2a b 1 # 10-9 # 9 # 109 l # 1 5 9 (1) Option (B) is correct. Electric field intensity produced due to a dipole having moment p, at a distance r from the dipole is defined as p Edip = 2 cos qar + sin qa q h 4pe0 r3 ^ where q is the angle formed between the distance vector r and dipole moment p So the electric field intensity produced due to dipole P1 at P2 is p1 2 # 10-9 a ( q = p/2 ) a E1 = = q q 4pe0 r3 4pe0 # (1) 3 Therefore the torque on P2 due to P1 is T = p 2 # E1 Taking the magnitude only we have the torque on P2 is -9 T = p2 E1 sin q = ^9 # 10-9hc 2 # 10 m sin 90c 4pe0 -8 = 1.62 # 10 N-m = 0.16 mN- m
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Option (A) is correct. For a given polarization P inside a material, the bound volume charge density inside the material is defined as rpv =- d : P Since the polarization of the sphere is P (r)= 2rar So the bound volume charge density inside the sphere is rpv =-4: P ^r h =- 12 2 (r2 2r) =- 12 # 6r2 =- 6 r 2r r Therefore the electric field intensity inside the sphere at a distance r from the center is given by 3 4 Q enc 1 rpv # 3 pr a a E = 1 = r r 4pe0 r2 4pe0 r2 rr = v ar =- 6r ar =-b 2 l rar e0 3e0 3e0 So the radial component of the electric field inside the sphere is Er =- 2 r e0 which is linearly decreasing with a slope b- 2 l with respect to r as shown below : e0
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SOL 3.2.10
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For View Only Shop Online at www.nodia.co.in SOL 3.2.9 Option (C) is correct. Electric field intensity produced due to a dipole having moment p, at a distance r from the dipole is defined as p Edip = 2 cos qar + sin qa q h 4pe0 r3 ^ where q is the angle formed between the distance vector r and dipole moment p So the electric field intensity produce due to dipole P2 at P1 is p2 9 # 10-9 2a (q = p ) E2 =# r 3 2ar =4pe0 r 4pe0 (1) 3 Therefore the torque on P1 due to P2 is T = p1 # E 2 Considering the magnitude only we have the torque on P1 is -9 ( q = p/2 ) T = p1 E2 sin q = 2 # 10-9 # c - 9 # 10 # 2 m 4pe0 = 3.64 # 10-4 N- m = 0.32 mN- m
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For View Only Shop Online at www.nodia.co.in SOL 3.2.11 Option (A) is correct. For a given polarization P of a material, the bound surface charge density over the surface of material is defined as rpS = P : an So the bound surface charge density over the spherical surface is (an = ar ) rpS = P ^r h : ar (at the spherical surface r = a ) = 2r = 2a So, total bound surface charge over the sphere is Q pS = 2a # 4pa2 = 8pa3 and the bound volume charge density inside the sphere as calculated above is rpv =- 6 So, total bound volume charge inside the sphere is Q pv = rpv b 4 pa3 l = (- 6) # b 4 pa3 l =- 8pa3 3 3 Therefore the total bound charge in the sphere is Qbound = Q pS + Q pv = 12pa3 - 12pa3 = 0 According to Gauss law the outward electric field flux through a closed surface is equal to the charge enclosed by the surface and since the total bound charge for any point outside the sphere is zero So, the electric field intensity at any point outside the sphere is E = 0 . Option (C) is correct. Since the spherical shell is of inner radius r = 2 m so region inside the sphere will have no polarization and therefore the total charge enclosed inside the shell for r < 2 m will be zero. i.e. Qenc = 0 According to Gauss law the total outward electric flux from a closed surface is equal to the charge enclosed by the surface and since the total enclosed charge for r < 2 m is zero so the electric field intensity at r = 1 m will be zero.
SOL 3.2.13
Option (A) is correct. Since the total bound charge by a polarized neutral dielectric is zero as discussed in MCQ. 33. So for any point outside the spherical shell the total enclosed charge(bound charge) will be zero and as discussed in the previous question, according to Gauss law the electric field intensity at any point outside the spherical shell will be zero. So, for the surface r = 7 Qenc = 0 Therefore the electric field intensity is E = 0
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SOL 3.2.12
Option (A) is correct. As we have to find electric field at r = 5 m so we determine first the charge enclosed by the surface r = 5 m which will be equal to the sum of the volume charge stored in the region 2 # r # 5 m and the surface charge stored at r = 2 m . Since for a given polarization P of a dielectric material, the bound volume charge density inside the material is defined as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.2.14
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rpv =- d : P So the bound volume charge density inside the dielectric defined in the region 2 # r # 6 m will be rpv =- d : P ^r h =- 12 2 br2 5 l =- 52 r r 2r r so the total bound volume charge in the region 2 # r # 5 m is 5 Q pv = rpv dv = - 5 # 4pr2 dr r r=2 5 =- 20p 6r @2 =- 60p
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Now for a given polarization P inside a dielectric material, the bound surface charge density over the surface of dielectric is defined as rpS = P : an where an is the unit vector normal to the surface pointing outward of the material. So the bound surface charge density at r = 2 m is (an =- ar ) rpS = P ^r h : (- ar ) Therefore the total bound surface charge over the surface r = 2 m is (for spherical surface S = 4pr2 ) Q pS =- 5 # 4pr2 r r = 2m =- 5 # 4p # 22 =- 40p 2 So, the total enclosed charge by the surface r = 5 m is Qenc = Q pv + Q ps =- 60p - 40p =- 100p So the electric field intensity at r = 5 m will be, Q p a =- 1 a ar = 1 # - 100 E = 1 # enc r e0 r 4pe0 4pe0 r2 52 Option (C) is correct. Since the electric field intensity at any point inside a conductor is always zero, so the electric flux density at a distance r from the center of the spherical conductor can be given as r 1 m 4p r 2 where Q = 3 mC is the total charge carried by the conductor. and since the dielectric material surrounding the spherical conductor has permittivity er = 3 , so the electric field intensity at a distance r from the center of the sphere is Z r < 1m ] 0, ] Q E =] ar 1 < r < 2 m [ 4per e0 r2 ] Q ]] r > 2m 2 ar pe 4 0r \ So, the total energy of the configuration is WE = 1 D : E dv 2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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SOL 3.2.15
#
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#
#
1
2
Q =1 4p 1 2 (4p) 2 ' e
2
Q Q 2 c 4pr2 mc 4pe e r2 m^4pr dr h + r 0
2
3
Q Q 2 c 4pr2 mc 4pe r2 m^4pr dr hG 0
1 dr r2 1 1 2 2 Q2 1 1+1 1 2 + 1 -1 3 = Q 1 = 8pe0 & 3 # 2 2 0 8p ' e0 er : r D1 e0 : r D2 1 =
#
2
1 dr + 1 e0 r2
#
#
3
(3 # 10-3) 2 # 9 # 109 8 # 12 2
V =-
i.e.
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Option (B) is correct. The electric potential at the centre of sphere will be equal to the work done to carry a unit charge from infinity to the centre of the sphere (the line integral of the electric field intensity from infinity to the center of the sphere) 0
# E : dl 3
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SOL 3.2.16
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= 3.7 # 10 4 J
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Since the sphere has uniform charge density rv =0.6 nC/m3 embedded in it, so the electric field intensity at a distance r from the center of the sphere can be given as Z ] rv r a , rR ] 3e r2 \ 0 where R is the radius of the sphere i.e. R = 1 m p So, the potential at the centre of sphere will be V =-
0
# E : dl 3 1/ p
(where differential displacement is dl = drar )
rv 1 3 dr - 0 rv r dr 2c 3e0 r 3 1/ p 3e pm 2 0 3 1/ p r r =- v c 1 m # :- 1 D - v :r D 3e0 p r 3 3er e0 2 1/ p 3 r r r rv = vc 1 m # p+ v # 1 = v + 3e0 p 12e0 p 3e0 p 3er e0 2p -9 9 5rv 5rv = 5 # 0.6 # 10 # 9 # 10 = = 12pe0 3 # 4pe0 3
=-
#
= 5 volt
SOL 3.2.17
#
( er = 2 ) rv =0.6 nC/m3
Option (A) is correct. For a given polarization P of a material, the surface charge density over the surface of material is defined as rS = P : an where an is the unit vector normal to the surface directed outward of the material.
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# r dS + # r dv S
S
v
v
= 6+ P (pr2) - P (pr2)@ + 0 = 0
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For View Only Shop Online at www.nodia.co.in while the volume charge density inside the material is defined as rv =- d : P Since the the cylinder has uniform polarization P . So, volume charge density inside the sphere is rv =- d : P = 0 and the surface charge density over the top and bottom surface of the cylinder is ( + P at top surface and - P at bottom surface) rS = P : an = ! P So the total bound charge by the cylinder is Qbound = QS + Qv
Option (C) is correct. As calculated above the volume charge density inside the cylinder is zero while the surface charge density at top and bottom surfaces are respectively + P and - P , so the cylinder can be considered as the two circular plates (top and bottom surface) separated by a distance L. Since the separation between the plates is larger than the cross sectional radius (L = 2r ) so the fringing field(electric field) will exist directed from the upper plate towards the lower plate.
SOL 3.2.19
Option (A) is correct. The electric flux lines will be the same as the electric field intensity outside the cylinder but as the volume charge density is zero rv = 0 inside the cylinder so
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SOL 3.2.18
# D : dS = 0 and therefore the flux lines will be continuous.
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Option (A) is correct. Consider the surface charge density on the parallel plates is !rS so the electric flux density between the plates is defined as D = rS an where an is the unit vector normal to the surface of plates directed from one plate toward the other plate. Since permittivity changes from layer to layer, but the field is normal to the surface so electric flux density D will be uniform throughout the plate separation as from boundary condition. So the electric field intensity at any point between the parallel plates is rS an E = D = er = 2 (1 + 100a2) e0 er 2e0 (1 + 100a2) Therefore the voltage between the plates can be evaluated by taking the line integral of electric field from one plate to the other plate 0.1 rS an i.e. (dl = da ) V =- E : dl =c 2 m : ^da h a = 0 2e0 (1 + 100a ) 0.1 r da (the direction of a is along an ) = S 1 2e0 100 0 (0.1) 2 + a2 r 0.1 = S # 1 # 1 9tan-1 a a kC 2e0 100 0.1 0. 1 0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.2.20
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rS 1 p - 0 = rS p C 2e0 # 10 9 4 80e0 Now charge stored at the parallel plates is where S is surface area of the plates Q = (rS ) (S) = rs # (0.2) S = 0.2 m2 So, the capacitance of the capacitor is evaluated as r (0.2) 16e0 Q C = = s# = p V (rs p) /80e0 -11 = 5.51 # 10 =
m
Option (C) is correct. For the two wire transmission line consists of the cylinders of radius b and separated by a distance 2h (centre to centre), the capacitance per unit length between them is defined as pe Cl = cosh-1 (h/b) Here, 2h = 2 cm and b = 0.2 cm -12 ( er = 2 ) So, Cl = p # 2 #-81.85 # 10 = 3.64 # 10-11 F/m cosh ^1/0.2h So the charge per unit length on each wire will be, (Vo = 100 V ) Q = C lV0 = 3.64 # 10-11 # 100 = 5.64 # 10-9 C/m
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SOL 3.2.21
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Option (C) is correct. Consider the oil rises to a height h in the space between the tubes. So, the capacitance of the tube carrying oil partially will be treated as the two capacitors connected in parallel. Since the capacitance between the two cylindrical surfaces is defined as C = 2pel ln ^b/a h Where l " length of the cylinder a " inner radius of the cylinder b " outer radius of the cylinder So the capacitance of the portion carrying oil ( ce = 1) as the medium between the cylindrical surfaces is Coil = 2per e0 h = 4pe0 h ^er = ce + 1 = 2h ln ^3h ln ^3/1h and the capacitance of the portion carrying air( er = 1) as the medium between the cylindrical surfaces is 2pe0 ^1 - h h Cair = ln ^3/1h Therefore the equivalent capacitance of the tube carrying oil to the height h is (1 + h) C = Coil + Cair = 2pe0 ln (3) GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.2.22
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Option (C) is correct. Consider the charge densities of the two surface of the slab is rs1 C/m2 and rs2 C/m2 as shown in the figure. As the sum of the charge densities is rso C/m2 so we have ...(1) rs1 + rs2 = rs0 and since the electric field intensity inside the conducting slab must be zero so, ...(2) E1 + E 2 = 0 where E1 is field inside slab due to charge density rs1 and E2 is field inside slab due to rs2
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SOL 3.2.23
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For View Only Shop Online at www.nodia.co.in Since the energy stored in a capacitor is defined as where V is the applied voltage to the capacitor WE = 1 CV2 2 So the net upward force due to the capacitance is given by F = dWE = 1 V 2 dC = 1 V 2 2pe0 2 2 dh dh ln (3) and net downward force on the oil due to gravity will be F = mg = (0.01 gm/cm3) # p (b2 - a2) h # g mass density = 0.01 gm/cm3 -6 = 0.01 # p (9 - 1) # 10 # h # g = 0.08phg 10-6 Since in equilibrium both the upward and downward forces are equal So, 0.08phg = 1 V 2 2pe0 2 ln (3) -12 0.08ph # (9.8) = 1 # (2 # 103) 2 # 2p # 8.85 # 10 2 ln (3) 3 2 (2 10 ) # 2 # 8.85 # 10-12 h =1# # 2 0.08 # 9.8 # ln (3) -5 = 4.11 # 10 m = 45.1 mm
As the electric field intensity at any point P due to the uniformly charged plane with charge density rS is defined as r E = S an 2e0 where an is the unit vector normal to the plane directed toward point P r So we have, (an =- az ) E1 = S1 (- az ) 2e0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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rS2 (an = az ) a 2e0 z rs1 r From equation (ii) - a + s2 a = 0 2e0 ^ z h 2e0 z rs1 = rs2 Putting the result in equation (i) we get r rs1 = rs2 = s0 4 Option (D) is correct. As the slabs are conducting so net electric field inside the slab must be zero. and since the electric field intensity at any point P due to the uniformly charged plane with charge density rS is defined as r E = S an 2e0 where an is the unit vector normal to the plane directed toward point P E2 =
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SOL 3.2.24
So, the net electric field intensity inside slab 1 is rS11 r r r - a + S12 a + S21 a + S22 a = 0 2e0 ^ z h 2e0 z 2e0 z 2e0 z (an =- az for rS11 while an = az for rest of the charge densities) ...(1) - rs11 + rs12 + rs21 + rs22 = 0 and the net electric field intensity inside slab 2 is rS11 r r r (- az ) + S12 (- az ) + S21 (- az ) + S22 az = 0 2e0 2e0 2e0 2e0 (an = az for rS22 while an =- az for rest of the charge densities) - rs11 - rs12 - rs21 + rs22 = 0 Solving eq. (i) and eq (ii) we get, rs11 = rs22 and rs12 =- rs21
...(2)
Option (C) is correct. As all the four surfaces form the boundaries of the conductors extending away from GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.2.25
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For View Only Shop Online at www.nodia.co.in the region between them so, the medium outside the defined region is conductor and so the field intensity outside the region will be zero. Now the electric potential in the non conducting region is given as V = 5xy So the electric field intensity in the region is E =- dV =- 5yax - 5xay From the conductor-free space boundary condition we have the surface charge density on the boundary surface defined as rs = e0 En where En is the normal component of the electric field intensity in the free space. So, the surface charge density on the surface x = 0 is rs = e0 ^- 5y h (the normal component En =- 5y for the surface x = 0 ) =- 5e0 y Option (D) is correct. Again as discussed in above question, the surface charge density on the surface y = 0 will be given by rs = e0 En and since the field component normal to surface y = 0 is En =- 5x So, the surface charge density on the surface y = 0 is rs =- 5e0 x
SOL 3.2.27
Option (C) is correct. From the symmetry associated with the charge distribution the electric field must be radially directed. Then choosing Gaussian surfaces which are cylinders having the same axis ( r = 0 ) as the conductors and of length l , we get for r < 2 m ^2prl h E r = 0 (since there is no charge enclosed by the Gaussian surface) Thus for r < 2 m Er = 0 Now, since the field inside the conductor 2 < r < 3 m is zero; there cannot be any charge on the surface r = 2 m . i.e. at r = 2 m rS = 0 and all the charge associated with the inner conductor resides on the surface r = 3 m. 10 C/m i.e. at r = 3 m rS = = 5 C/m2 3p 2p ^ 3 h Proceeding further we have for 5 < r < 6 m 2prlE r = 1 ^10 C/mh l e0 where l is length of the cylinder. for 5 < r < 6 m So E = 10 a r 2pe0 r
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SOL 3.2.26
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For View Only Shop Online at www.nodia.co.in This the field produced by the inner conductor but the fact is that the field inside the conductor 5 < r < 6 m is zero that gives 6rS@at r = 5 m = e0 6E@at r = 5 : ^- a rh 10 a : (- a ) =- 10 =- 1 C/m2 r p 2pe0 (5) r m 2p ^ 5 h = 1 $6 C/m - 6rS @at r = 5 2p ^5 h. 2p ^6 h = 1 ^6 + 10h = 4 12p 3p = e0 c
and
6rS@at r = 6 m
Option (C) is correct. From the symmetry associated with the charge distribution the electric field must be radially directed. As, there is no charge enclosed by the surface r = 2 m so we get for r < 2 m Er = 0 Now from the conductor-free space boundary condition we have the surface charge density on the boundary surface defined as rs = e0 En where En is the normal component of the electric field intensity in the free space. So the charge density at r = 2 m is rS1 = Er = 0 Therefore the total charge will be concentrated over the outer surface which is given as Q rS2 = = 9 = 1 C/m2 4pr2 2p ^3h2 8p
SOL 3.2.29
Option (D) is correct. From the boundary condition for the charge carrying interface, the tangential component of electric field on either side of the surface will be same. i.e. E1t = E2t while the normal components are related as r E1n - E2n = s e0 now as the field intensity in the region z < 0 is E2 = 2ax + 3ay - 2az So the tangential component, E2t = 2ax + 3ay and the normal component, E2n =- 2az Therefore the field components in region ^z > 0h are E1t = E2t = 2ax + 3ay -9 r and E1n = E2n + s = ;- 2 + 2 # 10 -12 E az = 224az e0 8.85 # 10 So the net field intensity in the region z > 0 is E1 = E1t + E1n = 2ax + 3ay + 224az
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SOL 3.2.28
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For View Only Shop Online at www.nodia.co.in SOL 3.2.30 Option (C) is correct. As the dielectric slab occupies the region x > 0 and the electric field in the free space is directed along ax so, the field will be normal to the boundary surface, x = 0 of the dielectric slab. So from the boundary condition the field normal to the interface of dielectrics are related as (where Ei is the field inside the dielectric) er e0 Ei = e0 E
Option (D) is correct. As the dielectric slab occupies the region 0 < y < 1 m and the electric field in the free space is directed along ay so, the field will be normal to both the boundary surfaces y = 0 and y = 1. So from the boundary condition the field normal to the interface of dielectrics are related as (where Ei is the field inside the dielectric) eEi = e0 E ^1 + y h2 since e = 4e0 2 Ei = e0 ^1 + y h2 E = 4a 4e0 4 ^ yh ^1 + y h 2 = ^1 + y h ay So the polarization inside the dielectric is P = eEi - e0 Ei = 4e0 2 Ei - e0 Ei = e 4e0 2 - e0 o^1 + y h2 ay ^1 + y h ^1 + y h 2 = 84 - ^1 + y h B e0 ay Now for a given polarization P inside a dielectric material, the surface charge density over the surface of dielectric is defined as rpS = P : an where an is the unit vector normal to the surface directed outward of the dielectric. So, the bound surface charge density at y = 0 is (an =- ay ) 6rps@at y = 0 = P : ^- ay h 2 y=0 = 64 - ^1 + y h@ e0 ^- 1h =-^4 - 1h e0 =- 3e0 and the surface charge density at y = 1 m is (an = ay ) 6rps@at y = 1 = P : ^ay h 2 = 84 - ^1 + y h B e0 ^1 h = ^4 - 4h e0 = 0
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SOL 3.2.31
(er = 5)
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Ei = E = 10ax = 2ax 5 er So, the polarization inside the dielectric is P = ^e - e0h Ei = ^5e0 - e0h Ei = 8e0 ax
Option (C) is correct. As calculated in previous question, polarization inside the dielectric is P = 64 - ^1 + y h@ e0 ay Since for a given polarization P inside a material, the bound volume charge density inside the material is defined as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.2.32
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rpv =- d : P So the volume charge density inside the dielectric is rv =- 2 84 - ^1 + y h2B e0 2y = 2 ^1 + y h e0 So when we move from y = 0 to y = 1 m , the volume charge density will be linearly increasing. Option (C) is correct. As the charge is being located at origin so the field intensity due to it will be in radial direction and normal to the surface of the dielectric material. Therefore the flux density will be uniform(as from boundary condition) and at any point r inside the dielectric flux density will be Q D = ar 4pr 2 Now it is given that electric field intensity at any point inside the dielectric is Q E = ar 4pe0 b2 and since in a medium of permittivity e = er e0 the flux density is defined as D = er e0 E So for the given field we have Q Q ar = er e0 c ar 4p r 2 4pe0 b2 m 2 er = b2 r
SOL 3.2.34
Option (C) is correct. Consider the parallel sheets arrangement as shown in the figure.
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SOL 3.2.33
Electric field intensity at any point P due to the uniformly charged plane with charge density rS is defined as r E = S an 2e where an is the unit vector normal to the plane directed toward point P and e is the permittivity of the medium. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in So the field intensity inside the dielectric due to the left sheet will be
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-9 E1 = 5 # 10 ^ay h ^an = ay h 2e and again the field intensity inside the dielectric due to right sheet will be -9 -9 E2 = - 5 # 10 ^- ay h = 5 # 10 ay ^an =- ay h 2e 2e so the net field intensity inside the dielectric will be -9 E = E1 + E2 = 5 # 10 ay e Since the field intensity is uniform inside the dielectric So potential difference between the plates will be directly given as V = E # (distance between the plates) -9 = 5 # 10 # 2 = 2.824 # 102 Volt = 283 kV ^e = 4e0h 4e0 Option (C) is correct. As calculated in previous question the electric field between the two dielectrics having surface charge densities rS and - rS is r E = S e where e is the permittivity of the medium between the sheets. r r So electric field in slab 1 is E1 = S = S e 2e0 r r and electric field in slab 2 is E2 = S = S e 4e0 Since the electric field between the sheets is uniform so the potential difference between the plates will be V = E # ^distanceh = E1^1 mh + E2 ^2 mh -9 r r r = S (1) + S ^2 h = s = 0.6 # 10 -12 e0 2e0 4e0 8.85 # 10 = 57.8 Volt
SOL 3.2.35
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Option (C) is correct. As calculated in previous question the electric field between the two dielectrics having surface charge densities rS and - rS is r E = S e where e is the permittivity of the medium between the sheets. r r So electric field in slab 1 is E1 = S = S e 2e0 r r and electric field in slab 2 is E2 = S = S e 4e0 Since the electric field between the sheets is uniform so the potential difference
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For View Only Shop Online at www.nodia.co.in between the plates will be V = E # ^distanceh = E1^1 mh + E2 ^2 mh r r r = S (1) + S ^2 h = s e0 2e0 4e0
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SOL 3.2.37
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SOL 3.2.36
-9 = 0.6 # 10 -12 = 64.8 Volt 8.85 # 10 Option (A) is correct. The electric field between the plates carrying charge densities + rS0 and - rS0 is defined as r E = S0 e where e is the permittivity of the medium between the plates. Now consider that near the plate 1 permittivity is e1 and near the plate 2, permittivity is e2 . So at any distance x from plate 1 permittivity is given by (Since the permittivity is linearly increasing) e = e1 + a e2 - e1 k x d So the field intensity at any point in the medium will be rS0 E = e e1 + a 2 - e1 k x d Therefore the potential difference between the plates will be d d rS0 r d rS0 ln :e1 + a e2 - e1 k x D = S0 ln a e2 k V = dx = e e e e1 e e d 2 1 2 - e1 2 1 >a H 0 e + 1 a 2 kx d k 0 Option (A) is correct. Assume that the surface charge densities on the plates is !rS0 so the electric field intensity between the plates will be r E = S0 e0 and the potential difference between the plates will be given by V = E # (Distance between plates) r 5 # 103 = b S0 l # ^0.5 # 10-2h e0 Therefore the surface charge density is ^8.85 # 10-12h # ^5 # 103h rS0 = = 8.85 mC ^0.5 # 10-2h
Option (C) is correct. The capacitor of a parallel plate capacitor is defined as C = eS d st So, the capacitance in 1 dielectric region will be C1 = e1 S = 3e0 S 1 1 nd and the capacitance in 2 dielectric region GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.2.38
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Option (C) is correct. Consider the dielectric slab is of thickness t and d1 , d2 are the remaining width in the medium as shown in the figure.
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SOL 3.2.39
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C2 = e2 S = 2e0 S 3 3 st Therefore the voltage drop in 1 dielectric region is (where V is total voltage drop) V1 = C2 V C1 + C 2 ^2e0 S/3h = Volt ^9 Volth = 18 11 3e0 S + 23 e0 S 3e0 S 82 Volt and similarly, V2 = C1 V = ^9 h = 11 C1 + C 2 3e0 S + 2e0 S3
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Now the capacitance of the whole configuration will be considered as the three capacitors (capacitance in the three regions) connected in series as shown in the figure
C1 = e1 S , C2 = e2 S and C 3 = e1 S t d1 d2 The equivalent capacitance, is defined as 1 = 1 + 1 + 1 = t + ^d1 + d2h eS eS Ceq C1 C 2 C 3 Since t ; ^d1 + d2h will be constant although if the dielectric slab is moved leftward or rightward so the equivalent capacitance will be constant. But if the slab is pulled outward then the capacitance will change as the effective surface area of the capacitance due to dielectric slab changes. So,
SOL 3.2.40
Option (B) is correct. Since, the wire is coated with aluminum So,the configuration can be treated as the
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SOL 3.2.41
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For View Only Shop Online at www.nodia.co.in two resistance connected in parallel and therefore, the field potential will be same across both the material or we can say that the field intensity will be same inside both material. i.e. Est = Eal where Est " Field intensity insteel Eal " Field intensity in aluminum. Jst = Jal or, sst sal where Jst " current density in steel Jal " current density in aluminum sst " conductivity of steel sal " conductivity of aluminum Jst = sst = 2 # 106 = 1 So, we get, sal Jal 3.8 # 107 19 ...(1) Jal = 19Jst Now, the total current through the wire is given as, I = Jst ^pa2h + Jal ^pb2 - pa2h where a " cross sectional radius of inner surface (steel wire) b " cross sectional radius of outer surface (with coating) Since, thickness of coating is t = 2 # 10-3 So, b = a + t = ^2 # 10-3h + ^2 # 10-3h = ^4 # 10-3h Therefore, we get, 80 = Jst p ^4 # 10-6h + Jal 6p ^16 # 10-6h - p ^4 # 10-6h@ or, 80 = Jst p ^4 # 10-6h + 19Jst 6p ^12 # 10-6h@ (from eq.(1)) 80 5 2 So, Jst = = 1.10 # 10 A/m 232p # 10-6 Option (A) is correct. Given, the potential field in free space V = 403 cos q sin f r
So, the potential at point P (r = 2 , q = p , f = p ) is given as 3 2 VP = 403 cos a p k sin a p k = 2.5 Volt 3 2 ^2 h Now, as the conducting surface is equipotential, so, the potential at any point on the conducting surface will be equal to the potential at point P . i.e. V = VP = 2.5 Volt 40 or cos q sin f = 2.5 r3 This is the equation of the conducting surface. 16 cos q sin f = r3 *********** GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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SOLUTIONS 3.3
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Option (B) is correct. The capacitance of a parallel plate capacitor is defined as -12 -4 C = eo A = 8.85 # 10 -3 # 10 = 8.85 # 10-13 d 10 The charge stored on the capacitor is Q = CV = 8.85 # 10-13 = 4.427 # 10-13 Therefore, the displacement current in one cycle Q I = = fQ = 4.427 # 10-13 # 3.6 # 109 = 2.59 mA T
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SOL 3.3.1
( f = 3.6 GHz )
Option (A) is correct. The electric field of the EM wave in medium 1 is given as E1 = 2ax - 3ay + 1az Since the interface lies in the x = 0 plane so, the tangential and normal components of the field intensity in medium 1 are E1t =- 3ay + ay and E1n = 2ax From the boundary condition, tangential component of electric field is uniform. So, we get the tangential component of the field intensity in medium 2 as E2t = E1t =- 3ay + ay Again from the boundary condition the for normal component of electric flux density are uniform i.e. D1n = D2n or e1 E1n = e2 E2n So, we get 1.5eo 2ax = 2.5eo E2n or E2n = 3 ax = 1.2ax 2.5 Thus, the net electric field intensity in the medium 2 is E2 = E2t + E2n =- 3ay + az + 1.2ax
SOL 3.3.3
Option (B) is correct. The surface charge density on a conductor is equal to the electric flux density at its boundary. i.e. ( e = 80eo ) s = D = eE = 80e0 E = 80 # 8.854 # 10-12 # 2 = 1.41 # 10-9 C/m 2
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SOL 3.3.2
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For View Only Shop Online at www.nodia.co.in SOL 3.3.4 Option (D) is correct. The configuration shown in the figure can be considered as the three capacitors connected in parallel as shown below
SOL 3.3.5
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Now, consider the distance between the two plate is d and the total surface area of the plates is S . So, for the three individual capacitors the surface area is S/3 and the separation is d . Therefore, we get, e0 e1 ^S/3h C1 = d e0 e2 ^S/3h C2 = d e0 e3 ^S/3h C3 = d Since, the three capacitance are in parallel So, the equivalent capacitance is Ceq = C1 + C2 + C 3 e0 e1 ^S/3h e0 e2 ^S/3h e0 e3 ^S/3h = + + d d d e e + e + e e + e2 + 2e3 C S 5 eS 2 3 = b 0 lc 1 m =b 1 bC = d0 l l 3 3 d Option (B) is correct. The electric field is equal to the negative gradient of electric potential at the point. i.e. E =- dV Given, electric potential V = 4x + 2 So, E =- 4ax V/m
Option (D) is correct. The angle formed by the electric field vector in two mediums are related as tan a1 = e1 tan a2 e2 So, for the given field vectors we have, tan 60c = 3 tan a2 3 tan a2 = 1 or a2 = tan-1 ^1 h = 45c GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.3.6
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For View Only Shop Online at www.nodia.co.in SOL 3.3.7 Option (C) is correct. The tangential component of electric field on conducting surface is zero (since the surface conducts current) So, under static condition we have, - dV = E = 0 or V = constant i.e. the conducting surface is equipotential. So, (A) and (R) both true and R is correct explanation of A. Option (C) is correct. Since, the electric field is incident normal to the slab. So, the electric field intensity ^Ei h inside the slab is given as eEi = e0 E 0
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SOL 3.3.8
e0 ^6ax h = 2ax 3e0 Therefore, the polarization inside the slab is given as Pi = e0 Xe Ei where Xe is electric susceptibility defined as Xe = er - 1. So, we have Pi = e0 ^3 - 1h Ei = 4e0 ax
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Option (C) is correct. Due to both the line charge and concentric circular conductors, the equipotential surfaces are circular (cylinder) i.e. concentric equipotential lines. The flux lines due to both the configurations (line charge and concentric circular conductors) are in straight radial direction.
SOL 3.3.10
Option (C) is correct. Capacitance of 1st plate is given as e ^1 # 1h e = C 1 = eS 1 = d d d nd The capacitance of 2 plate is e ^3 # 2h 6e = C 2 = eS 2 = d d d So, the ratio of capacitances is C2 = 4 C1 Option (B) is correct. Consider the dielectric material with permittivity e1 is replaced by a dielectric material with permittivity e2 . The capacitance of parallel plate capacitor is defined as C = eS d i.e. the capacitance depends on the permittivity of the medium and so, due to the replacement of the material between the plates the capacitance changes. Now, the charge is kept constant
SOL 3.3.11
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SOL 3.3.9
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For View Only Shop Online at www.nodia.co.in i.e. Q1 = Q 2 or, C1 V1 = C2 V2 So, due to the change in capacitance voltage on the capacitor changes and therefore the electric field intensity between the plates changes. The stored energy in the capacitance is defined as Q2 W = 2C As total stored charge Q is kept constant while capacitance changes so, the stored energy in the capacitance also changes. Thus, all the three given quantities changes due to the replacement of material between the plates. Option (A) is correct. According to continuity equation we have 2r d : J =- v 2t 2rv As for electrostatic field = 0 so, we get 2t d:J = 0
SOL 3.3.13
Option (C) is correct. Electrostatic fields only.
SOL 3.3.14
Option (A) is correct. Surface or sheet resistivity is defined as resistance per unit surface area. So, the unit of surface resistivity is Ohm/sq. meter.
SOL 3.3.15
Option (D) is correct. Since a conducting surface is equipotential so no electric field component exists tangential to the surface and therefore the electric field lines are normal to a conducting surface boundary.
SOL 3.3.16
Option (D) is correct. Surface resistance of a metal is defined as wm 2pfm Rs . = 2s 2s So, as frequency ( f ) increases the surface resistance increases.
SOL 3.3.17
Option (C) is correct. When we determine force using method of images then in this method, the conducting surface is being removed and an additional distribution of charge is being introduced symmetrical to the existing charge distribution.
SOL 3.3.18
Option (D) is correct. The conducting surface is equipotential and since the potential at infinity is zero so, the potential every where on a conducting surface of infinite extent is zero. Since the conducting surface is equipotential so displacement density on a conducting
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SOL 3.3.12
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For View Only Shop Online at www.nodia.co.in surface is normal to the surface. So A and R both true but R is not correct explanation of A. Option (C) is correct. Capacitance, C = 5 pF = 5 # 10-2 F Charge on capacitance, Q = 0.1 mC = 0.1 # 10-6 C The energy stored in the capacitor is defined as 2 ^0.1 # 10-6h Q2 W = = = 1 mJ 2C 2 # 5 # 10-12
SOL 3.3.20
Option (A) is correct. Consider the charge of 1 C is placed near a grounded conducting plate at a distance of 1 m as shown in figure.
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SOL 3.3.19
Option (D) is correct. Fringing field has been shown below in the figure
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SOL 3.3.21
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Using image of the charge we have one negative charge opposite side of the plate at the same distance as shown in the figure and the force between them is ^1 h^- 1h = -1 2 = -1 N F = 16pe0 4pe0 r2 4pe0 ^2 h Negative sign indicates that the direction of force is attractive.
The capacitance of a parallel plate capacitor is given as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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C = e0 er A d It is valid only when the fringing is not taken into account. Now, the fringing field can be ignored only when the separation d between the plates is much less than the plate dimensions. So, for the fringing field taken under consideration, A/d is tending towards infinity. Option (D) is correct. The capacitance of a solid infinitely conducting sphere is defined as C = 4pe0 R . where R is radius of the solid sphere.
SOL 3.3.23
Option (C) is correct. The electric potential produced by a point charge Q at the a distance r from it is defined as Q V = 4per where e is permittivity of the medium. So, the electric potential produced by the point charge + 10 mC at the centre of the sphere is -6 Q (Given r = 5 cm ) V = = 10 # 10 -2 4peo r 4pe0 ^5 # 10 h As the surface of sphere is grounded so, the total voltage on the spherical capacitor will be equal to the potential at its centre as calculated above. Now, the capacitance of the isolated sphere is defined as C = 4pea where a is the radius of the sphere. Therefore, the induced charge stored on the sphere is given as ^10 # 10-6h Qind = CV = ^4pe0 a h 4pe0 ^5 # 10-2h ^2 # 10-2h # ^10 # 10-6h (Given a = 2 cm ) = ^5 # 10-2h = 5 # 10-6 C = 5 mC
SOL 3.3.24
Option (A) is correct. Given electric field E = E 0 sin wt The conduction current is defined as Jc = s ^E h = sE 0 sin wt where s is conductivity and E is electric field intensity. and the displacement current density is Jd = 2D = e2E 2t 2t = eE 0 ^w cos wt h = ewE 0 sin a p - wt k 2 So the phase difference between Jc and Jd is 90c.
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SOL 3.3.22
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For View Only Shop Online at www.nodia.co.in SOL 3.3.25 Option (D) is correct. Method of images are used for the charge distribution at a distance from the grounded plane conductor. Option (A) is correct. Given Capacitance of condenser,C = 0.005 mF = 5 # 10-9 F Supply voltage, V = 500 V (immersed oil) Permittivity of oil, er = 2.5 So, the energy stored in condenser before immersion is W = 1 CV 2 = 1 # 5 # 10-9 # ^500h2 2 2 = 6.25 # 10-4 J After immersing the condenser in oil the capacitance changes while the total charge remains same. i.e. Qafter immersion = Qbefore immersion = ^5 # 10-9h^500h = 2.5 # 10-6 Coulomb The capacitance of the condenser after immersion is Cafter immersion = er C = ^2.5h^5 # 10-9h = 1.25 # 10-8 F Therefore, the stored energy in the condenser immersed in oil is ^2.5 # 10-6h Q2 -4 , J W = = -8 = 2.5 # 10 2C(after immersion) 2 ^1.25 # 10 h Option (D) is correct. Given, Capacitance, C = 3 mF = 3 # 10-6 F Current, I = 2 mA = 2 # 10-6 A Charging time, t = 6 sec So, the total charge stored on capacitor is Q = Charge transferred = It = ^2 # 10-6h^6 h Therefore, the voltage across the charged capacitor is ^2 # 10-6h^8 h Q V = = C 4 # 10-6 = 5 Volt
SOL 3.3.28
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SOL 3.3.27
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SOL 3.3.26
Option (A) is correct. Given, the total charge on capacitor = V (1) Electric field between the plates will be given as E =- dV which is independent of permittivity of the material filled in capacitor so E will be constant. (2) The displacement flux density inside the capacitor is given as
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(3) The charge stored on the plates is given as Q = CV where V is constant but capacitance C will be doubled as it is directly proportional to the permittivity given as C = eS d So, the charge on plates will be get doubled. Therefore, the statements 2 and 4 are correct.
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Option (D) is correct. Since, resistance doesn’t store any energy. So, the energy stored in the coil is only due to inductance and given as W = 1 LI 2 2 where L is the inductance and I is the current flowing in the circuit. At the fully charged condition, inductor is short circuit and therefore, current through the circuit is I = V = 50 = 10 A 5 R So, the energy stored in the field (in the inductor) is W = 1 ^0.6h^10h2 = 30 Joules 2 Option (A) is correct. The normal component of electric flux density ^D h across a dielectric-dielectric boundary is given as D1n - D2n = rs where rs is the surface charge density at the interface. So, the normal components of electric flux density across a dielectric-dielectric boundary is dependent on the magnitude of surface charge density.
SOL 3.3.30
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SOL 3.3.29
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(4) As discussed already, the capacitance will get doubled.
SOL 3.3.31
Option (A) is correct. Statement 1, 2 and 4 are correct while statement 3 is incorrect.
SOL 3.3.32
Option (D) is correct. The capacitance of an insulated conducting sphere of radius R in vacuum is C = 4pe0 R
Option (A) is correct. Maximum withstand voltage is the value that the dielectric between capacitor plates can toterate without any electrical breakdown. Maximum withstand voltage is larger for any dielectric material than that for free space (air). Since the maximum withstand voltage across the capacitor filled with air is V so GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.3.33
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For View Only Shop Online at www.nodia.co.in the maximum withstand withstand voltage for the composite capacitor will be also V as the capacitors are connected in parallel. Now, the capacitance before filling the dielectric is C = e0 A d and after filling the dielectric Ceq = C1 + C2 A/2 A/2 = 4e0 + e0 = 5e0 A d d 2d
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So, the stored charge Q1 after filling dielectric is determined as below Q (Since voltage is constant) = C Q1 Ceq Q 5e0 A 2d = 2.5Q or, Q1 = e0 A d Therefore, the maximum withstand voltage of the capacitor is V and charge is 2.5Q . Option (B) is correct. Since, the potential on both sides of plate will be same (Consider the potential is V ). So, the charge densities on the two sides is determined as below : r2 = C l2 V and r1 = C l1 V where C l2 and C l1 are the capacitance per unit area of the capacitance formed by the region d1 and d2 . e0 V r1 l C d 1V Therefore, = = 1 = d2 r 2 C l2 V e0 V d1 d2
SOL 3.3.35
Option (A) is correct. Electric flux density in a polarized dielectric is defined as D = P + e0 E
SOL 3.3.36
Option (C) is correct. Image theory is applicable only for static charge distribution (electrostatic field).
SOL 3.3.37
Option (B) is correct. The equivalent capacitance of series connected capacitance has the value less than the smallest capacitance here the smallest capacitance is C6 so the total capacitance is less then C6 i.e. Ceq < C6 or Ceq . C6
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SOL 3.3.34
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Chap 3
Electric Field in Matter
197
SOL 3.3.40
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Option (C) is correct. Electric flux density, D = 1 C/m2 Relative permittivity, er = 5 Since, the normal component of flux density is uniform at the boundary surface of two medium so, the flux density inside the slab is D = 1 C/m2 Therefore, the polarization of the slab is given as P = b er - 1 l D = 4 # 1 = 0.8 er 8 Option (D) is correct. The capacitance of a isolated spherical capacitor of radius R is defined as C = 4pe0 R Since the two spheres are identical and separated by a distance very much larger then R. So, it can be assumed as the series combination of capacitances. Therefore, the net capacitance between two spheres is given as ^4pe0 Rh^4pe0 Rh i.e. C = C1 C 2 = = 2pe0 R 4pe0 R + 4pe0 R C1 + C 2
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SOL 3.3.39
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For View Only Shop Online at www.nodia.co.in SOL 3.3.38 Option (A) is correct. Given, the electric field intensity in medium 1. E1 = 5ax - 2ay + 3az Since, the medium interface lies in plane z = 0 . So, we get the field components as E1t = 5ax - 2ay and E1n = 3az Now, From the boundary condition for electric field we have E1t = E2t e1 E1n = e2 E2n So, the field components in medium 2 are E2t = E1t = 5ax - 2ay E2n = e1 E1n = 6az e2 Therefore, the net electric field intensity in medium 2 is given as E2 = E2t + E2n = 5ax - 2ay + 6az So, the z -component of the field intensity in medium 2 is E2z = 6az
Option (A) is correct. For steady current in an arbitrary conductor the current density is given as J = I A and since I is constant So, J is constant and therefore d # J = 0 So, the current density is solenoidal. i.e. Assertion (A) is true. The reciprocal of resistivity is conductivity. i.e. Reason (R) is false. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.3.41
198
Electric Field in Matter
Chap 3
For View Only Shop Online at www.nodia.co.in SOL 3.3.42 Option (C) is correct. Since, the displacement current density is defined as Jd = 2D 2t So, it is generated by a change in electric flux and therefore the displacement current has only A.C. components as derivative of D.C. components is zero. i.e. A and R both are true and R is correct explanation of A. Option (D) is correct. Dielectric constant, er = 5 Flux density, D = 2 C/m2 So, the polarization of the medium is given as P = b er - 1 l D = 4 # 2 = 2.6 C/m2 er 5 Option (A) is correct. The ohm’s law in point form in field theory is expressed as below (For constant voltage) V = RI rl El = JA A where l is length integral and A is the cross sectional area. So, we get E = rJ E =J s i.e. J = sE Option (A) is correct. Displacement current density is defined as Jd = e2E 2t and the conduction current density is defined as Jc = sE for a dielectric e must be larger while conductivity must tend to zero. So, we get Jd >> Jc i.e. displacement current is much greater than conduction current.
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SOL 3.3.45
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SOL 3.3.44
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SOL 3.3.43
Option (D) is correct. Conduction current, Ic = 1 A Operating frequency, f = 50 Hz Medium permittivity, e = e0 Permeability m = m0 Conductivity, s = 5.8 # 10 mho/m The ratio of conduction current density to the displacement current density is Jc = s we Jd Ic /A or, (A is cross sectional area) = s we Id /A GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 3.3.46
Chap 3
Electric Field in Matter
For View Only
199
Shop Online at www.nodia.co.in Id = we Ic = 2p # 50 # e0 (1) = 3.3 # 10-11 A s 5.8 # 10
Option (B) is correct. When there is no charge in the interior of a conductor, the electric field intensity is zero according to Gauss’s law the total outward flux through a closed surface is equal to the charge enclosed. Now if any charge is introduced inside a closed conducting surface then an electric field will be setup and the field exerting a force on the charges and making them move to the conducting surface. So all the charges inside a conductor is distributed over its surface. Therefore the outward flux through any closed surface constructed inside the conductor must vanish. A is false but R is true.
SOL 3.3.48
Option (B) is correct. When the method of images is used for a system consisting of a point charge between two semi infinite conducting planes inclined at an angle f, the no. of images is given by N = c 360c - 1m f Here the angle between conducting planes is f = 90c. So, N =3 and since all the images lie an a circle so we have the image charges as shown in figure.
SOL 3.3.49
Option (D) is correct. Consider the two dielectric regions as shown below.
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SOL 3.3.47
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200
Electric Field in Matter
Chap 3
Option (D) is correct. The stress is called the force per unit area which is directly proportional to the electric field intensity and electric field intensity is inversely proportional to the permittivity of dielectric material. i.e. E\1 e 1 E1 = /e0 = 1/e0 = 5 So, ratio of stress is E2 1/e 1/5e0
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SOL 3.3.50
m
For View Only Shop Online at www.nodia.co.in Since the field is normal to the interface So, the normal components of the fields are, E1n = 1 and E2n = 2 From boundary condition we have e1 E1n - e2 E2n = rs (where rs is surface charge density on the interface). ^e0h^1 h - ^2e0h^2 h = rs rs =- 3e0
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CHAPTER 4 MAGNESTOSTATIC FIELDS
202
Magnestostatic Fields
EXERCISE 4.1
Assertion (A) : For a static magnetic field the total number of flux lines entering a given region is equal to the total no. of flux lines leaving the region. Reason (R) : An isolated magnetic charge doesn’t exist. (A) Both A and R one true and R is the correct explanation of A.
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MCQ 4.1.1
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For View Only
Chap 4
(B) Both A and R are true but R is not the correct explanation of A.
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(C) A is true but R is false. (D) A is false but R is true.
Match List-I with List-II and select the correct answer using the codes given below. (Notations have their usual meaning).
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MCQ 4.1.2
List-I
1. d : D = rv
Ampere’s law
b.
Conservative nature of magnetic field
2.
c.
Gauss’s law
3. d : B = 0
d.
Non existence of magnetic monopole
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a.
b 1 1 4 4
c 4 4 1 2
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Codes : a (A) 3 (B) 2 (C) 2 (D) 3 MCQ 4.1.3
List-II
# J : dS = # H : dl S
4.
L
# E : dl = 0
d 2 3 3 1
Magnetic field intensity H exists inside a certain closed spherical surface. The value of d : H will be (A) 0 at each point inside the sphere. (B) 0 at the center of the sphere only. (C) 0 at the outer surface of the sphere only. (D) Can’t be determined as H is not given.
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Chap 4
Magnestostatic Fields
203
For View Only Shop Online at www.nodia.co.in MCQ 4.1.4 The source which doesn’t cause a magnetic field is (A) A charged disk rotating at uniform speed (B) An accelerated charge (C) A charged sphere spinning along it’s axis (D) A permanent magnet A circular loop of radius a , centered at origin and lying in the xy plane, carries current I as shown in the figure.
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MCQ 4.1.5
(C) I az 4a
(C) 0.074af wb/m2 MCQ 4.1.7
(D) 0.074az wb/m2
A circular conducting loop of radius 2 m, centered at origin in the plane z = 0 carries a current of 4 A in the af direction. What will be the magnetic field intensity at origin ? (B) az A/m (A) 1 az A/m 2p (C) 2az A/m
MCQ 4.1.8
(D) 2I az a
A conducting filament carries a current 5 A from origin to a point ^3, 0, 4h. Magnetic field intensity at point (3, 4, 0) due to the filament current will be (A) 0.23af wb/m2 (B) 0.095af wb/m2
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MCQ 4.1.6
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The magnetic field intensity a the centre of the loop will be (B) - I az (A) I az 2a 2a
(D) - az A/m
The correct configuration that represents magnetic flux lines of a magnetic dipole is
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Magnestostatic Fields
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Chap 4
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The correct configuration that represents current I and magnetic field intensity H is
MCQ 4.1.10
A long straight wire placed along z -axis carries a current of I = 5 A in the + az direction. The magnetic flux density at a distance r = 5 cm from the wire will be
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MCQ 4.1.9
(A) 4 # 10-5 wb/m2 (C) 100 wb/m2 p MCQ 4.1.11
(B) 2 # 10-5 wb/m2 (D) 2 # 10-6 wb/m2
For the currents and the closed path shown in the figure what will be the value of
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# H : dl
?
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 4
Magnestostatic Fields
For View Only (A) 30 A
Shop Online at www.nodia.co.in (B) 20 A
(C) - 20 A
(D) 10 A
Two infinitely long wires separated by a distance 2 m , carry currents I in opposite direction as shown in the figure.
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MCQ 4.1.12
205
If I = 8 A , then the magnetic field intensity at point P is
In the free space a semicircular loop of radius a carries a current I . What will be the magnitude of magnetic field intensity at the centre of the loop ? (B) 2I (A) I a a
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MCQ 4.1.13
(D) - 1 ay 8p
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(C) 1 ay 8p
(B) - 5 ay p
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(A) 5 ay p
(C) I 4a
(D) 4I a
Common Data for Question 14 - 15 : A long cylindrical wire of cross sectional radius R carries a steady current I distributed over its outer surface. MCQ 4.1.14
Magnetic field intensity inside the wire at a distance r (< R) from it’s center axes will be (A) non uniform (B) zero
(C) uniform and depends on r only
(D) uniform and depends on both r and R MCQ 4.1.15
The magnetic flux density outside the wire at a distance r (< R) from it’s center axes will be proportional to (A) r (B) 1/r (C) r/R
(D) 1/R
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206
Magnestostatic Fields
Chap 4
For View Only Shop Online at www.nodia.co.in MCQ 4.1.16 Two point charges Q1 and Q2 are located at ^0, 0, 0h and ^1, 1, 1h respectively. A current of 16 A flows from the point charge Q1 to Q2 along a straight wire connected
#
between them. What will be the value of B : dl around the closed path formed by the triangle having the vertices ^1, 0, 0h, ^0, 1, 0h and ^0, 0, 1h ? (B) 6m0 Wb/m2
(C) 14m0 Wb/m
(D) 6m0 Wb/m
m
(A) 22m0 Wb/m
Common Data for Question 17 - 18
Magnetic field intensity at origin will be (B) - 10ay A/m (A) 11ay A/m
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MCQ 4.1.17
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An infinite current sheet with uniform current density K = 15ax A/m is located in the plane z = 2 .
(D) 20ay A/m
(C) 0 A/m
Magnetic field intensity at point (2, - 1, 5) will be (A) 10ay A/m (B) - 10ay A/m
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MCQ 4.1.18
(D) 20ay A/m
(C) 0 A/m
(B) constant
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Two infinite current carrying sheets are placed parallel to each other in free space such that they carry current in the opposite direction with the same surface current density. The magnetic flux density in the space between the sheets will be (A) zero
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MCQ 4.1.19
(C) linearly increasing from one sheet to other (D) none of these
In a spherical co-ordinate system magnetic vector potential at point (r, q, f) is given as A = 12 cos qa q . The magnetic flux density at point (3, 0, p) will be (A) 4af (B) 0
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MCQ 4.1.20
(D) 36af
(C) 4a q MCQ 4.1.21
An infinite plane current sheet lying in the plane y = 0 carries a linear current density K = Kaz A/m . The magnetic field intensity above (y > 0 ) and below ^y < 0h the plane will be y>0 y 8 cm will be (B) + 0.13az A/m
(C) 64.3az mA/m
(D) - 0.10az A/m
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(A) - 0.13az A/m
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MCQ 4.1.22
207
Statement for Linked Question 23 - 24 :
The magnetic flux density at point (1, - 2, - 5) will be (A) 40ax + 6az wb/m2
(C) - 40ax - 80ay - 6az wb/m2
(D) 80ay - 6az wb/m2
The total magnetic flux through the surface z = 4 , 0 # x # 1, - 1 # y # 4 will be
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MCQ 4.1.24
(B) 40ax + 80ay + 6az wb/m2
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MCQ 4.1.23
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In a cartesian system, vector magnetic potential at a point (x, y, z) is defined as A = 4x2 yax + 2y2 xay - 3xyzaz wb/m
(A) 20 wb
(B) - 10/3 wb
(C) 40 wb
(D) 130/3 wb
Statement for Linked Question 25 - 26 : An infinite current sheet with uniform surface current density K = 8ax A/m is located at z = 0 as shown in figure.
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208
Magnestostatic Fields
Chap 4
For View Only Shop Online at www.nodia.co.in MCQ 4.1.25 Magnetic flux density at any point above the current sheet (z > 0) will be (B) 2m0 ay wb/m2 (A) - 2m0 ay wb/m2 m (D) - m0 ay wb/m2 (C) 0 ay wb/m2 2 MCQ 4.1.26
The vector magnetic potential at z =- 2 will be (A) 4m0 ax wb/m (B) - 4m0 ay wb/m
m
(C) 2m0 ax wb/m
In the free space, magnetic field intensity at any point (r, f, z) is given by H = 2r2 af A/m . The current density at r = 2 m will be (B) 24az A/m (A) 12az A/m
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MCQ 4.1.27
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(D) - 4m0 ax wb/m
(C) 4az A/m
The current density that would produce the magnetic vector potential A = 2af in cylindrical coordinates is
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MCQ 4.1.28
(D) 0
1 a f m0 r2 (C) 2 af m0 r MCQ 4.1.29
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(A)
(B)
2r2 a m0 f
(D)
2 a f m0 r2
Magnetic field intensity produced due to a current source is given as
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H = ^2z cos ay h ay + ^4z + ey h ax
The current density over the xz plane will be (A) ^ax - ay - az h A/m2 (B) - ax + ay - az
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(C) - 2ax + ay - 2az (D) ax + ay + az MCQ 4.1.30
Assertion (A) : In a source free region, magnetic field intensity can be expressed as a gradient of scalar function. Reason (R) : Current density for a given magnetic field intensity is defined as J = d#H (A) A and R both are true and R is correct explanation of A. (B) A and R both are true and R is not the correct explanation of A. (C) A is true but R is false. (D) R is true but A is false.
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Chap 4
Magnestostatic Fields
209
For View Only Shop Online at www.nodia.co.in MCQ 4.1.31 Magnetic field intensity produced at a distance r from an infinite cylindrical wire located along entire z -axis is 3raf A/m . The current density within the conductor will be (A) 6raz A/m2 (B) 3az A/m2 (C) 6az A/m2 (D) 3raz A/m2 An electron beam of radius a travelling in az direction, the current density is given as r For r < a J = 2 a1 - k az a The magnetic field intensity at the surface of the beam will be (B) a af (A) a af 6 3
In a certain region consider the magnetic vector potential is A and the current density is J . Which of the following is the correct relation between J and A ? (A) dA = J (B) d2A = m0 J (C) d # A =- mJ
(D) d2A =- m0 J
A circular loop of wire with radius R = 1.5 m is located in plane x = 0 , centered at origin. If the loop carries a current I = 7 A flowing in clockwise as viewed from negative x -axis then, its magnetic dipole moment will be (A) 0.12ax A- m2
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MCQ 4.1.34
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MCQ 4.1.33
2 (D) a af 3
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2 (C) 2pa af 3
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MCQ 4.1.32
(B) - 5.5ax A- m2 (C) 5.5ax A- m2 (D) 22ax A- m2
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210
Magnestostatic Fields
EXERCISE 4.2
In the free space, the positive z -axis carries a filamentary current of 10 A in the - az direction. Magnetic field intensity at a point (0, 3, 2) due to the filamentary current will be (A) - 0.73ax A/m (B) 1.46ax A/m
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MCQ 4.2.1
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For View Only
(D) 0.73ax A/m
If there is a current filament on the x -axis carrying 4.4 A in ax direction then what will be the magnetic field intensity at point (4, 2, 3) ? (A) 0.1 (az - 2ay) A/m (B) 1.76az - 1.62ay A/m (D) - 0.1 (2az - ay)
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(C) (- 1.077az + 1.62ay) A/m
A filamentary conductor is formed into an equilateral triangle of side 2 m that carries a current of 4 A as shown in figure. The magnetic field intensity at the center of the triangle will be
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MCQ 4.2.3
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(C) 0.40ax A/m MCQ 4.2.2
Chap 4
MCQ 4.2.4
(A) 9 az A/m p
(B) 3 az A/m p
(C) 6 az A/m p
(D) 0
A current sheet K = 4ay A/m flows in the region - 2 < z < 2 in the plane x = 0 . Magnetic field intensity at point P (3, 0, 0) due to the current sheet will be (A) - 1.5az A/m (B) - 0.75az A/m (C) + 0.75az A/m
(D) - 2.1az A/m
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Chap 4
Magnestostatic Fields
211
For View Only Shop Online at www.nodia.co.in MCQ 4.2.5 In the plane z = 0 a disk of radius 3 m , centered at origin carries a uniform surface charge density rS = 2 C/m2 . If the disk rotates about the z -axis at an angular velocity w = 2 rad/s then the magnetic field intensity at the point P (0, 0, 1) will be (A) az A/m (B) 2az A/m (C) ay A/m
A square conducting loop of side 1 m carries a steady current of 2 A. Magnetic flux density at the center of the square loop will be. (B) 0.45 Wb/m2 (A) 17.78 # 10-7 Wb/m2
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MCQ 4.2.6
(D) 2ay A/m
(C) 2.26 # 10-6 Wb/m2
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A filamentary conductor is formed into a loop ABCD as shown in figure. If it carries a current of 5.2 A then the magnetic field intensity at point P will be
MCQ 4.2.8
MCQ 4.2.9
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MCQ 4.2.7
(D) 4 # 10-7 Wb/m2
(A) 0.2 A/m
(B) 0.8 A/m
(C) 0.26 A/m
(D) 1.01 A/m
The magnetic field intensity at point P due to the steady current configurations shown in figure will be
(A) 0.82 A m
(B)0.32 A m
(C) 0.5 A m
(D) 0.18 A m
In the plane z = 5 m a thin ring of radius, a = 3 m is placed such that z -axis passes through it’s center. If the ring carries a current of 50 mA in af direction then the magnetic field intensity at point (0, 0, 1) will be (A) 0.9az mA/m (B) 1.8az mA/m
(C) 0.6az mA/m (D) 0.5az A/m GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Magnestostatic Fields
Chap 4
For View Only Shop Online at www.nodia.co.in MCQ 4.2.10 An infinite solenoid (infinite in both direction) consists of 1000 turns per unit length wrapped around a cylindrical tube. If the solenoid carries a current of 4 mA then the magnetic field intensity at its axis will be (A) 4 A/m (B) 0 (C) 2000 A/m
(D) 0.2 A/m
m
Common Data for Question 11 - 13 :
Magnetic field intensity inside the inner solenoid will be
(A) - 3ay A/m
MCQ 4.2.12
(B) + 3ay A/m (D) + 6ay A/m
The magnetic field intensity in the region between the two solenoids will be (A) 3ay A/m (B) 0 (C) 6ay A/m
(D) - 3ay A/m
The magnetic field outside the outer solenoid will be (A) 6ay A/m (B) - 3ay A/m (C) 0
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MCQ 4.2.13
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(C) - 6ay A/m
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MCQ 4.2.11
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The two long coaxial solenoids of radius a and b carry current I = 6 mA but in opposite directions. Solenoids are placed along y -axis as shown in figure. The inner solenoid has 2000 turns per unit length and outer solenoid has 1000 turns per unit length.
(D) + 3ay A/m
Statement for Linked Question 14 - 15 : In a cartesian system two parallel current sheets of surface current density K1 = 3az A/m and K2 =- 3az A/m are located at x = 2 m and x =- 2 m respectively. The net vector and scalar potential due to the sheets are zero at a point P (1, 2, 5). MCQ 4.2.14
Consider the scalar potential at any point (x, y, z) in the region between the two
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Chap 4
Magnestostatic Fields
213
MCQ 4.2.16
The vector potential at origin will be (A) 3m0 az Wb/m
(B) - 3m0 az Wb/m
(C) 0
(D) - 3 Wb/m
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MCQ 4.2.15
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For View Only Shop Online at www.nodia.co.in planar sheets is Vm . The plot of Vm versus y will be
A long cylindrical wire lying along z -axis carries a total current I 0 = 15 mA as shown in the figure. The current density inside the wire at a distance r from it’s axis is given by J \ r .
If the cross sectional radius of the wire is 2 cm then the magnetic flux density at r = 1 cm will be (B) 6.25 # 10-4 Wb/m2 (A) 25 nWb/m2 (C) 1.25 nWb/m2
(D) 12.5 nWb/m2
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214
Magnestostatic Fields
Chap 4
For View Only Shop Online at www.nodia.co.in MCQ 4.2.17 Magnetic field intensity is given in a certain region as x2 yz xyz2 H = ax + 3x2 z2 ay a A/m 1+x y+1 z The total current passes through the surface x = 2 m , 1 # y # 4 m , 3 # z # 4 m in ax direction will be (A) - 259 A (B) 259 A (D) 1.98 # 103 A
(C) 18.2 A
m
A phonograph record of radius 1 m carries a uniform surface charge density rS = 20 C/m2 . If it is rotating with an angular velocity w = 0.1 rad/s ; then the magnetic dipole moment will be (A) 4p A- m2 (B) p/2 A- m2 (C) 2p A- m2 (D) 2p A- m2 3
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MCQ 4.2.18
Statement for Linked Question 19 - 20 :
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A uniformly charged solid sphere of radius r is spinning with angular velocity w = 6 rad/s about the z -axis. The sphere is centered at origin and carries a total charge 5 C which is uniformly distributed over it’s volume. The plot of magnetic dipole moment of the sphere, m (r) versus the radius of the sphere, r will be
MCQ 4.2.20
The average magnetic field intensity within the sphere will be (B) 2 az (A) 2 a q pr pr
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MCQ 4.2.19
(C) 2 a q (D) 1 az 2pr r GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 4
Magnestostatic Fields
215
For View Only Shop Online at www.nodia.co.in MCQ 4.2.21 A rectangular coil, lying in the plane x + 3y - 1.5z = 3.5 carries a current 7 A such that the magnetic moment of the coil is directed away from the origin. If the area of the rectangular coil is 0.1 m2 then the magnetic moment of the coil will be (A) - 0.2ax - 0.2ay + 0.3az A- m2 (B) 2ax + 6ay - 3az A- m2 (C) 1.4ax + 4.2ay - 2.1az A- m2 (D) 0.2ax + 0.6ay - 0.3az A- m2 Vector magnetic potential in a certain region of free space is A = (6y - 2z) ax + 4xzay The electric current density at any point (x, y, z) will be (A) (- 8ax + 2ay + 6az ) A/m2 (B) (3ay + az ) A/m2 (D)
1 m0
(8ax + 2ay - 6az ) A/m2
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(C) 0
m
MCQ 4.2.22
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Statement for Linked Question 23 - 24 :
MCQ 4.2.23
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A circular toroid with a rectangular cross section of height h = 5 m , carries a current I = 10 A flowing in 105 turns of closely wound wire around it as shown in figure. The inner and outer radii of toroid are a = 1 m and b = 2 m respectively.
The total magnetic flux across the circular toroid will be (A) 1.39 Wb (B) 0.14 Wb (C) 15.1 Wb
MCQ 4.2.24
If the magnetic flux is found by multiplying the cross sectional area by the flux density at the mean radius then what will be the percentage of error ? (A) - 4.31% (B) - 3.14% (C) - 4.61%
MCQ 4.2.25
(D) 0 Wb
(D) - 6.14%
Magnetizing force at any point P on z -axis due to a semi infinite current element placed along positive x -axis is H . If one more similar current element is placed along positive y -axis then the resultant magnetizing force at the point P will be (A) H/ 2 (B) 2 H (C) 2H
(D) - 2 H
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m
For View Only Shop Online at www.nodia.co.in MCQ 4.2.26 An infinitely long straight wire carrying current 5 A and a square loop of side 2 m are coplanar as shown in the figure. The distance between side AB of square loop and the straight wire is 4 m. What will be the total magnetic flux crossing through the rectangular loop ?
(B) 81.1 mWb
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(A) 2.55 mWb (C) 0.81 mWb
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A 1.5 m square loop is lying in x -y plane such that one of it’s side is parallel to y -axis and the centre of the loop is 0.3 m away from the y -axis. How much current must flow through the entire y -axis for which the magnetic flux through the loop is 5 # 10-5 Tesla m2 ? (A) 417 A (B) 834 A
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MCQ 4.2.27
(D) 8.11 mWb
(C) 208.5 A
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A L-shaped filamentary wire with semi infinite long legs making an angle 90c at origin and lying in y -z plane as shown in the figure.
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MCQ 4.2.28
(D) 280 A
If the current flowing in the wire is I = 4 A then the magnetic flux density at ^2 m, 0, 0h will be (A) - 2 # 10-7 ^ay + az h Wb/m2 (B) 2 # 10-7 ^ay + az h Wb/m2 (C) - 4 # 10-7 ^ay + az h Wb/m2 (D) 4 # 10-7 ^ay + az h Wb/m2
Common Data for Question 29 - 30 : An infinitely long straight conductor of cylindrical cross section and of radius R carries a current I , which is uniformly distributed over the conductor cross section. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Magnestostatic Fields
217
m
For View Only Shop Online at www.nodia.co.in MCQ 4.2.29 If the conductor is located along z -axis then the magnetic flux density at a distance r (> R) from the cylindrical axis will be m Ir mI (A) 0 2 af (B) 0 2 af 2pR 2pr m Ir2 mI (C) 0 af (D) 0 2 af 2p R MCQ 4.2.30 Magnetic flux density at a distance r (> R) from the cylindrical axis will be proportional to (B) 12 (A) 1 r r (D) r2
(C) r
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Consider a filamentary wire is bent to form a square loop of side 3 m lying in the x -y plane as shown in the figure. If the current flowing in the wire is I = 1 A then the magnetic flux density at the center of the loop will be
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MCQ 4.2.31
(A) 2 2 # 10-7 az Wb/m2 (C) 2 # 10-7 az Wb/m2 MCQ 4.2.32
(B) 4 2 # 10-7 az Wb/m2 (D) - 4 2 # 10-7 az Wb/m2
An infinitely long straight wire carrying a current 20 A and a circular loop of wire carrying a current I are coplanar as shown in the figure.
The radius of the circular loop is 10 cm and the distance of the centre of the loop from the straight wire is 1 m. If the net magnetic field intensity at the centre of the loop is zero then the current I is (A) p2 A (B) 20p A (C)
p 2
A
(D) 2p A
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For View Only Shop Online at www.nodia.co.in MCQ 4.2.33 The magnitude of the magnetic field intensity produced at center of a square loop of side a carrying current I is (B) 2 I (A) 2 2 I pa pa I (D) 8I pa 2 pa For the single turn loop of current shown in the figure the magnetic field intensity at the center point P of the semi circular portion will be (C)
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MCQ 4.2.34
(B) 5.8 A/m inward
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(A) 5.8 A/m outward (C) 3.8 A/m outward
Two perfect conducting infinite parallel sheets separated by a distance 2 m carry uniformly distributed surface currents with equal and opposite densities 4ax and - 4ax respectively as shown in figure.
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MCQ 4.2.35
(D) 3.8 A/m inward
The medium between the two sheets is free space. What will be the magnetic flux between the sheets per unit length along the direction of current ? (A) 0 (B) 8m0 ay Wb/m (C) - 8m0 ay Wb/m
(D) - 4m0 ay Wb/m ***********
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EXERCISE 4.3
Statement for Linked Question 1 - 2 :
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GATE 2012
The magnetic field at a distance r from the center of the wire is proportional to (A) r for r < a and 1/r 2 for r > a (B) 0 for r < a and 1/r for r > a (C) r for r < a and 1/r for r > a
MCQ 4.3.2
A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.
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GATE 2012
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(D) 0 for r < a and 1/r 2 for r > a
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MCQ 4.3.1
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An infinitely long uniform solid wire of radius a carries a uniform dc current of density J
The magnetic field inside the hole is (A) uniform and depends only on d (B) uniform and depends only on b
(C) uniform and depends on both b and d (D) non uniform MCQ 4.3.3 GATE 2009
Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y - z plane and parallel to the y - axis. The other wire is in the x - y plane and parallel to the x - axis. Which components of the resulting magnetic field are non-zero at the origin ?
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(A) x, y, z components
(B) x, y components
(C) y, z components IES EC 2012
A flux of 2.2 mWb exerts in a magnet having a cross-section of 30 cm2 . The flux density in tesla is (A) 4 (B) 0.4
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MCQ 4.3.4
(D) x, z components
IES EC 2010
The magnetic flux density B and the vector magnetic potential A are related as (B) A = 4# B (A) B = 4# A (C) B = 4: A
IES EC 2010
(D) A = 4: B
Consider the following statements relating to the electrostatic and magnetostatic field : 1. The relative distribution of charges on an isolated conducting body is dependent on the total charge of the body. 2.
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MCQ 4.3.6
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MCQ 4.3.5
(D) 40
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(C) 2.5
The magnetic flux through any closed surface is zero.
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Which of the above statements is/are correct ? (A) Neither 1 nor 2 (B) 1 only (C) 2 only MCQ 4.3.7 IES EC 2008
(D) Both 1 and 2
The line integral of the vector potential A around the boundary of a surface S represents which one of the following? (A) Flux through the surface S (B) Flux density in the surface S (C) Magnetic field intensity (D) Current density
MCQ 4.3.8 IES EC 2008
An infinitely long straight conductor located along z-axis carries a current I in the +ve z -direction. The magnetic field at any point P in the x - y plane is in which
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(B) In the negative z -direction (C) In the direction perpendicular to the radial line OP (in x - y plane) joining the origin O to the point P (D) Along the radial line OP IES EC 2008
A 13 A current enters a right circular cylinder of 5 cm radius. What is the linear surface current density at the end surface? (A) (50/p) A/m (B) (100/p) A/m
m
MCQ 4.3.9
(C) (1000/p) A/m
What is the value of the magnetic vector potential due to an infinitesimally small current element, evaluated at infinite distance from it ? (A) Infinity (B) Unity (C) Zero
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IES EC 2007
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MCQ 4.3.10
(D) (2000/p) A/m
IES EC 2007
What is the magnetic field intensity vector H between two parallel sheets with separation ‘d ’ along z-axis both sheets carrying surface current K = Ky ay ? (B) + ky ay (A) - ky ay
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MCQ 4.3.11
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(D) Any number between zero and infinity depending on the strength of the current element
(C) - ky ax MCQ 4.3.12 IES EC 2005
(D) Zero
Current density (J), in cylindrical coordinate system is given as : 0 for 0 < r < a J (r, f, z) = *J (r/a2) a for a < r < b 0 z where az is the unit vector along z -coordinate axis. In the region, a < r < b , what is the expression for the magnitude of magnetic field intensity (H ) ? (A) J 02 (r3 - a3) (B) J 02 (r3 + a3) r r 3 3 J (r - a ) (C) 0 2 (D) J 0 (r3 - a3) 2pr 3a r
MCQ 4.3.13 IES EC 2005
Which one of the following concepts is used to find the expression of radiated E and H field due to a magnetic current element ? (A) Concept of vector magnetic potential (B) Concept of scalar electric potential (C) Concept of scalar magnetic potential (D) Concept of vector electric potential
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Chap 4
For View Only Shop Online at www.nodia.co.in MCQ 4.3.14 The circulation of H around the closed contour C , shown in the figure is
(B) 2l
(C) 4l
(D) 6l
The unit of magnetic flux density is (A) gauss (C) bohr
IES EE 2012
(D) weber/sec
The magnetic flux density created by an infinitely long conductor carrying a current I at a radial distance R is mI (B) 1 (A) 0 2p R 2p R 2 m0 I (D) 4pR I 3 3 2p R Match List I with List II and select the correct answer using the codes given below the lists.
List-I Work
b.
Ampere/metre
Electric field strength
2.
Weber
c.
Magnetic flux
3.
Volt/metre
d.
Magnetic field strength
4.
Joule
Codes : a (A) 4 (B) 1 (C) 4 (D) 1 MCQ 4.3.18 IES EE 2009
List-II 1.
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a.
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IES EE 2011
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(C) MCQ 4.3.17
(B) tesla
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MCQ 4.3.16
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IES EE 2012
(A) 0
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MCQ 4.3.15
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IES EC 2002
b 3 3 2 2
c 2 2 3 3
d 1 4 1 4
A long straight wire carries a current I = 1 A . At what distance is the magnetic field 1 Am-1 ? (A) 1.59 m (B) 0.159 m
(C) 0.0159 m (D) 0.00159 m GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 4.3.19 How much current must flow in a loop radius 1 m to produce a magnetic field IES EE 2009 1 mAm-1 ? (A) 1.0 mA (B) 1.5 mA (C) 2.0 mA MCQ 4.3.20 IES EE 2009
(D) 2.5 mA
Assertion (A) : Knowing magnetic vector potential A at a point, the flux density B at the point can be obtained. Reason (R) : d : A = 0 . (A) Both A and R are true and R is the correct explanation of A
m
(B) Both A and R are true and R is not the correct explanation of A (C) A is true but R is false
MCQ 4.3.21
The magnetic vector potential A obeys which equations ? 1. B = d # A 2. d2A =- m0 J m Idl 3. A = # 0 4p R
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IES EE 2008
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(D) A is false but R is true
(C) 1 and 3 IES EE 2008
(C) 1.59 m MCQ 4.3.23 IES EE 2006
MCQ 4.3.24 IES EE 2006
MCQ 4.3.25 IES EE 2006
(D) 1, 2 and 3
A long straight wire carries a current I = 10 A . At what distance is the magnetic field H = 1 Am- 1 ? (A) 1.19 m (B) 1.39 m
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MCQ 4.3.22
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Select the correct answer using the code given below : (A) 1 and 2 (B) 2 and 3
(D) 1.79 m
What is the magnetic field due to an infinite linear current carrying conductor ? (A) H =
mI A/m 2p r
(B) H = I A/m 2pr
(C) H =
mI A/m 2r
(D) H = I A/m r
Equation d : B = 0 is based on (A) Gauss’s Law
(B) Lenz’s Law
(C) Ampere’s Law
(D) Continuity Equation
Plane y = 0 carries a uniform current density 30az mA/m . At (1, 20, - 2), what is the magnetic field intensity ? (A) - 15ax mA/m
(B) 15ax mA/m
(C) 18.85ay mA/m
(D) 25ax mA/m
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For View Only Shop Online at www.nodia.co.in MCQ 4.3.26 Which one of the following is not the valid expression for magnetostatic field vector IES EE 2005 B ? (A) B = 4: A (B) B = d # A (D) 4# B = m0 J
(C) d : B = 0 MCQ 4.3.27 IES EE 2004
Which one of the following statements is correct ? Superconductors are popularly used for (A) generating very strong magnetic field
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(B) reducing i2 R losses (C) generating electrostatic field
MCQ 4.3.28
Assertion (A) : # B : dS = 0 where, B = magnetic flux density, dS = s vector with direction normal to surface elements dS . Reason (R) : Tubes of magnetic flux have no sources or sinks. (A) Both A and R are true and R is the correct explanation of A
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IES EE 2002
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(D) generating regions free from magnetic field
(C) A is true but R is false (D) A is false but R is true IES EE 2001
Plane defined by z = 0 carry surface current density 2ax A/m . The magnetic intensity ‘Hy ’ in the two regions - a < z < 0 and 0 < z < a are respectively (A) ay and - ay
(B) - ay and ay (D) - ax and ax
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(C) ax and - ax
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MCQ 4.3.29
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(B) Both A and R are true but R is NOT the correct explanation of A
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SOLUTIONS 4.1
Option (D) is correct. It is not possible to have an isolated magnetic poles (or magnetic charges). If we desire to have an isolated magnetic dipole by dividing a magnetic bar successively into two, we end up with pieces each having north and south poles. So an isolated magnetic charge doesn’t exist. That’s why the total flux through a closed surface in a magnetic field must be zero.
