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Electromagnetic Theory For

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Syllabus

Syllabus for Electromagnetic Theory Coulomb's Law, Electric Field Intensity, Electric Flux Density, Gauss's Law, Divergence, Electric Field and Potential due to Point, Line, Plane and Spherical Charge Distributions, Effect of Dielectric Medium, Capacitance of Simple Configurations, Biot‐Savart’s Law, Ampere’s Law, Curl, Faraday’s Law, Lorentz Force, Inductance, Magnetomotive Force, Reluctance, Magnetic Circuits, Self and Mutual Inductance of Simple Configurations.

Analysis of GATE Papers Year

Percentage of Marks

2015

4.00

2014

3.60

2013

4.00

2012

2.00

2011

2.00

2010

0.00

2009

0.00

2008

4.00

2007

6.00

2006

0.67

Overall Percentage

2.627%

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Contents

Contents

#1.

Chapters Electromagnetic Field                

Introduction Operators Material and Physical Constants Electromagnetic (EM Field) Electric Field Intensity Electric Dipole Divergence of Current Density and Relaxation Boundary Conditions The Magnetic Vector Potential Faraday’s Law Maxwell’s Equation’s Magnetic Field Solved Examples Assignment 1 Assignment 2 Answer Keys & Explanations

Page No. 1 – 46 1 2–7 7–8 8–9 9 – 12 12 – 17 18 19 – 21 21 – 25 25 – 27 27 – 28 28 – 32 32 – 38 39 – 41 41 – 42 43 – 46

Module Test

47 – 50

 

47 – 48 49 – 50

Test Questions Answer Keys & Explanations

Reference Books

51

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i

“Picture yourself vividly as winning and that alone will contribute immeasurably to success."

CHAPTER

…Harry Fosdick

1

Electromagnetic Field

Learning Objectives After reading this chapter, you will know: 1. Elements of Vector Calculus 2. Operators, Curl, Divergence 3. Electromagnetic Coulombs’ law, Electric Field Intensity, Electric Dipole, Electric Flux Density 4. Gauss's Law, Electric Potential 5. Divergence of Current Density and Relaxation 6. Boundary Conditions 7. Biot-Savart’s Law, Ampere Circuit Law, Continuity Equation 8. Magnetic Vector Potential, Energy Density of Electric & Magnetic Fields, Stored Energy in Inductance 9. Faraday’s Law, Motional EMF, Induced EMF Approach 10. Maxwell’s Equations

Introduction Cartesian coordinates (x, y, z), −∞ < x < ∞, −∞ < y < ∞, −∞ < z < ∞ Cylindrical coordinates (ρ , ϕ, z), 0 ≤ ρ < ∞, 0 ≤ ϕ < 2π, −∞ < z < ∞ Spherical coordinates (r, θ, ϕ ) , 0 ≤ r < ∞, 0 ≤ θ ≤ π, 0 ≤ ϕ < 2π Other valid alternative range of θ and ϕ are----(i) 0 ≤ θ < 2π, 0 ≤ ϕ ≤ π (ii) −π ≤ θ ≤ π, 0 ≤ ϕ ≤ π π (iii) − 2 ≤ θ ≤ π⁄2 , 0 ≤ ϕ < 2π (iv) 0 < θ ≤ π, −π ≤ ϕ < π Vector Calculus Formula SL. No (a) (b)

(c)

Cartesian Coordinates Differential Displacement dl = dx ax + dy ay + dz az Differential Area dS = dy dz ax = dx dz ay = dx dy az Differential Volume dv = dx dy dz

Cylindrical Coordinates dl = dρaρ + ρdϕaϕ +dzaz dS = ρ dϕ dz aρ = d ρ dz aϕ = ρdρd ϕ az

Spherical Coordinates dl = drar + rdθaθ + r sin θdϕaϕ ds = r 2 sin θ dθ dϕ ar = r sin θ dr dϕ aθ = r dr dθ aϕ

dv = ρ dρ dϕ dz

dv = r 2 sin θ dθ dϕ dr

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1

Electromagnetic Field

Operators 1) ∇ V – Gradient, of a Scalar V 2) ∇ V – Divergence, of a Vector V 3) ∇ × V – Curl, of a Vector V 4) ∇2 V – Laplacian, of a Scalar V DEL Operator: ∂ ∂ ∂ ∇ = ax + ay + a (Cartesian) ∂x ∂y ∂z z ∂ 1 ∂ ∂ = aρ + aϕ + a (Cylindrical) ∂ρ ρ ∂ϕ ∂z z ∂ 1 ∂ 1 ∂ = ar + aθ + a (Spherical) ∂r r ∂θ rsi n θ ∂ϕ ϕ Gradient of a Scalar field V is a vector that represents both the magnitude and the direction of maximum space rate of increase of V. ∂V ∂V ∂V ∇V = ax + ay + a For Cartisian Coordinates ∂x ∂y ∂z z ∂V 1 ∂V ∂V = aρ + aϕ + a For Spherical Coordinates ∂ρ ρ ∂ϕ ∂z z ∂V 1 ∂V 1 ∂V = ar + aθ + a For Cylindrical Coordinates ∂r r ∂θ rsi n θ ∂ϕ ϕ The following are the fundamental properties of the gradient of a scalar field V 1. The magnitude of ∇V equals the maximum rate of change in V per unit distance. 2. ∇V points in the direction of the maximum rate of change in V. 3. ∇V at any point is perpendicular to the constant V surface that passes through that point. 4. If A = ∇V, V is said to be the scalar potential of A. 5. The projection of ∇V in the direction of a unit vector a is ∇V. a and is called the directional derivative of V along a. This is the rate of change of V in direction of a. Example: Find the Gradient of the following scalar fields: (a) V = e−z sin 2x cosh y (b) U = ρ2 z cos 2ϕ (c) W = 10r sin2θ cos ϕ Solution: (a) ∇V =

∂V

a ∂x x −z

= 2e (b) ∇U =

∂U

a ∂ρ ρ

+

∂V

a ∂y y

+

∂V ∂z

az

cos 2x cosh y ax + e−z sin 2x sinh y ay − e−z sin 2x cosh y az 1 ∂U

∂U

+ ρ ∂ϕ aϕ + ∂z az

= 2ρz cos 2ϕ aρ − 2ρz sin 2ϕ aϕ + ρ2 cos 2 ϕ az (c) ∇W =

∂W ∂r

ar +

1 ∂W r ∂θ

aθ +

1

∂W

r sin θ ∂ϕ

aϕ

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2

Electromagnetic Field

Divergence of a Vector Statement: Divergence of A at a given point P is the outward flux per unit volume as the volume shrinks about P. Hence, ∮ A . ds DivA = ∇. A = lim S … … … … … … … … … … … … … … … (1) ∆v→0 ∆v Where, ∆v is the volume enclosed by the closed surface S in which P is located. Physically, we may regard the divergence of the vector field A at a given point as a measure of how much the field diverges or emanates from that point. ∂Ax ∂Ay ∂Az ∇. A = + Cartisian System ∂x ∂y ∂z P 1 ∂ 1 ∂Aϕ ∂A z (ρAρ ) + + Cylindrical System = ∂z ρ ∂ρ ρ ∂ϕ 1 ∂ 2 ∂ 1 ∂Aϕ 1 = 2 (r Ar ) + (Aθ sin θ) + Sphearical System r ∂r r sin θ ∂ϕ r sin θ ∂θ From equation (1), ∮ A . dS = ∫ ∇ . A dv S

V

This is called divergence theorem which states that the total outward flux of the vector field A through a closed surface S is same as the volume integral of the divergence of A. Example: Determine the divergence of these vector field (a) P = x 2 yzax + xzaz (b) Q = ρ sin ϕ aρ + ρ2 zaϕ + z cos ϕ az 1

(c) T = r2 cos θ ar + r sin θ cos ϕ aθ + cos θ aϕ Solution: ∂

∂

∂

(a) ∇. P = ∂x Px + ∂y Py + ∂z Pz ∂ 2 ∂ ∂ (x yz) + (0) + (xz) ∂x ∂y ∂z = 2xyz + x

=

1 ∂

1 ∂

∂

(b) ∇ . Q = ρ ∂ρ (ρQρ ) + ρ ∂ϕ Qϕ + ∂z Qz 1 ∂ 2 1 ∂ 2 ∂ (ρ sin ϕ) + (ρ z) + (z cos ϕ) ρ ∂ρ ρ ∂ϕ ∂z = 2 sin ϕ + cos ϕ

= (c) ∇. T =

1 ∂ r2 ∂r

(r 2 Tr ) +

1

∂

r sin θ ∂θ

(Tθ sin θ) +

1

∂

r sin θ ∂ϕ

(Tϕ )

1 ∂ 1 ∂ 1 ∂ (cos θ) + (r sin 2 θ cos ϕ) + (cos θ) 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ 1 =0+ 2r sin θ cos θ cos ϕ + 0 r sin θ = 2 cos θ cos ϕ =

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Electromagnetic Field

Curl of a Vector Curl of a Vector field provides the maximum value of the circulation of the field per unit area and indicates the direction along which this maximum value occurs. That is, ∮ A . dl ) an … … … … … … … . . (2) Curl A = ∇ × A = lim ( L ΔS→0 ∆S max ax ay az ∂ ∂ ∂ | ∇×A = | ∂x ∂y ∂z Ax Ay Az aρ ρaϕ az ∂ ∂ 1 ∂ | = | ρ ∂ρ ∂ϕ ∂z Aρ ρA ϕ Az aρ raθ r sin θ aϕ ∂ 1 ∂ ∂ || | = 2 r sin θ ∂r ∂θ ∂ϕ | A r rAθ r sin θ Aϕ From equation (2) we may expect that ∮ A dl = ∫(∇ × A) . ds L

S

This is called stoke’s theorem, which states that the circulation of a vector field A around a (closed) path L is equal to the surface integral of the curl of A over the open surface S bounded by L, Provided A and Δ × A are continuous no s. Example: Determine the curl of each of the vector fields. (a) P = x 2 yz ax + xzaz (b) Q = ρ sin ϕaρ + ρ2 zaϕ + z cos ϕaz 1

