Electromagnetism 1
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Success through conceptual learning 4A, Nandanavanam, Plot No.23A, S R Nagar X Road, Hyderabad-38
Electromagnetism-I In this module we are going study about •
Magnetic fields produced by currents
•
Ampere’s Law and Biot-Savart law
•
Motion of a charged particle in magnetic field
•
The force on a current in a magnetic field
•
The torque on a current carrying coil
•
Galvanometer, Ammeter and Voltmeter
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Electromagnetism-I
ELECTROMAGNETISM - I Sources of magnetic field : 1. The first magnetic phenomena observed were those associated with naturally occurring magnets fragments of iron ore found near the ancient city of magnesia (whence the term “magnet”). These natural magnets attract unmagnetized iron, the effect is most pronounced at certain regions of the magnet known as its poles. 2. In the year 1819, Danish scientist H.C. Oersted observed that a pivoted magnet (a compass needle) was deflected when in the neighborhood of a wire carrying current. Twelve years later Michael Faraday found that a momentary current existed in a circuit while the current in a nearly circuit was being started or stopped. Shortly afterward it was discovered that motion of magnet toward or away from the circuit would produce the same effect. 3. The work of Oersted demonstrated that magnetic effect could be produced by moving electric charges and that of Faraday and Henry showed that currents could be produced by moving magnets. 4. The magnetic field : a) An electric charge sets up or creates an electric field E in the space surrounding it. b) The electric field E exerts a force F = Eq on a charge q placed in the field.
5. 6. 7.
Similarly a) A moving charge or a current sets up or creates a magnetic field in the space surrounding it. b) The magnetic field exerts a force on the moving charge or current carrying conductor in the field. A static charge produces only electric field but moving charge produces both electric field and magnetic field in space. Moving charge current is the source of magnetic field. Magnetic field of a moving charge : v µ qv sin θ . The magnetic field due to moving charge q is B = 0 4π r 2 q
θ Where v is velocity of charge, r is distance from charge, θ is angle between r v and r . B µ q v × rˆ B= 0 4π r 2 The direction of B is perpendicular to plane containing v and r . 8. Ampere established relationship between the current in the conductor and strength of the magnetic field around the conductor. 9. Oersted’s experiment : A magnetic compass needle placed in the vicinity of a conductor carrying conductor aligned perpendicular to the conductor. 10. The direction of deflection of north pole of magnetic needle is given by Ampere’s swimming rule. 11. Ampere’s swimming rule : Imagine that a man is swimming along the conductor in the direction of the current facing a magnetic needle the north pole of the needle will deflect towards his left hand. i
S
N
1
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Electromagnetism-I
12. The magnetic lines of force around a current carrying conductor are concentric circles with their center lying on the conductor.
i
13. The direction of magnetic field around the current carrying conductor is given by a) Maxwell’s cork screw rule b) Right hand thumb rule i
B ↑
(a) The direction of lines of induction around a current carrying conductor (Right Hand Screw Rule)
i
(b) Right hand thumb rule. Thumb shows the direction of current and curled fingers show the direction of lines of induction
Maxwell’s cork screw rule : Imagine a right hand screw is advancing in the direction of current in a conductor. Then the direction of rotation of the screw gives the direction of magnetic lines of induction. Right hand thumb rule : Imagine that a current carrying conductor is held in the right hand palm such that the direction of current is indicated by the thumb. Then the other fingers indicate the direction of magnetic lines of induction. B 14. Ampere’s law : Ampere’s law states that the line integral of B.dl along a closed r i path round the current carrying conductor is equal to µ0i where i is the current dl through the surface bounded by the closed path and µ0 permeability of free space.