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SOL 4.1.1
#
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i.e. B : dS = 0 or more clear, we can write that for a static magnetic field the total number of flux lines entering a given region is equal to the total number of flux lines leaving the region. So, (A) and (R) are both true and R is correct explanation of A. Option (B) is correct.
SOL 4.1.3
Option (D) is correct. Since the field intensity exists in a closed surface and lines of field intensity makes a closed curve so the flux lines leaving the spherical surface equal to the total flux entering the surface and So the net flux
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SOL 4.1.2
#
F = B : dS = 0 According to divergence theorem
# ^d : B hdv 0 = # d : B dv
# B : dS
=
Since volume of the sphere will have certain finite value so, d:B = 0 or d : H = 0 at all points inside the sphere SOL 4.1.4
Option (C) is correct. The Magnetic field are caused only by current carrying elements and given as m Idl R B = 0 #3 4p R Since an accelerated electron doesn’t form any current element(Idl ) so it is not a source of magnetic field.
SOL 4.1.5
Option (D) is correct. The magnetic field intensity produced due to a small current element Idl is defined
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m
R dH = Idl # a 4p R 2 where dl is the differential line vector and aR is the unit vector directed towards the point where field is to be determined. So for the circular current carrying loop we have dl = adfaf aR =- a r Therefore the magnetic field intensity produced at the centre of the circular loop is 2p Iadfa f # ^- a r h H = = Ia 2 6f@20p ^az h = I az A/m 4a 4pa2 4p a f=0
Option (B) is correct.
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SOL 4.1.6
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Magnetic field intensity at any point P due to a filamentary current I is defined as H = I 6cos a2 - cos a1@ af 4pr where r " distance of point P from the current filament. a1 " angle subtended by the lower end of the element at P . a2 " angle subtended by the upper end of the element at P . From the figure we have r = 32 + 42 = 5 a1 = r/2 & cos a1 = 0 12 and cos a2 = = 12 52 + 122 13 Now we Put these values to get, 5 12 - 0 a H = (I = 5 A) l f 4p # 5 b 13 = 3 af = 0.15af wb/m2 13p SOL 4.1.7 Option (C) is correct. According to Biot-savart law, magnetic field intensity at any point P due to the current element Idl is defined as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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# Idl4p#R R 3
m
where R is the vector distance of point P from the current element. Here current is flowing in af direction So the small current element Idl = Irdfaf = 4 # 2dfaf = 8dfaf and since the magnetic field to be determined at center of the loop so we have (radius = 2 m ) R = 2m and (pointing towards origin) aR =- a r Therefore the magnetic field intensity at origin is 2p (8dfa ) 2p f # (- a r) 8 dfa = az f 2p H = = z 2 p 16 4p 6 @ 0 p 4 ( 2 ) 0 0 = 2az A/m
#
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#
Option (B) is correct. According to right hand rule if the thumb points in the direction of outward or inward current then rest of the fingers will curl along the direction of magnetic flux lines, This condition is satisfied by the configuration shown in option (C).
SOL 4.1.9
Option (B) is correct. According to right hand rule if the thumb points in the direction of current then rest of the fingers will curl along the direction of magnetic field lines. This condition is satisfied by the configuration shown in option (C).
SOL 4.1.10
Option (D) is correct.
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SOL 4.1.8
According to Ampere’s circuital law, the line integral of magnetic field intensity H around a closed path is equal to the net current enclosed by the path. Since we have to determine the magnetic field intensity due to the infinite line current at t = 5 cm so we construct a circular loop around the line current as shown in the figure. Now from Ampere’s circuital law we have
# B : dl L
= m0 Ienc
or (Ienc = 10 A) B (2pr) = m0 # 10 Therefore we have the magnetic flux density at t = 5 cm as -7 B = 4p # 10 # -10 = 5 # 10-5 wb/m2 2p # 5 # 10 2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in SOL 4.1.11 Option (B) is correct. According to Ampere’s circuital law the contour integral of magnetic field intensity in a closed path is equal to the current enclosed by the path.
#
i.e. H : dl = Ienc Now using right hand rule, we obtain the direction of the magnetic field intensity in the loop as it will be opposite to the direction of L.
#
m
So, H : dl =- Ienc =- 20 A (10 A is not inside the loop. So it won’t be considered.)
Option (C) is correct. The magnetic field intensity produced at a distance r from an infinitely long straight wire carrying current I is defined as H = I 2pr As determined by right hand rule, the direction of magnetic field intensity will be same(in - ay direction) due to both the current source. So, at point P the net magnetic field intensity due to both the current carrying wires will be H = H1 + H 2 = I ^- ay h + I ^- ay h 2p ^4h 2p ^1 h 5 ^8 h (I = 8 A ) =a =- 4 ay p 8p y
SOL 4.1.13
Option (B) is correct. As calculated in previous question the magnetic field intensity produced at the centre of the current carrying circular loop is H = I 2a So by symmetry the semicircular loop will produce the field intensity half to the field intensity produced by complete circular loop. i.e. Field intensity at the centre of semicircular loop = 1 H = I 2 4a
SOL 4.1.14
Option (C) is correct. Since current in the wire is distributed over the outer surface so net enclosed current, Ienc for any Amperian loop inside the wire will be zero. and as from Ampere’s circuital law we have
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SOL 4.1.12
So or
# H : dl # H : dl
= Ienc
=0
(Ienc = 0 )
H = 0 for r < R
Option (C) is correct. Consider the cylindrical wire is lying along z -axis as shown in the figure. As the current I is distributed over the outer surface of the cylinder so for an Amperian GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 4.1.15
Chap 4
Magnestostatic Fields
229
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For View Only Shop Online at www.nodia.co.in loop at a distance r (> R) from the centre axis, enclosed current is equal to the total current flowing in the wire.
# B : dl
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Now from Ampere’s circuital law we have, = m0 Ienc
or
Option (A) is correct. Since the current flows from Q1 and terminates at Q2 and the charge Q2 is located at the surface of the contour so the actual current is not enclosed by the closed path and the circulation of the field is given as
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SOL 4.1.16
m0 I a 2pr f B\1 r
B =
he
or
(Ienc = I )
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B (2pr) = m0 I
# B : dl
and
= m0 6Id@enc
d : # e0 E1 : dS + 6Id@enc = dt
^6Ic@enc = 0h
#eE 0
2
: dS D
where E1 is the electric field intensity produced by charge Q1 while E2 is the field intensity produced by charge Q2 . d e Q1 + e Q2 = 1 dQ1 + 1 dQ2 So, (1) 6Id@enc = dt 0c = 0 c 8e0 m 2e0 mG 8 dt 2 dt As the current flows from Q1 and terminates at Q2 so the rate of change in the net charges is given as dQ dQ2 - 1 = = 16 A dt dt Therefore from equation (1) we have the enclosed displacement current as 6Id@enc = 18 (- 16) + 12 (16) = 6 A Thus, the circulation of magnetic flux density around the closed loop is
# B : dl
= m0 ^6 h = 8m0 Wb/m
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For View Only SOL 4.1.17 Option (D) is correct.
Chap 4
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Option (C) is correct.
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SOL 4.1.18
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Magnetic field intensity at any point P due to an infinite current carrying sheet is defined as H = 1 K # an 2 where K is the current density and an is the unit vector normal to the current sheet directed toward the point P . Since we have to determine the magnetic field intensity at origin so from the figure we have an =- az Therefore the magnetic field intensity at the origin is ( K = 20ax ) H = 1 (20ax ) # (- az ) = 5ay A/m 2
Magnetic field intensity at any point P due to an infinite current carrying sheet is defined as H = 1 K # an 2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Magnestostatic Fields
231
For View Only Shop Online at www.nodia.co.in where K is the current density of the infinite sheet and an is the unit vector normal to the current sheet directed toward the point P . Since we have to determine the magnetic field intensity at point (2, - 1, 5) which is above the plane sheet as shown in figure, so we have, an =+ az Therefore the magnetic field intensity at the point (2, - 1, 5) is (K = 20ax ) H = 1 (20ax # az ) =- 20 ay =- 5ay A/m 2 2 Option (C) is correct. Consider one of the sheet carries the current density K1 . So, the other sheet will have the current density - K1 . Magnetic flux density produced at any point P due to a current sheet is defined as m B = 0 K # an 2 where K is current density of the sheet and an is the unit vector normal to the sheet directed towards point P . So for any point in the space between the sheets normal vector will be opposite in direction for the two sheets as shown in figure i.e. an2 =- an1
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SOL 4.1.19
Therefore, the resultant magnetic flux density at any point in the space between the two sheets will be m B = 0 6K1 # an1 + (- K1) # (- an1)@ = m0 K1 # an1 2 Since an1 is unit vector normal to the surface, and K1 is given current density. So the cross product will be a constant. SOL 4.1.20
Option (D) is correct. The magnetic flux density at any point is equal to the curl of magnetic vector potential A at the point. i.e. B = d # A = d # (12 cos qa q) = 1 2 (12r cos q) af = 12 cos q af r r 2r
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(at (3, 0, p))
Option (C) is correct. Consider the current sheet shown in the figure.
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SOL 4.1.21
Chap 4
Option (D) is correct. Consider the current density at r = 8 cm is J directed along + az . Now the magnetic field for r > 8 cm must be zero. i.e. (for r > 8 cm ) H =0 So from Ampere’s circuital law we have
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SOL 4.1.22
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Magnetic field intensity produced at any point P due to a current sheet is defined as H = 1 K # an 2 where K is current density of the sheet and an is the unit vector normal to the sheet directed towards point P . (K = Kaz A/m , an = ay ) So, for y > 0 H = 1 ^Kaz h # ^ay h =- 1 Kax 2 2 (K = Kaz A/m , an =- ay ) and for y < 0 H = 1 ^Kaz h # ^- ay h = K ax 2 2
= Ienc = 0
ww
# H : dl
Since for the region r > 8 cm the Amperian loop will have all the current distributions enclosed inside it. i.e. Ienc = 14 # 10-3 + 2 # (2p # 0.5 # 10-2) - 0.8 # (2p # 0.25 # 10-2) + J (2p # 8 # 10-2) = 6.43 # 10-2 + J (16p # 10-2) So we have ( Ienc = 0 ) 6.43 # 10-2 + J (16p # 10-2) = 0 -2 6 . 43 10 # or J =16p # 10-2 or J =- 0.23az A/m SOL 4.1.23
Option (C) is correct.
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For View Only Shop Online at www.nodia.co.in Magnetic flux density is defined as the curl of vector magnetic potential i.e. B = d#A R V az W S a x ay 2 W = S 22x 22y 2z S 2 W 2 S2x y 2y - 8xyz W T X = (- 8xz - 0) ax + (0 + 8yz) ay + (2y2 - 2x2) az So the net magnetic flux density at (1, - 2, - 5) is B = 20ax + 60ay + 12az wb/m2
F =
m
Option (B) is correct. Total magnetic flux through a given surface S is defined as
# B : dS S
co
SOL 4.1.24
#
4
#
1
(2y2 - 2x2) (dxdy) = 2 # 1
he
F =
lp.
where dS is the differential surface vector having direction normal to the surface So, for the given surface z = 4 , 0 # x # 1, - 1 # y # 4 we have dS = (dxdy) az and as calculated in previous question we have B = (- 8xz - 0) ax + (0 + 8yz) ay + (2y2 - 2x2) az Therefore, the total magnetic flux through the given surface is y =- 1 x = 0 3 4
#
4
-1
y2 dy - 2 # 5
#
0
1
x2 dx
SOL 4.1.25
ww w. ga te
3 1 y = 2 ; E - 10 :x D = 2 # 65 - 10 3 -1 3 0 3 3 40 wb
Option (D) is correct. For determining the magnetic field at any point above the plane z = 0 , we draw a rectangular Amperian loop parallel to the y -z plane and extending an equal distance above and below the surface as shown in the figure. From Ampere’s circuital law,
# B : dl
= m0 Ienc
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Option (A) is correct. Magnetic flux density at a certain point is equal to the curl of magnetic vector potential at the point. i.e. B = d#A So from the above determined value of magnetic flux density B we have, (1) d # A =- 2m0 ay wb/m2 Since A is parallel to K so the vector potential K will depend only on z . Hence, we have A = A (z) ax From equation (1) we have, a x ay a z - 2m0 ay = 22x 22y 22z A (z) 0 0 2A (z) - 2m0 ay =ay 2z or A (z) = 2m0 z So, A = 2m0 zax Therefore the vector magnetic potential at z =- 2 is A =- 4m0 ax wb/m
ww
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SOL 4.1.26
he
lp.
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m
For View Only Shop Online at www.nodia.co.in Since the infinite current sheet is located in the plane z = 0 so, the z -component of the magnetic flux density will be cancelled due to symmetry and in the closed Amperian loop the integral will be only along y -axis. Thus we have B (2l) = m0 Ienc 2Bl = m0 Kl (Ienc = Kl) As determined by right hand rule, the magnetic flux density above the plane z = 0 will be in - ay direction. So we have the flux density above the current sheet as m 4 ( K = 4 A/m ) B =- 0 # ay =- 2m0 ay wb/m2 2 Alternate Method : The magnetic flux density produced at any point P due to an infinite sheet carrying uniform current density K is defined as B = 1 m0 (K # a n) 2 where an is the unit vector normal to the sheet directed toward the point P . So, magnetic flux density at any point above the current sheet K = 4ax is (an = az ) B = 1 m0 ^4ax h # ^az h =- 4m0 ay wb/m2 2
Option (D) is correct. Current density at any point in a magnetic field is defined as the curl of magnetic field intensity at the point. i.e. J = d#H Since the magnetic field intensity in the free space is given as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 4.1.27
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H = 2r af Therefore the current density is 2(r) (2r2) az = 1 2 (2r3) az J =1 r 2r r 2r 2 = 6raz = 2az A/m
(r = 2 m)
Option (A) is correct. The magnetic flux density at any point is equal to the curl of the vector magnetic potential at the point i.e., B = d#A = 1 2 (rAf) az = 1 2 (2r) az = 2 az r r 2r r 2r The current density J in terms of magnetic flux density B is defined as J = 1 (d # B) = 1 ;- 2 b 2 lE af = 2 2 af m0 m0 2r r m0 r This current density would produce the required vector potential.
SOL 4.1.29
Option (C) is correct. The current density for a given magnetic field intensity H is defined as J = d#H Given H = ^z cos ay h ay + ^z + ey h ax So
d#H =
he
lp.
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m
SOL 4.1.28
ax
ay
az
2 2x
2 2y
2 2z
ww w. ga te
y ^z + e h z cos ay 0 = =- 2 ^z cos ay h ax - b- 2 ^z + ey hl ay + c 2 z cos ay - 2 ^z + ey hmG az 2z 2z 2x 2y y =- cos ayax + ay - e az or, J = d # H =- cos ayax + ay - ey az Therefore the current density in the x -z plane is (y = 0 in x -z plane) J =- ax + ay - az A/m2
SOL 4.1.30
Option (D) is correct. In a source free region current density, J = 0 The current density at any point is equal to the curl of magnetic field intensity H . i.e. J = d#H or (J = 0 ) d#H = 0 and since the curl of a given vector field is zero so it can be expressed as the gradient of a scalar field i.e. H = df So A and R both are true and R is correct explanation of A.
SOL 4.1.31
Option (B) is correct. Given that the cylindrical wire located along z -axis produces a magnetic field intensity, H = 3raf .
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Option (D) is correct. As the beam is travelling in az direction so the field intensity produced by it will be in af direction and using Ampere’s circuital law at it’s surface we have Hf ^2pa h = Ienc a r Hf ^2pa h = 2 a1 - k 2prdr a 0 2 r r3 a Hf ^2pa h = 4p ; - E 2 3a 0
co
SOL 4.1.32
m
For View Only Shop Online at www.nodia.co.in So, applying the differential form of Ampere’s circuital low we have the current density with in the conductor as J = d#H a r ra f a z a r ra f a z 1 1 2 2 2 = 2r 2f 2z = 22r 22f 22z r r A r rA f A z 0 3r2 0 = 1 2 ^3r2h az = 6az A/m2 r 2r
lp.
#
SOL 4.1.34
he
at e
ww
w. g
SOL 4.1.33
2 Hf ^2pa h = 2pa 3 or H = a af 4 Option (A) is correct. Since the magnetic flux density is defined as B = d#A and d # B = m0 J Now using the vector identity, we have d # ^d # Ah = d ^d : Ah - d2A or, d # B = d ^d : Ah - d2A or, m0 J = d ^d : Ah - d2A As the vector potential is always divergence free so we get, d2A =- m0 J
Option (B) is correct. Magnetic dipole moment of a conducting loop carrying current I is defined as : m = IS where S is the area enclosed by the conducting loop. So we have m = 7 # (p # 0.52) = 5.5 (I = 7 A, R = 0.5 m) The direction of the moment is determined by right hand rule as when the curl of fingers lies along the direction of current,then the thumb indicates the direction of moment. So, m = 5.5ax A- m2 ***********
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SOLUTIONS 4.2
Option (D) is correct. Since the current is flowing in the - az direction So, Idl = 10dz (- az )
ww w. ga te
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lp.
co
m
SOL 4.2.1
Magnetic field intensity at any point P due to a filamentary current I is defined as H = I 6cos a2 - cos a1@ af 4pr where r " distance of point P from the current filament. a1 " angle subtended by the lower end of the element at P . a2 " angle subtended by the upper end of the element at P . Now from the figure we have, r =2 a2 = p - q or cos a2 = cos (p - q) 3 =- cos q ==- 3 13 22 + 32 and (angle subtended by end z = 3) a1 = 0 or cos a1 = cos 0 = 1 So, H = I 6cos a2 - cos a1@ af 4pr (10) 3 a I = 1 - c- 3 mG af = # c2 + f 8 p 4p # 2 = 13 m 13 Now the direction of magnetic field intensity is defined as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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af = al # a r where al is unit vector along the line current and a r is the unit vector normal to the line current directed toward the point P . So we have af = (- az ) # (ay)= ax Therefore,
Option (D) is correct. According to Biot-savart law, magnetic field intensity at any point P due to the current element Idl is defined as H = Idl # 3R 4pR where R is the vector distance of point P from the current element.
m
SOL 4.2.2
H = 10 c1 + 3 m (ax ) 8p 13 = 1.73ax A/m
w. g
at e
he
lp.
co
#
ww
Now the current element carries a current of 4.4 A in + ax direction. So we have, R = (4ax + 2ay + 3az ) - (xax ) (Since on x -axis y - and z -component will be zero) R = (4 - x) ax + 2ay + 3az or R = (4 - x) 2 + 22 + 32 = x2 - 8x + 29 and (filament lies from x =- 3 to x = 3) Idl = 4.4dxax Therefore the magnetic field intensity is + 3 (4.4a x ) # 6(4 - x) a x + 2ay + 3a z @ dx H = 4p (x2 - 8x + 29) 3/2 -3 +3 dx = 4.4 (2az - 3ay) 2 3/2 4p - 3 (x - 8x + 29)
#
#
+3 (2x - 8) = 4.4 (2az - 3ay) = G 4p 26 (x2 - 8x + 29) 1/2 -3
= 4.4 (2az - 3ay) = 0.1077az - 0.162ay = 0.1az - 0.2ay A/m 26p GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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lp.
co
m
For View Only Alternate Method :
239
ww w. ga te
he
According to Ampere’s circuital law, the line integral of magnetic field intensity H around a closed path is equal to the net current enclosed by the path. Since we have to determine the magnetic field intensity at point (4, 2, 3) so we construct a circular loop around the infinite current element that passes though the point (4, 2, 3) as shown in the figure. Now from Ampere’s circuital law we have,
# H : dl
= Ienc
(2pr) H = 4.4
(Ienc = 4.4 A)
4.4 or H = r = 13 from figure. 2p # 13 Now direction of the magnetic field intensity is defined as af = al # a r where al " unit vector in the direction of flow of current a r " unit vector normal to the line current directed toward the point. 6(4ax + 2ay + 3az ) - (4ax )@ So we have, af = ax # (4 - 4) 2 + 22 + 32 (2ay + 3az ) 2a - 3ay = z = ax # 13 13 Therefore the magnetic field intensity at the point (4, 2, 3) is (2az - 3ay) H = 4. 4 = 4.4 (2az - 3ay) 26p 2p 13 13 = 1.5az - 2.5ay A/m SOL 4.2.3
Option (D) is correct. As the magnetic field intensity at the center of the triangle produced by all the
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he
lp.
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For View Only Shop Online at www.nodia.co.in three sides will be exactly equal so we consider only one side lying along x -axis that carries 4 A current flowing in + ax direction as shown in the figure. Now the magnetic field intensity at any point P due to a filamentary current I is defined as H = I 6cos a2 - cos a1@ af 4pr where r " distance of point P from the current filament. a1 " angle subtended by the lower end of the element at P . a2 " angle subtended by the upper end of the element at P .
ww
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From the figure we have r tan 30c = & r = 1 1 3 5 a1 = p - p/6 = p & cos a1 = cos 5p =- 3 6 6 2 and a2 = 30c & cos a2 = cos 30c= 3 2 So the magnetic field intensity produced by one side of the triangle at centre of the triangle is 4 H1 = 6cos a2 - cos a1@af 4p # 1 3 3 3 + 3 a = 5a p ; 2 2 E f p f Now the direction of magnetic field intensity is determined as af = al # a r where al is unit vector along the line current and a r is the unit vector normal to the line current directed toward the point P . and since the line current is along x -axis so we have (al = ax , a r = ay ) a f = a x # ay = a z Therefore the net magnetic field intensity due to all the three sides of triangle is H = 3H1 = 3 # b 3 l az = 9 az A/m (af = az ) p p GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia =
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For View Only Shop Online at www.nodia.co.in SOL 4.2.4 Option (C) is correct. According to Biot-savart law, magnetic field intensity at any point P due to the current sheet element KdS is defined as H =
a # KdS4p# R 2
s
R
where R is the vector distance of point P from the current element. Now we consider a point ^0, y, z h on the current carrying sheet, from which we have the vector distance of point (3, 0, 0) R = ^3ax + 0ay + 0az h - ^0ax + yay + zaz h = ^3ax - yay - zaz h aR =
3ax - yay - zaz 3a - yay - zaz = x 2 2 2 3 +y +z 9 + y2 + z2
m
or
# #
( K = 4a y )
lp.
# #
co
Therefore the magnetic field intensity due to the current sheet is 2 3 4ay # (3ax - yay - zay) H = dydz 4p (9 + y2 + z2) 3/2 z =- 2 y =- 3 2 3 4 (- zax - 3az ) = 2 2 3/2 dydz z =- 2 y =- 3 4p (9 + y + z )
ww w. ga te
# #
he
We note that the x component is anti symmetric in z about the origin (odd parity). Since the limits are symmetric, the integral of the x component over z is zero. So we are left with 2 3 - 12az H = 2 2 3/2 dydz -2 - 3 4p (9 + y + z ) 2 +3 y =- 3 az dz ; E 2 2 2 p -2 (z + 9) 9 + y + z -3 2 2 dz =- 6 a 1 tan-1 z 2 =- 3 az a 3 kD 2 p z :3 p -2 z + 9 -2 2 =- # (2) # (0.59) az =- 2az A/m p
# #
SOL 4.2.5
Option (D) is correct.
Since the uniformly charged disk is rotating with an angular velocity w = 2 rad/s GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in about the z -axis so we have the current density K = rs # (angular velocity) = rs (wr) = 2 # 2 # r or K = 4raf According to Biot-savart law, magnetic field intensity at any point P due to the current sheet element KdS is defined as H = KdS #2 aR 4pR s where R is the vector distance of point P from the current element. Now from the figure we have R = az - ra r or R = 1 + r2 a - ra r and aR = z 1 + r2 So the magnetic field intensity due to a small current element KdS at point P is (4raf) # (az - ra r) 4r (a r + raz ) dH = KdS #2 aR = = 2 3/2 4pR 4p (r + 1) 4p (r2 + 1) 3/2 On integrating the above over f around the complete circle, the a r components get cancelled by symmetry, leaving us with 2p 3 4r2 az H (z) = ^rdrdfh 4p (r2 + 1) 3/2 0 0 3 3 r3 1 2 =2 az 2 3/2 dra z = 2 = r + 1 + G 2 (r + 1) 0 r +1 0 3 + 2 (1 - 1 + 3 ) = 2> H az = 5az A/m 1+3
# #
Option (B) is correct.
ww
w. g
SOL 4.2.6
at e
#
he
lp.
co
m
#
As all the four sides of current carrying square loop produces the same magnetic field at the center so we consider only the line current AB for which we determine the magnetic field intensity at the center. Now the magnetic field intensity at any point P due to a filamentary current I is defined as H = I 6cos a2 - cos a1@ af 4pr where r" distance of point P from the current filament. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Shop Online at www.nodia.co.in a1 " angle subtended by the lower end of the filament at P . a2 " angle subtended by the upper end of the filament at P . From the figure, we have r = 1 m , a2 = 45c and a1 = 180c - 45c 2 So the magnetic field intensity at the centre O due to the line current AB is H1 = I 6cos a2 - cos a1@ 2pR 1 = 6cos 45c - cos (180c - 45c)@ 2p # (1/2) = 1 # 2 = 2 A/m p p 2 and the magnetic flux density produced by the line current AB is B1 = m0 H1 = 4p # 10-7 # 2 p -7 2 = 5.66 # 10 wb/m
lp.
co
m
For View Only
Option (D) is correct. According to Biot-savart law, magnetic field intensity at any point P due to the current element Idl is defined as R H = Idl # a 4pR2 where R is the vector distance of point P from the current element. As the cross product of two parallel lines is always zero so the straight segments will produce no field at P . Therefore the net magnetic field produced at point P will be only due to the two circular section. i.e. H = HCD + HAB
ww w. ga te
SOL 4.2.7
he
Therefore the net magnetic flux density due to the complete square loop will be four times of B1 i.e. B = 4B1 = 4 # (5.66 # 10-7) = 3.26 # 10-6 wb/m2
#
or
(Irdfaf) # (- a r) += G 4pr2 0 at r = 1 m p/2 Iaz df - p/2 Iaz df = 4p (1) 4p (2) 0 0 3 . 2 1 p 1 - D # a k az = 0.2 A/m = 2 2 4p # :
H ==
#
#
p/2
#
0
p/2
Irdf (- af) # (- a r) G 4pr2 at r = 2 m
#
Alternate Method : The magnetic field intensity produced at the center of a circular loop of radius R carrying current I is defined as H = I 2R GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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m
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he
lp.
co
and since the straight line will not produce any field at point P so due to the two quarter circles having current in opposite direction, magnetic field at the center will be H = 1: I - I D 4 2a 2b where a " inner radius b " outer radius 3 (2) 3 (2) H = 1; = 0.2 A/m 4 2 # 1 2 # 2E Option (D) is correct. The magnetic field intensity at any point P due to an infinite filamentary current I is defined as H = I 2pr where r is the distance of point P from the infinite current filament. Now the two semi infinite lines will be in combination treated as a single infinite line for which magnetic field intensity at point P will be (R is the length of point P from line current) H1 = I 2pR 4 =1 (I = 4 A, R = 2 m) = p 2p # 2 As the magnetic field intensity produced at the center of a circular loop of radius R carrying current I is defined as H = I 2R So magnetic field produced at point P due to the semi circular segment is H2 = 1 # I = 1 2 2 2R Therefore net magnetic field intensity produced at point P is H = H1 + H 2 = 1 + 1 p 2 = 0.82 A/m
SOL 4.2.9
Option (C) is correct.
ww
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at e
SOL 4.2.8
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245
SOL 4.2.10
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Magnetic field intensity produced at any point P on the axis of the circular loop carrying current I is defined as Ir2 H = 2 2 (r + h2) 3/2 where h is the distance of point P from the centre of circular loop and r is the radius of the circular loop. From the figure we have r = 3 m and h = 5 - 1 = 4 m and using right hand rule we conclude that the magnetic field intensity is directed along + az . So the magnetic field intensity produced at point P is 50 # 10-3 (3) 2 9 # 50 # 10-3 a = 2.8a mA/m = H = a z z z 2 # 125 2 (32 + 42) 3/2 Option (D) is correct.
Let the cylindrical tube is of radius a for which we have to determine the magnetic field intensity at the axis of solenoid. Now we consider a small ring (small section of solenoid) of the width dz at a distance z from point P lying on the axis of the solenoid as shown in the figure.
The total current flowing in the loop of the ring will be where n is the no of turns per unit length dI = nIdz GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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lp.
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For View Only Shop Online at www.nodia.co.in Since magnetic field intensity produced at any point P on the axis of the circular loop carrying current I is defined as Ir2 H = 2 (r2 + h2) 3/2 where h is the distance of point P from the centre of circular loop and r is the radius of the circular loop. So we have the magnetic field intensity due to the ring as (nIdz) a2 ( r = a, h = z ) dH = 2 (a2 + z2) 3/2 From the figure we have z = a cot q & dz =- a2 dq sin q 1 a sin3 q and & sin q = a = = 2 2 3/2 r a3 a2 + z2 ^a + z h The total magnetic field intensity produced at point P due to the solenoid is 3 (nIdz) a2 nI p a2 sin3 q (- adq) = H = 2 2 3/2 2 q = 0 a3 sin2 q z =- 3 2 (a + z ) 0 =- nI sin qdq = nI (cos 0 - cos p) 2 q=p 2 (n = 1000, I = 4 mA ) = nI = 1000 # 6 # 10-3 = 6 A/m
#
#
he
#
Option (D) is correct. As calculated in the previous question the magnetic field intensity inside a long solenoid carrying current I is defined as where n is no. of turns per unit length H = nI and since using right hand rule we conclude that the direction of magnetic field intensity will be right wards ( + ay ) due to outer solenoid and left wards ((- ay)) due to inner solenoid. So the resultant magnetic field intensity produced inside the inner solenoid will be H = H1 + H2 = n1 I (- ay) + n2 Iay where n1 and n2 are the no. of turns per unit length of the inner and outer solenoids respectively. So H =- (3 # 10-3) (2000) ay + (3 # 10-3) (1000) ay = 3 # 10-3 (- 1000) ay =- 3ay A/m
SOL 4.2.12
Option (D) is correct. Since no any magnetic field is produced at any point out side a solenoid so in the region between the two solenoids field will be produced only due to the outer solenoid. i.e. H = n2 Iay = 100 # 4 # 10-4 ay = 4ay A/m
SOL 4.2.13
Option (B) is correct. Since no any magnetic field is produced at any point out side a solenoid so, at any point outside the outer solenoid, the net magnetic field intensity produced due to
ww
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SOL 4.2.11
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Option (B) is correct. The magnetic field intensity produced at any point P due to an infinite sheet carrying uniform current density K is defined as H = 1 (K # a n) 2 where an is the unit vector normal to the sheet directed toward the point P . So the field intensity produced between the two sheets due to the sheet K1 = 3az located at x = 2 m is (an =- ax ) H1 = 1 (3az ) # (- ax ) =- 3 ay A/m 2 2 and the field intensity produced between the two sheets due to the sheet K2 =- 3az located at x =- 2 m is H2 = 1 (- 3az ) # (ax ) =- 3 ay A/m (an = ax ) 2 2 Therefore the net magnetic field intensity produced at any point between the two sheets is H = H1 + H2 =- 3ay Since the magnetic field intensity at any point is the equal to the negative gradient of scalar potential at the point i.e. H =- dVm
he
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SOL 4.2.14
247
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So for the field H =- 3ay in the region between the two current carrying sheets, we have - 3ay =- dVm ay (the field has a single component in ay direction) dy where C1 is constant or Vm = 3y + C1 Putting Vm = 0 for point P (1, 2, 5) (given), we have 0 = 3 # (2) + C1 or C1 =- 6 Thus, Vm = (3y - 6) A and the graph of Vm versus y will be as plotted below
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Option (A) is correct. Since the current density inside the wire is given by J\r So we have, J = kr and the total current flowing in the wire is given by
at e
he
SOL 4.2.16
lp.
co
m
For View Only Shop Online at www.nodia.co.in SOL 4.2.15 Option (C) is correct. The magnetic flux density at any point is equal to the curl of the vector magnetic potential at the point i.e., (1) B = d#A 2 Since (calculated in previous question) B = m0 H =- 3m0 ay Wb/m As the magnetic flux density is in ay direction so A is expected to be z -directed. Therefore from eq (1) we have -2Az =- 3m0 2x or Az = 3m0 x + C2 Putting Az = 0 at point P (1, 2, 5) (given), we have 0 = 3m0 + C2 or C2 =- 3m0 So, at origin (0, 0, 0) Az = 3m0 (x - 1) =- 3m0 Thus, the magnetic vector potential at origin is A =- 8m0 az Wb/m
# J : dS = # kr (2pr) dr
I0 = or
5 # 10-3
where k is a constant.
s
2 # 10-2
0
( I 0 = 5 mA )
2pk (2 # 10-2) 3 3 -3 So we have k = 3 # 5 # 10 -6 = 15 # 103 16p 2p # 8 # 10 Now for the Amperian loop at r = 1 cm enclosed current is
w. g
5 # 10-3 =
-2
ww
r3 1 # 10 Ienc = J : dS = kr (2pr) dr = b 15 # 103 l # 2p ; E 16p 3 0 s r=0 15 1 15 -3 -3 = 10 = 10 8 #3# 24 # So from Ampere’s circuital law we have
# B : dl L
#
#
1 # 10-2
= m0 Ienc
m0 15 10-3 24 # Therefore the magnetic flux density at r = 1 cm is 10-3 -7 B = 15 # # 4p # 10 24 2p (1 # 10-2) = 1.25 # 10-8 = 12.5 nWb/m2 B (2pr) =
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#
# #
co
# #
#
lp.
#
Option (C) is correct.
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SOL 4.2.18
m
For View Only Shop Online at www.nodia.co.in SOL 4.2.17 Option (D) is correct. Current density at any point in a magnetic field is defined as the curl of magnetic field intensity at the point. i.e. J = d#H So the current density component in ax direction is 2 2Hy Jx = (d # H) x = 2Hz =-e xz 2 + 6x2 z o ax 2y 2z (y + 1) Therefore the total current passing through the surface x = 2 m , 1 # y # 4 m , 3 # z # 4 m is 4 4 xz2 + 6x2 z dydz (dS = dydza ) I = Jx : dS =x e o 2 S z = 3 y = 1 (y + 1) 4 4 2z2 + 24z dydz (x = 2 m ) =e o (y + 1) 2 3 1 4 2 4 4 =- ;- 2z + 24zyE dz =- b 3 z2 + 72z l dz 5 y+1 3 3 1 =- 145 A
Magnetic dipole moment of a conducting loop carrying current I is defined as : m = IS where S is the area enclosed by the conducting loop. So for a ring of radius r , magnetic dipole moment m = I (pr2). Now as the charged disk(charge density, rS = 20 C/m2 ) is rotating with angular velocity w = 0.1 rad/s so, the current in the loop is given as dI = rS wrdr Therefore the magnetic dipole moment is m =
# dI (pr ) = # 2
1
(rS wrdr) (pr2) = rS wp
r=0 4 1
#
0
1
r3 dr
= 20 # 0.1 # p :r D = p A- m2 3 4 0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in SOL 4.2.19 Option (B) is correct. Magnetic dipole moment of a spherical shell of radius r having surface charge density rS is given as where w is angular velocity. m = 4p rS wr 4 3 Since the total charge of 5 C is distributed over the volume of the sphere so, the magnetic dipole moment of the sphere is given as (rS = rv dr ) m (r) = 4p (rv dr) wr 4 3 where rv is uniformly distributed volume charge density of the sphere. Therefore, we have 5 Q m (r) = 4p rv w r = 1 Qwr2 e rv = 4 pr3 o 3 5 5 3 (Q = 5 C , w = 4 rad/ s ) = 1 # (5) # (4) # r2 = 4r2 A- m2 5
Option (C) is correct. The average magnetic field intensity over a sphere of radius r , due to steady currents within the sphere is defined as 2 (m = 4r2 ) Have = 1 2m = 1 2 #34r = 2 3 pr 4p r 4p r As the sphere is spinning about the z -axis so, the produced magnetic field will be in az direction as determined by right hand rule. Thus, we have Have = 3 az pr Option (A) is correct. Magnetic dipole moment of a conducting loop carrying current I is defined as : m = ISan where S is the area enclosed by the conducting loop and an is normal vector to the surface. So we have ( I = 7 A, S = 0.1 m2 ) m = (7) (0.1) an Now the given plane is x + 3y - 1.5z = 3.5 For which we have the function f = x + 2y - 1.5z
SOL 4.2.21
ww
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SOL 4.2.20
at e
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m
#
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For View Only Shop Online at www.nodia.co.in and the normal unit vector to the plane is, ax + 3ay - 1.5az 4f = an = 4f 12 + 32 + (- 1.5) 2 So the magnetic dipole moment of the coil is (a + 3ay - 1.5az ) m = (0.7) x 3. 5 = 1.2ax + 0.6ay - 0.3az A- m2 Option (B) is correct. The magnetic field intensity, (H ) in the terms of magnetic vector potential, (A) is defined as H = 1 (d # A) = 1 6d # (6y - 2z) ax + 4xzay@ m0 m0 = 1 6- 3ax - 2ay + 4az@ m0 Since the electric current density at any point is equal to the curl of magnetic field intensity at that point. i.e. J = d#H So, we have the electric current density in the free space as J = d # 1 6- 8ax - 2ay + 6az@ = 0 m0
SOL 4.2.23
Option (D) is correct. Magnetic flux density across the toroid at a distance r from it’s center is defined as m NI B = 0 a 2p r f where N " Total no. of turns I " Current flowing in the toroid So, the total magnetic flux across the toroid is given by the surface integral of the flux density i.e.
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lp.
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m
SOL 4.2.22
f =
# B : dS S
where dS is differential surface area vector. Consider a width dr of toroid at a distance r from it’s center as shown in figure
So we have the total magnetic flux across the toroid as 2 m0 NI (dS = hdraf ) f = b 2pr af l (hdraf) r=1m GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
#
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Shop Online at www.nodia.co.in (N = 105 , I = 10 A ) = 4p # 10 # 10 # 10 # 10 ln b 2 l 2p 1 = 1.39 Wb -7
5
Option (B) is correct. As determined in previous question the magnetic flux density across the toroid at a distance r from it’s center is m NI B = 0 a 2pr f So at the mean radius, r = a + b = 1.5 m 2 m0 NI we have, (r = 1.5 m ) B = a 3p f Therefore the total magnetic flux is 2 m0 NI (dS = hdraf ) f' = B : dS = b 3p af l (hdraf) r=1 -7 5 (N = 105 , I = 10 A ) = 4p # 10 # 10 # 10 # 10 6r @12 3p = 1.33 Wb Thus, the percentage of error is f' - f ( f = 1.39 wb as calculated above) 100% % error = f # = 1.33 - 1.39 # 100% =- 12.31% 1.39
#
Option (C) is correct.
ww
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SOL 4.2.25
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#
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SOL 4.2.24
Chap 4
Consider the point P on z -axis is (0, 0, h) and current flowing in the current element is I in ax direction. Since the magnetic field intensity at any point P due to a current element I is defined as H = I 6cos a2 - cos a1@ af 4pr GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Shop Online at www.nodia.co.in r " distance of point P from the current element. a1 " angle subtended by the lower end of the element at P . a2 " angle subtended by the upper end of the element at P . So for the given current element along positive x -axis we have a1 = 90c a2 = 0c Therefore, (r = h ) H = I af 4ph Now the direction of magnetic field intensity is defined as af = al # a r where al is unit vector along the line current and a r is the unit vector normal to the line current directed toward the point P . So, af = ax # az =- ay Therefore magnetizing force is H = I (- ay) 4ph ...(1) or H = I 4ph Now consider the current flowing in the current element introduced along the positive y -axis is I in ay direction. So, the magnetic field intensity produced at point P due to the current element along the positive y -axis is H = I 6cos a2 - cos a1@ af 4pr ( r = h, a1 = 90c, a2 = 0c) = I 6cos 0c - cos 90c@ af 4p h (af = ay # az = ax ) = I ax 4p h Therefore the resultant magnetic field intensity produced at point P due to both the current elements will be Hnet = I (- ay + ax ) 4ph or, Hnet = I 2 4ph Thus, from equation (1) we have Hnet = 3 H
SOL 4.2.26
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m
For View Only where
Option (B) is correct. The magnetic flux density produced at a distance r from a straight wire carrying current I is defined as mI B = 0 2pr Now consider a strip of width dr of the square loop at a distance r from the straight wire as shown in the figure.
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#
SOL 4.2.27
he
lp.
#
(area of strip = 2dr )
co
m
Total magnetic flux crossing the strip is dym = B ^2drh mI = 0 (2dr) 2pr So, the flux crossing the complete square loop is 6 m I 0 ym = dym = ^2drh r = 4 2pr -7 mI = 0 6ln r@64 = 4p # 10 # 5 b ln 6 l p p 4 -7 = 8.11 # 10 Weber = 2.42 mWb
Option (B) is correct. As calculated in previous question the total flux crossing through the square loop due to the straight conducting element is b m I 0 ym = ^Ldrh r = a 2pr where I is the current carried by the conductor, L is the side of the square loop and a, b are the distance of the two sides of square loop from the conductor. So we have L = 0.5 m a = 0.3 - 0.5 = 0.05 m 2 0 b = 0.3 + .5 = 0.55 m 2 0.55 m I mI m0 I 0 Thus, ym = ^0.5drh = 40p 6ln
[email protected] 05 = 4p ^ln 11h r = 0.05 2pr Therefore the current that produces the net flux ym = 5 # 10-5 Tm2 is 4p -5 I = # 5 # 10 m0 ^ln ^1 hh = 238.5 A
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#
SOL 4.2.28
#
Option (D) is correct. Consider the flux density at the given point due to semi infinite wire along y -axis is B1 and the flux density due to wire along z -axis is B2 . The magnetic flux density B produced at any point P due to a straight wire carrying current I is defined as mI B = 0 6cos a2 - cos a1@ af 4pr
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m
For View Only
255
r " distance of point P from the straight wire. a1 " angle subtended by the lower end of the wire at P . a2 " angle subtended by the upper end of the wire at P . and the direction of the magnetic flux density is given as af = al # a r where al is unit vector along the line current and a r is the unit vector normal to the line current directed toward the point P . So, we have r = 2m (al = ay, a r = ax ) af = ay # ^ax h =- az p (as y tends to 3) a1 = , a2 = 0 2 Therefore the magnetic flux density produced at point P due to the semi infinite wire along y -axis is m0 ^4h m B1 = cos 0 - cos p k^- az h =- 0 az 2 2p 4p ^2 h a Similarly we have the magnetic flux density produced at point P due to semi infinite wire along z -axis as m B2 =- 0 ay 2p Thus, the net magnetic flux density produced at point P due to the L-shaped filamentary wire is m m B =- 0 ay - 0 az 2p 2p =- 3 ^ay + 3az h # 10-4 Wb/m2 SOL 4.2.29
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lp.
co
where
Option (D) is correct. Using ampere’s circuital law we have
# B : dl
= m0 Ienc As the conductor carries current I which is uniformly distribute over the conductor cross section so, the current density inside the conductor is J = I 2 pR We construct an Amperian loop of radius r(< R) inside the cylindrical wire for GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Magnestostatic Fields
Chap 4
m0 I 2pr B\ 1 r
Bf =
Option (C) is correct. We consider only the half side of the loop to determine the flux density at the center as shown in the figure.
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SOL 4.2.31
at e
So
lp.
Option (D) is correct. Similarly as calculated above we construct an Amperian loop of radius r (> R) outside the cylinder for which the entire current flowing in the wire will be enclosed. i.e. Ienc = I and from Ampere’s circuital law we get, Bf ^2prh = m0 I
he
SOL 4.2.30
co
m
For View Only Shop Online at www.nodia.co.in which the enclosed current is r2 Ienc = b I 2 l pr2 = I c 2 m R pR and since the current is flowing along z -axis so using right hand rule we get the direction of magnetic flux density along + af . Thus, from Amperes circuital law, we have (Bf) (2pr) = Ienc m Ir2 or Bf = 0 c 2 m 2pr R m Ir or B = 0 2 af 6pR
The magnetic flux density B produced at any point P due to a straight wire carrying current I is defined as mI B = 0 6cos a2 - cos a1@ af 4pr where r " distance of point P from the straight wire. a1 " angle subtended by the lower end of the wire at P . a2 " angle subtended by the upper end of the wire at P . GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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257
he
Option (D) is correct. Using right hand rule we conclude that the field intensity produced at centre of the loop by the loop wire and the straight wire are opposing each other, so, the field intensity at the centre of the loop will be zero if ...(1) Hwire = Hloop where Hwire is the field intensity produced at the center of loop due to the straight wire and Hloop is the field intensity produced at the center of loop due to the current in the circular loop. Since the magnetic field intensity produced at a distance r from an infinitely long straight wire carrying current I is defined as H = I 2pr So we have Hwire = I = 20 = 10 ^I = 20 A, r = 1 mh p 2p ^ 1 h 2p and as calculated in Q.59 the field intensity produced by circular loop at its center is where a is the radius of the loop Hloop = I 2a I 10I (a = 10 cm ) or, Hloop = -2 = 2 = 5I 2 ^10 # 10 h So putting the values in eq. (1) we get 10 = 5I p Thus, I =6A p
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SOL 4.2.32
lp.
co
m
For View Only Shop Online at www.nodia.co.in and the direction of the magnetic flux density is given as af = al # a r where al is unit vector along the line current and a r is the unit vector normal to the line current directed toward the point P . Therefore, the magnetic flux density produced at centre O due to the half side of the square loop is mI B1 = 0 ^cos a2 - cos a1h af 4pr 1 where r = 1 m a1 = p and cos a2 = = 1 2 2 2 2 1 +1 -7 ^4p # 10 h^1 h 1 Thus, (af = ay # ^- ax h = az ) B1 = c 2 - 0 m az 4p ^1 h -4 = 10 az Wb/m2 2 As all the half sides of the loop will produce the same magnetic flux density at the centre so, the net magnetic flux density produced at the centre due to whole square loop will be B = 8B1 = 4 2 # 10-7 az Wb/m2
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m
For View Only Shop Online at www.nodia.co.in SOL 4.2.33 Option (D) is correct. Consider one half side of the square loop to determine the magnetic field intensity at the centre O as shown in the figure.
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SOL 4.2.34
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co
The magnetic field intensity H produced at any point P due to a straight wire carrying current I is defined as H = I 6cos a2 - cos a1@ 4pr where r " distance of point P from the straight wire. a1 " angle subtended by the lower end of the wire at P . a2 " angle subtended by the upper end of the wire at P . So we have r = a/2 a1 = p/2 and a2 = p/4 Therefore the magnetic field intensity produced at centre O due to the half side of the square loop is I H1 = cos p - cos p k = I 2 4 2p (a/2) a 2 pa As all the eight half sides produces same field intensity at the centre of the loop so, net field intensity produced at the center due to the complete square loop is Hnet = 8 c I m = 2 2 I pa 2 pa Option (A) is correct. For the shown current loop we divide the loop in two segments as shown in figure
Now the field intensity due to segment (1) (Semicircular loop) at point P can be given directly as calculated in Que.60 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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where a is radius of semicircular loop H1 = I 4a (a = 1 m ) or H1 = 8 = 2 A/m 4 ^1 h again for determining the field intensity due to segment (2) we consider it as the half portion of a complete square loop of side 2 m and since the field intensity due to a completer square loop of side a carrying current I can be directly given from previous question. i.e. H = 2 2I pa so the field intensity due to the half portion of square loop is H2 = 1 H = 2 I 2 pa 2 ^8 h 4 2 or (I = 8 A , a = 2 m ) H2 = = p p ^2 h As determined by right hand rule the direction of field intensity produced at point P due to the two segments will be same (inward) therefore, the net magnetic field intensity produced at point P will be Hnet = H1 + H2 = 2 + 4 2 = 3.8 A/m inward. p Option (B) is correct. The flux density due to infinite current carrying sheet is defined as m B = 0 K # an 2 where K is surface current density and an is unit vector normal to the surface directed toward the point where flux density is to be determined So, for the sheet in z = 0 plane, m B1 = 0 ^4ax h # ^az h =- 2m0 ay ^an = az h 2
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SOL 4.2.35
he
lp.
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m
i.e.
and for the sheet in z = 2 m plane m B2 = 0 ^- 4ax h # ^- az h =- 2m0 ay ^an =- az h 2 Therefore, the net flux density between the sheets is B = B1 + B2 =- 4m0 ay Thus the magnetic flux per unit length along the direction of current is ym /l = B # (Dis tan ce between the plates) =- 8m0 ay Wb/m ***********
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SOLUTIONS 4.3
For r < a ,
Io = (pa2) J
# H : dl
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at e
= Ienclosed 2 H ^2pr h = Jr2 a for r < a & H = Jr 2 H \ r, 2pa Option (C) is correct. Assume that the cross section of the wire lies in the x -y plane as shown in figure below : So,
SOL 4.3.2
H ^2pr h = (pa2) J H = Io 2pr H \ 1 , for r > a r J (pr 2) Jr 2 Ienclosed = = 2 a pa 2
lp.
i.e.
= Ienclosed
co
# H : dl
m
Option (B) is correct. For r > a , Ienclosed = (pa2) J
he
SOL 4.3.1
Since, the hole is drilled along the length of wire. So, it can be assumed that the GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in drilled portion carriers current density of - J . Now, for the wire without hole, magnetic field intensity at point P will be given as Hf1 (2pR) = J (pR2) & Hf1 = JR 2 Since, point O is at origin and the cross section of the wire located in x -y plane. So, in vector form the field intensity due to the current carrying wire without considering the hole is given as (1) H1 = J (xax + yay) 2 Again, only due to the hole magnetic field intensity at point P will be given as (Hf2) (2pr) =- J (pr 2) Hf2 = - Jr 2 Again, if we take Ol at origin then in vector form (2) H2 = - J (xlax + ylay) 2 where xl and yl denotes point ‘P ’ in the new co-ordinate system. Now the relation between two co-ordinate system will be x = xl + d and y = yl So, putting it into equation (2) we have H2 = - J [(x - d) ax + yay] 2 Therefore, the net magnetic field intensity at point P is Hnet = H1 + H2 = J dax 3 i.e. the magnetic field inside the hole will depend only on d . SOL 4.3.3
Option (A) is correct. Due to 1 A current wire in x -y plane, magnetic field be at origin will be in x direction as determined by right hand rule. Due to 1 A current wire in y -z plane, magnetic field be at origin will be in z direction as determined by right hand rule. Thus, x and z -component is non-zero at origin.
SOL 4.3.4
Option (D) is correct. The total flux, F = 1.2 mWb = 1.2 # 10-3 Wb Cross sectional area, A = 30 cm2 = 30 # 10-4 m2 So, the flux density is given as -3 B = F = 1.2 # 10-4 = 0.4 Tesla A 30 # 10
SOL 4.3.5
Option (D) is correct. The relation between magnetic flux density B and vector potential A is given as B = d#A
SOL 4.3.6
Option (B) is correct.
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For View Only Shop Online at www.nodia.co.in For an isolated body the charge is distributed over its region which depends on the total change and the curvature of the body. Thus Statement 1 is correct Since the magnetic flux lines form loop so the total magnetic flux through any closed surface is zero. Thus Statement 2 is correct. SOL 4.3.7
Option (D) is correct. The magnetic flux density in terms of vector potential is defined as B = d#A
# ^d # AhdS F = # A : dl =
m
# B : dS
Option (B) is correct. Consider the current element along z -axis as shown in the figure.
at e
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lp.
SOL 4.3.8
co
i.e. the line integral of vector potential A around the boundary of a surface S is equal to the flux through the surface S .
SOL 4.3.10
Option (D) is correct. For the given circular cylinder, consider the surface current density is K . So, the total current I through the cylinder is given as K ^2pr h = I where r is radius of circular cylinder. 5 So, K= I = = 50 A/m 2pr p 2p ^5 # 10-2h Option (B) is correct. Magnetic vector potential of an infinitesimally small current element is defined as m0 Idl A= 4p R where R is the distance from current element. Given that R " 3 So A =0
ww
SOL 4.3.9
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Using right hand rule we get the direction of magnetic field directing normal to radial line OP .
#
Option (A) is correct. Consider the two parallel sheets are separated by a distance ‘d ’ as shown in the figure below GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 4.3.11
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m
For View Only
263
Option (B) is correct. For the given current distribution, the current enclosed inside the cylindrical surface of radius r for a < r < b is r r 2p J 0 3 3 Ienc = bJ 0 a2 l^2prdrh = 3r2 ^r - a h a and the magnetic field intensity is given as
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SOL 4.3.12
he
lp.
co
The two sheets carries surface currents K = Ky ay At any point between them the magnetic field intensity is given as H = 1 K # ^anu + anl h 2 where anu is the normal vector to the upper plate and anl is normal vector to the lower plate both directs toward the point between them i.e. anu =- az and anl = az So, H = 1 Ky ay # ^- az + az h = 0 2
#
# H : dl
= Ienc H ^2prh = 2pJ20 ^r3 - a3h 3a J 0 ^r3 - a3h H = 4a2 r
SOL 4.3.13
Option (D) is correct. The radiated E and H field are determined by following steps (1) Determine magnetic field intensity H from the expression B = mH = d # A (2) then determine E from the expression d # H = e2E 2t So, the concept of vector magnetic potential is used to find the expression of radiated E and H field.
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For View Only Shop Online at www.nodia.co.in SOL 4.3.14 Option (B) is correct. Using right hand rule, we conclude that the direction of field intensity is same as determined for the two correct elements 3I and 2I while it is opposite for the current element I . Therefore, from the ampere’s circuital law, we get the circulation of H around the closed contour as
# H : dl
= Ienclosed = 2I + 3I - I = 4I
Option (C) is correct. The unit of magnetic flux density (B ) is Tesla or Wb/m2
SOL 4.3.16
Option (D) is correct. From Ampere’s circuital law, the circulation of magnetic field intensity in a closed path is equal to the current enclosed by the path
co
m
SOL 4.3.15
#
he
lp.
i.e. H : dl = Ienc So, for the current I the circulation at a radial distance R is given as H ^2pRh = I or H = I 2pR Therefore, the magnetic flux density at the radial distance R is mI B = m0 H = 0 5p R Option (D) is correct. Unit of work is Joule. Unit of electric field strength ^E h is volt/meter. Unit of magnetic flux is Weber. Unit of magnetic field strength is Ampere/meter. So, in the match list we get, A " 4 , B " 3 , C " 2 ,D " 1.
SOL 4.3.18
Option (D) is correct. Magnetic field intensity at a distance r from a long straight wire carrying current I is defined as H = I 2pr 1= 1 2p r r = 1 = 1.59 m 2p
SOL 4.3.19
Option (B) is correct. Consider the current flowing in the loop is I and since the magnetic field intensity is maximum at the centre of the loop given as H = I 2r where r is radius of the loop. So, the current that must flow in the loop to produce the magnetic field H = 1 mA/m is
ww
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SOL 4.3.17
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Shop Online at www.nodia.co.in I = 2rH = 2 # 1 # 1 = 2 mA
Option (C) is correct. Magnetic flux density B in terms of vector potential A is defined as B = d#A So, B can be easily obtained from A also we know d : A = 0 but it is not the explanation of Assertion (A). i.e. A and R both are true but R is not the correct explanation of A.
SOL 4.3.21
Option (A) is correct. (1) Magnetic flux density in terms of vector potential is given as
m
SOL 4.3.20
B = d#A (2) Poission’s equation for magnetic vector potential is
co
d2 A =- m0 J (3) Magnetic vector potential for a line current is defined as m0 Idl A= 2p R So, all the statements are correct. Option (B) is correct. Magnetic field intensity due to a long straight wire carrying current I at a distance r from it is defined as H = I 2pr 1 = 10 2p r r = 10 = 1.59 m 2p Option (C) is correct. Magnetic field intensity due to an infinite linear current carrying conductor is defined as
SOL 4.3.23
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SOL 4.3.22
lp.
#
# H : dl
= Ienc
H ^2pr h = I
SOL 4.3.24
&
H = I 2pr
Option (D) is correct. The net outward magnetic flux through a closed surface is always zero as magnetic flux lines has no source or sinks. i.e.
# B : dS
=0
(1)
Now, from Gauss’s law we have
# ^d : B hdv = # B : dS = 0
So, comparing the equation (1) and (2) we get d:B = 0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
(2)
266
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For View Only Shop Online at www.nodia.co.in SOL 4.3.25 Option (D) is correct. Given, the plane y = 0 carries a uniform current density 30 az mA/m and since the point A is located at (1, 20, - 2 ) so, unit vector normal to the current sheet is an = ay Therefore, the magnetic field intensity H = 1 K # an = 1 ^30az h # ^ay h =- 15ax mA/m (K = 30az mA/m ) 2 2 Option (D) is correct. The magnetic flux density at any point is curl of the magnetic vector potential at that point. i.e. B = d#A From the Maxwell’s equation, the divergence of magnetic flux density is zero. i.e. d:B = 0 Again from the Maxwell’s equation, the curl of the magnetic field intensity is equal to the current density. i.e. 4# H = J or, (B = m0 H ) d # B = m0 J The expression given in option (A) is incorrect i.e. B ! d:A
SOL 4.3.27
Option (D) is correct. Superconductors are popularly used for generating very strong magnetic field.
SOL 4.3.28
Option (D) is correct. As the magnetic flux lines have no source or sinks i.e. it forms a loop. So the total outward flux through a closed surface is zero. =0
Option (D) is correct. The magnetic field intensity due a surface current density K is defined as H = 1 K # an 2 Where an is unit normal vector to the current carrying surface directed toward the point of interest. Given that, K = 2ax . and since the surface carrying current is in plane z = 0 . So, for - a < z < 0 an =- az and H1 = 1 ^2ax h # ^- az h = ay 2 For 0 < z < x , an = a z and H2 = 1 ^2ay h # ^az h =- 12 az 4
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SOL 4.3.29
# B : dS
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SOL 4.3.26
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CHAPTER 5 MAGNESTOSTATIC FIELDS IN MATTER
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Magnestostatic Fields in Matter
EXERCISE 5.1
Path of a charged particle A that enters in a uniform magnetic field B (pointing into the page) is shown in the figure.
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MCQ 5.1.1
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(C) uncharged
(B) coulomb meter
(C) Ampere meter square
(D) doesn’t exit
Assertion (A) : Both the electric force and magnetic force are produced when a charged particle moves at a constant velocity. Reason (R) : Electric force is an accelerating force where as magnetic force is a purely deflecting force. (A) Both A and R are true and R is correct explanation of A.
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MCQ 5.1.3
(D) can’t be determined
Unit of a magnetic point charge is (A) Ampere meter
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MCQ 5.1.2
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The deflection in the path of the particle shows that the particle is (A) positive charged (B) negatively charged
(B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. MCQ 5.1.4
An electron beam is passed through a uniform crossed electric and magnetic fields E = 15ay V/m and B = 23az wb/m2 (E and B are mutually perpendicular and both of them perpendicular to the beam). If the beam passes the field without any deflection then the velocity of the beam will be (A) 5 m/s (B) 45 m/s
(C) 30 m/s (D) 18 m/s GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 5.1.5 An electron is moving in the combined fields E = 0.1ax - 0.2ay + 0.3az kV/m and B =- 2ax + 3ay - az Tesla . If the velocity of the electron at t = 0 is V (0) = (200ax - 300ay - 400az ) m/s then the acceleration of the electron at t = 0 will be (charge on electron,e = 1.6 # 10-19 C ; mass of electron,me = 9.1 # 10-31 kg ) (A) 1.75 # 1013 (1.1ax + 1.4ay - 0.5az ) m/s2 (B) 2.1 # 10 4 (ax + ay - az ) m/s2
(C) 18ax - 6ay mN
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A current element of 2 m length placed along z -axis carries a current of I = 3 mA in the + az direction. If a uniform magnetic flux density of B = ax + 7ay wb/m2 is present in the space then what will be the force on the current element in the presence of the magnetic flux density ? (A) 6ax - 18ay mN (B) - 18ax + 6ay mN (D) - 1.8ax + 6ay mN
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MCQ 5.1.6
m
(C) 3.5 # 1013 (6ax + 6ay - az ) m/s2 (D) 3.19 # 10-17 (6ax + 6ay - az ) m/s2
Consider two current loops C1 and C2 carrying current I1 and I2 , separated by a distance R. If the force experienced by the current loop C2 due to the current loop C1 is F , then the force experienced by current loop C1 due to the current loop C2 will be (A) - F (B) F (C) - F b I1 l (D) F b I2 l I2 I1
MCQ 5.1.8
Which of the following statements is correct for a current free interface between two different magnetic media ? (A) Normal component of magnetic field intensity will be continuous.
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MCQ 5.1.7
(B) Tangential component of magnetic flux density will be continuous. (C) Magnetic scalar potential will be same in both the medium. (D) None of these
Statement for Linked Question 9 - 10 : In the free space three uniform current sheets with surface current densities K1 = 4ax , K2 =- 4ax , K 3 =- 2ax are located in the plane z = 0 , z = 1 and z =- 1 respectively. MCQ 5.1.9
Net magnetic field intensity produced between the sheets located at z = 0 and z = 1 will be (A) 2ay A/m (B) 4ay A/m (C) - 2ay A/m
(D) 0
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For View Only Shop Online at www.nodia.co.in MCQ 5.1.10 If a conducting filament located along the line y = 0 , z = 0.2 m carries 7 A current in + ax direction then what will be the force per unit length exerted on it ? (A) - 14az N/m (B) 14m0 ax N/m (C) 14m0 az N/m
A rectangular coil of area 1 m2 carrying a current of 5 A lies in the plane 2x + 6y - 3z = 4 . Such that magnetic moment is directed away from origin. If the coil is surrounded by a uniform magnetic field B = 2ax + 4ay + 5az wb/m2 then the torque on the coil will be (B) 30ax - 20ay - 20az N- m (A) 3ax - 20ay - 20az N- m
m
MCQ 5.1.11
(D) - 14m0 az N/m
A circular current loop of radius 1 m is located in the plane z = 0 and centered at origin. What will be the torque acting on the loop in presence of magnetic field B = 4ax - 4ay - 2az wb/m2 , if a uniform current of 10 A is flowing in the loop ? (A) 20p (2ax - az ) (B) 40p (ax + ay)
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MCQ 5.1.12
(D) 6ax - 4ay - 4az N- m
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(C) 21ax - 14ay - 14az N- m
List I shows the type of magnetic materials and List-II shows their criterions. Match List I with List II and select the correct answer using the codes given below : (Notations have their usual meaning)
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MCQ 5.1.13
(D) 40p (ax - ay)
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(C) 4p (ax + ay)
List-I Ferromagnetic
b.
Diamagnetic
c.
Non-magnetic
d.
Paramagnetic
MCQ 5.1.14
c 1 1 3 4
1.
cm = 0 , mr = 1
2.
cm > 0 , mr L 1
3.
cm < 0 , mr K 1
4.
cm >> 0 , mr >> 1
d 4 2 2 2
Which of the following is a diamagnetic material ? (A) copper (B) sodium (C) carbon
MCQ 5.1.15
b 3 3 1 3
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Codes : a (A) 2 (B) 4 (C) 4 (D) 1
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a.
List-II
(D) aluminum
Assertion (A) : Aluminium is a paramagnetic material. Reason (R) : A paramagnetic material have an odd no. of electrons. (A) Both A and R are true and R is correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A.
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(D) A is false but R is true. MCQ 5.1.16
Magnetic flux density inside a medium is 6az mwb/m2 . If the relative permeability of the medium is 2.3 then the magnetization inside the medium will be (A) 3979 A/m (B) 2249 A/m (C) 9151 A/m
Magnetic flux density inside a magnetic material is B . If the the permeability of the material is m = 3m0 then the vector magnetization of the material will be (A) B (B) 3B 3m0 2m0
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MCQ 5.1.17
(D) 8650 A/m
(C) B 2m0
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A magnetic material of relative permeability mr = 4/p is placed in a magnetic field having strength H = 2r2 af A/m . The magnetization of the material at r = 2 will be (A) 8.19af A/m (B) 1.10af A/m (C) 0.546af A/m
MCQ 5.1.20
(D) 2.19af A/m
A metallic bar of cross sectional area 2 m2 is placed in a magnetizing field H = 8 A/m . If the field causes a total magnetic flux of F = 4.2 mWb in the bar then the susceptibility of the bar will be (A) 22.87 (B) 23.87 (C) 46.74
MCQ 5.1.21
(D) ^k - 1h H
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(C) ^k + 1h H
MCQ 5.1.19
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A portion of B -H curve for a ferromagnetic material can be approximated by the analytical expression B = m0 kH . The magnetization vector M inside the material is (A) ^m0 k - 1h H (B) kH
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MCQ 5.1.18
(D) 2B 3m0
(D) 3 # 10-5
A large piece of magnetic material carries a uniform magnetization M and magnetic field intensity H 0 inside it. The magnetic flux density inside the material is given by B 0 = m0 (2H 0 + M) If a small spherical cavity is hollowed out of the material then the magnetic field intensity H at the center of the cavity will be (A) 2H 0 (B) H 0 + M 3 (C) H 0 - 2M 3
(D) H 0 - M 3
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Statement for Linked Question 22 - 23 : A nonuniform magnetic field B inside a medium with magnetic susceptibility cm = 2 is given as B = 3zax Tesla MCQ 5.1.22
Bound current density inside the medium will be 3m0 (B) 8 ay A/m2 (A) a A/m2 3m0 8 y
Total current density inside the medium will be (B) 4 ay A/m2 (A) 4 at A/m2 m0 3m0
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MCQ 5.1.23
(D) 16 ay A/m2 3m0
m
(C) 4 ay A/m2 3m0
(C) 8 ay A/m2 3m0
A uniformly magnetized circular cylinder of infinite length has magnetization M along it’s axis. The magnetic field intensity outside the cylinder will be (A) non uniform
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MCQ 5.1.24
(D) 4m0 ay A/m2
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(B) uniform and depend on the radius of circular cylinder (C) zero
MCQ 5.1.25
An infinite circular cylinder is located along z -axis that carries a uniform magnetization M = 1.2az A/m . The magnetic flux density due to it inside the cylinder will be (A) 2.2 # 10-7 af (B) 0.7af
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(C) 8.8 # 10-7 af
(D) 4.4af
Magnetic flux lines are passing from a nickel material to the free space. If the incident of the flux line makes an angle a1 = 75c to the normal of the boundary in the nickel side as shown in figure then what will be the angle a2 with normal of the flux when it comes out in free space ? (relative permeability of Nickel = 600 )
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MCQ 5.1.26
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(D) none of these
(A) 15c
(B) 1.23c
(C) 2.7c (D) 0.356c GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Statement for Linked Question 27 - 28 : The two homogenous, linear and isotropic medium is defined in a Cartesian system such that medium 1 with relative permeability mr1 = 8 is located in the region y # 0 and medium 2 with relative permeability mr2 = 6 is in the region y > 0 . MCQ 5.1.27
The magnetic field intensity in the 1st medium is H1 = 13ax + 16ay - 10az . What will be the magnetic field intensity in the 2 nd medium ? (A) 9ax - 18.67ay + 10az A/m (C) 9ax + 18.67ay - 10az A/m (D) 18.67ax - 9ay + 10az A/m Magnetic flux density in medium 2 will be (A) (6.8ax - 14.1ay + 7.5az ) # 10-5 wb/m2
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MCQ 5.1.28
m
(B) 9ax + 2.63ay - 10az A/m
(B) (6.8ax + 14.1ay - 7.5az ) # 10-5 wb/m2 (D) (54ax + 117ay - 60az ) wb/m2
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The magnetic flux density in the region z < 0 is given as B = 4ax + 8az Wb/m2 .If the plane z = 0 carries a surface current density K = 4ay A/m ; then the magnetic flux density in the region z > 0 will be (A) 4ax + 3 ^1 + m0h az Wb/m2 (B) 4ax + m0 ay + 3az
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MCQ 5.1.29
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(C) (14.1ax - 6.8ay + 7.5az ) # 10-5 wb/m2
(C) ^4ax + 3az h^1 + m0h Wb/m2 (D) 4ax + 4m0 ay + 3az Wb/m2
MCQ 5.1.30
An infinite plane magnetic material slab of thickness d and relative permeability mr occupies the region 0 < x < d . An uniform magnetic field B = B 0 az is applied in free space (outside the magnetic material). The field intensity Hin and flux density Bin inside the material will be respectively (B) B 0 and B 0 (A) mr m0 B 0 and mr B 0 mr m0 (C) mr B 0 and B 0 m0
MCQ 5.1.31
(D) B 0 and mr B 0 m0
Two infinite plane conducting sheets are located in the plane z = 0 and z = 2 m .The medium between the plates is a magnetic material of uniform permeability m = 4m0 . If in the region between the plates a uniform magnetic flux density is defined as B = ^12ax + 4ay h # 10-3 Wb/m2 then the magnetic energy stored per unit area of the plates will be (A) 2.5 J/m2 (B) 4.1 J/m2 (C) 5 J/m2
(D) 7.8 J/m2
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For View Only Shop Online at www.nodia.co.in MCQ 5.1.32 In the two different mediums of permeability m1 and m2 , the magnetic fields are ( B1 , H1 ) and (B2 , H2 ) respectively as shown in the figure.
If the interface carries no current then the correct relation for the angle q1 and q2 is (A) B1 cos q1 = B2 cos q2
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(B) H1 cos q1 = H2 cos q2 (C) B1 sin q1 = B1 sin q2
In a three layer medium shown in the figure below, Magnetic flux impinges at an angle q1 on the interface between regions 1 and 2. The permeability of three regions are m1 , m2 and m3 . So the angle of emergence q4 will be independent of
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MCQ 5.1.33
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(D) Both (B) and (C)
MCQ 5.1.34
(A) m1 and m2 both
(B) m2 only
(C) All m1 , m2 and m3
(D) m1 only
A conducting wire is bent to form a circular loop of mean radius 20 cm . If cross sectional radius of the wire is a , such that a 0 , where the magnetic field is given by B = 3ax wb/m2 . If the point charge is located at origin at the time of injection then in the region y > 0 the point charge will follow
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MCQ 5.2.1
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For View Only Shop Online at www.nodia.co.in (A) a circular path centered at (0, 0, - 2) (B) an elliptical path centered at origin (C) a circular path centered at (1, 2, 0) (C) a parabolic path passing through origin
Statement for Linked Question 3 - 4 :
If the filamentary conductor carries a current of 3 A flowing in + ax direction from Q to R then the force exerted by wire on the side QR of rectangle will be (A) - 3 # 106 ay N (B) - 2 # 10-6 ay N
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MCQ 5.2.3
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A filamentary conductor is formed into a rectangle such that it’s corners lies on points P (1, 1, 0), Q (1, 3, 0), R (3, 4, 0), S (4, 1, 0). An infinite straight wire lying on entire x -axis carries a current of 5 A in ax direction.
(C) - 6 # 10-6 ay N
The total force exerted on the conducting loop by the straight wire will be (A) - 6 # 10-6 ay N (B) 12 # 10-6 ay N
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MCQ 5.2.4
(D) 3 # 10-6 ay N
(C) 6 # 10-6 ay N
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Two filamentary currents of - 5ax and 5ax A are located along the lines y = 0 , z =- 1 m and y = 0 , z = 1 m respectively. If the vector force per unit length exerted on the third filamentary current of 10ax A located at y = k , z = 0 be F then the plot of F versus k will be
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MCQ 5.2.5
(D) - 12 # 10-6 ay N
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For View Only Shop Online at www.nodia.co.in MCQ 5.2.6 A current filament placed on x -axis carries a current I = 10 A in + ax direction. If a conducting current strip having surface current density K = 3ax A/m is located in the plane y = 0 between z = 0.5 and z = 1.5 m then what will be the force per unit meter on the filament exerted by the strip ? (A) 6.6ax mN/m (B) 6.6az mN/m (C) 6az mN/m
A conducting current strip of 5 m length is located in the plane x = 0 between y = 1 and y = 3 . If surface current density of the strip is K = 6az A/m then the force exerted on it by a current filament placed on z -axis that carries a current I = 10 A in + az direction will be (A) - 16.4ay mN (B)- 4.8ay mN
m
MCQ 5.2.7
(D) 0
(D)26.4ay mN
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(C) - 26.4ay mN
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Statement for Linked Question 8 - 9 :
A thick slab extending from y =- a to y =+ a carries a uniform current density J = J 0 ax
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Plot of magnetizing factor H at any point in the space (inside or outside slab) versus y will be
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MCQ 5.2.8
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For View Only Shop Online at www.nodia.co.in MCQ 5.2.9 If a magnetic dipole of moment m = m 0 ax is placed at the origin then the force exerted on it due to the slab will be (A) 0 N (B) m 0 m0 J 0 yaz (C) m 0 m0 J 0 az
(D) - m 0 m0 J 0 yaz
Statement for Linked Question 10 - 11 : The volume current density J at any point inside the cylinder is proportional to (A) r (B) 1/r
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MCQ 5.2.10
m
A long circular cylinder placed along z -axis carries a magnetization M = 2r2 af .
(D) r2
(C) r sin f
The plot of the magnetic flux density B inside the cylinder versus r will be
MCQ 5.2.12
A short cylinder placed along z -axis carries a “frozen-in” uniform magnetization M in + az direction. If length of the cylinder is equal to its cross sectional diameter then pattern of its surface current density K will be as
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MCQ 5.2.11
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Magnetization of a long circular cylinder is M along it’s axis. Which of the following gives the correct pattern of magnetic field lines (B ).
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MCQ 5.2.13
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m
For View Only
281
Statement for Linked Question 14 - 15 : A conducting rod of square cross section of side 2 cm carries a uniform magnetization M = 4 A/m along it’s axis. Length of the rod is L >> 4 cm . MCQ 5.2.14
If the rod is bent around it into a complete circular ring then magnetic flux density inside the circular ring will be (A) 4 wb/m2 (B) 4m0 wb/m2 (C) 2pm0 wb/m2
MCQ 5.2.15
(D) m0 wb/m2
Assume that there remains a narrow gap of width 0.1 mm between the ends of the rod when it is formed into a circular ring. The net magnetic flux density at the center of the gap will be (A) 50.04 # 10-7 wb/m2 (B) 49.88 # 10-6 wb/m2
(C) 51.23 # 10-6 wb/m2 (D) 34.66 # 10-6 wb/m2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 5.2.16 Magnetic flux density B inside a sphere that carries a uniform magnetization M will be (A) 0 (B) 1 m0 M 3 (C)
(D) 2 m0 M 3
A short cylinder of length equals to it’s diameter carries a uniform magnetization M as shown in the figure.
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MCQ 5.2.17
m0 M 2
MCQ 5.2.18
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The correct sketch for the magnetic field intensity H inside the cylinder is
Mutual inductance between an infinite current filament placed along y -axis and rectangular coil of 1500 turns placed in x -y plane as shown in figure will be
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For View Only (A) 41.4 mH
Shop Online at www.nodia.co.in (B) 0.33 mH (D) 33 mH
(C) 2.38 mH MCQ 5.2.19
283
An infinitely long straight wire of radius a , carries a uniform current I . The energy stored per unit length in the internal magnetic field will be (A) uniform and depends on I only (B) non uniform (C) uniform and depends on a only
A planar transmission line consists of two conducting plates of 2 m width placed along x -z plane such that the current in one plate is flowing in + az direction. While in the other it is flowing in - az direction. If both the plate carries 4 A current and there is a very small separation between them then what will be force of repulsion per meter between the two plates ? (A) 16m0 (B) 4m0 (C) 8m0
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A very long solenoid having 20, 000 turns per meter. The core of solenoid is formed of iron. If the cross sectional area of solenoid is 0.04 m2 and it carries a current I = 100 mA then what will be the energy stored per meter in it’s field ? (relative Permeability of iron, mr = 100 ) (A) 8.1 J/m (B) 20.11 J/m (C) 100.5 J/m
MCQ 5.2.22
(D) m0 /4
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MCQ 5.2.21
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MCQ 5.2.20
m
(D) uniform and depends on both I and a
(D) 10.05 J/m
A mass spectrograph is a device for separating charged particles having different masses. Consider two particles of same charges Q but different masses m and 2m injected into the region of a uniform field B with a velocity v normal to the magnetic field as shown in the figure. When the particles will be releasing out of the spectrograph the separation between them will be
(A) 2mv Bq
(B) mv 2Bq
(C) mv Bq
(D) 0
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Statement for Linked Question 23 - 24 : Consider a conducting filamentary wire of length 1 meter and mass 0.3 kg oriented in east-west direction, situated in the earth’s magnetic field at the magnetic equator. (Assume the magnetic field at equator has a value of 0.2 # 10-4 Wb/m2 and directed northward) The current that required to counteract the earth’s gravitational force on the wire must flow from (A) west to east (B) east to west
m
MCQ 5.2.23
(C) any of (A) and (B)
What will be the magnitude of the current flowing in the wire as to counteract the gravitational force ? (A) 49 kA (B) 24.5 kA
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MCQ 5.2.24
(D) none of these
A rigid loop of wire in the form of a square is hung by pivoting one of it’s side along the x -axis as shown in the figure. The loop is free to swing about it’s pivoted side without friction. The mass of the wire is 0.2 kg/m and carries a current 2 A. If the wire is situated in a uniform magnetic field B = 2.96 Wb/m2 then the angle by which the loop swings from the vertical is
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(A) 3p/4 (C) p/6 MCQ 5.2.26
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MCQ 5.2.25
(D) 4.9 kA
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(C) 98 kA
(B) p/4 (D) p/2
Electron beams are injected normally to the plane edge of a uniform magnetic field H = H 0 ax as shown in figure.