(c) T = r2 cos θ ar + r sinθ cos ϕaθ + cos ϕ aϕ Solution: ∂Py ∂Px ∂P ∂Py ∂Px ∂Pz (a) ∇ × P = ( z − ) ax + ( − ) ay + ( − )a ∂y ∂z ∂z ∂x ∂x ∂y z = (0 − 0)ax + (x 2 y − z)ay + (0 − x 2 z)az = (x 2 y − z)ay − x 2 zaz ∂Qρ ∂Qz ∂Qρ 1 ∂Qz ∂Qϕ 1 ∂ ] aρ + [ ] aϕ + [ (ρQϕ ) − ]a − − ρ ∂ϕ ∂z ∂z ∂ρ ρ ∂ρ ∂ϕ z −z 1 = ( sin ϕ − ρ2 ) aρ + (0 − 0)aϕ + (3ρ2 z − ρ cos ϕ)az ρ ρ 1 = − (z sin ϕ + ρ3 )aρ + (3ρz − cos ϕ)az ρ

(b) ∇ × Q = [

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4

Electromagnetic Field

∂ ∂ 1 [ (Tϕ sin θ) − T ]a ∂ϕ θ r r sin θ ∂θ 1 1 ∂ ∂ 1 ∂ ∂ + [ Tr − (rTϕ )] aθ + [ (rTθ ) − Tr ] aϕ r sin θ ∂ϕ ∂r r ∂r ∂θ ∂ ∂ 1 [ (cos θ sin θ) − = (r sin θ cos ϕ)] ar ∂ϕ r sin θ ∂θ 1 1 ∂ (cos θ) ∂ + [ − (r cos θ)] aθ r sin θ ∂ϕ r 2 ∂r 1 ∂ 2 ∂ (cos θ) ] aϕ + [ (r sin θ cos ϕ) − r ∂r ∂θ r 2 1 1 (cos 2θ + r sin θ sin ϕ)ar + (0 − cos θ)aθ = r sin θ r 1 sin θ + (2r sin θ cos ϕ + 2 ) aϕ r r cos 2θ cos θ 1 =( + sin ϕ) ar − aθ + (2 cos ϕ + 3 ) sin θ aϕ r sin θ r r

(c) ∇ × T =

Laplacian (a) Laplacian of a scalar field V, is the divergence of the gradient of V and is written as ∇2 V. ∂2 V ∂2 V ∂2 V ∇2 V = 2 + 2 + 2 → For Cartisian Coordinates ∂x ∂y ∂z 1 ∂ ∂V 1 ∂2 V ∂2 V ∇2 V = (ρ ) + 2 2 + 2 → For Cylindrical Coordinates ρ ∂ρ ∂ρ ρ ∂ϕ ∂z ∂ 1 ∂ ∂V 1 ∂V 1 ∂2 V = 2 (r 2 ) + 2 (sin θ ) + 2 → For Spherical Coordinates r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ2 If ∇2 V = 0, V is said to be harmonic in the region. A vector field is solenoid if ∇.A = 0; it is irrotational or conservative if ∇ × A = 0 ∇. (∇ × A) = 0 ∇ × (∇V) = 0 ̅ (b) Laplacian of Vector A 2⃗ ∇ A = ⋯ is always a vector quantity ⃗ = (∇2 Ax )âx + (∇2 Ay )ây + (∇2 Az )âz ∇2 A ∇2 Ax → Scalar quantity ∇2 Ay → Scalar quantity ∇2 Az → Scalar quantity −p ∇2 V = ϵ ........Poission’s E.q.

∇2 V = 0 ........Laplace E.q. ⃗ ∂E ∂2 E ∇2 E = μσ + μE 2 . . . . . . . wave E. q. ∂t ∂t

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5

Electromagnetic Field

Example: The potential (scalar) distribution in free space is given as V = 10y 4 + 20x 3 . If ε0 : permittivity of free space what is the charge density ρ at the point (2,0)? ρ 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Poission’s Equation ∇2 V = − ε ∂2 ∂2 −ρ ∂2 ( 2 + 2 + 2 ) (10 y 4 + 20x 3 ) = ∂y ∂z ∂x ε0 ∵ ε = εr ε0 [ε = ε0 as εr = 1] −ρ 20 × 3 × 2x + 10 × 4 × 3y 2 = ε 0 −ρ At pt(2, 10) ⇒ 20 × 3 × 2 × 2 = ρ = −240ε0 ε0 Example: Find the Laplacian of the following scalar fields (a) V = e−z sin 2x cosh y (b) U = ρ2 z cos 2ϕ (c) W = 10r sin2 θ cos ϕ Solution: The Laplacian in the Cartesian system can be found by taking the first derivative and later the second derivative. ∂2 V ∂2 V ∂2 V (a) ∇2 V = 2 + 2 + 2 ∂x ∂y ∂z ∂ ∂ ∂ = (2e−z cos 2x cosh y) + (e−z sin 2x sinh y) + (−e−z sin 2x cosh y) ∂y ∂x ∂z = −4e−z sin 2x cosh y + e−z sin 2x cosh y + e−z sin 2x cosh y = −2e−z sin 2x cosh y ∂U 1 ∂2 U ∂2 U 1 ∂ (b) ∇2 U = (ρ ) + 2 2 + 2 ρ ∂ϕ ∂z ρ ∂ρ ∂ρ 1 ∂ 1 (2ρ2 z cos 2ϕ) − 2 4ρ2 z cos 2ϕ + 0 = ρ ∂ρ ρ = 4z cos 2ϕ − 4z cos 2ϕ =0 ∂ ∂2 W 1 ∂ ∂W 1 ∂W 1 (c) ∇2 W = 2 (r 2 )+ 2 (sin θ )+ 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin2 θ ∂ϕ2 ∂ 1 ∂ 1 10r sin2 θ cos ϕ (10r sin 2θ sin θ cos ϕ) − = 2 (10 r 2 sin2 θ cos ϕ) + 2 r ∂r r sinθ ∂θ r 2 sin2 θ 2 20 sin θ cos ϕ 20r cos 2θ sin θ cos ϕ 10r sin 2θ cos θ cos ϕ 10 cos ϕ = + + − r r 2 sin θ r 2 sin θ r 10 cos ϕ = (2 sin2 θ + 2 cos 2θ + 2 cos2 θ − 1) r 10 cos ϕ = (1 + 2 cos 2θ) r Stoke’s Theorem ⃗ integrated over any closed curve C is always equal to Statement: Closed line integral of any vector A ⃗ integrated over the surface area ‘s’ which is enclosed by the the surface integral of curl of vector A closed curve ‘c’. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

6

Electromagnetic Field

S

C

⃗ = ∫ ∫(∇ × ⃗A) dS⃗ ∮ ⃗A . dL S

The theorem is valid irrespective of (i) Shape of closed curve ‘C’ (ii) Type of vector ‘A’ (iii) Type of co-ordinate system Divergence Theorem S

V ⃗ dv ∯ ⃗A dS⃗ = ∭ ⃗V. A S

V

⃗ integrated over any closed surface area. S is Statement: Closed surface integral of any vector A ⃗ integrated over the volume V always equal to the volume integral of the divergence of vector A which is enclosed by the closed surface are ‘S’ the theorem holds good, irrespective (i) Shape of closed surface (ii) Type of coordinate system (iii) Type of vector ⃗A

Material & Physical Constants (a) Material Constants Material Air Aluminum Bakelite Brass Carbon Copper Glass Graphite Mica Paper Paraffin

Conductivity (σ ) S/m 0 3.186 × 107 10−14 2.564 × 107 3 × 104 5.8 × 107 10−13 105 10−15 − 10−15

Relative Permittivity (εr) 1.0006 1.0 5 1 − 1 6 − 6 3 2.1

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Electromagnetic Field

Plexiglas Polystyrene PVC Porcelain Quartz Rubber Rutile Soil(Clay) (Sandy) Urban ground Vaseline Terflon Water (Distilled) (Fresh) (Sea) Wood Transformer oil Ebonite Epoxy

− 10−16 − − 10−17 10−13 − 5 × 10−3 2 × 10−3 2 × 10−4 − 10−15 10−4 10−2 to 10−3 4 to 5 − − − −

3.4 2.7 2.7 5 5 5 100 14 10 4 2.2 2.1 80 80 80 2 2 to 3 2.6 4

(b) Physical Constants Permittivity of free space, ∊0 = 8.854×10-12 F/m = (1/36 π) 10-9 F/m Permeability of free space, μ0 = 4π×10-7 H/m Impedance of free space, η0 = 120 π Ohms = 377 Ohms Velocity of free space, c = 3 ×108 m/sec = 3 ×1010 cm/sec Charge of an Electron, q = 1.602 × 10-19 C Mass of electron, m = 9.107 × 10 -31 kg Boltzman’s constant, k = 1.38×10-23 J/0K Planck’s constant, h = 1.054× 10-34 J-s Base of natural logarithm, e = 2.718

Electromagnetic (EM Field) 



  

In general, electromagnetic field is regarded as interplay between time varying electric and magnetic fields. The study of electromagnetic can be accomplished with study of electrostatics, magneto statics and time varying electric and magnetic fields. Electrostatics deals with field related to stationary charge(s). The charge can be positive or negative. The unit of charge is called a coulomb. The charge of an electron is e = −1.6019 × 10−19 Coulombs q = |e| Charge may be distributed in space or may be concentrated in a small volume at a point. A charge that occupies a volume in space may be considered to be a point charge for analysis purposes if this volume is small compared to the surrounding dimensions. A charge density defines charge distribution on a line (or) over a surface (or) throughout a volume. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

8

Electromagnetic Field

Coulomb’s Law Statement: “The force between any two point charges Q1 and Q2 is proportional to the product of the two charges, inversely proportional to square of the distance between the two charges, and directed along the line connecting the two charges”. The Mathematical expression of Coulomb’s law is Q1 Q2 ̂ F= R 4π ε R2 Where (1/4π ε) is proportionality factor R is the distance between the two charges. ̂ is the unit vector pointing form Q1 to Q2 (or) Q2 to Q1. R The proportionality factor depends on the material in which the charges are located. ‘∊’ is a material constant and is called the permittivity of the material and its units are “Farad/meter”. Force is measured in “Newton”. ε = ε0 εr , where εr is relative permittivity and ε0 is permittivity of free space.