∫ B.dl = µ o i B 2π r = µ o i µ i B= o 2πr
15. Biot-Savart’s law : The magnetic induction at a point near a current carrying conductor is directly proportional to the length of the conductor, the strength of the current and sine of the angle made by the line joining the point with the conductor and inversely proportional to the square of the distance of the point µ i.dl.sin θ µ i dB = o . ; dB = o . 3 d lxr 2 4π 4π r r The direction of magnetic induction is given by the vector (d i × rc) . This law is valid only for small current segments. 2
P i
θ
d C
Q
r
dB
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Electromagnetism-I
16. Ampere's law and Biot-Savart law are equivalent but Ampere's law is more useful in some symmetrical conditions. A
17. The magnetic induction at a point P due to a conductor of finite length is µ i B = 0 (sin α + sin β) 4πr
α β
r
i
P
B
18. The magnetic field induction due to the current carrying conductor of infinite µ 2i µ i length is given by B = 0 ⋅ or B = 0 4π r 2πr µ i Magnetic field at one end of infinite long conductor is B = 0 4πr 19. The magnetic field induction at a point along the axis of a circular coil is µ 2πnir 2 B= 0 ⋅ 2 where n = number of turns, i = current in 4π (r + x 2 )3 / 2 the coil, r = radius and x = the distance of the point from the centre of the coil. µ 2niA B= 0 ⋅ 2 4π (r + x 2 )3 / 2 µ 2niA If x >> r, then B = 0 ⋅ 4π x 3 µ 2M i.e., B = 0 ⋅ where M = magnetic moment of the current loop. 4π x 3
i
r
B
i
r 2 + x2
r
B
x
i
20. (a) The magnetic field induction at the centre of a circular coil of n turns is given by relation µ ni B= 0 ⋅ 2 r µ iα (b) Magnetic field induction at the centre of an arc is B = 0 4πr
α
Ex-1. An equilateral triangular loop of edge d carries a current i. Determine the magnetic field induction at its centroid. Sol. The loop can be considered to be made up of three straight wires AB, BC and CA, each of length d and carrying a current i as shown in figure.
Let us first find the magnetic induction B1 due to the wire AB at the centroid O. 1 1 d ⎛π⎞ Here, a = OM = (CM) = dcos ⎜ ⎟ = 3 3 ⎝6⎠ 2 3 and θ1 = θ2 =
I
π . 3
π/3 π/3 O
µ0 i(sin θ1 + sin θ2 ) , 4π a µ i π π⎞ ⎛ B= B= 0 ⎜ sin + sin ⎟ 4π (d / 2 3 ) ⎝ 3 3⎠
Using equation B =
=
A
B
µ0 6i directed into the plane of the page. 4π d
3
I d
I
π/6
C
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Electromagnetism-I
µ0 6i each directed 4π d
Similarly, the magnetic inductions at O due to the wires BC and CA are also along the inward normal to the plane of the page. ∴ The net magnetic field induction at O is 9µ0i . B = 3B1 = 2πd
Ex-2. For the wire PQRST shown in figure, find the magnetic induction B at the point O. R d
i
i
–∞ P
i Q
S
+∞ T
O
Sol. The entire wire can be divided into four parts as PQ, QR, RS and ST. Obviously, the magnetic fields due to each portion, at O is directed toward the inward normal to the plane of page. It remains for us to determine the field inductions B1, B2, B3 and B4 corresponding to the four parts respectively and add them to get the resultant field induction B. µ i(1 − sin θ) Making use of equation B = 0 4π a µ0 i π ⎡ ⎤ 1 − sin ⎥ B1 = B4 = 4π (d / 2) ⎢⎣ 4⎦ µ0 isin θ 4π a µ isin( π / 4) µ i B2 = B3 = B = 0 = 0 4π d 4π d 2
–∞
Making use of equation B =
∴ B = 2(B1 + B2) =
M P
Q d/√2 π/4
µ0 2i ⎡ 1 ⎤ 2+ ⎢ ⎥ 4π d ⎣ 2⎦
O
3µ0i
. 2 2πd 21. For a solenoid of finite length at any point on the axis B=
R
µ Ni B = 0 [sin α + sin β] 2
α β L
N is number of turns per unit length. 22. A solenoid consists of closely wounded helical coil. Inside the solenoid the field is almost uniform. µ ni The magnetic induction at the centre of the solenoid is B = 0 where l is the axial length and n l is total number of turns in length l of the solenoid. The equation holds good only when the radius of the turns is very small when compared with the length. 23. When current is passed through a helical spring, it contracts due to mutual attraction between consecutive turns. 24. Magnetic field at a point due to a cylindrical current carrying wire Case (A) : When the wire is solid Consider a solid cylindrical wire of radius R carrying a current I distributed uniformly across its cross-section. Let P be a point distant r from the axis, where the magnetic field induction B is to be determined. 4
I d B
P
O r R
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We consider a circular arc of radius r coaxial with the wire, such that the former passes through point P. From symmetry B must be tangential at each point on the curve and should also be equal in magnitude. Application of Ampere’s law B ⋅ d I = µ0i yields,
∫
B(2πr) = µ0 i where i is the current embraced by the closed circular arc. The following two cases arise. i) When P lies within the cross-section of the wire, i.e., r < R : Iπr 2
⎛r ⎞ = I⎜ ⎟ Obviously, i = 2 ⎝R⎠ πR ⎛r ⎞ ∴ B(2πr) = µ0 I ⎜ ⎟ ⎝R⎠ µ rI or, B = i = 0 2 . 2πR
2
2
…(i)
ii) When P lies outside the cross-section of the wire, i.e., r > R : Obviously, i in this case will be I. ∴ B(2πr) = µ0I µ I …(ii) or, B= 0 . 2πr It is evident that, at the surface of the wire r = R. Both equations (i) as well as (ii) yield consistent and expected results, viz., µ I B= 0 2πr Figure displays graphically the variation of B with distance r from the axis.