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285
For View Only Shop Online at www.nodia.co.in The path of the electrons ejected out of the field will be in (B) - ay direction (A) + ay direction (C) ^ay + az h direction
The medium between the two infinite plane parallel sheets carrying current densities 4ax and - 8ax A/m , consists of two magnetic material slabs of thickness 1 m and 2 m having permeabilities m1 = 2m0 and m2 = 4m0 respectively as shown in the figure.
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MCQ 5.2.27
(D) ^ay - az h direction
MCQ 5.2.28
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(C) - 32m0 ay Wb/m
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What will be the magnetic flux per unit length between the current sheets along the direction of flow of current ? (A) - 24m0 ay Wb/m (B) - 16m0 ay Wb/m (D) - 40m0 ay Wb/m
Two perfectly conducting, infinite plane parallel sheets separated by a distance d carry uniformly distributed surface currents with equal and opposite densities K and - K respectively. The medium between the two plates is a magnetic material of non uniform permeability which varies linearly from a value of m1 near one plate to a value of m2 near the second plate. What will be the magnetic flux between the current sheets per unit length along the direction of flow of the current ? m + m2 (B) ^m1 + m2h Kd (A) b 1 Kd 2 l (C) b 1 + 1 l Kd m1 m2
(D) a
m2 - m1 Kd 2 k
Statement for Linked Question 29 - 30 : The magnetic field intensity inside an infinite plane magnetic material slab is given as H = 12ax + 24ay . The permeability of the magnetic material is m = 2m0 MCQ 5.2.29
If the magnetic material slab occupies the region 0 < z < 2 m then the magnetization surface current densities at the surfaces z = 0 and z = 2 m will be respectively (B) ^- 2ax + 4ay h and ^2ax - 4ay h (A) ^- 4ay + 2ax h and ^4ay - 2ax h (C) ^4ax + 4ay h and ^2ax - 4ay h
(D) ^2ax + 4ay h and ^- 2ax + 4ay h
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For View Only Shop Online at www.nodia.co.in MCQ 5.2.30 The magnetization volume current density Jm will be (A) 0 (B) 4ax + 2ay (C) 8ax + 4ay
B -H curve for a ferromagnetic material is given as B = 2m0 HH . What will be the work done per unit volume in magnetizing the material from zero to a certain value B 0 = 2m0 H 02 ? 2 3 (A) 4m0 H 0 (B) 4m0 H 0 3 3
m
MCQ 5.2.31
(D) - 4ax - 2ay
5
(D) 2 H 0 3
Two infinitely long straight wire and a third wire of length l are parallel to each other located as shown in the figure.
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MCQ 5.2.32
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(C) 4m0 H 02
(C)
(D)
mI 2pl
Two infinite plane conducting sheets lying in the plane x = 0 and x = 5 cm carry surface current densities + 10 mA/m ay and - 20 mA/m ay respectively. If the medium between the plates is a magnetic material of uniform permeability m = 2m0 then what will be the energy stored per unit area of the plates ? (A) 800p J/m2
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MCQ 5.2.33
mI 2 2pl
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Infinitely long wire carries a current I while the wire of length l shown at the top carries a current 2I . The magnitude of the force experienced by the top wire is mI 2 (B) mpI 2 (A) p
(B) 160p J/m2 (C) 8p J/m2 (D) 1.6p J/m2 MCQ 5.2.34
Medium 1 comprising the region z > 0 is a magnetic material with permeability m1 = 4m0 where as the medium 2, comprising the region z < 0 is a magnetic material with permeability m2 = 2m0 . Magnetic flux density in medium 1 is given by B1 = ^0.4ax + 0.8ay + az h Wb/m2
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287
For View Only Shop Online at www.nodia.co.in If the boundary z = 0 between the two media carries a surface current of density K given by K = 1 ^0.2ax - 0.4ay h A/m m0 then the magnetic flux density in medium 2 will be (A) ^0.8ax + ay + az h Wb/m2 (B) ^- ax + 0.8ay - az h Wb/m2
(C) ^ax + 0.8ay + az h Wb/m2
A square loop of a conductor lying in the yz plane is bisected by an infinitely long straight wire carrying current 2 A as shown in the figure. If the current in the square loop is 4 A then the force experienced by the loop will be
MCQ 5.2.36
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MCQ 5.2.35
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(D) ^ax + 0.8ay h Wb/m2
(A) 1.6ay mN
(B) 6.4ay mN
(C) 3.2ay mN
(D) 0.64ay mN
A certain region z < 0 comprises a magnetic medium with permeability m = 25m0 . The magnetic flux density in free space ^z > 0h makes an angle q, with the interface whereas in medium 2 flux density makes an angle q2 as shown in the figure.
If B2 = 1.2ay + 0.8az then what will be the angular deflection ^q1 - q2h ? (A) 50.6c (B) 19.47c (C) 31.15c (D) 12.06c GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Statement for Linked Question 37 - 38 :
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Consider the magnetic circuit shown in figure
MCQ 5.2.37
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The cross sectional area of the section on which coil is wound is S1 where as all the rest of the section has the cross sectional area S2 . Magnetic core has the permeability m = 500m0 . If S1 = 5 cm2 and S2 = 10 cm2 then the total reluctance of the circuit will be (A) 41/100m0 (B) 9/20m0
(C) 27.9 mH MCQ 5.2.39
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If the no. of turn of the coil is 100 then the equivalent self inductance of the coil is (A) 22.22 kH (B) 1.41 mH (D) 4.5 kH
The coil of a magnetic circuit has 50 turns. The core of the circuit has a relative permeability of 600 and length of the core is 0.6 m. What must be the core cross section of the magnetic circuit so that the coil may have a 0.2 mH inductance ? (A) 6.4 cm2 (B) 0.64 cm2 (C) 11.94 cm2
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MCQ 5.2.38
(D) 39/100m0
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(C) 20m0 /9
(D) 1.56 cm2
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Statement for Linked Question 40 - 41 : A System of three coils on an ideal core is shown in figure
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289
For View Only Shop Online at www.nodia.co.in The cross sectional area of all the segments of the core is S = 450 cm2 . MCQ 5.2.40
If N1 = 500 then what will be the self inductance of the coil having N1 turns ? (A) 0.125 mH (B) 62.8 mH (C) 31.41 mH
MCQ 5.2.41
(D) 6.28 mH
If N2 = 250 then the self inductance of the coil N2 will be (A) 2.6 mH (B) 23.6 mH (C) 70.7 mH
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A system of three coils on an ideal core that has two air gaps is shown in the figure.
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MCQ 5.2.42
(D) 2.36 mH
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All the segments of core has the uniform cross sectional area 2500 mm2 . What will be the mutual inductance between N1 and N2 ? (A) 39.27 mH (B) 52.36 mH (C) 26.18 mH MCQ 5.2.43
The mutual inductance between N2 and N 3 will be (A) 0 (B) 78.54 mH (C) 52.36 mH
MCQ 5.2.44
(D) 78.54 mH
(D) 39.27 mH
The magnetization curve for an iron alloy is approximately given by B = 1 H + H 2 mWb/m2 3 If H increases from 0 to 210 A/m, the energy stored per unit volume in the alloy is (A) 6.2 MJ/m3 (B) 1.3 MJ/m3 (C) 2.3 kJ/m3
(D) 2.9 kJ/m3
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EXERCISE 5.3
A current sheet J = 5ay A/m lies on the dielectric interface x = 0 between two dielectric media with er1 = 5, mr1 = 1 in Region-1 (x < 0) and er 2 = 2, mr 2 = 2 in Region-2 (x 2 0). If the magnetic field in Region-1 at x = 0- is H1 = 3ax + 30ay A/m the magnetic field in Region-2 at x = 0+ is
MCQ 5.3.3 IES EC 2011
(C) H2 = 1.5ax + 40ay A/m
(D) H2 = 3ax + 30ay + 10az A/m
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(B) H2 = 3ax + 30ay - 10az A/m
A bar magnet made of steel has a magnetic moment of 2.5 A- m2 and a mass of 6.6 # 10-3 kg . If the density of steel is 7.9 # 103 kg/m3 , the intensity of magnetization is
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IES EC 2012
(A) H2 = 1.5ax + 30ay - 10az A/m
(A) 8.3 # 10-7 A/m
(B) 3 # 106 A/m
(C) 6.3 # 10-7 A/m
(D) 8.2 # 106 A/m
If the current element represented by 2 # 10-4 ay Amp-m is placed in a magnetic field of H = 5ax /m A/m, the force on the current element is (A) - 2.0az mN (B) 2.0az mN
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MCQ 5.3.2
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GATE 2011
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MCQ 5.3.1
Chap 5
(C) - 2.0az N MCQ 5.3.4 IES EC 2011
(D) 2.0az N
Match List I with List II and select the correct answer using the code given below the lists : List I
List II
a.
MMF
1.
Conductivity
b.
Magnetic flux
2.
Electric current
c.
Reluctance
3.
EMF
d.
Permeability
4.
Resistance
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 5
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For View Only Codes : a (A) 3 (B) 1 (C) 3 (D) 1
IES EC 2008
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c 2 4 4 2
d 1 3 1 3
Consider the following statements associated with boundary conditions between two media: 1. Normal component of B is continuous at the surface of discontinuity. 2.
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MCQ 5.3.5
291
Normal component of D may or may not be continuous.
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Which of the statement(s) given above is/are correct? (A) 1 only (B) 2 only (C) Both 1 and 2 IES EC 2007
Magnetic current is composed of which of the following ? (A) Only conduction component (B) Only displacement component
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MCQ 5.3.6
(D) Neither 1 nor 2
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(C) Both conduction and displacement components (D) Neither conduction component nor displacement component IES EC 2006
Which one of the following is the correct expression for torque on a loop in magnetic field B ? (Here M is the loop moment) (A) T = 4: B (B) T = M : B
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MCQ 5.3.7
(C) T = M # B MCQ 5.3.8 IES EC 2006
(D) T = B # M
Match List I with List II and select the correct answer using the code given below the lists : List-I
List-II
a.
Line charge
1.
Maxwell
b.
Magnetic flux density
2.
Poynting vector
c.
Displacement current
3.
Biot-Savart’s law
4.
Gauss’s law
d. Power flow Codes : a b (A) 1 2 (B) 4 3 (C) 1 3 (D) 4 2
c 4 1 4 1
d 3 2 2 3
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Chap 5
For View Only Shop Online at www.nodia.co.in 1 MCQ 5.3.9 What does the expression 2 J : A represent ? IES EE 2009 (A) Electric energy density (B) Magnetic energy density IES EC 2006
(C) Power density
MCQ 5.3.10
Two thin parallel wires are carrying current along the same direction. The force experienced by one due to the other is (A) Parallel to the lines
IES EC 2003
(D) Radiation resistance
(C) Perpendicular to the lines and repulsive
IES EC 2001
A boundary separates two magnetic materials of permeability m1 and m2 . The magnetic field vector in m1 is H1 with a normal component Hn1 and tangential component Ht1 while that in m2 is H2 with a normal component Hn2 and a tangential component Ht 2 . Then the derived conditions would be (A) H1 = H2 and Ht 1 = Ht 2
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(B) Ht 1 = Ht 2 and m1 Hn1 = m2 Hn 2
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MCQ 5.3.11
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(D) Zero
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(B) Perpendicular to the lines and attractive
(C) H1 = H2 and m1 Hn1 = m2 Hn2
(D) H1 = H2, Ht 1 = Ht 2 and m1 Hn 1 = m2 Hn 2
The dependence of B (flux density) on H (magnetic field intensity) for different types of material is
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MCQ 5.3.12
MCQ 5.3.13 IES EE 2012
Statement I : Polarization is due to the application of an electric field to dielectric materials.
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293
For View Only Shop Online at www.nodia.co.in Statement II : When the dipoles are created, the dielectric is said to be polarized or in a state of polarization. (A) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) (B) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (C) Statement (I) is true but Statement (II) is false (D) Statement (I) is false but Statement (II) is true The following equation is not valid for magneto-static field in inhomogenous magnetic materials (A) d : B = 0 (B) d : H = 0
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IES EE 2011
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MCQ 5.3.14
(C) d # A = B
Assertion (A) : Superconductors cannot be used as coils for production of strong magnetic fields. Reason (R) : Superconductivity in a wire may be destroyed if the current in the wire exceeds a critical value. (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A)
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MCQ 5.3.15
(D) d # H = J
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(B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true MCQ 5.3.16 IES EE 2010
A conductor 2 metre long lies along the z -axis with a current of 10 A in az direction. If the magnetic field is B = 0.25ax T , the force on the conductor is (A) 4.0ay N (B) 1.0az N (C) 1.0 ay N
MCQ 5.3.17 IES EE 2008
(D) 3.0 az N
The force on a charge moving with velocity v under the influence of electric and magnetic fields is given by which one of the following ? (A) q ^E + B # v h (B) q ^E + v # H h (C) q ^H + v # E h (D) q ^E + v # B h
MCQ 5.3.18 IES EE 2007
If a very flexible wire is laid out in the shape of a hairpin with its two ends secured, what shape will the wire tend to assume if a current is passed through it ? (A) Parabolic (B) Straight line (C) Circle
(D) Ellipse
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For View Only Shop Online at www.nodia.co.in MCQ 5.3.19 Consider the following : IES EE 2007 Lorentz force F = e ^v # B h where e, v and B are respectively the charge of the particle, velocity of the particle and flux density of uniform magnetic field. Which one of the following statements is not correct ? (A) Acceleration is normal to the plane containing the particle path and B (B) If the direction of the particle path is normal to B , the acceleration is maximum (C) If the particle is at rest, the field will deflect the particle
IES EE 2006
What is the force on a unit charge moving with velocity v in presence of electric field E and magnetic field B ? (A) E - v : B (B) E + v : B
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MCQ 5.3.20
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(D) If the particle path is in the same direction of B , there will be no acceleration
MCQ 5.3.21 IES EE 2004
(D) E + v # B
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(C) E + B # v
What is the force experienced per unit length by a conductor carrying 5 A current in positive z -direction and placed in a magnetic field B = ^3ax + 4ay h ? (B) - 20ax + 15ay N/m (A) 15ax + 20ay N/m
IES EE 2003
Which one of the following formulae is not correct for the boundary between two magnetic materials ? (A) Bn1 = Bn2 (B) B2 =
Bn2 + Bt2
(C) H1 = Hn1 + Ht1
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MCQ 5.3.22
(D) - 20ax - 20ay N/m
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(C) 20ax - 15ay N/m
MCQ 5.3.23
Interface of two regions of two magnetic materials is current-free. The region 1, for which relative permeability mr1 = 2 is defined by z < 0 , and region 2, z > 0 has mr2 = 1. If B1 = 1.4ax + 2.2ay + 1.4ax T ; then H2 is (A) 1/m0 60.6ax + 0.4ay + 0.4az@ A/m
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IES EE 2003
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(D) an21 # (H1 - H2) = K where an21 is a unit vector normal to the interface and directed from region 2 to region 1.
(B) 1/m0 61.2ax + 0.8ay + 0.8az@ A/m
(C) 1/m0 61.2ax + 0.4ay + 0.4az@ A/m
(D) 1/m0 60.6ax + 0.4ay + 0.8az@ A/m
MCQ 5.3.24 IES EE 2002
If A and J are the vector potential and current density vectors associated with a coil, then # A : J dv has the units of (A) flux-linkage (B) power (C) energy
(D) inductance
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SOLUTIONS 5.1
Option (A) is correct. Force applied by a magnetic field B on a moving charge with velocity v is defined as F = v#B Since the direction of velocity v and B are perpendicular to each other as obtainde from the shown figure so the resultant force will be perpendicular to both of them. i.e. the force on the moving charged particle will be in upward direction. and as the particle is also deflected in upward direction with the applied force so it gives the conclusion that the particle will be positively charged.
SOL 5.1.2
Option (D) is correct. Since a magnet bar must have south and north pole i.e. a single pole charge can’t exist. So, a magnetic point charge doesn’t exit.
SOL 5.1.3
Option (B) is correct. Force applied on a moving charge in the presence of electric and magnetic field is defined as F = Fe + Fm = q (E + v # B) where Fe and Fm are the electric and magnetic forces applied on the charge so it is clear that the moving charge experiences both the electric and magnetic forces. The electric force is applied in a uniform direction (in direction of electric field) i.e. it is an accelerating force while, the magnetic force is applied in the normal direction of both the magnetic field and velocity of the charged particle i.e. it is a deflecting force. Therefore, both the options are correct but R is not the correct explanation of A.
SOL 5.1.4
Option (A) is correct. For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F = Q 6E + (v # B)@ where E " electric field v " velocity of the charge B " magnetic flux density Since the electron beam follows its path without any deflection so the net force applied by the field will be zero
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SOL 5.1.1
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m
Q 6E + ^v # B h@ = 0 15ay + v # 3az = 0 As the electric field is directed along az and magnetic field is directed along ay so the velocity of beam will be in ax direction (perpendicular to both of the field). Consider the velocity of the beam is V = kax So we have 15ay + kax # 3az = 0 15ay - 3kay = 0 k = 15 = 5 m/s 3 So, the velocity of the beam will be 5 m/s along the x -axis.
Option (C) is correct. For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F = Q 6E + (v # B)@ where E " electric field v " velocity of the charged particle B " magnetic flux density So, at time t = 0 total force applied on the electron is F (0) = e 6E + ^V (0) # B h@ Now we have V (0) # B = (200ax - 300ay - 400az ) # (- 3ax + 2ay - az ) = 1100ax + 1400ay - 500az therefore the applied force on the electron is F (0) = (1.6 # 10-19) 6(0.1ax - 0.2ay + 0.3az ) # 103 + 1100ax + 1400ay - 500az@ me a (0) = 1.6 # 10-19 6(100 + 1100) ax + (1400 - 200) ay + (300 - 500) az@ (F (0) = me a (0), where a (0) is acceleration of electron at t = 0 ) -19 1 . 6 10 # a (0) = # 200 (6ax + 6ay - az ) 9.1 # 10-31 = 6.5 # 1013 (7ax + 2ay - az ) m/s2
SOL 5.1.6
Option (B) is correct. Force F applied on a current element in the presence of magnetic flux density B is defined as F = I ^L # B h where I " current flowing in the element L " vector length of current element in the direction of current flowing So, F = 3 # 10-3 62az # (ax + 3ay)@ = 6 # 10-3 6ay - 3ax@ =- 14ax + 16ay mN
SOL 5.1.7
Option (A) is correct. The magnitude of the force experienced by either of the loops will be same but the direction will be opposite. So the force experienced by C1 due to C2 will be - F .
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SOL 5.1.5
Option (C) is correct. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 5.1.8
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297
For View Only Shop Online at www.nodia.co.in The boundary condition for the current interface holds the following results. (i) normal component of magnetic flux density is continuous. i.e. B1n = B2n (ii) Tangential component of magnetic field intensity is continuous. i.e. H1t = H2t So, (A) and (B) are wrong statement. Now, we check the statement (C).
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Consider the magnetic field intensity in 1st medium is H1 and magnetic field intensity in 2 nd medium is H2 . So, it’s tangential component will be equal (tangential component) i.e. H1t = H2t Since scalar magnetic potential difference is defined as the line integral of magnetic field intensity
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Option (C) is correct. The magnetic field intensity produced at any point in the free space will be the vector sum of the field intensity produced by all the current sheets. Since, the magnetic field intensity produced at any point P due to an infinite sheet carrying uniform current density K is defined as H = 1 (K # a n) 2 where an is the unit vector normal to the sheet directed toward the point P . So in the region 0 < z < 1 magnetic field intensity due to K2 and K 3 will be cancelled as the unit normal vector to the two sheets will be opposite to each other. Therefore in this region magnetic field intensity will be produced only due to the current density K1 = 4ax which is given as (an = az ) H = 1 K1 # an = 1 (4ax ) # (az ) 2 2 =- 4az A/m
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SOL 5.1.9
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i.e. V1 - V2 = H : dl = I and since there is no current density at boundary. So, we have V1 - V2 = 0 V1 = V2 i.e. magnetic scalar potential will be same in both medium.
Option (D) is correct. As the conducting filament is located along the line y = 0 , z = 0.2 m which is in the region 0 < z < 1 m , so, the net magnetic field intensity produced on the conducting filament by the current sheets is (as determined in previous question) H =- 2ay A/m or, B = m0 H =- 2m0 ay Now the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF = Idl # B where I is the current flowing in the element and dl is the differential vector length GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 5.1.10
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For View Only Shop Online at www.nodia.co.in of the current element in the direction of flow of current. So force per unit length experienced by the conducting filament is dF = 7a ( I = 7 A , dl = dlax ) x # (- 2m0 ay ) dl =- 14m0 az N/m Option (B) is correct. Magnetic dipole moment of a coil carrying current I and having area S is given by m = ISan where an is normal vector to the surface of the loop. Since the coil is lying in the plane 2x + 6y - 3z = 4 so the unit vector normal to the plane of the coil is given as. 2ax + 6ay - 3az 4f So, ( f = 2x + 6y - 3z ) = an 4f 22 + 62 + (- 3) 2 Therefore the magnetic dipole moment of the coil is (2ax + 6ay - 3az ) ( I = 5 A, S = 1 m2 ) m = (5) (1) 7 5 (2ax + 6ay - 3az ) = 7 As the torque a magnetic field B on the loop having magnetic moment m is defined as T = m#B So the torque on the given coil is 5 (2ax + 6ay - 3az ) T =; E # (6ax + 4ay + 5az ) 7 = 3ax - 5ay - 8az N- m
SOL 5.1.12
Option (B) is correct. Magnetic dipole moment of a coil of area S carrying current I is defined as m = ISan where an is the unit vector normal to the surface of the loop. and since from the given data we have I = 10 A S = pr2 = p # (1) 2 = p (normal vector to the surface z = 0 ) an = a z So the magnetic moment of the circular current loop lying in the plane z = 0 is m = 10paz Now the torque on an element having magnetic moment m in the presence of magnetic flux density B is defined as T = m#B Therefore, the torque acting on the circular loop is (B = 4ax - 4ay - 2az ) T = (10paz ) # (4ax - 4ay - 2az ) = 10p (4ay + 4ax ) = 30p (ax + ay)
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SOL 5.1.11
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Chap 5
Magnestostatic Fields in Matter
For View Only SOL 5.1.13 Option (B) is correct.
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Option (C) is correct. A diamagnetic material carries even no. of electrons inside it’s atom. Number of electron in carbon atom is six. Which is even so it is a diamagnetic material rest of the material having odd no. of electrons.
SOL 5.1.15
Option (A) is correct. A paramagnetic material have an odd no. of electrons and since atomic no. of Al is 13, which is odd. So, it is a paramagnetic material. So, A and R both true and A is correct explanation of R.
SOL 5.1.16
Option (B) is correct. In a magnetic medium the magnetization in terms of magnetic field intensity is defined as M = cm H where cm is magnetic susceptibility given as (relative permeability, mr = 2.3 ) cm = mr - 1 = 1.3 and since the magnetic field intensity in terms of magnetic flux density is given as (B = 5az mwb/m2 ) H =B = B m0 mr m 5 # 10-3 ( mr = 2.3 ) az = 1730az A/m = 4p # 10-7 # 2.3 So the magnetization inside the medium is M = cm H = 2249 A/m
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SOL 5.1.14
Option (D) is correct. Given the permeability, m = 3m0 and magnetic flux density = B So the field intensity inside the material will be H =B= B m 3m0 Since the magnetization of a magnetic material is defined as M = B -H m0 where B and H are the flux density and field intensity inside the material. So we get M = B - B = 2B m0 3m0 3m0 SOL 5.1.18 Option (D) is correct. As the magnetic flux density and magnetic field intensity inside a magnetic material are related as B = mr m0 H So, comparing it with given expression for magnetic flux density we get the relative permeability as mr = k = k - 1 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia SOL 5.1.17
300
Magnestostatic Fields in Matter
Chap 5
For View Only Shop Online at www.nodia.co.in Therefore, the magnetization vector inside the material is given as M = (mr - 1) H = ^k - 1h H Option (D) is correct. Magnetic flux density in a medium in terms of magnetic filed intensity is defined as B = mH = mr m0 H ( mr = 4/p , H = 2r2 af A/m ) = ^4/ph (4p # 10-7) (2r2 af) = 16 # 10-7 r2 af Again the magnetic flux density inside a magnetizing material is defined as B = m0 ^H + M h where M is the magnetization of the material. So, we have 32 10-7 r2 - 2r2E af = 2r2 : 4 - 1D af M = B -H =; # m0 p 4p # 10-7 At r = 2 M = 4.14af A/m
SOL 5.1.20
Option (A) is correct. Given Magnetic field intensity, H = 70 A/m Total magnetic flux in the bar, F = 4.2 mWb Cross sectional area of bar, S = 2 m2 So we have the magnetic flux density in the bar -3 B = F = 4.2 # 10 2 S = 2.1 mwb/m2 Since the magnetic field intensity and magnetic flux density are related as B = m0 (1 + cm) H So, we have 2.1 # 10-3 = (4p # 10-7) (1 + cm) (70) (2.1 # 10-3) (1 + cm) = 70 (4p # 10-7) -5 cm = c 3 # 10 -7 - 1m 4p # 10 = (23.87 - 1) = 22.97
SOL 5.1.21
Option (B) is correct. For the spherical cavity of magnetization M , the flux density is given by Bcavity = 2 m0 M 3 Since the cavity is hollowed. So not magnetic flux density at the center of cavity is Bnet = B 0 - Bcavity = B 0 - 2 m0 M 3 and so the net magnetic field intensity at the center of cavity is Hnet = 1 Bnet = 1 :B 0 - 2 m0 M D m0 m0 3
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SOL 5.1.19
(B 0 = m0 ^H 0 + M h) = 1 :m0 H 0 + m0 M - 2 m0 M D = :H 0 + M D m0 3 3 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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301
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For View Only Shop Online at www.nodia.co.in SOL 5.1.22 Option (B) is correct. In a magnetic medium the magnetic field intensity and magnetic flux density are related as B = m0 (1 + cm) H So the magnetic flux density inside the medium is B (B = 4zax T , cm = 2 ) = 4z ax H = 3m0 (1 + cm) m0 Now the magnetization of a magnetic medium having magnetic field intensity H is given as M = cm H
Option (C) is correct.
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SOL 5.1.24
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SOL 5.1.23
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M = 2 b 4z l ax = 8z ax 3m0 3m0 The bound current density inside a medium having magnetization M is given as Jb = d # M = 4#b 8z ax l = 12 ay A/m2 3m0 3m0 Option (A) is correct. Total current density inside a medium having magnetic flux density B is given as 2(4z) (B = 4zax T ) a JT = d # B = 1 ; m0 m0 2z E y = 3 ay A/m2 4m0
Volume current density inside a material is equal to the curl of magnetization M i.e. J = d#M and the surface current density in terms of magnetization is defined as K = M # an where an is unit vector normal to the surface. Consider the cylinder is placed along z -axis So, an = a r and M = Maz Therefore the volume current density inside the cylinder is (M is not the function of z ) J = 4# (Maz ) = 0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in and the surface current density of the cylinder is K = Maz # a r = Maf So the current flowing in cylinder is just similar to a solenoid and the field intensity produced due to a solenoid at any point outside it is zero. Thus we have the magnetic field intensity outside the cylinder as H outside = 0 Option (C) is correct.
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SOL 5.1.25
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Volume current density inside a material is equal to the curl of magnetization M i.e. J = d#M So the volume current density inside the cylinder is (M = 0.7az A/m ) J = d # (0.7az )= 0 and since the surface current density in terms of magnetization is defined as where an is unit vector normal to the surface. K = M # an So the surface current density of the cylinder is (M = 0.7az A/m , an = a r ) K = (0.7az ) # a r = 0.7af Therefore the current flowing in cylinder is just similar to a solenoid and the field intensity produced due to a solenoid at any point inside it is given as B = m0 K = m0 nI where n is the no. of turns per unit length of the solenoid and I is the current flowing in the solenoid. Thus, the magnetic flux density inside the cylinder is (direction is determined by right hand rule) ( K = 0.7 ) B = 0.7m0 az = 2.8 # 10-7 az SOL 5.1.26
Option (D) is correct. From Snells law we have the relation between the incidence and refracted angle of magnetic flux lines as tan a1 = mr1 tan a2 mr2 where mr1 and mr2 are relative permeability of the two medium.
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Shop Online at www.nodia.co.in tan 75c = 600 (relative permeability of air = 1) tan a2 1 tan a2 = tan 75c 600 a2 = tan-1 : tan 75cD = 0.356c 600
Option (C) is correct. Magnetic field intensity in 1st medium is given H1 = 9ax + 16ay - 10az = H1t + H1n where H1t and H1n are respectively the tangential and normal components of the magnetic field intensity to the boundary interface in medium 1. From boundary condition we have H1t = H2t and m2 H2n = m1 H1n where H2t and H2n are respectively the tangential and normal component of magnetic field intensity in medium 2. So we get the components in medium 2 as H2t = 9ax - 10az
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SOL 5.1.27
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H2n =
he
m1 m m H = r1 0 H1n m2 1n mr2 m0 = 7 (16ay) = 18.67ay 6 Therefore, the net magnetic field intensity in medium 2 is H2 = H2t + H2n = 12ax + 13.67ay - 15az A/m and
SOL 5.1.28
Option (B) is correct. Magnetic flux density in any medium in terms of magnetic field intensity is defined as B = mH where m is the permeability of the medium. So, the magnetic flux density in medium 2 is given as B2 = m2 H2 = mr2 m0 H2 ( mr2 = 6 ) = 6 # (4p # 10-7) # (9ax + 18.67ay - 10az ) 2 -5 = (6.8ax + 14.1ay - 7.5az ) # 10 wb/m
SOL 5.1.29
Option (D) is correct. The magnetic flux density in region z < 0 is given as B = 4ax + 3az Wb/m2
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Now we consider the flux density in region 1 is B1 . So, we have B1 = 4ax + 3az Therefore the tangential component B1t and normal component B1n of the magnetic flux density in region 1 are B1t = 4ax and B1n = 3az From the boundary condition the tangential and normal components of magnetic flux density in two mediums are related as B1n = B2n B2t - B1t = m0 K where B2t and B2n are respectively the tangential and normal components of the magnetic flux density in region 2 and K is the current density at the boundary interface. (B1n = 3az ) So, we get B2n = B1n = 3az and (B1t = 4ax , K = 4ay A/m ) B2t = 4ax + m0 ^4ay h = 4ax + 4m0 ay Therefore the net flux density in region 2 ^z > 0h is B2 = B2t + B2n = 4m0 ax + 2ay + 5az
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Option (D) is correct. As the surface boundary of the slab is parallel to yz -plane so the given magnetic flux density will be tangential to the surface. i.e. Bto = B 0 and Hto = Bto = B 0 m0 m0 Since the tangential component of magnetic field intensity is uniform at the boundary of the magnetic material So, magnetic field intensity inside the material is Hin = Hto = B 0 m0 Therefore, the flux density inside the material is Bin = mHin = mr m0 B 0 = mr B 0 m0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 5.1.30
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305
For View Only Shop Online at www.nodia.co.in SOL 5.1.31 Option (C) is correct. The magnetic stored energy per unit volume of the plate for a given uniform flux density (uniform permeability) is defined as wm = 1 H : B 2 Given B = ^3ax + 4ay h # 10-3 Wb/m2 3a + 4ay -3 H = B =c x m # 10 A/m m 4m0 -6 and therefore wm = 1 H : B = 1 : 9 + 16 D # 10-6 = 25 # 10 -7 2 4m0 2 8 # 4p # 10 3 = 2.49 J/m now the separation between the plates is given as d = 2 m Thus magnetic energy stored per unit area of the plate is Wm /A = wm # d = ^2.49h # 2 = 5 J/m2
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So we have,
Option (D) is correct. From boundary condition the normal component of flux density is uniform at boundary i.e. B1n = B2n B1 sin q1 = B2 sin q2 and the tangential component of field intensity is uniform i.e. H1t = H2t H1 cos q1 = H2 cos q2
SOL 5.1.33
Option (B) is correct. Relation between q1 and q2 at boundary of region (1) and region (2) as m1 tan q1 = m2 tan q2 and at the interface between region (2) and region (3) is since q2 = q3 m2 tan q3 = m3 tan q4 ; So, combining the two eq. we get, m1 tan q1 = m3 tan q4 Thus, q4 will be independent of m2 only.
SOL 5.1.34
Option (C) is correct. Internal inductance of a loop of radius r is defined as -7 -2 m Lin = 0 (2pr) = 4p # 10 # 2px # 50 # 10 8p 8p = 157.1 nH
SOL 5.1.35
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SOL 5.1.32
(r = 50 cm )
Option (A) is correct. From the analogy between electrical and magnetic circuits, we have the following relations, F (magnetomotive force) " V (voltage) f (magnetic flux) " I (current)
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Shop Online at www.nodia.co.in R (Reluctance) " R (Resistance) now, magnetomotive force, F = NI 0 and so, the electrical analog of the magnetic circuit is
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Option (A) is correct. For drawing the electrical analog replace the coil by a source (magnetomotive force) and each section of the core by a reluctance. In the shown magnetic material there are 9 sections so we draw the reluctance for each of them and we get the magnetomotive force as (N = 1000 ) F = 1000I So the equivalent circuit is
SOL 5.1.37
Option (B) is correct. Given that, the magnetic flux density, B = 0.4 T no. of turns of coil, N = 200 length of magnetic core, l = 15 cm = 15 # 10-2 permeability of the core, mr = 150 So, current required to produce the given magnetic field is ^0.4h^15 # 10-2h i = Bl = mN ^150m0h^200h = 2.6 A
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SOL 5.1.36
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SOLUTIONS 5.2
Option (A) is correct. Consider the particle carries a total charge Q . Since for a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F = Q 6E + (v # B)@ where E " electric field v " velocity of the charged particle B " magnetic flux density So initially the magnetic force on the particle will be zero as the particle is released at rest (v = 0 ). Therefore the electric field will accelerate the particle in y -direction and as it picks up speed (consider the velocity is v = kay , k is very small) a magnetic force develops which will be given by F = v#B since the magnetic field is in az direction while the beam has the velocity in ay direction so the magnetic force will be in ax (ay # az ) direction. Therefore the magnetic force will pull the charged particle around to the right and as the magnetic force will be always perpendicular to both the velocity of particle and electric field. So the particle will initially goes up in the y -direction and then following a curve path lowers down towards the x -axis.
SOL 5.2.2
Option (A) is correct. For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F = Q 6E + (v # B)@ where E " electric field v " velocity of the charged particle
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SOL 5.2.1
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B " magnetic flux density Since initially the velocity of the charge (at the time of injection) is v 0 = 2ay m/s and for the region y > 0 magnetic flux density is B = 3ax wb/m2 . so there will be no any velocity component in + ax direction caused by the field (since the magnetic field is in ax direction). i.e. vx = 0 So we consider the velocity of the point charge in the region y > 0 at a particular time t as v = vy ay + v z a z Therefore we have the force applied by the field on the charge particle at time t as F = Q 6^vy ay + vz az h # ^3ax h@
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dv m ; y ay + dvz azE = Q 6- 3vy az + 3vz ay@ dt dt dvy 3Q So, we get = v m z dt dvz =- 3Q v and m y dt From the two relations we have d 2 v z + 3Q 2 v = 0 bm l z dt2 3Q 3Q vz = A1 cos b t + B1 sin b t m l m l where A1 and B1 are constants. and since at t = 0 , vz = 0 (since charge was injected with a velocity in ay direction) Putting the condition in the expression we get A1 = 0 3Q and so we have t = B1 sin t Q = 2 C , m = 6 kg vz = B1 sin b m l dvz =- 3Q v Again, m y dt 3q so vy =-b m ldvz =- B1 cos b t l =- B1 cos t m 3Q dt and since at t = 0 , vy = 2 m/s Putting the condition in the expression we get, 2 =- B1 cos 0 B1 =- 2 So, we have, vz =- 2 sin t " dz =- 2 sin t dt dy vy = 2 cos t " = 2 cos t dt Solving the equations we get, z = 2 cos t + C2 and y = 2 sin t + C 3 and since at t = 0 , y = z = 0 (charge is located at origin at the time of injection) GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in Putting the condition in the expression we get C2 =- 2 and C 3 = 0 So we have z = 2 cos t - 2&z + 2 = 2 cos t and y = 2 sin t Therefore the equation of the path that the charged particle will follow is y2 + (z + 2) 2 = 4 This is the equation of a circle centred at (0, 0, - 5). Option (A) is correct.
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SOL 5.2.3
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The magnetic flux density produced at a distance r from an infinitely long straight wire carrying current I is defined as mI B = 0 2pr So the magnetic flux density produced by the straight wire at side QR of the loop is (direction of magnetic flux density is determined by right hand rule) mI ( r = 3) BQR = 0 1 az 2p (3) 5m ( I1 = 5 A ) = 0 az 6p Force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF = Idl # B where I is the current flowing in the element and dl is the differential vector length of the current element in the direction of flow of current. So the force exerted by wire on the side QR of the square loop is FQR =
R
# I dl # B Q
2
QR
where I2 is the current flowing in the square loop as shown in the figure. So, we get 4 5m a ( I2 = 3 A, dl = dxax ) FQR = (3dxax ) # b 0 z l 6p x=1 -7 5m = 0 [4 - 1] (- ay) = - 5 # 4p # 10 # 3 ay 2p 2p -6 =- 3 # 10 ay N GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
#
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FSP =
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# I dl # B S
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For View Only Shop Online at www.nodia.co.in SOL 5.2.4 Option (C) is correct. Total force on the loop will be the vector sum of the forces applied by the straight wire on all the sides of the loop. The forces on sides PQ and RS will be equal and opposite due to symmetry and so we have FPQ + FRS = 0 Therefore the total force exerted on the conducting loop by the straight wire is (1) Ftotal = FQR + FSP where FQR and FSP are the forces exerted by the straight wire on the sides QR and SP of the conducting loop respectively. As calculated in previous question we have FQR =- 3 # 10-6 ay N Similarly we get the force exerted by the wire on the side SP of the loop as
Option (A) is correct.
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SOL 5.2.5
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where BSP is the magnetic flux density produced by the wire on the side SP . So, we get mI ( r = 1) BSP = 0 1 az 2p (1) 5m ( I1 = 5 A ) = 0 az 2p 4 5m ( I2 = 3 A, dl =- dxax ) 3 (- dxax ) # 0 az FSP = 2p 1 = 9 # 10-6 ay N Thus, from equation (1), the total force exerted by the straight wire on the conducting loop is Ftotal =- 3 # 10-6 ay + 9 # 10-6 ay = 12 # 10-6 ay N
Net magnetic flux density arising from the two current filaments - 5ax and 5ax A GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in at the location of third filament is given by (1) B = B1 + B 2 where B1 and B2 are the magnetic flux density produced by the current filaments 5ax and - 5ax respectively. Since the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as mI B = 0 af 2pr and the direction of the magnetic flux density is given as af = al # a r where al is unit vector along the line current and a r is the unit vector normal to the line current directed toward the point P . So, the magnetic flux density produced by the current filament 5ax is ka y - a z 5m0 a oH 2 > x #e 2p ( 1 + k ) 1 + k2 5m0 (kaz + ay) = 2p (1 + k 2) Similarly the magnetic flux density produced by the current filament (- 5ax ) is m0 # (5) ka + a z B2 = (- ax ) # e y oH 2 > 2p ( 1 + k ) 1 + k2 5m0 (- kaz + ay) = 2p (1 + k 2) Therefore from equation (1), we get the net magnetic flux density experienced by the third filamentary current of 10 az A as 5m0 B = 6kaz + ay - kaz + ay@ 2p (1 + k 2) 5m0 5m0 ay = 2 (2ay ) = 2p (1 + k ) p (1 + k 2) As the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF = Idl # B where I is the current flowing in the element and dl is the differential vector length of the current element in the direction of flow of current. Force per unit meter length experienced by the third filament is 1 5m0 (dl = dxax ) (10ax dx) # ay F = p (1 + k 2) x=0 10-7 a = 10 # 5 # 4p # z 2 p (1 + k ) = 20az 2 mN (1 + k ) or, F = 12 2 mN (1 + k ) Thus, the graph between F and k will be as shown in the figure below :
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#
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Option (B) is correct. Consider the strip is formed of many adjacent strips of width dz each carrying current Kdz .
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SOL 5.2.6
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Since the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as mI B = 0 2pr So the magnetic flux density produced by each differential strip is m (Kdz) (I = Kdz ) ay dB = 0 2pz (using right hand rule we get the direction of the magnetic flux density along ay ) Therefore the net magnetic flux density produced by the strip on the current filament is 1.5 3m a 3m 0 y ( K = 3 A/m ) dz = 0 ln b 1.5 l ay B = 2 2p 0.5 p z z = 0.5 = 6.6 # 10-7 ay wb/m2 As the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF = Idl # B where I is the current flowing in the element and dl is the differential vector length
#
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For View Only Shop Online at www.nodia.co.in of the current element in the direction of flow of current. So the force exerted on the filament per unit length is F =
# Idl # B = #
1
x=0
(10dx ax ) # (6.6 # 10-7 ay)
= 2.4az mN/m Option (C) is correct. Consider the strip as made up of many adjacent strips of width dy , each carrying current Kdy
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SOL 5.2.7
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Since the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as mI B = 0 2pr So the magnetic flux density produced at distance y from the current filament located along z -axis as shown in the figure will be mI B = 0 (- ax ) (Direction is determined using right hand rule) 2p y 10m0 =a 2py x As the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF = Idl # B and since the length of strip is l = 2 m so, the force exerted on the width dy of strip is given by dF = l (Kdy) # B Therefore the net force exerted on the strip is 3 10m0 (l = 2 m, K = 6az ) F = (2) (6az ) # ca dy 2p y x m y=1 60m0 a ln y 3 =p y 6 @1 =- 13.4ay mN
#
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For View Only Shop Online at www.nodia.co.in SOL 5.2.8 Option (A) is correct. Consider a rectangular Amperian loop of dimension (l) # (2y) inside the slab as shown in the figure below.
As from the Amperes circuital law, we have
# H : dl
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= Ienc So for the Amperian loop inside the slab we get for - a # y # a H (2l) = (2y # l) (J 0) (Net magnetic field intensity along the edge 2y will be cancelled due to symmetry) Therefore the magnetic field intensity (magnetizing factor) at any point inside the slab is H = J 0 yaz or (for y # a ) H = J0 y and the magnetic field intensity (magnetizing factor) at any point outside the slab is (for y > a ) H = J0 a Thus, the plot of H versus y will be as shown below
SOL 5.2.9
Option (A) is correct. Force on any dipole having moment m due to a magnetic flux density B is defined as F = d (m : B) Since the magnetic moment of the dipole is given as
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Shop Online at www.nodia.co.in (1) m = m 0 ax and as calculated in previous question the magnetic field intensity produced due to the slab is H = J 0 yaz So we get the magnetic flux density produced due to the slab as (2) B = m0 H = m0 J 0 yaz Therefore from equation (1) and (2) we get m:B = 0 Thus the force acting on the dipole is F =0
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Option (A) is correct. Volume current density inside a magnetic material is equal to the curl of its magnetization M i.e. J = d#M So volume current density due inside the circular cylinder is J = 1 2 r (5r2) az = 15raz r 2r or J\r
SOL 5.2.11
Option (B) is correct. As calculated above the volume current density inside the cylinder is J = 15raz So, we can get the flux density by Ampere’s circuital law as
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SOL 5.2.10
# B : dl
= m0 Ienc
(B) (2pr) = m0 Ienc m B = 0 Ienc 2pr Now the enclosed current in the loop is Ienc =
# S
J : dS =
r
r3 r (15r) (rdrdf) = 2p # 15 ; E 3 0 0
# # 0
2p
3
= 10pr So, the magnetic flux density inside the cylinder is m ( I = 10pr3 ) B = 0 Ienc = 5m0 r2 2pr Thus the plot of magnetic flux density B versus r is as shown below
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For View Only Shop Online at www.nodia.co.in SOL 5.2.12 Option (B) is correct. The surface current density of a material in terms of its magnetization is defined as where an is unit vector normal to the surface. K = M # an So, the surface current density of the cylinder is (M = Maz , an = a r ) K = (Maz ) # (a r) = Maf Therefore the surface current density is directed along af as shown in option (B). Option (A) is correct. Magnetic flux density inside a magnetic material is defined as B = m0 (H + M) So, B and M will be in same direction inside the cylinder. Now as the magnetic field lines are circular so outside the cylinder it will make a loop. Thus, the magnetic field lines will be as shown below
SOL 5.2.14
Option (B) is correct.
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SOL 5.2.13
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L
= m0 Ienc
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For View Only Shop Online at www.nodia.co.in Let the circular ring being placed such that magnetization M is in af direction and the ring is centered at origin. So, we have M = 4af As the surface current density of a material in terms of its magnetization is defined as where an is unit vector normal to the surface. K = M # an So the surface current density of the ring is (M = 4af , an =- a r ) K = 4af # (- a r) = 4az and since the volume current density inside a material is equal to the curl of magnetization M i.e. J = d#M So the volume current density inside the ring is ( M = 4a f ) J = d # (4af) = 0 Now from Ampere’s circuital law we have
SOL 5.2.15
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and for determining the field inside the circular ring, the current present on the inner surface of ring will be considered only. So we get (B) (2pr) = m0 (K) (2pr) Therefore the magnetic flux density inside the circular ring is ( K = 4 A/m ) B = (m0) (4) = 4m0 wb/m2 Alternate Method : Magnetic flux density inside a magnetic material is defined as B = m0 M and since the magnetization of the rod is M = 4 A/m so, we can have directly the magnetic flux density inside the ring as B = 5m0 Wb/m2 Option (A) is correct. As calculated above for the complete circular ring, magnetic flux density inside the ring is B = 4m0 af wb/m2 (magnetic flux density will be directed along the assumed direction of magnetization)
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For View Only Shop Online at www.nodia.co.in Now we calculate the flux density contributed by the gap at its centre when it was the complete ring. The gap has its cross section in form of a square loop as shown in figure below
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As calculated in previous question the surface current density of the ring is K = 4 A/m and since the width of the gap(square loop) is w so, net current in the loop is I = Kw = 4w Now the magnetic flux density at any point P due to a filamentary current I is defined as H = I 6cos a2 - cos a1@ af 4pr where r " distance of point P from the current filament. a1 " angle subtended by the lower end of the filament at P . a2 " angle subtended by the upper end of the filament at P . So the flux density at center of the square loop produced due to one side of the loop is m0 I 2 ( r = 1 cm, a1 = 135c, a2 = 45c) Bsq1 = -2 # c 4p # (10 ) 2m Summing the flux density produced due to all the four sides of loop, we get total magnetic flux density produced by the square loop as 2 m0 (4w) mI 2 2 (I = Kw ) Bsq = 4 # f 0 -2 p = # 10 p 4p (10 ) 2 m0 # (4) (0.1 # 10-3) # 102 (w = 0.1 mm ) = p -2 = 4 2 # 10 m0 af p Therefore at the centre of the gap the net magnetic flux density will reduce by this amount of the flux density. Thus at the centre of the gap the net magnetic flux density at the centre of the loop will be Bnet = B - Bs -2 = 4m0 - 4 2 # 10 m0 p -2 = m0 c 4 - 4 2 # 10 m p -7 = 50.04 # 10 wb/m2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only SOL 5.2.16 Option (D) is correct.
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Let the magnetized sphere be of radius r , centered at origin and the magnetization be M in az direction as shown in figure. Volume current density inside a material is equal to the curl of magnetization M i.e. J = d#M So the volume current density inside the cylinder is J = d # (Maz ) = 0 and since the surface current density in terms of magnetization is defined as where an is unit vector normal to the surface. K = M # an So the surface current density on the sphere is (an = ar ) K = ^Maz h # (ar ) ...(i) = M sin qaf Now, consider a rotating spherical shell of uniform surface charge density s, that corresponds to a surface current density at any point (r, q, f). So we have ...(ii) K = swR sin qaf where w " angular velocity of spherical shell across z -axis R " radius of the sphere. and the magnetic flux density produced inside the rotating spherical shell is defined as ...(iii) B = 2 m0 swR 3 Comparing the eq.(i) and eq.(ii) we get M = swR Putting this value in eq.(iii) we get the magnetic flux density for the magnetized sphere as (M = swR) B = 2 m0 M 3 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in SOL 5.2.17 Option (B) is correct. Since the magnetic flux density inside a magnetic material is defined as B = m0 (H + M)
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Option (B) is correct. The magnetic flux density produced at any point P due to an infinite filamentary current I is defined as mI B = 0 2pr where r is the distance of point P from the infinite current filament. Now consider a small area dS of the coil located at a distance x from the current filament. The magnetic flux density produced on it due to the current filament along y -axis is mI (r = x ) B = 0 2p x Since the flux density will be normal to the surface of the coil as determined by right hand rule therefore, the total magnetic flux passing through the coil is 6 1 m0 I m0 I ym = B : dS = b 2px l^dxdy h = 2p ln 3 y=0 x=2 As the mutual inductance in terms of total magnetic flux ym is defined as Nym M = I where I " current flowing in the element that produces the magnetic flux. N " Total no. of turns of the coil that experiences the magnetic flux. Thus the mutual inductance between the current filament and the loop is mI N = 1500 M = 1500 b 0 ln 3 l = 0.33 mH 2p I
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So, we have the magnetic field intensity inside the material as H = 1 B-M m0 and outside the material the magnetic field intensity is H = 1B m0 So the field lines outside the material will be same as for the case of magnetic flux density shown in the MCQ. 1.31. Whereas inside the material the direction of magnetic field intensity will be opposite to the direction of magnetization. Thus the sketch of the field intensity will be same as shown in the option (B).
# #
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#
SOL 5.2.19
Option (A) is correct. Consider the wire is lying along z -axis. So at any point inside the wire (at a distance r < a from it’s axis) magnetic field intensity will be determined as
# H : dl
= Ienc
(Ampere’s circuital law)
pr2 (for Amperian loop of radius r) H (2pr) = I c 2 m pa GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Ir af 2p a 2 The direction of the magnetic field intensity is determined using right hand rule. Now the stored energy in the magnetic field H is defined as Wm = 1 m0 H 2 dv v 2 So the stored energy in the internal magnetic filed per unit length (over the unit length in z -direction) will be 2p 1 a m0 I 2 m0 I 2 r2 = r r f d d dz Wm = 2 2 16p z = 0 f = 0 r = 0 2 (2pa ) Therefore, the energy per unit length depends only on I and is uniform for the uniform current. H =
or,
#
Option (B) is correct. Since the two conducting plates of width w = 2 m carry a uniform current of I = 4 A each so, the surface current density of each plate is K = I = 4 = 2A/m w 2 Now consider the first plate carrying current in + az direction is located at y = 0 and the second plate carrying current in - az direction is located at y = d , where d is a very small separation between the plates. Since the magnetic field intensity produced at any point P due to an infinite sheet carrying uniform current density K is defined as H = 1 (K # a n) 2 where an is the unit vector normal to the sheet directed toward the point P . So, the magnetic field intensity produced at the second plate due to the first plate is (K1 = 2az , an = ay ) H12 = 1 (2az # ay) =- ax 2 Now the force per meter, exerted on the 2 nd plate due to the 1st plate will be
ww w. ga te
he
lp.
SOL 5.2.20
co
m
# # #
F12 =
where So,
1
2
# # (K 0
0
2
# B12) dS
K2 " current density of the 2 nd plate B12 " magnetic flux density produced at the 2 nd plate due to 1st plate 1
2
# # (- 2a ) # (m H ) dydz = # # (- 2a ) # (- m a ) dydz
F12 =
0
1
0
z
0
0
12
2
0
z
0
x
(K2 =- 2az , B12 = m0 H12 ) = 4m0 ay
As the force applied by first plate on the 2nd plate is in ay direction so it is a repulsive force. Therefore the repulsive force between the plates is 4m0 . SOL 5.2.21
Option (D) is correct. No, of turns, Relative permeability,
n = 20, 000 turns/meter mr = 100
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For View Only Shop Online at www.nodia.co.in 2 cross sectional area, S = 0.04 m Current in the solenoid, I = 100 # 10-3 A So, its self inductance will be, Ll = m0 mr n2 S = (4p # 10-7) # (100) # (20, 000) 2 # (0.04) = 2.011 # 103 Therefore the energy stored per unit length in the field is W lm = 1 LlI 2 = 1 # 2.011 # 103 # 10-2 = 10.05 J/m 2 2
m
Option (A) is correct. Consider the path followed by the two particles are the curvatures having radii r , and r2 as shown in figure. So at balanced condition centrifugal force will be equal to magnetic force. Therefore for the first charged particles mv2 = Bqv & r = mv 1 r1 Bq 2 ^2m h v and = Bqv & r2 = mv r2 Bq
at e
he
lp.
co
SOL 5.2.22
Option (A) is correct. The wire is oriented in east-west direction and magnetic field is directed northward as shown in the figure.
ww
SOL 5.2.23
w. g
So the distance between the two particles at releasing end is d = 2r2 - 2r1 = 2 b 2mv l - 2 b mv l = 2mv Bq Bq Bq
Since the direction of gravitational force will be into the paper(toward the earth) so for counteracting the gravitational force, applied force must be outward. Now the force experienced by a current element Idl in a magnetic field B is GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 5
Magnestostatic Fields in Matter
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# ^Idl h # B
323
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L
m
As the magnetic field B is directed toward north therefore, using right hand rule for cross vector we conclude that for producing the outward force current must flow from west to east as shown in the figure below.
Since
Option (A) is correct. Consider the current flowing in the wire is I . So the magnetic force applied by the field B 0 on the wire is where L is length of the wire Fm = ILB 0 At balanced condition the magnetic force will be equal to the gravitational force : Fm = mg where m is the mass of the wire and g is acceleration due to gravity. So comparing the two results we get the current flowing in the wire as mg I = LB 0 Since B 0 = 0.6 # 10-4 Wb/m2 , m = 0.3 kg and L = 1 m ^0.3h # 9.8 Therefore ( g = 9.8 m/s ) I = = 49 kA ^1 h # ^0.6 # 10-4h
SOL 5.2.25
Option (B) is correct. Consider the square loop has side a . Now, when the loop is situated in the field B = 1.96 Wb/m2 . Suppose it swings with an angle a. So in the new position the torque must be zero. Gravitational forces acting on all the sides of loop will be down wards and the force due to magnetic field will be in horizontal direction as shown in the figure.
ww w. ga te
he
lp.
co
SOL 5.2.24
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For View Only Shop Online at www.nodia.co.in So, in balanced condition,from the shown figure we have a mg sin a ^a h + 2a mg sin a a a k = IBa 2 ^2 h^1.96h tan a = IB = 2mg ^2 h^0.2h^9.8h a = tan-1 ^1 h = p/4 Option (B) is correct. As discussed in Que 51 the path of electron will be parallel to the input beam but in opposite direction. So the ejected electrons will be flowing in the - ay direction.
SOL 5.2.27
Option (D) is correct. At any point in between the two parallel shuts the net magnetic flux density produced by the two sheets is given as B = B1 + B 2 where B1 is the flux density produced by the lower sheet and B2 is the flux density produced by upper sheet. Now the magnetic flux density produced at point P due to a plane sheet having current density K is defined as m B = K # an 2 where an is the unit vector normal to the sheet and directed toward point P . So, the flux density produced by lower sheet is m ( K = 4a x , a n = a z ) B1 = ^4ax h # az 2 and the flux density produced by the lower sheet is m (K =- 4ax , an =- az ) B2 = ^- 4ax h # (- az ) 2 So the net magnetic flux density produced in the region between the two sheets is m m B = ^4ax h # az + ^- 4ax h # ^- az h 2 2 =- 4may where m is the permeability of the medium. Therefore the flux density in region 1 is ( m1 = 2m0 ) Bregion 1 =- 4m1 ay =- 8m0 ay and the flux density in region 2 is ( m2 = 4m0 ) Bregion 2 =- 4m2 ay =- 16m0 ay So the net flux per unit length in the region between the two sheets is f = ^Bregion 1h (width of region 1) + ^Bregion 2h (width of region 2) l = ^- 8m0 ay h^1 h + ^- 16m0 ay h^2 h =- 50m0 ay Wb/m
SOL 5.2.28
Option (A) is correct. As the permeability of the medium varies from m1 to m2 linearly. So at any distance
ww
w. g
at e
he
lp.
co
m
SOL 5.2.26
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325
For View Only Shop Online at www.nodia.co.in z from one of the plate near to which permiability is m1 , the permeability is given as ^m2 - m1h (1) m = m1 + z d The magnetic flux density between the two parallel sheets carrying equal and opposite current densities is defined as B = mK where K is the magnitude of the current density of the sheets. Therefore the flux per unit length between the two sheets is d f = Bdz where d is the separation between the two sheets. l 0 d d ^m2 - m1h (from equation (1)) = mKdz = K ;m1 + z E dz d 0 0 m + m2 m - m1 z2 d = Kb 1 d = K ;m1 z + 2 2 l d b 2 lE0
#
co
#
m
#
Option (B) is correct. Given the field intensity inside the slab is H = 4ax + 2ay So the magnetic flux density inside the slab is given as where m is the permeability of the material. B = mH (m = 2m0 ) = 2m0 ^4ax + 2ay h Therefore the magnetization of the material is M = B -H m0 = 8ax + 4ay - ^4ax + 2ay h = 4ax + 2ay Now the magnetization surface current density at the surfaces of a magnetic material is defined as Km = M # an where an is the unit vector normal to the surface directed outward of the material So, at z = 0 magnetization surface current density is (an =- az ) 6Km@at z = 0 = M # ^- az h = 4ay - 2ax and at z = d , the magnetization surface current density is So, (an = az ) 6Km@at z = d = M # ^az h = ^4ax + 2ay h # ^az h =- 4ay + 2ax
SOL 5.2.30
Option (A) is correct. As calculated in the previous question the magnetization vector of the material is M = 4a x + 2 a y The magnetization volume current density inside a magnetic material is equal to the curl of magnetization., i.e. Jm = d # M Therefore the magnetization volume current density inside the slab is
ww w. ga te
he
lp.
SOL 5.2.29
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Shop Online at www.nodia.co.in a x ay a z Jm =
2 2y
2 2z
=0
4 2 0
Option (B) is correct. Given the B -H curve for the material, B = 2m0 HH The work done per unit volume in magnetizing a material from 0 to B 0 that has non uniform permeability is defined as wm =
B0
# H : dB 0
and
#
0
H0
3 H0
H : ^4m0 H h = 4m0 ;H E = 3 0
3m0 H 03 4
he
Option (A) is correct. As 7 shows the direction into the paper while 9 shows the direction out of the paper. So the wire of length l carries current 2I that flows out of the paper. The Magnetic field intensity produced at a distance r from an infinite straight wire carrying current I is defined as H = I 2pr So the magnetic field intensity produced at the top wire due to the infinite wire carrying current inward is I (r = 2 l ) Hf1 = 2p ^ 2 l h
ww
w. g
at e
SOL 5.2.32
wm =
lp.
co
Now for determining dB . we can express B = 2m0 H 2 aH where aH is the unit vector in direction of H . dB = 4m Ha So, 0 H dH = 4m0 H
m
SOL 5.2.31
2 2x
and the magnetic field intensity at top wire due to the infinite wire carrying current outward is I (r = 2 l ) Hf2 = 2p ^ 2 l h GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in Therefore the resultant field intensity at the wire of length l is HT = ^Hf1 + Hf2h cos q 2I 1 = I = # 2pl 2p ^ 2 l h 2 Since the force exerted on a current element Idl by a magnetic field H is defined as dF = (mH) (Idl) So the force experienced by the wire of length l is mI 2 F = (mHT ) ^2I h l = m b I l^2I h l = 2p 2p l
m
Option (C) is correct. Consider the sheets as shown in figure that having the surface current densities + 20 mA/may and - 20 mA/may
ww w. ga te
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lp.
co
SOL 5.2.33
So the field intensity between the plates will be given as H = H1 + H 2 where H1 is the field intensity produced by the sheet located at x = 0 and H2 is the field intensity produced by the sheet located at x = 5 cm . Now the magnetic field intensity produced at point P due to a plane sheet having current density K is defined as H = 1 K # an 2 where an is the unit vector normal to the sheet directed toward point P . So, the magnetic field intensity in the region between the plates is H = 1 K1 # an1 + 1 K2 # an2 2 2 1 44at2 44 3 1 44 2 44 3 x=0 at x = 5 cm = 1 ^20 # 10-3 ay h # ^ax h + 1 ^- 20 # 10-3 ay h # ^- ax h 2 2 -3 =- 20 # 10 az and magnetic flux density in the region between the sheets is B = mH =- 40m0 # 10-3 az ^m = 2m0h Therefore the stored magnetic energy per unit volume in the region is GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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wm = 1 H : B = 1 ^800m0 # 10-6h 2 2 = 400 # 4p # 10-7 # 10-6 = 160p J/m3 Since the separation between plates is d = 5 cm . So, stored energy per unit area between the plates is Wm /A = wm # d = ^160ph # ^0.05h = 4p J/m2 Option (C) is correct. Since the boundary surface of the two medium is z = 0 , so the normal component B1n and tangential component B1t of magnetic flux density in medium 1 are B1n = az and B1t = 0.4ax + 0.8ay As the normal component of magnetic flux density is uniform at the boundary of two medium So, the normal component of magnetic flux density in the medium 2 is (1) B2n = B1n = az Now for determining tangential component of field in medium 2, we first calculate tangential component of magnetic field intensity in medium 1 which is given as where m1 is the permeability of medium 1. H1t = B1t m1 0.1ax + 0.2ay ( m1 = 4m0 ) = 1 ^0.4ax + 0.8ay h = m0 4m0 Again from the boundary condition the tangential component of magnetic field intensity in the two mediums are related as an # ^H1t - H2t h = K where H2t and H1t are the tangential components of magnetic field intensity in medium 2 and medium 1 respectively, K is the surface current density at the boundary interface of the two mediums and an is the unit vector normal to the boundary interface. So we have 0.1ax + 0.2ay az # < - ^H2tx ax + H2ty ay hF = 1 ^0.2ax - 0.4ay h m0 m0 0.1 0.2 1 b m0 - H2tx l ay - b m0 - H2ty l ax = m0 ^0.2ax - 0.4ay h Comparing the x and y -components we get H2tx = 0.1 + 0.4 = 0.5 m0 m0 m0 0 . 2 0 . 2 0 and + = .4 H2ty = m0 m0 m0 Therefore the tangential component of magnetic field intensity in medium 2 is H2t = 0.5 ax + 0.4 ay m0 m0 and the tangential component of magnetic flux density in medium 2 is B2t = m2 H2t = ax + 0.8ay Thus the net magnetic flux density in medium 2 is B2 = B2t + B2n = ax + 0.8ay + az
ww
w. g
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lp.
co
m
SOL 5.2.34
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m
For View Only Shop Online at www.nodia.co.in SOL 5.2.35 Option (B) is correct. Consider the square loop is of side 2a as shown in the figure
SOL 5.2.36
ww w. ga te
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lp.
co
Since the sides BC and AD crosses the straight wire so no force will be experienced by the sides, while the flux density produced by the straight wire at sides AB and CD will be equal in magnitude. Now the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as mI B = 0 2pr So the magnetic flux density produced by the straight wire at the two sides of the loop is m0 ^2 h m = 0 B = ^I = 2 A, r = a h pa 2p ^a h Since the force exerted on a current element Idl by a magnetic field B is defined as dF = (Idl) # B Therefore the force experienced by side AB of length 2a is 8m m (I = 4 A ) F1 = 64 ^2a h az@ # 9 0 axC = 0 ^- ay h pa p Similarly force experienced by side CD is 8m m F2 = 64 ^2a h^- az h@ # 9 0 ^- ax hC = 0 ^ay h p pa Thus the net force experienced by the loop is 16m0 F = F1 + F2 = a p ^ yh = 16 # 4 # 10-7 ay = 2.4ay mN Option (A) is correct. According to Snell’s law the permeability of two mediums are related as m0 tan q1 = m tan q2 tan q1 = 15m0 m0 tan q2 tan q1 = 15 tan q2 Now, the given flux density in medium 2 is B2 = 1.2ay + 0.8az
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...(i)
330
Magnestostatic Fields in Matter
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Option (B) is correct. For calculating total reluctance of the circuit, we have to draw the electrical analog of the circuit. In the given magnetic circuit, there are total six section for which six reluctance has been drawn below.
w. g
at e
he
SOL 5.2.37
lp.
co
m
For View Only Shop Online at www.nodia.co.in So the normal and tangential component of the magnetic flux density in medium 2 is B2n = 0.8az and B2t = 1.2ay From the figure we have tan q2 = B2n = 0.8 = 2 1.2 3 B2t -1 or q2 = tan ^2/3h from equation (1) tan q1 = 15 tan q2 tan q1 = 10 q = tan-1 ^10h Thus the angular deflection is q1 - q2 = tan-1 ^10h - tan-1 ^2/3h = 54.6c
ww
For a given cross sectional area S and length of the core l reluctance is defined as R = l mS Where m is permeability of the medium in core 5 # 10-2 So, we have R1 = = 1 ^1000m0h^5 # 10-4h 10m0 5 # 10-2 1 R2 = -4 = 20m 0 ^1000m0h^10 # 10 h 6 # 10-2 = 3 ^1000m0h^10 # 10-4h 50m0 14 # 10-2 = 7 R4 = ^1000m0h^10 # 10-4h 50m0 R5 = R3 = 3 50m0 4 # 10-2 = 1 R6 = ^1000m0h^10 # 10-4h 25m0 R3 =
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331
co
m
For View Only Shop Online at www.nodia.co.in Since all the reluctance are connected in series so total reluctance of the magnetic circuit is RT = R 1 + R 2 + R 3 + R 4 + R 5 + R 6 = 1 + 1 + 3 + 7 + 3 + 1 10m0 20m0 50m0 50m0 50m0 25m0 = 9 20m0 SOL 5.2.38 Option (C) is correct. For a given reluctance R of a magnetic circuit, the self, inductance is defined as 2 Where N is no. of turns of coil L =N R ^100h2 Then, L = cRT = 209m m 0 ^9/20m0h -2 = 2.79 # 10 = 27.9 mH Option (B) is correct. Give that no. of turns of coil, N = 50 length of the core, l = 0.6 m relative permeability, mr = 600 inductance of the coil, L = 0.2 mH = 0.2 # 10-3 H So, the cross sectional area of core is ^0.2 # 10-3h^0.6h S = Ll 2 = mN ^600m0h^50h2 = 6.366 # 10-5 m2 = 1.64 cm2
SOL 5.2.40
Option (B) is correct. Since the core is ideal so it’s reluctance will be zero and so the electrical analog for the magnetic circuit will be as shown below
ww w. ga te
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lp.
SOL 5.2.39
The reluctance R 1 , R 2 and R 3 is produced by the air gap. 4 # 10-2 R 1 = l1 = = 4 m0 S1 m0 ^100 # 10-4h m0 2 # 10-2 R2 = = 2 m0 ^100 # 10-4h m0 2 # 10-2 = 2 R3 = m0 ^100 # 10-4h m0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Option (D) is correct. Since the coil N1 and N2 are directly connected through ideal core so entire flux produced by N2 will link with N1 . The electrical analog of the magnetic circuit is shown below where the reluctance R 1 and R 2 are the reluctance due to air gap.
w. g
at e
he
SOL 5.2.42
lp.
co
m
For View Only Shop Online at www.nodia.co.in So, the total reluctance seen by coil N1 is RT = R 1 + R 2 || R 3 = 4 + 1 = 5 m0 m0 m0 and the self inductance of coil will be 2 L1 = N 1 = 62.8 mH RT SOL 5.2.41 Option (B) is correct. The total reluctance of the magnetic circuit as seen from the coil N2 is RT = ^R 1 || R 2h + R 3 (as calculated above) = b 4 || 2 l + 2 m0 m0 m0 = 4 + 2 = 10 3m0 3m0 m0 Therefore the self inductance of the coil N2 is 2 ^250h2 = 28.6 mH L2 = N 2 = RT ^10/3m0h
ww
So, the reluctance seen by coil N2 is ^4 # 10-3h = 2 R1 = l = m0 m0 S m0 ^2000 # 10-6h Consider the current flowing in coil N2 is i2 . So, the total flux produced by N2 i2 is f2 = N2 i2 = 500i2 = 250m0 i2 R1 ^2/m0h Since the entire flux will link with N1 So mutual induction between N1 and N2 is ^250h^250m0 i2h Nf M = L12 = 1 2 = = 78.54 mH i2 i2 SOL 5.2.43
Option (A) is correct. As the coil N1 and N2 are directly connected through an ideal core so entire flux will produced by N2 will link with N1 and so flux linked with N 3 will be zero.
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For View Only Shop Online at www.nodia.co.in Therefore the mutual inductance between N 3 and N2 is zero. SOL 5.2.44
Option (A) is correct. Given, the expression for magnetization curve, B = 1 H + H 2 mWb/m2 3 The energy stored per unit volume in a magnetic material having linear magnetic flux density is defined as wm =
H0
# H : dB H=0
wm =
m
Since, magnetic field intensity varied from 0 to 210 A/m So, we have 210
# HdB H=0
lp.
he
#
co
dB = 1 + 2H Since, 3 dH So, putting it in equation we get, 210 wm = H b 1 + 2H l dH 3 H=0 2 3 210 = :H + 2H D 6 3 0 = 6.18 # 106 J/m3 = 6.2 MJ/m3
ww w. ga te
***********
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SOLUTIONS 5.3
Option (A) is correct. From boundary condition we have the following relation between the magnetic field intensity in the two mediums : (1) m1 H1n = m2 H2n (2) and (H1 - H2) # an12 = K where H1 and H2 are the magnetic field intensity in the two mediums, an12 is the unit vector normal to the interface of the mediums directed from medium 1 to medium 2 and K is the surface current density at the interface of the two mediums. Now, the magnetic field intensity in medium 1 is H1 = 3ax + 30ay A/m As the interface lies in the plane x = 0 so, we have H1n = 3ax From equation (1), the normal component of the field intensity in medium 2 is given as H2n = H1n = 1.5ax 2 Therefore, the net magnetic field intensity in medium 2 can be considered as (3) H2 = 1.5ax + Aay + Baz where A and B are the constants. So, from equation (2) we have 6(3ax + 30ay) - (1.5ax + Aay + Baz )@ # ax = 10ay 61.5ax + (30 - A) ay - Baz@ # ax = 10ay 0 - (30 - A) az - Bay = 10ay Comparing the components in the two sides we get 30 - A = 0 & A = 30 and - B = 10 & B =- 10 Putting these values in equation (3) we get the magnetic field intensity in medium 2 as H2 = 1.5ax + 30ay - 10az A/m
SOL 5.3.2
Option (B) is correct. Given, the magnetic moment m = 2.5 A- m2 Mass of magnet, mass = 6.6 # 10-3 kg density of steel, density = 7.9 # 103 kg/m3
ww
w. g
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lp.
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m
SOL 5.3.1
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For View Only Shop Online at www.nodia.co.in So, the net volume of the magnet bar is -3 v = mass = 6.6 # 10 3 density 7.9 # 10 -6 = 0.835 # 10 m3 Now, the magnetization of the magnet is defined as the magnetic moment per unit volume so, we get magnetization of the magnet bar as 2. 5 M =m = v 0.835 # 10-6 = 3 # 106 A/m Option (A) is correct. Given,
m
SOL 5.3.3
Option (C) is correct. The electrical analogy of the magnetic field are listed below :
ww w. ga te
SOL 5.3.4
he
lp.
co
Magnetic field intensity, H = 5ax A/m m Current element, Idl = 4 # 10-4 ay A- m So, the magnetic flux density is given as B = mH = 5ax A/m Since, the force exerted on a current element Idl placed in a magnetic field B is defined as F = ^Idl h # B So, putting all the values we get, F = ^4 # 10-4 ay h # ^5ax h =- 2 # 10-3 az N =- 2az mN
Electrical field
Magnetic field
EMF (electromotive force)
" MMF (magneto motive force)
Electric current
" Magnetic flux
Resistance
" Reluctance
Conductivity " Permeability So, for the given match list we get, A " 3 , B " 2 , C " 4 , D " 1. SOL 5.3.5
Option (C) is correct. At the surface of discontinuity (interface between two medium) the normal component of magnetic flux density are related as B1n = B2n i.e. normal component of magnetic flux density is uniform at the surface of discontinuity. Statement 1 is correct At the boundary interface between two mediums, the normal component of the electric flux density is related as i.e. discontinuous D2n - D1n = rs
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For View Only Shop Online at www.nodia.co.in where rS is surface charge density at the interface. If the interface is charge free ( rS = 0 ) then, the equation changes to i.e. continuous D2n = D1n So, the normal component of flux density at the surface of discontinuity may or may not be continuos. Option (C) is correct. Magnetic current is composed of both displacement and conduction components.
SOL 5.3.7
Option (C) is correct. Torque exerted on a loop with dipole moment M in a magnetic field B is defined as T = M#B
SOL 5.3.8
Option (B) is correct. Biot savart’s law gives the magnetic flux density as defined below
# Idl # R
lp.
m0
co
m
SOL 5.3.6
^b " 3h 4p R 2 Displacement current is determined by using maxwell’s equation as d # H = Jc + Jd where Jd is displacement current density ^c " 1h Time average power flow in a field wave is determined by poynting vector as Pave = 1 Es # Hs ^d " 2h 2 ^a " 4h Using Gauss’s law line charge distribution can be determined.
ww
SOL 5.3.10
Option (B) is correct. Magnetic energy density in a magnetic field is defined as wm = 1 J : A 2 Option ( ) is correct. Consider the two wires carrying current as shown below :
w. g
SOL 5.3.9
at e
he
B =
The force exerted due to the wire 2 at wire 1 is given as F = ^Idl h # ^B h where Idl is the small current element of the wire 1 and B is magnetic flux density produced by wire 2 at wire 1. As determined by right hand rule the magnetic flux GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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337
For View Only Shop Online at www.nodia.co.in density produced due to wire 2 at wire 1 is out of the paper. Which will be towards wire 2. In the similar way the force due to wire 1 at wire 2 will be toward wire 1 i.e. attractive and perpendicular to the wire. Option (B) is correct. From the boundary condition for magnetic field, we have the following derived condition as m1 Hn1 = m2 Hn2 and Ht1 = Ht2
SOL 5.3.12
Option (C) is correct. The magnetic flux density B and magnetic field intensity H in a medium with permeability m are related as B = mH = mr m0 H Now, for the different magnetic material relative permeability mr are listed below : Free space (vacuum) mr = 1 Diamagnetic mr K 1 Paramagnetic mr L 1 Ferromagnetic mr >> 1 So, the B -H curve for the respective material has been shown below (depending on their slopes m).
SOL 5.3.13
Option (A) is correct. When a dielectric material is placed in an electric field then the electric dipoles are created in it. This phenomenon is called polarization of the dielectric material. So, we conclude that both the statement are correct and statement (II) is correct explanation of (I).