Electric Field Intensity Electric Field Intensity is defined as the force per unit charge, when placed in an electric field and its unit is Newton/Coulomb (or) Volt/meter. The electric field can be viewed as starting at a positive charge and ending at a negative charge. In the electric field E of a charge (say Q1), if we introduce another charge (say Q2), there will be a force acting on this charge Q2 . i.e., F = Q2 E â r

r

Q

Q â . . . . . . . . . . . . due to pt. chrage 4πε0 R2 r δ dl ⃗E = ∫ L â . . . . . . . . . . . . (Line Charge) 4πε0 R2 r

⃗ = E

C

⃗ = ∫∫ E S

ps ds â . . . . . . . . . . . . (Surface Charge) 4πεR2 r

⃗E = ∫ ∫ ∫ V

ρv dv â . . . . . . . . . . . . (Volume Charge) 4πε0 R2 r

General Form of Coulomb’s Law ⃗ = r − ⃗⃗r ′ R ⃗ | = |r − r⃗⃗′ | R = |R ⃗⃗⃗⃗⃗ R r − ⃗⃗r ′ ̂= R = ⃗ | |r − ⃗⃗r ′ | |R

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9

Electromagnetic Field

Electrical field y

̂R Q r

⃗ = E

Q

⃗⃗r ′

O ⃗E =

⃗R = r − ⃗⃗r ′ x

Q Q ⃗R × â R = 2 4πε0 R 4πε0 R3 Q(r − r⃗⃗′ ) 3

4πε0 |r − r⃗⃗′ | Equipotential surface An equipotential surface has its every point at equal potential. Properties:  The movement of charge over such a surface would require no work.  Tangential to such surface is zero electric field.  Electric field is always perpendicular to an equipotential surface for static fields, a conductor surface is always an equipotential surface. Electric Field Intensity of a Finite Line Charge Consider a line charge with uniform charge density ρL extending from A to B along the z – axis as shown in For a finite line charge ρL [−(sin α2 − sin α1 )ar + (cos α2 − cosα1 )az ] E= 4πε0 ρ Where ρL is line charge density and r is the perpendicular distance from the line to point of interest. +π −π As a special case, for an infinite line charge α1 = and α2 = 2 2 ρL And E= a 2πε0 ρ r z r α2

α1

P (x, y, z)

B

A 0

Y

x

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10

Electromagnetic Field

For an infinite sheet of charge, ρs E= a 2ε0 n Where ρs is surface charge density of sheet and an is a unit vector normal to sheet. Example: Point charges 1 mC and −2 mC are located at (3, 2, −1) and (−1, −1, 4), respectively. Calculate the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that point. Solution: QQk QQk (r − rk ) ∑ F= ∑ a = R 4πε0 R2 4πε0 |r − rk |3 k=1,2

k=1,2

Q 10−3 [(0, 3, 1) − (3, 2, −1)] 2 × 10−3 [(0, 3, 1) − (−1, −1, 4)] { } = − |(0, 3, 1) − (3, 2, −1)|3 |(0, 3, 1) − (−1, −1, 4)|3 4πε0 10−3 × 10 × 10−9 (−3, 1, 2) 2(1, 4, −3) [ ] = − −9 3/2 10 (9 + 1 + 4) (1 + 16 + 9)3/2 4π × 36π (−3, 1, 2) (−2, −8, 6) ] = 9 × 10−2 [ + 14√14 26√26 F = −6.507 ax − 3.817 ay + 7.506az mN At that point, F E= Q 10−3 = (−6.507 − 3.817, 7.506) × 10 × 10−9 E = −650 × 7ax − 381 × 7ay + 750 × 6az kV/m Example: The finite sheet 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 on the z = 0 plane has a charge density ρs = xy(x 2 + y 2 + 25)3⁄2 n C/m2 . Find (a) The total charge on the sheet (b) The electric field at (0, 0, 5) (c) The force experienced by a −1 mC charge located at (0, 0, 5) Solution: 1

1

(a) Q = ∫S ρS dS = ∫0 ∫0 xy(x 2 + y 2 + 25)3/2 dx dy nC

Since x dx = 1/2 d(x 2 ), we now integrate with respect to x 2 (or change variable: x 2 = u so that x dx = du⁄2. 1

1

1 Q = ∫ y ∫(x 2 + y 2 + 25)3⁄2 d(x 2 )dy nC 2 0 1

0

1 2 1 = ∫ y (x 2 + y 2 + 25)5⁄2 |0 dy 2 5 0 1

1 1 = ∫ [(y 2 + 26)5⁄2 − (y 2 + 25)5⁄2 ]d(y 2 ) 5 2 0

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11

Electromagnetic Field

1 2 2 1 × [(y + 26)7⁄2 − (y 2 + 25)7/2 ]|0 10 7 7 1 [(27)7⁄2 + (25)2 − 2(26)7⁄2 ] = 35 Q = 33.15 nC =

(b) E = ∫ S

ρS dS (r − r ′ ) ρS dS aR ∫ = 4πε0 |r − r ′ |3 4πε0 r 2 S

Where r − r ′ = (0, 0, 5) − (x, y, 0) = (−x, −y, 5). Hence, 1 1

E = ∫∫ 0 0

10−9 xy(x 2 + y 2 + 25)3⁄2 (−xax − yay + 5az )dx dy 10−9 4π × 36π (x 2 + y 2 + 25)3⁄2 1

1

1

1

1

1

= 9 [− ∫ x 2 dx ∫ y dy ax − ∫ x dx ∫ y 2 dy ay + 5 ∫ x dx ∫ y dy az ] 0

0

0

0

0

0

−1 −1 5 = 9( , , ) 6 6 4 = (−1.5, −1.5,11.25) V/m (c) F = qE = (1.5, 1.5, −11.25) mN

Electric Dipole Two equal and opposite electric charges, separated by a very short distance is called electric dipole and is shown below. P +Q θ d

R R>>d

Q Electric Dipole    

The electric dipole moment, p = Q d The dipole moment is directed from –Q to + Q The electric field intensity of a dipole varies as 1/R3 , where as the electric field intensity of a point charge varies as 1/R2 The electric field due to the dipole is given by p (2 cos θ ar + sin θ aθ ) E= 4πε0 r 3

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12

Electromagnetic Field

Electric Flux Density (D) D = ε0 E  This vector has the same direction as E, but it is independent of ‘∊’ and therefore of material properties.  The unit of ‘D’ is Coulomb/meter2.  Electric Flux Ψ in terms of D is defined as Ψ = ∫ D. ds S

Gauss’s Law 



Gauss’s law states that the total electric flux Ψ through any closed surface is equal to the total charge enclosed by that surface. Thus Ψ = Qenc Ψ = Qenc i. e. , Ψ = ∮ dΨ = ∮ D ∙ ds S

Total charge enclosed Q = ∫ ρv dv v

or Q = ∮ D ∙ dS = ∫ ρv dv S

V

By applying divergence theorem, ∮ D ∙ dS = ∫ ∇ ∙ D dv v

So, ρv = ∇ ∙ D Which is one of the four Maxwell’s equation and it states that the volume charge density is the same as the divergence of the electric flux density.  When, at any point if charge density is zero, then divergence of electric flux density and divergence of electric field intensity is zero.  Curl of static electric field intensity is zero. Mathematically, ∇ × E = 0  Thus, electrostatic field is Irrotational (curl free) and Non-Solenoidal (non zero divergence).  Gauss’s law is an alternate form of Coulomb‘s law.  Gauss’s law may be used either to calculate the equivalent charges from known electric fields or electric fields due to known charges. Example: Find the Total flux in a cylinder of radius r and length L placed in a uniform electric field E parallel to the axis of cylinder. dA B E E ⃗ dA dA L A C : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

13

Electromagnetic Field

Solution: The total flux ϕE is the sum of flux coming out of surface A, B and C ϕE = ∫ E dA cos θ + ∫ E dA cos θ + ∫ E dA cos θ A

B

C

θA = 0° θB = 90° θC = 180° dAA = πr 2 dAB = 2πrl dAC = πr 2 ϕE = E. A + 0 − E. A = 0 As ϕE = 0, charge enclosed Qenclosed = 0 and hence D and E are 0 at all point inside the plane of uniformly charged circular ring. Example: Find the electric flux inside and outside a symmetrically charged sphere of radius ‘A’ R

0

A r

Solution: By Guss’s Law ∫ E. dA = surface

1 Q ε0 enclosed

∫ E dA = E (4πR2 ) =

1 Q ε0

1 Q 4πε0 R2 Where Q is the charge in the sphere of radius A and enclosed by sphere of radius R. Suppose there is a point charge Q at the origin O. Then the electric field at the distance R will be 1 Q E= 4πε0 R2 E.g.: Find the field for a long st. wire of charge. E=

E Gaussian h Surface

E E

∫ EdA = E=

λh λh , E(2πrh) = ε0 ε0

λ × r, Where, λ = Line Change / Unit Length 2πε0

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14

Electromagnetic Field

Example: If A = ρ sin ϕ μp + ρ2 μp , and L is the contour of fig. given below, the circulation y

∮A. dL

2

c

L

1

C −2

d −1

b

a 1

2

x

Solution: ∮ A. dL = ( ∫ + ∫ + ∫ + ∫ ) A. dL C

ab

bc

cd

da

Along ab, dϕ = 0, ϕ = 0 b

A. dL = 0 ∫ A. dL = 0 a

Along bc, dρ = 0 A.dL = ρ3 dϕ c

π

∫ A. dL = ∫ ρ3 dϕ = 23 π = 8π b

0

Along cd, dϕ = 0, ϕ = π, A.dL = 0 d

∫ A. dL = 0 C

Along da, dρ = 0, A.dL = ρ3dϕ a

0

∫ A. dL = ρ3 ∫ dϕ = (1)3 (−π) = −π d a

𝜋

∫ A. dL = 0 + 8π + 0 − π = 7π d

Example: Let J =

800 sin θ r2+4

× Ur A/m2 the total current flowing in a portion of spherical surface having

radius r = 0.8 bounded by 0.1 π < θ < 0.3π, 0 < ϕ < 2π will be Solution: I = ∫ ∫ J. n d s 2π