B
µ0I 2 πR
Case (B) : When the wire is hollow and thin walled As discussed above, with the same reasoning and using Ampere’s law it is easy to see that B=0 (within the wire) µ0 I (outside the wire) and B= 2πr
R
O
R
r
r
I
P d
R B
µ0I 2 πR
The variation of B and r is shown graphically in figure.
R
O
R
r
r
Ex-3. A straight wire of circular cross-section of radius R carries a total current I distributed in such a manner that j varies directly as distance form the axis. Find the magnetic field induction B at a distance r from the axis (where r < R). Sol. Let P be the point where B is to be determined. We draw a circle of radius r coaxial with the wire so that P lies on it. Next, we consider a circular strip of infinitesimally small thickness dr. Current flowing through this strip is I dI = j(2πrdr) = (kr) (2πrdr) where k is a constant R
or, Total current I =
∫ 0
∴k=
3I 2πR
3
2πkR3 . (k2π)r dr = 3
d
2
B
. 5
dr P
r
O R
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Electromagnetism-I
Applying Ampere’s law, yields B ⋅ dl = µ0i
∫
r
i=
⎛ 3Ir ⎞ ⎛r ⎞ 2πdr = I ⎜ ⎟ ⎜ 3 ⎟ ⎝R⎠ ⎝ 2πR ⎠ 0
∫
⎛r ⎞ ∴ B(2πr) = µ0 I ⎜ ⎟ ⎝R⎠
B=
µ0r 2I 2πR3
3
3
.
25. The magnetic field induction due to the long current carrying cylindrical conductor is µ ir B= 0 ⋅ 2π (R + r ) 2 Where R is the radius of the conductor and r is the distance of the point from the surface of the conductor at which the value of B is given. However if r >> R, then µ i B= 0 2πr 26. Force on a moving charge :
v q
B
Fmax = q v B
v
B
+q
27. An electric charge moving in a uniform magnetic field experiences a force (F). F = q(v x B) or F = qvB sin θ 28. The direction of force is obtained by right hand grip rule. When the charge enters into a uniform magnetic field perpendicular to its direction, then F = qvB. If it enters along the direction of the field, F = 0. 29. The force can only change the direction in which the charge is moving but not its speed. Hence no work is done by it. 30. When a charged particle moves in a uniform magnetic field at right angles to the direction of the field. i) the trajectory of motion changes and ii) the speed and energy of a particle remain the same. 31. When a charged particle moves in an electric field, work is done and hence its kinetic energy increases. 32. If we assume that the earth’s magnetic field is due to a long circular loop of current in the interior of the earth, then the plane of the loop will be east-west and the current passes in clockwise direction when seen from earth’s north pole. 33. An electron is moving vertically downwards at a certain place. The direction of force on it due to the horizontal component of the earth’s magnetic field is towards west.