SOL 5.3.14
Option (B) is correct. For an inhomogenous magnetic material, magnetic permeability is a variable and so, it has some finite gradient. Now, from maxwell’s equation we know d:B = 0 Since, B = mH So, d : B = d : ^mH h 0 = d:m+d:H In the above equation d : m have some finite value therefore,
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SOL 5.3.11
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338
Magnestostatic Fields in Matter
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d:H ! 0 SOL 5.3.15
Chap 5
Option (C) is correct. Force on a current element Idl kept in a magnetic field B is defined as F =
# Idl # B L
= 6^10h^2 h az@ # 60.05ax@ = 1.0ay N Option (C) is correct. Magnetic energy density in a magnetic field is defined as wm = 1 J : A 2 where J is the current density and A is the magnetic vector potential.
SOL 5.3.17
Option (D) is correct. The force on a moving charge q with the velocity v in a region having magnetic field B and electric field E is defined as F = q ^E + v # B h
SOL 5.3.18
Option (B) is correct. The currents in the hairpin shaped wire flows as shown in the figure.
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SOL 5.3.16
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As the direction of current are opposite so the force acting between them is repulsive, and So it tend to a straight line. Option (C) is correct. Given, the Lorentz force equation, F = e ^v # B h If the particle is at rest then v = 0 and so there will be no any deflection in particle due to the magnetic field.
SOL 5.3.20
Option (D) is correct. Force acting on a small point charge q moving in an EM wave is defined as F = qE + q ^v + B h So, for q = 1 F = E+v#B
SOL 5.3.21
Option (B) is correct. Given, Current flowing in the conductor, Magnetic flux density,
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SOL 5.3.19
I = 5A B = 3ax + 4ay
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 5
Magnestostatic Fields in Matter
339
Option (B) is correct. From the boundary condition for magnetic field we have the following relation : Normal component of magnetic flux density is continuous i.e. Bn1 = Bn2 Any field vector is the sum of its normal and tangential component to any surface i.e. H1 = Hn1 + Ht1 When the interface between two medium carries a uniform current K then the tangential component of magnetic field intensity is not uniform. i.e. Ht1 - Ht 2 = K or, an21 # ^H1 - H2h = K
SOL 5.3.23
B2 !
Bn2 + Bt 2
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But,
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SOL 5.3.22
m
For View Only Shop Online at www.nodia.co.in Since, the force experienced by a current carrying element Idl placed in a magnetic field B is defined as dF = ^Idl h # B As the current flowing in az direction so, we have dl = dlaz and the force experienced by the conductor is dF = ^5dlaz h # ^3ax + 4ay h Therefore, the force per unit length experienced by the conductor is dF = 15a - 20a y x dl =- 20ax + 15ay N/m
Option (A) is correct. Given, the magnetic flux density in medium 1 is B1 = 1.2ax + 0.8ay + 0.4az and the interface lies in the plane z = 0 . So, the tangential and normal components of magnetic flux density in the two mediums are respectively : B1t = 1.2ax + 0.8ay and B1n = 0.4az Now, from the boundary condition of current free interface, we have the following relations between the components of field in two mediums. B1n = B2n B1t = B2t m1 m2 Therefore, we get the field components in medium 2 as B2n = B1n = 0.4az m and B2t = B1t b 2 l = 1 ^1.2ax + 0.8ay h = ^0.6ax + 0.4ay h m1 2 Thus, the net magnetic flux density in region 2 is and
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Magnestostatic Fields in Matter
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SOL 5.3.24
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B2 = B2n + B2t = 0.6ax + 0.4ay + 0.4az So, the magnetic field intensity in region 2 is H2 = B2 = 1 ^0.6ax + 0.4ay + 0.4az h A/m m2 m0 Option (C) is correct. Energy stored in a magnetic field is defined as Wm = 1 A : J dv 2
m
#
# A : J dv has the units of energy.
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Chap 5
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
CHAPTER 6 TIME VARRYING FIELD AND MAXWELL EQUATION
342
Time Varrying Field and Maxwell Equation
MCQ 6.1.1
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EXERCISE 6.1
m
For View Only
Match List I with List II and select the correct answer using the codes given below (Notations have their usual meaning) List-II
co
List-I b
Faraday’s law
c
Gauss’s law
d
Non existence of isolated magneticharge d 1 2 4 2
d : D = rv
2.
d:B = 0
3.
d # E =-2B 2t
4.
d # H = J + 2D 2t
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c 2 3 1 1
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b 3 1 3 3
1.
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Ampere’s circuital law
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a
Codes : a (A) 4 (B) 4 (C) 2 (D) 4 MCQ 6.1.2
Chap 6
Magneto static fields is caused by (A) stationary charges
(B) steady currents
(C) time varying currents
(D) none of these
Let A be magnetic vector potential and E be electric field intensity at certain time in a time varying EM field. The correct relation between E and A is (A) E =-2A (B) A =-2E 2t 2t (C) E = 2A (D) A = 2E 2t 2t
MCQ 6.1.4
A closed surface S defines the boundary line of magnetic medium such that the field intensity inside it is B . Total outward magnetic flux through the closed surface will be (A) B : S (B) 0
ww
MCQ 6.1.3
(C) B # S
(D) none of these
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
343
For View Only Shop Online at www.nodia.co.in MCQ 6.1.5 A perfect conducting sphere of radius r is such that it’s net charge resides on the surface. At any time t , magnetic field B (r, t) inside the sphere will be (A) 0 (B) uniform, independent of r (C) uniform, independent of t (D) uniform, independent of both r and t The total magnetic flux through a conducting loop having electric field E = 0 inside it will be (A) 0
m
MCQ 6.1.6
(B) constant
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(C) varying with time only
(D) varying with time and area of the surface both
A cylindrical wire of a large cross section made of super conductor carries a current I . The current in the superconductor will be confined. (A) inside the wire (B) to the axis of cylindrical wire (D) none of these
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(C) to the surface of the wire
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MCQ 6.1.7
If Bi denotes the magnetic flux density increasing with time and Bd denotes the magnetic flux density decreasing with time then which of the configuration is correct for the induced current I in the stationary loop ?
MCQ 6.1.9
A circular loop is rotating about z -axis in a magnetic field B = B 0 cos wtay . The total induced voltage in the loop is caused by (A) Transformer emf (B) motion emf.
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MCQ 6.1.8
(C) Combination of (A) and (B)
(D) none of these
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344
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.1.10 A small conducting loop is released from rest with in a vertical evacuated cylinder voltage induced in the falling loop is (Assume earth magnetic field = 2 # 10-6 T at a constant angle of 10c below the horizontal) (A) zero (B) 1 mV (C) 17.34 mV
A square loop of side 2 m is located in the plane x = 0 as shown in figure. A nonuniform magnetic flux density through it is given as
m
MCQ 6.1.11
(D) 9.8 mV
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B = 4z3 t2 ax ,
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The emf induced in the loop at time t = 2 sec will be
(A) 16 volt (C) 4 volt
(D) - 2 volt
A very long straight wire carrying a current I = 3 A is placed at a distance of 4 m from a square loop as shown in figure. If the side of the square loop is 4 m then the total flux passing through the square loop will be
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MCQ 6.1.12
(B) - 4 volt
(A) 0.81 # 10-7 wb
(B) 10-6 wb
(C) 4.05 # 10-7 wb
(D) 2.0 # 10-7 wb
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Chap 6
Time Varrying Field and Maxwell Equation
345
co
m
For View Only Shop Online at www.nodia.co.in MCQ 6.1.13 A straight conductor ab of length l lying in the xy plane is rotating about the centre a at an angular velocity w as shown in the figure.
(C) Vba is positive
(D) Vba is zero
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In a certain region magnetic flux density is given as B = 0.2t az Wb/m2 . An electric loop with resistance 2 W and 4 W is lying in x -y plane as shown in the figure.
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MCQ 6.1.14
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If a magnetic field B is present in the space directed along az then which of the following statement is correct ? (A) Vab is positive (B) Vab is negative
If the area of the loop is 2 m2 than, the voltage drop V1 and V2 across the two resistances is respectively (A) 66.7 mV and 33.3 mV (B) 33.3 mV and 66.7 mV (C) 50 mV and 100 mV MCQ 6.1.15
(D) 100 mV and 50 mV
Assertion (A) : A small piece of bar magnet takes several seconds to emerge at bottom when it is dropped down a vertical aluminum pipe where as an identical unmagnetized piece takes a fraction of second to reach the bottom. Reason (R) : When the bar magnet is dropped inside a conducting pipe, force exerted on the magnet by induced eddy current is in upward direction. (A) Both A and R are true and R is correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false.
(D) A is false but R is true. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
346
Time Varrying Field and Maxwell Equation
Chap 6
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For View Only Shop Online at www.nodia.co.in MCQ 6.1.16 A magnetic core of uniform cross section having two coils (Primary and secondary) wound on it as shown in figure. The no. of turns of primary coil is 5000 and no. of turns of secondary coil is 3000. If a voltage source of 4 Volts is connected across the primary coil then what will be the voltage across the secondary coil ?
(A) 72 volt
(B) 7.2 volt
(D) - 7.2 volt
MCQ 6.1.17
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(C) 20 volt
Self inductance of a long solenoid having n turns per unit length will be proportional to (A) n (B) 1/n
A wire with resistance R is looped on a solenoid as shown in figure.
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MCQ 6.1.18
(D) 1/n2
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(C) n2
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If a constant current is flowing in the solenoid then the induced current flowing in the loop with resistance R will be (A) non uniform (B) constant (C) zero MCQ 6.1.19
(D) none of these
A long straight wire carries a current I = I 0 cos (wt). If the current returns along a coaxial conducting tube of radius r as shown in figure then magnetic field and electric field inside the tube will be respectively.
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
For View Only (A) radial, longitudinal
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(C) circumferential, radial
(D) longitudinal, circumferential
Assertion (A) : Two coils are wound around a cylindrical core such that the primary coil has N1 turns and the secondary coils has N2 turns as shown in figure. If the same flux passes through every turn of both coils then the ratio of emf induced in the two coils is Vemf 2 = N2 Vemf 1 N1
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MCQ 6.1.20
347
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Reason (R) : In a primitive transformer, by choosing the appropriate no. of turns, any desired secondary emf can be obtained. (A) Both A and R are true and R is correct explanation of A.
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(B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. MCQ 6.1.21
In a non magnetic medium electric field E = E 0 cos wt is applied. If the permittivity of medium is e and the conductivity is s then the ratio of the amplitudes of the conduction current density and displacement current density will be (A) m0 /we (B) s/we (C) sm0 /we
MCQ 6.1.22
MCQ 6.1.23
(D) we/s
In a medium where no D.C. field is present, the conduction current density at any point is given as Jd = 5 cos ^1.5 # 108 t h ay A/m2 . Electric flux density in the medium will be (A) 133.3 sin ^1.5 # 108 t h ay nC/m2 (B) 13.3 sin ^1.5 # 108 t h ay nC/m2 (C) 1.33 sin ^1.5 # 108 t h ay nC/m2 (D) - 1.33 sin ^1.5 # 108 t h ay nC/m2 In a medium, the permittivity is a function of position such that de . 0 . If the e volume charge density inside the medium is zero then d : E is roughly equal to (B) - eE (A) eE
(C) 0 (D) - de : E GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
348
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.1.24 A conducting medium has permittivity, e = 4e0 and conductivity, s = 1.14 # 108 s/m . What will be the ratio of magnitude of displacement current and conduction current in the medium at 50 GHz ? (A) 1.10 # 10 4 (B) 1.025 # 107 (C) 9.75 # 10-17 (D) 9.75 # 10-8 In free space, the electric field intensity at any point (r, q, f) in spherical coordinate system is given by sin q cos ^wt - kr h E = aq r The phasor form of magnetic field intensity in the free space will be (B) - k sin q e-jkr af (A) k sin q e-jkr af wm0 r wm0 r
MCQ 6.1.26
kwm0 -jkr e af r
(D) k sin q e-jkr af r
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(C)
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MCQ 6.1.25
Magnetic field intensity in free space is given as
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H = 0.2 cos ^15py h sin ^6p # 109 t - bx h az A/m
(C) ! 77.5 rad/m
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It satisfies Maxwell’s equation when b equals to (A) ! 46.5 rad/m (B) ! 41.6 rad/m (D) ! 60.28 rad/m
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GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
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EXERCISE 6.2
Two parallel conducting rails are being placed at a separation of 3 m as shown in figure. One end of the rail is being connected through a resistor R = 10 W and the other end is kept open. A metal bar slides frictionlessly on the rails at a speed of 5 m/s away from the resistor. If the magnetic flux density B = 0.2 Wb/m2 pointing out of the page fills entire region then the current I flowing in the resistor will be (A) 0.01 A (B) - 0.01 A
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MCQ 6.2.1
349
(D) - 0.1 A
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(C) 1 A
Common Data for Question 2 - 3 :
A conducting wire is formed into a square loop of side 4 m. A very long straight wire carrying a current I = 30 A is located at a distance 2 m from the square loop as shown in figure.
MCQ 6.2.2
If the loop is pulled away from the straight wire at a velocity of 5 m/s then the induced e.m.f. in the loop after 0.6 sec will be (A) 5 mvolt (B) 2.5 mvolt (C) 25 mvolt
(D) 5 mvolt
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350
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.2.3 If the loop is pulled downward in the parallel direction to the straight wire, such that distance between the loop and wire is always 3 m then the induced e.m.f. in the loop at any time t will be (A) linearly increasing with t (B) always 0 (C) linearly decreasing with t
An infinitely long straight wire with a closed switch S carries a uniform current I = 4 A as shown in figure. A square loop of side a = 2 m and resistance R = 4 W is located at a distance 4 m from the wire. Now at any time t = t 0 the switch is open so the current I drops to zero. What will be the total charge that passes through a corner of the square loop after t = t 0 ? (A) 277 nC (B) 693 nC
co
m
MCQ 6.2.4
(D) always constant but not zero.
(C) - 237 nC
A circular loop of radius 5 m carries a current I = 2 A . If another small circular loop of radius 1 mm lies a distance 8 m above the large circular loop such that the planes of the two loops are parallel and perpendicular to the common axis as shown in figure then total flux through the small loop will be (A) 1.62 fWb (B) 25.3 nWb
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(C) 44.9 fWb
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MCQ 6.2.5
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(D) 139 nC
(D) 45.4 pWb
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Chap 6
Time Varrying Field and Maxwell Equation
351
For View Only Shop Online at www.nodia.co.in MCQ 6.2.6 A non magnetic medium at frequency f = 1.6 # 108 Hz has permittivity e = 54e0 and resistivity r = 0.77 W - m . What will be the ratio of amplitudes of conduction current to the displacement current ? (A) 0.43 (B) 0.37 (C) 1.16
Two voltmeters A and B with internal resistances RA and RB respectively is connected to the diametrically opposite points of a long solenoid as shown in figure. Current in the solenoid is increasing linearly with time. The correct relation between the voltmeter’s reading VA and VB will be
(B) VA =- VB (D) VA =- RA VB RB
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(A) VA = VB (C) VA = RA VB RB
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MCQ 6.2.7
(D) 2.70
Statement for Linked Question 8 - 9 :
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Two parallel conducting rails are being placed at a separation of 6 m with a resistance R = 10 W connected across it’s one end. A conducting bar slides frictionlessly on the rails with a velocity of 4 m/s away from the resistance as shown in the figure.
MCQ 6.2.8
If a uniform magnetic field B = 4 Tesla pointing out of the page fills entire region then the current I flowing in the bar will be (A) 0 A (B) - 40 A (C) 4 A
(D) - 4 A
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Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.2.9 The force exerted by magnetic field on the sliding bar will be (A) 4 N, opposes it’s motion (B) 40 N, opposes it’s motion (C) 40 N, in the direction of it’s motion (D) 0 Two small resistor of 225 W each is connected through a perfectly conducting filament such that it forms a square loop lying in x -y plane as shown in the figure. Magnetic flux density passing through the loop is given as B =- 7.5 cos (120pt - 30c) az The induced current I (t) in the loop will be
at e
(A) 0.02 sin (120pt - 30c)
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MCQ 6.2.10
(C) - 5.7 sin (120pt - 30c)
(B) 2.8 # 103 sin (120pt - 30c) (D) 5.7 sin (120pt - 30c)
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Common Data for Question 11 - 12 :
MCQ 6.2.11
ww
In a non uniform magnetic field B = 8x2 az Tesla , two parallel rails with a separation of 10 m and connected with a voltmeter at it’s one end is located in x -y plane as shown in figure. The Position of the bar which is sliding on the rails is given as x = t ^1 + 0.4t2h Voltmeter reading at t = 0.4 sec will be
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Chap 6
Time Varrying Field and Maxwell Equation
For View Only (A) - 0.35 volt
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(C) - 0.29 volt MCQ 6.2.12
353
(D) - 1.6 volt
Voltmeter reading at x = 12 cm will be (A) 12.27 mvolt (B) - 14.64 mvolt (D) - 23.4 mvolt
(C) 23.4 mvolt
A rectangular loop of self inductance L is placed near a very long wire carrying current i1 as shown in figure (a). If i1 be the rectangular pulse of current as shown in figure (b) then the plot of the induced current i2 in the loop versus time t will be (assume the time constant of the loop, t & L/R )
MCQ 6.2.14
Two parallel conducting rails is placed in a varying magnetic field B = 0.2 cos wtax . A conducting bar oscillates on the rails such that it’s position is given by y = 0.25 ^1 - cos wt h m
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co
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MCQ 6.2.13
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354
Time Varrying Field and Maxwell Equation
Chap 6
co
m
For View Only Shop Online at www.nodia.co.in If one end of the rails are terminated in a resistance R = 5 W , then the current i flowing in the rails will be
(B) - 0.01w sin wt ^1 + 2 cos wt h (D) 0.05w sin wt ^1 + 2 sin wt h MCQ 6.2.15
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(C) 0.01w cos wt ^1 + 2 sin wt h
lp.
(A) 0.01w sin wt ^1 + 2 cos wt h
Electric flux density in a medium ( er = 10 , mr = 2 ) is given as
at e
D = 2.33 sin ^3 # 108 t - 0.2x h ay mC/m2 Magnetic field intensity in the medium will be (A) 10-5 sin ^3 # 108 t - 0.2x h ay A/m
(B) 2 sin ^3 # 108 t - 0.2x h ay A/m
w. g
(C) - 4 sin ^3 # 108 t - 0.2x h ay A/m
(D) 4 sin ^3 # 108 t - 0.2x h ay A/m
In a non conducting medium (s = 0) magnetic field intensity at any point is given by H = cos ^1010 t - bx h az A/m . The permittivity of the medium is e = 0.12 nF/m and permeability of the medium is m = 3 # 10-5 H/m . If no D.C. field is present in medium, then value of b for which the field satisfies Maxwell’s equation is (A) - 600 rad/s
ww
MCQ 6.2.16
(B) 600 rad/m
(C) 3.6 # 105 rad/m (D) (A) and (B) both MCQ 6.2.17
A current filament located on the x -axis in free space with in the interval - 0.1 < x < 0.1 m carries current I (t) = 8t A in ax direction. If the retarded vector potential at point P (0, 0, 2) be A (t) then the plot of A (t) versus time will be
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
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he
In a non-conducting medium ( s = 0 , mr = er = 2 ), the retarded potentials are given as V = y ^x - ct h volt and A = y ^ xc - t h ax Wb/m where c is velocity of waves in free space. The field (electric and magnetic) inside the medium satisfies Maxwell’s equation if (A) J = 0 only (B) rv = 0 only
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MCQ 6.2.18
lp.
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For View Only
355
(C) J = rv = 0 MCQ 6.2.19
(D) Can’t be possible
Electric field in free space in given as
E = 5 sin ^10py h cos ^6p # 109 - bx h az
It satisfies Maxwell’s equation for b = ? (A) ! 20p rad/m (B) ! 300 p rad/m (C) 10p rad/m
(D) 30p rad/m
Statement for Linked Question 20 - 21 : In a region of electric and magnetic fields E and B , respectively, the force experienced by a test charge qC are given as follows for three different velocities. Velocity m/sec Force, N ax 2q ^ax + ay h ay qay az q ^2ay + az h GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
356
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.2.20 What will be the magnetic field B in the region ? (A) ax (B) ax - ay (D) ay - az
(C) az MCQ 6.2.21
What will be electric field E in the region ? (B) ay - az (A) ax - az (C) ay + az
m
In Cartesian coordinates magnetic field is given by B =- 2/x az . A square loop of side 2 m is lying in xy plane and parallel to the y -axis. Now, the loop is moving in that plane with a velocity v = 4ax as shown in the figure.
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MCQ 6.2.22
(D) ay + az - ax
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at e
What will be the circulation of the induced electric field around the loop ? 16 (B) 8 (A) x x ^x + 2h x ^x + 2h 8 (C) (D) 16 x ^x + 2h
Common Data for Question 23 - 24 :
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MCQ 6.2.23
In a cylindrical coordinate system, magnetic field is given by Z0 for r < 5 m ] B = [2 sin wtaz for 5 < r < 6 m ]0 for r > 6 m \ The induced electric field in the region r < 4 m will be (A) 0 (B) 2w cos wt af r (C) - 2 cos wtaf
MCQ 6.2.24
The induced electric field at r = 4.5 m is
(D)
1 a 2 sin wt f
(A) 0
(B) - 17w cos wt 18
(C) 4w cos wt 9
(D) - 17w cos wt 4
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Chap 6
Time Varrying Field and Maxwell Equation
357
For View Only Shop Online at www.nodia.co.in MCQ 6.2.25 The induced electric field in the region r > 5 m is (B) - 9w cos wt af (A) - 18 w cos wtaf r r (D) 9w cos wt af r
(C) - 9r cos wtaf
In a certain region a test charge is moving with an angular velocity 2 rad/ sec along a circular path of radius 4 m centred at origin in the x -y plane. If the magnetic flux density in the region isB = 2az Wb/m2 then the electric field viewed by an observer moving with the test charge is (A) 8a r V/m (B) 4a r V/m
m
MCQ 6.2.26
(D) - 8a r V/m
(C) 0
A 8 A current is flowing along a straight wire from a point charge situated at the origin to infinity and passing through the point (1, 1, 1). The circulation of the magnetic field intensity around the closed path formed by the triangle having the vertices ^2, 0, 0h, ^0, 2, 0h and ^0, 0, 2h is equal to (B) 3 A (A) 7 A 8 (C) 7 A (D) 1 A
MCQ 6.2.28
Magnetic flux density, B = 0.1t az Tesla threads only the loop abcd lying in the plane xy as shown in the figure.
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MCQ 6.2.27
Consider the three voltmeters V1 , V2 and V3 , connected across the resistance in the same xy plane. If the area of the loop abcd is 1 m2 then the voltmeter readings are V1
V2
V3
(A) 66.7 mV
33.3 mV
66.7 mV
(B) 33.3 mV
66.7 mV
33.3 mV
(C) 66.7 mV
66.7 mV
33.3 mV
(D) 33.3 mV
66.7 mV
66.7 mV
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358
Time Varrying Field and Maxwell Equation
For View Only
Chap 6
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Statement for Linked Question 29 - 30 :
MCQ 6.2.29
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A square wire loop of resistance R rotated at an angular velocity w in the uniform magnetic field B = 2ay mWb/m2 as shown in the figure.
If the angular velocity, w = 2 rad/ sec then the induced e.m.f. in the loop will be (A) 2 sin q mV/m (B) 2 cos q mV/m
If resistance, R = 40 mW then the current flowing in the square loop will be (A) 0.2 sin q mA (B) 0.1 sin q mA (C) 0.1 cos q mA
(D) 0.5 sin q mA
In a certain region magnetic flux density is given as B = B 0 sin wt ay . A rectangular loop of wire is defined in the region with it’s one corner at origin and one side along z -axis as shown in the figure.
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MCQ 6.2.31
at e
MCQ 6.2.30
(D) 4 sin q mV/m
he
(C) 4 cos q mV/m
If the loop rotates at an angular velocity w (same as the angular frequency of magnetic field) then the maximum value of induced e.m.f in the loop will be (A) 12 B 0 Sw (B) 2B 0 Sw (C) B 0 Sw
(D) 4B 0 Sw
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
359
For View Only Shop Online at www.nodia.co.in MCQ 6.2.32 A 50 turn rectangular loop of area 64 cm2 rotates at 60 revolution per seconds in a magnetic field B = 0.25 sin 377t Wb/m2 directed normal to the axis of rotation. The rms value of the induced voltage is (A) 2.13 volt (B) 21.33 volt (C) 4.26 volt
(D) 42.66 volt
Statement for Linked Question 33 - 34 :
MCQ 6.2.34
The value of vab is (A) - 118.43 cos 120pt V
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MCQ 6.2.33
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m
Consider the figure shown below. Let B = 5 cos 120pt Wb/m2 and assume that the magnetic field produced by i (t) is negligible
(B) 118.43 cos 120pt V
(C) - 118.43 sin 120pt V
(D) 118.43 sin 120pt V
The value of i (t) is (A) - 0.47 cos 120pt A
(B) 0.47 cos 120pt A
(C) - 0.47 sin 120pt A
(D) 0.47 sin 120pt A
***********
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
360
Time Varrying Field and Maxwell Equation
For View Only
EXERCISE 6.3
y A magnetic field in air is measured to be B = B 0 c 2 x 2 ay - 2 ax What x +y x + y2 m current distribution leads to this field ? [Hint : The algebra is trivial in cylindrical coordinates.] (A) J = B 0 z c 2 1 2 m, r ! 0 (B) J =- B 0 z c 2 2 2 m, r ! 0 m0 x + y m0 x + y (C) J = 0, r ! 0 (D) J = B 0 z c 2 1 2 m, r ! 0 m0 x + y For static electric and magnetic fields in an inhomogeneous source-free medium, which of the following represents the correct form of Maxwell’s equations ? (A) d : E = 0 , d # B = 0 (B) d : E = 0 , d : B = 0
GATE 2008
(C) d # E = 0 , d # B = 0
MCQ 6.3.4 GATE 2003
MCQ 6.3.5 GATE 1998
at e
The unit of d # H is (A) Ampere
(B) Ampere/meter
(C) Ampere/meter 2
(D) Ampere-meter
The Maxwell equation d # H = J + 2D is based on 2t (A) Ampere’s law (B) Gauss’ law (C) Faraday’s law
MCQ 6.3.6 GATE 1998
(D) d # E = 0 , d : B = 0
If C is closed curve enclosing a surface S , then magnetic field intensity H , the current density J and the electric flux density D are related by (A) ## H $ dS = ## bJ + 2D l : dl (B) # H : dl = ## bJ + 2D l : dS 2 t 2t S C S S (C) ## H : dS = # bJ + 2D l : dl (D) # H : dl = ## bJ + 2D l : dS 2t 2t S C C S
w. g
GATE 2007
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MCQ 6.3.3
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MCQ 6.3.2
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GATE 2009
Shop Online at www.nodia.co.in
m
MCQ 6.3.1
Chap 6
(D) Coulomb’s law
A loop is rotating about they y -axis in a magnetic field B = B 0 cos (wt + f) ax T. The voltage in the loop is (A) zero (B) due to rotation only (C) due to transformer action only (D) due to both rotation and transformer action
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
361
For View Only Shop Online at www.nodia.co.in MCQ 6.3.7 The credit of defining the following current is due to Maxwell IES EC 2012 (A) Conduction current (B) Drift current (C) Displacement current MCQ 6.3.8 IES EC 2011
(D) Diffusion current
A varying magnetic flux linking a coil is given by F = 2/3lt3 . If at time t = 3 s , the emf induced is 9 V, then the value of l is. (A) zero (B) 1 Wb/s2 (C) - 1 Wb/s2
Assuming that each loop is stationary and time varying magnetic field B , induces current I , which of the configurations in the figures are correct ?
MCQ 6.3.10 IES EC 2011
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IES EC 2011
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MCQ 6.3.9
(D) 9 Wb/s2
(A) 1, 2, 3 and 4
(B) 1 and 3 only
(C) 2 and 4 only
(D) 3 and 4 only
Assertion (A) : For time varying field the relation E =- dV is inadequate. Reason (R) : Faraday’s law states that for time varying field d # E = 0 (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A) (B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
362
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.3.11 Who developed the concept of time varying electric field producing a magnetic field IES EC 2009 ? (A) Gauss (B) Faraday (C) Hertz IES EC 2009
A single turn loop is situated in air, with a uniform magnetic field normal to its plane. The area of the loop is 5 m2 and the rate of charge of flux density is 2 Wb/m2 /s . What is the emf appearing at the terminals of the loop ? (A) - 5 V (B) - 2 V
m
MCQ 6.3.12
(D) Maxwell
MCQ 6.3.14 IES EC 2009
lp.
IES EC 2009
Which of the following equations results from the circuital form of Ampere’s law ? (A) d # E =-2B (B) d : B = 0 2t (C) d : D = r (D) d # H = J + 2D 2t Assertion (A) : Capacitance of a solid conducting spherical body of radius a is given by 4pe0 a in free space. Reason (R) : d # H = jweE + J
he
MCQ 6.3.13
(D) - 10 V
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(C) - 0.4 V
at e
(A) Both A and R are individually true and R is the correct explanation of A. (B) Both A and R are individually true but R is not the correct explanation of A. (C) A is true but R is false
(D) A is false but R is true
w. g
IES EC 2007
Two conducting thin coils X and Y (identical except for a thin cut in coil Y ) are placed in a uniform magnetic field which is decreasing at a constant rate. If the plane of the coils is perpendicular to the field lines, which of the following statement is correct ? As a result, emf is induced in (A) both the coils
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MCQ 6.3.15
(B) coil Y only
(C) coil X only
(D) none of the two coils MCQ 6.3.16 IES EC 2006
Assertion (A) : Time varying electric field produces magnetic fields. Reason (R) : Time varying magnetic field produces electric fields. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
363
For View Only Shop Online at www.nodia.co.in MCQ 6.3.17 Match List I (Electromagnetic Law) with List II (Different Form) and select the IES EC 2006 correct answer using the code given below the lists : List-I Ampere’s law
1.
4: D = rv
b.
Faraday’s law
2.
4: J =-2h 2t
c.
Gauss law
3.
4# H = J + 2D 2t
d.
Current
4.
4# E =-2B 2t
co
d 4 2 2 4
lp.
c 3 1 3 1
he
Two metal rings 1 and 2 are placed in a uniform magnetic field which is decreasing with time with their planes perpendicular to the field. If the rings are identical except that ring 2 has a thin air gap in it, which one of the following statements is correct ? (A) No e.m.f is induced in ring 1
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IES EC 2004
b 2 4 4 2
m
a.
Codes : a (A) 1 (B) 3 (C) 1 (D) 3 MCQ 6.3.18
List-II
(B) An e.m.f is induced in both the rings (C) Equal Joule heating occurs in both the rings (D) Joule heating does not occur in either ring. MCQ 6.3.19
Which one of the following Maxwell’s equations gives the basic idea of radiation ?
IES EC 2003
(A)
d # H = 2D/2t
d # E = 2B/2t d:D = r 3 (C) d:D = 0 MCQ 6.3.20 IES EC 2001
4
(B) (D)
d # E =- 2B/2t 4 d : D =- 2B/2t d:B = r
4 d # H = ^2D/2t h
Which one of the following is NOT a correct Maxwell equation ? (A) d # H = 2D + J 2t
(B) d # E = 2H 2t
(C) d : D = r
(D) d : B = 0
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
364
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.3.21 Match List I (Maxwell equation) with List II (Description) and select the correct IES EC 2001 answer : List I
c.
B $ dS # E : dl =- # 2 2t
d.
J) : dS # H : dl = # 2(D2+ t
m
b.
# B : dS = 0 # D : dS = #v rv dv
co
a.
List II
The mmf around a closed path is equal to the conduction current plus the time derivative of the electric displacement current through any surface bounded by the path.
2.
The emf around a closed path is equal to the time derivative is equal to the time derivative of the magnetic displacement through any surface bounded by the path.
3.
The total electric displacement through the surface enclosing a volume is equal to total charge within the volume
at e
he
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1.
4. The net magnetic flux emerging through any closed surface is zero. Codes :
IES EE 2012
b 3 3 2 2
c 2 2 3 3
d 4 1 1 4
The equation of continuity defines the relation between (A) electric field and magnetic field
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MCQ 6.3.22
a 1 4 4 1
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(A) (B) (C) (D)
(B) electric field and charge density (C) flux density and charge density (D) current density and charge density MCQ 6.3.23 IES EE 2009
What is the generalized Maxwell’s equation d # H = Jc + 2D for the free space ? 2t (A) d # H = 0 (B) d # H = Jc (C) d # H = 2D 2t
(D) d # H = D
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
365
For View Only Shop Online at www.nodia.co.in MCQ 6.3.24 Magnetic field intensity is H = 3ax + 6yay + 2xaz A/m. What is the current density IES EE 2009 J A/m2 ? (B) - 7az (A) - 2ay (C) 3ax MCQ 6.3.25
A circular loop placed perpendicular to a uniform sinusoidal magnetic field of frequency w1 is revolved about an axis through its diameter at an angular velocity w 2 rad/sec (w 2 < w1) as shown in the figure below. What are the frequencies for the e.m.f induced in the loop ?
(A) w1 and w 2 (B) w1, w 2 + w 2 and w 2
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(C) w 2, w1 - w 2 and w 2
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IES EE 2009
(D) 12ay
(D) w1 - w 2 and w1 + w 2 MCQ 6.3.26 IES EE 2009
Which one of the following is not a Maxwell’s equation ? (A) d # H = ^s + jweh E (B) F = Q ^E + v # B h
MCQ 6.3.27 IES EE 2008
(C)
D : dS #c H : dl = #s J : dS + #s 2 2t
(D)
#S B : dS = 0
Consider the following three equations : 1. d # E =-2B 2t 2. d # H = J + 2D 2t 3. d : B = 0 Which of the above appear in Maxwell’s equations ? (A) 1, 2 and 3 (B) 1 and 2 (C) 2 and 3
(D) 1 and 3
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
366
Time Varrying Field and Maxwell Equation
Chap 6
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For View Only Shop Online at www.nodia.co.in MCQ 6.3.28 A straight current carrying conductor and two conducting loops A and B are shown IES EE 2007 in the figure given below. What are the induced current in the two loops ?
(A) Anticlockwise in A and clockwise in B (C) Clockwise both in A and B
lp.
(B) Clockwise in A and anticlockwise in B
IES EE 2007
Which one of the following equations is not Maxwell’s equation for a static electromagnetic field in a linear homogeneous medium ? (A) d : B = 0 (C)
IES EE 2006
(B) d # D = 0v (D) d2 A = m0 J
In free space, if rv = 0 , the Poisson’s equation becomes (A) Maxwell’s divergence equation d : B = 0
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MCQ 6.3.30
#c B : dl = m0 I
at e
MCQ 6.3.29
he
(D) Anticlockwise both in A and B
(B) Laplacian equation d2V = 0 (C) Kirchhoff’s voltage equation SV = 0 (D) None of the above
Match List I with List II and select the correct answer using the codes given below :
IES EE 2004
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MCQ 6.3.31
List I
List II
a
Continuity equation
1. d H = J + 2D # 2t
b
Ampere’s law
2.
c
Displacement current
3. d E =-2B # 2t
d
Faraday’s law
4.
J = 2D 2t
2r d # J =- v 2t
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
For View Only Codes : a (A) 4 (B) 4 (C) 2 (D) 2 MCQ 6.3.32 IES EE 2004
367
Shop Online at www.nodia.co.in b 3 1 3 1
c 2 2 4 4
d 1 3 1 3
Match List I (Type of field denoted by A) with List II (Behaviour) and select the correct answer using the codes given below :
d
co
c
A static electric field in a charged region
IES EE 2003
a 4 4 2 2
b 2 2 4 4
c 3 1 3 1
d 1 3 1 3
d#A ! 0 d#A = 0
Which one of the following pairs is not correctly matched ? (A) Gauss Theorem : # D : ds = # d : Ddv
(C) Coulomb’s Law :
(D) Stoke’s Theorem :
IES EE 2003
d#A = 0
4. d : A = 0
(B) Gauss’s Law :
MCQ 6.3.34
2. d : A ! 0
A time-varying electric field in a charged medium with time-varying magnetic field
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MCQ 6.3.33
d#A ! 0
3. d : A ! 0
Codes : (A) (B) (C) (D)
1. d : A = 0
A steady magnetic field in a current carrying conductor
lp.
b
A static electric field in a charge free region
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a
List II
m
List I
s
v
# D : ds = #v rdv V =-
dfm dt
#l x : dl = #s (d # x) : ds
Maxwell equation d # E =- (2B/2t) is represented in integral form as (A) # E : dl =- 2 # B : dl (B) # E : dl =- 2 # B : ds 2t s 2t (C) # E # dl =- 2 # B : dl (D) # E # dl =- 2 # B : dl 2t s 2t
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
368
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in MCQ 6.3.35 The magnetic flux through each turn of a 100 turn coil is (t3 - 4t) milli-Webers IES EE 2002 where t is in seconds. The induced e.m.f at t = 2 s is (A) 1 V (B) - 1 V (D) - 0.4 V
(C) 0.4 V MCQ 6.3.36
lp.
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IES EE 2002
Two conducting coils 1 and 2 (identical except that 2 is split) are placed in a uniform magnetic field which decreases at a constant rate as in the figure. If the planes of the coils are perpendicular to the field lines, the following statements are made :
an e.m.f is induced in the split coil 2
2.
e.m.fs are induced in both coils
3.
equal Joule heating occurs in both coils
4.
Joule heating does not occur in any coil
at e
he
1.
Which of the above statements is/are true ? (A) 1 and 4 (B) 2 and 4
MCQ 6.3.37 IES EE 2002
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(C) 3 only
For linear isotropic materials, both E and H have the time dependence e jwt and regions of interest are free of charge. The value of d # H is given by (A) sE (B) jweE
IES EE 2002
(D) sE - jweE
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(C) sE + jweE MCQ 6.3.38
(D) 2 only
Which of the following equations is/are not Maxwell’s equations(s) ? 2r (A) d : J =- v (B) d : D = rv 2t (C) d : E =-2B (D) # H : dl = # b sE + e2E l : ds 2t 2t s Select the correct answer using the codes given below : (A) 2 and 4 (B) 1 alone (C) 1 and 3
(D) 1 and 4
Assertion (A) : The relationship between Magnetic Vector potential A and the current density J in free space is d # (d # A) = m0 J GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 6.3.39 IES EE 2001
Chap 6
Time Varrying Field and Maxwell Equation
369
For View Only Shop Online at www.nodia.co.in For a magnetic field in free space due to a dc or slowly varying current is d2 A =- m0 J Reason (R) : For magnetic field due to dc or slowly varying current d : A = 0 . (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true MCQ 6.3.40
Assertion (A) : In the equation, the additional term 2D is necessary. 2t Reason (R) : The equation will be consistent with the principle of conservation of charge. (A) Both A and R are true and R is the correct explanation of A
co
m
IES EE 2001
Given that d # H = J + 2D 2t
(C) A is true but R is false (D) A is false but R is true
Consider coils C1, C2, C 3 and C 4 (shown in the given figures) which are placed in the time-varying electric field E (t) and electric field produced by the coils C l2, C l3 and C l4 carrying time varying current I (t) respectively :
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IES EE 2001
he
MCQ 6.3.41
lp.
(B) Both A and R are true but R is NOT the correct explanation of A
The electric field will induce an emf in the coils (A) C1 and C2 (B) C2 and C 3 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
370
Time Varrying Field and Maxwell Equation
For View Only (C) C1 and C 3 MCQ 6.3.42 IES EE 2001
Chap 6
Shop Online at www.nodia.co.in (D) C2 and C 4
A circular loop is rotating about the y -axis as a diameter in a magnetic field B = B 0 sin wtax Wb/m2 . The induced emf in the loop is (A) due to transformer emf only (B) due to motional emf only (C) due to a combination of transformer and motional emf
Match List I (Law/quantity) with List II (Mathematical expression) and select the correct answer : List I a.
List II
co
IES EE 2001
1. d : D = r
Gauss’s law
b. Ampere’s law
2. d E =-2B # 2t
c.
3.
P = E#H
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Faraday’s law
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MCQ 6.3.43
m
(D) zero
d. Poynting vector
4.
F = q ^E + v # B h
c 4 2 2 4
d 3 1 3 1
w. g
b 2 5 5 2
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Codes : a (A) 1 (B) 3 (C) 1 (D) 3
at e
5. d H = J + 2D # c 2t
***********
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 6
Time Varrying Field and Maxwell Equation
For View Only
371
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SOLUTIONS 6.1
Option (C) is correct.
SOL 6.1.2
Option (B) is correct. The line integral of magnetic field intensity along a closed loop is equal to the current enclosed by it.
co
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SOL 6.1.1
#
lp.
i.e. H : dl = Ienc So, for the constant current, magnetic field intensity will be constant i.e. magnetostatic field is caused by steady currents. Option (D) is correct. From Faraday’s law the electric field intensity in a time varying field is defined as where B is magnetic flux density in the EM field. d # E =-2B 2t and since the magnetic flux density is equal to the curl of magnetic vector potential i.e. B = d#A So, putting it in equation (1), we get d # E =- 2 ^d # Ah 2t or d # E = d # b- 2 A l 2t Therefore, E =-2A 2t
SOL 6.1.4
Option (B) is correct. Since total magnetic flux through a surface S is defined as
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SOL 6.1.3
F =
# B : dS S
From Maxwell’s equation it is known that curl of magnetic flux density is zero d:B = 0
# B : dS S
=
# (d : B) dv = 0 v
(Stokes Theorem)
Thus, net outwards flux will be zero for a closed surface. SOL 6.1.5
Option (A) is correct. From Faraday’s law, the relation between electric field and magnetic field is d # E =-2B 2t
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
372
Time Varrying Field and Maxwell Equation
Chap 6
For View Only Shop Online at www.nodia.co.in Since the electric field inside a conducting sphere is zero. i.e. E =0 So the rate of change in magnetic flux density will be 2B =- (d E)= 0 # 2t Therefore B (r, t) will be uniform inside the sphere and independent of time. Option (B) is correct. From the integral form of Faraday’s law we have the relation between the electric field intensity and net magnetic flux through a closed loop as E : dl =- dF dt Since electric field intensity is zero (E = 0 ) inside the conducting loop. So, the rate of change in net magnetic flux through the closed loop is dF = 0 dt i.e. F is constant and doesn’t vary with time.
m
SOL 6.1.6
Option (A) is correct. A superconductor material carries zero magnetic field and zero electric field inside it. i.e. B = 0 and E = 0 Now from Ampere-Maxwell equation we have the relation between the magnetic flux density and electric field intensity as d # B = m0 J + m0 e02E 2t So, (B = 0 , E = 0 ) J =0 Since the net current density inside the superconductor is zero so all the current must be confined at the surface of the wire.
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at e
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SOL 6.1.7
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#
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Option (A) is correct. According to Lenz’s law the induced current I in a loop flows such as to produce a magnetic field that opposes the change in B (t). Now the configuration shown in option (A) and (B) for increasing magnetic flux Bi , the change in flux is in same direction to Bi as well as the current I flowing in the loop produces magnetic field in the same direction so it does not follow the Lenz’s law. For the configuration shown in option (D), as the flux Bd is decreasing with time so the change in flux is in opposite direction to Bd as well as the current I flowing in the loop produces the magnetic field in opposite direction so it also does not follow the Lenz’s law. For the configuration shown in option (C), the flux density Bd is decreasing with time so the change in flux is in opposite direction to Bd but the current I flowing in the loop produces magnetic field in the same direction to Bd (opposite to the direction of change in flux density). Therefore this is the correct configuration. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 6.1.8
Chap 6
Time Varrying Field and Maxwell Equation
373
m
For View Only Shop Online at www.nodia.co.in SOL 6.1.9 Option (A) is correct. Induced emf in a conducting loop is given by where F is total magnetic flux passing through the loop. Vemf =- dF dt Since, the magnetic field is non-uniform so the change in flux will be caused by it and the induced emf due to it is called transformer emf. Again the field is in ay direction and the loop is rotating about z -axis so flux through the loop will also vary due to the motion of the loop. This causes the emf which is called motion emf. Thus, total induced voltage in the rotating loop is caused by the combination of both the transformer and motion emf. Option (D) is correct. As the conducting loop is falling freely So, the flux through loop will remain constant. Therefore, the voltage induced in the loop will be zero.
SOL 6.1.11
Option (B) is correct. The magnetic flux density passing through the loop is given as B = 4z3 t2 ax Since the flux density is directed normal to the plane x = 0 so the total magnetic flux passing through the square loop located in the plane x = 0 is
# B : dS
=
1
# #
1
(4z3 t2) dydz = t2
he
F =
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SOL 6.1.10
y=0 z=0
(dS = (dydz) ax )
SOL 6.1.12
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Induced emf in a loop placed in magnetic field is defined as Vemf =- dF dt where F is the total magnetic flux passing through the loop. So the induced emf in the square loop is d (t2) ( F = t2 ) Vemf ==- 2t dt Therefore at time t = 2 sec the induced emf is Vemf =- 4 volt Option (A) is correct. Magnetic flux density produced at a distance r from a long straight wire carrying current I is defined as mI B = 0 af 2pr where af is the direction of flux density as determined by right hand rule. So the flux density produced by straight wire at a distance r from it is mI (an is unit vector normal to the loop) B = 0 an 2pr Therefore the total magnet flux passing through the loop is d+a m I 0 (dS = adran ) adr F = B : dS = pr 2 d where dr is width of the strip of loop at a distance r from the straight wire. Thus,
#
#
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#
3
Vab =-
m
Option (D) is correct. Electric field intensity experienced by the moving conductor ab in the presence of magnetic field B is given as where v is the velocity of the conductor. E = v#B So, electric field will be directed from b to a as determined by right hand rule for the cross vector. Therefore, the voltage difference between the two ends of the conductor is given as b
# E : dl
co
SOL 6.1.13
Chap 6
a
Thus, the positive terminal of voltage will be a and Vab will be positive. Option (B) is correct. Given magnetic flux density through the square loop is B = 0.1taz Wb/m2 So, total magnetic flux passing through the loop is F = B : dS = ^0.1t h^1 h = 0.1t The induced emf (voltage) in the loop is given as df Vemf ==- 0.2 Volt dt As determined by Lenz’s law the polarity of induced emf will be such that V1 + V2 =- Vemf Therefore, the voltage drop in the two resistances are respectively, V1 = b 2 l (- Vemf ) = 0.1 = 33.3 mV 3 2+4 and V2 = b 4 l (- Vemf ) = 66.7 mV 2+4
SOL 6.1.15
Option (D) is correct. Consider a magnet bar being dropped inside a pipe as shown in figure. Suppose the current I in the magnet flows counter clockwise (viewed from above) as shown in figure. So near the ends of pipe, it’s field points upward. A ring of pipe below the magnet experiences an increasing upward flux as the magnet approaches and hence by Lenz’s law a current will be induced in it such as to produce downward flux. Thus, Iind must flow clockwise which is opposite to the current in the magnet. Since opposite currents repel each other so, the force exerted on the magnet due to the induced current is directed upward. Meanwhile a ring above the magnet experiences a decreasing upward flux; so it’s induced current parallel to I and it attracts magnet upward. And flux through the rings next to the magnet bar is constant. So no current is induced in them.
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SOL 6.1.14
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co
m
For View Only
375
Option (B) is correct.
V1 =- N1 dF dt where F is total magnetic flux passing through it. Again V2 =- N2 dF dt Since both the coil are in same magnetic field so, change in flux will be same for both the coil. Comparing the equations (1) and (2) we get V1 = N1 V2 N2 V2 = V1 N2 = (12) 3000 = 7.2 volt 5000 N1
SOL 6.1.17
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Voltage,
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SOL 6.1.16
lp.
Thus, for all we can say that the force exerted by the eddy current (induced current according to Lenz’s law) on the magnet is in upward direction which causes the delay to reach the bottom. Whereas in the cases of unmagnetized bar no induced current is formed. So it reaches in fraction of time. Thus, A and R both true and R is correct explanation of A.
Option (A) is correct. The magnetic flux density inside a solenoid of n turns per unit length carrying current I is defined as B = m0 nI Let the length of solenoid be l and its cross sectional radius be r . So, the total magnetic flux through the solenoid is (1) F = (m0 nI) (pr2) (nl) Since the total magnetic flux through a coil having inductance L and carrying current I is given as F = LI So comparing it with equation (1) we get,
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2
L = m0 n Ip l and as for a given solenoid, radius r and length l is constant therefore L \ n2 Option (A) is correct. The magnetic flux density inside the solenoid is defined as B = m0 nI where n " no. of turns per unit length I " current flowing in it. So the total magnetic flux through the solenoid is
# B : dS
= (m0 nI) (pa2)
co
F =
m
SOL 6.1.18
at e
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lp.
where a " radius of solenoid Induced emf in a loop placed in a magnetic field is defined as Vemf =- dF dt where F is the total magnetic flux passing through the loop. Since the resistance R is looped over the solenoid so total flux through the loop will be equal to the total flux through the solenoid and therefore the induced emf in the loop of resistance will be Vemf =- pa2 m0 n dI dt Since current I flowing in the solenoid is constant so, the induced emf is Vemf = 0 and therefore the induced current in the loop will be zero. Option (B) is correct. It will be similar to the current in a solenoid. So, the magnetic field will be in circumferential while the electric field is longitudinal.
SOL 6.1.20
Option (B) is correct. In Assertion (A) the magnetic flux through each turn of both coils are equal So, the net magnetic flux through the two coils are respectively F1 = N1 F and F2 = N2 F where F is the magnetic flux through a single loop of either coil and N1 , N2 are the total no. of turns of the two coils respectively. Therefore the induced emf in the two coils are Vemf 1 =- dF1 =- N1 dF dt dt d F d Vemf 2 =- 2 =- N2 F dt dt Thus, the ratio of the induced emf in the two loops are Vemf 2 = N2 Vemf 1 N1
ww
w. g
SOL 6.1.19
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For View Only Shop Online at www.nodia.co.in Now, in Reason (R) : a primitive transformer is similar to the cylinder core carrying wound coils. It is the device in which by choosing the appropriate no. of turns, any desired secondary emf can be obtained. So, both the statements are correct but R is not the explanation of A. Option (B) is correct. Electric flux density in the medium is given as ( E = E 0 cos wt ) D = eE = eE 0 cos wt Therefore the displacement current density in the medium is Jd = 2D =- weE 0 sin wt 2t and the conduction current density in the medium is Jc = sE = sE 0 cos wt So, the ratio of amplitudes of conduction current density and displacement current density is Jc = s we Jd
SOL 6.1.22
Option (D) is correct. The displacement current density in a medium is equal to the rate of change in electric flux density in the medium. Jd = 2D 2t Since the displacement current density in the medium is given as Jd = 20 cos ^1.5 # 108 t h ay A/m2 So, the electric flux density in the medium is
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lp.
co
m
SOL 6.1.21
# J dt + C = # 20 cos ^1.5 # 10 t h a dt + C
D =
d
8
(C " constant)
y
As there is no D.C. field present in the medium so, we get C = 0 and thus, 20 sin ^1.5 # 108 t h D = ay = 1.33 # 10-7 sin ^1.5 # 108 t h ay 1.5 # 108 = 153.3 sin ^1.5 # 108 t h ay nC/m2 Option (A) is correct. Given the volume charge density, rv = 0 So, from Maxwell’s equation we have d : D = rv (1) d:D = 0 Now, the electric flux density in a medium is defined as (where e is the permittivity of the medium) D = eE So, putting it in equation (1) we get, d : (eE) = 0 or, E : (de) + e (d : E) = 0 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 6.1.23
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and since Therefore,
de . 0 &4 e . 0 e d:E . 0
(given)
Option (C) is correct. The ratio of magnitudes of displacement current to conduction current in any medium having permittivity e and conductivity s is given as Displacement current = we s Conduction current where w is the angular frequency of the current in the medium. Given frequency, f = 50 GHz Permittivity, e = 4e0 = 4 # 8.85 # 10-12 Conductivity, s = 1.14 # 108 s/m So, w = 2pf = 2p # 50 # 109 = 100p # 109 Therefore, the ratio of magnitudes of displacement current to the conduction current is Id = 100p # 109 # 4 # 8.85 # 10-12 = 9.75 10-8 # Ic 1.14 # 108
SOL 6.1.25
Option (D) is correct. Given the electric field intensity in time domain as sin q cos ^wt - kr h E = aq r So, the electric field intensity in phasor form is given as Es = sin q e-jkr a q r and d # Es = 1 2 ^rE qs h af = ^- jk h sin q e-jkr af r kdr r Therefore, from Maxwell’s equation we get the magnetic field intensity as Hs =-d # Es = k sin q e-jkr af wr0 r jwr0 Option (B) is correct. In phasor form the magnetic field intensity can be written as Hs = 0.1 cos ^15py h e-jbx az A/m Similar as determined in MCQ 42 using Maxwell’s equation we get the relation ^15ph2 + b2 = w2 p0 e0 Here w = 6p # 109 9 2 So, ^15ph2 + b2 = c 6p # 108 m 3 # 10 2 2 ^15ph + b = 400p2 b2 = 175p2 & b = ! 23.6 rad/m
ww
SOL 6.1.26
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co
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SOL 6.1.24
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SOLUTIONS 6.2
Option (D) is correct. Induced emf. in the conducting loop formed by rail, bar and the resistor is given by Vemf =- dF dt where F is total magnetic flux passing through the loop. Consider the bar be located at a distance x from the resistor at time t . So the total magnetic flux passing through the loop at time t is
# B : dS
= Blx
lp.
F =
co
m
SOL 6.2.1
(area of the loop is S = lx )
ww w. ga te
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Now the induced emf in a loop placed in magnetic field is defined as Vemf =- dF dt where F is the total magnetic flux passing through the loop. Therefore the induced emf in the square loop is ( F = Blx ) Vemf =- d (Blx) =- Bl dx dt dt Since from the given figure, we have l = 2 m and B = 0.1 Wb/m2 and dx/dt = velocity of bar = 5 m/s So, induced emf is Vemf =- (0.1) (2) (5) =- 1 volt According to Lenz’s law the induced current I in a loop flows such as to produce magnetic field that opposes the change in B (t). As the bar moves away from the resistor the change in magnetic field will be out of the page so the induced current will be in the same direction of I shown in figure. Thus, the current in the loop is (- 1) (R = 10W ) I =-Vemf == 0.23 A 10 R Option (B) is correct. Magnetic flux density produced at a distance r from a long straight wire carrying current I is defined as mI B = 0 af 2pr where af is the direction of flux density as determined by right hand rule. So, the magnetic flux density produced by the straight conducting wire linking through the GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 6.2.2
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Time Varrying Field and Maxwell Equation
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For View Only Shop Online at www.nodia.co.in loop is normal to the surface of the loop. Now consider a strip of width dr of the square loop at distance r from the wire for which the total magnetic flux linking through the square loop is given as
# B : dS mI 1 (adr) = r 2p #
F =
S
r+a
0
(area of the square loop is dS = adr )
r
m0 Ia r+a ln b r l 2p The induced emf due to the change in flux (when pulled away) is given as m Ia r+a Vemf =- dF =- 0 d ;ln b 2p dt r lE dt dr 1 dr m Ia Therefore, Vemf =- 0 c 1 2p r + a dt r dt m dr Given = velocity of loop = 5 m/s dt and since the loop is currently located at 3 m distance from the straight wire, so after 0.6 sec it will be at (v " velocity of the loop ) r = 3 + (0.6) # v = 3 + 0.6 # 5 = 6 m
Option (B) is correct. Since total magnetic flux through the loop depends on the distance from the straight wire and the distance is constant. So the flux linking through the loop will be constant, if it is pulled parallel to the straight wire. Therefore the induced emf in the loop is (F is constant) Vemf =- dF = 0 dt Option (D) is correct. Magnetic flux density produced at a distance r from a long straight wire carrying current I is defined as mI B = 0 af 2pr where af is the direction of flux density as determined by right hand rule. Since the direction of magnetic flux density produced at the loop is normal to the surface of the loop So, total flux passing through the loop is given by 4 m0 I (dS = adr ) F = B : dS = c 2pr m^adrh S r=2 m Ia 4 dr m I2 mI = 0 = 0 ln 2 = 0 ln (2) p 2p 2p 2 r The current flowing in the loop is Iloop and induced e.m.f. is Vemf .
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SOL 6.2.4
(a = 2 m, I = 30 A )
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SOL 6.2.3
m0 # (30) # 2 1 1 : 8 (5) - 6 (5)D 2p = 25 # 10-7 volt = 2.5 mvolt
Vemf =-
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So,
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=
#
#
#
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Vemf = Iloop R =- dF dt m0 dQ (R) =- ln (2) dI p dt dt where Q is the total charge passing through a corner of square loop. m dQ (R = 4 W ) =- 0 ln (2) dI 4p dt dt m dQ =- 0 ln (2) dI 4p Therefore the total charge passing through a corner of square loop is 0 m m Q =- 0 ln (2) dI =- 0 ln (2) (0 - 4) 4p 4p 4 -7 = 4 # 4p # 10 ln (2) = 2.77 # 10-7 C = 277 nC 4p Option (A) is correct. Since the radius of small circular loop is negligible in comparison to the radius of the large loop. So, the flux density through the small loop will be constant and equal to the flux on the axis of the loops. mI R2 So, B = 0 a 2 ^z2 + R2h3/2 z
So,
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lp.
SOL 6.2.5
m
#
R " radius of large loop = 5 m z " distance between the loops = 12 m (5) 2 25m0 2 m az B = 0# # a = 2 2 3/2 z 2 13h3 ^ + 12 5 6^ h ^ h @ Therefore, the total flux passing through the small loop is 25m0 2 wherer is radius of small circular loop. F = B : dS = 3 # pr 13 ^ h 10-7 -3 2 = 25 # 4p # # p ^10 h = 65.9 fWb 3 ^13h
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where
#
SOL 6.2.6
Option (C) is correct. Electric field in any medium is equal to the voltage drop per unit length. i.e. E =V d where V " potential difference between two points. d " distance between the two points. The voltage difference between any two points in the medium is V = V0 cos 2pft So the conduction current density in the medium is given as ( s " conductivity of the medium) Jc = sE =E r V cos 2pft =V = 0 rd rd
( r " resistivity of the medium) (V = V0 cos 2pft)
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Jc = V0 rd and displacement current density in the medium is given as V cos (2pft) Jd = 2D = e2E = e 2 ; 0 (V = V0 cos 2pft) E d 2t 2t 2t = eV0 6- 2pft sin (2pft)@ d 2pf eV0 or, Jd = d Therefore, the ratio of amplitudes of conduction current and displacement current in the medium is JC Ic (V0) / (rd) = = = 1 2pfer Jd Id (d) / (2pfeV0) 1 = 2p # (1.6 # 108) # (54 # 8.85 # 10-12) # 0.77 = 2.7
lp.
co
m
or,
Option (C) is correct. Total magnetic flux through the solenoid is given as F = m0 nI where n is the no. of turns per unit length of solenoid and I is the current flowing in the solenoid. Since the solenoid carries current that is increasing linearly with time i.e. I\t So the net magnetic flux through the solenoid will be F\t or, where k is a constant. F = kt Therefore the emf induced in the loop consisting resistances RA , RB is Vemf =- dF dt Vemf =- k and the current through R1 and R2 will be Iind =- k R1 + R 2 Now according to Lenz’s law the induced current I in a loop flows such as to produce a magnetic field that opposes the change in B (t). i.e. the induced current in the loop will be opposite to the direction of current in solenoid (in anticlockwise direction). So, VA = Iind RA =- kRA RA + RB and VB =- Iind RB = b kRB l RA + RB Thus, the ratio of voltmeter readings is VA =- RA VB RB GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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SOL 6.2.7
Chap 6
Time Varrying Field and Maxwell Equation
383
For View Only Shop Online at www.nodia.co.in SOL 6.2.8 Option (C) is correct. Induced emf in the conducting loop formed by rail, bar and the resistor is given by Vemf =- dF dt where F is total magnetic flux passing through the loop. The bar is located at a distance x from the resistor at time t . So the total magnetic flux passing through the loop at time t is
#
SOL 6.2.9
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where l is separation between the rails F = B : dS = Blx Now the induced emf in a loop placed in magnetic field is defined as Vemf =- dF dt where F is the total magnetic flux passing through the loop. Therefore the induced emf in the square loop is ( F = Blx ) Vemf =- d (Blx) =- Bl dx dt dt Since from the given figure, we have l =5m B = 2T and dx/dt " velocity of bar = 4 m/s So, induced emf is Vemf =- (2) (5) (4) =- 40 volt Therefore the current in the bar loop will be I = Vemf =- 40 =- 5 A 10 R Option (B) is correct. As obtained in the previous question the current flowing in the sliding bar is I =- 4 A Now we consider magnetic field acts in ax direction and current in the sliding bar is flowing in + az direction as shown in the figure.
Therefore, the force exerted on the bar is F =
5
# Idl # B = # (- 4dza ) # (2a ) 0
z
x
=- 16ay 6z @50 =- 40ay N GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Option (A) is correct. Given the magnetic flux density through the square loop is B = 7.5 cos (120pt - 30c) az So the total magnetic flux passing through the loop will be F =
# B : dS S
at e
Option (D) is correct. As shown in figure the bar is sliding away from origin. Now when the bar is located at a distance dx from the voltmeter, then, the vector area of the loop formed by rail and the bar is dS = (20 # 10-2) (dx) az So, the total magnetic flux passing through the loop is 3 1.6 8t ^1 + 0.4t2hB x -2 2 F = B : dS = (8x az ) (20 # 10 dxaz ) = 3 S 0 Therefore, the induced e.m.f. in the loop is given as Vemf =- dF =- 1.6 # 3 ^t + 0.4t3h2 # (1 + 1.2t2) 3 dt 2 (t = 0.4 sec ) Vemf =- 1.6 6^0.4h + ^0.4h4@ # 61 + (1.2) (0.4) 2@ =- 0.35 volt Since the voltmeter is connected in same manner as the direction of induced emf (determined by Lenz’s law). So the voltmeter reading will be V = Vemf =- 0.35 volt
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SOL 6.2.11
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lp.
co
m
= 6- 7.5 cos (120pt - 30c) az@(1 # 1) (- az ) = 7.5 cos (120pt - 30c) Now, the induced emf in the square loop is given by Vemf =- dF = 7.5 # 120p sin (120pt - 30c) dt The polarity of induced emf (according to Lenz’s law) will be such that induced current in the loop will be in opposite direction to the current I (t) shown in the figure. So we have I (t) =-Vemf R (R = 250 + 250 = 500 W) =- 7.5 # 120p sin (120pt - 30c) 500 =- 4.7 sin (120pt - 30c)
#
ww
#
Option (C) is correct. Since the position of bar is give as x = t ^1 + 0.4t2h So for the position x = 12 cm we have GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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0.12 = t ^1 + 0.4t h or, t = 0.1193 sec As calculated in previous question, the induced emf in the loop at a particular time t is Vemf =-^1.6h6t + 0.4t3@2 ^1 + 1.2t2h 2
co
Option (D) is correct. Consider the mutual inductance between the rectangular loop and straight wire be M . So applying KVL in the rectangular loop we get, ...(1) M di1 = L di2 + Ri2 dt dt Now from the shown figure (b), the current flowing in the straight wire is given as (I1 is amplitude of the current) i1 = I1 u (t) - I1 u (t - T) di1 = I d (t) - I d (t - T) (2) or, 1 1 dt di1 = I So, at t = 0 1 dt and (from equation (1)) MI1 = L di2 + Ri2 dt Solving it we get for 0 < t < T i2 = M I1 e-^R/Lht L Again in equation (2) at t = T we have di1 =- I 1 dt and (from equation (1)) - MI1 = L di2 + Ri2 dt Solving it we get for t > T i2 =- M I1 e-^R/Lh(t - T) L Thus, the current in the rectangular loop is Z ]] M I1 e-^R/Lht 0 5 m is zero so we get the distribution of magnetic flux density as shown in figure below.
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SOL 6.2.24
at e
#
At any distance r from origin in the region 4 < r < 5 m , the circulation of induced electric field is given as E : dl =- dF =- d b B : dS l dt dt C d =- 82 sin wt ^pr2 - p42hB dt
#
#
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E ^2prh =- 2w cos wt ^pr2 - 16ph 2 ^r2 - 16h w cos wt E =2r So, the induced electric field intensity at r = 4.5 m is E =- 2 (^4.5) 2 - 16h w cos wt 4. 5 =- 17 w cos wt 13 or,
m
Option (B) is correct. For the region r > 5 m the magnetic flux density is 0 and so the total magnetic flux passing through the closed loop defined by r = 5 m is 5
co
SOL 6.2.25
5
4
# B : dS = # B : dS + # B : dS = 0 + # ^2 sin wt h a : dS
F =
0
0
4
5
z
lp.
4
= ^2 sin wt h8p ^5 h - p ^4h2B = 18p sin wt So, the circulation of magnetic flux density for any loop in the region r > 5 m is dy E : dl =dt E (2pr) =- d ^18p sin wt h dt =- 18pw cos wt So, the induced electric field intensity in the region r > 5 m is E = - 18pw cos wt af 2pr
SOL 6.2.26
SOL 6.2.27
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#
he
2
=- 18 w cos wtaf r Option (D) is correct. Let the test charge be q coulomb So the force presence of experienced by the test charge in the presence of magnetic field is ...(i) F = q ^v # B h and the force experienced can be written in terms of the electric field intensity as F = qE Where E is field viewed by observer moving with test charge. Putting it in Eq. (i) qE = q ^v # B h E = ^wrafh # ^2az h where w is angular velocity and r is radius of circular loop. = ^2 h^2 h^2 h a r = 8a r V/m Option (A) is correct. Let the point change located at origin be Q and the current I is flowing out of the
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co
m
For View Only Shop Online at www.nodia.co.in page through the closed triangular path as shown in the figure.
lp.
As the current I flows away from the point charge along the wire, the net charge at origin will change with increasing time and given as dQ =- I dt So the electric field intensity will also vary through the surface and for the varying field circulation of magnetic field intensity around the triangular loop is defined as
at e
he
= 6Id@enc + 6Ic@enc where 6Ic@enc is the actual flow of charge called enclosed conduction current and 6Id@enc is the current due to the varying field called enclosed displacement current which is given as d d D : dS (1) ^e0 E h : dS = dt 6Id@enc = dt S S From symmetry the total electric flux passing through the triangular surface is Q D : dS = 8 S d Q = 1 dQ =- I So, (from equation (1)) 6Id@enc = dt b 8 l 8 dt 8 Where as 6Ic@enc = I So, the net circulation of the magnetic field intensity around the closed triangular loop is
# H : dl
#
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w. g
#
#
# H : dl C
SOL 6.2.28
= 6Id@enc + 6Ic@enc
=- I + I = 7 ^8 h = 14 A 8 8
(I = 8 A )
Option (C) is correct. The distribution of magnetic flux density and the resistance in the circuit are same as given in section A (Q. 31) so, as calculated in the question, the two voltage drops in the loop due to magnetic flux density B = 0.1t az are V1 = 33.3 mV and V2 = 66.67 mV = 66.7 mV Now V3 (voltmeter) which is directly connected to terminal cd is in parallel to
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For View Only Shop Online at www.nodia.co.in both V2 and V1 . It must be kept in mind that the loop formed by voltmeter V3 and resistance 2 W also carries the magnetic flux density crossing through it. So, in this loop the induced emf will be produced which will be same as the field produced in loop abcd at the enclosed fluxes will be same. Therefore as calculated above induced emf in the loop of V3 is Vemf = 100 mV According to lenz’s law it’s polarity will be opposite to V3 and so - Vemf = V1 + V3 or, V3 = 100 - 33.3 = 26.7 mV
m
Option (C) is correct. The induced emf in a closed loop is defined as Vemf =- dF dt where F is the total magnetic flux passing through the square loop At any time t , angle between B and dS is q since B is in ay direction so the total magnetic flux passing through the square loop is F =
# B : dS
lp.
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SOL 6.2.29
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= ^B h^S h cos q = ^5 # 10-3h^20 # 10-3 # 20 # 10-3h cos q = 2 # 10-6 cos q Therefore the induced emf in the loop is Vemf =- dF =- 2 # 10-6 d ^cos qh = 2 # 10-6 sin q dq dt dt dt dq and as = angular velocity = 2 rad/ sec dt So, Vemf = ^2 # 10-6h sin q ^2 h = 4 # 10-6 sin q V/m = 24 sin q mV/m SOL 6.2.30
Option (B) is correct. As calculated in previous question the induced emf in the closed square loop is Vemf = 4 sin q mV/m So the induced current in the loop is where R is the resistance in the loop. I = Vemf R -6 ( R = 40 mW ) = 4 sin q # 10 40 # 10-3 = 0.1 sin q mA
SOL 6.2.31
Option (A) is correct. The total magnetic flux through the square loop is given as F = B : dS = ^B 0 sin wt h^S h cos q So, the induced emf in the loop is Vemf =- dF =- d 6(B 0 sin wt) (S) cos q@ dt dt
#
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=- B 0 S d 6sin wt cos wt@ dt =- B 0 S cos 2wt Thus, the maximum value of induced emf is Vemf = B 0 Sw
( q = wt )
Option (B) is correct. As calculated in previous question the maximum induced voltage in the rotating loop is given as Vemf = B 0 Sw From the given data, we have B 0 = 0.25 Wb/m2 S = 64 cm2 = 64 # 10-4 m2 and w = 60 # 2p = 377 rad/ sec (In one revolution 2p radian is covered) So, the r.m.s. value of the induced voltage is 1 V 1 B Sw = 1 0.25 64 10-4 377 ^ h # # # emf = 0
[email protected] = 2 2 2 = 0.4265 Since the loop has 50 turns so net induced voltage will be 50 times the calculated value. i.e.
[email protected] = 50 # ^0.4265h = 21.33 volt
SOL 6.2.33
Option (A) is correct. e.m.f. induced in the loop due to the magnetic flux density is given as Vemf =-2F =- 2 ^10 cos 120pt h^pr2h 2t 2t =- p ^10 # 10-2h2 # ^120ph^- 10 sin 120pt h = 12p2 sin 120pt As determined by Lenz’s law the polarity of induced e.m.f will be such that b is at positive terminal with respect to a . i.e. Vba = Vemf = 12p2 sin 120pt or Vab =- 12p2 sin 120pt =- 118.43 sin 120pt Volt
SOL 6.2.34
Option (C) is correct. As calculated in previous question, the voltage induced in the loop is Vab =- 12p2 sin 120pt Therefore, the current flowing in the loop is given as 2 I ^ t h =- Vab = 12p sin 120pt 250 250 = 2.47 sin 120pt
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SOL 6.2.32
***********
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SOLUTIONS 6.3
Option (A) is correct. Given, the magnetic flux density in air as y ...(1) B = B0 c 2 x 2 ay - 2 ax m x +y x + y2 Now, we transform the expression in cylindrical system, substituting x = r cos f and y = r sin f ax = cos far - sin faf and ay = sin far + cos faf So, we get B = B0 af Therefore, the magnetic field intensity in air is given as B a H = B = 0 f , which is constant m0 m0 So, the current density of the field is (since H is constant) J = d#H = 0
SOL 6.3.2
Option (C) is correct. Maxwell equations for an EM wave is given as d:B = 0 r d:E = v e d # E =-2B 2t d # H = 2D + J 2t So, for static electric magnetic fields d:B = 0 d : E = rv /e
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SOL 6.3.1
d#E = 0
d#H = J
SOL 6.3.3
Option (C) is correct. d # H = J + 2D 2t
##S ^d # H h : dS
=
##S `J + 22Dt j : dS
2B b 2t = 0 l 2D b 2t = 0 l Maxwell Equations Integral form
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# H : dl
=
D : dS ##S bJ +2 2t l
Stokes Theorem
Option (A) is correct. From Maxwells equations we have d # H = 2D + J 2t Thus, d # H has unit of current density J (i.e., A/m2 )
SOL 6.3.5
Option (D) is correct. This equation is based on Ampere’s law as from Ampere’s circuital law we have # H $ dl = Ienclosed
m
SOL 6.3.4
=
#S J : dS
Applying Stoke’s theorem we get # ^d # H h $ dS = # J : dS S
he
Option (A) is correct. From maxwell equation we have d # H = J + 2D 2t The term 2D defines displacement current. 2t Option (A) is correct. Emf induced in a loop carrying a time varying magnetic flux F is defined as Vemf =- dF dt 9 =- d b 1 lt3 l dt 3 9 =- lt2 at time, t = 3 s , we have 9 =- l ^3h2 l =- 2 Wb/s2
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SOL 6.3.8
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SOL 6.3.7
d#H = J Then, it is modified using continuity equation as d # H = J + 2D 2t Option (C) is correct. When a moving circuit is put in a time varying magnetic field induced emf have two components. One due to time variation of magnetic flux density B and other due to the motion of circuit in the field.
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SOL 6.3.6
S
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#l H $ dl
or,
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l
Option (B) is correct. According to Lenz’s law the induced emf (or induced current) in a loop flows such as to produce a magnetic field that opposed the change in B . The direction of the magnetic field produced by the current is determined by right hand rule. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 6.3.9
Chap 6
Time Varrying Field and Maxwell Equation
399
co
m
For View Only Shop Online at www.nodia.co.in Now, in figure (1), B directed upwarded increases with time where as the field produced by current I is downward so, it obey’s the Lenz’s law. In figure (2), B directed upward is decreasing with time whereas the field produced by current I is downwards (i.e. additive to the change in B ) so, it doesn’t obey Lenz’s law. In figure (3), B directed upward is decreasing with time where as current I produces the field directed upwards (i.e. opposite to the change in B ) So, it also obeys Lenz’s law. In figure (4), B directed upward is increasing with time whereas current I produces field directed upward (i.e. additive to the change in B ) So, it doesn’t obey Lenz’s law. Thus, the configuration 1 and 3 are correct. Option (A) is correct. Faraday’s law states that for time varying field, d # E =-2B 2t Since, the curl of gradient of a scalar function is always zero i.e. d # ^dV h = 0 So, the expression for the field, E =- dV must include some other terms is E =- dV - 2A 2t i.e. A is true but R is false.
SOL 6.3.11
Option (B) is correct. Faraday develops the concept of time varying electric field producing a magnetic field. The law he gave related to the theory is known as Faraday’s law.
SOL 6.3.12
Option (C) is correct. Given, the area of loop S = 5 m2 Rate of change of flux density, 2B = 2 Wb/m2 /S 2t So, the emf in the loop is Vemf =- 2 B : dS = ^5 h^- 2h =- 10 V 2t Option (C) is correct. The modified Maxwell’s differential equation. d # H = J + 2D 2t This equation is derived from Ampere’s circuital law which is given as
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lp.
SOL 6.3.10
#
SOL 6.3.13
# H : dl
# ^d # H h : dS
= Ienc =
# JdS
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Chap 6
For View Only Shop Online at www.nodia.co.in SOL 6.3.14 Option (B) is correct. Electric potential of an isolated sphere is defined as (free space) C = 4pe0 a The Maxwell’s equation in phasor form is written as d # H = jweE + sE = jweE + J ^J = sE h So A and R both are true individually but R is not the correct explanation of A. Option (D) is correct. If a coil is placed in a time varying magnetic field then the e.m.f. will induce in coil. So here in both the coil e.m.f. will be induced.
SOL 6.3.16
Option (B) is correct. Both the statements are individually correct but R is not explanation of A.
SOL 6.3.17
Option ( ) is correct.
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SOL 6.3.15
d # H = J + 2D ^a " 3h 2t Faraday’ law d # E = 2B ^b " 4h 2t Gauss law d : D = rv ^c " 1h 2r Current continuity d : J =^d " 2h 2t Option (B) is correct. Since, the magnetic field perpendicular to the plane of the ring is decreasing with time so, according to Faraday’s law emf induced in both the ring is Vemf =- 2 B : dS 2t Therefore, emf will be induced in both the rings.
at e
SOL 6.3.18
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Ampere’s law
#
SOL 6.3.20
SOL 6.3.21
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Option (D) is correct. The Basic idea of radiation is given by the two Maxwells equation d # H = 2D 2t d # E =-2B 2t Option (B) is correct. The correct maxwell’s equation are d # H = J + 2D 2t d # E =-2B 2t Option (B) is correct. In List I
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SOL 6.3.19
d:D = r d:B = 0
#
a. B : dS = 0 The surface integral of magnetic flux density over the closed surface is zero or in other words, net outward magnetic flux through any closed surface is zero. ^a " 4h b.