S 3π

∫ ∫ 0

0.1π

800 sin θ × (0.8)2 sin θ dθ dϕ = 154.8 A (0.8)2 + 4

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15

Electromagnetic Field

Example: In a certain region where the relative permitivity is 2.4, D = 2ux − 4uy + 5uz nc⁄m2 . Polarization = ? Solution: D = ε0 E + P ⇒ where D = ε0 εr E ε0 E = D/εr D D p = D − = (εr − 1) εr εr 2.4 − 1 p = (2ux − 4uy + 5uz ) 2.4 p = 1.2ux − 2.3uy + 2.9uz nc⁄m2 Electric Potential The scalar electric potential is defined using fundamental ideas of force and work related to the electric field. When a charge is allowed to move due to force in the electric field, work is said to be done as expressed below. Work, W = (Force) (Displacement) dW = −F. dI = (QE) (Displacement) = QE. dl The total work done in moving a point charge Q from A to B is, B

W = −Q ∫ E. dl A

Where negative sign indicates that the work is being done by an external agent. Work done per unit charge is potential difference. Potential difference between two points is difference of absolute potentials at the two points. Absolute Potential The potential at any point is the work per unit charge required to bring a unit charge form infinity to the point. In simple, it is potential difference between any point and a reference point at infinity. Due to a point charge, Q Absolute potential at a point r = a is Va = Q/4π ε a Absolute potential at a point r = b is Vb = Q / 4π ε b Potential difference between ‘a’ and ‘b’ is Vab = Va – Vb Q 1 1 = ( − ) 4πε a b The potential, only depends on the distance between points ‘a’ and ‘b’ and the point charge, regardless of path between ‘a’ and ‘b’. In moving a charge along a closed path in electrostatic field, total work done is zero. Thus, ∮ E ∙ dl = 0 L

By applying stoke’s theorem ∮ E ∙ dl = ∫(∇ × E). ds = 0 L

S

∇×E =0 It is one of the four Maxwell’s equations. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

16

Electromagnetic Field

  

Any surface on which the potential is same throughout is known an equipotential surface. The direction of E is everywhere normal to the equipotential surface. No work is done in moving a charge from one point to other point along equipotential surface.

The lines of constant potential are always perpendicular to the electric field intensity. Electric scalar potential is scalar. Electric field intensity and electric potential are related as, E = − ∇ V. The negative sign shows that the direction of E is opposite to the direction in which V increases; E is directed from higher to lower level of V. An electric flux line is an imaginary path or line drawn in such a way that its direction at any point is the direction of the electric field at that point. Example: Two point charges −4 μC and 5 μC are located at (2, −1, 3) and (0, 4, −2), respectively. Find the potential at (1, 0, 1), assuming zero potential at infinity. Solution: Let, Q1 = −4 μC, Q2 = 5 μC Q1 Q2 V(r) = + + C0 4πε0 |r − r1 | 4πε0 |r − r2 | If V(∞) = 0, C0 = 0, |r − r1 | = |(1, 0, 1) – (2, − 1, 3)| = |(−1, 1, −2)| = √6 |r − r2 | = |(1, 0, 1) – (0, 4, −2,)| = |(1, −4, 3)| = √26 Hence, −4 5 10−6 [ ] + V(1, 0, 1) = −9 10 4π × 36 π √6 √26 = 9 × 103 (−1.633 + 0.9806) = − 5.872 kV In electrostatics, charges are considered to be stationary. This certainly does not mean that charges cannot move. A conductor is a material that allows free movement of charge within its volume. In other words, if a charge is introduced into a conductor, it can move freely until something prevents it from moving. This something may be an electric field or the surface of the conductor. The movement of charges is merely a mechanism to reach the steady state. After charges have reached their final state, the conductor has no effect on the charges.  Conductors in electrostatic field are said to be perfect conductors. A perfect conductor (σ = ∞) can not contain an electrostatic field within it. E = 0, ρv = 0, Vab = 0 inside a conductor.  Unlike conductors, dielectrics are materials in which charges are not free to move.  A perfect dielectric is a material which has bound charges but no free charges.  A material is linear if a particular property like permittivity does not change when the fields are changed.  A homogeneous material is a material whose physical properties do not vary from point to point in space.  An isotropic material is one whose properties are independent of direction in space.  A linear, homogeneous, isotropic material is called simple material. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

17

Electromagnetic Field

Divergence of Current Density and Relaxation Charges in conductors under static conditions are distributed on the surface of the conductor in such a way that the potential energy in the system is minimum. If charges are placed in the interior of a conductor, they will move to the surface. This motion of charges constitutes a current. Since the current ceases, once the charges are static on the surface, the current is a transient current. Example: A block of silicon made in the form of a sphere of radius 100 mm is given. The conductivity of silicon is 4 × 10-4 S/m, its relative permittivity is 12 and both are constant. Suppose that by some means a volume charge density ρv = 10-6 C/m3 is placed in the interior of the sphere at t = 0; calculate (a) The current produced by the charge as they move to the surface (b) The time constant of the charge decay in the silicon (c) The divergence of the current density during the transient Solution: At time t = 0, the charges start moving toward the surface. The charge density must satisfy the continuity equation at all times, ∇. J = −

∂𝛅𝐯 ∂t

In addition, the current in the conducting material must satisfy Ohm’s law, J = σE and ρv Gauss’s law ∇. E = ℰ σρv ∂ρv ∴ ∆. σE = =− ε ∂t ∂δv σ ⇒ + ρv = 0 ∂t ε This is a homogenous linear differential equation. By separately variables, ∂ρv −σ σt = ∂t and integrating ln ρv = − + ln ρvo ρv ε ε −t/τ ρv = ρ0 e where τ = ε/σ This equation shows that the introduction of charge at some interior point of material results in a decay of volume charge density 𝛒𝐯 . At a radius ‘R’, the total current crossing the surface defined by the sphere of radius ‘R’ is d(QR ) 4πR3 d −t ( ) [ρv (t)] = 15.8 R3 exp [ ] I R, t = − =− dt 3 dt 2.65 × 10−7 The current depends on the location and increases with the radius. Therefore, it is not constant in space or time. The time constant of the charge decay is ε/σ. This time constant depends on material alone and is called the relaxation time. A long time constant (poor conductors) means charges take longer to ‘relax’ or to reach the surface. A short time constant (good conductor) means the charges quickly reach their static state (at the surface). In the present case, τ = ε/σ = 2.65 × 10−7 sec the divergence of the current density is − ∂ρv −∂ −t [ρ0 e ϵ ] = 3.77 exp (−t⁄ ) ∇.J = = − 2.65 × 10−7 ( ⁄σ) ∂t ∂t The divergence of the current density is clearly not zero, but decays with time. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

18

Electromagnetic Field

Boundary Conditions If the field exists in a region consisting of two different media, the condition that the field must satisfy at the interface separating the media are called “Boundary Conditions”. (a) Dielectric – Dielectric Boundary Conditions: The tangential component Et of field undergoes no change on the boundary and it is said to be continuous across the boundary, E1t = E2t D1t D2t ⇒ = ε1 ε2 Hence Dt undergoes some change across the boundary and said to be discontinuous. For normal component, D1n − D2n = ρS Where ρS is the free charge density placed at the boundary. If ρS = 0 then D1n = D2n ε1 E1n = ε2 E2n So the normal component of E is discontinuous across the interface. tanθ1 εr1 Law of refraction, = tan θ2 εr2 Where θ1 and θ2 are angles which E1 and E2 makes with the normal at the interface. (b) Conductor – Dielectric Boundary Condition: No electric field, E may not exist inside conductor that is E = 0, ρV = 0 Et = 0 = Dt Dn = ρS ε2 En = ρS Thus, an electric field must be external to the conductor and must be normal to the surface. (c) Conductor – Free Space Boundary Condition: Dt = εo Et = 0 Dn = εo En = ρS Biot–Savart’s Law The Biot – Savart’s Law is used to compute the magnetic field generated by a steady current, i.e., a continuous flow of charges, for example through a wire, which is constant in time and in which charge is neither building up nor depleting at any point. The equation is as follows: μ0 Idl × ̂r μ0 Idl × r B=∫ or (equivalently) B = ∫ (in SI units) 2 4π r 4π r 3 dl

I

α r

P dB (Inside)

Magnetic Field dB at P Due to Current Element I dl : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

19

Electromagnetic Field

Where, I is the current, dl is a vector, whose magnitude is the length of the differential element of the wire, and whose direction is the direction of conventional current, B is the net magnetic field, μ0 is the magnetic constant, ̂r is the displacement unit vector in the direction pointing from the wire element towards the point at which the field is being computed, r = rr̂ is the full displacement vector from the wire element to the point at which the field is being computed, the symbols in boldface denote vector quantities. Magnetic field B, at point P due to a straight line conductor, μo I (cos α2 − cos α1 )aϕ B= 4πρ Where I is current of conductor and ρ is perpendicular distance. B α2 I A α1 ρ

P Field at P Due to a Straight Filamentary Conductor When the conductor is semi infinite, α1 = 90 and α2 = 0 μo I B= a 4πρ ϕ For a infinite conductor α1 = 180 and α2 = 0 μo I B= a 2πρ ϕ Ampere's Circuit Law In classical electromagnetism, Ampère's circuit law, discovered by André-Marie Ampère in 1826, relates the integrated magnetic field around a closed loop to the electric current passing through the loop. It states that the line integral of B around a closed path is the same as the net current Ienc enclosed by the path multiplied by μ permeability. In SI units (the version in cgs units is in a later section), the "integral form" of the original Ampère's circuital law is: ∮ B. dl = μ0 Ienc C

By applying stoke’s Theorem to the right side of equation,

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20

Electromagnetic Field

∮ B. dl = ∫(∇ × B). ds = μ0 Ienc C

S

But Ienc = ∫ J. ds S

∇ × B = μo J

Where  ∮C is the closed line integral around the closed curve C.  B is the magnetic field in tesla.  ". " is the vector dot product.  dl is an infinitesimal element (differential) of the curve C (i.e., a vector with magnitude equal to

   

  

the length of the infinitesimal line element, and direction given by the tangent to the curve C, see below), ∬ denotes an integral over the surface S enclosed by the curve C (see below). The double integral sign is meant simply to denote that the integral is two-dimensional in nature. μ0 is the magnetic constant also called the absolute permeability of free space. Jf is the free current density through the surface S enclosed by the curve C dS is the vector area of an infinitesimal element of surface S (that is, a vector with magnitude equal to the area of the infinitesimal surface element, and direction normal to surface S. The direction of the normal must correspond with the orientation of C by the right hand rule, see below for further discussion), Ienc is the net free current that penetrates through the surface S. ∇ × H = J ≠ 0, that is a magnetostatic field is not conservative. This is one of the four Maxwell’s equation.