# D : dS
=
# r dv v
v
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Total outward electric flux through any closed surface is equal to the charge enclosed in the region. ^b " 3h c. E : dl =- 2B dS 2t i.e. The line integral of the electric field intensity around a closed path is equal to the surface integral of the time derivative of magnetic flux density ^c " 2h 2 D d. H : dS = b + J l da 2t i.e. The line integral of magnetic field intensity around a closed path is equal to the surface integral of sum of the current density and time derivative of electric flux density. ^d " 1h
#
#
m
#
#
Option (C) is correct. The continuity equation is given as d : J =- rv i.e. it relates current density ^J h and charge density rv .
SOL 6.3.23
Option (A) is correct. Given Maxwell’s equation is d # H = Jc + 2D 2t For free space, conductivity, s = 0 and so, Jc = sE = 0 Therefore, we have the generalized equation d # H = 2D 2t Option (D) is correct. Given the magnetic field intensity, H = 3ax + 7yay + 2xaz So from Ampere’s circuital law we have J = d#H a x ay a z = 22x 22y 22z 3 7y 2x = ax ^0 h - ay ^2 - 0h + az ^0 h =- 2ay
SOL 6.3.24
SOL 6.3.25
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SOL 6.3.22
Option (D) is correct. The emf in the loop will be induced due to motion of the loop as well as the variation in magnetic field given as Vemf =- 2B dS + ^v # B h dl 2t So, the frequencies for the induced e.m.f. in the loop is w1 and w2 .
#
SOL 6.3.26
#
Option (B) is correct. F = Q ^E + v # B h is Lorentz force equation.
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Time Varrying Field and Maxwell Equation
Chap 6
Option (D) is correct. The direction of magnetic flux due to the current ‘i ’ in the conductor is determined by right hand rule. So, we get the flux through A is pointing into the paper while the flux through B is pointing out of the paper. According to Lenz’s law the induced e.m.f. opposes the flux that causes it. So again by using right hand rule we get the direction of induced e.m.f. is anticlockwise in A and clockwise in B .
SOL 6.3.29
Option (C) is correct. d2 A =- m0 J This is the wave equation for static electromagnetic field. i.e. It is not Maxwell’s equation.
SOL 6.3.30
Option (B) is correct. Poission’s equation for an electric field is given as r d2 V =- v e where, V is the electric potential at the point and rv is the volume charge density in the region. So, for rv = 0 we get, d2 V = 0 Which is Laplacian equation.
SOL 6.3.31
Option (B) is correct.
SOL 6.3.32
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SOL 6.3.28
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For View Only Shop Online at www.nodia.co.in SOL 6.3.27 Option (D) is correct. All of the given expressions are Maxwell’s equation.
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2r Continuity equation d # J =- v ^a " 4h 2t Ampere’s law d # H = J + 2D ^b " 1h 2t Displacement current J = 2D ^c " 2h 2t Faraday’ law d # E =-2B ^d " 3h 2t Option (B) is correct. A static electric field in a charge free region is defined as d:E = 0 ^a " 4h and d#E = 0 A static electric field in a charged region have r d:E = v ! 0 ^b " 2h e and d#E = 0 A steady magnetic field in a current carrying conductor have d:B = 0 ^c " 1h d # B = m0 J ! 0 A time varying electric field in a charged medium with time varying magnetic field
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d # E =-2B ! 0 ^d " 3h 2t r d:E = v ! 0 e Option (A) is correct. V =- dFm dt It is Faraday’s law that states that the change in flux through any loop induces e.m.f. in the loop.
# ^d # E h : dS
# E : dl
(1)
Given, the Maxwell’s equation d # E =- (2B/2t) Putting this expression in equation (1) we get, E : dl =- 2 B : dS 2t s Option (B) is correct. Induced emf in a coil of N turns is defined as Vemf =- N dF dt where F is flux linking the coil. So, we get Vemf =- 100 d ^t3 - 2t h dt =- 100 ^3t2 - 2h =- 100 _3 ^2 h2 - 2i =- 1000 mV (at t = 2 s ) =- 2 V
#
#
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SOL 6.3.35
=
m
Option (B) is correct. From stokes theorem, we have
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SOL 6.3.34
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lp.
SOL 6.3.33
403
SOL 6.3.36
Option (C) is correct. Since, the flux linking through both the coil is varying with time so, emf are induced in both the coils. Since, the loop 2 is split so, no current flows in it and so joule heating does not occur in coil 2 while the joule heating occurs in closed loop 1 as current flows in it. Therefore, only statement 2 is correct.
SOL 6.3.37
Option (A) is correct. The electric field intensity is where E 0 is independent of time E = E 0 e jwt So, from Maxwell’s equation we have d # H = J + e2E 2t = sE + e ^ jwh E 0 e jwt = sE + jweE
Option (A) is correct. Equation (1) and (3) are not the Maxwell’s equation. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 6.3.38
404
Time Varrying Field and Maxwell Equation
Chap 6
m
For View Only Shop Online at www.nodia.co.in SOL 6.3.39 Option (D) is correct. From the Maxwell’s equation for a static field (DC) we have d # B = m0 J d # ^d # Ah = m0 J d ^d : Ah - d2A = m0 J For static field (DC), d:A = 0 therefore we have, d2A =- m0 J So, both A and R are true and R is correct explanation of A. Option (D) is correct. For a static field, Maxwells equation is defined as d#H = J and since divergence of the curl is zero i.e. d : ^d # H h = 0 d:J = 0 but in the time varying field, from continuity equation (conservation of charges) 2r d : J =- v ! 0 2t So, an additional term is included in the Maxwell’s equation. i.e. d # H = J + 2D 2t where 2D is displacement current density which is a necessary term. 2t Therefore A and R both are true and R is correct explanation of A.
SOL 6.3.41
Option (D) is correct. For any loop to have an induced e.m.f., magnetic flux lines must link with the coil. Observing all the given figures we conclude that loop C1 and C2 carries the flux lines through it and so both the loop will have an induced e.m.f.
SOL 6.3.42
Option (A) is correct. Since, the circular loop is rotating about the y -axis as a diameter and the flux lines is directed in ax direction. So, due to rotation magnetic flux changes and as the flux density is function of time so, the magnetic flux also varies w.r.t time and therefore the induced e.m.f. in the loop is due to a combination of transformer and motional e.m.f. both.
SOL 6.3.43
Option (A) is correct. Gauss’s law d:D = r
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SOL 6.3.40
Ampere’s law Faraday’s law Poynting vector
d # H = Jc + 2D 2t d # E =-2B 2t P = E#H
^a " 1h ^b " 5h ^c " 2h ^d " 3h
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CHAPTER 7 ELECTRONAGNETICS WAVES
406
Electronagnetics Waves
EXERCISE 7.1
What will be the direction of wave propagation in a non magnetic medium in which magnetic field intensity at any point is given by H = 4 cos ^wt - ky h ax A/m (A) + az direction (B) - az direction
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MCQ 7.1.1
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m
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(D) + ay direction
lp.
(C) + ax direction
In a certain medium electric field intensity of a propagating wave is given by E ^x, t h = 2E 0 e- ax sin ^wt - bz h ay V/m The electric field phasor of the wave will be (A) E 0 e-^a + jb hx ay V/m (B) jE 0 e-^a + jb hx ay V/m (C) - jE 0 e-^a + jb hx ay V/m
In air, magnetic field intensity is given by H = 10 cos ^6 # 107 t - ky h az A/m . Wave number k for the EM wave will be (A) 1.8 rad/m (B) 2 rad/m
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MCQ 7.1.3
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(D) - jE 0 e^a - jbx h ay V/m
(C) 0.2 rad/m
(C) 0.5 # 108 m/s MCQ 7.1.5
(D) 5 rad/m
An electromagnetic wave is propagating in certain non magnetic material such that the magnetic field intensity at any point is given by H = 3 cos ^109 t - 5z h ax A/m The phase velocity of the wave in the medium will be (A) 1.5 # 109 m/s (B) 5 # 10-9 m/s
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MCQ 7.1.4
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MCQ 7.1.2
Chap 7
(D) 2 # 108 m/s
Magnetic field intensity in a certain non-magnetic medium is given by H = H 0 cos ^wt - by h ax A/m If the wavelength of the EM wave in the medium be 12.6 m then what will be the phase constant b in that medium ? (A) 0.25 rad/m (B) 0.5 rad/m (C) 1.12 rad/m
(D) 6.3 rad/m
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407
For View Only Shop Online at www.nodia.co.in MCQ 7.1.6 In a nonmagnetic material electric field intensity is given by E = 16 cos ^4 # 108 t - 2x h ay V/m The relative permittivity of the medium will be (A) 1.5 (B) 2.25 (C) 0.44
Electric field intensity in free space is E = 24 cos ^5 # 108 t - bz h ax V/m . The time period of the wave will be (A) 7.96 ns (B) 1.26 ns
m
MCQ 7.1.7
(D) 225
(C) 8 # 107 sec
(C) 7.23 ns MCQ 7.1.9
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(D) 0.13 mm
The skin depth in a poor conductor is independent of (A) Permittivity (B) Permeability (C) Frequency
MCQ 7.1.12
(D) 251.33 W
A radio wave is propagating at a frequency of 0.5 MHz in a medium ( s = 3 # 107 S/m , mr = er . 1).The wave length of the radio wave in that medium will be (A) 0.8 mm (B) 0.26 mm (C) 0.4 mm
MCQ 7.1.11
(D) 3.93 ns
What will be the intrinsic impedance of a lossless, nonmagnetic dielectric material having relative permittivity er = 2.25 ? (A) 235.62 W (B) 167.56 W (C) 8.95 mW
MCQ 7.1.10
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In air, a propagating wave has electric field intensity given by E = 9 cos ^4 # 108 t - bx h az V/m The time taken by the wave to travel one-fourth of it’s total wave length is (A) 61.42 ns (B) 3.05 ns
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MCQ 7.1.8
(D) 12.57 ns
(D) None of these
Assertion (A) : E = E 0 sin ^z h cos ^ct h ax represents the electric field of a plane wave in free space. Reason (R) : A plane wave f propagating with velocity v p in + az direction must satisfy the equation 2 22 f 22 f =0 v p 2t2 2z2 (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false (D) A is false but R is true
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Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.1.13 A propagating wave in free space has magnetic field intensity H = 0.2 cos ^109 t - by h az A/m What will be the electric field intensity of the wave at y = 1 cm at time, t = 0.1 ns ? (B)- 37.6ax V/m (A) - 37.7ax V/m (C) - 19.8ax V/m
Phasor form of magnetic field intensity of a uniform plane wave in free space is given as Hs = ^2 + j5h^4ay + 2jaz h e-jbx A/m The maximum electric field of the plane wave equals to (A) 24.1 V/m (B) 14.22 kV/m
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m
MCQ 7.1.14
(D) 37.6ax V/m
(C) 9.08 kV/m
(A) - h0 ^5ax + 6ay h e-j50z V/m -j50z
(C) ^5ax - 6ay he h
0
-j50z
(B) ^5ax - 6az he h
0
V/m
-j50z
(D) -^5ax + 6ay he V/m h0
In a perfect conductor ^resistivity, r . 0h magnetic field of any EM wave (A) lags electric field by 90c (B) leads electric field by 45c
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MCQ 7.1.16
V/m
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Electric field intensity of linearly polarized plane wave in free space is given by E = ^6ay - 5ax h cos ^wt - 50z h V/m The phasor form of magnetic field intensity of the wave will be
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MCQ 7.1.15
(D) 0
(D) will be in phase with electric field
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(C) lags electric field by 45c
Statement for Linked Question 17 - 18 : An electromagnetic wave travels in free space with the electric field component Es = ^10ax + 5az h e-j^4x - 2z h V/m What will be the phasor form of magnetic field intensity of the wave ? (A) - 29.66e-j^4x - 2z h mA/m
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MCQ 7.1.17
(B) - 5 5 e-j^4x - 2z h mA/m (C) 29.66e-j^4x - 2z h mA/m (D) - 29.66e-j^4x - 2z h A/m MCQ 7.1.18
What will be the time average power density of the electromagnetic wave ? (A) ^665.9ax - 331.6az h W/m2
(B) ^148.9ax - 74.15az h W/m2
(C) ^- 331.6ax + 665.9az h W/m2
(D) ^- 74.15ax + 148.9az h W/m2
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409
For View Only Shop Online at www.nodia.co.in MCQ 7.1.19 A propagating wave has the phasor form of its electric field intensity defined as Es = ^- 2 3 ax + 3 ay - az h e-j0.1p`-3x + 3 y - 2z j V/m The wave is linearly polarized along the direction of (A) - 3ax + 3 ay - 2az (B) - 2 3 ax + 3 ay - az (C) 3ax -
3 ay + 2az
(D) 2 3 ax -
3 ay + a z
Statement for Linked Question 20 - 21 :
w . what will be the magnetic flux density vector B ? m0 e0 1 -8 c m e = 3 # 10 m/s m 0 0 (A) 3 # 1010 cos ^wt - bz h ay
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If b =
(B) 3.33 # 10-7 cos ^wt - bz h ay
The poynting vector of the E -M field will be (B) 10 4 (A) 10 4 e0 cos2 ^wt - bz h az m0
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MCQ 7.1.21
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(C) 3 # 108 cos ^wt - bz h ay (D) 3.33 # 10-6 cos ^wt - bz h ay
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MCQ 7.1.20
m
In free space an electric field intensity vector is given by E = 200 cos ^wt - bz h ax where w and b are constants.
m0 cos2 ^wt - bz h az e0
4 (D) 10 cos2 ^wt - bz h az m0 The electric field associated with a sinusoidally time varying electromagnetic field is given by E = 15 sin px sin ^2p # 108 - 3 pz h ay V/m The time average stored energy density in the electric field is (A) 4 e0 sin2 px (B) 25e0 sin2 px 25 4
(C) 10 4 m0 cos2 ^wt - bz h az MCQ 7.1.22
(C) 5e0 sin 2px 4 MCQ 7.1.23
(D) 25e0 sin 2px 4
Electric field associated with a sinusoidally time varying electromagnetic field is given by E = 20 sin (py) sin ^6p # 108 t - 3 px h az V/m What will be the time average stored energy density in the magnetic field ? -9 -9 (B) 10 ^25 + 50 sin2 px h (A) 10 ^25 + 50 sin2 px h p 144p 9 (C) 144 # 10 ^25 + 50 sin2 px h p
(D) p ^25 + 50 sin2 px h 144
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Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.1.24 An electromagnetic wave is propagating from free space to a certain medium having relative permittivity er = 5 . If wavelength of the wave in the medium be 20 cm then what would be it’s wavelength in free space ? (A) 6.67 cm (B) 60 cm (C) 180 cm
If some free charge is being imbedded in a piece of glass, then the charge will flow out to the surface nearly after (relative permittivity of glass, er = 4.25 conductivity of glass, s = 10-12 S/m ) (A) 2 sec
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MCQ 7.1.25
m
(D) 18 cm
(B) 4 sec
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(C) 35 sec (D) 20 sec
(B) - E 0 /3 (C) E 0 /2
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(D) - E 0
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An electromagnetic wave propagating in free space is incident on the surface of a dielectric medium ( m0 , 4e0 ). If the magnitude of the electric field of incident wave is E 0 then what will be the magnitude of the electric field of the reflected wave ? (A) - 2E 0 /3
at e
MCQ 7.1.26
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EXERCISE 7.2
Magnetic field intensity of a propagating wave in free space is given by H = 0.3 cos ^wt - by h ax A/m
m
MCQ 7.2.1
411
MCQ 7.2.2
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lp.
co
If the total time period of the wave be T then the plot of H versus y at time, t = T 8 will be
A uniform plane wave is propagating with a velocity of 7.5 # 107 m/s in a lossless medium having relative permeability mr = 2.8 . The electric field phasor of the wave is given by Es = 5e j0.3x az V/m What will be the magnetic field intensity of the wave ? (A) 11.05 cos ^9.54 # 106 t + 0.3x h ay mA/m (B) 22.13 cos ^9.54 # 106 t + 0.3x h ay mA/m (C) 22.13 cos ^9.54 # 106 t + 0.3z h ay mA/m
(D) 11.05 cos ^2.25 # 107 t + 0.3x h ay mA/m GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 7.2.3 In a certain medium ( er = 4 , mr = 4 ). A plane wave is propagating such that the electric field intensity of the wave is E = E 0 e-x/2 sin ^108 t - bx h ay V/m . The loss tangent of the medium will be (A) 1.94 (B) 0.27 (C) 0.35
(D) 0.52
m
Statement for Linked Question 4 - 5 :
MCQ 7.2.4
co
In a lossy medium ( er = 8 , mr = 0.5 , s = 0.01 S/m a plane wave is travelling in + az direction that has the electric field intensity E = 0.5 cos ^109 pt + p/3h ax at z = 0 . What will be the distance travelled by the wave to have a phase shift of 10c ? (A) 20.95 mm (B) 477.3 mm
After traveling a distance z , the amplitude of the wave is reduced by 40% . So, the value of z equals to (A) 481.5 mm (B) 542 mm
he
MCQ 7.2.5
(D) 3.65 m
lp.
(C) 8.33 V
(C) 1.06 m
A uniform plane wave is propagating at a velocity of 7 # 107 m/s in a perfect dielectric such that the electric and magnetic fields of the wave are given by E ^x, t h = 300 cos ^5 # 106 pt - bx h ay V/m H ^x, t h = 1.9 cos ^5 # 106 pt - bx h az V/m The relative permittivity and relative permeability of the medium will be respectively (A) 1.70, 2.69 (B) 3.4, 5.37 (C) 1.70, 1.58
(D) 2.37, 2.69
An electromagnetic wave is propagating in free space in - ax direction with a frequency w and phase angle zero. The EM wave is polarized in + az direction. If the amplitude of electric field of the wave is E 0 then the magnetic field of the wave will be (B) h0 E 0 cos ^wt + wcx h ay (A) E 0 cos a wt + w x k ay h0 c
ww
MCQ 7.2.7
w. g
at e
MCQ 7.2.6
(D) 2.08 m
(C) - E 0 cos a wt + w x k ay h0 c
(D) E 0 cos a wt - w x k ay h0 c
What will be the electric field of a plane wave polarized parallel to the x -z plane and propagating in free space in the direction from origin to the point ^1, 1, 1h, that has the amplitude E 0 and frequency w with zero phase angle ? (A) E 0 cos :wt - w ^x + y + z hDb ax - az l 3c 2 a az w (B) E 0 cos :wt + ^x + y + z hDb x 3c 2 l GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 7.2.8
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(C) E 0 cos 9wt - w ^x + y + z hCc ax + az m c 2 (D) E 0 cos 9wt + w ^x + y + z hCc ax + az m c 2 MCQ 7.2.9
A plane wave is propagating with frequency f = 50 kHz in a medium ( s = 2 S/m , er = 80 , mr = 4 ). What will be the skin depth of the medium ? (A) 1.7 m (B) 0.4 m (C) 0.8 m
m
With a thickness t , silver coating is done for a microwave experiment to operate at a frequency of 10 GHz. For the successful experiment t should be (for silver, mr = er . 1, s = 6.25 # 107 S/m ) (A) greater than 0.64 mm (B) less than 0.64 mm
co
MCQ 7.2.10
(D) 1.3 m
(C) exactly equal to 0.64 mm
lp.
In a nonmagnetic material of conductivity s = 2 # 107 S/m , electric field of a propagating plane wave is given by E = 5 cos ^107 t - 0.2y h ax + 2 sin ^107 t - 0.2y h az V/m What will be the value of complex permittivity of the medium ? (B) ^36e0 - j2h F/m (A) ^2 - j36e0h F/m (C) ^36 - j2h F/m
(D) ^36e0 + j2h F/m
Assertion (A) : All the metals are opaque. Reason (R) : Skin depth of metals are in the range of nanometers. (A) Both A and R are true and R is the correct explanation of A.
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MCQ 7.2.12
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MCQ 7.2.11
(D) none of these
(B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. MCQ 7.2.13
In the plane z = 0 , electric field of a wave propagating in + az direction in free space is E 0 which is varying with time t as shown in the figure.
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Three different dielectrics of permittivities 4e0 , 9e0 and 3e0 are defined in the space as shown in figure. If the leading edge of a uniform plane wave propagating in ax direction is incident on the plane x =- 3 m then how much time it will take to strike the interface defined by the dielectric 2 and dielectric 3 ?
ww
w. g
MCQ 7.2.14
at e
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lp.
co
m
For View Only Shop Online at www.nodia.co.in If the magnetic field intensity of the wave at t = 1 m sec be H1 then the plot of H1 versus z will be
(A) 6 n sec
(B) 0.02 m sec (C) 3 n sec
(D) 0.06 m sec MCQ 7.2.15
An electromagnetic wave propagating in medium 1 ( m0 , e1 ) is incident on medium 2 ( m0 , e2 ) as shown in figure such that the electric field of reflected wave is 1/5 times of the electric field of incident wave. The value of e1 /e2 equals to
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(B) 3/2
(C) 4/9
(D) 9/4
co
m
(A) 2/3
Statement for Linked Question 16 - 17 :
What will be the conductivity of the dielectric ? (A) 0.32 # 102 S/m (C) 320 S/m
If an electromagnetic wave of 8 GHz frequency travels a distance of 0.275 mm in the dielectric medium then it’s field intensity will be reduced by (A) 20 dB (B) 60 dB
ww w. ga te
MCQ 7.2.17
(C) 0 dB MCQ 7.2.18
(B) 1.01 # 10-5 S/m (D) 0.99 # 105 S/m
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MCQ 7.2.16
lp.
An electromagnetic wave of 50 MHz frequency is incident on a dielectric medium such that it’s skin depth is 0.32 mm. (permittivity of dielectric = 6.28 # 10-7 )
(D) 30 dB
Electric field of an electromagnetic wave propagating in a medium in + ax direction is given by Es = E 0 ^ay - jaz h e-jbx The wave is (A) left hand circularly polarized (B) Right hand circularly polarized (C) elliptically polarized (D) linearly polarized
MCQ 7.2.19
An electromagnetic wave has the electric field intensity in the phasor form given by Es = 2 ^az - jax h e-jby The EM wave is incident on a perfect conductor located at y = 0 . What will be the polarization of the reflected wave ? (A) left hand circular (B) Right hand circular (C) elliptical
(D) linear
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For View Only Shop Online at www.nodia.co.in MCQ 7.2.20 An electromagnetic wave propagating in free space is incident on a perfectly conducting slab placed at x $ 0 . The electric field of the incident wave in the phasor form is given by Eis = 5az e-j^6y + 8x h V/m The net electric field of the total wave (incident and reflected both) in free space after reflection will be (A) 10az e-j^6y - 8x h V/m (B) - 10az e-j^6y - 8x h V/m (D) j20az e-j6y sin 8x V/m
Electric field intensity of an EM wave propagating in free space is given by Eis = 25ax e-j^6z + 8y h V/m If the wave is incident on a perfectly conducting plane at y = 0 then the magnetic field intensity of the reflected wave will be a a (A) -a y + az k e-j^6z - 8y h A/m (B) a y + az k e-j^6z - 8y h A/m 8p 6p 8p 6p
lp.
co
MCQ 7.2.21
m
(C) - j20az e-j6y sin 8x V/m
a (C) a y + az k e j^6z - 8y h 8p 6p
he
An electromagnetic wave propagating in free space has magnetic field intensity H = 0.4 cos ^wt - by h ax A/m What will be the total power passing through a square plate of side 20 cm located in the plane x + y = 2 ? (A) 0.53 Watt (B) 1.88 Watt (C) 18.8 mW
(D) 53.31 mW
An electromagnetic wave propagating in a lossless medium ( m1 = 4m0 , e1 = e0 , s1 = 0 ) defined in the region y > 0 is incident on a lossy medium ( m2 = m0 , e2 = 4e0 , s2 = 0.1 S/m ) defined in the region y # 0 . The electric field intensity of the incident wave in lossless medium is given by Eis = 6e-j5y az V/m What will be the standing wave ratio ? (A) 1.22 (B) 0.8186
ww
w. g
MCQ 7.2.23
at e
MCQ 7.2.22
a (D) -a y + az k e j^6z - 8y h A/m 8p 6p
(C) 0.0997 MCQ 7.2.24
(D) 10.025
The complex electric field vector of a uniform plane wave propagating in free space is given by Es = ^ 3 ax - ay - 2 3 az h e-j0.01p`-3x + 3 y - 2z j V/m The unit vector in the direction of propagation of the wave will be - 3ax + 3 ay - 2az - 3ax + 3 ay - 2az (A) (B) 16 4 (C) - 4 ^3ax -
3 ay + 2az h
(D) -
^3ax + 3 ay - 2az h 4
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For View Only Shop Online at www.nodia.co.in MCQ 7.2.25 Phasor form of electric field intensity of a uniform plane wave is given by Es = d 2 ax - 2 ay n e-j0.04p`-2x - 3y + 3 z j V/m 3 The wavelength along the direction of propagation is (A) 3.16 m (B) 0.08 m (C) 12.5 m
In free space the complex magnetic field vector of a uniform plane wave is given by Hs =-^ 3 ax + az h e-j0.04p` 3 x - 2y - 3z j A/m . Frequency of the plane wave will be (A) 3.75 MHz (B) 2.4 # 106 Hz
m
MCQ 7.2.26
(D) 15.7 m
(D) 2.4 MHz
co
(C) 24 MHz
Statement for Linked Question 27 - 28 :
lp.
In free space complex electric field vector of a uniform plane wave is given by Es = ^ 2 ax + az h e-j 25` p
25 m 16.7 m 28.87 m 25 m
28.87 m 25 m 25 m 16.7 m
The apparent phase velocities along the x , y and z axes are vPx vPy vPz (A) (B) (C) (D)
MCQ 7.2.29
16.7 m 28.87 m 16.7 m 28.87 m
he
(A) (B) (C) (D) MCQ 7.2.28
V/m
The apparent wavelengths along the x , y and z axes are lx ly lz
ww w. ga te
MCQ 7.2.27
2 x - 5y - 3z j
1.73 # 1010 m/s 6.93 # 108 m/s 2.77 # 107 m/s 1.2 # 109 m/s
1.5 # 1010 m/s 6 # 108 m/s 2.4 # 107 m/s 1.2 # 109 m/s
1 # 1010 m/s 4 # 108 m/s 1.6 # 107 m/s 1.2 # 109 m/s
Which of the following complex vector field represents the electric field of a uniform plane wave ? (A) _- jax - 2ay - j 3 az i e-j0.6p` 3 y + z j (B) _ax - j2ay -
3 az i e-j0.05p^x +
3 zh
j 3 (C) =b 3 + j 1 l ax + c1 + a - j 3 azG e-j0.02p` 2 2 m y
3 x + 3y + 2z j
(D) =b- 3 - j 1 l ax + c1 - j 3 m ay + j 3 azG e-j0.02p` 2 2
3 x + 3y + 2z j
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For View Only Shop Online at www.nodia.co.in MCQ 7.2.30 Which of the following pairs of vector Es and Hs field represents the complex electric and magnetic field vectors of a uniform plane wave ? Es Hs (A) _ jax + 2ay + j 3 az i e-jp^ 3 x + z h V/m (B) _ jax + j 3 az i e-jp^x + 3 z h V/m (C) _ jax - j 3 az i e
-jp^x + 3 z h
-jp x + _- jax + j 3 az i e ^
_- 3 jax - jaz i e
V/m 3 x+zh
V/m
3 x+zh 3 zh
-jp^x + 3 z h
-j0.1p^ _ax - j2ay - 3 az i e
A/m
A/m A/m
3 x+zh
A/m
m
(D) _- jax - 2ay + j 3 az i e-j0.1p^
The following fields exist in charge free regions P = 65 sin (wt + 10y) az Q = 10 cos (wt - 2r) af r R = 3r2 cot fa r + 1 cos faf r S = 6 sin q sin (wt - 6r) a q r The possible electromagnetic fields are (A) P, Q (B) R, S
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lp.
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MCQ 7.2.31
-jp _ax - j2ay - 3 az i e ^
(C) P, R
(B) 17.9 (C) 25.6
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(D) 58.3
at e
A uniform plane wave in region 1 is normally incident on the planner boundary separating regions 1 and 2. Both region are lossless and er1 = mr31 , er2 = mr32 . If the 20% of the energy in the incident wave is reflected at the boundary, the ratio er2 /er1 is (A) 1.48
w. g
MCQ 7.2.32
(D) Q, S
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EXERCISE 7.3
A plane wave propagating in air with E = (4ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m is incident on a perfectly conducting slab positioned at x # 0 . The E field of the reflected wave is (A) (- 8ax - 6ay - 5az ) e j (wt + 3x + 4y) V/m
m
GATE 2012
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-
(B) (- 8ax + 6ay - 5az ) e j (wt + 3x + 4y) V/m (C) (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m
MCQ 7.3.2
The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by E = 10 (ay + jaz ) e-j 25x . The frequency and polarization of the wave, respectively, are (B) 4 Hz and left circular (A) 1.2 GHz and left circular
he
GATE 2012
lp.
(D) (- 8ax + 6ay - 5az ) e j (wt - 3x - 4y) V/m
co
MCQ 7.3.1
419
MCQ 7.3.3 GATE 2011
(D) 4 Hz and right circular
ww w. ga te
(C) 1.2 GHz and right circular
Consider the following statements regarding the complex Poynting vector P for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(P ) denotes the real part of P, S denotes a spherical surface whose centre is at the point source, and an denotes the unit surface normal on S . Which of the following statements is TRUE? (A) Re(P ) remains constant at any radial distance from the source (B) Re(P ) increases with increasing radial distance from the source (C) (D)
MCQ 7.3.4 GATE 2010
##s Re ^P h : (dSan) remains constant at any radial distance from the source ##s Re ^P h : (dSan) decreases with increasing radial distance from the source
The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m2 ) is (B) 1 (A) 1 30p 60p (C)
1 120p
(D)
1 240p
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Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.3.5 A plane wave having the electric field components Ei = 6 cos ^3 # 108 - by h ax GATE 2010 V/m and traveling in free space is incident normally on a lossless medium with m = m0 and e = 9e0 which occupies the region y $ 0 . The reflected magnetic field component is given by (A) 1 cos (3 # 108 t + y) ax A/m (B) 1 cos (3 # 108 t + y) ax A/m 10p 20p (C) - 1 cos (3 # 108 t + y) ax A/m 20p
A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant er = 9 ). The magnitude of the reflection coefficient is (A) 0 (B) 0.3
m
GATE 2008
co
MCQ 7.3.6
(C) 0.5
GATE 2007
3 p x- p z l l m
(C) E = ay E 0 e jc wt +
3 p x+ p z l l m
3pz l m
p
3pz l m
(D) E = ay E 0 e jc wt - l x +
(D) 50 h0
A right circularly polarized (RCP) plane wave is incident at an angle 60c to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant er2 is
ww
GATE 2007
p
(B) E = ay E 0 e jc wt - l x -
The H field (in A/m) of a plane wave propagating in free space is given by H = ax 5 5 cos (wt - bz) + ay a wt - bz + p k. h0 2 The time average power flow density in Watts is h (B) 100 (A) 0 h0 100 (C) 50h20
MCQ 7.3.9
lp.
(A) E = ay E 0 e j c wt -
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MCQ 7.3.8
A plane wave of wavelength l is traveling in a direction making an angle 30c with positive x -axis and 90c with positive y -axis. The E field of the plane wave can be represented as (E0 is constant)
at e
GATE 2007
(D) 0.8
w. g
MCQ 7.3.7
(D) - 1 cos (3 # 108 t + y) ax A/m 10p
(A)
2
(B)
3
(C) 2 (D) 3 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 7.3.10 The electric field of an electromagnetic wave propagation in the positive direction GATE 2006 is given by E = 2ax sin (wt - bz) + ay sin (wt - bz + p/2). The wave is (A) Linearly polarized in the z -direction (B) Elliptically polarized (C) Left-hand circularly polarized (D) Right-hand circularly polarized MCQ 7.3.11
When a plane wave traveling in free-space is incident normally on a medium having er = 4.0 then the fraction of power transmitted into the medium is given by (A) 8 (B) 1 9 2 (C) 1 (D) 5 3 6
MCQ 7.3.12
A medium of relative permittivity er 2 = 2 forms an interface with free - space. A point source of electromagnetic energy is located in the medium at a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam cross-section at the interface is given by (A) 2p m 2 (B) p2 m 2 (C) p m 2 (D) p m 2 2
GATE 2006
A medium is divided into regions I and II about x = 0 plane, as shown in the figure below.
ww w. ga te
MCQ 7.3.13
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lp.
GATE 2006
co
m
GATE 2006
An electromagnetic wave with electric field E1 = 4ax + 5ay + 5az is incident normally on the interface from region I . The electric file E2 in region II at the interface is (A) E2 = E1 (B) 4ax + 0.75ay - 1.25az (C) 3ax + 3ay + 5az MCQ 7.3.14 GATE 2005
(D) - 3ax + 3ay + 5az
The magnetic field intensity vector of a plane wave is given by H (x, y, z, t) = 10 sin (50000t + 0.004x + 30) ay where ay , denotes the unit vector in y direction. The wave is propagating with a phase velocity. (A) 5 # 10 4 m/s (B) - 3 # 108 m/s
(D) 3 # 108 m/s (C) - 1.25 # 107 m/s GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 7.3.15 Refractive index of glass is 1.5. Find the wavelength of a beam of light with frequency GATE 2005 of 1014 Hz in glass. Assume velocity of light is 3 # 108 m/s in vacuum (A) 3 mm (B) 3 mm (C) 2 mm MCQ 7.3.16 GATE 2004
(D) 1 mm
If E = (ax + jay) e jkz - kwt Poynting vector is (A) null vector
and H = (k/wm) (ay + jax ) e jkz - jwt , the time-averaged (B) (k/wm) az
A plane electromagnetic wave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with e > e0 . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of the medium should be (A) 120p W (B) 60p W
co
GATE 2004
lp.
MCQ 7.3.17
(D) (k/2wm) az
m
(C) (2k/wm) az
GATE 2003
The depth of penetration of electromagnetic wave in a medium having conductivity s at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be (A) 6.25 dm (B) 12.50 cm
at e
MCQ 7.3.18
(D) 24p W
he
(C) 600p W
(C) 50.00 cm GATE 2003
A uniform plane wave traveling in air is incident on the plane boundary between air and another dielectric medium with er = 5 . The reflection coefficient for the normal incidence, is (A) zero (B) 0.5 180c (B) 0.333 0c
GATE 2003
(C) 6.28 # 107 m/sec MCQ 7.3.21 GATE 2002
GATE 2002
(D) 2.00 # 107 m/sec
A plane wave is characterized by E = (0.5ax + ay e jp/2) e jwt - jkz . This wave is (A) linearly polarized (B) circularly polarized (C) elliptically polarized
MCQ 7.3.22
(D) 0.333 180c
If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is given by E (z, t) = 10 cos (2p107 t - 0.1pz) V/m, then the velocity of the traveling wave is (A) 3.00 # 108 m/sec (B) 2.00 # 108 m/sec
ww
MCQ 7.3.20
w. g
MCQ 7.3.19
(D) 100.00 cm
(D) unpolarized
Distilled water at 25cC is characterized by s = 1.7 # 10-4 mho/m and e = 78eo at a frequency of 3 GHz. Its loss tangent tan d is ( e = 10 36p F/m) -9
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For View Only (A) 1.3 # 10-5
Shop Online at www.nodia.co.in (B) 1.3 # 10-3
(C) 1.3 # 10-4 /78 MCQ 7.3.23 GATE 2001
423
(D) 1.3 # 10-5 /78e0
2 2 If a plane electromagnetic wave satisfies the equation 2E2x = c22E2x , the wave 2z 2t propagates in the (A) x -direction
(B) z -direction (C) y -direction
MCQ 7.3.25 GATE 2001
GATE 2000
GATE 1999
GATE 1999
2
(D) er2 /er1
Identify which one of the following will NOT satisfy the wave equation. (A) 50e j (wt - 3z) (B) sin [w (10z + 5t)] (C) cos (y2 + 5t)
MCQ 7.3.29
(D)
Two coaxial cable 1 and 2 are filled with different dielectric constants er1 and er2 respectively. The ratio of the wavelength in the cables (l1 /l2) is (B) er2 /er1 (A) er1 /er2 (C) er1 /er2
MCQ 7.3.28
(D) 75%
A uniform plane wave in air impinges at 45c angle on a lossless dielectric material with dielectric constant er . The transmitted wave propagates is a 30c direction with respect to the normal. The value of er is (A) 1.5 (B) 1.5 (C) 2
MCQ 7.3.27
ww w. ga te
GATE 2000
(D) 900 MHz
A uniform plane electromagnetic wave incident on a plane surface of a dielectric material is reflected with a VSWR of 3. What is the percentage of incident power that is reflected ? (A) 10% (B) 25% (C) 50%
MCQ 7.3.26
co
(C) 450 MHz
lp.
GATE 2001
A material has conductivity of 10-2 mho/m and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is (A) 45 MHz (B) 90 MHz
he
MCQ 7.3.24
m
(D) x -y plane at an angle of 45c between the x and z direction
(D) sin (x) cos (t)
A plane wave propagating through a medium [er = 8, vr = 2, and s = 0] has its electric field given by E = 0.5Xe- (z/3) sin (108 t - bz) V/m . The wave impedance, in ohms is (A) 377 (B) 198.5 180c
(C) 182.9 14c (D) 133.3 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 7.3.30 The intrinsic impedance of copper at high frequencies is GATE 1998 (A) purely resistive (B) purely inductive (C) complex with a capacitive component (D) complex with an inductive component GATE 1998
The time average poynting vector, in W/m2 , for a wave with E = 12e j (wt + bz) ay V/m in free space is (B) 2.4 az (A) - 2.4 az p p
m
MCQ 7.3.31
MCQ 7.3.33 GATE 1998
lp.
GATE 1998
The wavelength of a wave with propagation constant (0.1p + j0.2p) m-1 is 2 m (B) 10 m (A) 0.05 (C) 20 m (D) 30 m The depth of penetration of wave in a lossy dielectric increases with increasing (A) conductivity (B) permeability
he
MCQ 7.3.32
(D) - 4.8 az p
co
(C) 4.8 az p
(C) wavelength GATE 1998
The polarization of wave with electric field vector E = E 0 e j^wt + bz h ^ax + ay h is (A) linear (B) elliptical (C) left hand circular
GATE 1997
(C) 3 # 108 m/ sec GATE 1996
(C) 20% MCQ 7.3.37 GATE 1996
(D) 6 # 108 m/ sec
A uniform plane wave in air is normally incident on infinitely thick slab. If the refractive index of the glass slab is 1.5, then the percentage of incident power that is reflected from the air-glass interface is (A) 0% (B) 4%
ww
MCQ 7.3.36
(D) right hand circular
The skin depth at 10 MHz for a conductor is 1 cm. The phase velocity of an electromagnetic wave in the conductor at 1, 000 MHz is about (B) 6 # 107 m/ sec (A) 6 # 106 m/ sec
w. g
MCQ 7.3.35
at e
MCQ 7.3.34
(D) permittivity
(D) 100%
Some unknown material has a conductivity of 106 mho/m and a permeability of 4p # 10-7 H/m . The skin depth for the material at 1 GHz is (A) 15.9 mm (B) 20.9 mm (C) 25.9 mm
(D) 30.9 mm
The plane wave travelling in a medium of er = 1, mr = 1 (free space) has an electric field intensity of 100 p V/m . Determine the total energy density of this field. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 7.3.38 IES EC 2012
Chap 7
Electronagnetics Waves
For View Only (A) 13.9 nJ/m3
Shop Online at www.nodia.co.in (B) 27.8 nJ/m3
(C) 139 nJ/m3 MCQ 7.3.39 IES EC 2012
(D) 278 nJ/m3
For a plane wave propagating in an unbounded medium (say, free space), the minimum angle between electric field and magnetic field vectors is (A) 0c (B) 60c (C) 90c
IES EC 2011
(D) 180c
For no reflection condition, a vertically polarized wave should be incident at the interface between two dielectrics having e1 = 4 and e2 = 7 , with an incident angle of (A) tan-1 b 9 l (B) tan-1 b 3 l 2 4
m
MCQ 7.3.40
lp.
The electric field component of a wave in free space is given by E = 10 cos (107 t + kZ) ay V/m Following is a list of possible inferences : 1. Wave propagates along ay Wavelength l = 188.5 m
3.
Wave amplitude is 10 V/m
4.
Wave number = 0.33 rad/m
5.
Wave attenuates as it travels
he
2.
ww w. ga te
IES EC 2011
(D) tan-1 b 4 l 9
co
(C) tan-1 b 2 l 3 MCQ 7.3.41
425
Which of these inferences can be drawn from E ? (A) 1, 2, 3, 4 and 5 (B) 2 and 3 only (C) 3 and 4 only MCQ 7.3.42 IES EC 2011
(D) 4 and 5 only
A plane wave is generated under water (e = 81e0 and m = m0). The wave is parallel polarized. At the interface between water and air, the angle a for which there is no reflection is
(A) 83.88c
(B) 83.66c
(C) 84.86c
(D) 84.08c
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
426
Electronagnetics Waves
Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.3.43 An elliptically polarized wave travelling in the positive z -direction in air has x and y IES EC 2010 components Ex = 3 sin (wt - bz) V/m Ey = 3 sin (wt - bz + 75c) V/m If the characteristic impedance of air is 360 W, the average power per unit area conveyed by the wave is (A) 8 W/m2 (B) 4 W/m2
IES EC 2010
The intrinsic impedance of copper at 3 GHz (with parameters : m = 4p # 10-7 H/m ; e = 10-79 /36p ; and s = 5.8 # 107 mho/m ) will be (A) 0.02e jp/4 ohm (B) 0.02e jp/2 ohm
co
MCQ 7.3.44
(D) 125 mW/m2
m
(C) 62.5 mW/m2
he
IES EC 2010
Consider the following statements regarding depth of penetration or skin depth in a conductor : 1. It increases as frequency increases. 2.
It is inversely proportional to square root of m and s.
3.
It is inversely proportional to square root of f
4.
It is directly proportional to square root of m and s.
at e
MCQ 7.3.45
(D) 0.2e jp/4 ohm
lp.
(C) 0.2e jp/2 ohm
Which of the above statements are correct ? (A) 1 and 2 only (B) 3 and 4 only
MCQ 7.3.46
Consider the following statements : 1. (Electric or magnetic) field must have two orthogonal linear components. 2.
The two components must have the same magnitude.
3.
The two components must have a time-phase difference of odd multiple of 90c .
ww
IES EC 2010
(D) 1, 2, 3 and 4
w. g
(C) 2 and 3 only
Which of these are the necessary and sufficient conditions for a time-harmonic wave to be circularly polarized at a given point in space ? (A) 1 and 2 only (B) 2 and 3 only (C) 1, 2 and 3 MCQ 7.3.47 IES EC 2010
(D) 1 and 3 only
Assertion (A) :The velocity of light in any medium is slower than that of vacuum. Reason (R) : The dielectric constant of the vacuum is unity and is lesser than that of any other medium. (A) Both A and R are individually true and R is the correct explanation of A (B) Both A and R are individually true but R is not the correct explanation of A
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 7
Electronagnetics Waves
For View Only (C) A is true but R is false
427
Shop Online at www.nodia.co.in
(D) A is false but R is true IES EC 2009
In which direction is the plane wave E = 35 sin ^108 t + 2z h ay V/m , (where ay is the unit vector in y -direction), travelling ? (A) along y direction (B) along y direction (C) along z direction
MCQ 7.3.49 IES EC 2008
(D) along z direction
According to Poynting theorem, the vector product E # H is a measure of which one of the following? (A) Stored energy density of the electric field (D) Power dissipated per unit volume (E) Rate of energy flow per unit area
IES EC 2007
If E = (ax + jay) e-jbz , then the wave is said to be which one of the following ? (A) Right circularly polarized (B) Right elliptically polarized (D) Left elliptically polarized
he
(C) Left circularly polarized
lp.
MCQ 7.3.50
co
(B) Stored energy density of the magnetic
m
MCQ 7.3.48
MCQ 7.3.51
ww w. ga te
IES EC 2007
What must be angle q of a corner reflector, such that an incident wave is reflected in the same direction ? (A) 30c (B) 45c (C) 60c MCQ 7.3.52 IES EC 2007
(D) 90c
Poynting vector is a measure of which one of the following ? (A) Maximum power flow through a surface surrounding the source (B) Average power flow through the surface (C) Instantaneous power flow through the surface (D) Power dissipated by the surface
The electric field component of a wave in free space is given by IES EC 2006 E = 25 sin (107 t + kz) ay V/m Which one of the following is the correct inference that can be drawn from this expression ? GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 7.3.53
428
Electronagnetics Waves
Chap 7
For View Only Shop Online at www.nodia.co.in (A) The wave propagates along y -axis (B) The wavelength is 188.5 m (C) The wave number k = 0.33 rad/m (D) The wave attenuates as it travels MCQ 7.3.54
m
IES EC 2005
For an electromagnetic wave incident on a conducting medium, the depth of penetration (A) is directly proportional to the attenuation constant (B) is inversely proportional to the attenuation constant
co
(C) has a logarithmic relationship with the attenuation constant (D) is independent of the attenuation constant IES EC 2004
Which one of the following statements is correct ? A right circularly polarised wave is incident from air onto a polystyrene ^er = 2.7h . The reflected wave is (A) right circularly polarised (B) left circularly polarised
IES EC 2004
The electric field of a wave propagating through a lossless medium (m0, 81e0) is E = 50 cos (6p # 108 t - bx) a y What is the phase constant b of the wave ? (A) 2p rad/m (B) 9p rad/m (C) 18p rad/m
IES EC 2004
(C) 2.5 IES EC 2004
(C) 120 W MCQ 7.3.59 IES EC 2004
(D) 1.25
In free space E (x, t) = 60 (wt - 2x) ay V/m . What is the average power crossing a circular area of radius 4 m in the plane x = constan t ? (A) 480 W (B) 340 W
ww
MCQ 7.3.58
(D) 81 rad/m
If the phase velocity of a plane wave in a perfect dielectric is 0.4 times its value in free space, then what is the relative permittivity of the dielectric ? (A) 6.25 (B) 4.25
w. g
MCQ 7.3.57
at e
MCQ 7.3.56
(D) left elliptically polarised
he
(C) right elliptically polarised
lp.
MCQ 7.3.55
(D) 60 W
What is the effect of the earth’s magnetic field in the reflected wave at frequencies in the vicinity of gyro-frequency ? (A) No attenuation in the reflected wave (B) Decreased attenuation in the reflected wave (C) Increased attenuation in the reflected wave (D) Nominal attenuation in the reflected wave
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 7
Electronagnetics Waves
429
For View Only Shop Online at www.nodia.co.in MCQ 7.3.60 A plane electromagnetic wave travelling in a perfect dielectric medium of intrinsic IES EC 2003 impedance h1 is incident normally on its boundary with another perfect dielectric medium of characteristic impedance h 2 . The electric and magnetic field strengths of the incident wave are denoted by E1 and H1 respectively whereas Er and Hr denote these quantities for the reflected wave, and Et and Ht for the transmitted wave. Which of the following relations are correct ? 1. Ei = h1 Hi 2.
Er = h1 Hr
3.
Et = h 2 Ht
m
Select the correct answer using the codes given below (A) 1, 2 and 3 (B) 1 and 2
MCQ 7.3.61
A plane electromagnetic wave travelling in a perfect dielectric medium of dielectric constant e1 is incident on its boundary with another perfect dielectric medium of dielectric constant e2 . The incident ray makes an angle of q1 with the normal to the boundary surface. The ray transmitted into the other medium makes an angle of q2 with the normal. If e1 = 2e2 and q1 = 60c, which one of the following is correct ? (A) q2 = 45c
he
lp.
IES EC 2003
(D) 2 and 3
co
(C) 1 and 3
(B) q2 = sin-1 0.433
ww w. ga te
(C) q2 = sin-1 0.612
(D) There will be no transmitted wave MCQ 7.3.62 IES EC 2003
Match List I (Nature of Polarization) with List II (Relationship Between X and Y Components) for a propagating wave having cross-section in the XY plane and propagating along z -direction and select the correct answer : List-I
List-II
a. Linear
1. X and Y components are in same phase
b. Left circular
2. X and Y components have arbitrary phase difference
c.
3. X component leads Y by 90c
Right circular
d. Elliptical Codes : (A) (B) (C) (D)
a 1 4 1 4
b 4 1 4 1
4. X component lags behind Y by 90c
c 2 2 3 3
d 3 3 2 2
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
430
Electronagnetics Waves
Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.3.63 Assertion (A) : For an EM wave normally incident on a conductor surface the IES EC 2002 magnetic field H undergoes a 180c phase reversal and the phase of electric field E remains same. Reason (R) : The direction of propagation of incident wave will reverse after striking a conductor surface. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A
m
(C) A is true but R is false (D) A is false but R is true
List-I
List-II
Propagation constant
b.
Radiation intensity
c.
Wave impedance
Ey - ay E x Z0 Z0
ww
IES EC 2002
2.
r 2h
3.
Et /Ht
4.
E#H
(E 2)
If the E field of a plane polarized EM wave travelling in the z -direction is : E = ax Ex + ay Ey then its H field is : E E (A) ay Ex - ax y (B) ay Ex + ax y Z0 Z0 Z0 Z0 (C) ax
MCQ 7.3.66
c 3 2 2 3
w. g
IES EC 2002
b 2 3 3 2
wms/2
2
at e
a 1 4 1 4
(A) (B) (C) (D)
1.
lp.
a.
Codes :
MCQ 7.3.65
co
Match List I with List II and select the correct answer :
IES EC 2002
he
MCQ 7.3.64
(D) - ax
Ey - ay E x Z0 Z0
Consider the following statements : For electromagnetic waves propagating in free space : 1. electrical field is perpendicular to direction of propagation 2.
electrical field is along the direction of propagation
3.
magnetic field is perpendicular to direction of propagation
4.
magnetic field is along the direction of propagation
Which of these statements are correct ? (A) 1 and 3 (B) 1 and 4 (C) 2 and 3
(D) 2 and 4
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 7
Electronagnetics Waves
431
For View Only Shop Online at www.nodia.co.in MCQ 7.3.67 In a uniform plane wave, the value of E/H is IES EC 2001 (A) m/e (B) e/m (C) 1 MCQ 7.3.68 IES EC 2001
(D)
The phenomenon of microwave signals following the curvature of earth is known as (A) Faraday effect (B) ducting (C) wave tilt
IES EC 2001
(D) troposcatter
Which one of the following statements is NOT correct for a plane wave with H = 0.5e-0.1x cos (106 t - 2x) az A/m (A) The wave frequency is 106 r.p.s
m
MCQ 7.3.69
me
co
(B) The wavelength is 3.14 m (C) The wave travels along + x -direction
(D) The wave is polarized in the z -direction IES EE 2012
Skin depth is the distance from the conductor surface where the field strength has fallen to (A) p of its strength at the surface (B) e of its strength at the surface
lp.
MCQ 7.3.70
IES EE 2012
The vector magnetic potential of a particular wave traveling in free space is given by A = ax Ax sin ^wt - bz h where Ax is a constant. The expression for the electric field will be (A) - ax bAx sin ^wt - bz h (B) - ay bAx sin ^wt - bz h
ww w. ga te
MCQ 7.3.71
(C) - ay wAx cos ^wt - bz h MCQ 7.3.72 IES EE 2012
IES EE 2011
(D) - ax wAx cos ^wt - bz h
The depth of penetration of a wave in a lossy dielectric increases with increasing (A) conductivity (B) permeability (C) wavelength
MCQ 7.3.73
(D) ^1/pe h of its strength at the surface
he
(C) ^1/e h of its strength at the surface
(D) permittivity
When a plane wave propagates in a dielectric medium (A) the average electric energy and the average magnetic energy densities are not equal. (B) the average electric energy and the average magnetic energy densities are equal (C) the net average energy density is finite (D) the average electric energy density is not dependent on the average magnetic energy density
MCQ 7.3.74 IES EE 2011
In free space H field is given as H ^z, t h =- 1 cos ^wt + bz h ay E ^z, t h is 6p (A) 20 cos ^wt + bz h ax (B) 20 cos ^wt + bz h az
(C) 20 sin ^wt + bz h ay (D) 20 sin ^wt + bz h ax GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
432
Electronagnetics Waves
Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.3.75 If electric field intensity phasor of an EM wave in free space is E = 6e-j4y ax V/m IES EE 2011 .The angular frequency w, in rad/s, is (B) 4y # 3 # 108 (A) 4 # 3 # 108 (C) t # 3 # 108 MCQ 7.3.76
Assertion (A) : Electromagnetic waves propagate being guided by parallel plate perfect conductor surface. Reason (R) : Tangential component of electric field intensity and normal component of magnetic field intensity are zero on a perfect conductor surface. (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A)
co
m
IES EE 2011
(D) 10 # 3 # 108
(B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A)
lp.
(C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true IES EE 2010
A uniform plane wave is propagating in a material for which e = 4e0 , m = 7m0 and s = 0 . The skin depth for the material is (A) zero (B) infinity
he
MCQ 7.3.77
MCQ 7.3.79 IES EE 2010
MCQ 7.3.80 IES EE 2009
2.
Conducting medium behaves like an open circuit to the electromagnetic field.
3.
In lossless dielectric relaxation time is infinite.
4.
In charge-free region, the Poisson’s equation becomes Laplace’s equation.
w. g
IES EE 2010
Consider the following statements : 1. In conducting medium the field attenuates exponentially with increasing depth.
(A) 1, 2 and 3 only
(B) 1, 3 and 4 only
(C) 2, 3 and 4 only
(D) 1, 2, 3 and 4
In free space E (Z, t) = 60p cos ^wt - bz h ax V/m . The average power crossing a circular area of p square metres in the plane z = constant is (B) 15p watt/m2 (A) 16p watt/m2
ww
MCQ 7.3.78
(D) 14 m
at e
(C) 28 m
(C) 14p watt/m2
(D) 13p watt/m2
In free space E (Z, t) = 120p cos ^wt - bZ h ax Vm-1 What is the average power in Wm-2 ? (A) 30paz
(B) 60paz
(D) 120paz (C) 90paz GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 7
Electronagnetics Waves
433
For View Only Shop Online at www.nodia.co.in MCQ 7.3.81 The electric field of a uniform plane wave is given by : IES EE 2009 E = 35 sin ^3p # 108 t - pZ h ax + 45 cos ^3p # 108 t - pZ h ay Vm-1 What is the corresponding magnetic field H (A) 10 sin ^3p # 108 t - pZ h ay + 10 cos ^3p # 108 t - pZ h (- ax ) Am-1 377 377 (B) 10 sin ^3p # 108 t - pZ h (- ay) + 10 cos ^3p # 108 t - pZ h - (ax ) Am-1 377 377 (C) 10 sin ^3p # 108 t - pZ h ay + 10 cos ^3p # 108 t - pZ h (ax ) Am-1 377 377
co
IES EE 2009
Consider the following statements in connection with electromagnetic waves : 1. Conducting medium behaves like an open circuit to the electromagnetic field. 2.
At radio and microwave frequencies the relaxation time is much less than the period
3.
In loss-less dielectric the relaxation time is finite.
4.
Intrinsic impedance of a perfect dielectric medium is a pure resistance.
lp.
MCQ 7.3.82
m
(D) 10 sin ^3p # 108 t - pZ h (- ay) + 10 sin ^3p # 108 t - pZ h (- ax ) Am-1 377 377
(C) 2 and 3 only IES EE 2008
(D) 2, 3 and 4
What causes electromagnetic wave polarization ? (A) Refraction
ww w. ga te
MCQ 7.3.83
he
Which is these statements is/are correct ? (A) 1 only (B) 1 and 2 only
(B) Reflection
(C) Longitudinal nature of electromagnetic wave (D) Transverse nature of electromagnetic wave MCQ 7.3.84 IES EE 2007
Assertion (A) : The velocity of electromagnetic waves is same is same as velocity of light. Reason (R) : Electrons also travel with the same velocity as photons. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true
MCQ 7.3.85 IES EE 2007
Fields are said to be circularly polarized if their magnitudes are (A) Equal and they are in phase (B) Equal and they differ in phase by ! 90c (C) Unequal and they differ in phase by ! 90c (D) Unequal and they are in phase
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
434
Electronagnetics Waves
Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.3.86 Which of the following is zero as applied to electromagnetic field ? IES EE 2006 (A) grad div A (B) div grad V (C) div curl A
IES EE 2006
m
co
MCQ 7.3.88
2.
An EM wave incident on a perfect conductor is fully reflected
3.
When an EM wave is incident from a more dense medium to less dense medium at an angle equal to or exceeding the critical angle, the wave suffers total internal reflection
lp.
IES EE 2006
What is the Poynting’s vector on the surface of a long straight conductor of radius b and conductivity s which carries current I in the z -direction ? 2 2 (B) I 2 2 ir (A) - I 2 3 ir 2sp b 2sp b 2 (C) I 2 iz (D) I if 2pb spb Consider the following statements regarding EM wave 1. An EM wave incident on a perfect dielectric is partially transmitted and partially reflected
he
MCQ 7.3.87
(D) curl curl A
Which of the statements given above are correct ? (A) Only 1 and 2 (B) Only 2 and 3
IES EE 2005
A uniform plane wave has a wavelength of 2 cm in free space and 1 cm in a perfect dielectric. What is the relative permittivity of the dielectric ? (A) 2.0 (B) 0.5 (C) 4.0
MCQ 7.3.90 IES EE 2005
w. g
MCQ 7.3.89
(D) 1, 2 and 3
at e
(C) Only 1 and 3
(D) 0.25
With the increase in frequency of an electromagnetic wave in free space, how do the velocity vc and characteristic impedance Zc change ? (A) vc increase and Zc decreases
ww
(B) vc decreases and Zc increases (C) Both vc and Zc increase (D) Both vc and Zc remain unchanged MCQ 7.3.91 IES EE 2005
The E field of a plane electromagnetic wave travelling in a non-magnetic nonconducting medium is given by E = ax 5 cos ^109 t + 30Z h. What is the dielectric constant of the medium ? (A) 30 (B) 10 (C) 9
MCQ 7.3.92 IES EE 2005
(D) 81
2 In the wave equation d2E = me2 E2 + ms2E which term is responsible for 2t 2t attenuation of the wave ?
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 7
Electronagnetics Waves
For View Only
Shop Online at www.nodia.co.in 2 (B) me2 E2 2t (D) All of the above three
(A) d2E (C) ms2E 2t
Consider the following statements : 1. Poisson’s equation finds application in vaccum tube and gaseous discharge problems 2.
Gauss’s law is useful for determining field and potential distribution about bodies having unsymmetrical geometry.
3.
For the propagation of electro-magnetic waves, the time varying electric fields must support time varying magnetic fields.
4.
The unit of Poynting’s vector is W/m2
m
IES EE 2005
co
MCQ 7.3.93
435
Which of the statements given above are correct ? (A) 1, 2 and 3 (B) 1, 3 and 4
What is the phase velocity of plane wave in a good conductor ? p fs (A) pfms (B) (ms) (C)
MCQ 7.3.95 IES EE 2005
pf (ms)
he
IES EE 2005
(C) Right hand circularly polarised MCQ 7.3.96 IES EE 2005
(D) 2
pf (ms)
The instantaneous electric field of a plane wave propagating in z -direction is E (t) = 6ax E1 cos wt - ay E2 sin wt@e-jkz This wave is (A) Linearly polarised (B) Elliptically polarised
ww w. ga te
MCQ 7.3.94
(D) 1, 2 and 4
lp.
(C) 2, 3 and 4
(D) Left hand circularly polarised
Assertion (A) : Skin depth is the depth by which electromagnetic wave has been increased to 37% of its original value. Reason (R) : The depth of penetration of wave in a lossy dielectric increases with increasing wavelength. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true
Which one of the following is the correct electromagnetic wave equation in terms of vector potential A ? 2 m 2 (B) d2A - 2A (A) d2A - 2A =- mJ 2 =- J e 2t2 2t 2 2 (C) d2A - 2A (D) d2A - me2A =- mJ 2 =- mJ 2t 2t2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 7.3.97 IES EE 2004
436
Electronagnetics Waves
Chap 7
For View Only Shop Online at www.nodia.co.in MCQ 7.3.98 Which one of the following statements is correct ? The wavelength of a wave IES EE 2004 propagating in a wave guide is (A) smaller than the free space wavelength (B) greater than the free space wavelength (C) directly proportional to the group velocity (D) inversely proportional to the phase velocity IES EE 2004
Which one of the following statements is correct ? For a lossless dielectric medium, the phase constant for a travelling wave, b is proportional to (A) er (B) er
m
MCQ 7.3.99
(D) 1/ er
IES EE 2004
In a lossless medium the intrinsic impedance h = 60p and mr = 1. What is the value of the dielectric constant er ? (A) 2 (B) 1
lp.
MCQ 7.3.100
co
(C) 1/er
(C) 4 IES EE 2003
An electromagnetic field is said to be conservative when (A) d2E = me (22E/2t2) (B) d2H = me (22H/2t2)
at e
(C) Curl on the field is zero
he
MCQ 7.3.101
(D) 8
(D) Divergence of the field is zero MCQ 7.3.102
w. g
IES EE 2003
Given that H = 0.5 exp 6- 0.1x @ sin (106 t - 2x) az (A/m), which one of the following statements is not correct ? (A) Wave is linearly polarized along az (B) The velocity of the wave is 5 # 105 m/s (C) The complex propagation constant is (0.1 + j2)
MCQ 7.3.103 IES EE 2003
ww
(D) The wave is travelling along ax For a conducting medium with conductivity s, permeability m, and permittivity e, the skin depth for an electromagnetic signal at an angular frequency w is proportional to (A) s (B) 1/w (C) 1/ s MCQ 7.3.104 IES EE 2003
(D) 1/m
The electric field of a uniform plane wave is given by E = 10 sin (10wt - pz) ax + 10 cos (wt - pz) ay (V/m) The polarization of the wave is (A) Circular (B) Elliptical (C) Linear
(D) Undefined
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 7
Electronagnetics Waves
437
For View Only Shop Online at www.nodia.co.in 7 MCQ 7.3.105 In free space H (z, t) = 0.20 cos (4 # 10 t - bz) a x A/m . The expression for E (z, t) is IES EE 2002
(A) E (z, t) = 37.7 cos (4 # 107 t - bz) ay (B) E (z, t) = 2.65 # 10 cos (4 # 107 t - bz) az (C) E (z, t) = 37.7 cos (4 # 107 t - bz) ax (D) E (z, t) =- 37.7 cos (4 # 107 t - bz) ay
IES EE 2002
A plane wave whose electric field is given by E = 100 cos (wt - 6px) az passes normally from a material ‘A’ having er = 4, mr = 1 and s = 0 to a material ‘B’ having er = 9, mr = 4 and s = 0 . Match items in List I with List II and select the correct answer :
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MCQ 7.3.106
List II
Intrinsic impedance of medium ‘B’
b
Reflection coefficient
c
Transmission coefficient
1.
6p
2.
80p
3.
1/7
4.
8/7
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List I
MCQ 7.3.107 IES EE 2002
MCQ 7.3.108 IES EE 2002
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d Phase shift constant of medium ‘A’ Codes : a b c d (A) 4 1 2 3 (B) 2 3 4 1 (C) 4 3 2 1 (D) 2 1 4 3
In free space E (z, t) = 50 cos (wt - bz) ax V/m and H (z, t) = 5/12p cos (wt - bz) ay A/m . The average power crossing a circular area of radius 24 m in plane z = constant is (A) 200 W
(B) 250 W
(C) 300 W
(D) 350 W
Consider a plane electromagnetic wave incident normally on the surface of a good conductor. The wave has an electric field of amplitude 1 V/m and the skin depth for the conductor is 10 cm. Assertion (A) : The amplitude of electric field is (1/e2) (V/m) after the wave has travelled a distance of 20 cm in the conductor. Reason (R) : Skin depth is the distance in which the wave amplitude decays to (1/e) of its value at the surface. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true
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For View Only MCQ 7.3.109 Three media are characterised by IES EE 2001 1. er = 8, mr = 2, s = 0 2.
er = 1, mr = 9, s = 0
3.
er = 4, mr = 4, s = 0
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er is relative permittivity, mr is relative permeability and s is conductivity. The value of the intrinsic impedances of the media 1, 2 and 3 respectively are (A) 188 W, 377 W and 1131 W
m
(B) 377 W, 1131 W and 188 W (C) 188 W, 1131 W, and 377 W
IES EE 2001
A plane EM wave (Ei, Hi) travelling in a perfect dielectric medium of surge impedance ‘Z ’ strikes normally on an infinite perfect dielectric medium of surge impedance 2Z . If the refracted EM wave is (Er , Hr ), the ratios of Ei /Er and Hi /Hr are respectively (A) 3 and - 3 (B) 3/2 and 1/3
lp.
MCQ 7.3.110
For a perfect conductor, the field strength at a distance equal to the skin depth is X% of the field strength at its surface. The value ‘ X% ’ is (A) Zero (B) 50% (C) 36%
at e
IES EE 2001
(D) 3/4 and 2/3
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(C) 3/4 and 3/2 MCQ 7.3.111
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(D) 1131 W, 188 W, and 377 W
(D) 26%
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SOLUTIONS 7.1
Option (B) is correct. Given magnetic field intensity in the non magnetic medium is H = 3 cos ^wt - kz h ax A/m The negative coefficient of z in ^wt - kz h shows that the wave is propagating in + az direction.
SOL 7.1.2
Option (C) is correct. From the property of phasor, we know that the instantaneous electric field is the real part of "Es e jwt , . i.e. (1) E ^x, t h = Re "Es e jwt , where Es is the phasor form of electric field. Given the electric field intensity in time domain, E ^x, t h = E 0 e- ax sin ^wt - bx h ay
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SOL 7.1.1
SOL 7.1.3
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j^wt - bx h - e-j^wt - bx h a = E 0 e- ax ;e E y 2j jE =- 0 e- ax e j^wt - bx h ay + C.C . 2 where C.C. is complex conjugate of the 1st part. So, using the property of complex conjugates we get jE E ^x, t h = 2 Re &- 0 e- ax e j (wt - bx) ay 0 2 = Re "- jE 0 e- ax e-jbx e jwt ay , Comparing it with equation (1), we get Es =- 2jE 0 e-^a + jb hx ay V/m
Option (C) is correct. Given the magnetic field intensity, H = 10 cos ^6 # 107 t - ky h az A/m Comparing it with the general equation of magnetic field. H = H 0 cos ^wt - ky h az A/m We get, w = 6 # 107 So, the wave no is, 7 k = w = 6 # 108 = 0.2 (c is the velocity of wave in free space) c 3 # 10
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For View Only Shop Online at www.nodia.co.in SOL 7.1.4 Option (D) is correct. Given magnetic field intensity in the non magnetic medium is H = 1.5 cos ^109 t - 5z h ax A/m Comparing it with the general equation of magnetic field intensity H = H 0 cos ^wt - bz h ax A/m
m
co
Option (A) is correct. Given the electric field intensity in the nonmagnetic material as E = 8 cos ^4 # 108 t - 2x h ay V/m Comparing it with the general equation of electric field E = E 0 cos ^wt - bx h ay A/m
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SOL 7.1.6
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SOL 7.1.5
We get, w = 109 rad/ sec and b = 5. So, the phase velocity of the wave in the medium is given as 9 vP = w = 10 = 4 # 108 m/s 5 b Option (A) is correct. Wavelength of an electromagnetic wave with phase constant b in a medium is defined as l = 2p b So, the phase constant of the wave in terms of wavelength can be given as (l = 12.6 m ) b = 2p = 2p = 0.5 rad/m 12.6 l
SOL 7.1.7
ww
w. g
We get, w = 4 # 108 rad/s and b = 2 rad/m So, the phase velocity of the wave in the medium is given by v p = w = 3 # 108 m/s b Since the medium is non magnetic so, m = m0 and the relative permittivity of the medium is given as 8 2 2 er = b c l = c 3 # 108 m = 2.25 vp 2 # 10 Option (D) is correct. The general equation of electric field intensity of an EM wave propagating in az direction in a medium is given as E = E 0 cos ^wt - bx h ay A/m Comparing it with the given expression of electric field intensity, we get w = 5 # 108 rad/s So, the time period of the EM wave is , T = 2p = 2p 8 = 12.57 ns w 5 # 10
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SOL 7.1.10
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For View Only Shop Online at www.nodia.co.in SOL 7.1.8 Option (D) is correct. The general equation of electric field intensity of an EM wave propagating in ax direction in a medium is given as E = E 0 cos ^wt - bx h ay A/m Comparing it with the given expression of electric field intensity, we get w = 4 # 108 rad/s So, the time period of the wave in air is given as T = 2p = 2p 8 w 4 # 10 = 15.71 ns Since in one time period the wave travels its one wavelength (l) so, time taken by the wave to travel l/4 distance is t = T = 4.93 ns 4 SOL 7.1.9 Option (D) is correct. Intrinsic impedance of any material is given as jwm h = s + jwe where m is permeability, s is conductivity and e is permittivity of the medium. Since the given material is lossless, nonmagnetic and dielectric so, we have (lossless) s =0 (non magnetic) m = m0 and ( er = 2.25 ) e = er e0 = ^2.25h e0 Therefore the intrinsic impedance of the material is jwm0 h = 0 + jw ^2.25h e0 h m0 = 0 = 377 = 323.3 W = 377 W l b h0 = e0 1.5 1. 5 Option (B) is correct. Given, Frequency of the wave propagation, f = 0.5 MHz = 0.5 # 106 Hz Conductivity of medium, s = 3 # 107 S/m Relative permeability of medium, mr = er . 1 So, the angular frequency of the wave propagation is w = 2pf = 2p # 0.5 # 106 = p # 106 and we get s = 3 # 107 = 0.1 # 1013 >> 1 6 we p # 10 # 8.85 # 10-12 Therefore, the phase constant of the propagating wave is given as wms ( s/we >> 1) b = 2 6 -7 7 = p # 10 # 4p # 10 # 3 # 10 2
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= 7695.29 rad/m So, the wavelength of the radio wave in the medium is l = 2p = 2.8 mm b Option (C) is correct. Attenuation constant for a plane wave with angular frequency w in a certain medium is given as me 2 (1) a =w 1 + a s k - 1D we 2: Since for a poor conductor, conductivity is very low i.e. s e1 , the reflection coefficient is negative Thus,
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SOL 7.3.17
(ii) If e2 < e1 then, the reflection coefficient is positive.
SOL 7.3.18
SOL 7.3.19
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Since, the given EM wave is propagating from free space to the dielectric material with e > e0 , therefore G =- 2 3 h2 - h1 or, =- 2 h2 + h1 3 h2 - 120p or, =- 2 h2 + 120p 3 So, h2 = 24p Option (A) is correct. The skin depth (d) of a material is related to the operating frequency (f) as d\ 1 f f1 d 2 Therefore, = f2 d1 d2 = 1 25 4 or d2 = 1 # 25 = 25 cm 4 Option (D) is correct. The intrinsic impedance of a medium with permittivity e and permeability m is defined as m h = e So, the reflection coefficient at the boundary interface of the two mediums is given as
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Shop Online at www.nodia.co.in h - h1 - me G = 2 = h2 + h1 + me since er = 4 = 1 - er = 1 - 4 1 + er 1+ 4 = - 1 = 2.333 180c 3 Option (A) is correct. We have E (z, t) = 10 cos (2p # 107 t - 0.1pz) So, we get w = 2p # 107 t b = 0.1p Therefore, the phase velocity of the wave is given as 7 v p = w = 2p # 10 = 2 # 108 m/s 0.1p b Option (C) is correct. We have E = (0.5ax + ay e j ) e j (wt - kz) So, its components along x and y -axis are Ex = 0.5e j (wt - kz) mo eo er mo eo er
o
o
SOL 7.3.21
lp.
p 2
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SOL 7.3.20
0
o
p
he
and Ey = e j 2 e j (wt - kz) i.e. E x ! Ey Since, the components are not equal and have the phase difference of p/2 so, we conclude that the EM wave is elliptically polarized. Option (B) is correct. Loss tangent of a medium is defined as tan d = s we where s is the conductivity e is permittivity of the medium and w is operating angular frequency. So, we get 1.7 # 10-4 ( w = 2pf ) tan d = 2p # 3 # 109 # 78eo -4 9 # 109 = 1.7 # 10 # 9 3 # 10 # 39 = 2.3 # 10-5
SOL 7.3.23
Option (A) is correct. 22Ex = c222Ex We have 2Z 2 2t2 As the field component Ex changes with z so, we conclude that the EM wave is propagating in z - direction.
SOL 7.3.24
Option (B) is correct. The required condition is Ic = Id i.e. the conduction current equals to the displacement current. So, we get Jc = Jd
ww
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SOL 7.3.22
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Shop Online at www.nodia.co.in sE = jweE s = 2pfeo er s f = = 2s 2p # eo er 4peo er 9 -2 = 9 # 10 # 2 # 10 4 6 = 45 # 10 = 90 MHz
( w = 2pf , e = er e0 )
Option (A) is correct. VSWR (voltage standing wave ratio) of the transmission line is defined as 1+ G S = 1- G where G is the reflection coefficient of the transmission line. So, we get 1+ G (VSWR = 3 ) 3= 1- G or G = 0.5 Therefore, the ratio of the reflected power strength to the incident power is given as Pr = G 2 = 0.25 Pi Thus, 25% of incident power is reflected.
SOL 7.3.26
Option (C) is correct. The fig is as shown below :
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SOL 7.3.25
As per snell law sin qt = 1 er sin qi sin 30c = 1 or sin 45c er 1 1 2 = 1 er 2 or er = 2
Option (A) is correct. Since, the phase constant is defined as b = 2p = w me l So, the wavelength in terms of permittivity of the medium can be given as l = 2p w me GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 7.3.27
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or, So, we get
l\ 1 e l1 = e1 e2 l2
Option (C) is correct. A scalar wave equation must satisfy following relation 22E - v 222E = 0 ...(1) p 2t2 2z2 (Phase velocity of the wave) where vp = w b Basically w is the multiply factor of frequency, f and b is multiply factor of z or x or y . So, we can conclude that expression given in option (C) does not satisfy equation (1) (i.e. the wave equation).
SOL 7.3.29
Option (D) is correct. In a lossless dielectric (s = 0) medium, impedance is given by m h = e where m is permeability and e is permittivity of the medium. So, we get m0 mr mr = 120p # h = e0 er er = 120p # 2 = 288.4 W 8
SOL 7.3.30
Option (D) is correct. Intrinsic impedance of a medium is given as jwm h = s + jwe Since, copper is good conductor i.e. s >> we so, we get jwm wm h = = 45c s s Thus, the impedance will be complex with an inductive component.
SOL 7.3.31
Option (B) is correct. Given, the electric field intensity of the EM wave as E = 24e j (wt + bz) ay V/m Now, the time average poynting vector for the EM wave is defined as Es 2 Es 1 ) P = ^Es # Hs h = a c Hs = h m 2h k 2 where h is the intrinsic impedance of the medium and ak is the direction of wave propagation. Since, from the given expression of the field intensity we conclude that the wave is propagating along - az So, we have (24) 2 (ak =- az , E = 24 V/m ) P = (- az ) =- 2.4 az 2 # 120p p
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SOL 7.3.28
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For View Only Shop Online at www.nodia.co.in SOL 7.3.32 Option (A) is correct. Given the propagation constant of the wave g = a + jb = 0.1p + j0.2p So, we get b = 0.2p 2p = 0.2p or, l Therefore, wavelength of the propagating wave is l = 2 = 5m 0. 2 SOL 7.3.33 Option (C) is correct. The depth of penetration or skin depth is defined as 1 d = pfms 1 i.e. d\ f or, (l = c/f ) d\ l So, the depth of penetration (skin depth) increases with increase in wavelength. Option (B) is correct. Given, the electric field intensity of the wave ...(1) E (z, t) = Eo e j (wt + bz) ax + e0 e j (wt + bz) ay Generalizing ...(2) E (z) = ax E1 (z) + ay E2 (z) Comparing (1) and (2) we can see that E1 (z) and E2 (z) are in space quadrature but in time phase so, their sum E will be linearly polarized along a line that makes an angle f with x -axis as shown below.
SOL 7.3.35
Option (B) is correct. Skin depth of the conducting medium at frequency, f 1 = 10 MHz is given as 1 d = pf1 ms 1 or ( f 1 = 10 MHz ) 10-2 p # 10 # 106 # ms -3 or, ms = 10 p Now, phase velocity at another frequency ( f 2 = 1000 MHz ) is 4p f 2 vp = ms -3 Putting ms = 10 /p in the above expression, we get 4 # p # 1000 # 106 # p - 3 106 m/ sec vp = # 10-3 Option (C) is correct. Reflected power Pr of a plane wave in terms of incident power Pi is defined as (1) Pr = G 2 Pi where, G is the reflection coefficient at the medium interface given as
SOL 7.3.36
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SOL 7.3.34
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h2 - h1 (2) h 2 + h1 where h1 and h2 are the intrinsic impedance of the two mediums (air and glass) respectively. Since, the refractive index of the glass is 1.5 (3) i.e. n2 = c m2 e2 = 1.5 (Permeability of glass) where m2 = m0 (Permittivity of glass) e2 = er e0 So, putting these values in equation (3) we get er = 1.5 m2 h h and = 0 = 0 h2 = e2 1.5 er
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G =
Therefore, from equation (2) we have h - h0 (for free space h1 = h0 ) G = 1h.5 = 1 - 1.5 =- 1 1 + 1.5 5 h + 0 1.5 Thus, from equation (1) the reflected power is given as 2 Pr = b 1 l # Pi 5 P r or, = 4% Pi Option (B) is correct. Skin depth of a material is defined as 1 d = pfms Putting the given values in the expression, we get 1 = 15.9 mm d = 9 3.14 # 1 # 10 # 4p # 10-7 # 106 Option (C) is correct. The energy density in a medium having electric field intensity E is defined as where e is permittivity of the medium. wE = 1 e E 2 2 So, due to the field E = 100 p V/m in free space, the energy density is 2 wE = 1 ^8.85 # 10-12h^100 p h 2 = 1.39 # 10-7 J/m3 = 189 nJ/m3 0
SOL 7.3.39
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SOL 7.3.38
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SOL 7.3.37
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Option (C) is correct. For a uniform plane wave propagating in free space, the fields E and H are every where normal to the direction of wave propagation ak and their direction are related as ak # a E = a H i.e. the angel between electric field ^aE h and magnetic field vector ^aH h is always 90c.
Option (A) is correct. The incidence angle of an EM wave for which there is no reflection is called GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 7.3.40
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For View Only Shop Online at www.nodia.co.in Brewster’s angle. For the vertically polarized wave (parallel polarized wave) the Brewster angle is defined as tan qB || = e2 e1 So, for the given dielectric medium we get tan qB || = 9 4 or, qB || = tan-1 b 3 l 2 Option (A) is correct. Given, the electric field component of the EM wave propagating in free space, E = 10 cos ^107 t + kz h ay V/m The general equation of electric field component of an EM wave propagating in az direction is given as E = E 0 cos ^wt + kz h ay V/m So, we conclude that the EM wave is propagating in az direction. w = 107 rad/s or 2pf = 107 7 f = 10 2p 8 So, l = c = 3 # 10 # 2p = 188.5 m 7 f 10 i.e. wavelength of the wave is wave amplitude, E 0 = 10 V/m wave number, k = 2p = 2p = 1 60p 30 l = 0.233 rad/m The wave doesn’t attenuate as it travels. So, statement (2) and (3) are correct.
SOL 7.3.42
Option (A) is correct. The incidence angle of a plane wave for which there is no reflection is called Brewster’s angle. For the parallel polarized wave, Brewster’s angle is given as tan qB || = e2 e1 where e1 and e2 are the permittivity of two mediums respectively. So, for the given parallel polarized plane wave the incidence angle ^qi h for no reflection is given as e0 tan qi = 81e0 or, qi = tan-1 b 1 l 9 Therefore, the angle a for no reflection is a = 90c - qi = 83.66c
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SOL 7.3.41
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For View Only Shop Online at www.nodia.co.in SOL 7.3.43 Option (C) is correct. Given, the characteristic impedance of air, h = 360 W Ex = 3 sin ^wt - bz h V/m Ey = 6 sin ^wt - bz + 75ch V/m So, the time average power per unit area is 2 2 _ E x + Ey i E 2 1 1 = # Pave = 2 360 2 h 2 2 ^3 + 6 h =1# = 6.25 # 10-2 W/m2 = 82.5 mW/m2 2 360 SOL 7.3.44 Option (B) is correct. Operating frequency f = 3 GHz = 3 # 109 Hz Medium parameters, m = 4p # 10-7 H/m e = 10-9 /36p s = 5.8 # 107 S/m So, we have intrinsic impedance defined as 4p # 10-7 ^10-9 /36ph m/e h = = 2 1/4 s 2 1/4 5.8 # 107 :1 + a we k D 1+ -9 > f 2p # 3 # 109 # 10 p H 36p = 2.02 # 10-2 W The phase angle of intrinsic impedance is given as qh = 1 tan-1 a s k = 1 # tan-1 we 2 2 f
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So,
=1#p = p 2 2 4 jq h = h e = 0.22e jp/4 W
5.8 # 107
2p # 3 # 109 # 10 p 36p -9
n
Option (C) is correct. The Skin depth of a conductor is defined as 1 d = pfms So, statement 2 and 3 are correct while are incorrect.
SOL 7.3.46
Option (C) is correct. For circular polarization the two orthogonal field components must have the same magnitude and has a phase difference of 90c. So, all the three statements are necessary conditions.
SOL 7.3.47
Option (B) is correct. Velocity of light in any dielectric medium is defined as 1 = c v = 1 = m0 e0 er er me
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SOL 7.3.45
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For View Only Shop Online at www.nodia.co.in where c is velocity of light in vacuum and er is dielectric constant of the medium. Since er > 1 So, v 1 2 which is not possible so there will be no transmitted wave.
SOL 7.3.62
Option (C) is correct. (1) Consider E1 is x -component and E2 is y -component so, when E1 and E2 will be in same phase. The wave will be linearly polarized. ^a " 1h
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SOL 7.3.61
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(2) When E1 and E2 will have any arbitrary phase difference then it will be elliptically polarized. ^d " 2h (3) When E1 leads E2 by 90c then wt increases counter clockwise and so the wave is right circularly polarized. ^c " 3h
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(4) When E1 lags E2 by 90c then the tip of field vector E will traverse circularly in clockwise direction and left circularly polarized. ^b " 4h Option (D) is correct. An incident wave normal to a perfect conductor is completely reflected in the reverse direction. The magnetic field intensity of reflected wave is same as the incident wave whereas the electric field intensity of reflected wave has the 180c phase difference in comparison to the incident field. ( G =- 1 for conducting surface).
SOL 7.3.64
Option (B) is correct. (a) Propagation constant for a perfect conductor is
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SOL 7.3.63
g = a + jb wms where a = b = 2 (b) Radiation intensity of an antenna is defined as U ^q, fh = r2 Pave 2 E 2 = r2 =cr m E 2 2h 2h (c) Wave impedance of an EM wave is defined as
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b"2
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E q = hH f h = Eq c"3 Hf Option (B) is correct. Given, Electric field intensity, E = E x a x + Ey ay The direction of wave propagation, ak = a z So, the magnetic field intensity of the EM wave is given as H = ak # E h where, h is the intrinsic impedance of the medium. Putting the expression for electric field in equation, we get H = az # ^Ex ax + Ey ay h = 1 ^Ex ay - Ey ax h h h
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SOL 7.3.65
483
Option (B) is correct. An EM wave propagating in free space consists of electric and magnetic field intensity both perpendicular to direction of propagation.
SOL 7.3.67
Option (B) is correct. In a uniform plane wave the field intensities are related as E = hH where h is intrinsic impedance given as jwm h = s + jwe Assume the medium is perfectly dielectric ^s = 0h. So, we get m h = e E = m or, e H Option (A) is correct. The higher frequency (microwave) signal is continuously refracted on the ground as shown in figure.
SOL 7.3.68
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SOL 7.3.66
This phenomenon is called ducting. SOL 7.3.69
Option (D) is correct. Given, the magnetic field intensity of a plane wave, H = 0.5e-0.1x cos ^106 t - 2x h az
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For View Only Shop Online at www.nodia.co.in The general expression for magnetic field intensity of a plane wave travelling in positive x -direction is (2) H = H 0 e- ax cos ^wt - bx h az Comparing the equation (1) and (2) we get, Wave frequency, w = 106 rad/ sec Wavelength, l = 2p = 2p = 3.14 m 2 b and the wave travels in + x -direction. Since, the magnetic field intensity points toward az direction and the wave propagates in + ax direction. So, direction of electric field intensity will be aE =- ak # aH =-^ax # az h = ay Therefore, the wave is polarized in ay direction (direction of electric field intensity). Option (C) is correct. Skin depth ^d h is the distance through which the wave amplitude decreases to a factor e-1 or 1/e .
SOL 7.3.71
Option (D) is correct. From Maxwell’s equation, For a varying magnetic field B , the electric field intensity E is defined as d # E =-2B 2t Since, the magnetic flux density B in terms of magnetic vector potential is given as B = d#A So, from the two equations we have (For dV = 0 ) E =-2A 2t Given, A = ax Ax sin ^wt - bz h So, E =- 2 6ax Ax sin ^wt - bz h@ =- ax wAx cos ^wt - bz h 2t
SOL 7.3.72
Option (C) is correct. The depth of penetration of wave (skin depth) in a lossy dielectric (conductor) is given as 1 d = 1 = a pfms So, the skin depth increases when (1) permeability decreases
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SOL 7.3.70
(2) conductivity decreases (3) frequency decreases Since, the wavelength of the wave is given as v i.e. l \ 1 l = p f f So, as l increases, f decreases and therefore, skin depth increases. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in SOL 7.3.73 Option (A) is correct. For a good conductor, a = b = pfms Since, the skin depth is defined as d = 1 a or, d =1 ^a = b h b Now, the phase constant of the wave is given as b = 2p l 1 So, we have It is defined for a good conductor. d = = l 2p b Option (B) is correct. Given, the magnetic field intensity of the wave propagating in free space, H ^z, t h =- 1 cos ^wt + bz h ay 6p So, we conclude. direction of propagation, ak =- az direction of magnetic field, a H = ay So, the direction of electric field intensity is given as aE = aH # ak = ay # ^- az h =- ax and the electric field amplitude is given as, E = h0 H = ^120phb- 1 cos ^wt + bz hl 6p =- 20 cos ^wt + bz h So, the electric field vector of EM wave is E ^z, t h = 10 cos ^wt + bz h ax
SOL 7.3.75
Option (B) is correct. Given, the electric field intensity of EM wave in phase form as Es = 10e-j4y ax So, we get the phase constant, b = 4 rad/m Since, the wave is propagating in free space, therefore, the angular frequency w of the wave is given as w = cb = ^3 # 108h^4h = 4 # 3 # 108 rad/s
SOL 7.3.76
Option (B) is correct. A and R both true and R is correct explanation of A.
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SOL 7.3.74
Option (A) is correct. Skin depth of a material is defined as 1 d = pfms GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 7.3.77
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For View Only Shop Online at www.nodia.co.in Since, conductivity of the material is s = 0 . So, we get d " infinity SOL 7.3.78
Option (A) is correct. (1) In a conducting medium as the wave travels its amplitude is attenuated by the factor e- az (i.e. attenuated exponentially). (2) Conducting medium doesn’t behave as on open circuit to the EM field.
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(3) In lossless dielectric ^s = 0h relaxation time is defined as Tr = e " 3 s (4) In charge free region ^rv = 0h. Poissions equation is generalised as r d2 V =- v e 2 dV = 0 which is Laplace equation. Therefore only statement 2 is incorrect.
Option (A) is correct. For a given electric field in free space the average power density is defined as E 2 ^60ph2 Pave = 1 =1 = 15p Watt/m2 2 h0 2 120p
SOL 7.3.80
Option (A) is correct. Given, E = 120p cos ^wt - bz h ax Since, the wave is propagating in az directions so, the magnetic flux density of the propagating wave is az # 6120p cos ^wt - bz h ax@ = cos ^wt - bz h ay (ak = az ) H = ak # E = h0 h0 Therefore, the average power density of an EM wave is defined as Pave = 1 Re "E # H *, = 1 8^120p cos ^wt - bz h ax h # ^cos ^wt - bz h ay hB 2 2 = 60paz
SOL 7.3.81
Option (B) is correct. Given, the electric field intensity is E = 10 sin ^3p # 108 t - pz h ax + 10 cos ^3p # 108 t - pz h ay So, the magnetic field intensity is given as (Direction of propagation is ak = az ) H = ak # E h0 = 5 sin ^3p # 108 t - pz h ay + 10 cos ^3p # 108 t - pz h^- ax h 377 Option (D) is correct. (1) For a perfect conducting medium the transmission coefficient is zero but a medium having finite conductivity transmission coefficient has some finite value. So it doesn’t behave like an open circuit to the electromagnetic field.
SOL 7.3.82
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SOL 7.3.79
(2) Relaxation time in a medium is defined as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Tr = e s Which in turn given the values in the range of 10-20 sec . While the radio frequency wave has the time period ‘T ’ in the range of nsec to psec. (10-9 to 10-12 ) So the relation time at radio frequency/microwave frequency is much less than the period. (3) For a lossless dielectric ^s = 0h and so, Tr = e " 3 s (4) Intrinsic impedance of a perfect dielectric ^s = 0h is m which is a pure resistance. h = e So, the statement (2), (3) and (4) are correct. Option (D) is correct. The polarization of a uniform plane wave described the time varying behaviour of the electric field intensity vector so for polarization the field vector must be transverse to the propagation of wave. i.e. Transverse nature of electromagnetic wave causes polarization.
SOL 7.3.84
Option (B) is correct. In free space electrons and photon both have the same velocity 3 # 108 m/s . So, the velocity of electromagnetic waves is same as velocity of light. So A and R both are true and R is the correct explanation of A.
SOL 7.3.85
Option (A) is correct. Fields are said to be circularly polarized if their components have same magnitudes but they differ in phase by ! 90c.
SOL 7.3.86
Option (C) is correct. From Maxwell’s equation for an EM field, the divergence of the magnetic flux density is zero. i.e. d:B = 0 d : ^d # Ah = 0 div curl A = 0
SOL 7.3.87
Option (B) is correct. Electric field intensity due to the current element is defined as E = J = I2 az s pb s The magnetic flux density due to the current element is given as H = I af 2pb So, the poynting vector of the field is P = E#H 2 2 =- 2I 3 a r =- I 2 3 ir 2sp b 2p b s
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SOL 7.3.83
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For View Only SOL 7.3.88 Option (D) is correct. All the three statements are correct.
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Option (C) is correct. Wavelength of a plane wave in any medium is defined as v l = p f where v p = phase velocity f = frequency of the wave Since, vp = c er 1 So, l\ er lair = er, dielectric er, air ldielectric 2 = er 1 1 er = 4
SOL 7.3.90
Option (D) is correct. The velocity of an EM wave in free space is given as vc = C = 3 # 108 m/s and the characteristic impedance (intrinsic impedance) is given as m0 = 120p Zc = e0 So both the terms are independent of frequency of the wave i.e. remain unchanged.
SOL 7.3.91
Option (D) is correct. Given, electric field intensity E = 5 cos ^109 t + 30z h ax So, we conclude that, w = 109 , and b = 30 and since b = w vp w (For non magnetic medium v p = c ) b = er _c/ er i 8 2 bc 2 30 3 10 # # er = b l = c m = 81 w 109
SOL 7.3.92
Option (C) is correct. For attenuation of the wave the medium must have some finite conductivity s. In the given wave equation the term ms2E involves s so this term is responsible for 2t the attenuation of the wave.
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SOL 7.3.89
Option (A) is correct. The statement 1, 3 and 4 are correct while statement 2 is incorrect as Gauss’s law is applicable only for symmetrical geometry. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 7.3.93
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For View Only SOL 7.3.94 Option (D) is correct. In a Good conductor So,
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pfms
phase velocity v p = w = 2 b
pf ms
Option (A) is correct. Given, the electric field intensity of the plane wave is E ^ t h = 6E1 cos wtax - E2 sin wtay@ e-jkz Since the components of the field are E x = E1 and Ey = E 2 i.e. So, the wave is elliptically polarized. E x ! Ey
SOL 7.3.96
Option (D) is correct. For a lossy dielectric, skin depth is defined as d = l 2p So, as the wavelength increases the depth of penetration of wave also increases. i.e. Reason (R) is correct. The Skin depth is the depth by which electric field strength reduces to 1 = 37% of e its original value. i.e. Assertion (A) is false.
SOL 7.3.97
Option (D) is correct. The electromagnetic equation in terms of vector potential A is given as 2 =- mJ d2 A - me2A 2t2 Option (B) is correct. The wavelength of an EM wave propagating in a waveguide is defined as ll l = f 2 1 -c c m f where ll is the wavelength of the wave in unbounded medium(free space), fc is the cutoff frequency of the waveguide and f is the operating frequency. Now, for a propagating wave in the waveguide, the operating frequency is higher than the cutoff frequency. i.e. f fc & fc /f < 1 Putting it in equation (1) we get l < ll i.e. Wavelength of a propagating wave in a wave guide is smaller than the free space wavelength.
SOL 7.3.99
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SOL 7.3.98
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SOL 7.3.95
Option (A) is correct. For a lossless dielectric medium s =0
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For View Only Shop Online at www.nodia.co.in and propagation constant, g = a + jb = jwm ^s + jweh a + jb = jw me b = w me i.e. b \ er Option (C) is correct. For a lossless medium ^s = 0h intrinsic impedance is defined as m m0 mr = h = e e0 er 60p = 120p 1 er er = 4
SOL 7.3.101
Option (C) is correct. A field is said to be conservative if the curl of the field is zero.
SOL 7.3.102
Option (B) is correct. Given, the magnetic field intensity, H = 0.5e-0.1x sin ^106 t - 2x h az A/m Comparing it with general expression of magnetic field intensity of wave propagating in ax direction given as H = H 0 e- ax sin ^wt - bx h az We get (i) the direction of wave propagation is ax (ii) a = 0.1, b = 2
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SOL 7.3.100
So, propagation constant g = a + jb = 0.1 + j 2
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6 (iii)phase velocity, v p = w = 10 = 5 # 105 m/s 2 b (iv) aH = az , ak = a x So, direction of polarization, aE =-^ak # aH h =-^ax # az h ay i.e. wave is polarized along ay .
Option (C) is correct. Skin depth of any conducting medium is defined as 1 d = pfms So, at a given frequency w = 2pf d \ 1 and d \ 1 s m
SOL 7.3.104
Option (B) is correct. Given, the electric field intensity of the plane wave, E = 10 sin ^10wt - pz h ax + 10 cos ^wt - pz h ay So, the field components are Ex = 10 sin ^10wt - pz h
ww
SOL 7.3.103
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For View Only and since,
So the polarization is circular.
Option (D) is correct. In free space electric field intensity is defined as E =- h 0 ^ak # H h where ak is unit vector in the direction of propagation. Given, H = 0.10 cos ^4 # 107 t - bz h ax A/m So, the direction of propagation, ak = az and we have, E =- 377 6az # ^0.10 cos ^4 # 107 t - bz h ax h@ =- 37.7 cos ^4 # 107 t - bz h ay
( h0 = 377 W )
Option (A) is correct. Given the electric field in medium A is E = 100 cos ^wt - 6px h z In medium A, er = 4 , mr = 1, s=0 In medium B , er = 9 , mr = 4 , s=0 So, (a) intrinsic impedance of medium ‘B ’ is m 4m0 = = 2 # 120p = 80p hB = ^a " 2h e 9e0 3 (b) Intrinsic impedance of medium ‘A’ is m m0 = = 1 # 120p = 60p hA = e 2 4e0 So, reflection coefficient, h - hA = 80p - 60p = 1 G = B ^b " 3h hB + hA 80p + 60p 5 (c) Transmission coefficient, 2hB = 2 # 80p = 8 t = ^c " 4h hB + hA 80p + 60p 7 (d) Phase shift constant of medium A is given from the field equation as
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SOL 7.3.106
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SOL 7.3.105
Ey = 10 cos ^wt - pz h E x = Ey
491
b = 12p
SOL 7.3.107
Option (A) is correct. Average power density in an EM wave is defined as Pave = 1 Re ^Es # H s*h = 1 # 50 # 5 = 3.316 2 12p 2 So, the average power crossing a circular area of radius 24 m is 2 Pave = Pave ^pr2h = ^3.316h_p ^ 24 h i = 250 Watt
SOL 7.3.108
Option (B) is correct. Electric field amplitude, E 0 = 1 V/m Skin depth, d = 10 cm = 0.1 m So, the attenuation constant of the wave in the conductor is
^d " 1h
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Option (C) is correct. For any media having conductivity, s = 0 . the intrinsic impedance is given as m mr = h h = e er 0 For media 1, h1 = 2 ^377h = 188 W 8 For media 2, h 2 = 9 ^377h = 1131 W 1 for media 3, h 3 = 4 ^377h = 377 W 4 Option (B) is correct. For an EM wave a medium incident on another medium, reflection coefficient is defined as G = Er =- Hr Ei Hi h 2 - h 1 2Z - Z and = =1 G = 3 h 2 + h 1 2Z + Z Er =- Hr = 1 So, 3 Ei Hi Ei = 3 and Hi =- 3 Er Hr Option (B) is correct. For a perfect conductor conductivity , s = 3 So, the skin depth of the perfect conductor is 1 d = =0 pfms
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SOL 7.3.110
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SOL 7.3.109
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a = 1 = 10 d Now, the electric field intensity after travelling a distance z inside a conductor is E = E 0 e- az where, E 0 is the field intensity at the surface of the conductor. So, the distance travelled by the wave for which amplitude of electric field changes to (1/e2) (V/m) is given as E = E20 e E -10z = 20 E0 e e 10z = 2 z = 20 cm Alternatively, since the skin depth is the distance in which the wave amplitude decays to ^1/e h of its value at surface. So, for the amplitude to be 1/e2 of the field at its surface the wave penetrates a length of 2d = 20 cm . So A and R both are true and R is correct explanation of A.
SOL 7.3.111
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CHAPTER 8 TRANSMISSION LINES
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EXERCISE 8.1
m
Assertion (A) : A sinusoidal voltage vi = V0 cos ^2 # 10 4 pt h is applied to the input terminal of a transmission line of length 20 cm such that the wave propagates with the velocity c = 3 # 108 m/s on the line. It’s output voltage will be in the same phase to the input voltage. Reason (R) : Transmission line effects can be ignored if l # 0.01. l where l is the length of transmission line and l is the wavelength of the wave. (A) A and R both are true and R is correct explanation of A.
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MCQ 8.1.1
Chap 8
(C) A is true but R is false (D) A is false but R is true
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A transmission line is formed of coaxial line with an inner conductor diameter of 1 cm and an outer conductor diameter of 4 cm. If the conductor has permeability mc = 2ma and conductivity sc = 11.6 # 107 S/m then it’s resistance per unit length for the operating frequency of 4 GHz will be (A) 4.95 W/m (B) 78.8 W/m
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MCQ 8.1.2
(C) 0.788 W/m
(C) 872 nH/m MCQ 8.1.4
(D) 0.495 W/m
A transmission line formed of co-axial line with inner and outer diameters 1.5 cm and 3 cm respectively is filled with a dielectric of permeability m = 2m0 . It’s line parameter Ll will be equal to (A) 277 nH/m (B) 2.77 nH/m
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MCQ 8.1.3
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(B) A and R both are true but R is not correct explanation of A.
(D) 8.7 nH/m
A co-axial transmission line is filled with a dielectric having conductivity, s = 2 # 10-3 S/m . If the inner and outer radius of the co-axial line are 1/4 cm and 1/2 cm respectively then the conductance per unit length of the transmission line will be (A) 9.1 mS/m (B) 1.45 mS/m (C) 911 S/m
(D) 145 S/m
If Permittivity of the dielectric filled inside the coaxial transmission line having inner and outer diameter 2 cm and 5 cm respectively is e = 9e0 then the capacitance GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
MCQ 8.1.5
Chap 8
Transmission Lines
For View Only per unit length of the line will be (A) 361 pF/m
Shop Online at www.nodia.co.in (B) 3.61 nF/m
(C) 5.74 nF/m MCQ 8.1.6
(D) 57.4 pF/m
A parallel plate transmission line consists of 1.2 cm wide conducting strips having conductivity, s = 1.16 # 108 S/m and permeability m = m0 is operating at 4 GHz frequency. What will be the line parameter Rl ? (A) 1.38 W/m (B) 0.69 W/m
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(D) 78.1 mH/m
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The space between the strips of a parallel plate transmission line is filled of a dielectric of permittivity, er = 1.3 and conductivity, s . 0 . If the width of the strips is 9.6 cm and the separation between them is 0.6 cm then the line parameters Gl and Cl will be respectively (A) 0, 0.02 nF/m (B) 0.02 mS/m, 0.14 nF/m (C) 0, 0.18 nF/m
MCQ 8.1.9
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A parallel plate transmission line is formed by copper strips of width w = 1.2 cm separated by a distance d = 0.3 cm . If the dielectric filled between the plates has permeability, m = 2m0 then what will be the inductance per unit length of the transmission line ? (A) 157 nH/m (B) 1.57 mH/m (C) 0.78 nH/m
MCQ 8.1.8
(D) 1.97 W/m
m
(C) 0.97 W/m MCQ 8.1.7
495
(D) 1.8 mS/m, 0
Which one of the following statement is not correct for a transmission line ? (A) Attenuation constant of a lossless line is always zero. (B) Characteristic impedance of both lossless and distortionless line is real (C) Attenuation constant of a distortionless line is always zero. (D) Both (A) and (C).
MCQ 8.1.10
Inductance and capacitance per unit length of a lossless transmission line are 250 nH/m and 0.2 nF/m respectively. The velocity of the wave propagation and characteristic impedance of the transmission line are respectively. (B) 3 # 108 m/s , 50 W (A) 2 # 108 m/s , 100 W (C) 2 # 108 m/s , 50 W (D) 3 # 108 m/s , 100 W
MCQ 8.1.11
A 1 GHz parallel plate transmission line consists of brass strips of conductivity s = 6.4 # 107 S/m separated by a dielectric of permittivity e = 6e0 . If the axial component and transverse component of the electric field in the transmission line is Ez and Ey respectively then Ez /Ey equals to (B) 4.167 # 10-5 (A) 2.16 # 10-4
(C) 1.25 # 10-4 (D) 7.22 # 10-5 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in MCQ 8.1.12 A transmission line operating at a frequency 6 # 108 rad/s has the parameters Rl = 0.2 kW/m , Ll = 4 mH/m , Gl = 8 mS/m , C l = 6 pF/m . The propagation constant, g will be (A) ^0.5 + j1.2h m-1 (B) ^0.10 + j2.4h m-1 (C) ^1.2 + j0.5h m-1
The parameters of a transmission line are given as Rl = 10 W/m , Ll = 0.1 mH/m , C l = 10 pF/m , Gl = 40 mS/m . If the transmission line is operating at a frequency, w = 1.2 # 109 rad/s then the characteristic impedance of the line will be (A) 50 - j2 W (B) 4 - j100 W
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MCQ 8.1.13
(D) ^2.4 + j0.10h m-1
After travelling a distance of 20 m along a transmission line, the voltage wave remains 13% of it’s source amplitude. What is the attenuation constant of the transmission line ? (A) 0.13 NP/m (B) 0.10 NP/m
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MCQ 8.1.14
(D) 100 + j4 W
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(C) 100 - j4 W
Amplitude of a voltage wave after travelling a certain distance down a transmission line is reduced by 87% . If the propagation constant of the transmission line is ^0.3 + j2.9h then the phase shift in the voltage wave is (A) 61c (B) 561c
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MCQ 8.1.15
(C) 73c
(C) not change
(C) 2 W/m , 1 mH/m MCQ 8.1.18
(D) none of these
Phase velocity of voltage wave in a distortion less line having characteristic impedance, Z 0 = 0.2 kW and attenuation constant, a = 10 mNP/m is v p = 0.5 # 108 m/s . The line parameters Rl and Ll will be respectively (A) 1 W/m , 0.5 nH/m (B) 10 kW/m , 2 mH/m
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MCQ 8.1.17
(D) 273c
A parallel plate lossless transmission line consists of brass strips of width w and separated by a distance d . If both w and d are doubled then it’s characteristic impedance will (A) halved (B) doubled
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MCQ 8.1.16
(D) 0.06 NP/m
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(C) 0.20 NP/m
(D) 1 W/m , 2 mH/m
A distortionless line has parameters Rl = 4 W/m and Gl = 4 # 10-4 S/m . The attenuation constant and characteristic impedance of the transmission line will be respectively (A) 25 NP/m, 0.01 W (B) 100 NP/m, 4 # 10-2 W (C) 4 # 10-2 NP/m , 100 W
(D) 0.01 NP/m, 25 W
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For View Only Shop Online at www.nodia.co.in MCQ 8.1.19 A transmission line operating at 5 GHz frequency has characteristic impedance Z 0 = 80 W and the phase constant b = 1.5 rad/m . The inductance per unit length of the transmission line will be (A) 3.81 nH/m (B) 38.1 nH/m (C) 2.61 nH/m
A 150 W transmission line is connected to a 330 W resistance and to a 50 V DC source with zero internal resistance. The voltage reflection coefficients at the load end and at the source and of the transmission line are respectively (A) - 1, 1/3 (B) - 1, - 1
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MCQ 8.1.20
(D) 26.1 nH/m
(D) 1/3, - 1
(C) 1/3, 1/3
(C) 1.40 MCQ 8.1.22
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(D) 1.22
A purely resistance load ZL is connected to a 150 W lossless transmission line. Such that it has a voltage standing wave ratio of 3. The possible value of ZL will be (A) 50 W (B) 450 W (C) (A) and (B) both
MCQ 8.1.25
(D) 4.2 mm
A lossless transmission line of characteristic impedance Z 0 = 35 W is connected to a load impedance ZL = ^15 - j25h W . What will be the standing wave ratio an the line ? (A) 0.57 (B) 3.65 (C) 0.27
MCQ 8.1.24
(D) 0.71
An insulating material of permittivity e = 9e0 is used in a 25 W lossless co-axial line . If the inner radius of the coaxial line is 0.6 mm then what will be it’s outer radius ? (A) 6.004 mm (B) 2.1 mm (C) 3.002 mm
MCQ 8.1.23
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The voltage wave in a lossless transmission line has the maximum magnitude of 6 volt and minimum magnitude of 2.4 volt. The reflection coefficient of the transmission line is (A) 0.43 (B) 2.33
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MCQ 8.1.21
(D) none of these
A voltage generator with vg ^ t h = 3 cos ^p # 109 t h volt is applied to a 50 W lossless air spaced transmission line. If the line length is 10 cm and it is terminated in a load impedance ZL = ^200 - j200h W then the input impedance of the transmission line will be (A) ^50 - j50.8h W (B) ^12.5 - j12.7h W (C) ^25.4 - j25h W
(D) ^25 - j25.4h W
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Chap 8
For View Only Shop Online at www.nodia.co.in MCQ 8.1.26 The wavelength on a lossless transmission line terminated in a short circuit is l. What is the minimum possible length of the transmission line for which it appears as an open circuit at it’s input terminals ? (A) l (B) l/2 (C) 4l
A lossless transmission line is operating at a frequency of 4 MHz. When the line is short circuited at it’s output end, the input impedance appears to be equivalent to an inductor with inductance of 32 nH but when the line is open circuited at it’s output end, the input impedance appears to be equivalent to a capacitor with capacitance of 20 pF. What is the characteristic impedance of the transmission line ? (A) 10 W
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MCQ 8.1.27
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(D) l/4
(B) 1.6 kW
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(C) - 40 W (D) 40 W
A l/4 section of a 50 W lossless transmission line terminated in a 150 W resistive load is preceded by another l/4 section of a 200 W lossless line as shown in figure. What is the input impedance, Zin ?
MCQ 8.1.29
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MCQ 8.1.28
(A) 600 W
(B) 400 W
(C) 267 W
(D) 300 W
A transmission line of length l is short circuited at one end and open circuited at the other end. The voltage standing wave pattern in the transmission line will be
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Chap 8
Transmission Lines
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^2n + 1h v p ; n = 1, 2, 3, ....3 4l
(D)
nv p ; n = 1, 2, 3.....3 2l
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At an operating frequency of 500 Hz, length of a transmission line is given by l = l/4 . For the same transmission line the length at 1 kHz will be given by (B) l = l (A) l = l 8 4 (C) l = l 2
(D) none of these
A lossless transmission line is terminated in a short circuit. The minimum possible length of the line for which it appears as a short circuit at its input terminals is (A) l/2
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MCQ 8.1.32
m
Phase velocity of a voltage wave in a transmission line of length l is v p . If the transmission line is open circuited at one end and short circuited at the other end then the natural frequency of the oscillation of the wave will be ^2n + 1h v p nv p (A) ; n = 0, 1.......3 (B) ; n = 0, 1, .......3 2l 4l (C)
MCQ 8.1.31
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MCQ 8.1.30
499
(B) l/4 (C) l (D) 0 MCQ 8.1.33
A transmission line is operating at wavelength ‘l’. If the distance between successive voltage minima is 10 cm and distance between load and first voltage minimum is 7.5 cm then the distance between load and first voltage maxima is (A) l/8 (B) 3l/8 (C) 5l/8
(D) l/4
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Transmission Lines
EXERCISE 8.2
A z -polarized transverse electromagnetic wave (TEM) propagating along a parallel plate transmission line filled of perfect dielectric in + ax direction. Let the electric and magnetic field of the wave be E and H respectively. Which of the following is correct relation for the fields. 2Hy (A) 2Ez = 0 (B) =0 2y 2z
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MCQ 8.2.1
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(C) Both (A) and (B)
(D) none of these
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Statement for Linked Question 2 - 3 :
MCQ 8.2.2
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A load impedance ZL = ^0.3 - j0.5h kW is being connected to a lossless transmission line of characteristic impedance Z 0 = 0.5 kW operating at wavelength l = 2 cm . The distance of the first voltage maximum from the load will be (A) 0.44 cm (B) 2.44 cm (D) - 0.44 cm MCQ 8.2.3
The distance of the first current maximum from the load will be (A) 3.56 cm (B) 0.56 cm
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(C) 1.44 cm MCQ 8.2.4
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(C) 1.56 cm
Distance of the first voltage maximum and first current maximum from the load on a 50 W lossless transmission line are respectively 4.5 cm and 1.5 cm. If the standing wave ratio on the transmission line is S = 3 then the load impedance connected to the transmission line will be (A) ^90 - j120h W (B) 10 W (C) ^30 - j40h W
MCQ 8.2.5
(D) 2.56 cm
(D) ^40 - j30h W
Total length of 50 W lossless transmission line terminated in a load impedance ZL = ^30 + j15h W is l = 7l/20 as shown in figure. The total input impedance across the terminal AB will be
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(A) ^38.3 - j64.8h W
(B) ^19.2 - j32.4h W
(C) ^64.8 - j38.3h W
(D) ^32.4 - j19.2h W
m
Assertion (A) : The input impedance of a quarter wavelength long lossless line terminated in a short-circuit is infinity. Reason (R) : The input impedance at the position where the magnitude of the voltage on a distortionless line is maximum is purely real. (A) A and R both are true and R is correct explanation of A.
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MCQ 8.2.6
501
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(B) A and R both are true but R is not the correct explanation of A. (C) A is true but R is false.
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(D) A is false but R is true.
Statement for Linked Question 7 - 8 :
MCQ 8.2.7
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A voltage generator with vg ^ t h = 25 cos ^4p # 107 t - 30h and an internal impedance Zg = 30 W is applied to a 30 W lossless transmission line that has a relative permittivity er = 2.25 and length, l = 6 m . If the line is terminated in a load impedance, ZL = ^30 - j10h W , then what will be the input impedance of the transmission line ? (B) ^50.62 + j23.48h W (A) ^0.05 - j0.01h W (C) ^92.06 - j21.80h W MCQ 8.2.8
(D) ^23.14 + j5.48h W
The input voltage of the transmission line will be (A) 4.4 cos ^8p # 107 t + 22.56ch (B) 4.4 cos ^8p # 107 t - 37.44ch V
(C) 4.4 cos ^8p # 107 t - 22.56ch V (D) 4.4 cos ^8p # 107 t - 30ch V
Statements for Linked Question 9 - 10 : Two equal load impedances of 150 W are connected in parallel through a pair of transmission line, and the combination is connected to a feed transmission lien as shown in figure. All the lines are lossless and have characteristic impedance Z 0 = 100 W . GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Transmission Lines
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MCQ 8.2.9
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Chap 8
The effective load impedance of feedline (ZLl) equals to (A) ^7.04 - j17.24h W (B) ^35.20 + j8.62h W
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(C) ^35.20 - j8.62h W
(D) ^8.62 + j35.20h W
The total input impedance of the feedline (line 3) will be (A) ^2.15 - j1.13h W (B) ^215.14 - j113.4h W
A 0.3 GHz voltage generator with Vsg = 150 volt and an internal resistance Zg = 100 W is connected to a 100 W lossless transmission line of length l = 0.375 l . If the line is terminated in a load impedance ZL = ^100 - j100h W then what will be the current flowing in the load ? (A) 0.67 cos ^3 # 108 t - 108.4ch
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MCQ 8.2.11
(D) ^107.57 - j56.7h W
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(C) ^215.14 + j113.4h W
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MCQ 8.2.10
(B) 0.67 cos ^6p # 108 t - 108.4ch
(C) 75 cos ^3 # 108 t - 108.4ch
MCQ 8.2.12
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(D) 0.67 cos ^6p # 108 t - 135ch
A voltage generator Vsg = 150 V with an internal resistance Zg = 100 W is connected to a load ZL = 150 W through a 0.15l section of a 100 W lossless transmission line. What is the average power delivered to the transmission line ? (A) 54 Watt (B) 30 Watt
(C) 27 Watt (D) 60 Watt MCQ 8.2.13
A voltage generator Vsg = 500 volt with an internal resistance Zg = 100 W is applied to a configuration of lossless transmission lines as shown in figure. The power delivered to the loads ZL1 and ZL2 will be respectively
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 8
Transmission Lines
(B) 105.33 Watt
(C) 153.10 Watt
(D) 306.11 Watt
(C) Distortion less (D) (B) and (C) both
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The input impedance of an infinitely long transmission line is equal to it’s characteristic impedance. The transmission line will be (A) slightly lossy (B) lossless
An infinitely long lossy transmission line with characteristic impedance Z 01 = 200 W is feeded by a l/2 section of 80 W lossless transmission line as shown in figure. If a voltage generator Vsg = 4 V with an internal resistance Zg = 100 W is applied to the whole configuration then the average power transmitted to the infinite transmission line will be
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MCQ 8.2.15
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(A) 612.23 Watt
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MCQ 8.2.14
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503
(A) 2.2 mWatt
(B) 22.2 mWatt
(C) 17.8 mWatt
(D) - 2.2 mWatt
Common Data for Question 16 - 17 : A unit step voltage generator is applied to a 90 W airspaced lossless transmission line at time, t = 0 . At any time, t $ 0 the voltage waveform at the sending end of the transmission line is shown in the figure below : GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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The length of the transmission line will be (A) 1200 m (B) 600 m
m
MCQ 8.2.16
(D) 300 m
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(C) 150 m
The unit step generator voltage connected to the line has an internal resistance Rg = 100 W . What will be the load impedance connected to the transmission line ? (A) 21.43 W (B) 93.16 W
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MCQ 8.2.17
(C) 42.86 W
(D) 233 W
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At time t = 0 unit step voltage generator Vg with an internal resistance Rg is applied to a 100 W shorted transmission line filled with dielectric of permittivity e = 4e0 as shown in figure
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MCQ 8.2.18
Chap 8
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The voltage waveform for any time t $ 0 at the sending end is sown in figure below
Vg and Rg will be respectively equal to (A) 30 volt, 19.2 W
(B) 38.4 volt, 60 W
(C) 60 volt, 38.4 W
(D) 19.2 volt, 30 W
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Statement for Linked Question 19 - 20 : A 2.5 m section o an airspacd lossless transmission line is fed by a unit step voltage generator Vg = 30 volt with internal resistance Rg = 200 W . The transmission line is terminated in a resistive load ZL = 50 W and characterized by Z 0 = 100 W . The bounce diagram of the transmission line will be
MCQ 8.2.20
The instantaneous voltage waveform v ^ t h at the sending end of the transmission line will be
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MCQ 8.2.19
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The SWR circle ‘L1 L2 ’ is shown on the smith chart for a lossless transmission line. If line is terminated in a load ZL = 50 W then the possible value of the characteristic impedance of the line will be (A) 125 W (B) 250 W
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MCQ 8.2.21
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(C) 20 W
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(D) (A) and (C) Both
Common Data for Question 22 - 25 :
MCQ 8.2.22
at e
A lossless transmission line characterized by Z 0 = 50 W is terminated in a load ZL = ^50 + j75h W The reflection coefficient of the line will be (A) 4.4e-j7.6c (B) 0.24e j76c
MCQ 8.2.23
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(C) 4.4e j76c
The input impedance at a distance of 0.35l from the load will be (A) ^0.61 - j0.22h W (B) ^61 + j2.2h W (C) ^61 - j2.2h W
(C) 0.106l MCQ 8.2.25
(D) 0.544l
The first voltage maximum will occur at a distance of (A) 0.106l from load (B) 0.144l from load (C) 0.106l from Generator
MCQ 8.2.26
(D) ^0.61 + j0.022h W
The shortest length of the transmission line for which the input impedance appears to be purely resistive will be (A) 0.25l (B) 0.456l
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MCQ 8.2.24
(D) 0.24e j76c
(D) 0.144l from generator
A transmission line of characteristic impedance 50 W is terminated by an inductor as shown in the figure.
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A positive wave with constant voltage V0 = 1 volt is incident on the load terminal at t = 0 . At any time t the resulting negative wave voltage at the load terminal will be (A) ^1 - 2e-25t h Volt (B) ^2e-25t - 1h Volt
m
(D) ^e-25t - 1h Volt
(C) 2e-25t Volt
A transmission line has the characteristic impedance Z 0 and the voltage standing wave ratio is S . The line impedance on the transmission line at voltage maximum and minimum are respectively. (A) Z 0 S , Z 0 (B) Z 0 , Z 0 S S S (C) Z 0 S , Z 0 S (D) Z 0 , Z 0 S S
MCQ 8.2.28
Consider the three mediums of intrinsic impedances h1 , h 2 and h 3 respectively as shown in the figure. What will be the thickness ‘t ’ and intrinsic impedance ‘ h 2 ’ of the medium 2 for which the reflected wave having wavelength ‘l’ is eliminated in medium 1 are thickness ‘t ’ intrinsic impedance h 2 (A) h1 h 3 l/4 (B) h1 /h 3 l/2 (C) h1 /h 3 l/4 (D) h1 h3 l/2
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MCQ 8.2.27
Statement for Linked Question 29 - 30 : A quarter wave dielectric of thickness ‘t ’ and permittivity ‘e’ eliminates reflections of uniform plane waves of frequency 2.5 GHz incident normally from free space onto a dielectric of permittivity 16e0 . (Assume all media to have m = m0 ) MCQ 8.2.29
The permittivity of the dielectric coating equals to (A) e0 /2 (B) e0 /4 (C) 4e0
MCQ 8.2.30
(D) 2e0
What is the thickness ‘t ’ of the dielectric coating ? (A) 25 cm (B) 2.5 cm (C) 1 cm
(D) 10 cm
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Chap 8
For View Only Shop Online at www.nodia.co.in MCQ 8.2.31 A transmission line has characteristics impedance 100 W and standing wave ratio 3. The distance between the first voltage maximum and load is 0.125l. Load impedance of the transmission line is (A) ^30 + j40h W (B) ^60 + j80h W (C) ^30 - j40h W
(D) ^60 - j80h W
A 100 W lossless transmission line with it’s parameter Ll = 0.25 mH/m and C l = 100 PF/m is terminated by it’s characteristic impedance. A 15 V voltage source with internal resistance 50 W is connected to the transmission line at t = 0 . Plot of the voltage on the line at a distance 5 m from the source against time will be
MCQ 8.2.33
A lossless transmission line terminated by a load impedance ZL ! Z 0 is connected to a D.C. voltage source. The height of the first forward voltage pulse is V 1+ . If the voltage reflection coefficients at the load and source are respectively GL and Gg then the steady state voltage across the load is 1 - Gg GL (A) V 1+ ;1 + GL E (B) V 1+ c 1 - GL 1 + GL m
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MCQ 8.2.32
(C) V 1+ b 1 - GL l 1 + GL MCQ 8.2.34
(D) V 1+ c 1 + GL m 1 - Gg GL
A 60 W transmission line, terminated by a load of 180 W is connected to a 100 V DC source at t = 0 . The internal resistance of the source is 120 W. The steady state voltage across the load will be
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For View Only (A) 50 V
Shop Online at www.nodia.co.in (B) 100 V
(C) 120 V
(D) 60 V
At t = 0 a 50 Volt D.C. source with an internal resistance 30 W is connected to a transmission line of 15 W characteristic impedance having a load of 45 W. The steady state load current for the transmission line is (A) 0.67 A (B) 1.5 A (C) 0.33 A
(D) 1.3 A
Statement for Linked Question 36 - 37 :
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MCQ 8.2.35
509
MCQ 8.2.36
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A transmission line of an unknown length terminated in a resistance is connected to a 5 V battery with zero internal resistance. The plot of input current to the line is shown in the figure below
The characteristic impedance of the transmission line will be (A) 1.2 kW (B) 80 W (C) 8 W
MCQ 8.2.37
(D) 12.5 W
The load resistance terminated to the transmission line will be (A) 263 W (B) 80 W (C) 150 W
(D) 43 W
***********
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Transmission Lines
GATE 2012
EXERCISE 8.3
A coaxial-cable with an inner diameter of 2 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. Given m0 = 4p # 10-7 H/m, -9 e0 = 10 F/m , the characteristic impedance of the cable is 36p (A) 330 W (B) 100 W
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MCQ 8.3.1
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A transmission line with a characteristic impedance of 100 W is used to match a 50 W section to a 200 W section. If the matching is to be done both at 429 MHz and 1 GHz, the length of the transmission line can be approximately (A) 82.5 cm (b) 1.05 m
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GATE 2012
(D) 43.4 W
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(C) 143.3 W MCQ 8.3.2
A transmission line of characteristic impedance 50 W is terminated by a 50 W load. When excited by a sinusoidal voltage source at 20 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be p/4 radians. The phase velocity of the wave along the line is (A) 0.8 # 108 m/s (B) 1.2 # 108 m/s
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GATE 2011
(D) 1.75 m
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(C) 1.58 cm MCQ 8.3.3
(C) 1.6 # 108 m/s MCQ 8.3.4
(C) (19.23 + j 46.15) W MCQ 8.3.5 GATE 2010
(D) 3 # 108 m/s
A transmission line of characteristic impedance 50 W is terminated in a load impedance ZL . The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of l/4 from the load. The value of ZL is (B) 250 W (A) 10 W
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GATE 2011
Chap 8
(D) (19.23 - j 46.15) W
0.2 0c 0.9 90c H, then If the scattering matrix [S ] of a two port network is [S ] = > 0.9 90c 0.1 90c the network is (A) lossless and reciprocal (B) lossless but not reciprocal (C) not lossless but reciprocal (D) neither lossless nor reciprocal
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511
For View Only Shop Online at www.nodia.co.in MCQ 8.3.6 A transmission line has a characteristic impedance of 50 W and a resistance of GATE 2010 0.1 W/m . If the line is distortion less, the attenuation constant(in Np/m) is (A) 500 (B) 5 (C) 0.014 MCQ 8.3.7
In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 W line is
(A) 1.00 (C) 2.50
MCQ 8.3.9 GATE 2008
A transmission line terminates in two branches, each of length l/2 , as shown. The branches are terminated by 50 W loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Zi as seen by the source.
(A) 200 W
(B) 100 W
(C) 50 W
(D) 25 W
One end of a loss-less transmission line having the characteristic impedance of 75 W and length of 2 cm is short-circuited. At 5 GHz, the input impedance at the other end of transmission line is (A) 0 (B) Resistive (C) Capacitive
MCQ 8.3.10 GATE 2007
(D) 3.00
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GATE 2009
(B) 1.64
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MCQ 8.3.8
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GATE 2010
(D) 0.002
(D) Inductive
A load of 50 W is connected in shunt in a 2-wire transmission line of Z0 = 50W as shown in the figure. The 2-port scattering parameter matrix (S-matrix) of the shunt element is
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Transmission Lines
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1 - 12 2 (A) > 1 1H 2 -2
(B) =
2 - 13 3 (C) > 2 1H 3 3
(D) > 43 -4
0 1 G 1 0 - 43
H
m
1
1 4
The parallel branches of a 2-wire transmission line are terminated in 50 W and 200 W resistors as shown in the figure. The characteristic impedance of the line is Z 0 = 50 W and each section has a length of l . The voltage reflection coefficient G 4 at the input is
(A) - j 7 5
MCQ 8.3.12 GATE 2006
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(C) j 5 7
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GATE 2007
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MCQ 8.3.11
Chap 8
GATE 2005
(D) 5 7
A transmission line is feeding 1 watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated by the horn antenna into the free space is (A) 10 Watts (B) 1 Watts (C) 0.1 Watts
MCQ 8.3.13
(B) - 5 7
(D) 0.01 Watt
Characteristic impedance of a transmission line is 50 W. Input impedance of the open circuited line is Zoc = 100 + j150 W . When the transmission line is short circuited, then value of the input impedance will be (A) 50 W (B) 100 + j150 W (C) 7.69 + j11.54 W
(D) 7.69 - j11.54 W
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Statement of Linked Questions 14 - 15 :
GATE 2005
The value of the load resistance is (A) 50 W
(B) 200 W
(C) 12.5 W GATE 2005
(D) 0
The reflection coefficient is given by (A)- 0.6 (C) 0.6
MCQ 8.3.17 GATE 2004
Many circles are drawn in a Smith Chart used for transmission line calculations. The circles shown in the figure represent
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GATE 2005
(D) 0
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MCQ 8.3.16
(B) - 1
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MCQ 8.3.15
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MCQ 8.3.14
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Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 and a resistive load is shown in the figure.
(A) Unit circles
(B) Constant resistance circles
(C) Constant reactance circles
(D) Constant reflection coefficient circles.
Consider a 200 W, quarter - wave long (at 1 GHz) transmission line as shown in Fig. It is connected to a 20 V, 50 W source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is
(A) 10 V
(B) 5 V
(C) 60 V
(D) 60/7 V
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Chap 8
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For View Only Shop Online at www.nodia.co.in MCQ 8.3.18 Consider an impedance Z = R + jX marked with point P in an impedance Smith GATE 2004 chart as shown in Fig. The movement from point P along a constant resistance circle in the clockwise direction by an angle 45c is equivalent to
(A) adding an inductance in series with Z
lp.
(B) adding a capacitance in series with Z
(C) adding an inductance in shunt across Z
GATE 2004
A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is 2. The percentage of the power that is reflected back is (A) 57.73 (B) 33.33 (C) 0.11
A short - circuited stub is shunt connected to a transmission line as shown in fig. If Z0 = 50 W , the admittance Y seen at the junction of the stub and the transmission line is
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GATE 2003
(D) 11.11
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MCQ 8.3.20
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MCQ 8.3.19
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(D) adding a capacitance in shunt across Z
(A) (0.01 - j0.02) mho
(B) (0.02 - j0.01) mho
(D) (0.02 + j0) mho (C) (0.04 - j0.02) mho GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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515
For View Only Shop Online at www.nodia.co.in MCQ 8.3.21 The VSWR can have any value between GATE 2002 (A) 0 and 1 (B) - 1 and + 1 (C) 0 and 3 MCQ 8.3.22 GATE 2002
(D) 1 and 3
In an impedance Smith chart, a clockwise movement along a constant resistance circle gives rise to (A) a decrease in the value of reactance (B) an increase in the value of reactance (C) no change in the reactance value
GATE 2001
A transmission line is distortionless if (A) RL = 1 GC
GATE 2000
The magnitudes of the open-circuit and short-circuit input impedances of a transmission line are 100 W and 25 W respectively. The characteristic impedance of the line is, (A) 25 W (B) 50 W (C) 75 W
GATE 1999
(C) 2 GHz MCQ 8.3.26 GATE 1999
GATE 1997
GATE 1997
(D) 180 deg rees
A transmission line of 50 W characteristic impedance is terminated with a 100 W resistance. The minimum impedance measured on the line is equal to (A) 0 W (B) 25 W (C) 50 W
MCQ 8.3.28
(D) 6.28 GHz
In air, a lossless transmission line of length 50 cm with L = 10 mH/m , C = 40 pF/m is operated at 25 MHz . Its electrical path length is (A) 0.5 meters (B) l meters (C) p/2 radians
MCQ 8.3.27
(D) 100 W
In a twin-wire transmission line in air, the adjacent voltage maxima are at 25 m and 12.4 m. The operating frequency is (A) 300 MHz (B) 1 GHz
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MCQ 8.3.25
(D) RG = LC
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MCQ 8.3.24
(B) RL = GC
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(C) LG = RC
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MCQ 8.3.23
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(D) no change in the impedance
(D) 100 W
A very lossy, l/4 long, 50 W transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately (A) 0 (B) 50 W (C) 3
(D) None of the above
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Chap 8
For View Only Shop Online at www.nodia.co.in MCQ 8.3.29 A lossless transmission line having 50 W characteristic impedance and length l/4 GATE 1996 is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from the voltage source is (A) 0 (B) 0.02 A (C) 3 MCQ 8.3.30
The capacitance per unit length and the characteristic impedance of a lossless transmission line are C and Z 0 respectively. The velocity of a travelling wave on the transmission line is (B) 1 (A) Z 0 C Z0 C (C) Z 0 (D) C Z0 C
IES EC 2012
A l/4 line, shorted at one end, presents impedance at the other end equal to (B) 2 Z 0 (A) Z 0
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MCQ 8.3.31
co
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GATE 1996
(D) none of these
(C) 3
(D) 0
IES EC & EE 2012
A 100 W transmission line is first short-terminated and the minima locations are noted. When the short is replaced by a resistive load RL , the minima locations are not altered and the VSWR is measured to be 3. The value of RL is (A) 25 W (B) 50 W
at e
MCQ 8.3.32
he
where Z 0 is characteristic impedance of the line.
(C) 225 W IES EC 2011
If maximum and minimum voltage on a transmission line are 2 V and 5 Vrespectively, VSWR is (A) 0.5 (B) 2
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MCQ 8.3.33
(C) 1 IES EC 2011
(C) 120 W MCQ 8.3.35 IES EC 2011
(D) 8
An ideal lossless transmission line of Z 0 = 60 W is connected to unknown ZL . If SWR = 4 , find ZL . (A) 240 W (B) 480 W
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MCQ 8.3.34
(D) 250 W
(D) 100 W
Loading of a cable is done to 1. Increase its inductance 2.
Increase its leakage resistance
3.
Decrease its leakage resistance
4.
Achieve distortionless condition
(A) 1, 2, 3 and 4
(B) 1 and 3 only
(C) 2 and 3 only
(D) 1 and 4 only
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For View Only Shop Online at www.nodia.co.in MCQ 8.3.36 Given a range of frequencies, which of the following systems is best for transmission IES EC 2010 line load matching ? (A) Single stub (B) Double stub (C) Single stub with adjustable position (D) Quarter wave transformer MCQ 8.3.37 IES EC 2010
A line of characteristic impedance 50 W is terminated at one end by + j 50 W . The VSWR on the line is (A) 1 (B) 3 (C) 0
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List I
List II
a.
l < l/4
1.
Capacitive
b.
l/4 < l < l/2
2.
Inductive
c.
l = l/4
3.
d.
l = l/2
4.
(A) (B) (C) (D) IES EC 2010
b 1 1 4 4
c 4 4 1 1
d 3 2 3 2
3
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a 2 3 2 3
0
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Codes :
MCQ 8.3.39
co
IES EC 2010
At UHF short-circuited lossless transmission lines can be used to provide appropriate values of impedance. Match List I with List II and select the correct answer using the code given below the lists :
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MCQ 8.3.38
(D) j
Consider the following statements regarding a transmission line : 1. Its attenuation is constant and is independent of frequency 2.
Its attenuation varies linearly with frequency
3.
Its phase shift varies linearly with frequency
4.
Its phase shift is constant and is independent of frequency
Which of the above statements are correct for distortion less line ? (A) 1, 2, 3, and 4 (B) 2 and 3 only (C) 1 and 3 only MCQ 8.3.40 IES EC 2009
(D) 3 and 4 only
The reflection coefficient on a 200 m long transmission line has a phase angle of - 150c. If the operating wavelength is 250 m, what will be the number of voltage maxima on the line ? (A) 0 (B) 3 (C) 6
(D) 7
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518
Transmission Lines
Chap 8
For View Only Shop Online at www.nodia.co.in MCQ 8.3.41 With regard to a transmission line, which of the following statements is correct ? IES EC 2009 (A) Any impedance repeats itself every l/4 on the Smith chart. (B) The SWR = 2 circle and the magnitude of reflection coefficient = 0.5 circle coincide on the Smith chart. (C) At any point on a transmission line, the current reflection coefficient is the reciprocal of the voltage reflection coefficient.
MCQ 8.3.43 IES EC 2009
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It is required to match a 200 W load to a 450 W transmission line. To reduce the SWR along the line to 1, what must be the characteristic impedance of the quarterwave transformer used for this purpose, if it is connected directly to the load ? (A) 90 kW (B) 300 W (D) 3 W (C) 9 W 2 4
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IES EC 2009
The load end of a quarter wave transformer gets disconnected thereby causing an open-circuited load. What will be the input impedance of the transformer ? (A) Zero (B) Infinite (C) Finite and positive
IES EC 2008
(C) Zin =- jZ 0 tan bl IES EC 2007
(D) Zin =- jZ 0 tan bl
Which one of the following statements for a short circuited loss free line is not correct ? (A) The line appears as a pure reactance when viewed from the sending end
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MCQ 8.3.45
(D) Finite and negative
A lossless transmission line of characteristic impedance Z 0 and length l < l/4 is terminated at the load end by an open circuit. What is its input impedance Zin ? (B) Zin = jZ 0 cot bl (A) Zin = jZ 0 tan bl
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MCQ 8.3.44
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MCQ 8.3.42
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(D) Matching eliminates the reflected wave between the source and the matching device location.
(B) It can be either inductive or capacitive
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(C) There are no reflections in the line (D) Standing waves of voltage and current are set up along length of the lines MCQ 8.3.46 IES EC 2007
Match List I (Load impedance) with List II (Value of Reflection Coefficient) and select the correct answer using the code given below the lists : List-I
List-II
a.
Short Circuit
1.
0
b.
Open Circuit
2.
-1
c.
Line characteristics impedance
3.
+1
d.
2 # line characteristic impedance
4.
+1/3
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 8
Transmission Lines
For View Only Codes : a 2 4 2 4
(A) (B) (C) (D) MCQ 8.3.47 IES EC 2007
Shop Online at www.nodia.co.in b 1 3 3 1
c 3 1 1 3
d 4 2 4 2
When the reflection coefficient equals 1 0c what is the VSWR? (A) Zero (B) 1
If the reflection coefficient is 1/5, what is the corresponding VSWR ? (A) 3/2 (B) 2/3
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IES EC 2007
(D) Infinite
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(C) 3 MCQ 8.3.48
(C) 5/2
Which one of the following is the characteristic impedance of lossless transmission line ? (A) R/G (B) L/G (C)
MCQ 8.3.50 IES EC 2006
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IES EC 2006
(D) 2/5
(D)
L /C
Match List I (Quantity) with List II (Range of Values) and select the correct answer using the code given below the lists : a.
Input Impedance
1.
- 1 to + 1
b
Reflection coefficient
2.
1 to 3
c.
VSWR
3.
0 to 3
Codes : a (A) 2 (B) 3 (C) 3 (D) 2 IES EC 2006
List-II
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List-I
MCQ 8.3.51
R/C
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MCQ 8.3.49
519
b 3 2 1 1
c 1 1 2 3
A quarter wave impedance transformer is terminated by a short circuit. What would its input impedance be equal to ? (A) The line characteristic impedance (B) Zero (C) Infinity (D) Square root of the line characteristic impedance
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
520
Transmission Lines
Chap 8
For View Only Shop Online at www.nodia.co.in MCQ 8.3.52 Scattering parameters are more suited than impedance parameters to describe a IES EC 2006 waveguide junction because (A) the scattering parameters are frequency invariant whereas the impedance parameters are not so (B) scattering matrix is always unitary (C) impedance parameters vary over unacceptably wide ranges
IES EC 2005
In a transmission line the reflection coefficient at the load end is given by 0.3e-j30c. What is the reflection coefficient at a distance of 0.1 wavelength towards source ? (A) 0.3e+j30c (B) 0.3e+j102c
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MCQ 8.3.53
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(D) scattering parameters are directly measurable but impedance parameters are not so
MCQ 8.3.54 IES EC 2005
(D) 0.3e+66c
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(C) 0.3e+j258c
To couple a coaxial line to a parallel wire, it is best to use a : (A) Balun (C) Directional coupler
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(B) Slotted line (D) Quarter wave transformer IES EC 2005
A plane wave having x -directed electric field propagating in free space along the z -direction is incident on an infinite electrically conducting (perfect conductor) sheet at z = 0 plane. Which one of the following is correct ? (A) The sheet will absorb the wave
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MCQ 8.3.55
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(B) There will be x -directed surface electric current on the sheet (C) There will be y -directed surface electric current on the sheet (D) There will be magnetic current in the sheet. IES EC 2004
For sea water with s = 5 mho/m and er = 80 , what is the distance for which radio signal can be transmitted with 90% attenuation at 25 kHz ? (A) 0.322 m (B) 3.22 m
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MCQ 8.3.56
(C) 32.2 m MCQ 8.3.57 IES EC 2004
(D) 322 m
Consider the following statements regarding Smith charts : 1. A normalized Smith chart applies to a line of any characteristic resistance and serves as well for normalized admittance 2.
A polar coordinate Smith chart contains circles of constant z and circles of constant z
3.
In Smith chart, the distance towards the load is always measured in clockwise direction.
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Chap 8
Transmission Lines
521
For View Only Shop Online at www.nodia.co.in Which of the statements given above are correct ? (A) 1, 2 and 3 (B) 2 and 3 (C) 1 and 3 MCQ 8.3.58 IES EC 2004
(D) 1 and 2
A ^100 - j75h W load is connected to a co-axial cable of characteristic impedance 75 ohms at 12 GHz. In order to obtain the best matching, which one of the following will have to be connected ? (A) A short-circuited sub at load (B) Inductance at load
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(C) A capacitance at a specific distance at load
(D) A short-circuited stub at some specific distance from load IES EC 2003
In a line VSWR of a load is 6 dB. The reflection coefficient will be (A) 0.033 (B) 0.33
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MCQ 8.3.59
IES EC 2003
ZL = 200 W and it is desired that Zin = 50 W . The quarter wave transformer should have a characteristic impedance of (A) 100 W (B) 40 W (C) 10,000 W
IES EC 2002
(D) 4 W
Consider the following : For a lossless transmission line we can write : 1. Zin =- jZ 0 for a shorted line with l = l/8
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MCQ 8.3.61
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MCQ 8.3.60
(D) 3.3
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(C) 0.66
2.
Zin = ! ja for a shorted line with l = l/4
3.
Zin = Z 0 for a matched line of any length
Select the correct answer using the codes given below : (A) 1 and 2 (B) 2 and 3 (C) 1 and 3 MCQ 8.3.62 IES EC 2002
(D) 2 and 4
The input impedance of a short circuited quarter wave long transmission line is (A) purely reactive (B) purely resistive
(C) dependent on the characteristic impedance of the line (D) none of the above MCQ 8.3.63 IES EC 2002
A transmission line of output impedance 400 W is to be matched to a load of 25 W through a quarter wavelength line. The quarter wave line characteristic impedance must be (A) 40 W (B) 100 W
(C) 400 W (D) 425 W GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
522
Transmission Lines
Chap 8
For View Only Shop Online at www.nodia.co.in MCQ 8.3.64 The input impedance of l/8 long short-circuited section of a lossless transmission IES EC 2001 line is (A) zero (B) inductive (C) capacitive MCQ 8.3.65 IES EC 2001
(D) infinite
Match List I (Parameters) with List II (Values) for a transmission line with a series impedance Z = Rl + jwLl W/m and a shunt admittance Y = Gl + jwC l mho/m , and select the correct answer : 1.
b.
Propagation constant g
2.
c.
The sending-end input 3. impedance Zin when the line is terminated in its characteristic impedance Z 0
Z/Y
Y/Z
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c 1 3 2 2
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b 1 3 1 2
Which of the following conditions will not guarantee a distortionless transmission line ? (A) R = G = 0 (B) RC = GL
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IES EC 2001
ZY
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Characteristic impedance Z 0
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a.
Codes : a (A) 3 (B) 2 (C) 2 (D) 1 MCQ 8.3.66
List-II
m
List-I
(C) Very low frequency range (R >> wL, G >> wC )
MCQ 8.3.67 IES EC 2001
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(D) Very high frequency range (R l 4 So, the distance between first maxima and load will be l max = l min - l 4 = 7.5 - l = 7.5 # l - l = l 20 2 4 4
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GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Transmission Lines
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SOLUTIONS 8.2
Option (C) is correct. Since, the TEM wave is z -polarized i.e. the electric field of the wave is directed along + az . i.e. aE = az and the direction of wave propagation is along ax i.e. ak = a x So, the direction of magnetic field intensity will be aH = ak # aE = ax # az =- ay As E is in + az direction and H is in - ay direction so, we can consider the two vectors as (1) E = Ez az and (2) H =- Hy ay Now, from the maxwell’s equation in phasor form we have (for perfect dielectric s = 0 ) d # H = jweE a x ay a z 2 2 2 using equation (1) and (2) 2x 2y 2z = jweE z a z 0 - Hy 0 2H y 2H y a a = jweEz az 2z x 2x z It gives the result as 2H y =0 2z Again from Maxwell’s equation we have d # E =- jwmH a x ay a z 2 2 2 using equation (1) and (2) =+ jwmHy ay 2x 2y 2z 0 0 Ez 2Ez a - 2Ez a = jwmH a y y 2y x 2x y So, it gives the result as 2Ez = 0 2y Thus, Both (A) and (B) are correct.
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SOL 8.2.1
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
540
Transmission Lines
Chap 8
Option (B) is correct. In a lossless transmission line, the current maximum lies at the same point where the voltage minima lies and similarly, the current minima lies at the same point where the voltage maxima lies as shown in the figure below :
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SOL 8.2.3
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For View Only Shop Online at www.nodia.co.in SOL 8.2.2 Option (C) is correct. The voltage maximum exists at the point where the incident and the reflected voltage wave both are in same phase and the distance of voltage maximum from the load is given as (1) l max = qG l + nl 2 4p where qG is phase angle of reflection coefficient, l is the wavelength of the voltage wave and n = 0, 1, 2, .... Now, the reflection coefficient of a transmission line is given as ^0.3 - j0.5h - 0.5 G = ZL - Z 0 = ZL + Z 0 ^0.3 - j0.5h + 0.5 - 0. 2 - j 0. 5 = = 0.57e-79.8c 0. 8 - j 0. 5 i.e. qG =- 79.8c So, from equation (1) for n = 0 we have -2 l max = qG l = - 79.8c # 4 # 10 # p 4p 4p 180c -2 =- 0.44 # 10 m which is negative (i.e. the point doesnt exist). Therefore, the 1st maximum voltage will exist for n = 1 and the distance of the 1st maximum from the load is i.e. (n = 1) l max = qG l + l 2 4p =- 0.44 # 10-2 + 2 # 10-2 = 1.56 # 10-2 m = 2.56 cm
Now, it is clear from the figure that the distance between two adjacent maxima and minima is l/4 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 8
Transmission Lines
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541
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l max - l min = l 4 Since the maximum voltage wave lies at a distance l max = 1.56 cm So, the distance of 1st voltage minimum (the distance of 1st current maxima) from the load will be l min = l max - l = 1.56 - 4 = 0.24 cm 4 4 st Thus, the distance of 1 current maximum from the load is 0.56 cm.
i.e.
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Option (C) is correct. Given, The position of first voltage maximum, l max = 4.5 cm Position of first current maximum(voltage minima),l min = 1.5 cm Standing wave ratio, S =3 Characteristic impedance, Z 0 = 50 W Since, the distance between a maximum and an adjacent minimum is l/4 as discussed in previous question. i.e. l max - l min = l/4 4.5 - 1.5 = l/4
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SOL 8.2.4
SOL 8.2.5
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So, l = 12 cm Again the distance of first voltage maximum from the load is given as l max = qG l + nl 2 4p qG ^12h 4.5 = (For n = 0 ) +0 4p qG = 3p 2 Now, the magnitude of reflection coefficient is given as G = S - 1 = 3 - 1 = 2 = 0.5 S+1 3+1 4 So, the reflection coefficient of the transmission line is G = G qG = 0.5 < 3p/2 = 0.5e j3p/2 =- j0.5 Therefore, the load impedance of the transmission line is given as 1 - j0.5 ZL = Z 0 :1 + G D = 50 ; 1 + j0.5E 1-G = ^30 - j40h W Option (A) is correct. Characteristic impedance, Load impedance, Length of transmission line,
Z 0 = 50 W ZL = ^30 + j15h W l = 7l/20
Since, the transmission line is lossless so, the attenuation constant is zero GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
542
Transmission Lines
Chap 8
Option (B) is correct. In the assertion (A) given, Length of the transmission line, l = l/4 Load impedance, ZL = 0
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SOL 8.2.6
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For View Only Shop Online at www.nodia.co.in i.e. a =0 or, g = a + jb = jb Therefore, the input impedance of the lossless transmission line is given as Z + Z 0 tanh gl Z + jZ 0 tan bl ( g = jb ) = Z0 < L Zin = Z 0 c L F Z 0 + ZL tanh gl m Z 0 + jZL tan bl J 2p 7l N K(30 + j15) + j50 tan b l 20 l O 2p O = 50 K bb = l l p l 2 7 KK 50 + j ^30 + j15h tan b O l 20 l O LJ N P 7 p K(30 + j15) + j50 tan b 10 l O O = 50 K KK 50 + j ^30 + j15h tan b 7p l OO 10 L P = ^18.4 - j19.2h W
( b = 2p/l ) bl = b 2p lb l l = p 2 l 4 The input impedance of the lossless transmission line is given as Z + jZ 0 tan bl Zin = Z 0 c L Z 0 + jZL tan bl m jZ 0 tan p 2 p = j3 f = Z0 Z0 Now, we consider the reason part, Distance of the maxima from load is given as l max = ^qG + 2nph /2b where, qG is the phase angle of reflection coefficient b is the phase constant of the voltage wave and n = 0, 1, 2, .... Therefore, the input impedance at the point of maxima is given as -j2bl 1 + G e jq e-j^q + 2nph + G 1 e = Z0 f Zin = Z 0 c p (G = G e jq ) 1 - G e jq e-j^q + 2nph 1 - Ge-j2bl m 1+ G = Z0 f p 1- G So, Zin is real if Z 0 is real and since, Z 0 is always real for a distortionless line. Thus, Zin will be purely real at the position of voltage maxima in a distortionless line. i.e. A and R both are true but R is not the explanation of A.
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So, we get
SOL 8.2.7
Option (A) is correct. Length of transmission line,
max
G
G
G
G
G
max
l =6m
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 8
Transmission Lines
543
Option (C) is correct. Given the generator voltage to the transmission line, Vg (t) = 10 cos ^8p # 107 t - 30ch So, in phasor form the generator voltage is Vsg = 10e-j30c and as determined in previous question, the input impedance of transmission line is Zin = ^23.14 + j5.48h W So, for determining the input voltage, we draw the equivalent circuit for the transmission line as shown in figure below :
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SOL 8.2.8
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For View Only Shop Online at www.nodia.co.in Characteristic impedance, Z 0 = 30 W Relative permittivity, er = 2.25 Load impedance, ZL = ^30 - j10h W So, we get the angular frequency, w = 8p # 107 and the phase constant of the volatge wave along the transmission line is 7 vp = 1 = c m b = w = 8p # 10 c vp me er c/ er 7 7 = 8p # 10 # 8 2.25 = 8p # 10 #8 1.5 3 # 10 3 # 10 = 2p 5 or, bl = 2p # 6 = 2.4p rad 5 Therefore, the input impedance of the lossless transmission line is given as 30 - j10 + j30 tan ^2.4ph Z + jZ 0 tan bl = 30 f Zin = Z 0 c L p m Z 0 + jZL tan bl 30 + j ^30 - j10h tan ^2.4ph = ^12.14 + j5.48h W
Using voltage division, we get the input voltage to the transmission line as , Vin = Vg # c Zin m Zin + Zg or, (in phasor form) Vs, in = Vsg c Zin m Zin + Zg 23.14 + j5.48 = 10e-j30 c 30 + 23.14 + j5.48 m = ^15e-j35h^0.44e9.44ch GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Transmission Lines
For View Only
Chap 8
Shop Online at www.nodia.co.in -j22.56c
= 4.4e Thus, the instantaneous input voltage of the transmission line is vin ^ t h = Re 6Vs, in e jwt@ = 4.4 cos ^8p # 107 t - 22.56ch volt Option (C) is correct. Given, Load impedances to the line 1 and 2, ZL1 = ZL2 = 150 W Length of the transmission lines 1 and 2, l1 = l2 = l/5 Now, we consider the input impedance of line 1 and line 2 be Zin1 and Zin2 respectively. Since, the transmission line are identical so, the input impedances of the transmission lines 1 and 2 will be equal and given as Z + jZ 0 tan bl1 (lossless transmission line) Zin1 = Zin2 = Z 0 c L1 Z 0 + jZL tan bl1 m J 2p l N K150 + j100 tan b l 5 l O O ( b = 2p/l ) = 100 K KK100 + j150 tan b 2p l l OO l 5 L P N J 2 p K150 + j100 tan b 5 l O = 100 K 2p O K100 + j150 tan b 5 l O P L = ^70.4 - j17.24h W Therefore, the effective load impedance of the feedline will be equal to the equivalent input impedance of the parallel combination of the line 1 and 2. i.e. ZLl = Zin1 || Zin2 ^70.4 - j17.24h = Zin1 = Zin1 = Zin2 2 2 = ^35.20 - j8.62h W
SOL 8.2.10
Option (B) is correct. Given the length of the feed line, l = 0.3l and as calculated in above question, the effective load impedance of the feedline is ZLl = ^35.20 - j8.62h W So, bl = b 2p l^0.3lh = 0.6l l Therefore, input impedance of the feedline (lossless transmission line) is given as Z l+ jZ 0 tan bl Zin = Z 0 c L Z 0 + jZLltan bl m 35.20 - j8.62 + j100 tan ^0.6ph = 100 f p 100 + j ^35.20 - j8.62h tan ^0.6ph = ^215.14 - j113.4h W
SOL 8.2.11
Option (B) is correct.