An isolated magnetic charge does not exist. Thus the total flux through a closed surface in a magnetic field must be zero; that is, ∮ B. ds = 0 By applying divergence theorem, ∮ B. ds = ∫ ∇. B dv = 0 S

V

or ∇. B = 0 This is the fourth Maxwell’s equation.

The Magnetic Vector Potential (A) The magnetic vector potential is defined based on the divergence free condition of ‘B’, the magnetic flux density. The definition of ‘A’ is based entirely on the mathematical properties of the vector ‘B’, not on its physical characteristics. ∴‘A’ is viewed as an auxiliary function rather than fundamental field quantity. Since the magnetic vector potential is a vector function, both its curl and divergence must be specified. The magnetic vector potential does not have a simple physical meaning in the sense that it is not a measurable physical quantity like B or H. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

21

Electromagnetic Field

Because the magnetic vector potential relates to the magnetic flux density through ‘A’, it is at right angles to the magnetic flux density ‘B’. The magnetic vector potential is always in the direction of the current (and perpendicular to B). It is a sort of current distributed in space. It is sometimes even called a “fuzzy current” since it is spread around the current, as shown in below figure A I A

∇ . B = 0, ∴ B = ∇ × A ∇. A = 0 and is called the coulomb guage for static fields. Ampere’s law is ∇ × H = J ∴ ∇ × B = μJ Since, B = ∇ × A, ∇ × ∇ × A = μ J ∇(∇. A). ∇2 A = μ J Taking ∇. A = 0 −∇2 A = μ J i.e., ∇2 A = − μ J The above equation is a vector Poisson’s equation. Energy Density of Electric and Magnetic Fields Electric and magnetic fields store energy. In a vacuum, the (volumetric) energy density (in SI units) is given by ε0 1 2 U = E2 + B 2μ0 2 The electric (Potential) energy in an electrostatic field is given by 1 1 WE = ∫ D. E dv = ∫ εE2 dv 2 2 The energy in an magnetostatic field in given by 1 1 WM = ∫ B. H dv = ∫ μH 2 dv 2 2 Capacitance of capacitor (Parallel Plate) is defined as the ratio of the magnitude of the charge on one of the plate to the potential difference between them. Q C= V Q

q 1 Q2 1 2 1 Wcharging = ∫ dq = = CV = Wstored = QV C 2 C 2 2 0

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22

Electromagnetic Field

Capacitance of Simple Systems Type Capacitance Resistance Comment d Parallel-Plate Capacitor εA/d A: Area d: Distance σA a 1 2πεl Coaxial Cable / a1 : Inner Radius ln ⁄a1 Cylindrical Capacitor a2 : Outer Radius ln(a2 /a1 ) 2πσl l : Length 1 1 4πεa a Concentric Spheres a1 : Inner Radius 1 2 a−b a2 : Outer Radius a2 − a1 4πσ 1 Sphere 4πεa a : Radius 4πσa Circular Disc 8εa a : Radius Example: The point charges −1 nC, 4 nC, and 3 nC are located at (0, 0, 0), (0, 0, 1), and (1, 0, 0), respectively. Find the energy in the system Solution: W = W1 + W2 + W3 = 0 + Q2 V21 + Q3 (V31 + V32 ) Q1 Q3 Q1 Q2 [ ] = Q2 + + 4πε0 |(0, 0, 1) − (0, 0, 0)| 4πε0 |(1, 0, 0) − (0, 0, 0)| |(1, 0, 0) − (0, 0, 1)| Q2 Q3 1 (Q1 Q2 + Q1 Q3 + ) = 4πε0 √2 1 12 12 = (−4 − 3 + ) 10−18 = 9 ( − 7) nJ = 13.37 nJ −9 10 √2 √2 4π 36π Alternatively, 3

1 1 W = ∑ Qk Vk = (Q1 V1 + Q2 V2 + Q3 V3 ) 2 2 k=1

Q2 Q1 Q1 Q1 Q3 Q2 Q3 Q3 Q2 [ ]+ [ ] + [ ] = + + + 2 4πε0 (1) 4πε0 (1) 2 4πε0 (1) 4πε0 (√2) 2 4πε0 (1) 4πε0 (√2) Q2 Q3 12 1 = (Q1 Q2 + Q1 Q3 + ) = 9 ( − 7) nJ = 13.37 nJ 4πε0 √2 √2 As obtained in the first solution. Stored Energy in Inductance The energy (measured in joules, in SI) stored by an inductor is equal to the amount of work required to establish the current through the inductor, and therefore the magnetic field. This is given by 1 Estored = LI2 2 The quantitative definition of the self inductance of a wire loop in SI units (weber per ampere known as henries) is Nϕ L = I Where ϕ denotes the magnetic flux through the area spanned by the loop, and N is the number of wire turns. The flux linkage thus is Nϕ = LI : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

23

Electromagnetic Field

Inductance of a Solenoid A solenoid is a long, thin coil, i.e., a coil whose length is much greater than the diameter. Under these conditions, and without any magnetic material used, the magnetic flux density B within the coil is practically constant and is given by B = μ0 NI/ l Where μ0 is the magnetic constant, N the number of turns, I the current and l the length of the coil. Ignoring end effects the total magnetic flux through the coil is obtained by multiplying the flux density B by the cross-section area A and the number of turns N ϕ = μ0 N2 IA/l From which it follows that the inductance of a solenoid is given by L = μ0 N2 A/l

Inductance of Simple Systems Coaxial Cable, High Frequency

Toroidal Core (Circular crosssection)

l a1 ln μ 2π a L = μ0 μr

a1 : Outer radius a : Inner radius l : Length L = Inductance (H) μ0 = Permeability of free space = 4π × 10−7 H/m μr = Relative permeability of core material N = Number of turns r = Radius of coil winding (m) D = Overall diameter of toroid (m)

N2 r 2 D

The Continuity Equation Consider figure below which shows an isolated volume, charged with a charge density ρv. No charge leaves or enters the volume and, therefore, the charge is conserved. This is a trivial example of conversation of charge.

ρv Now, consider figure below where we connect the volume through wire and allow the charge to flow through the wire to some other body (not shown) ds I

ρv The rate of decrease of charge in volume v is the current out of the volume. Because charges flow there is a current I in the wire. At the same time, the charge that flows out of the volume over a time dt (i.e., the charge that flows through the wire) is dQ. The time rate of decrease of charge in the volume ‘v’ is = dQ/dt. ∴ The rate of decrease of the charge in volume ‘v’ must equal the current out of the volume. I = −dQ/dt : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

24

Electromagnetic Field

This is the basic definition of current. The current is considered to be positive because it flows through the surface of the volume ‘v’ in the direction of ds. The total charge in volume ‘v’ is Q =∫v ρv dv dρv ∴ I = − d⁄dt ∫ ρv dv = ∫ dv v v dt The current I flowing out of the volume ‘v’ in terms of the current density J that flows through the surface enclosing the volume ‘v’ is I = ∮ J. ds S

∴ ∮ J. ds = ∫ − dρv ⁄dt dv v

The surface ‘s’ is a closed surface (encloses a volume). Thus, we can apply the divergence theorem to the LHS of the above equation to convert the closed surface integral to a volume integral. ∫ ∇ ∙ J dv = ∫ (− dρv ⁄dt) dv v

v

Since both integrals are taken over the same volume, we get ∇. J = − dρv ⁄dt This is the general form of the continuity equation. This expression holds at any point in space and is not limited to conductors. In the particular case of steady currents, the charge density ρ v does not vary with time. The rate of change of charge with time is zero and the charge decreases from volume ‘v’ must be replenished constantly to maintain a steady current. ∇. J = 0, If ρv = Constant The above equation means that a steady current must flow in closed circuits, it cannot end in a point because the divergence at that point would not be zero, invalidating the requirement of steady current. This also means that the total current entering any volume must equal the total leaving this volume. The steady current density is conservative, since ∮ E . dI = 0 i. e. , ∮ J⁄σ . dI = 0 ∴∇ ×J=0 The steady current density is solenoidal, Since, ∮ J . ds = 0 i. e. , ∇ . J = 0 ∴ The current density is an irrotational, solenoidal field.