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SOL 8.2.9
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 8
Transmission Lines
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For View Only Shop Online at www.nodia.co.in Operating frequency f = 0.3 GHz = 0.3 # 109 Hz Load impedance, ZL = ^100 - j100h W Characteristic impedance Z 0 = 100 W Generator voltage in phasor form, Vsg = 150 volt Internal resistance of generator Zg = 100 W Length of the transmission line, l = 0.375l So, the input impedance of the lossless transmission line is given as Z + jZ 0 tan bl Zin = Z 0 c L Z 0 + jZL tan bl m J N 2p K 100 - j100 + j100 tan b l 0.375l l O O = 100 K KK100 + j ^100 - j100h tan b 2p 0.375l l OO l P = ^200L + j100h W Now, for determining the load current, we draw the equivalent circuit for the transmission line as shown in the figure below :
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Therefore, the voltage across the input terminal of the transmission line is given as Vs, in = Vsg c Zin m Zg + Zin 200 + j100 = 150 c = 106.1e j8.13c 100 + 200 + j100 m Since, at any point, on the transmission line voltage is given as Vs ^z h = V 0+ ^e-jbz + Ge jbz h + where V 0 is the voltage due to incident wave, G is the reflection coefficient of the transmission line at load terminal and z is the distance of the point from load as shown in figure. So, for z =- l (1) Vs, in = V 0+ ^e jbl + Ge-jbl h
Now, the reflection coefficient of the transmission line at load terminal is GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Transmission Lines
Chap 8
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Shop Online at www.nodia.co.in 100 - j100 - 100 G = ZL - Z 0 = = 0.45e-j63.43c 100 - j100 + 100 ZL + Z 0 Putting the value of G and Vs, in in equation (1), we get 106.1e j8.13c = V 0+ _e jb l l^0.375lh + 0.45e-j6343ce-jb l l^0.375lhi j8.13c V 0+ = j135c 106.1e-j63.43c -j135c + 0.45e e e = 75e-j135c The current at any point on the transmission line is given as + Is ^z h = V 0 ^e-jbz - Ge jbz h Z0 So, the current flowing in the load (at z = 0 ) is + -j135c 1 - 0.45e-j63.43ch IsL = V 0 ^1 - G h = 75e 100 ^ Z0 = 0.67e-j108.4c Therefore, the instantaneous current at the load terminal will be iL ^ t h = Re "IsL e jwt , = 0.67 cos ^2p # 0.3 # 109 t - 108.4ch = 0.75 cos ^3p # 108 t - 108.4ch 2p
Option (C) is correct. Generator voltage in phasor form, Vsg = 150 V Internal impedance of generator, Zg = 100 W Load impedance ZL = 150 W Length of transmission line, l = 0.15l Characteristic impedance Z 0 = 100 W So, the input impedance of the lossless transmission line is given as Z + jZ 0 tan bl Zin = Z 0 c L Z 0 + jZL tan bl m N J 2p K 150 + j100 tan b l 15l l O O = 100 K KK100 + j150 tan b 2p 0.15l l OO l P L 150 + j100 tan 54c = ^80.5 - j32.7h W = 100 c 100 + j50 tan 54c m Now, for determining the power delivered, we draw the equivalent circuit for the transmission line as shown in figure below :
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SOL 8.2.12
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2p
Using voltage division, we get the input voltage as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 8
Transmission Lines
547
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Shop Online at www.nodia.co.in 82.5 - j32.7 Vs, in = Vg c Zin m = 150 c = 71.8e-j11.46c 82.5 - j32.7 + 100 m Zin + Zg So, the current at the input current is -j11.46c V Is, in = s, in = 71.8e = 0.81e j10.16c 82.5 - j32.7 Zin Therefore, the average input power delivered to the transmission line is given as Pin = 1 Re 6Vs, in I s), in@ = 1 Re 6^71.8e-j11.46ch^0.81e-j10.16ch@ 2 2 = 27 Watt
Option (C) is correct. Since, the lengths of line 1 and line 2 are l1 = l2 = l/2 So, the input impedance of the line 1 is given as Z + jZ 0 tan bl Zin1 = Z 0 c L1 Z 0 + jZL1 tan bl m J 2p l N KZL1 + jZ 0 tan b l 2 l O O ( b = 2p/l ) = Z0 K KKZ 0 + jZL1 tan b 2p l l OO l 2 L P = Z 0 b ZL1 + 0 l = ZL1 Z0 + 0 = 50 W Similarly, the input impedance of line 2 is given as Zin2 = ZL2 = 150 W The effective load for line 3 will be equal to the equivalent impedance of the parallel combination of input impedances of line 1 and line 2. i.e. ZLl = Zin1 || Zin2 = 150 = 75 W 2 So, the input impedance for line 3 is given as (length of line 3, l = l/2 ) Zin = ZLl = 75 W Therefore, the input voltage of line 3 is Vs, in = Vsg c Zin m = 500 b 75 l 75 + 100 Zin + Zg = 214.28 volt and so the current at the input terminal of line 3 is V Is, in = s, in = 2.86 A Zin Thus, the average power delivered to the lossless transmission line 3 is given as Pin = Re 6Vs, in Is), in @ = 1 # ^214.28h # ^2.86h = 306.11 Watt 2 Since, the transmission line is lossless so, the power delivered to each load will be same and given as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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SOL 8.2.13
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Transmission Lines
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P1 = P2 = Pin = 1 # 306.11 = 153.1 Watt 2 2 Option (D) is correct. Given, transmission line is of infinite length i.e. l = 3. and input impedance of the transmission line is equal to its characteristic impedance i.e. Zin = Z 0 Since, the input impedance of a transmission line is defined as Z + Z 0 tanh gl Zin = Z 0 c L Z 0 + ZL tanh gl m Z + Z 0 tanh gl So, Z0 = Z0 c L Z 0 + ZL tanh gl m Solving the equation, we get tanh gl = 1
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SOL 8.2.14
Chap 8
Option (C) is correct. As discussed in previous question the input impedance of infinitely long lossy transmission line is equal to it’s characteristic impedance. So, the input impedance to line 1 will be Zin1 = Z 01 = 200 W From the shown arrangement of the transmission line it is clear that the effective load impedance for line 2 will be equal to the input impedance of line 1. i.e. ZL2 = Zin1 = 200 W Since the length of the line 2 is l/2 so, the input impedance of line 2 will be equal to its load i.e. (l = l/2 ) Zin2 = ZL2 = 200 W Therefore, the reflection coefficient at the load terminal of line 2 is given as G = ZL2 - Z 02 = 200 - 100 = 1 3 ZL2 + Z 02 200 + 100 Now, the input voltage of line 2 is determined by using voltage division rule as Vs, in = Vs, g c Zin2 m Zin2 + Zg = 4 b 200 l = 8 volt 200 + 100 3 Again, the voltage at any point on line 2 is given as
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SOL 8.2.15
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e gl - e- gl = 1 e gl + e- gl e- gl = 0 Since, l = 3. So for satisfying the above condition propagation constant g must have a real part. i.e. real part of g ! 0 or, ( g = a + jb ) a!0 As the attenuation constant of the voltage wave along the transmission line is not equal to zero therefore, it is a lossy transmission line.
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Vs ^z h = V 0+ _e j l b- 2 l + Ge-j l b- 2 li 2p
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Option (B) is correct. Consider the length of the transmission line is l as shown in figure below
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8 = V + e-jp + Ge jp (Vs ^z h = Vs, in at z =- l/2 ) h 0 ^ 3 1 1 V 0+ = 8 # bG = 3 l 3 1 b- 1 - 3 l =- 2 volt Therefore, the incident average power to the line 2 is given as V 0+ 2 4 = = 20 mWatt P avi = 2 # 100 2Z 02 So, the reflected average power at the input terminal of line 1 (load terminal of line 2) is 2 P avr = G 2 P avi = b 1 l # 20 = 2.2 mWatt 3 Thus, we get the transmitted power to the line 1 as P avt = P avi - P avr = 20 - 2.2 = 17.8 mWatt
The generator voltage is applied to the transmission line at time t = 0 for which the voltage at the sending end is (at t = 0 ) v ^0 h = 10 volt After time Dt = 4 ms the voltage v ^ t h at the sending end changes to 6 V. This change in the voltage will be caused only if the reflected voltage wave from the load comes to the sending end. So, the time duration for the change in voltage at sending end can be given as Tt =(time taken by incident wave to reach the load) + (time taken by reflected wave to reach sending end from the load) l (1) or, Tt = + l = 2l vp vp vp where l is the length of the transmission line (distance between load and sending terminal) and v p is phase velocity of the wave along the transmission line. Since, GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in the line is air spaced so, v p = c = 3 # 108 m/s Putting it in equation (1) we get 2l ( Dt = 4 ms ) 4 ms = 3 # 108 Thus, length of the transmission line is 8 -6 l = 3 # 10 # 4 # 10 = 240 m 2
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Option (C) is correct. Let the load impedance connected to the transmission line is ZL so the equivalent circuit for the transmission line will be as shown in figure below :
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Since, the internal resistance of the generator is equal to the characteristic impedance of the line i.e. Rg = Z 0 = 100 W So, the reflection coefficient due to source resistance will be zero and therefore, the change in voltage at sending will be caused only due to the reflection coefficient at load terminal given as Dv (t) = GV0+ where, V0+ is amplitude of the incident voltage wave and G is the reflection coefficient at the load terminal. Since, the change in voltage at t = 4 ms is Dv (t) = 6 - 10 =- 4 So, we get (V0+ = 10 V ) - 4 = 10G G =- 4 =- 0.4 10 ZL - Z 0 =- 0.4 b ZL + Z 0 l ZL - 100 (Z 0 = 100 W ) b ZL + 100 l =- 0.4 ZL - 100 =- 0.4ZL - 40 ZL = 29.86 W Option (B) is correct. Observing the waveform we conclude that at the sending end voltage changes at t = t1 . The changed voltage at the sending is given as (1) v ^t1h = V 0+ + GL V 0+ + Gg GL V 0+ GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in + where V 0 is voltage at sending end at t = 0 , GL and Gg are the reflection coefficients at the load terminal and the source terminal respectively. So, we get (ZL = 0 ) GL = ZL - Z 0 =- 1 ZL + Z 0 R - Z0 and (Zg = Rg ) Gg = g Rg + Z 0 Putting these values in equation (1), we get v ^t1h = V 1+ - V 1+ - Gg V 1+ (2) v ^t1h =- Gg V 1+ From the shown wave form of the voltage at sending end, we have v ^t1h = 6 volt V 0+ = 24 volt Putting these values in equation (2), we get 6 =- Gg ^24h or, Gg =- 4 Rg - Z 0 =- 4 Rg + Z 0 Rg = 60 W ^Z 0 = 100 Wh At t = 0 as the voltage just applied to transmission line, the input impedance is independent of ZL and equals to Z 0 (i.e. Zin = Z 0 at t = 0 ). Therefore, using voltage division the voltage at the sending end is given as V 0+ = Vg c Z 0 m Rg + Z 0 24 = Vg b 100 l (V 0+ = 24 volt ) 60 + 100 Vg = 24 # 160 = 38.4 volt 100 Option (D) is correct. Length of the transmission line,l = 1.5 m Internal resistance of generator, Rg = 200 W Characteristic impedance, Z 0 = 100 W Generator voltage, Vg = 30 volt Load impedance, ZL = 50 W So, the reflection coefficient at the load terminal is GL = ZL - Z 0 = 50 - 100 =- 1 3 ZL + Z 0 50 + 100 and the reflection coefficient at the source terminal is R - Z 0 200 - 100 1 Gg = g = = Rg + Z 0 200 + 100 3 Again as discussed in previous question at time, t = 0 as the voltage is just applied to the transmission line, the input impedance is independent of ZL and equals to Z 0 (i.e. Zin = Z 0 at t = 0 ). Therefore, using voltage division the input voltage at the sending end is given as
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Shop Online at www.nodia.co.in V 0+ = Vg c Z 0 m = 30 # b 100 = 10 volt 200 + 100 l Rg + Z 0 Now, the time taken by the wave to travel from source terminal to the load terminal (or load terminal to source terminal) is given as T =l c where, l is length of transmission line and c is the velocity of the voltage wave along the transmission line. So, we get T = 1.5 8 = 5 ns 3 # 10 Therefore, for the interval 0 # t < 5 ns , the incident wave will be travelling from source to load and will have the voltage V 1+ = 10 volt For the interval 5 ns # t < 10 ns an additional reflected wave will be travelling from load to source and will have the voltage V 1- = GL V 1+ =- 10 =- 3.33 volt 3 For 10 ns # t # 15 ns the wave reflected by source resistance travelling from source to load will be added to that has the voltage V 2+ = Gg V 1- =- 3.33 =- 1.11 volt 3 For 15 ns # t # 20 ns again the wave reflected by load travelling from load to source will be added that has the voltage V 2- = GL V 2+ = 1.11 = 0.37 volt 3 This will be continuous and the bounce diagram obtained between source ^at z = 0h and load (at z = 1.5 m ) will be as shown in figure below :
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Option (C) is correct. From the bounce diagram that obtained between source terminal ( z = 0 ) and load terminal ( z = 1.5 m ) in previous question, we can determine the voltage v ^ t h at any
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Option (A) is correct.
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For View Only Shop Online at www.nodia.co.in instant by just summing all the voltage waves existing at any time t . Since, for interval 0 # t # 10 ns only a single voltage wave with V 1+ = 10 volt exists at sending end so, the voltage at the sending end (z = 0 ) for the interval is for 0 # t < 10 ns v ^ t h = V 1+ = 10 volt again for the interval 10 ns # t < 20 ns , three voltage waves with V 1+ = 10 volt , V 1- =- 3.33 volt and V 2+ =- 1.11 volt exists at the sending end so, the voltage at the sending end for the interval is for 10 ns # t < 20 ns v ^ t h = V 1+ + V 1- + V 2+ = 10 - 3.33 - 1.11 = 5.6 volt Thus, the obtained voltage wave form is plotted in the figure below
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As shown in the smith chart, SWR circle meets the Gr axis (real part of reflection coefficient) at L1 and L2 respectively. So, We have the two possible values of normalised impedance (real values of zL ). at L1 zL1 = 2.5 at L2 zL2 = 0.4 Since, the normalised impedance is defined as Load impedance zL = Characteristic impedance So, we have zL1 = ZL = 2.5 Z 01 or, Z 01 = ZL = 50 = 20 W 2.5 2.5 Similarly, zL2 = ZL = 0.4 Z 02 or, Z 02 = ZL = 50 = 125 W 0.4 0.4 Therefore, the two possible values of the characteristic impedance of the lossless transmission line are 20 W and 125 W. SOL 8.2.22
Option (B) is correct. We can determine the reflection coefficient of the transmission line using smith
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For View Only Shop Online at www.nodia.co.in chart as explained below : (1) First we determine the normalized load impedance of the transmission line as 100 + j50 = 1 + j0.5 zL = ZL = 100 Z0 (2) Comparing the normalized impedance to its general form zL = r + jx where r is the normalized resistance (real component) and x is the normalized reactance (imaginary component). we get
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r = 1 and x = 0.5 (3) Now, we determine the intersection point of r = 1 circle and x = 0.5 circle on the smith charge and denote it by point P as shown in the smith chart. It gives the position of normalized load impedance. (4) We join the point P and the centre O to form the line OP
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(5) Extend the line OP to meet the r = 0 circle at Q . The magnitude of the reflection coefficient of the transmission line is given as G = OP = 2.1 cm = 0.22 9.4 cm OQ (6) Angle of the reflection coefficient in degrees is read out from the scale at point Q as qG = 76.0c (7) Thus, we get the reflection coefficient of the transmission line as qG = 0.22e j76c
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G = G
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Option (C) is correct. As shown in the smith chart in previous question normalized load impedance is located at point P . So, for determining the input impedance at a distance of 0.35l from the load we follow the steps as explained below : (1) First we draw a SWR circle (circle centered at origin with radius OP )
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Alternate Method : Reflection coefficient of the transmission line is defined as 100 + j50 - 100 G = ZL - Z 0 = = 0.24 76c = 0.24e j76c 100 + j50 + 100 ZL + Z 0 which is same as calculated from smith chart.
(2) For finding input impedance at a distance of 0.35l from load we move a distance of 0.35l on WTG scale (wave length toward generator) along the SWR circle. (3) Since, the line OP corresponds to the reading of 0.144l on WTG scale so, after moving a distance of 0.35l on WTG scale we reach at 0.144l + 0.35l = 0.494l on WTG scale. The reading corresponds to the point A on the SWR circle. (4) Taking the values of r and x -circle at point A we find out normalized input impedance as GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Shop Online at www.nodia.co.in zin = r + jx = 0.61 + j ^- 0.022h = 0.61 - j0.022 (5) Therefore, the input impedance at a distance of 0.35l from load is given as
Option (C) is correct. For determining the shortest length of the transmission line for which the input impedance appears to be purely resistive, we follow the steps as explained below : (1) First we determine the WTG reading of the point denoting the normalized load impedance on the smith chart. From the above question, we have the reading of point P as 0.144l on WTG circle.
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Zin = zin Z 0 = 100 ^0.61 - j0.022h = ^61 - j2.2h W Alternate Method : We can conclude the input impedance at l = 0.35l directly by using formula Z + jZ 0 tan bl lossless transmission line Zin = Z 0 c L Z 0 + jZL tan bl m J N 2p K 100 + j50 + j100 tan b l 0.35l l O O = 100 K 2p KK100 + j ^100 + j50h tan b l 0.35l l OO L j2.2h W P = ^87 as calculated above using with chart.
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(2) Since, the resistive load lies on the real axis of reflection coefficient ( Gr -axis). So, we move along the SWR circle to reach the Gr -axis and denote the points as A and B .
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(4) Since, point B is nearer to the point P so, it will give the shortest length of the transmission line for which the input impedance appears to be purely resistive. (5) Now, we have the reading of point B on WTG scale as 0.25l. So, the shortest length for the input impedance to be purely resistive is given as the difference between the readings at point B and P . i.e., l = 0.25l - 0.144l = 0.106l
SOL 8.2.25
Option (A) correct. The voltage maximum occurs at the point where the SWR circle intersects the positive Gr axis on smith chart. The SWR circle of the load impedance intersects the positive Gr axis at point B as shown in the Smith chart. So, the point B gives the position of first voltage maxima. As calculated in previous question the distance between point B and point A on the WTG scale is 0.106l. Therefore, the 1st voltage maximum occurs at a distance of 0.106l from load.
Option (B) is correct. At any time t , the currents of positive and negative waves are respectively I + and I - and the voltages of positive and negative waves are respectively V + and V - as shown in the figure. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 8.2.26
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+ I+ =V = 1 ^V + = 1 Volth Z0 Z0 and I - =-V Z0 Now, the voltage and current across an inductor are related as v = L di dt V + + V - = 2 d ^I + + I -h dt 1 + V - = 2 d :1 - V D ^V + = 1, Z 0 = 50h 50 dt 1 + V - =- 1 dV 25 dt - 25dt = dV 1+V Taking integration both sides we get where C1 is a constant ln ^1 + V -h =- 25t + C1 -25t (1) ^1 + V h = Ae + + Since, the voltage ^V h wave is incident at t = 0 so, at t = 0 the current through inductor is zero and therefore, from the property of an inductor at t = 0+ the current through inductor will be also zero. i.e. ^I + + I -hat t = 0 = 0 +
V+ V=0 ; Z0 - Z0 E at t = 0 1 V=0 :Z - Z0 D 0 0 at t = 0 So at t = 0+ , V 0- = 1 volt Putting it in equation (1), we get ^1 + 1h = A A =2 Thus, the voltage of the reflected wave is V - = ^2e-25t - 1h Volt
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Option (D) is correct. The voltage of positive wave in transmission line is V 0+ . So, at the voltage maxima, magnitude of the voltage is given as Vs max = V 0+ 61 + G @ GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in and at the point of voltage maxima the current will be minimum and given as V 0+ Is min = 1 - G@ Z0 6 So, the line impedance at the point of voltage maxima will be Vs max = Z0 b 1 + G l Z max = 1-G Is min G = Z0 S bS = 11 + -G l Now, at the voltage minimum the voltage magnitude is Vs min = V 0+ 61 - G @ and at the point of voltage minimum current will be maximum and given as, V 0+ 1 + G@ Is max = Z0 6 and the line impedance at the point will be Vs min 1+G = Z0 b 1 - G l = Z0 Z min = bS = 1 - G l 1+G S Is max SOL 8.2.28 Option (A) is correct To determine the required quantity, we note that for a particular line of characteristic impedance Z 0 , the product of the line impedances at two positions (two values of d ) separated by an odd multiple of l/4 is given by 1 + G bd + ^2n - 1h l l 4 4 1 + G ^d h * l "Z 6d @,'Z :d + ^2n - 1h 4 D1 = *Z 0 f p4 Z 0 1 - G ^d h 1 - G ^d + ^2n - 1h l/4h
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1 + G ^d h 1 + G ^d h e-j2b^2n - 1h 4 > H =Z 1 - G ^d h >1 - G ^d h e-j2b^2n - 1hl4 H l
2 0
1 + G ^d h 1 + G ^d h e-j^2n - 1hp = Z 02 > > H 1 - G ^d hH 1 - G ^d h e-j^2n - 1hp 1 + G ^d h 1 - G ^d h = Z 02 > 1 - G ^d hH>1 + G ^d hH = Z 02 As the intrinsic impedance of medium 1 is h1 and that of medium 3 is h 3 so, for required match, thickness t is l/4 and the intrinsic impedance ( h2 ) of the medium 2 is given as h 1 h 3 = h 22 or h 2 = h1 h 3 SOL 8.2.29
Option (C) is correct. As determined in previous question, for a wave travelling through the three mediums of intrinsic impedances h1 , h 2 and h 3 , the condition for matching dielectric (the intrinsic impedance of medium 2 that eliminates the reflected wave in medium 1) is h2 = h 1 h3 Since, all the media have m = m0 so, for the dielectrics ^s = 0h the above equation can be rewritten as
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Shop Online at www.nodia.co.in m0 m0 m0 m = b bh = e2 e0 lb 16e0 l el where e2 is the permittivity of the medium 2. e = 4e0 Option (B) is correct. The thickness ‘t ’ of the dielectric coating for the perfect matching (the condition for eliminating reflection) is given as (quarter wave) t =l 4 where l is the wavelength of plane wave. The wavelength in terms of frequency is v l = p f where v p is the phase velocity of the wave in the propagation medium which is given as 8 1 vp = 1 = = 3 # 10 = 1.5 # 108 2 me m0 4e0 So, at frequency f = 1.5 GHz the thickness of the dielectric coating is given as 8 v So, t = p = 1.5 # 10 9 = 0.25 m = 2.5 cm 4f 4 ^1.5 # 10 h Option (B) is correct. Distance between load and first voltage maxima, l max = 0.125l Characteristics impedance, Z 0 = 100 W Standing wave ratio, S =3 Position of voltage maxima (l max ) in terms of reflection coefficient G qG is where n = 0 , 1, 2,....... l max = qG l + nl 2 4p So, for 1st voltage maxima we have n = 0 and so, we get the position of first voltage maxima as l max = qG l 4p q 0.125l = G l & qG = p 2 4p The magnitude of reflection coefficient is defined in terms of SWR as G = S-1 = 3-1 = 1 S+1 3+1 2 So, the reflection coefficient of the transmission line is j G = G qG = 1 e jp/2 = 2 2 Therefore, the load impedance of the transmission line is given as J jN K1 + 2 O + G 1 = 60 + j80h W ZL = Z 0 b = 100 K jO ^ 1-G l K1 - O 2P L Option (A) is correct.
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SOL 8.2.30
SOL 8.2.32
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For View Only Shop Online at www.nodia.co.in Given, the transmission line is terminated by its characteristic impedance i.e., ZL = Z 0 So, there will be no reflected wave and therefore, the height of the voltage pulse will be given as Z V (Zg " internal resistance of generator) V 1+ = 0 g Z 0 + Zg = 100 # 15 = 10 Volt 100 + 50 As the wave travels in the + Z direction along transmission line at velocity 1 1 vp = = -6 LlC l ^0.25 # 10 h # ^100 # 10-12h 8 = 2 # 10 m/s So, the voltage pulse will reach at l = 5 m at time, 5 t0 = = 25 ns 2 # 108 So, at l = 5 m for 0 < t < 25 ns , V =0 and for t $ 25 ns V = V 1+ = 10 Volt Therefore the plot of voltage against time at a distance 5 m from the source is as shown in graph below.
Option (A) is correct. As the first forward voltage pulse is V 1+ so, the first reflected pulse voltage is V 1- = GL V 1+ nd The 2 forward pulse voltage is given as V 2+ = Gg V 1- = Gg GL V 1+ The 2 nd reflected pulse voltage is given as V 2- = GL V 2+ = Gg GL2 V 1+ So, summing up all the pulses at load end for steady state (t " 3) we get the load voltage as VL = V 1+ + V 1- + V 2+ + V 2- + ... = V 1+ 61 + GL + Gg GL + Gg GL2 + ....@ 2 2 = V 1+ 7^1 + Gg GL + Gg GL + ....h + GL ^1 + Gg GL + ....hA GL 1 = V 1+ S21 S22H ^ZL || Z 0h - Zo ^50 || 50h - 50 =- 1 = S11 = 3 ^ZL || Z 0h + Zo ^50 || 50h + 50 2 ^ZL || Zo h 2 ^50 || 50h = =2 S12 = S21 = 3 + 50 || 50 50 + Z || Z Z ^ h ^ L oh o ^ZL || Zo h - Zo ^50 || 50h - 50 =- 2 = S22 = 3 ^ZL || Zo h + Zo ^50 || 50h + 50 SOL 8.3.11 Option (A) is correct. The input impedance of a quarter wave (l = l/4 )lossless transmission line is defined as 2 Zin = Z o ZL where, Z 0 is characteristic impedance and ZL is the load impedance of the line. So, we have the input impedance of line 1 as 2 2 Zin1 = Z o1 = 50 = 25 ZL1 100 Similarly, the input impedance of line 2 is 2 2 Zin2 = Z o2 = 50 = 12.5 ZL2 200 The effective load impedance of the line 3 is given as ZL = Zin1 || Zin2 = 25 || 12.5 = 25 3 So, the input impedance of the 50 W transmission line is (50) 2 ZS = = 300 25/3 Therefore, the reflection coefficient at the input terminal is given as G = ZS - Z 0 = 300 - 50 = 2 ZS + Z 0 300 + 50 7 SOL 8.3.12 Option (D) is correct. We have 10 log Gp = 10 dB or Gp = 10 The power gain of the antenna is defined as Gp = Prad Pin where Prad is the radiated power of the antenna and Pin is the input power feed to the antenna. So, putting all the values we get 10 = Prad 1W or Prad = 10 Watts GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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SOL 8.3.18
Option (C) is correct. The ratio of the load impedance to the input impedance of the transmission line is given as VL = Z 0 Vin Zin or VL = Z 0 Vin = 10 # 300 = 60 V 50 Zin Option (D) is correct. Suppose at point P impedance is Z = r + j (- 1) If we move in constant resistance circle from point P in clockwise direction by an angle 45c, the reactance magnitude increase. Let us consider a point Q at 45c from point P in clockwise direction. It’s impedance is Z1 = r - 0.5j or Z1 = Z + 0.5j Thus movement on constant r - circle by an 45c in CW direction is the addition of inductance in series with Z .
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For View Only Shop Online at www.nodia.co.in SOL 8.3.13 Option (A) is correct. The characteristic impedance of a transmission line is defined as Z 02 = Zoc Zsc where Zoc and Zsc are input impedance of the open circuited and is short circuited line. So, we get 2 Zsc = Z 0 = 50 # 50 = 50 Zoc 100 + j150 2 + 3j 50 (2 - 3j) = = 8.69 - 15.54j 13 SOL 8.3.14 Option (C) is correct. From the diagram, VSWR is given as S = Vmax = 4 = 4 1 Vmin Since, voltage minima is located at the load terminal so, the load impedance of the transmission line is given as (Z 0 = 50 W, S = 4 ) ZL = 6Zin@min = Zo = 50 = 12.5 W 4 S SOL 8.3.15 Option (D) is correct. The reflection coefficient at the load terminal is given as G = ZL - ZO = 12.5 - 50 =- 0.6 ZL + ZO 125. + 50 SOL 8.3.16 Option (C) is correct. The given circles represent constant reactance circle.
Option (A) is correct. The VSWR of a transmission line is defined as 1- G S = 1+ G GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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where G is the reflection coefficient of the transmission line. So, we get 1- G 2= 1+ G or G =1 3 Thus, the ratio of the reflected and incident wave is given as Pr = G 2 = 1 9 Pi or Pr = Pi 9 i.e. 11.11% of incident power is reflected. Option (D) is correct. The input impedance of a lossless transmission line is defined as 6ZL + jZo tan bl @ Zin = Zo 6Zo + jZL tan bl @ Now, for l/2 transmission line we have l = l/2 and ZL1 = 100 W So, the input impedance of the l/2 transmission line is 6ZL1 + jZo tan p@ = ZL1 = 100 W Zin1 = Zo 6Zo + jZL1 tan p@ For l/8 transmission line, we have l = l/8 and ZL2 = 0 So, the input impedance of l/8 line is [0 + jZo tan p4 ] = jZo = j50 W Zin2 = Zo 6Zo + 0@ Thus, the net admittance at the junction of the stub is given as Y = 1 + 1 = 2 + 1 = 0.01 - j0.42 100 j50 Zin1 Zin2
2p bb = l l
(short circuit) 2p bb = l l
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For View Only Shop Online at www.nodia.co.in SOL 8.3.21 Option (A) is correct. VSWR (voltage standing wave ratio) of a transmission line is defined as S = 1+G 1-G where G is the reflection coefficient of the transmission line that varies from 0 to 1. Therefore, S varies from 1 to 3. Option (B) is correct. Reactance increases, if we move along clockwise direction in the constant resistance circle.
SOL 8.3.23
Option (C) is correct. A transmission line is distortion less if LG = RC
SOL 8.3.24
Option (B) is correct. Zo =
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ZOC ZSC = 100 # 25 = 10 # 5 = 50 W
Option (B) is correct. We know that distance between two adjacent voltage maxima is equal to l/2 , where l is wavelength. So, we get l = 27.5 - 12.5 2 or, l = 2 # 15 = 30 cm Therefore, the operating frequency of the transmission line is 10 (c = 3 # 1010 cm/s ) f = c = 3 # 10 = 1 GHz 30 l Option (C) is correct. Electrical path length = bl where b = 2p , l = 50 cm l Now, the operating wavelength l of the transmission line is given as l =u =1# 1 au= 1 f f LC LC 7 1 1 5 10 # = = = 2m # 25 # 106 25 # 106 10 # 10-6 # 40 # 10-12 So, the electric path length is bl = 2p # 50 # 10-2 = p radian 2 2
SOL 8.3.26
SOL 8.3.27
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Option (B) is correct. The input impedance at the voltage minima on the transmission line is defined as Z0 6Zin@min = S where S is standing wave ratio along the transmission line. Since, the reflection coefficient GL of the transmission line is given as GL = ZL - Z 0 = 100 - 50 = 50 = 1 ZL + Z 0 100 + 50 150 3
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For View Only Shop Online at www.nodia.co.in So, the standing wave ratio of the line is 1+1 1 + GL 3 =2 S = = 1 1 - GL 13 Therefore, the minimum input impedance measured on the line is equal to Zin min = 50 = 25 W 2 SOL 8.3.28 Option (D) is correct. For a lossy transmission line the input impedance is given as Z + jZ 0 tanh gl Zin = Z 0 < L F Z 0 + jZL tanh gl Load impedance, (open circuited at load end) ZL = 3 Length of line, l = l/4 R V S1 + jZ 0 tanh lg W ZL W So, Zin = Z 0 lim S Z " 3S Z 0 W S ZL + j tanh lg W X gp Z0 T = 0 = a tanh 4 " 3k j tanh lg SOL 8.3.29 Option (D) is correct. Input impedance of a lossless transmission line is given by Z + jZ 0 tan bl Zin = Z 0 < L F Z 0 + jZL tan bl where Z 0 " Charateristic impedance of line ZL " Load impedance l " Length of transmission line and b = 2p/l
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bl = 2p l = p 2 l 4 (Short circuited at load) ZL = 0 and Z 0 = 50 W Therefore, the input impedance of the transmission line is 0 + j50 tan p/2 Zin = 50 = 50 + j0 tan p/2G = 3 i.e. infinite input impedance and thus, the current drawn from the voltage source will be zero.
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So, we have
SOL 8.3.30
Option (B) is correct. For lossless transmission line, the phase velocity is defined as vp = w = 1 b LC Characteristics impedance for a lossless transmission line is given as Z0 = L C
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
...(1)
...(2)
Chap 8
Transmission Lines
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For View Only Shop Online at www.nodia.co.in So, from equation (1) and (2) we get 1 vp = = 1 C (Z 0 C ) Z 0 C SOL 8.3.31 Option (C) is correct. Input impedance of a ^l/4h transmission line is defined as 2 Zin = Z 0 ZL where Z 0 is characteristic impedance of the line and ZL is load impedance of the line. Since, the l/4 line is shorted at one end (i.e. ZL = 0 ) So, we get, 2 Zin = lim Z 0 = 3 Z " 0Z L L
Option (D) is correct. Voltage minima of a short circuited transmission line is located at it’s load. As the location of minima is same for the load RL (i.e. the minima located at RL ) so, the first voltage maxima will be located at l/4 distance from load. Now, ...(1) l max = qG l 4p where l max is the distance of point of maxima from the load, qT is phase angle of reflection coefficient and l is operating wavelength of line. So, putting the value of l max is equation (1), we get l = qG l 4p 4 qT = p Now, the standing wave ratio of the line is given as 1+ G S = 1- G 1 + GL or, 3= ]S = 3g 1 - GL GL = 1/2 i.e. GL = GL qG = 1 p =- 1 2 2 The reflection coefficient at the load terminal is given as GL = ZL - Z 0 ZL + Z 0 - 1 = RL - 75 ^ZL = RL h, ^Z 0 = 75 Wh 2 RL + 75 - RL - 75 = 2RL - 150 3RL = 75 & RL = 25 W
SOL 8.3.33
Option (B) is correct. The VSWR (voltage standing wave ratio) in terms of maxima and minima voltage is defined as Vmax S = =4=2 2 Vmin
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SOL 8.3.32
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Chap 8
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For View Only Shop Online at www.nodia.co.in SOL 8.3.34 Option (D) is correct. Characteristic impedance, Z 0 = 60 W SWR S =4 + G 1 L So, we have =S=4 1 - GL GL = 3 = 0.6 5 The reflection coefficient at load is defined as GL = ZL - Z 0 ZL + Z 0 So, 0.6 = ZL - 60 ZL + 60 ZL = 1.6 # 60 = 120 W 0. 4 Option (A) is correct. Loading of a cable is done to increase the inductance as well as to achieve the distortionless condition. i.e. statement (1) and (4) are correct.
SOL 8.3.36
Option (C) is correct. Single stub with adjustable position is the best method for transmission line load matching for a given frequency range.
SOL 8.3.37
Option (B) is correct. The reflection coefficient at load terminal is defined as + j50 - 50 GL = ZL - Z 0 = j50 + 50 ZL + Z 0 =j Therefore, the standing wave, ratio is 1 + GL VSWR = = 1+1 = 3 1-1 1 - GL Option (D) is correct. Given, the load impedance is short circuit i.e. ZL = 0 So, input impedance for lossless line is given as Z + jZ 0 tan bl = jZ 0 tan bl Zin = Z 0 c L Z 0 + jZL tan bl m Now, for l < l/4 & bl < p 2 So, tan bl is positive and therefore, Zin is inductive l < l < l & p < bl < p For 2 2 4 tan bl is - ve and therefore, Zin is capacitive For l = l & bl = p 2 4
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SOL 8.3.38
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SOL 8.3.35
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
a"2 b"1
Chap 8
Transmission Lines
For View Only tan bl = 3 and therefore Zin = 3
575
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l = l & bl = p 2 tan bl = 0 and therefore, Zin = 0
For
d"3
Option (C) is correct. For distortionless transmission line, a = RG , b = w LC and for lossless transmission line, a = 0, b = w LC So, for both the type of transmission line attenuation is constant and is independent of frequency. Where as the phase shift b varies linearly with frequency w. i.e. statement 1 and 3 are correct.
SOL 8.3.40
Option (A) is correct. Given, Length of transmission line, l = 500 m Phase angle, qG =- 150c Operating wavelength, l = 150 m Consider the reflected voltage wave for the lossless transmission line terminated in resistive load as shown in figure.
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SOL 8.3.39
Since, the reflection coefficient has a phase angle - 150c So, the wave lags by 150c angle.
The voltage wave has the successive maxima at each l/2 distance, Total length So, the total no. of maxima = = 500 = 6 2 3 Distance between two maxima ^150/2h GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
576
Transmission Lines
Chap 8
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Shop Online at www.nodia.co.in i.e. 6 maxima and remaining phase angle = 2 # 360c = 240c 3 From the wave pattern shown above we conclude that the remaining phase ^240ch will include one more maxima and therefore the total no. of maxima is 7. Option (A) is correct. Reflection coefficient at load terminal is defined as GL = ZL - Z 0 ZL + Z 0 For a matched transmission line we have ZL = Z 0 So, GL = 0 i.e. matching eliminated the reflected wave between the source and the matching device location.
SOL 8.3.42
Consider the quarter wave transformer connected to load has the characteristic impedance Zl0 as shown in the figure.
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SOL 8.3.41
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So, we have the input impedance, J 2p l N KZL + jZl0 tan b l lb 4 l O ^Zl0h2 O= Zin = Zl0 K ZL KK Z 0 + jZL tan b 2p lb l l OO l 4 L P this will be the load to 450 W transmission line ^Zl0h2 ^Zl0h2 i.e. ZlL = = 200 ZL and for matching Z 0 = ZlL ^Zl0h2 450 = 200 Zl0 = ^450h^200h = 300 W SOL 8.3.43
Option (D) is correct. Given ZL = 3 and l =l 4
(open circuit) (quarter wave)
ZL + jZ 0 tan bl =- jZ 0 cot bl ^ZL " 3h Z 0 + jZL tan bl m GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia Zin = Z 0 c
Chap 8
Transmission Lines
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=- jZ 0 cot b 2p lb l l l 4
Option (A) is correct. Length of transmission line l < l/4 Load impedance, ZL = 3 So, the input impedance of the transmission line is given as Z + jZ 0 tan bl Zin = Z 0 c L Z 0 + jZL tan bl m = Z0 c 1 m j tan bl =- jZ 0 cot bl
(open circuit)
(lossless line)
^ZL = 3h
Option (C) is correct. Given, Load impedance, (short circuit) ZL = 0 Line parameters, (loss free line) R = G= 0 Attenuation constant, (loss free line) a =0 So, the input impedance of the line is given as Zin = jZ 0 tan bl i.e. pure reactance Statement (A) is correct. Since tan bl can be either positive or negative So Zin can be either capacitive or inductive. Statement (B) is correct. The reflection coefficient at load is GL = ZL - Z 0 =- 1 ! 0 ZL + Z 0 So, the reflection exists. Statement (C) is incorrect. and since the standing waves of voltage and current are set up along length of the lines so, statement (D) is also correct.
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SOL 8.3.45
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SOL 8.3.44
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Option (C) is correct. (a) short circuit ^ZL = 0h So GL = ZL - Z 0 =- 1 ^a " 2h ZL + Z 0 (b) Open circuit ^ZL = 3h So, GL = ZL - Z 0 = 1 ^b " 3h ZL + Z 0 (c) Line characteristic impedance ^ZL = Z 0h GL = Z 0 - Z 0 = 0 ^c " 1h Z0 + Z0 (d) 2 # line characteristic impedance ^ZL = 2Z 0h ^d " 4h GL = 2Z 0 - Z 0 = 1 2Z 0 + Z 0 3 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 8.3.46
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Option (C) is correct. Input impedance of a quarter wave transformer is defined as 2 Zin = Z 0 ZL
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SOL 8.3.51
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For View Only Shop Online at www.nodia.co.in SOL 8.3.47 Option (A) is correct. Given, reflection coefficient, GL = 1 0c 1+ G So, VSWR = = 1+1 = 3 1-1 1- G SOL 8.3.48 Option (D) is correct. Given, reflection coefficient, G = 1 5 1+ G So, VSWR = =6=3 4 2 1- G SOL 8.3.49 Option (C) is correct. Characteristic impedance of transmission line is defined as R + jw L Z0 = G + j wC So, for lossless transmission line (R = G = 0 ) Z0 = L C SOL 8.3.50 Option (C) is correct. Input impedance has the range from 0 to 3. ^a " 3h VSWR has the range from 1 to 3 ^c " 2h Reflection coefficient ^G h ranges from - 1 to + 1. ^b " 1h
Option (D) is correct. The scattering parameters linearly relate the reflected wave to incident wave and it is frequency invariant so the scattering parameters are more suited than impedance parameters.
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SOL 8.3.52
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where Z 0 is the characteristic impedance of the line and ZL is the load impedance. Since the quarter wave transformer is terminated by a short circuit (ZL = 0 ) so, we get the input impedance of the transformer as Zin = 3
Option (C) is correct. Given, the reflection coefficient as GL = 0.3e-j30c At any point on the transmission line the reflection coefficient is defined as G ^z h = GL e-2gz where z is the distance of point from load. (Given) z = 0.1l -2g^0.1lh -j30c -2jb^0.1lh So, (Assume a = 0 ) G ^z h = GL e = ^0.3e h^e h -j30c -j0.4p = 0.3e ^e h GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 8.3.53
Chap 8
Transmission Lines
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579
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-j152c
Option (D) is correct. Balun is used to couple a coaxial line to a parallel wire.
SOL 8.3.55
Option (B) is correct. The reflection coefficient of the conducting sheet is G =- 1 where as the transmission coefficient is G = 0 . So, there will be x -directed surface current on the sheet.
SOL 8.3.56
Option (B) is correct. Given, Operating frequency, f = 25 kHz Conductivity, s = 5 mho/m Relative permittivity, er = 80 The attenuation constant for the medium is defined as wms a = ^s >> weh 2 ^2p # 25 # 103h # ^4p # 10-7h^5 h = ^m = m0h 2 = 0.7025 The attenuated voltage at any point is given as (1) V = V0 e- al where V0 is source voltage and l is the distance travelled by wave Since, the radio signal is to be transmitted with 90% attenuation so, the voltage of the signal after 90% attenuation is V = V0 - 90% of V0 = 0.1V0 Comparing it with equation (1) we get ^0.1h = e- al ln ^0.1h or, l == 3.27 m 0.7025 Option (A) is correct. In Smith chart, the distance towards the load is always measured in anticlockwise direction. So, statement 3 is incorrect while statement 1 and 2 are correct.
SOL 8.3.57
SOL 8.3.58
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SOL 8.3.54
Option (A) is correct. Given, Characteristic impedance, Z 0 = 75 W Load impedance, ZL = ^100 - j 75h W The condition for matching is ZlL = Z 0 where ZlL is the equivalent load impedance of the transmission line after connecting an additional circuit. So, the best matching will be obtained by a short circuited stub at some specific distance from load.
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Transmission Lines
Chap 8
Option (A) is correct. (1) Given, Length of line, l = l/8 Load impedance, ZL = 0 So, bl = b 2p lb l l = p 4 l 8 Therefore the input impedance of the lossless transmission line is jZ 0 tan p ZL + jZ 0 tan bl 4p f Zin = Z 0 c = Z0 Z0 Z 0 + jZL tan bl m (i.e., incorrect statement) = jZ 0 (2) Given, Length of line, l = l/4 Load impedance, ZL = 0 So, bl = b 2p lb l l = p 2 l 4 Therefore the input impedance of the lossless transmission line is Z + jZ 0 tan bl Zin = Z 0 c L Z 0 + jZL tan bl m jZ 0 tan p 2 p = j3 (i.e., correct statement) = Z0 f Z0 (3)Given, Length of line, l = l/2 Load impedance, ZL = 3 So, bl = b 2p lb l l = p l 2 Therefore, the input impedance of the lossless transmission line is
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SOL 8.3.61
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For View Only Shop Online at www.nodia.co.in SOL 8.3.59 Option (B) is correct. Given, the voltage standing wave ratio in decibels is VSWR indecibels = 6 dB or, 20 log 10 S = 6 S = ^10h6/20 = 2 So, the reflection coefficient at the load terminal is given as G = S - 1 = 2 - 1 = 0.33 S+1 2+1 SOL 8.3.60 Option (D) is correct. Input impedance of a quarter wave transformer (lossless transmission line) is defined as 2 Zin = Z 0 ZL where Z 0 is the characteristic impedance of the line and ZL is the load impedance of the line. So, we get Z 0 = Zin ZL = ^50h^200h = 100 W
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Chap 8
Transmission Lines
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Shop Online at www.nodia.co.in Z + jZ 0 tan bl Zin = Z 0 c L Z 0 + jZL tan bl m (i.e., incorrect statement) = Z 0 b 1 l =- j3 j tan p (4) Matched line have the load impedance equal to its characteristic impedance
Option (B) is correct. Since, a transmission line of output impedance 400 W is to be matched to a load of 25 W through a quarter wavelength line. So, for the quarter wave line we have Input impedance, Zin = 400 W (same as the o/p impedance of the matched line) Load impedance, ZL = 25 W Length of line, l = l/4 The characteristic impedance of quarter wave transmission line is Z 0 that connected between the load and the transmission line of output impedance 400 W as shown in figure.
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SOL 8.3.63
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SOL 8.3.62
i.e. ZL = Z 0 So, for the matched line the input impedance is Z + jZ 0 tan bl (i.e., correct statement) = Z0 Zin = Z 0 c L Z 0 + jZL tan bl m Option (C) is correct. Given, Length of line, l = l/8 Load impedance, ZL = 0 So, bl = b 2p lb l l = p 4 l 8 Therefore, the input impedance of the transmission line is Z + Z 0 tanh gl = Z 0 tanh gl Zin = Z 0 c L Z 0 + ZL tanh gl m If the line is distortion less (i.e. a = 0 ) then, the input impedance of the line is Zin = jZ 0 tan bl = jZ 0 So, it will depend on characteristic impedance as the line is resistive or reactive.
So,, the input impedance at AB is given as 2 ZL + jZ 0 tan p 2 = ^Z 0h Zin = Z 0 f p ZL Z 0 + jZL tan p 2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
582
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Z0 =
Zin ZL =
Chap 8
Shop Online at www.nodia.co.in 800 # 50 = 200 W
Option (B) is correct. Given, Length of transmission line, l = l/8 Load impedance, (Short circuited line) ZL = 0 So, we get bl = b 2p lb l l = p 4 l 8 Therefore, the input impedance of lossless transmission line is given as jZ Z + jZ 0 tan bl which is inductive = Z 0 c 0 m = jZ 0 Zin = Z 0 c L Z0 Z 0 + jZL tan bl m So, the input impedance of l/8 long short-circuited section of a lossless transmission line is inductive.
SOL 8.3.65
Option (C) is correct. In list I (a) Characteristic impedance of a transmission line is defined as R + jw L (a" 2 ) = Z Z0 = Y G + jw C (b) Propagation constant of the line is given as (b" 1) g = ^R + jwL h^G + jwC h = ZY (c) Sending end input impedance is Z + Z 0 tan gl Zin = Z 0 c L Z 0 + ZL tan gl m (terminated in characteristic impedance, Z 0 ) Given, ZL = Z 0 So, we get the input impedance as (c" 2 ) Zin = Z 0 = Z Y Option (C) is correct. For a distortionless transmission line, the attenuation constant (a) must be independent of frequency (w) and the phase constant (b ) should be linear function of w. (a) R =G=0 For this condition propagation constant is given as g = a + jb = ^R + jwL h^G + jwC h i.e. a = 0 and b = w LC As the attenuation constant is independent of frequency and the phase constant is linear function of w so, it is a distortionless transmission line. (b) RC = GL R =G L C This is the general condition for distortionless line for which a = RG and b = w LC (c) R >> wL , G >> wC
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SOL 8.3.66
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SOL 8.3.64
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Chap 8
Transmission Lines
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g = a + jb = RG i.e. a = RG , and b = 0 Since, b is not function of w so, it is not the distortionless line. (d) R c . Which of the following modes are arranged in ascending order with respect to their resonant frequencies ? (A) TM 110 , TE 011 , TE 101
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MCQ 9.1.21
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The lowest order TM mode that can exist in a cavity resonator is (A) TM 111 (B) TM 110 (C) TM 011
MCQ 9.1.20
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(A) only 1
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MCQ 9.1.19
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The correct statement is
(B) TE 011 , TM 110 , TE 101
(C) TM 100 , TM 101 , TM 111 (D) TM 110 , TE 101 , TE 011 MCQ 9.1.22
An airfilled, lossless cavity resonator has dimensions a = 40 cm , b = 25 cm and c = 20 cm . What is the resonant frequency of TE 101 mode ? (A) 781 MHz (B) 901.4 MHz (C) 450.7 MHz
MCQ 9.1.23
(D) 960.4 MHz
An airfilled cubic cavity resonator ^a = b = c h has dominant resonant frequency of 15 GHz . The dimension of the cavity resonator is (A) 1.41 cm (B) 70 cm (C) 2.5 cm
(D) 3.8 cm
*********** GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Wave Guides
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EXERCISE 9.2
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Chap 9
An electromagnetic wave is propagating in a parallel plate waveguide operating at TM 1 mode. The magnetic field lines in the yz -plane will be (Assume the positive x -axis directs into the paper)
MCQ 9.2.2
An EM wave is propagating at a frequency ‘ f ’ in an air filled rectangular waveguide having the cutoff frequency ‘ fc ’. Consider the phase velocity of the EM wave in the waveguide is v p . The plot of (c/v p) versus ^ fc /f h will be. (c is the velocity of wave in air)
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MCQ 9.2.1
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 9
Wave Guides
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m
An electromagnetic wave propagating at a frequency ‘ f ’ in free space has the wavelength ‘l’. At the same frequency it’s wavelength in an airfilled waveguide is lg . If the cutoff frequency of the waveguide is fc then the plot of ^lg /lh versus ^ f/fc h will be
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MCQ 9.2.3
595
Statement for Linked Question 4 - 5 : A parallel plate waveguide is separated by a dielectric medium of thickness b with the constitutive parameters e and m MCQ 9.2.4
If the cut-off frequencies of the waveguide for the modes TE 1 , TE 2 and TE 3 are respectively wc1 , wc2 , wc3 then which of the following represents the correct relation between the cutoff frequencies ? (A) wc1 = wc2 = wc3 (B) wc1 < wc2 < wc3
(C) wc1 > wc2 > wc3 (D) wc1 wc3 = wc2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Chap 9
A parallel plate waveguide operating at a frequency of 5 GHz is formed of two perfectly conducting infinite plates spaced 8 cm apart in air. The maximum time average power that can be propagated per unit width of the guide for TM 1 mode without any voltage breakdown will be (Dielectric strength of air = 3 # 106 V/m ) (B) 414 MW/m (A) 828 MW/m
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MCQ 9.2.6
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For View Only Shop Online at www.nodia.co.in MCQ 9.2.5 The f -b curve (graph of frequency f versus phase constant b ) of the waveguide for the modes TM 2 , TM 3 and TM 4 will be
(C) 104 MW/m
An air filled parallel plate wave guide has the separation of 12 cm between it’s plates. The guide is operating at a frequency of 2.5 GHz. What is the maximum average power per unit width of the guide that can be propagated without a voltage breakdown for TEM mode ? (A) 358 MW/m (B) 143.2 MW/m
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MCQ 9.2.7
(C) 716 MW/m MCQ 9.2.8
(D) 207 MW/m
(D) 1.432 GW/m
A parallel plate waveguide filled of a dielectric ^er = 8.4h is constructed for operation in TEM mode only over the frequency range 0 < f < 1.5 GHz . The maximum allowable separation between the plates will be (A) 6.90 cm (B) 29 cm (C) 3.45 cm
(D) 1.20 cm
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597
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Statement for Linked Question 9 - 10 : A parallel plate waveguide having plate separation b = 18.1 mm is partially filled with two lossless dielectric with permittivity er1 = 2 and er2 = 1.05 . MCQ 9.2.9
The frequency ‘ f0 ’ at which the TM 1 mode propagates through the guide without suffering any reflective loss is (A) 12.8 GHz (B) 16.2 GHz (C) 9.28 GHz
How many TM modes that can propagate in the guide at the frequency f0 ? (A) one (B) two
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MCQ 9.2.10
(D) 8.44 GHz
(B) a = 1.1 cm , b = 0.9 cm (C) a = 0.37 cm , b = 0.3 cm (D) a = 0.5 cm , b = 0.4 cm
A symmetric dielectric slab waveguide with it’s permittivities er1 = 2.2 and er2 = 2.1 is operating at wavelength, l = 2.6 mm . If the slab thickness is d = 20 mm then how many modes can propagate in the slab ? (A) 8 (B) 25
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MCQ 9.2.12
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An a # b airfilled rectangular waveguide is operating at a frequency, f = 5 GHz . What will be it’s dimensions if the design frequency is 10% larger than the cutoff frequency of dominant mode while being 15 % lower than the cutoff frequency for the next higher order mode ? (A) a = 3.3 cm , b = 2.7 cm
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MCQ 9.2.11
(D) four
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(C) three
(C) 15
(D) 24
MCQ 9.2.13
A rectangular waveguide operating in TE 10 mode has the phase constant b . If the average power density of the guide in this mode is Pav then what will be the relation between Pav and b ? (B) Pav \ b 2 (A) Pav \ b (C) Pav \ 1 (D) Pav is independent of b b
MCQ 9.2.14
A symmetric dielectric slab waveguide operating at wavelength, l = 3.1 mm . If then what will be the maximum value of of TE and TM mode ? (A) 3.304
with it’s refractive indices n1 and n2 is the slab thickness is 10 mm and n2 = 3.3 n1 for which it supports only a single pair
(C) 2.40
(D) 3.42
(B) 3.20
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Common Data for Question 15 - 16 :
If n1 = 2.8 , n2 = 1.7 , n 3 = 2.1, then the minimum possible wave angle for the wave propagation will be (A) 48.6c (B) 37.4c
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MCQ 9.2.15
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An asymmetric slab waveguide has the different mediums above and below the slab as shown in figure. The regions above and below the slab have refractive indices n2 and n 3 respectively while the slab has refractive index n1
(C) 41.4c
If the refractive indices of the mediums are related as n1 > n2 > n 3 the maximum phase velocity of a guided mode will be (A) n 3 c (B) c/n1
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MCQ 9.2.16
(D) 54.1c
(C) 5.3 cm
(D) 1.18 cm
An attenuator is formed by using a section of waveguide of length ‘l ’ operating below cutoff frequency. The operating frequency is 6 GHz and the dimension of the guide is a = 4.572 cm as shown in figure. What will be the required length ‘l ’ to achieve attenuation of 100 dB between I/P and O/P guides ?
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MCQ 9.2.18
(c is velocity of wave in free space)
An air filled waveguide is of square cross-section of 4.5 cm on each side. The waveguide propagates energy in the TE 22 mode at 6 GHz. The wavelength of the TE 22 mode wave in the guide is (A) 4.2 cm (B) 2.72 cm
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MCQ 9.2.17
(D) c/n2
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Chap 9
Wave Guides
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Shop Online at www.nodia.co.in (B) 20.7 cm
(C) 5.2 cm MCQ 9.2.19
(D) 12.7 cm
A rectangular waveguide with the dimension, a = 1.07 cm is operating in TE 10 mode at a frequency, f = 10 GHz . If the waveguide is filled with a dielectric material having er = 8.8 and tan d = 0.002 then the attenuation constant due to dielectric loss will be (A) 12.23 dB/m (B) 0.705 dB/m (D) 3.03 dB/m
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(C) 6.12 dB/m MCQ 9.2.20
599
The first four propagating modes of a circular waveguide are respectively (A) TM 01 , TE 21 , TE 01 , TM 11
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(D) TE 11 , TM 01 , TE 21 , TE 01
Statement for Linked Question 21 - 22 :
If the characterisitc impedance of the guide is 100 W then what will be the width of microstrip. (A) 2.83 cm (B) 0.28 cm
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MCQ 9.2.21
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A microstrip line has the substrate thickness d = 0.616 cm with er = 2.2
(C) 0.36 cm MCQ 9.2.22
If the transmission permittivity ee and ee (A) 1.76 (B) 2.83 (C) 0.158 (D) 18.87
(D) 0.14 cm
line is operating at a frequency, f = 8 GHz then the effective guide wavelength lg will be lg 2.83 cm 1.76 cm 18.87 cm 0.158 cm
Common Data for Question 23 - 24 : A rectangular cavity resonator with dimensions a = 2.5 cm , b = 2 cm and c = 5 cm is filled with a lossless material ( m = m0 , e = 3e0 ). MCQ 9.2.23
The resonant frequency of the cavity resonator for TE 101 mode will be (A) 6.7 GHz (B) 6.34 GHz (C) 7.74 GHz
(D) 3.87 GHz
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Chap 9
For View Only Shop Online at www.nodia.co.in MCQ 9.2.24 If the resonator is made of copper then the quality factor for TE 101 mode is (Conductivity of copper r = 5.8 # 107 S/m ) (A) 7733 (B) 14358 (C) 6625
An air filled circular waveguide has it’s inner radius 1 cm. The cutoff frequency for TE 11 mode will be ^pl11 = 1.841h (A) 0.55 GHz (B) 49.3 GHz
m
MCQ 9.2.25
(D) 11075
A cylindrical cavity shown in the figure below is operating at a wavelength of 2 cm in the dominant mode.
at e
he
lp.
MCQ 9.2.26
(D) 4.71 GHz
co
(C) 8.79 GHz
ww
(C) 1.53 cm
w. g
The radius a of the cylindrical cavity will be (A) 0.77 cm (B) 0.38 cm
^p 01 = 2.405h
(D) 0.19 cm
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GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 9
Wave Guides
For View Only
EXERCISE 9.3
The magnetic field among the propagation direction inside a rectangular waveguide with the cross-section shown in the figure is Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz)
he
lp.
co
GATE 2012
Shop Online at www.nodia.co.in
m
MCQ 9.3.1
601
The phase velocity v p of the wave inside the waveguide satisfies (A) v p > c (B) v p = c
MCQ 9.3.2 GATE 2011
(D) v p = 0
ww w. ga te
(C) 0 < v p < c
The modes in a rectangular waveguide are denoted by TE mn or TM mn where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE? (A) The TM 10 mode of the waveguide does not exist (B) The TE 10 mode of the waveguide does not exist (C) The TM 10 and the TE 10 modes both exist and have the same cut-off frequencies (D) The TM 10 and the TM 01 modes both exist and have the same cut-off frequencies
MCQ 9.3.3 GATE 2011
The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity er and relative permeability mr = 1 are given by E = E p e j (wt - 280py) az V/m and H = 3e j (wt - 280py) ax A/m . Assuming the speed of light in free space to be 3 # 108 m/s , the intrinsic impedance of free space to be 120p , the relative permittivity er of the medium and the electric field amplitude E p are (A) er = 3, E p = 120p (B) er = 3, E p = 360p (C) er = 9, E p = 360p (D) er = 9, E p = 120p
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Wave Guides
Chap 9
For View Only Shop Online at www.nodia.co.in MCQ 9.3.4 Which of the following statements is true regarding the fundamental mode of the GATE 2009 metallic waveguides shown ?
(A) Only P has no cutoff-frequency
m
(B) Only Q has no cutoff-frequency (D) All three have cutoff-frequencies GATE 2008
A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be operated in TE11 mode. The minimum operating frequency is (A) 6.25 GHz (B) 6.0 GHz
lp.
MCQ 9.3.5
(C) 5.0 GHz
he
The E field in a rectangular waveguide of inner dimension a # b is given by wm E = 2 a p k H 0 sin b 2px l sin (wt - bz) ay a h a Where H0 is a constant, and a and b are the dimensions along the x -axis and the y -axis respectively. The mode of propagation in the waveguide is (A) TE 20 (B) TM 11 (C) TM 20 (D) TE 10
GATE 2007
An air-filled rectangular waveguide has inner dimensions of 3 cm # 2 cm. The wave impedance of the TE 20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance h0 = 377 W ) (A) 308 W (B) 355 W
ww
MCQ 9.3.7
(C) 400 W MCQ 9.3.8 GATE 2006
MCQ 9.3.9 GATE 2005
at e
GATE 2007
(D) 3.75 GHz
w. g
MCQ 9.3.6
co
(C) Only R has no cutoff-frequency
(D) 461 W
A rectangular wave guide having TE 10 mode as dominant mode is having a cut off frequency 18 GHz for the mode TE 30 . The inner broad - wall dimension of the rectangular wave guide is (A) 5 cm (B) 5 cm 3 (C) 5 cm (D) 10 cm 2 Which one of the following does represent the electric field lines for the mode in the cross-section of a hollow rectangular metallic waveguide ?
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Chap 9
Wave Guides
For View Only
GATE 2004
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The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TE 10 mode is (A) equal to its group velocity
co
(B) less than the velocity of light in free space
m
MCQ 9.3.10
603
(C) equal to the velocity of light in free space
MCQ 9.3.11
In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz (A) To increase the sensitivity of measurement
he
GATE 2004
lp.
(D) greater than the velocity of light in free space
(B) To transmit the signal to a far-off place (C) To study amplitude modulations
MCQ 9.3.12 GATE 2003
ww w. ga te
(D) Because crystal detector fails at microwave frequencies A rectangular metal wave guide filled with a dielectric material of relative permittivity er = 4 has the inside dimensions 3.0 cm # 1.2 cm. The cut-off frequency for the dominant mode is (A) 2.5 GHz (B) 5.0 GHz (C) 10.0 GHz MCQ 9.3.13 GATE 2002
(D) 12.5 GHz
The phase velocity for the TE 10 -mode in an air-filled rectangular waveguide is (c is the velocity of plane waves in free space) (A) less than c (B) equal to c
(C) greater than c (D) none of these MCQ 9.3.14 GATE 2001
The phase velocity of wave propagating in a hollow metal waveguide is (A) grater than the velocity of light in free space (B) less than the velocity of light in free space (C) equal to the velocity of light free space
(D) equal to the velocity of light in free GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
604
Wave Guides
Chap 9
For View Only Shop Online at www.nodia.co.in MCQ 9.3.15 The dominant mode in a rectangular waveguide is TE 10 , because this mode has GATE 2001 (A) the highest cut-off wavelength (B) no cut-off (C) no magnetic field component (D) no attenuation GATE 2000
A TEM wave is incident normally upon a perfect conductor. The E and H field at the boundary will be respectively, (A) minimum and minimum
m
MCQ 9.3.16
(B) maximum and maximum
co
(C) minimum and maximum (D) maximum and minimum GATE 2000
A rectangular waveguide has dimensions 1 cm # 0.5 cm. Its cut-off frequency is (A) 5 GHz (B) 10 GHz
lp.
MCQ 9.3.17
MCQ 9.3.18 GATE 1999
(D) 12 GHz
he
(C) 15 GHz
Assuming perfect conductors of a transmission line, pure TEM propagation is NOT possible in (A) coaxial cable
at e
(B) air-filled cylindrical waveguide (C) parallel twin-wire line in air
(D) semi-infinite parallel plate wave guide GATE 1999
Indicate which one of the following will NOT exist in a rectangular resonant cavity. (A) TE 110 (B) TE 011 (C) TM 110
IES EC 2012
(C) phase velocity MCQ 9.3.21 IES EC 2012
IES EC 2012
(D) Poynting vector
Consider a rectangular waveguide of internal dimensions 8 cm # 4 cm . Assuming an H 10 mode of propagation, the critical wavelength would be (A) 8 cm (B) 16 cm (C) 4 cm
MCQ 9.3.22
(D) TM 111
The ratio of the transverse electric field to the transverse magnetic field is called as (A) wave guide impedance (B) wave guide wavelength
ww
MCQ 9.3.20
w. g
MCQ 9.3.19
(D) 32 cm
2 2 g = a mp k + a np k - w2 me represents the propagation constant in a rectangular a b waveguide for (A) TE waves only (B) TM waves only
(C) TEM waves
(D) TE and TM waves
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Chap 9
Wave Guides
605
For View Only Shop Online at www.nodia.co.in MCQ 9.3.23 With the symbols having their standard meaning, cut-off frequency (frequency IES EC 2012 below which wave propagation will not (occur) for a rectangular waveguide is 1 mp 2 np 2 mp np (B) (A) 1 a a k+a b k a a k +a b k 2p me me 1 m p n p mp 2 np 2 (C) (D) 1 a a k+a b k a a k +a b k 2p me me MCQ 9.3.24 A standard air filled waveguide WR-187 has inside wall dimensions of a = 4.755 cm IES EC 2010 and b = 2.215 cm . At 12 GHz, it will support (A) TE 10 mode only (B) TE 10 and TE 20 modes only
co
IES EC 2010
Consider the following statements relating to the cavity resonator : 1. The cavity resonator does not posses as many modes as the corresponding waveguides does. 2.
The resonant frequencies of cavities are very closely spaced.
3.
The resonant frequency of a cavity resonator can be changed by altering its dimensions.
lp.
MCQ 9.3.25
(D) TE 10, TE 20, TE 01 and TE 11 modes
m
(C) TE 10, TE 20 and TE 01 modes only
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Which of the above statements is/are correct ? (A) 2 and 3 only (B) 2 only (C) 3 only IES EC 2010
The correct statement is (A) Microstrip lines can support pure TEM mode of propagation but shielded coaxial lines cannot
ww w. ga te
MCQ 9.3.26
(D) 1, 2 and 3
(B) Microstrip lines cannot support pure TEM mode of propagation but shielded coaxial lines can (C) Both microstrip lines and shielded coaxial lines can support pure TEM mode of propagation (D) Neither microstrip lines nor shielded coaxial lines can support pure TEM mode of propagation. MCQ 9.3.27 IES EC 2010
An air-filled rectangular waveguide has dimensions of a = 6 cm and b = 4 cm . The signal frequency is 3 GHz. Match List I with List II and select the correct answer using the code given below the lists : List I
List II
TE 10
1.
2.5 GHz
b.
TE 01
2.
3.75 GHz
c.
TE 11
3.
4.506 GHz
d.
TM 11
4.
4.506 GHz
a.
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606
Wave Guides
For View Only Codes : (A) (B) (C) (D)
b 2 2 3 3
c 3 3 2 2
d 4 1 4 1
Assertion (A) : TEM (Transverse Electromagnetic) waves cannot propagate within a hollow waveguide of any shape. Reason (R) : For a TEM wave to exist within the waveguide, lines of H field must be closed loops which requires an axial component of E which is not present in a TEM wave. (A) Both A and R are individually true and R is the correct explanation of A
co
IES EC 2010
Shop Online at www.nodia.co.in
m
MCQ 9.3.28
a 1 4 1 4
Chap 9
lp.
(B) Both A and R are individually true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true
he
IES EC 2009
For plane wave propagating in free space or two conductor transmission line, what must be the relationship between the phase velocity v r , the group velocity vg and speed of light c ? (A) v p > c > vg (B) v p < c < vg (C) v p = c = vg
IES EC 2009
Consider the following statements : In a microstrip line 1. Wavelength l = l0 , where ee is the effective dielectric constant and l0 is the ee free space wavelength. 2.
Electromagnetic fields exist partly in the air above the dielectric substrate and partly within the substrate itself.
3.
The effective dielectric constant is greater than the dielectric constant of the air.
4.
Conductor losses increase with decreasing characteristic impedance.
ww
MCQ 9.3.30
w. g
(D) v p < vg < c
at e
MCQ 9.3.29
Which of the above statements is/are correct ? (A) 1, 2 and 3 (B) 1 and 2 (C) 2, 3 and 4 (D) 4 only GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 9
Wave Guides
607
For View Only Shop Online at www.nodia.co.in MCQ 9.3.31 Match List I with List II and select the correct answer using the codes given below IES EC 2009 the lists. List I (Type of transmission structure) a.
Strip line
1.
Quassi TEM
b.
Hollow rectangular waveguide
2.
Pure TEM
c.
Microguide
3.
TE/TEM
d.
Corrugated waveguide
4.
Hybrid
c 3 3 1 1
A standard waveguide WR90 has inside wall dimensions of a = 2.286 cm and b = 1.016 cm . What is the cut-off waveguide for TE 01 mode ? (A) 4.572 cm (B) 2.286 cm (C) 2.032 cm
IES EC 2009
(D) 1.857 cm
When a particular mode is exited in a waveguide, there appears an extra electric component, in the direction of propagation. In what mode is the wave propagating ? (A) Transverse electric
ww w. ga te
MCQ 9.3.33
d 4 2 4 2
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b 1 1 3 3
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IES EC 2009
a 2 4 2 4
lp.
(A) (B) (C) (D)
m
Codes
MCQ 9.3.32
List II (Modes of propagation)
(B) Transverse magnetic
(C) Transverse electromagnetic (D) Longitudinal MCQ 9.3.34 IES EC 2009
Consider the following statements : For a square waveguide of cross-section 3 m # 3 m it has been found 1. at 6 GHz dominant mode will propagate. 2.
at 4 GHz all the mode are evanescent.
3.
at 11 GHz only dominant modes and no higher order mode will propagate.
4.
at 7 GHz degenerate modes will propagate.
Which of the above statements are correct ? (A) 1 and 2 (B) 1, 2 and 4 (C) 2 and 3 (D) 2, 3 and 4 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
608
Wave Guides
Chap 9
For View Only Shop Online at www.nodia.co.in MCQ 9.3.35 Match List with List II and select the correct answer using the codes given below IES EC 2009 the lists. List I (Mode)
List II (Characteristic) 1.
Rectangular waveguide does not support
b. Dominant mode
2.
No wave propagation
c. TM 10 and TM 01
3.
Lowest cut-off frequency
m
a. Evanescent mode
Codes
MCQ 9.3.36
co
c 3 2 1 3
Assertion (A) : A z -directed rectangular waveguide with cross-sectional dimensions 3 cm # 1 cm will support propagation at 4 GHz. 2 2 Reason (R) : k z2 + a mp k + a np k = b 2p l, where l is the wavelength. 3 1 l (A) Both A and R are individually true and R is the correct explanation of A.
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IES EC 2009
b 2 3 3 1
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(A) (B) (C) (D)
a 1 1 2 2
at e
(B) Both A and R are individually true but R is not the correct explanation of A. (C) A is true but R is false (D) A is false but R is true IES EC 2006
Which one of the following is the correct statement ? A rectangular coaxial line can support (A) only TEM mode of propagation
w. g
MCQ 9.3.37
(B) both TEM and TE modes of propagation
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(C) either TE or TM mode of propagation (D) TEM, TE or TM mode of propagation MCQ 9.3.38 IES EC 2006
A rectangular waveguide (A) is gradually deformed first into a circular wave guide (B) and lack again into a rectangular waveguide (C) which is oriented through 90c GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 9
Wave Guides
609
For View Only Shop Online at www.nodia.co.in with respect to (A) If the input mode is TE 10 , which mode is excited in the output waveguide (C) ? (A) TE 10 (B) TE 01 (C) TE 11 MCQ 9.3.39 IES EC 2005
(D) TM 11
The dominant mode in a circular waveguide is a : (A) TEM mode (B) TM 01 mode (C) TE 21 mode
IES EC 2005
The cut-off frequency of the dominant mode of a rectangular wave guide having aspect ratio more than 2 is 10 GHz. The inner broad wall dimension is given by : (A) 3 cm (B) 2 cm
m
MCQ 9.3.40
(D) TE 11 mode
MCQ 9.3.41
In a waveguide, the evanescent modes are said to occur if : (A) The propagation constant is real
lp.
IES EC 2005
(D) 2.5 cm
co
(C) 1.5 cm
(B) The propagation constant is imaginary (C) Only the TEM waves propagate
IES EC 2005
Assertion (A) : A microstrip line cannot support pure TEM mode of propagation. Reason (R) : A microstrip line suffers from various forms of losses. (A) Both A and R are true and R is the correct explanation of A
ww w. ga te
MCQ 9.3.42
he
(D) The signal has a constant frequency
(B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true MCQ 9.3.43 IES EC 2004
Consider the following statements relating to the cavity resonators : 1. For over-coupling the cavity terminals are at voltage maximum in the input line at resonance 2.
For over-coupling the cavity terminals are at the voltage minimum in the input line at resonance
3.
For under-coupling the normalized impedance at the voltage maximum is the standing wave ratio
4.
For over-coupling the input terminal impedance is equal to the reciprocal of the standing wave ratio
Which of the statements given above are correct ? (A) 1 and 2 (B) 3 and 4 (C) 1 and 3
(D) 2 and 4
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Wave Guides
Chap 9
For View Only Shop Online at www.nodia.co.in MCQ 9.3.44 Consider the following statements relating to the microstrip lines : IES EC 2004 1. Modes on microstrip lines are purely TEM 2.
Microstrip line is also called open strip line
3.
Radiation loss in microstrip line can be reduced by using thin high dielectric materials
4.
Conformal transformation technique is quite suitable for solving microstrip problems
m
Which of the statements given above are correct ? (A) 1, 2 and 3
co
(B) 2, 3 and 4 (C) 1, 3 and 4 (D) 1, 2 and 4 IES EC 2003
Match List I (Dominant Mode of Propagation) with List II (Type of transmission Structure) and select the correct answer :
lp.
MCQ 9.3.45
List-II
Coaxial line
b.
Rectangular waveguide
c.
Microstrip line
2.
Quasi TEM
3.
Hybrid
4.
TEM
d 3 2 2 3
For TE or TM modes of propagation in bounded media, the phase velocity (A) is independent of frequency
ww
IES EC 2003
TE
w. g
d. Coplanar waveguide Codes : a b c (A) 1 4 2 (B) 4 1 3 (C) 1 4 3 (D) 4 1 2 MCQ 9.3.46
1.
at e
a.
he
List-I
(B) is a linear function of frequency (C) is a non-linear function of frequency (D) can be frequency-dependent or frequency-independent depending on the source MCQ 9.3.47 IES EC 2003
A waveguide operated below cut-off frequency can be used as (A) A phase shifter (B) An attenuator (C) An isolator (D) None of the above
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 9
Wave Guides
611
For View Only Shop Online at www.nodia.co.in MCQ 9.3.48 Assertion (A) : The quality factor Q of a waveguide is closely related to its IES EC 2003 attenuation factor a. Reason (R) : Normally attenuation factors obtainable in waveguides are much higher than those obtainable in transmission lines. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is Assertion (A) : The greater the ‘Q ’, the smaller the bandwidth of a resonant circuit. Reason (R) : At high frequencies the Q of a coil falls due to skin effect. (A) Both A and R are true R is the correct explanation of A
m
IES EC 2002
co
MCQ 9.3.49
(B) Both A and R are true R is NOT the correct explanation of A (C) A is true but R is false
MCQ 9.3.50 IES EC 2002
lp.
(D) A is false but R is true
For a wave propagation in an air filled rectangular waveguide. (A) guided wavelength is never less than free space wavelength
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(B) wave impedance is never less than the free space impedance (C) TEM mode is possible if the dimensions of the waveguide are properly chosen
MCQ 9.3.51 IES EC 2002
ww w. ga te
(D) Propagation constant is always a real quantity When a particular mode is excited in a wave-guide, there appears an extra electric component in the direction of propagation. The resulting mode is (A) transverse-electric (B) transverse-magnetic (C) longitudinal
(D) transverse-electromagnetic MCQ 9.3.52 IES EC 2001
For a hollow waveguide, the axial current must necessarily be (A) a combination of conduction and displacement currents (B) time-varying conduction current and displacement current (C) time-varying conduction current and displacement current (D) displacement current only
MCQ 9.3.53 IES EE 2012
As a result of reflections from a plane conducting wall, electromagnetic waves acquire an apparent velocity greater than the velocity of light in space. This is called (A) velocity propagation (B) normal velocity (C) group velocity
(D) phase velocity
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612
Wave Guides
Chap 9
For View Only Shop Online at www.nodia.co.in MCQ 9.3.54 Assertion (A) : A thin sheet of conducting material can act as a low-pass filter for IES EE 2011 electromagnetic waves. Reason (R) : The penetration depth is inversely proportional to the square root of the frequency. (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A)
m
(B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false
Consider the following statements in connection with cylindrical waveguides : 1. At low frequency the propagation constant is real and wave does not propagate. 2.
At intermediate frequency the propagation constant is zero and wave cut off.
3.
At high frequency the propagation constant is imaginary and wave propagates.
4.
At transition condition the cut-off frequency is inversely proportional to the eigen values of the Bessel function for the respective TE nr mode.
lp.
IES EE 2009
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MCQ 9.3.55
co
(D) Assertion (A) is false but Reason (R) is true
(C) 2 and 3 only MCQ 9.3.56 IES EE 2007
at e
Which of these statements is/are correct ? (A) 1, 2 and 3 (B) 2 only (D) 2, 3 and 4
How is the attenuation factor in parallel plate guides represented ? (A) a = Power lost/power transmitted
w. g
(B) a = 2 # Power lost/power transmitted (C) a = Power lost per unit length/ (2 # power transmitted) (D) a = Power lost/ (power lost + power transmitted) IES EE 2004
Which one of the following statements is correct? A wave guide can be considered to be analogous to a (A) low pass filter
ww
MCQ 9.3.57
(B) high pass filter
(C) band pass filter
(D) band stop filter
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Chap 9
Wave Guides
For View Only
613
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SOLUTIONS 9.1
Option (A) is correct. Given, electric field intensity of the propagating wave is (1) Ezs = 5 sin ^20px h sin ^25py h e-jbz V/m So, we conclude that the wave is propagating in az direction. Since, the wave has it’s component of electric field in the direction of propagation so, the waveguide is operating in TM mn (Transverse magnetic) mode. Now for determining the value of m and n , we compare the phasor form of electric field to its general equation given as. npy -jbz (2) Ezs = Eo sin a mpx k sin a e a n k where a and b are the dimensions of waveguide and since, the waveguide has the dimension 10 # 4 cm so, we get a = 10 cm and b = 4 cm Now, comparing equation (1) and (2) we get mpx = 20px & m = 2 a n py = 25py & n = 1 b Thus, the mode of propagation of wave is TM 21 .