Faraday’s Law Faraday found that the electromotive force (EMF) produced around a closed path is proportional to the rate of change of the magnetic flux through any surface bounded by that path. In practice, this means that an electrical current will be induced in any closed circuit when the magnetic flux through a surface bounded by the conductor changes. This applies whether the field itself changes in strength or the conductor is moved through it. Electromagnetic induction underlies the operation of generators, all electric motors, transformers, induction motors, synchronous motors, solenoids, and most other electrical machines. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

25

Electromagnetic Field

Faraday's law of electromagnetic induction states that: dΨ Vemf = − , dt Thus Vemf is the electromotive force (emf) in Volts ΨB is the magnetic flux in webers Thus a static magnetic field produces no current flow, but a time varying field, produces an induced voltage in a closed circuit and causes a flow of current. For the common but special case of a coil of wire, composed of N loops with the same area, Faraday's law of electromagnetic induction states that dΨ Vemf = −N dt Where Vemf is the electromotive force (emf) in Volts N is the number of turns of wire ΨB is the magnetic flux in Webers through a single loop An emf - produced field is non-conservative Motional EMF  When a conductor moves through a magnetic field an emf is produced in the conductor.  The charges in the conductor are carried along with the moving conductor and thus experience a magnetic force acting upon them which causes them to move inside the conductor.  As the conduction charges pile up at the end of conductor creating an electric field in the conductor.  The conduction electrons will stop piling up when the electric force on the interior conduction charges is equal to the magnetic force on those same charges so that the net force on the conduction charges is zero. B-Field + + + − L

Fr

Conductor v = Velocity of Conductor

Force Equilibrium +

Electric Force on FB Conduction Charge

E q−

v

v

− Built up of − − Conduction Electrons

−

FM

Magnetic Force on Conduction Charge

Motion in a Perpendicular Field



At Equilibrium, when the net force is zero. FE = FM qE = vB sin(θ)q E = vB sin(90o ) E = vB The charges that pile up create a voltage or emf across the length of the rod that is constant. E =

V L

, Vemf = EL = vBL

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26

Electromagnetic Field

Induced EMF Approach As the rod moves, it sweeps out area A. The change in magnetic flux can be found using Faraday's Law, L

Area A=Lx

x d d Vemf = − ϕM = − ∫ Bcos(θ′ )dA dt dt Bd − ∫ cos(180°)dA = Bd d dx A = B Lx = BL = BLv = dt dt dt dt The results are the same, a conductor moving in a perpendicular magnetic field produces an emf across its length Vemf = vLB  If the ends of the conductor are connected to an external circuit them the emf can act like a

battery. The end where the positive charges would build up would act like the positive pole of a battery. Alternately, the flow of current is in the same direction as the force that magnetic field exerts on positive conduction charges due to the conductors motion.

Maxwell’s Equations Static Integral Form Differential Form ∇ × H = Jc ∮ H. dl = ∫ J. ds L

S

∮ E. dl = 0

Time - Varying Differential Form Integral - Form ∇ × H = Jc + JD ∂D ∮ H. dl = ∫ (JC + ) . ds ∂t L

∇×E = 0

∇ × E = − ∂B⁄∂t

L

L

∮ D. dS = ∫ ρv dv

∇. D = ρv

∇. D = ρv

V

∮ B. dS = 0 S

S

−∂ ∮ E. dl = ∫ B. ds ∂t ∮ D. ds = ∫ ρv dv S

∇. B = 0

∇. B = 0

S

V

∮ B. ds = 0

∇ × H = Jc + JD is modified Ampere’s law ∇ × E = − ∂B⁄∂t is Faraday’s law ∇ × E = 0 conservative nature of electrostatic field ∇. D = ρv is Gauss’s law for electric field ∇. B = 0 is Gauss’s law for magnetic field (Non-existence of magnetic monopole) D = εE Electric flux density B = μH Magnetic flux density JC = σE Conduction current density, (this relation is referred to as Ohm’s law) JD = ∂D⁄∂t Displacement current density : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

27

Electromagnetic Field

σ is conductivity expressed in Siemens/ meter μ is permeability expressed in Henry / meter ε is permittivity expressed in Farad / meter  For a fields to “Qualify” as an electro magnetic field, it must satisfy all four Maxwell’s equations.  The ratio of conduction current density of displacement current density is referred to as loss tangent, i.e., loss tangent = (Jc / JD) = (σ / ωε).  If (σ / ωε) > > 1, the medium is referred to as high loss medium.  If (σ / ωε) < < 1, the medium is referred to as low loss medium.  If

σ

ωε

= 0, Loseless medium

 The frequency at which (Jc/ JD) is equal to one is referred to as Transition Frequency, fq i.e., When f = fq, Loss Tangent is one. fq = σ/2πε is called Transition Frequency. From continuity of current equation ∇. JC = − ∂ρv ⁄∂t And Gauss’s law, ∇. D = ρv And Ohms law, JC = σ E, We can find, ρv = ρV0 e−(σ⁄ε)t ρv = ρV0 e−t⁄τ Where τ = ε/σ and is called relaxation time constant.

The above relaxation indicates that at an interior point of any conductor if we place a charge, it decreases to (1/e) times of initial charge in one relaxation time constant.  When operating frequency ‘f’, in a medium with σ, μ, ε is greater than fq, then that medium is regarded as a Dielectric. f >> fq Dielectric  When operating frequency ‘f’, in a medium with σ, μ, ε is less than fq, then that medium is regarded as a Conductor. f 0, εr 1 = 4) has electric field 3ax̂ + 5aŷ + 2âz v/m. What will be the electric field in medium 2 (z < 0, εr = 16). (A) 3ax̂ + 5aŷ + 0.5aẑ (B) 3ax̂ + 5aŷ − 0.5aẑ (C) −2ax̂ + 5aŷ + 0.5aẑ (D) 2ax̂ + 5aŷ + 0.5aẑ A lossy material has μ = 5μ0 , ε = 2ε0 , The phase constant is 10 rad/m at 5 MHz. The loss tangent is (A) 2913 (C) 2468 (B) 1823 (D) 1374 Which one of the following is another form of Faraday’s law (A) ∇ − J = − ⃗⃗⃗⃗⃗

∂A (B) ⃗E = − ∂t

4.

Y 2.5

∂PV ∂t

5

X

⃗⃗⃗⃗⃗ ∂A

(C) ∇. ⃗A = −μE ∂t (D) None of these

If v = 10xyz − z + 5x V find the E field on V = 100 volt surface at the point x = 2, y=3 (A) −50.76ax̂ − 30.51aŷ − 59aẑv/m (B) −27.726ax̂ − 30.51aŷ − 59aẑv/m (C) −50.76ax̂ − 22.61aŷ − 59aẑv/m (D) −27.32ax̂ − 30.51aŷ − 11.22aẑv/m

μ0 2.25 ε0

(A) 1.564 (B) 2.468 8.

μ 0 ε0

Z

(C) −3.911 (D) 4.389

An electrostatic field is said to be conservative when (A) The divergence of the field is equal to zero (B) The curl of the field is equal to zero ∂B (C) The curl of the field is equal to − ∂t (D) The laplacian of the field is equal to με

∂2 E ∂t2

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41

Electromagnetic Field

9.

Two long parallel wires in free space are separated by a distance R and carry currents of equal magnitude but opposite in direction. At any general point, the Z – component of

12.

The electric field strength at a far – off point, P, due to a point charge, +q, located at the origin, O, is 100 mV/meter. The point charge is now enclosed by a perfectly conducting hollow metal sphere with its centre at the origin, 0. The electric field strength at the point, P, (A) Remains unchanged in its magnitude and direction (B) Remains unchanged in its magnitude but reverse in direction (C) Would be that due to a dipole formed by the charge, +q, at 0 and – q induced (D) Would be zero

13.

Which of the following field equations indicate that the free magnetic charges do not exist 1 (C) ∇ . H = 0 (A) H= ∇ × A

Z

I

I

d1

P

d2

Y

R

X

(A) The magnetic vector potential is μ0 I 4π

ln(d2 2 /d1 2 )

(B) The magnetic induction is μ0 I 2π

(d2 /d1 )

(C) The magnetic induction is zero (D) The magnetic vector potential is μ0 I 4π

10.

11.

(d2 2 /d1 2 )

On either side of a charge – free interface between two media (A) The normal components of the electric field are equal (B) The tangential components of the electric field are equal (C) The normal components of the electric flux density are equal (D) The tangential components of the electric flux density are equal Vector potential is a vector (A) Whose curl is equal to the magnetic flux density (B) Whose curl is equal to the electric field intensity (C) Whose divergence is equal to the electric potential (D) Which is equal to the vector product E x H

μ

(B) H = ∮

I dI×R 4πR2

(D) ∇ × H = J

14.

The incoming solar radiation at a place on the surface of the earth is 1.2 kW/m2 . The amplitude of the electric field corresponding to this incident power is nearly equal to (A) 80 mV/m (C) 30 V/m (B) 2.5 V/m (D) 950 V/m

15.

Given ⃗V = x cos2 y î + x 2 ez ĵ + z sin2 y k̂ and S the surface of a unit cube with one corner at the origin and edges parallel to the coordinate axes, the value of the integral ∫ ∫ ⃗V. n̂ ds is ________ s

16.

For a uniformly charged sphere of radius R and charged density ρ, the ratio of magnitude of electric fields at distances R/2 and 2R from the centre, i.e., E(r=R /2) E (r=2R)

is __________

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42

Electromagnetic Field

Answers Keys & Explanations Assignment 1 1. [Ans. A] Maxwell’s divergence equation, ρ ∇∙E = εo 2.

4.

5.

6.

[Ans. B] The new system has two capacitor in series, let the value of one capacitor is ‘C’. εA ′ εA 2ε A C = ,C = = = 2C t t/2 t New capacitance = 2C in series with 2C = C [Ans. D] ρs E = 2ε0

2 × 10−9 = −36π az V/m 2 × 8.854 × 10−12

8.

10.

[Ans. A] The E field vanishes everywhere inside the sphere.

11.

[Ans. B] Ampere’s law – Magnetic flux density at a point. Biosavart’s law – Force due to a current carrying conductor. Coulomb’s law – Force on a charge. Gauss’s law – Electric flux density at a point

12.

[Ans. C] H 2π r =

13.

14.

[Ans. C] Surface Charge Density = |D| = 2√1 + 3 = 4 C/m2

[Ans. B]

[Ans. C] Ndϕ = 100 × (3 t 2 − 2) × 10−3 dt [magnitude EMF] at t = 4 sec EMF = 100 × (48 − 2) × 10−3 ⇒ 100 × 46 × 10−3 = 4.6 V EMF = −

1 =1F 4πε0

[Ans. B] ∇∙B = 0 It does not pertain that magnetic field is perpendicular to the electric field.

Iπr 2 Ir ⇒H = 2 2πR2 πR

M = K √L1 L2 M 0.015 ⇒K= = = 0.5 √0.09 × 0.01 √L1 L2

[Ans. B] C = 4πε0 R = 4πε0 ×

7.

= 0.1 × π(10−3 )2 10−5 (2 ax + 2ay ) = 2π × 10−12 (ax + ay )Wb/m2

[Ans. C] Inductance of solenoid, μo N2 A L= l So L ∝ N2 When turns are double, inductance will be 4 times.

=

[Ans. B] T = m × B = IA az × B = IA az × (2 ax − 2 ay + az )10−5 = IA 10−5 (2ay + 2ax )

[Ans. B] 1 θ = sin−1 √ = sin−1 0.5 , θ = 30o 4

3.

9.

15.

[Ans. B] z A +Q C

−Q B

45°

45°

E2

E1

x

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43

Electromagnetic Field

E1 = Field due to A at point C E2 = Field due to B at point C both are equal in magnitude and make angle of 90°, their resultant lie in –z direction. 16.

4.

V = 10xyz − z + 5x ⃗E = −∇v 100 = 10xyz − z + 5x ⇒ −(10yz + 5)ax̅ − 10xzây + (1 − 10xy)aẑ 5x − 10 z= 1 − 210xy At point P(2,3, z), z = 1.52 ⃗E = −50.76 ax̂ − 30.51 aŷ − 59 aẑ V/m

[Ans. A] q2 r2 10−18 = 9 × 109 × = 9 × 10−3 N 10−6

F = 9 × 109 ×

17.

[Ans. B] q It 2 × 10−6 × 6 V = = = = 4V C C 3 × 10−6

18.

[Ans. B] 1 1 50 2 E = LI2 = × 0.4 × ( ) = 20 J 2 2 5

19.

5.

[Ans. D] For a < P < b & 0 < z < d d b μ0 A ϕ= ∫ ∫ sin(ωt) cos(β) dp dz P 0 a μ0 A b = ln ( ) β a −wμ0 A ln(b⁄a) sin(βd) cos(ωt) = β

6.

[Ans. C] At minimum

[Ans. C] By Ampere’c circuit law ∮ H. dl = Ienc ⇒ 3I + 2I − I = 4I

2π ⇒  = 6π λ S−1 5−1 4 |Γ | = = = = 0.67 S+1 5+1 6 η2 − η0 |Γ| = 67ej0.6π [∵ Γ = |Γ|ej ] η2 + η0 1 + 0.67ej0.6π η2 = η0 ( ) t − 0.67ej0.6π 0.67ej108° = −20 + j0.6365 1 − 20 + j 6365 η2 = 377 [ ] 1 + 20j0.6365 ≃ 377[0.3007 + j0.6899] η2 = 113.36 + j260.09 Ω

⃗⃗⃗⃗⃗⃗ D ̂ = ⃗⃗⃗⃗⃗⃗⃗ D2n 1n = 4 to (2az ) ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ E2n = 0.5aẑ, E2 = E2t + ⃗⃗⃗⃗⃗⃗ E2n [Ans. B] Loss Tangent

σ =x ωε

με σ 2 β = ω√ [√1 + ( ) + 1 ] 2 ωε 2π × 5 × 106 5 × 2 √ [√1 + x 2 + 1] 3 × 108 2 σ ⇒x= = 1823 ωε ⇒ 10 =

3.

[Ans. B] ⃗ ∂B ∂t ⃗⃗⃗⃗⃗ −∂ ∂A ⃗ = ∇×E [∂ × ⃗A], ⃗E = − ∂t ∂t ⃗ =− ∇×E

( + π) = 4λ 2β

Β=

Assignment 2 1. [Ans. A] ⃗⃗⃗⃗⃗ E1t = 3ax̂ + 5aŷ = ⃗⃗⃗⃗⃗ E2t

2.

[Ans. A]

7.

[Ans. A] Since a > b the dominant mode is TE10 C 3  108 In face space fc = = = 3 GHz 2a 2 0.05 η0 377 = = 406.7 Ω ⇒ η1 = 2 2 3 √1 − ( ) √1 − (fc ) 7 f0 c In dielectric medium fc = ≅ 2 GHz 2a√ϵr η0 377 η= = = 251.33 Ω √ϵr √2.25 251.33 η2 = = 259.23 Ω 2 2 √1 − ( ) 8

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44

Electromagnetic Field

Reflection Coefficient Γ =

η2 − η1 η2 + η1

μI [ln [L + √L2 + d12 ] − ln d1 ] 2π μI [ln[2L] − ln d1 ] , L >> d1 = 2π Vector magnetic potential A2 due to long wire RS: μI A2 = − [ln[2L] − ln d2 ] 2π Due to both wires, μI A = A1 + A2 = [ln d2 − ln d1 ] ⇒

259.23 − 406.7 = −0.22 259.23 + 406.7 1 + |Γ| 1 + 0.22 S = = = 1.564 1 + |Γ| 1 − 0.22 ⇒

8.

[Ans. B] An electrostatic field with electric field, ⃗E is said to be conservative, if the closed line integral of the field is zero. i.e., ⃗ . ⃗dl = 0………..(1) ∮E Applying stoke’s theorem, equation (1) becomes ⃗ = 0, i.e., the curl of the field, E ⃗ is ∇×E

2π

μI d2 μI d22 = ln [ ] = ln [ 2 ] 2π d1 2π d1 10.

[Ans. B, C] This question is regarding the boundary conditions at the interface between two media (1) and (2) For a charge free interface between two media with dielectric constant ε1 and ε2 (i) The tangential components of the electric field are equal. i.e., Etan1 = Etan2 or E is continuos at the boundary (ii) The normal components of electric flux density D are equal. i.e., DN1 = DN2 or DN is continuous at the boundary. Dtan1 ε1 EN2 It may be noted that = = Dtan2 ε2 EN1

11.

[Ans. A]

equal to zero. ⃗⃗ ∂B

Note that ∇ × ⃗E = − ∂t for time varying fields. Also note that laplacian of the field ⃗E is given by ⃗ ∂2 E

∇2 ⃗E = με ∂t2 , which is the wave equation of an electromagnetic wave. 9.

[Ans. A] The magnetic vector potential A, at point ‘P’ in the xy plane (z=o) is derived below (see Fig.) Taking an elemental current element I ⃗dz, Vector magnetic potential A1 due to the long wire PQ: Q z

M

I

Vector magnetic potential, ⃗A is a vector whose curl is equal to the magnetic flux ⃗ density, B

I

z

⃗⃗ ∇ × ⃗A = ⃗B = μ H

R o

y d1

Note that the unit of ⃗A is web/meter.

d2 P

12.

R

x P

N L

A1 =

μI dz ∫ , R = √d12 + z 2 4π R −L L

=

L

μI dz μI ∫ = ln [√z 2 + d12 ] 2π √d12 + z 2 2π 0

[Ans. D] E = 100 μ V/m is replaced by a 100 mV/m Note that the data either 100 μV/m has no effect on the field strength required at the point ‘P’. Therefore for the present question also the electric fields strength at ‘P’ would be zero.

0

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45

Electromagnetic Field

13.

[Ans. C] Gauss’s law for magnetic fields states that the electric flux flowing through the closed surface is equal to zero, as free magnetic charge do not exit and magnetic flux lines are closed. ⃗ =0 ∴ ∮ ⃗B. ⃗⃗⃗⃗ ds = 0 or ∇. B

⇒ ∫ sin2 y dy ∫ dx = ∫

sin 2y 1 −1 1 ] [x] = [ ] [y − 2 0 0 2 1 sin(2) 1 sin(2) = [1 − ] (−1) = − [1 − ] 2 2 2 2 I3 through OTSM = − ∬ Vy dx dz, y = 0, x = 0 to − 1, z = 0 to 1

s

⃗⃗ , ∇. μH ⃗⃗ = 0 As ⃗B = μH In homogeneous medias μ is independent of position. ∴ ∇ . ⃗⃗H = 0 14.

−1

x3 = −[x e dx dz] = − [ ] [ex ]10 3 0 1 1 = − [− ] [e1 − 1] = [e1 − 1] 3 3 I4 through PQRN = −I3 I5 through OTQP = ∬ xcos2 y dy dz , x = 0, y = 0 to − 1, z = 0 to 1 =0 I6 through MSRN = − ∬ xcos2 y dy dz , x = −1, y = 0 to − 1, z = 0 to 1 2 2

[Ans. D] Power density, P = 1.2 kW/m2, on the surface of the earth where η = η0 = (120π)Ω The relation between E and P is given by E2 =P 2η The amplitude, E of the electric field is given by

1

0

[Ans. *] Given ⃗V = x cos2 y î + x 2 ez ĵ + z sin2 y k̂

z

S

R Q

T

N

M O x

P

y

⃗ . n̂ dS The given integral : I = ∬s V I1 through OMNP = ∬ Vz dx dy, z = 0, x = 0 to − 1, y = 0 to 1 = 0 , as Vz = 0 at z = 0 I2 through TQRS = ∬ Vz dx dy, z = 1, x = 0 to − 1, y = 0 to 1 =∬ sin2 y dy ∫ dx

0

1 sin 2y 1 1 1 sin(2) = [y + ] [z]0 = [1 + ] [1] 2 2 0 2 2 1 sin(2) = [1 + ] 2 2 I = I1 + I2 + I3 + I4 + I5 + I6 1 sin(2) 1 sin(2) = − [1 − ] + [1 + ] 2 2 2 2 1 = sin(2) 2

= √2 × 120π × 1.2 × 103 V/m = 951 V/m ≈ 950 V/m

⃗V = x cos2 y ⃗ax + x 2 e2 ⃗ay + z sin2 y ⃗az = Vx ⃗ax + Vy ⃗ay + Vz ⃗az Vx = cos2 y, Vy = x 2 e2 and Vz = z sin2 y

1

1 + cos 2y ⇒ ∬ cos y dy dz = ∫ dy ∫ dz 2 2

E = √2η0 P

15.

1 − cos 2y dy ∫ dx 2

16.