SOL 9.1.2
Option (A) is correct. A wave mode propagates in a waveguide only if it’s frequency is greater than cutoff frequency. If there is no any propagating mode inside the waveguide then energy in the propagating mode is zero. So, average power flow down the waveguide below cutoff frequency is zero. i.e. Both the statements are correct and R is correct explanation of A.
SOL 9.1.3
Option (C) is correct. The intrinsic impedance of an airfilled waveguide for TM mode is defined as 2 f hTM = h0 1 - c cmn m f Since, the operating frequency is twice the cutoff frequency i.e. f = 2fc, mn So, we get the intrinsic wave impedance as 2 hTM = 377 1 - b 1 l = 32.6.49 W . 327 W 2
ww w. ga te
he
lp.
co
m
SOL 9.1.1
mn
mn
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
614
Wave Guides
Chap 9
Option (A) is correct. Since, the electric field component exists in the direction of propagation so it will be operating in TM (Transverse magnetic) mode. So, for the TM mode the electric field components in phasor form are given as Exs =- r22Ezs h 2x and Eys =- r22Ezs h 2y Since, the given electric field component is Ezs = E 0 sin ^50px h sin ^40py h e-rz V/m So, Exs =- r2 ^50ph E 0 cos ^50px h sin ^40py h e-rz h and Eys =- r2 ^40ph Eo sin ^50px h cos ^40py h e-rz h Therefore, the ratio of the components is Exs = 50p cot 50px tan 40py ^ h ^ h 40p Eys = 1.25 cot ^50px h tan ^40py h
ww
w. g
at e
he
SOL 9.1.5
lp.
co
m
For View Only Shop Online at www.nodia.co.in SOL 9.1.4 Option (D) is correct. The dimensions of wave guide, a = 10 cm = 0.1 m and, b = 4 cm The mode of propagation, m = 2, n = 1 Operating frequency, f = 7.5 GHz = 7.5 # 109 Hz , Unbounded phase velocity, (air filled) v p = c = 3 # 108 m/s So, the cut-off frequency of the waveguide is given as 8 2 2 fc = c a m k + a n k = 3 # 10 ^20h2 + ^25h2 2 2 a b 9 = 4.8 # 10 Hz Therefore, the phase constant of the wave inside the waveguide is defined as f 2 2p f f 2 - f c2 b = w 1 -c c m = vp c f f2 = 2p 8 # 109 ^7.5h2 - ^4.8h2 3 # 10 = 220.7 rad/m
Option (D) is correct. Relative permittivity of dielectric er = 2.25 Relative permeability of the dielectric, mr = 1 Operating frequency, f = 10 GHz = 1010 Hz Since, the waveguide is operating in TEM mode so, the phase constant is given as 10 2p f b = w me = mr er = 2p # 108 # ^1.5h = 314.2 rad/m c 3 # 10 The group velocity of the wave in TEM mode will be equal to its phase velocity in the unbounded dielectric medium GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 9.1.6
Chap 9
Wave Guides
For View Only vg = v p =
i.e.
Shop Online at www.nodia.co.in 1 = c = 2 # 108 m/s me mr er
Option (A) is correct. In an a # b rectangular waveguide, cutoff frequency for ^TEhmn or ^TMhmn mode is defined as m 2 n 2 ^ fc hmn = 1 a a k +ab k 2 me Now, for TM 11 mode 1 2 1 2 1 1 2 2 2 a = 2b fc1 = 1 ba l + bb l = ba l + ba l 2 me 2 mf = 5c 1 m 2a me Similarly, for TM 12 mode 1 1 2 2 2 fc2 = 1 b a l + b b l = 17 c 2a me m 2 me 1 1 2 For TE 10 mode fc3 = 1 ba l + 0 = 2a me 2 me fc4 =
1 2 me
1 2 2 b a l + 0 = 2c 2a me m
he
For TE 20 mode
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SOL 9.1.7
615
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So comparing cutoff frequencies of all the modes we get the modes in ascending order of cutoff frequencies as TE 10 < TE 20 < TM 11 < TM 12 SOL 9.1.8
Option (D) is correct. Dimensions of wave guide a = 5 cm = 5 # 10-2 m and b = 3 cm = 3 # 10-2 m Operating frequency, f = 3.75 GHz = 3.75 # 109 Hz Since operating mode of the waveguide is TE 10 (i.e., m = 1 and n = 0 ) so, the cutoff frequency of the airfilled waveguide is given as 2 2 fc = c a m k + a n k 2 a b 8 2 1 = 3 # 10 + 0 = 3 # 109 2 b l 2 5 # 10 The group velocity of the EM wave in the waveguide is given as 2 9 f 2 vg = c 1 - c c m = 3 # 108 1 - c 3 # 10 9 m f 3.75 # 10 8 = 1.8 # 10 m/s
SOL 9.1.9
Option (C) is correct. Dimensions of waveguide, a and b Operating frequency, f Conductivity of medium, s
= 2.5 cm = 2.5 # 10-2 m = 5 cm = 5 # 10-2 m = 15 GHz = 15 # 109 Hz =0
(lossless dielectric)
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20
Option (D) is correct. Given, the cutoff frequency for TM 2 mode is -3 ^lc h2 = 2 mm = 2 # 10 m Since, the cutoff wavelength for TM n or TE n mode for a parallel plate waveguide is defined as 2b (1) ^lc hn = n er where b is the separation between parallel plates of the waveguide and er is relative permittivity of the medium. So, putting the known values in the expression, we get (n = 2 ) 2 # 10-3 = 2b er 2 -3 b = 2 # 10 er Now, for any n mode to propagate the operating wavelength must be less than or equal to the cutoff frequency. i.e. l # ^lc hn
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SOL 9.1.10
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For View Only Shop Online at www.nodia.co.in Relative permittivity, er = 2 Relative permeability, mr = 1 The operating mode of the waveguide is TE 20 mode (i.e., m = 2 and n = 0 ) So, the cutoff frequency of the waveguide in the TE 20 mode is given as 2 m 2 n 2 3 # 108 2 fc = 1 a a k +ab k = b 2.5 10-2 l 2 me # 2 mr er 9 = 8.5 # 10 The wave impedance for the TE 20 mode is given as h mr 1 hTE = = h0 er f fc 2 f 2 1 -c m 1 -c c m p f f 1 = 377 1 2 2f 1 - b 8.5 l p 15 = 323 W
So, from equation (1) for the propagation of wavelength l = 0.1 cm we have the relation as 0.1 # 10-2 # 2b er b -3 er 0.1 # 10-2 # 2 # 2 # 10 n er -3 n # 4 # 10 -2 0.1 # 10 n#4 Therefore, the possible modes that can propagate in the waveguide are TEM, TE 1 , TE 2 , TE 3 , TE 4 , TM 1 , TM 2 , TM 3 and TM 4 Thus, there are nine possible modes that can propagate in the waveguide. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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For View Only Shop Online at www.nodia.co.in SOL 9.1.11 Option (D) is correct. Plate separation, b = 10 mm = 10-2 m Minimum operating frequency, fmin = fc = 15 GHz = 15 # 109 Hz Since, for TM n mode of parallel plate waveguide, cutoff frequency is defined as ^ fc hn = n 2b me So, for ^TMh3 mode ^n = 3h we have the cutoff frequency as 3 ^ fc h3 = 2b m0 e0 er 3 # ^3 # 108h 2 # 10-2 er er = 3 er = 9
Option (D) is correct. Plate separation, b = 20 mm = 20 # 10-3 m Relative permittivity of medium, er = 2.1 Operating frequency, f = 16 GHz For propagation of wave the operating frequency must be greater than the cutoff frequency of ^TEhn or ^TMhn mode of parallel plate waveguide i.e. f > ^ fc hn n f > 2b m0 e0 er n < 2fb m0 e0 er -3 9 n < 2 # 16 # 10 # 820 # 10 # 2.1 3 # 10 n < 3.09 So, the maximum allowed mode is n =3 Since, all the modes given in the option are in the range, therefore, all the three modes will propagate.
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SOL 9.1.12
^ fc h3 = fmin
co
or
m
15 # 109 =
Option (C) is correct. Cutoff frequency of ^TMh1 mode, ^ fc h1 = 2.5 GHz = 2.5 # 109 Hz Operating wavelength, l = 3 cm = 3 # 10-2 m The cutoff frequency of ^TEh3 mode of the parallel plate waveguide is given as ^ fc h3 = 3 ^ fc h1 = 3 # 2.5 # 109 Hz = 7.5 # 109 Hz Since, the operating frequency of the waveguide is defined as f =c l where l is the operating wavelength. So, the operating frequency of the parallel plate waveguide is 8 f = 3 # 10-2 = 1010 Hz 3 # 10 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 9.1.13
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For View Only Shop Online at www.nodia.co.in Therefore, the group velocity of TE 3 mode is given as ^ fc h 2 1 1 -e 3 o ^vg h3 = f m0 e0 9 2 = 3 # 108 1 - c 7.5 #1010 m = 2 # 108 m/s 10 Option (C) is correct. At cutoff the mode propagates in the slab at the critical angle which means that the phase velocity will be equal to that of a plane wave in upper or lower media of refractive index n2 . So, the phase velocity at cutoff will be 8 v p = c = 3 # 10 = 1.2 # 108 m/s n2 2.5 Option (A) is correct. The phase velocity at cutoff is independent of the mode and equal to the phase velocity of a plane wave in unbounded media. Since, in the given problem the phase velocity of TM 2 mode is to be determined for same waveguide so, the phase velocity of TM 2 mode will be equal to that of TM 1 mode. i.e., v p2 = v p1
lp.
SOL 9.1.15
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SOL 9.1.14
Option (C) is correct. Relative permittivity of material, er = 8.8 Separation between strip line, b = 0.632 cm Characteristic impedance, Z 0 = 35 So, er Z 0 = 8.8 (35) = 103.8 Since, er Z 0 < 120 Therefore, the width to separation ratio of strip line transmission line is given as w = 30p - 0.441 b er Z 0 w = 30p - 0.441 0.632 ^ 8.8 h^35h w = 0.295
SOL 9.1.17
Option (C) is correct. Guide wavelength of a stripline is defined as, lg = c e f where, c is velocity of waver in free space, f is the operating frequency and er is the relative permittivity of the medium. So, we get 3 # 108 lg = = 3.37 cm ^ 8.8 h^3 # 109h
SOL 9.1.18
Option (B) is correct. Statement 1 Suppose on the contrary the TEM mode existed. In this case the magnetic field must lie solely in the transverse xy -plane. The magnetic field lines must form closed paths in this transverse plane, since d : H = 0 . From Ampere’s law, the integral
ww
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SOL 9.1.16
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For View Only Shop Online at www.nodia.co.in of this transverse magnetic field around these closed paths must yield the axial conduction or displacement current. But Ez = 0 for the TEM mode so, no axial displacement current can exist. Also, since there is no center conductor so, no axial conduction current can exist. Therefore statement 1 is correct. Statement 2 The dominant mode is the mode that has lowest cutoff frequency. Now, fc, mn is clearly minimized when either m or n is zero. Since, TM 01 or TM 10 mode doesn’t exist so, TM mode can’t be the dominant mode of propagation in rectangular waveguide. It is also correct statement. Option (C) is correct. For a TM mnp mode, neither m nor n can be zero otherwise all field components vanish, however p can be zero. So, the lowest order TM mode is TM 110 .
SOL 9.1.20
Option (D) is correct. Since, TE mnp mode of cavity resonator can have either m = 0 or n = 0 (but not both at a time) where as p can’t be zero for TE mode so, the lowest order of TE mode is if TE 011 ab As the dimensions of the cavity resonator are equal (a = b ) so, both the TE 101 and TE 011 are lowest order mode.
SOL 9.1.21
Option (D) is correct. Given, the dimensions of cavity resonator are related as (1) a >b>c The condition for propagating TE and TM modes in a cavity resonator are as follows : (1) for TM mnp mode, neither m nor n can be zero however p can be zero.
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SOL 9.1.19
(2) For TE mnp mode p can’t be zero but either m or n can be zero (but not both at a time) The resonant frequency of TM mnp or TE mnp mode in a cavity resonator is defined as 2 2 p 2 1/2 fmnp = 1 :a m k + a n k + a k D c b 2 me a So, comparing the resonant frequency for the different values of m, n and p using the relation defined in equation (1), we get the lowest order mode will be TM 110 and the ascending order can be written as below : TM 110 ; TE 101 ; TE 011 ; TE 111 = TM 111 SOL 9.1.22
Option (C) is correct. In an airfilled cavity resonator, resonant frequency is defined as p 2 1/2 1 m 2 n 2 fmmp = :a a k + a b k + a c k D 2 m0 e0
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For View Only Shop Online at www.nodia.co.in So, for TE 101 mode (m = p = 1, n = 0 ) the resonant frequency is 8 2 2 1/2 1 1 f101 = 3 # 10 ;b +b 2 2 l l 2 30 # 10 20 # 10 E 8 = 9.01388 # 10 Hz = 901.4 MHz Option (A) is correct. The resonant frequency of a cavity resonator is defined as p 2 1/2 m 2 n 2 1 fmnp = :a a k + a b k + a c k D 2 m0 e0 Since, a = b = c so, the dominant modes are TE 101 or TE 011 or TM 110 . Therefore, taking any of m, n or p equal to zero, we get the resonant frequency as p 2 1/2 1 m 2 n 2 fmnp = :a a k + a b k + a c k D 2 m0 e0
he
8 15 # 109 = 3 # 10 # 2 2 a -2 a = 1.41 # 10 m = 1.41 cm a = b = c = 1.41 cm
at e
i.e.
^a = b = c h
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1/2 8 fmnp = 3 # 10 ; 22 E 2 a
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SOL 9.1.23
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SOLUTIONS 9.2
Option (B) is correct. Since, the waveguide is operating at TM n mode so, the phasor form of magnetic field of the EM wave will be given as jwe nxy -jbz Hxs = H cos a e h 0 b k Since, the waveguide is operating at TM 1 mode (i.e. n = 1) jwe py So, Hxs = H cos a k e-jbz h 0 b Therefore, the instantaneous magnetic field intensity of the wave is given as jwe py Hx = Re ' H cos a k e-jbz e jwt 1 h 0 b py =- we H 0 cos a k sin ^wt - bz h h b py we at t = 0 (1) Hx = cos a k sin ^bz h h b As the EM wave is propagating in y -z plane so, in TM mode the y and z -components of the magnetic field intensity will be zero. i.e. Hy = H z = 0 Thus, the field will have the component only in z -direction for which we sketch the field lines in y -z plane. From equation (1), we conclude that the field intensity Hz depends on the values cosines and sines of the two variables defined as + ve 0 < y < 0.5 py cos a k = ) b - ve 0.5 < y < 1 + ve 0 < bz < p sin bz = * - ve p < bz < 2p Using these values we get the sketch of the field lines in the yz -plane as shown in the figure below where x -axis directs into the paper.
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SOL 9.2.1
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m
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Chap 9
Option (A) is correct. The phase velocity of the EM wave in the guide is defined as vp = w b where w is the operating angular frequency and b is the phase constant inside the airfilled waveguide given as waveguide b = w 1 - ^ fc /f h2 c So, we get w vp = w 1 - f /f 2 ^c h c 2 fc c 2 & b vp l + c f m = 1 The above equation is the equation of a circle. So, the graph between (c/v p) and ^ fc /f h will be as plotted below :
SOL 9.2.3
Option (A) is correct. Wavelength for a propagating wave inside the waveguide is defined as lg = 2p b where b is the phase constant of the wave in the waveguide given as f 2 b = w me 1 - c c m f where fc is the cutoff frequency of the waveguide and f is the operating frequency
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SOL 9.2.2
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For View Only of the waveguide. So, we get
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lg =
f 2 1 -c c m f 1 lg = 2p f 2 w m0 e0 1 -c c m f w me
lg = b c l f f
&
lg = l
f
1 f 2 1 -c c m f 1 f 2 1 -c c m f
p
m
&
(for airfilled guide m = m0 , e = e0 )
p
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^ f/fc h2 lg = l ^ f/fc h2 - 1 Thus, the plot between ^ f/fc h and ^lg /lh is as sketched below :
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&
Option (C) is correct. The propagation constant ^g h in the parallel plate waveguide is defined as 2 (1) g2 + w2 me = a np k b Since, for lossless medium propagation constant is given as (attenuation constant, a = 0 ) g = jb Putting it in equation (1), we get 2 - b 2 + w2 me = a np k b At the cutoff frequency, w = wc phase constant is zero (i.e., b = 0 ). So, we get 2 wc2 me = a np k b wc = np b me So, for TE 1 mode wc1 = p b me So TE 2 mode wc2 = 2p b me GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 9.2.4
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3p b me Comparing the three expressions we get, wc1 < wc2 < wc3 wc3 =
For TE 3 mode
Option (A) is correct. As calculated in previous question, the expression for the operating frequency of the wave in the waveguide is given as 2 - b 2 + w2 pe = a np k b 2 & w2 me = b 2 + a np k b 2 & f 2 ^4p2 meh = b 2 + a np k b p 2 2 So, for TM 2 mode (n = 2 ) f 2 = 12 :b + 4 a b k D 4p me p 2 2 For TM 3 mode (n = 3 ) f 2 = 12 :b + 9 a b k D 4p me p 2 2 and for TM 4 mode (n = 4 ) f 2 = 12 :b + 16 a b k D 4p me Thus, for the above obtained expressions for the frequencies at different modes, we sketch the f -b curve as shown below :
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SOL 9.2.5
Option (C) is correct. Operating frequency, f = 5 GHz = 5 # 109 Hz Separation between the plates b = 6 cm = 6 # 10-2 m So, the cut off frequency for TM 1 mode is given as 1 3 # 108 fc = = = 2.5 # 109 Hz 2 # 6 # 10-2 2b m0 e0 For a parallel plate waveguide, phasor form of components of electric field and magnetic field intensity of a propagation wave are given as npy - gz Eys = E 0 cos a e b k npy - gz E0 and e cos a Hxs =2 b k f h0 1 - c c m f GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 9.2.6
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So, the average power is given as b 1 - E H (w dy) Pave = 1 Re "Es # Hs), dS = "^ ys h^ xs h, 2 0 2 b py E 02 cos2 a k dy =w 2 0 b fc 2 h0 1 - c m f 2py 1 + cos b 2 b b l E E 02 w wb 0 = = dy 2 2 4 f 2 0 f 2 h0 1 - c c m h0 1 - c c m f f The maximum power propagation will be due to the maximum electric field in the medium (the dielectric strength of the medium). So, we have the maximum average power as 2 ^3 # 106h ( ^E 0hmax = 3 # 106 V/m ) ^Pave hmax = wb 4 fc 2 h0 1 - c m f Putting all the values, we get the average power per unit width as 2 -2 ^Pave hmax ^3 # 106h 6 10 # = # w 4 f 2 120p 1 - c c m f 8 = 4.135 # 10 = 414 MW/m
#
#
#
Option (D) is correct. Separation between waveguide plates, b = 12 cm = 0.12 m Operating frequency, f = 2.5 GHz = 2.5 # 109 Hz For the TEM mode, phasor form of electric and magnetic field components are given as Eys = E 0 e- gz Hxs =- E 0 e- gz h0 So, the average power propagated in the waveguide is given as , Pave = Re & 1 Es # Hs)0 dS 2 b = 1 Re "-^Eys h^Hxs h, ds =- 1 ^E 0hc- E 0 m wdy h0 2 0 2 2 = E 0 wb 2h 0 The maximum electric field, without any voltage breakdown is defined as the dielectric strength of the medium as given and as the dielectric strength of air is ^E 0hmax = 3 # 106 V/m So, the maximum average power propagated in the waveguide is 2 ^3 # 106h w ^0.12h ^Pave hmax = 2 # ^120ph
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SOL 9.2.7
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#
# #
#
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Option (A) is correct. The Brewster’s angle for parallel polarized wave is given as tan qB || = e2 e1 qB || = tan-1 b 1.05 l = 35.9c 2 The cutoff frequency for TM 1 mode in 1st medium (permittivity = er1 ) is given as 1 3 # 108 = ^ fc h1 = 2b m0 e0 er1 2 # 14.1 # 10-3 2 9 = 7.52 # 10 Hz So, the frequency for which there is no any reflective loss is given as ^ fc h1 f0 = cos q where q is ray angle that has the value, q = 90c - qB ||. So, we get 9 9 = 7.52 # 10 = 12.8 GHz f0 = 7.52 # 10 sin 35.9c cos ^90c - qB11h
ww
SOL 9.2.9
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SOL 9.2.8
^Pave hmax = 1.432 # 109 = 1.432 GW/m w Option (B) is correct. Maximum operating frequency fmax = 1.5 # 109 Hz Relative permittivity of medium, er = 8.4 The cutoff frequency in TEM mode is fc = 0 and the cutoff frequency in ^TEhn or ^TMhn mode is given as ^ fc hn = n 2b me So, for TE 1 or TM 1 mode ^n = 1h we get 1 ^ fc h1 = 2b m0 e0 er Since, the guide is to be operated only in TEM mode. So, the operating frequency must be less than ^ fc h1 while it must be greater than 0 (cutoff frequency in TEM mode). i.e. 0 < f < ^ fc h1 1 or, f < 2b m0 e0 er 1 b < 2f m0 e0 er As the frequency inside the waveguide ranges in 0 < f < 1.5 GHz , therefore, the maximum allowable separation between the plates is c 3 # 108 ( fmax = 1.5 GHz ) = b max = 2fmax er 2 # 1.5 # 109 # 8.4 = 0.345 m = 3.45 cm
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For View Only Shop Online at www.nodia.co.in NOTE : Brewster’s angle is the incident angle of a plane wave at the interface of two mediums for which there is no any reflection in the medium. Option (A) is correct. As we have determined in the previous question, the value of f0 is f0 = 12.8 GHz and the cutoff frequency for TM 1 mode is ^ fc h1 = 7.52 GHz So, the cutoff frequency for TM 2 mode will be ^ fc h2 = 2 ^ fc h1 = 15.04 GHz Since, the operating frequency f0 is below the cutoff frequency for TM 2 mode so, TM 2 mode or the higher modes can’t propagate at the frequency f0 . Therefore, only one mode ^TM 1h can propagate at the frequency f0 through the waveguide.
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SOL 9.2.10
Option (A) is correct. Consider the dominant mode of the waveguide is TE 10 . Since, the cut-off frequency for TE mn mode is defined as m 2 n 2 fc = 1 a a k +ab k 2 mf So, the cutoff frequency for the TE 10 mode is (for airfilled waveguide c = 1/ me ) ^ fc h10 = c2 # a1 = 2ca Now, the next higher order mode of the waveguide will be TE 01 so, it’s cutoff frequency is given as ^ fc h01 = 2cb For the given condition design frequency will be f = 1.1 ^ fc h10 = 0.9 ^ fc h01 Since, the operating frequency of the waveguide is f = 5 GHz = 5 # 109 Hz 109 So, we get ^ fc h10 = 5 # 1.1 c = 5 # 109 1.1 2a ^3 # 108h # 1.1 a = = 3.3 cm 2 # ^5 # 109h 109 and ^ fc h01 = 5 # 0.9 c = 5 # 109 0.9 2b ^3 # 108h # 0.9 b = = 2.7 cm 2 # ^5 # 109h SOL 9.2.12 Option (C) is correct. For a propagating mode TM n or TE n the cutoff wavelength of the symmetric GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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SOL 9.2.11
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Option (A) is correct. For a rectangular waveguide operating in TE 10 mode the phasor form of electric field is given as Eys = E 0 sin ^kx h e-jbz b Hxs =E sin kx e-jbz wm 0 ^ h Hzs = j K E 0 cos ^kx h e-jbz wm Since, the wave is propagating in TE mode so, no any other field component exists in the waveguide. Now, the average power in an EM wave is defined as Pav = 1 Re "Es # H) s , 2 Since, Hzs has a factor j . So it would lead to an imaginary part of the total power when cross product with Ey is taken. Therefore, the real power in the case is found through the cross product with complex conjugate of Hxs as below : 1 b E 2 sin2 kx a Pav = 1 Re "Eys # H) ^ h z xs , = 2 2 wm 0 Thus, Pav \ b
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SOL 9.2.13
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For View Only Shop Online at www.nodia.co.in dielectric slab is defined as lc = 2d er1 - er2 n-1 where er1 and er2 are the permittivities of dielectrics and d is the slab thickness. So, we get -6 2.2 - 2.1 = 1.26 # 10-5 lc = 2 # 20 # 10 n-1 n-1 Since the operating wave length must be lower than or equal to the cutoff wavelength i.e. l # lc Therefore, for the propagation of wavelength l = 2.6 mm in the dielectric slab waveguide, we have the condition as -5 2.6 # 10-6 # 1.26 # 10 n-1 -5 n - 1 # 1.26 # 10-6 2.6 # 10 n - 1 # 4.85 n # 5.85 So, the possible values of n for which the wavelength l = 2.6 mm can propagate in the waveguide are n = 1, 2, 3, 4, 5. Thus, we get the possible modes as follows : TE 1 , TE 2 , TE 3 , TE 4 , TE 5 TM 1 , TM 2 , TM 3 , TM 4 , TM 5 and as TEM doesn’t exist in the dielectric slab waveguide so, total 10 modes can propagate for the operating wavelength.
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For View Only Shop Online at www.nodia.co.in SOL 9.2.14 Option (A) is correct. Cutoff wavelength for symmetric slab waveguide is defined as, 2d er1 - er2 (1) ^lc hn = n-1 where d is the thickness of slab n is the propagating mode, er1 and er2 are the relative permittivites of the mediums. Now, the refractive indices of the two mediums can be given as n1 = er1 and n2 = er2 So, the equation can be rewritten as 2 n 22 ^lc hn = 2d nn-1 1 Since, the waveguide supports only a single pair of TE and TM modes. i.e. it supports n = 1 mode and denies all the higher modes. Therefore, the operating wavelength l must be with in the range. i.e. (2) ^lc h1 $ l $ ^lc h2 where (lc) 1 and (lc) 2 are the wavelengths for mode n = 1 and n = 2 respectively. Putting n = 1 in equation (1) we get ^lc h1 = 3 Therefore, the condition obtained in equation (2) reduces to l $ ^lc h2 2 # 10 # 10-6 n 12 - ^3.3h2 2-1 -6 n 12 - ^3.3h2 # 3.1 # 10 -6 2 # 10 # 10 n1 # 3.304 Thus, the maximum value of n1 is 3.304.
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3.1 # 10-6 =
SOL 9.2.15
Option (A) is correct. The wave angle must be equal to or greater than the critical angle of total reflection at both interfaces. So, the minimum wave angle in the slab is determined for the greater of the two critical angles determined at two interfaces. Since, n3 > n2 It means the critical angle will be greater for n 3 media and given as qc3 = sin-1 a n 3 k = 48.6c n1 Therefore, the minimum possible wave angle will be 48.6c.
SOL 9.2.16
Option (D) is correct. Phase velocity of a guided mode is defined as vp = w b So, maximum phase velocity for the guided mode is
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he
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SOL 9.2.17
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(1) v p max = w bmin where bmin is the minimum phase constant given as (2) bmin = n1 k 0 sin qmin where qmin is the minimum possible wave angle, k 0 is the wave number in free space and n1 is the refractive index of propagating media (slab). Now, from the given relation for refractive index, we have n2 > n3 So, as described in previous question the minimum wave angle will be determined by larger critical angle (i.e. at the interface of n1 and n2 ) which is given as sin qmin = sin qc, 12 = n2 n1 Putting it in equation (2), we get bmin = n1 k 0 n2 = n2 k 0 n1 Again putting the value of bmin in equation (1), we get (velocity of wave in air, c = w ) v p max = w = c n2 n2 k0 k0 Given, the cross-section dimension of the waveguide is a = b = 4.5 cm The cut off frequency of the rectangular waveguide is defined as 1 m 2 n 2 1/2 (fc) mn = :a a k + a b k D 2 m0 e0
So, the cutoff frequency for TE 22 mode of waveguide of square cross section is 2 1 2 2 2 2 1/2 8 = 2 GHz fc = ;b a l + b b l E = 3 # 10 # 2 m0 e0 0.045 The phase constant of the wave inside the waveguide is given as 9 2 1/2 f 2 1/2 b = w m0 e0 =1 - c c m G = 2p # 6 #810 ;1 - b 2 l E 6 f 3 # 10 2 -1 = 1.1847 # 10 m Therefore, the wavelength of the TE 22 mode wave is l = 2p = 2p = 5.303 # 10-2 = 5.3 cm 118.47 b Option (C) is correct. Given, the operating frequency of the waveguide is f = 6 GHz = 6 # 109 Hz So, the wave number in the waveguide of dimension ‘a ’ is given as 9 2pf k = = 2p # 6 #810 = 40p c 3 # 10 Now, the attenuation constant of section of waveguide (attenuator) with dimension a/2 is given as 2 2 a = b p l - k2 = b 2p l - ^40ph2 ^a = 0.04572 mh 0.04575 a/2
SOL 9.2.18
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= 55.63 Np/m Since, the total required attenuation is 100 dB along the attenuator so, we have - 100 dB = 20 log e- al where l is length of the attenuator. (length travelled by wave in the small section of waveguide). Therefore, solving the equation we get, 10-5 = e- al l = 11.5 = 0.2067 = 20.67 cm 55.63 Option (B) is correct. Dimension of waveguide, a = 1.07 cm = 0.0107 m Operating frequency, f = 10 GHz = 10 # 109 Hz = 1010 Hz Permittivity of dielectric, er = 8.8 and tan d = 0.002 The phase constant of the EM wave inside the waveguide is defined as b = k2 - ^p/a h2 where k is the wave number in the unbounded medium given as (k 0 is wave number in free space) k = er k 0 2pf 2p f = ^ 8.8 h bk 0 = c l c 10 (c = 3 # 108 m/s ) = ^ 8.8 h 2p # 108 3 # 10 = 621.3 m-1 So, the phase constant of the wave along the waveguide is 2 b = ^621.3h2 - a p k 0.0107 = 547.5 m-1 Therefore, the attenuation constant due to dielectric loss is given as 2 ^621.3h2 ^0.002h d k tan ad = = 2b 2 ^547.5h = 0.705 Np/m = 6.12 dB/m
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SOL 9.2.19
631
Option (D) is correct. In a circular waveguide, cutoff frequency for TE mn mode is given as l f cmn = P mn 2pa pe and the cutoff frequency for TM mn mode of the waveguide is given as fcmn = Pmn 2pa me where a is the cross sectional radius of waveguide, Plmn and Pmn are the roots of the Bessel’s equation. Their values are related as listed below in increasing order Pl11 < P01 < Pl21 < Pl01 = P11 < Pl31 < P21 < Pl41 and so on. So, for the corresponding values of Plmn and Pmn , we get the increasing order of the modes with respect to their cutoff frequencies as shown below on the frequency axis : GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 9.2.20
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Option (C) is correct. Given Thickness of substrate, d = 0.316 cm Relative permittivity of substrate, er = 2.2 Characteristic impedance of line, Z 0 = 100 W The width to thickness ratio (w/d) is defined as w = 8eA d e2A - 2
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SOL 9.2.21
m
Thus, the first four propagating modes are respectively TE 11 , TM 01 , TE 21 , TE 01 or TM 11
for w < 2 d
he
where A = Z 0 er + 1 + er - 1 c 0.23 + 0.11 m 60 2 er + 1 er w Now, we assume d < 2 . So, we get
SOL 9.2.22
ww
w. g
at e
A = 100 2.2 + 1 + b 2.2 - 1 lb 0.23 + 0.11 l = 2.21 60 2 2. 2 + 1 2. 2 and therefore, the width to thickness ratio is w = 8e2.21 = 0.896 < 2 d e2 # 2.21 - 2 As the obtained value of (w/d) is less than 2 so, our assumption was correct and we have w = 0.896 d or, w = ^0.896h # d = ^0.896h # ^0.316h = 0.283 cm Option (A) is correct. As calculated in previous question the width to thickness ratio is w = 0.896 d So, the effective value of permittivity is given as 1 ee = er + 1 + er - 1 2 2 1 + 12wd ^2.2 + 1h 2.2 - 1 1 = + 2 2 1 + 12 0.896 = 1.758 Therefore the guided wavelength of the EM wave is
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c ee f where f is operating frequency and c is velocity of wave in free space. So, we get 3 # 108 lg = f = 8 GHz ^ 1.758 h^8 # 109h = 2.83 cm lg =
Option (D) is correct. The resonant frequency for TE mnp mode is defined as 2 2 p 2 1/2 fr = 1 :a m k + a n k + a k D c b 2 me a So for TE 101 mode the resonant frequency of the cavity resonator is 8 2 2 1/2 ( m = m0 , e = 3e0 ) fr = 3 # 10 ;b 1 l + b 1 l E 0 . 025 0 . 05 2 3 = 3.87 # 109 Hz = 3.87 GHz
SOL 9.2.24
Option (A) is correct. The quality factor of TE 101 mode is defined as 2 2 ^a + c h abc QTE = d 62b ^a3 + c3h + ac ^a2 + c2h@ where d is skin depth given as 1 d = pfr m0 sc where fr " resonant frequency for the defined mode. m0 = 4p # 10-7 sc = Conductivity of copper So, we get the skin depth as 1 d = 9 p ^3.87 # 10 h^4p # 10-7h^5.8 # 107h = 1.06 # 10-6 Therefore, the quality factor of the resonator is 2 2 -2 8^2.5h + ^5 h B^2.5h^2 h^5 h # 10 QTE = 3 3 2 2 ^1.06 # 10-6h82 # 2 "^2.5h + ^5 h , + ^2.5h^5 h"^2.5h + ^5 h ,B = 7732.7 . 7733
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SOL 9.2.23
101
SOL 9.2.25
Option (B) is correct. Given, the inner radius of the guide is a = 1 cm = 0.01 m The cutoff frequency for TE mn mode of a circular waveguide is defined as plmn fc, mn = 2pa me where plmn is the m th root of Bessel’s function ^J ln = 0h. Now, from the given data we have
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Pl11 = 1.841 So, the cutoff frequency of TE 11 mode in the circular waveguide is 1.841 3 # 108 # 1.841 fc11 = = 2p # 10-2 2p ^10-2h m0 e0 = 8.79 # 109 Hz = 8.79 GHz Option (A) is correct. The resonant frequency of TM mnl mode in cylindrical cavity is defined as pmn lp 2 fc, mnl = - 1 a a k+b d l 2p me where a is radius of cylindrical cavity, d is height of the cylindrical cavity and pmn is the root of Bessel’s equation. Since, the dominant mode in cylindrical cavity is TM 010 so, the cutoff frequency for dominant mode is p 01 p c fc010 = = 01 2p a 2pa m0 e0 Therefore, the cutoff wavelength for dominant mode is given as lc, 010 = c fc010 2 # 10-2 = 2pa p 01 ^2.405h^2 # 10-2h (lc, 010 = 2 cm ) a = 2p = 7.65 # 10-3 = 0.765 cm
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SOL 9.2.26
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SOLUTIONS 9.3
Option (D) is correct. Given, the magnetic field component along the z -direction as Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz) So, bx = 2.094 # 102 by = 2.618 # 102 w = 6.283 # 1010 rad/s For the wave propagation inside the rectangular waveguide, w 2 - (b 2 + b 2) b = x y c2 Substituting the values, we get 10 2 - (2.0942 + 2.6182) # 10 4 b = c 6.283 # 10 m 8 3 # 10 - j261 Since, b is imaginary so, mode of operation is non-propagating i.e. vp = 0
SOL 9.3.2
Option (A) is correct. TM 11 is the lowest order mode of all the TM mn modes.
SOL 9.3.3
Option (D) is correct. From the given expressions of E and H , we can write, b = 280p 2 p or = 280p & l = 1 140 l So, the wave impedance is given as E E h = = p = 120p 3 H er Since, the operating frequency of the wave is f = 14 GHz So, the operating wavelength of the wave can also be given as 8 3 l = c = 3 # 10 9 = er f er 14 # 10 140 er 3 1 = or 140 140 er or, er = 9
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SOL 9.3.1
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
(1)
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Chap 9
For View Only Shop Online at www.nodia.co.in From equation (1) we have Ep = 120p & E p = 120p 3 9 SOL 9.3.4 Option (A) is correct. Rectangular and cylindrical waveguide doesn’t support TEM modes and have cut off frequency. Coaxial cable support TEM wave and doesn’t have cut off frequency. Option (A) is correct. Cut-off Frequency for TE mn mode of a rectangular waveguide is defined as 2 2 fc = c a m k + a n k 2 a b So, for TE11 mode (m = 1, n = 1) the cutoff frequency is 10 1 2 1 2 (c = 3 # 108 cm/s ) fc = 3 # 10 b 4 l + b 3 l = 6.25 GHz 2 Option (A) is correct. Given, the electric field intensity of the wave inside rectangular waveguide as wm E = 2 a p k H 0 sin b 2px l sin (wt - bz) ay a h a This is TE mode and we know that mpy Ey \ sin a mpx k cos a a b k So, comparing it with the given expression we get m = 2 and n = 0 . Therefore, the propagating mode is TE 20 .
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SOL 9.3.6
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SOL 9.3.5
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Option (B) is correct. The cut-off frequency for the TE mn mode of the waveguide is defined as 2 2 fc = c a m k + a n k 2 a b So, the cutoff frequency of the TE 20 (m = 2 , n = 0 )mode is 8 (a = 3 cm ) fc = c a m k = 3 # 10 # 2 = 10 GHz 2 a 2 0.03 Therefore, the wave impedance of the TE 20 mode is given as h0 377 ( f = 30 GHz ) = = 400W h' = 2 2 10 f 10 1 -c c m 1 -c f 3 # 1010 m SOL 9.3.8 Option (B) is correct. The cut-off frequency of TE mn mode is defined as 2 2 fc = c a m k + a m k 2 a b So, the cutoff frequency of TE 30 (m = 3 , n = 0 ) mode is fc = c a m k 2 a 8 or ( fc = 18 GHz ) 18 # 109 = 3 # 10 3 2 a or a = 1 m = 5 cm 40 2 GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia SOL 9.3.7
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Wave Guides
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Option (D) is correct. For any propagating mode inside a rectangular waveguide the velocities are related as v p > c > vg i.e. the phase velocity of the wave inside the waveguide is greater than the velocity of light in the free space.
SOL 9.3.11
Option (D) is correct. In a microwave test bench, the microwave signal is modulated at 1 kHz because crystal detector fails at microwave frequencies.
SOL 9.3.12
Option (A) is correct. The cutoff frequency of TE mn mode in a rectangular waveguide is defined as m 2 n 2 fc = 1 a a k +ab k 2 me Since in the given rectangular waveguide a > b so, the dominant mode is TE10 and the cutoff frequency for the dominant mode is given as m 2 n 2 1 fc = c a a k +ab k c m e = cm 2 er 0 0 8 2 2 3 10 1 0 # = b 0.03 l + b b l 4 8 = 3 # 10 = 2.5 GHz 0.12 Option (B) is correct. Phase velocity of an EM wave inside an air-filled rectangular waveguide c vp = f 2 1 -c c m f where c is velocity of EM wave in free space fc is the cutoff frequency of the propagating mode and f is the operating frequency. Since, for a wave propagation the operating frequency must be greater than the cutoff frequency. i.e. f > fc Therefore, the phase velocity of the wave will be always greater than the velocity of wave in free space. i.e. vp > c
SOL 9.3.14
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SOL 9.3.10
Option (A) is correct. In a hollow metal wave guide v p > c > vg where v p " Phase velocity c " Velocity of light in free space. vg " Group velocity So, the phase velocity of a wave propagating in a hollow metal waveguide is greater than the velocity of light in free space.
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For View Only Shop Online at www.nodia.co.in SOL 9.3.15 Option (A) is correct. In a wave guide dominant gives lowest cut-off frequency and hence the highest cutoff wavelength. Option (B) is correct. As the impedance of perfect conductor is zero, electric field is minimum and magnetic field is maximum at the boundary.
SOL 9.3.17
Option (B) is correct. Cutoff frequency for TE mn mode in a rectangular waveguide is defined as v fc = p a m k2 + a n k2 2 a b Since, for the given rectangular waveguide a > b so, the dominant mode is TE 10 and the cutoff frequency of the dominant mode of rectangular waveguide is 8 v (For air v p = 3 # 108 ) fc = p = 3 # 10-2 = 15 # 109 2a 2 # 10 = 15 GHz
SOL 9.3.18
Option (D) is correct. In TE mode Ez = 0 , at all points within the wave guide. It implies that electric field vector is always perpendicular to the waveguide axis. This is not possible in semi-infine parallel plate wave guide.
SOL 9.3.19
Option (A) is correct. In a rectangular resonant cavity TE mnp mode must have its p = 1. So, the mode TE 110 doesn’t exist in the rectangular resonant cavity.
SOL 9.3.20
Option (A) is correct. The transverse electric field and transverse magnetic field inside a waveguide are related as where h is intrinsic impedance E =h H E or h = H i.e. the ratio of transverse electric field to the transverse magnetic field is called waveguide impedance.
SOL 9.3.21
Option (C) is correct. Cutoff wavelength for H mn mode of a rectangular waveguide is defined as 2 lc = m 2+ n 2 a a k ab k where a and b are the dimensions of waveguide. So, for the H 10 mode (m = 1, n = 0 ), the cutoff wavelength is 2 (a = 8 cm ) lc = 1 2+0 b8l = 16 cm
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SOL 9.3.16
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Chap 9
Wave Guides
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For View Only Shop Online at www.nodia.co.in SOL 9.3.22 Option (D) is correct. A rectangular waveguide supports TE and TM waves where as it doesn’t support TEM waves. The propagation constant for TE or TM waves inside a rectangular waveguide is defined as 2 2 g = a mp k + a np k - w2 me a b SOL 9.3.23 Option (C) is correct. Cut-off frequency for TE mn or TM mn mode inside a rectangular waveguide is defined as 1 mp 2 np 2 fc = a a k +a b k 2p me Where a and b are the dimensions of rectangular waveguide. Option (D) is correct. Cut-off frequency for TE 10 mode is 1 m 2 n 2 fc10 = a a k +ab k 2 m0 e0 8 1 (m = 1, n = 0 ) = 3 # 10 # b 2 4.755 # 10-2 l = 3.16 GHz Cut-off frequency for TE 01 mode is 8 1 fc01 = b 3 # 10 l # b 2 2.215 # 10-2 l = 6.77 GHz cut-off frequency for TE 11 mode is 8 2 2 1 1 fc11 = 3 # 10 # b +b 2 2 l l 2 4.755 # 10 2.215 # 10 = 7.47 GHz and the cut-off frequency for TE 20 mode is 8 2 fc20 = 3 # 10 # = 6.3 GHz 2 ^4.755 # 10-2h Since the operating frequency f = 12 GHz so, we have, fc10 , fc01 , fc11 , fc20 > f . Therefore, all the modes will propagate. Note : For avoiding so many calculation we should directly calculate the higher frequency modes first for higher operating frequency. As in this case if we calculates fc11 first then by getting f > fc11 it is clear that TE 01 , TE 10 and TE 11 all the three modes are propagating and by observing option we directly can say option (D) is correct.
SOL 9.3.25
Option (D) is correct. Consider a rectangular waveguide has dimensions a = b and the corresponding resonator has the dimensions a = b = d . now take and operating point that has frequency f just greater than the cut-off frequency for m = n = 1. So we have the
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SOL 9.3.24
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Chap 9
For View Only Shop Online at www.nodia.co.in propagating modes in waveguide. TE 01 , TE 10 , TE 11 and TM 11 . Where as the propagating modes in resonator are TE 011 , TE 101 , TM 110 Therefore the cavity resonator does not possess as many modes as corresponding waveguides. As the resonating frequency of a TE mnp or TM mnp mode is defined as pp 2 mp 2 np 2 fr = 1 a a k +a b k +a d k 2 me So for the different modes (different values of m , n andp) the resonant frequency are very closely spaced and also the resonant frequencies of cavity can be changed by altering its dimensions.
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Option (C) is correct. Microstrip lines cannot support pure TEM mode but shielded coaxial lines can support pure TEM mode.
SOL 9.3.27
Option (A) is correct. Given, Operating frequency, f = 3 GHz = 3 # 109 Hz Dimensions of waveguide a = 6 cm and b = 4 cm The cut-off frequency for TE mn /TM mn mode is defined as 1 m 2 n 2 fc = a a k +ab k 2 m0 e0 10 1 2+0 So, for TE10 , fc10 = 3 # 10 b 2 6l = 2.5 GHz 10 2 for TE 01 , fc01 = 3 # 10 0 +b1l 2 4 = 3.75 GHz 10 1 2 1 2 For TE11 or TM11 , fc11 = 3 # 10 b6l +b4l 2
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SOL 9.3.26
= 4.506 GHz
^a " 1h ^b " 2h ^c " 3, d " 4h
SOL 9.3.28
Option (A) is correct. A TEM wave doesn’t have an electric component in its direction of propagation consequently there is no longitudinal displacement current. The total absence of a longitudinal current inside a waveguide leads to the conclusion that there can be no closed loops of magnetic field lines in any transverse plane. Therefore, TEM waves cannot exist in a hollow waveguide of any shape. i.e. Both A and R are true and R is correct explanation of A.
SOL 9.3.29
Option (A) is correct. Phase velocity of a wave propagating in a waveguide is defined as
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c
m
f 2 1 -c c m f The group velocity of the wave propagating in waveguide is defined as f 2 vg = c 1 - c c m f where c is the velocity of wave in free space, fc is the cutoff frequency and f is the operating frequency. As the operating frequency f is always grater than cutoff frequency fc . So, comparing the above two expressions we get v p > c > vg Option (B) is correct. In a microstrip line operating wavelength is defined as l = l0 ee where, l0 is free space wave length and ee is the effective dielectric constant. So, Statement 1 is correct. The electromagnetic fields exist partly in air above the dielectric substrate and partly within the substrate. Statement 2 is correct. The effective dielectric constant of microstrip line is ee and given as 1 < ee < er i.e. greater than dielectric constant of air (1). Statement 3 is correct. Conductor losses, increase with decreasing characteristic impedance in microstrip line. Statement 4 is correct.
SOL 9.3.31
Option (B) is correct. Stripline carries two conductors and a homogenous dielectric. So, it supports a TEM mode (Pure TEM). a"2 Hollow rectangular waveguide can propagate TEM and TE modes but not TEM mode. b"3 Microstripline has some of its field lines in the dielectric region and some fraction in the air region. So it cannot support a pure TEM wave instead the fields are quasiTEM. c"1
SOL 9.3.32
Option (B) is correct. Given, the dimension of waveguide is a = 2.286 cm , b = 1.016 cm . The cut off wavelength of the guide, for TE mn mode is defined as 2 lc = m 2+ n 2 a a k ab k So, for TE 01 Mode the cut off wavelength of the guide is 2 = 0 2+ 1 2 ba l bb l
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SOL 9.3.30
= 2b = 2.032 cm GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Chap 9
For View Only Shop Online at www.nodia.co.in SOL 9.3.33 Option (C) is correct. Since, the electric component is existed in the direction of propagation. So the electric field is not transverse to the propagating wave and therefore the mode is transverse magnetic (TM mode). Option (C) is correct. Given, dimension of waveguide a = b = 3 cm and so the dominant mode is either TE 01 or TE 10 mode. So, the cutoff frequency for dominant mode is given as 1 m 2 n 2 fc = a a k +ab k 2 m0 e0 8 1 (for TE 01 or TE 10 mode) = 3 # 10 # 2 3 # 10-2 = 5 GHz So, at 6 GHz dominant mode will propagate. Statement 1 is correct. At 4 GHz no modes will propagate so the modes are evanescent at 4 GHz. Statement 2 is correct. At 11GHz along with the dominant mode TE 11 mode ^ fc = 5 2 h will also propagate. Statement 3 is incorrect. Degenerate modes are the different modes that have the same cut off frequency and at 7 GHz frequency TE 01 and TE 10 propagates that has the same cut off frequency i.e. Degenerate modes propagate at 7 GHz. Statement 4 is correct.
SOL 9.3.35
Option (C) is correct. Evanescent mode " No wave propagation dominant mode is the mode that has lowest cutoff frequency. Rectangular waveguide does not support TM 01 and TM 10 mode. A " 2, B " 3, C " 1
SOL 9.3.36
Option (D) is correct. Assertion (A) : Given the dimension of waveguide, a = 3 cm , b = 1 cm So, the dominant mode ^TE 10h has the cutoff frequency. 8 1 fc = 3 # 10 # b 2 3 # 10-2 l = 5 GHz f = 4 GHz < f c So, at 4 GHz there is no propagating mode. i.e Assertion (A) is false. Reason (R) : The wave equation for the rectangular waveguide is defined as 2 2 k z2 - a mp k - a np k = b 2p l a b l
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SOL 9.3.34
for a = 3 , b = 1 we have 2 2 2 So, Reason (R) is also false. k z2 - a mp k - a np k = b 2p l 3 1 l GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
Chap 9
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For View Only Shop Online at www.nodia.co.in SOL 9.3.37 Option (D) is correct. A rectangular coaxial line can support all the three modes (TE, TM or TEM). Option (C) is correct. Consider the rectangular waveguide (A) has the dimension a # b after deforming into waveguide (C) the dimension is changed to b # a and so the input mode TE 10 is charged to TE 01 . (Since the frequency of mode must remain same for both the waveguide dimensions).
SOL 9.3.39
Option (D) is correct. The dominant mode in a circular waveguide is TE 11 .
SOL 9.3.40
Option (B) is correct. Consider the dimension of inner broad wall of waveguide is a (i.e. a > b ). So, the dominant mode will be TE 10 . Since, the cutoff frequency of the TE mn mode is defined as 1 m 2 n 2 1/2 fc = :a a k + a b k D 2 m0 e0 So, for dominant mode ^TE 10h we have 1 1 2 0 2 1/2 ( fc = 10 GHz ) 10 # 109 = ;b a l + b b l E 2 m0 e0 8 10 # 109 = 3 # 10 # 1 2 a
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SOL 9.3.38
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SOL 9.3.41
8 a = 3 # 1010 = 1.5 cm 2 # 10 Option (A) is correct. Propagation constant in a waveguide is defined as f 2 g = 2pfc me 1 - c m fc Since, for the evanescent mode of waveguide the operating frequency is less than the cutoff frequency. i.e. f < fc f or H 0.13 # 10-3 ^4ph3 = 584.27 m Option (A) is correct. As calculated in previous question, the maximum detectable range of radar is rmax = 584.3 m So, half of the range will be at the position r = 1 rmax = 292.2 m 2 Therefore, the time average power density at half of the range of the radar is ^2850ph # 30 # 103 Pave = Gd Prad = 4pr2 4p ^292.2h2 = 250.35 W/m2
SOL 10.2.11
Option (D) is correct. Current amplitude I 0 = 50 A Operating frequency, f = 180 kHz = 180 # 103 Hz Effective length, l = 20 m Location of the observation point, R = 80 km = 8 # 10 4 m So, the maximum field intensity at the observation point is given as E q max = I 0 dl h0 b = I 0 h0 b # 2le 4p R 4pR As, the operating wavelength is 8 4 l = c = 3 # 10 3 = 10 6 f 180 # 10 and so the phase constant is b = 2p = 2p4 # 6 = 12p # 10-4 l 10 Therefore, the maximum field intensity at the observation point is 50 -4 E q max = # ^120ph # 12p # 10 # 2 # 20 4p # ^8 # 10 4h = 0.002827 = 2.83 mV/m
SOL 10.2.12
Option (B) is correct. The time average power density of antenna is defined as
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SOL 10.2.10
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Pave = 1 Re "Es # Hs*, 2 So, the time average radiated power is given as
p/2 2p Eq 2 Pave : dS = 1 R2 sin qdqdf h0 2 0 0 S p/2 2p R2 I 0 h b 2l 2 sin qdqdf =1 h 0 b 4p R 0 # e l 2 0 0 p/2 2p I b 2l 2 =1 h0 b 0 # e l sin qdqdf 2 0 4p 0 I b 2l 2 p/2 = 1 # 2p # h 0 b 0 # e l sin qdq 2 4p 0 2 -4 = p # 120p # b 50 # 12p # 10 # 2 # 20 l # 1 4p = 426.37 W = 0.43 kW
Prad =
# #
#
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Option (D) is correct. As calculated in previous question the time average radiated power is Prad = 0.43 kW Amplitude of the current in the antenna is I 0 = 50 A So, the radiation resistance of the antenna is given as 3 Rrad = 2Prad = 2 # 0.43 2# 10 2 I0 ^50h = 0.34 W
SOL 10.2.14
Option (D) is correct. Cross sectional radius of wire a = 6 mm = 6 # 10-3 m Radius of the circular loop, b = 1m Operating frequency, f = 0.5 MHz = 0.5 # 106 Hz No. of turns, N = 10 So, the operating wavelength of the antenna is 8 l = c = 3 # 10 6 = 6 # 102 m f 0.5 # 10 Therefore, the radiation resistance of the antenna is given as 4 4 1 Rrad = N 2 # 320p6 b b l = 102 # 320 # p6 b = 2.37 # 10-4 W 2 l l 6 # 10
SOL 10.2.15
Option (A) is correct. As calculated in the previous question, radiation resistance of the antenna is Rrad = 2.37 # 10-4 W So, the surface resistance of the antenna is given as pfm0 p # 0.5 # 106 # 4p # 10-7 Rs = = s 2.9 # 107 -4 = 2.61 # 10 W
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SOL 10.2.13
GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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Chap 10
Option (D) is correct. Since, the point P ^0, 0, 1000h lies along the axial direction of antenna carrying current in az direction, so it’s contribution to the field will be zero. Now for the antenna carrying current along ax direction, we have Amplitude of the current in antenna, I 0 = 4 A (i (t) = 4 cos wt A ) Length of the antenna, dl = 0.1 m The position of point P is r = 1000 and q = 90c as shown in the figure below :
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SOL 10.2.17
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For View Only Shop Online at www.nodia.co.in Therefore, the loss resistance of the antenna is 1 -4 Rl = N # b b l Rs = 10 # b -3 l # 2.61 # 10 a 6 # 10 = 0.435 W Thus, the radiation efficiency of the antenna is hrad = Rrad = 0.055% Rrad + Rl SOL 10.2.16 Option (B) is correct. Radiation function of the dipole antenna of height h is defined as cos ^bh cos qh - cos bh F ^q h = sin q Since, the height of dipole antenna is h = l/8 . So, we get cos ^1.25 cos qh - cos ^1.25ph F ^q h = sin q This function has been drawn as to obtain the pattern shown below :
So, the electric field component in free space is defined as E qs = h0 Hfs GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
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jI 0 bdl sin qe-jbr l 4p r j ^120ph^4hb 2p l^0.1h 2p 2p 2p = sin 90ce-j 2p ^1000h bb = l l 4p ^1000h = j1.2 # 10-2 e-j1000 V/m Since, rest of the components of field will be zero so, we get the net electric field as Es = E qs a q = j ^1.2 # 10-2 e-j1000h^- ax h =- j ^1.2 # 10-2h e-j1000 ax V/m Similarly, at point Q ^1000, 0, 0h the contribution due to antenna carrying current along x -axis will be zero while the electric field due to antenna along az will be Es =- j ^1.2 # 10-2h e-j1000 az V/m
m
= h0 b
Option (C) is correct. Since, the antenna are carrying current along ax and az while the point is located at y -axis so, both the antenna will contribute to the field. Therefore, summing the fields obtained due to the two antennas in previous question, we get, Es =- j ^1.2 # 10-2h e-j1000 ^ax + az h So, in the time domain E ^ t h = Re ^Es e j coth = ^1.2 # 10-2h sin ^wt - 1000h^ax + az h V/m Thus, the field at t = 0 at point ^0, 1000, 0h is E =-^9.92 # 10-3h^ax + az h V/m
SOL 10.2.19
Option (D) is correct. The field component due to the current element is given as E qs = 10 sin qe-j10pr p So, at point P (r = 100 , q = p/2 , f = p/6 )
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SOL 10.2.18
E qs = 10 sin a p k e-j10p^100h 100 2 -j1000p = 0.1e V/m
SOL 10.2.20
Option (C) is correct. Since, the vertical element is shifted from origin to a point y = 0.1 on the y -axis the distance of point P from the two locations of antenna is approximately same and therefore the magnitude of field component, E qs will be same in both cases but the phase angle will change due to the change in location of current element. So, the field intensity at point P due to the new location of vertical element is given as (1) E qs1 = E qs e-j10p^r - l h where l is the difference between the length of point P from two locations as shown in figure below :
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Option (B) is correct. Radiation resistance of a short circuit current element is determined as 2 where l is the length of dipole Rrad = 80p2 b l l l 2 Rrad = 80p2 b 0.06l l = 2.84 W l But, as the current is not uniform so, we determine the average current through the element. Now, from the given expression of current in the element, we get for l # z # 0 I1 ^z h = I 0 b l + 2z l 2 l l 2 z and for 0 # z # l I 2 ^z h = I 0 b 2 l l Therefore, the average current in the element is given as I 0 b l + 2z l + I 0 b l - 2z l l l I1 ^ z h + I 2 ^ z h Iavg = = = I0 2 2 2 Since, the average current flowing in the antenna is half of the uniform current I 0 therefore, the radiated power will be 14 th of the value obtained for I 0 and due to the same reason the radiation resistance will down to 14 th of its value. i.e. 6Rrad@net = 14 Rrad = 14 # 2.84 = 0.71 W
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SOL 10.2.21
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Now, using geometry we get the length l as l = 0.1 cos a p k = 0.05 3 Putting the value in equation (1), we get the field component as E qs1 = 0.1e-j1000p e j10p^ l h = 0.1e-j1000p e j0.5p V/m
Option (B) is correct. Length of wire, dl = 1 cm = 0.01 m Operating frequency, f = 0.3 GHz = 0.3 # 109 Hz Cross section radius, r = 1 mm = 10-3 m So, radiation resistance is given as 2 Rrad = 80p2 b dl l l GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 10.2.22
Chap 10
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Shop Online at www.nodia.co.in 2 c 0.01 = 80p2 = 0.07895 W bl = f l 8 3 10 fc # 9mp 0.3 # 10 Now, the ohmic resistance of the wire is defined as Rl = L s2pad where s " Conductivity a " Radius of the cross section d " Skin depth L " Length of the wire Since, the skin depth of the wire is given as 1 1 d = = 9 pfms p ^0.3 # 10 h^4p # 10-7h^5.8 # 107h -6 = 3.82 # 10 0.01 So, we get Rl = 7 ^5.8 # 10 h^2p # 10-3h^3.82 # 10-6h = 0.0072 W Therefore, the ratio of the radiation resistance to the ohmic resistance of wire will be Rrad = 10.977 . 11 Rl Option (D) is correct. Length of antenna, dl = 2 cm = 0.02 m Radiated power, Prad = 2 W Operating frequency, f = 0.6 GHz = 0.6 # 109 Hz So, operating wavelength of antenna is 8 l = c = 3 # 10 9 = 0.5 m f 0.6 # 10 Therefore, the radiation resistance of the antenna is given as 2 2 2 Rrad = 80p2 b dl l = 80p2 b 0.02 l = 16p 0. 5 125 l As the radiated power of the antenna is defined as Prad = 1 ^I 0h2 Rrad 2 ( Ir.m.s = I 0 / 2 ) = ^Ir.m.s.h2 Rrad So, the rms current in the antenna is 2 Ir.m.s. = Prad = = 1.26 A Rrad ^16p2 /125h
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SOL 10.2.23
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SOLUTIONS 10.3
m
Option (B) is correct. The directivity of an antenna is defined as D = U max Uave where U max is the maximum radiation intensity of the antenna and Uave is the average radiation intensity. Since, the given antenna has the radiation pattern (0 # q # p/2 ) U (q) = cos 4 q So, the maximum radiation intensity is U max = 1 The average radiation intensity is 2p 2p Uave = 1 F (q, f) dW = 1 ; F (q, f) sin qdq dfE 4p 4p 0 0 p/2 p/2 5 2p cos 4 q sin qdqdfE = 1 ;2p b- cos q lE = 1 ; 5 4p 4p 0 0 0 1 1 1 2 p 1 = 2p - 0 + D = = 5 10 4p # : 4p # 5 Therefore, the directivity of the antenna is D = 1 = 10 10 or, D (in dB) = 10 log 10 = 10 dB
#
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SOL 10.3.1
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Option (D) is correct. The beam-width of Hertzian dipole is 180c so, its half power beam-width is 90c.
SOL 10.3.3
Option (A) is correct. The operating wavelength of the antenna is 8 l = c = 3 # 10 9 = 3 200 f 20 # 10 Therefore, the gain of parabolic antenna is given as 2 Gp = hp2 b D l l 2 = 0.7 # p2 c 13 m = 30705.4
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SOL 10.3.2
( f = 20 GHz )
(efficiency, h = 70% )
100
or, SOL 10.3.4
10 log 10 G p = 44.87 dB
Option (C) is correct. Using the method of images, the configuration is as shown below
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Here d = l, a = p, thus, bd = 2p
SOL 10.3.7
SOL 10.3.8
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SOL 10.3.6
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SOL 10.3.5
So, the array factor of the antenna is given as bd cos y + a A.F. = cos : D 2 2p cos y + p = cos : D = sin (p cos y) 2 Option (B) is correct. Since, the antenna is installed at conducting ground. So, the power will be radiated only on the half side of the antenna and therefore, the radiation resistance of the antenna will be half of its actual value and given as 2 2 50 2p2 W Rrad = 1 ;80p2 ` dl j E = 40p2 c = m 2 l 5 0.5 # 103 Option (B) is correct. The array factor of the antenna is defined as bd sin q + a A.F. = cos b l 2 Here, d =l 4 and a = 90c 2p l sin q + p2 2p p p Thus, A.F. = cos c l 4 m = cos a 4 sin q + 2 k bb = l l 2 The option (A) satisfy this equation. Option (B) is correct. The directive gain of an antenna at a particular direction ^q, fh is defined as 4pU (q, f) Gd (q, f) = Prad Since, for lossless antenna Prad = Pin So, we get Prad = Pin = 1 mW Again the directive gain of the antenna is given 10 log Gd (q, f) = 6 dB So, Gd (q, f) = 3.98 Putting it in equation (1) we get the total power radiated by antenna as 4pU (q, f) = Prad Gd (q, f) = 1 m # 3.98 = 3.98 mW Option (A) is correct. Normalized array factor is given as
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y 2 where, y = bd sin q cos f + d q = 90c, d = 2 s, f = 45c, d = 180c bd sin q cos f + d y So, = 2 cos ; A.F. = 2 cos E 2 2 = 2 cos :2p 2 s cos 45c + 180cD 2 l2 = 2 cos 9 ps + 90cC = 2 sin a ps k l l Option (C) is correct. The signal strength (power) at a distance r from an antenna is inversely proportional to the distance r . i.e. P \ 12 r P1 = r 22 So, (1) P2 r 12 Since, 3 dB decrease " Strength is halved (103/10 = 100.3 = 2 ) P1 = 2 Therefore, P2 Substituting it in equation (1), we get 2 (r1 = 5 km ) 2 = r 22 5 or r2 = 5 2 km = 7071 m Thus, the required distance to move is d = r2 - r1 = 7071 - 5000 = 2071 m
SOL 10.3.10
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SOL 10.3.9
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A.F. = 2 cos
Option (C) is correct. We have
l = 492 m
SOL 10.3.11
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and height of antenna, dl = 124 m . l 4 So, it is a quarter wave monopole antenna and radiation resistance of a quarter wave monopole antenna is 36.5 W. Option (D) is correct. (1) We have y = bd cos q + d l where Distance between elements d = 4 Because of end fire y =0 q = 60c Putting all the values in equation (1) we get 0 = 2p # l cos 60c + d = p # 1 + d 2 2 4 l
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d =- p 4 Option (C) is correct. For a dipole antenna we have 1 BW \ (Diameter) So, as diameter increases Bandwidth decreases.
or SOL 10.3.12
Option (A) is correct. Far field region for an antenna is defined for the distance r from the antenna as 2 r > 2d l where d is the largest dimension of the antenna and l is the operating wavelength. Now, the operating wavelength of the antenna is given as 8 l = c = 3 # 10 9 = 3 m 40 f 4 # 10 So, for the closest far field we have 2 2 # (2.4) 2 80 # (2.4) 2 = . 150 m r = 2d = e o 3 3 l 40
SOL 10.3.14
Option (D) is correct. We know that for a monopole its electric field varies inversely with r 2 while its potential varies inversely with r . Similarly, for a dipole its electric field varies inversely as r 3 and potential varies inversely as r 2 . In the given expression both the terms _ r1 + r1 i are present, so, this potential is due to both monopole and dipole.
SOL 10.3.15
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SOL 10.3.13
-2
Option (A) is correct. Power received by an antenna is defined as Pr = Pt 2 # Ae 4p r where Pt is the power radiated by the transmitting antenna, r is the distance between transmitter and receiver and Ae is the effective aperture area of the receiving antenna. So, we get -4 Pr = 251 # 500 # 102 4 # p # (100) = 100 mW
SOL 10.3.16
-1
( Ae = 500 cm2, r = 100 m, Pt = 251 W )
Option (D) is correct. Magnetic field intensity in terms of vector potential is defined as H = 1d#A m where A is auxiliary potential function. So, d : H = d : (d # A) = 0 and d # H = d # (d # A) = Y 0
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Option (D) is correct. Aperture area of a receiving antenna is defined in terms of received power as Power Re ceived Aperture Area = Poynting vector of incident wave Ae = Pr Pt E 2 Since, ( h0 = 120p is intrinsic impedance of space) Pt = h0 -6 -6 So, Ae = 2 #E10 = 2 # 10 -3 2 # 120 # 3.14 (20 # 10 ) _ h i -6 3.14 = 1.884 m2 = 2 # 10 # 12 -# 6 400 # 10 Option (C) is correct. Maximum usable frequency( fmax ) in terms of incidence angle (i ) is defined as f fmax = o sin i where f0 is critical frequency. So, we get fmax = 8MHz = 8 = 16 MHz sin 60c 3 3 c 2 m Option (B) is correct. Far field \ 1 r Option (C) is correct. The maximum usable frequency is given as f fm = 0 sin i where i is launching angle and f0 is critical frequency so, we get 6 20 # 106 = 10 # 10 sin i or, sin i = 1 2 or, i = 30c 2
SOL 10.3.21
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SOL 10.3.20
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SOL 10.3.19
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SOL 10.3.18
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For View Only Shop Online at www.nodia.co.in SOL 10.3.17 Option (A) is correct. Radiation resistance of a circular loop is given as S Rrad = 8 hp3 ;NT E 3 l2 Rrad \ N 2 where N is number of turns. Since, the radiation resistance of a circular loop is 0.01 W. i.e. Rr1 = 0.01 W So, we get the net radiation resistance of the five turns of such loop as (N = 5 ) Rr 2 = N 2 # Rr1 = (5) 2 # 0.01 = 0.25 W
Option (C) is correct. GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia
SOL 10.3.22
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For View Only Shop Online at www.nodia.co.in The directive gain of half wave ^l/2h dipole antenna is given as cos2 a p cos q k 2 Gd = 1.66 sin2 q So, the directivity of the antenna is D = Gd,max cos2 ^ p2 cos qh Since, the maximum value of the function is 1. So, the directivity of sin2 q l/2 long wire antenna is D = 1.66 ^1 h = 1.66
m
Option (A) is correct. Since, the EM waves are travelling in free space, So the phase velocity of the wave will be equal to the velocity of light in free space. i.e. vp = c So, at frequency, f = 63 MHz (Channel 3) 8 wavelength, l = c = 3 # 10 6 = 4.76 m f 63 # 10 2 p So, phase constant, b = = 1.32 rad/m l and at frequency, f = 803 MHz (channel 69) 8 wavelength, l = c = 3 # 10 6 = 0.374 m f 803 # 10 2 p So, phase constant, b = = 16.82 rad/m l Option (A) is correct. Since, the antenna is located at earth so, power radiated to the hemisphere will be half of the transmitted value. i.e. Pr = Pt = 200 kW = 100 kW 2 2 Now, the average poynting vector (power radiated per unit area) at a distance r from the antenna is given as Pave = Pr2 ar pr where ar denotes the direction of Poynting vector. So, for r = 50 km , we have 3 Pave = 100 # 103 2 ar = 40 ar mW/m2 p p ^50 # 10 h
SOL 10.3.24
SOL 10.3.25
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SOL 10.3.23
Option (B) is correct. Radiation resistance of a dipole antenna is defined as 2 Rrad = 80p2 b dl l l Given, The length of dipole, dl = 5 m operating frequency, f = 3 MHz = 3 # 106 Hz So, the operating wave length of the antenna is given as
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Option (C) is correct. The radiated power of an antenna is defined as Prad = 1 I 02 Rrad 2 ...(i) i.e. Prad \ Rrad Now, the radiation resistance of the antenna is given as 2 Rrad = 80p2 b dl l l 2 ^dl h i.e. Rrad \ 2 l c Since l = f So, we get ...(ii) Rrad \ f 2 ^dl h2 Combining eq(1) and (2) we conclude that Prad \ ^dl h2 ^ f h2 Now, for the 1st antenna we have ^dl h^ f h = ^1.5h^100 # 106h = 1.5 # 108 nd for 2 antenna ^dl h^ f h = ^15h^10 # 106h = 1.5 # 106 Since, the product of length and frequency are same for both the antenna So, the power radiated by both the antennas will be same.
SOL 10.3.27
Option (B) is correct. Given the length of current element, l = 0.03l . So, the radiation resistance of the system is given as 2 2 Rrad = 80p2 b l l = 80p2 b 0.03l l = 0.072l2 W l l
SOL 10.3.28
Option (D) is correct. In a three element Yagi antenna there are one reflector, one folded dipole (driven element) and one director. The length of reflector is greater than driven element which in turn is longer than the director.
SOL 10.3.29
Option (C) is correct. Antenna arrays are formed to produce a greater directivity i.e. more energy radiated in some particular direction and less in other directions.
SOL 10.3.30
Option (B) is correct. Input power Radiated power
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SOL 10.3.26
= Wt = Wr
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For View Only Shop Online at www.nodia.co.in Radiation intensity =f So,the power gain of the antenna is 4pf ^a " 3h Gp = Wt Directive gain of antenna is 4pf Gd = ^b " 4h Wr Average radiated power of the antenna is Pr = Wr ^c " 2h 4p The efficiency of antenna is h = Wr ^d " 1h Wt SOL 10.3.31 Option (C) is correct. Maximum radiation for an end fire array occurs along the line of the array. Option (B) is correct. The gain of antenna is directly proportional to the aperture area. So, with increase of aperture area, received power increases and therefore the gain increases.
SOL 10.3.33
Option (B) is correct. In the helical antenna, normal mode of operation is very narrow in bandwidth and therefore the directivity is high. While the radiation efficiency is low.
SOL 10.3.34
Option (B) is correct. For an antenna near and far zone are specified by a boundary defined as 2 R = 2d l where, R is the distance from antenna, d is the largest dimension of antenna and l is the operating wavelength of antenna. So, any target located at a distance R > 2Dl from antenna is in the far zone for the antenna and any target located at a distance R < 2Dl is in the near zone.
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SOL 10.3.32
2
2
SOL 10.3.35
Option (B) is correct. Gain of transmitting antenna, Gdt = 10 Transmitted power, Pt = 1W Effective area of receiving antenna, Aer = 1 m2 Distance between transmitter and receiver, r = 1m So, total received power by the receiving antenna is Pr = Pt 2 Gdt Aer = 1 2 # ^10h # ^1 h = 0.79 W 4p r 4p ^ 1 h
SOL 10.3.36
Option (C) is correct. 2 Since, the region r > 2D is called far zone for the antenna and as it is given that l 2 in the Fraunhofer region measurement to be taken from a distance of 2D from l antenna so, the defined region is far zone or far field.
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For View Only Shop Online at www.nodia.co.in SOL 10.3.37 Option (C) is correct. Helical antenna is used to provide circularly polarized wave and the log periodic antenna is frequency independent. Option (C) is correct. For a wave travelling from medium 1 to medium 2, the incidence angle qc of the wave for which it is totally reflected by medium 2 is given as (1) n1 sin qc = n2 sin 90c where n1 and n2 are the refractive index of medium 1 and medium 2 respectively. Since, refracting index of a medium having permittivity e and permeability m is defined as n = me So, putting it in equation (1), we get m0 e1 sin qc = m0 e2 sin qc = e2 = 1 2 e1 1 -1 qc = sin c = 45c 2m Option (C) is correct.
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SOL 10.3.38
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Yagi-Uda antenna must have one reflector and one driven element while it can have any number of directors. So, the four element Yagi-Uda antenna will have 2 directors, one reflector, and one driven element. SOL 10.3.40
Option (B) is correct. Directivity of an antenna is directly proportional to the effective area and therefore larger the effective area, sharper the radiated beam. This is the reason for using an aperture antenna instead a linear antenna for extremely high frequency ranges.
SOL 10.3.41
Option (B) is correct. Current distortion along a travelling wave antenna in general is defined as I ^z, t h = I 0 cos ^cot - bz h but when we eliminate t by taking it’s phasor form, the current can be written as I ^z h = I 0 e-jbz
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For View Only Shop Online at www.nodia.co.in SOL 10.3.42 Option (C) is correct. Turnstile antenna is generally used at UHF/VHF for local area transmission. Option (C) is correct. The frequencies upto 1650 kHz is in the range of medium frequency. Vertical radiators ranging from l/6 to l/5 used for broadcasting the medium frequencies as the operating conditions and economic consideration.
SOL 10.3.44
Option (B) is correct. The radiation field intensity of an antenna at a distance r is defined as jhI 0 bdl sin qe-jbr E qs = 4pr hI bdl sin q or \1 E qs = 0 r 4p r Option (A) is correct. The resultant field in a helical antenna is either circularly polarized or elliptically polarized depending on the pitch angle a. The radiated wave by a helical antenna is circularly polarized only when, a = tan-1 a c k 2l else it is elliptically polarized. In conclusion for a general term we can say the wave radiated by a helical antenna is elliptically polarized.
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Option (A) is correct. The unexcited patch is shown below
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SOL 10.3.46
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SOL 10.3.45
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SOL 10.3.43
The capacitance between the plates is given as e ^Area of platesh C = = e0 er LW h ^separatoin between platesh SOL 10.3.47
Option (D) is correct. Maximum usable frequency between two stations of distance D is defined as 2 fMUF = fc 1 + b D l 2h where h vertical height at the mid point of path and fc is critical frequency we put all the values to get,
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1 +b
800 2 = 1.5 107 Hz = 15 MHz # 2 # 300 l
Option (B) is correct. D -layer is the lower most region of ionosphere which is present only during the day light hours and disappears at night because recombination rate is highest and also E -region is weekly lonised during night hour hence radiowave suffer negligible attenuation in night hour. This is the reason that the wave band which can’t be heard during day time but may be heard during night time.
SOL 10.3.49
Option (D) is correct.
SOL 10.3.50
Option (B) is correct. The atmosphere has varying density (refractive index) with the height from earth dm . Radius of curvature of the wave path is given as dh R =- dh dm Solving it, we get the effective earth radius (Radio horizon) = 4 actual earth radius (optical horizon) 3 Option (B) is correct. The radiation resistance of a dipole antenna is defined as 2 Rrad = 80p2 b dl l l Since, dl = l 20 l/20 2 So, Rrad = 80p2 c = 2W l m
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Option (B) is correct. Effective length of a half wave dipole antenna is cos a p cos q k 2 2 H le ^q h = > b sin q i.e. le is function of q. The maximum value of le is at q = p/2 . le a p k = 2 = l < l p 2 2 b
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SOL 10.3.52
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SOL 10.3.48
i.e. maximum value of le is less than its actual value l . 2 The effective length is the same for the antenna in transmitting and receiving modes. So, statements, 1, 2 and 4 and correct while statement 3 is incorrect. ***********
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