[Ans. *] Given the sphere of radius R with uniform surface charge density ρ C/m2 . Magnitude of the electric field at 2R from the centre of the sphere can be obtained by using ⃗⃗⃗⃗ = Charge Enclosed ⃗ . ds Gauss’s law: ∮ D D (at r = 2R) × 4π(2R)2 = ρ4πR2 ρ D(at r = 2R) = 4 D(at r = 2R) 1 ρ ∴ E(at r = 2R) = = ε ε4 R R But D (at r = ) = 0 or E (at r = ) = 0 2 2 Because no charge is enclosed within the imaginary spherical surface of radius R/2 E(r = R/2) ∴ =0 E(r = 2R)

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46

“People rarely succeed unless they have fun in what they are doing.” …. Dale Carnegie

Module Test 1.

If divergence is represented by D and curl is represented by C, which of the following is true?

(A) D = 0 , C = 0 (B) D ≠ 0 , C = 0 2.

3.

4.

A parallel plate capacitor has plate area of 5 cm2 and a separation of 3 mm and a voltage of 50 sin 103 t V applied to its plates. Assuming ε = 2 ε0 , the displacement current is, (A) 147.4 cos103 t nA (B) 150 cos 10 t A (C) 180 cos 103 A (D) 155 cos 10−3 mA Which of the following expression is not Maxwells’s equation for time varying fields? ∂B (A) ∇ × H = JC + JD (C) ∇.E= − ∂t

(D) ∮ B. ds = 0

If J = (zy ax + xz ay + z 3 az ) cos 104 t A/m2, then volume charge density ρv is, ( given ρv (x, y, 0, t)= 0) (A) ρv = −0.3z 2 sin 104 t mc/m3 (B) ρv = 0.6z 2 sin 104 t c/m3 (C) ρv = 0.3z 2 sin 104 t mc/m3 (D) ρv = 0.5z 2 sin 104 t mc/m3 Statement for Linked Answer Q.No. 6 & 7 A conducting bar can slide freely over two conducting rails a shown in figure.

(C) D = 0 , C ≠ 0 (D) D ≠ 0 , C ≠ 0

∇ × ⃗E = 0 implies that (A) E = 0 (B) E is a conservation field , and hence irrotational vector (C) E is a static and doesn’t change with time (D) E is a constant vector

(B) ∇.D = ρv

5.

P y

O V

B

6 cm x

Q

6.

If the bar is stationed at y = 8 cm and B = 4 cos 106 t az mWb/m2, the induced voltage in the bar is, (A) −19.2 sin 106 t V (B) 19.2 sin 106 t V (C) 19.2 cos 106 t V (D) −19.2 cos 106 t V

7.

If the bar slides at a velocity v = 20 ay m/s and B = 4 az m Wb/m2, the induced voltage in the bar is, (A) 4.8 mV (C) 48 mV (B) −4.8 mV (D) −48 mV

8.

Which law was given by Maxwell for the correction of the inconsistency of continuity equation for the time varying field δD (C) Faraday’s law (A) Ampere’s law δt

(B) Gauss’s law J

(D) None of these

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47

Module Test

9.

10.

11.

Maxwell’s Wave equation is −δB (A) ∇ × E = δt −δB δD (B) ∇ × E = ,∇ × H = δt δt 2 δ E (C) ∇. D = 0, ∇2 E = μ0 ε0 2 δt 2 δ E δ2 B (D) ∇2 E = μ0 ε0 2 , ∇2 B = μ0 ε0 2 δt δt In the source free wave equation ⃗ ⃗ ∂2 E ∂E ⃗ − μ0 ε0 μ ε ∇2 E − μ0 μσ =0 2 ∂t ∂t The term responsible for the attenuation of the wave is ⃗ ∂E (A) μ0 μσ ∂t ∂2 ⃗E (B) μ0 ε0 μ ε 2 ∂t ⃗ (C) ∇2 E ⃗ ⃗ ∂E ∂2 E (D) μ0 μ σ and μ0 ε0 μ0 ε 2 ∂t ∂t

Plane z = 10 m carries surface charge density 20 nc/m2 . What is the electric field at the origin? (A) −10âz V/m (C) 72 π âz V/m (B) −18 π âz V/m (D) −360 π âz V/m

15.

What is the value of total electric flux coming out of a closed surface? (A) Zero (B) Equal to volume charge density (C) Equal to the total charge enclosed by the surface (D) Equal to the surface charge density

16.

Two extensive homogenous isotropic dielectrics meet on a plane z = 0. For z ≥ 0, εr1 = 4 and for z ≤ 0, εr2 = 2. A uniform electric field exists at z ≥ 0 as E1 = 5âx − 2ây + 3â z kW/m. What is the value of E2 in the region z ≤ 0?

17.

Which one of the following is the poisson’s equation for a linear and isotropic but inhomogeneous medium? (A) ∇2 E = −ρ/ε ⃗ (εV) = −ρ (C) ⃗∇. ∇ ⃗ V) = −ρv (D) ∇2 V = −ρ/ε (B) ⃗∇. (ε∇

18.

⃗ = 0 is based on Equation ∇.B (A) Gauss’s Law (B) Lenz’s Law (C) Ampere’s Law (D) Continuity Equation

⃗ =0 (C) ∇ × B

L

⃗⃗⃗⃗ = 0 ⃗ . ds (B) ∮ B S

(D) ∇. ⃗B ≠ 0

15a⃗x + 20a⃗ y N/m −20a⃗x + 15a⃗y N/m 20a⃗x + 15a⃗ y N/m −20a⃗x − 20a⃗y N/m

14.

If a vector field B is solenoidal, which of these is true? ⃗⃗⃗ = 0 ⃗ . dl (A) ∮ B

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 ⃗ = (3a⃗x + 4a⃗ y )? B (A) (B) (C) (D)

A potential field is given by V = 3x 2 y − yz. Which of the following in NOT true? (A) At the point (1, 0, −1), V and the electric field E vanish (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 normal to the equipotential surface V = − 8 at P is (−0.83x̂ + 0.55 ŷ + 0.07 ẑ)

12.

13.

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48

Module Test

Answer Keys & Explanations 1.

[Ans. D] Both curl and divergence is not zero.

2.

[Ans. B] ∇ × E = 0 implies that E is a conservative field and hence it is irrotational vector.

3.

[Ans. A] V − D = εE = ε d ∂D ε ∂V Jd = = ∂t d ∂t εA ∂V εA dV ( )= Id = Jd . A = d dt d dt 10−9 5 × 10−4 =2× × × 103 36 π 3 × 10−3 × 50 cos 103 t A = 147.4 cos 103 t nA

4.

[Ans. A]

5.

[Ans. A]

8.

[Ans. A] δD⁄δt is called displacement current & behaves the same way as physical current & hence the continuity equation must also include this current.

9.

[Ans. D]

10.

[Ans. A] ⃗ ∂E ∂t involves the parameter σ i.e., conductivity of the lossy medium that correspond to losses in EM wave propagation. The term μ0 μ σ

11.

∂V ∂V ∂V â + â + â ] ∂x x ∂y y ∂z z E = −[6xyâx + (3x2 − z)ây − yây ] E = −∇V = − [

So, E at (+1, 0 − 1) = [0 + 3ây − 0] = −3ây ≠ 0

− ∂ρv ∇. J = (0 + 0 + 3z ) cos 10 t = ∂t 2

4

⇒ ρv = ∫ ∇. Jdt = ∫ 3z 2 cos 104 t dt

6.

−3z 2 = sin 104 t + C0 104 ρv /z = 0 ⇒ C0 =0 So, ρv = −0.3 z 2 sin 104 t mc/m3

[Ans. B] According to Gauss law for magnetic ⃗ = 0 i.e., no existence of field ∇.B monopoles. ⃗⃗⃗⃗ = 0 (∵ Divergence theorem) ⃗ ds ∮B

[Ans. B]

∴ Divergence of ⃗B = 0 i. e. , ⃗B is solenoidal.

Vemf = − ∫ 0.08 0.06

∂B . ds ∂t

∫ 4(10−3 ) (106 ) sin 106 t dx dy

⇒ ∫

y=0 x=0

= 19.2 sin 106 t V 7.

[Ans. A] V = 3x 2 y − yz

[Ans. B] Vemf = ∫(v × B). dl 0

= ∫(vay × Baz ). dx ax = −vBl

12.

s

13.

[Ans. B] Force on current carrying conductor due to magnetic field ⃗B = 3âx + 4ây ⃗ = l(la⃗z × ⃗B) F Force per unit length ⃗ 1 = l(âz × ⃗B) F = 5[âz × (3â x + 4ây )] = 5[3â y − 4âx ] ⃗F1 = −20âx + 15â y N/m

x=l

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Module Test

14.

[Ans. D] ρs (−âz ) E= 2 ε0 20 × 10−9 × 4π × 9 × 109 = (−âz ) 2×1 = −360π âz V/m

15.

[Ans. C] Net flux coming out from a closed ⃗⃗⃗⃗ = Qenclosed total ⃗ . ds surface i.e., ∮ D charge enclosed within the surface this is Gauss’s law for electric field.

16.

[Ans. C] From the boundary condition Dn1 = Dn2 (Charge-free) ∴ ε0 εr1 Ez1 = ε0 εr2 Ez2 ∴ 4 × 3 = 2 × Ez2 ∴ Ez2 = 6âz

17.

[Ans. B] ⃗ .D ⃗ = ρv We known that ∇ ⃗ But D = εE ⃗ . (εE ⃗ ) = ρv ∴ ∇ but ⃗E = −∇V V-potential (Scalar) ∇. (−ε∇V) = −ρv Medium is inhomogeneous i.e., ε is function of dimensions x, y, z so we cannot take it out side.

18.

[Ans. A] Magnetic monopoles are not physically ⃗ =0 possible as per Gauss’s Law i.e., ∇.B

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Reference Books

Reference Books 1.

Electromagnetic Waves and Radiating Systems By E.C. Jordan and K.G. Balsain

2.

Engg Electromagnetic – By William Hayt

3.

Antenna And Wave Propagation By KD Prasad

4.

Microwave devices & circuits By Lio

5.

Schaum series for problems

6.

Principles of Elctromagnetics By Matthew N.O. Sadiku

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