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34. The force acting on a charged particle when it enters a uniform magnetic field of induction B, with velocity v at right angles to the field will provide the necessary centripetal force and make the charged particle move along a circular path of radius ‘r’. mv 2 = qvB ; r mv v qB r= or = qB r m
F=
v
B
v
R F
F
2π 2πm , we have T = Since T = ω qB
+q F v
O F
F
v
v
Thus time period is independent of the particle-speed. Hence the faster particles move in bigger circles and slower particles move in smaller circles such that the period of revolution T is the same. qB and is called cyclotron frequency. 35. The frequency (ν ) = 2πm
v sin θ
36. If a charged particle enters a uniform magnetic field along the line of the field, it goes undeviated. If it enters at an angle of inclination other than 90° to the field, its path is helical. 37. Helical motion of a charged particle in a uniform magnetic field : Consider a uniform magnetic field B oriented along Y x-axis, and a charged particle having a charge q and mass m projected along the X-Y plane, making an B angle θ with the X-axis. The component of velocity v along the direction of B (i.e., v cos θ) does not θ contribute to the magnetic force, whereas the X v cosθ component, normal to the direction of B(i.e., v sin θ), p +q contributes for the magnetic force. Thus, B Z F = qvB sin θ This provides the necessary centripetal force, for rotational motion, whereas the component of velocity v cos θ, possessed by the particle imparts a movement along the magnetic field. The overall motion is said to be helical, wherein, the particle moves along a spring shaped path (figure). Notice, that the projection of the path of the particle, in a plane normal to the magnetic field, is a circle (Y-Z plane in the diagram). The particle rotates about the magnetic field (axis of the helix) with a period of revolution T 2πm given by, T = qB which is independent of the speed. The particle traverses along the magnetic field with a velocity v cos θ; the distance traveled along the helical path axis, in one period of revolution (T) is known as the pitch of the helix (p in the diagram) which is given by 2πmv cos θ p= . qB It should be noted that the particle started off from the X-Z plane and touches time and again, after each period of revolution (T). Thus, it can be concluded that, the charged particle touches the plane, from where it started, at times t = nT, where n ∈ I. 7
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Electromagnetism-I
Ex-4. A magnetic field of induction B = B0 ˆi exists along the X-axis. A
particle of charge q and mass m is projected from the origin O along the X-Y plane at an angle α to the X-axis with a speed v. (see figure). Find the position coordinates (x, y, z) and the velocity v of the particle after time t of the projection. Sol. The initial velocity v of the particle can be expressed as v 0 = v(cos α ˆi + sin αˆj) .
Y
v q
α
X
Z
B
The component v cos α ˆi along the magnetic field remains constant as no force acts along B (ie., X-axis). Therefore x = (v cos α) t.
Now, the projection of the helical path of the particle on the Y-Z plane (normal to B ) is a circle as shown in figure. The lorentz force (magnetic force) providing the necessary centripetal force is qvB sin α =
Y R
m(v sin α )2 R
O
where R is the radius of the helix. mv sin α . or R = qB
θ F
Z
C F
X
Also, the angular speed ω for the rotation of the particle about the pitch axis is 2π 2π(v sin α ) qB . ω= = = T 2πR m ∴ The angle rotated about the pitch axis (an imaginary line passing through the centre C and parallel to the X-axis) in time t, is Y qBt θ = ωt = P v sin α m R Figure shows the projection of the particle’s path; P is the θ O –Z M C projection of particle at any instant. X mv sin α ⎛ qBt ⎞ The y coordinate is y = PM = R sin θ = sin ⎜ ⎟. ⎝ m ⎠ qB where the sign of sin(qBt/m) automatically takes account all positions of the particle. The z coordinate is z = –OM = –(OC – MC) i.e., z = R(1 – cos θ) mv sin α ⎡ ⎛ qBt ⎞ ⎤ 1 − cos ⎜ = − ⎟ . ⎢ ⎝ m ⎠ ⎥⎦ qB ⎣ Thus, x = v cos αt, y = z= −
mv sin α ⎡ ⎛ qBt ⎞ ⎤ 1 − cos ⎜ ⎟ . ⎝ m ⎠ ⎥⎦ qB ⎢⎣
Coming to the velocity of the particle, the component parallel to X-axis remains unchanged. ∴ vx = v cos α. From figure, the velocity normal to the X-axis, (i.e., in the Y-Z plane) remains constant in magnitude, but has rotated by an angle θ from its initial direction (positive Y-axis). 8
Y v sin α cos θ
and
mv sin α ⎛ qBt ⎞ sin ⎜ ⎟ ⎝ m ⎠ qB
v sinα sinθ
θ
v sinα –Z
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Electromagnetism-I
∴ and ∴ or,
⎛ qBt ⎞ vy = v sin α cos θ = v sin α cos ⎜ ⎟ ⎝ m ⎠ ⎛ qBt ⎞ vz = – v sin α sin θ = –v sin α sin ⎜ ⎟ ⎝ m ⎠ v = v ˆi + v ˆj + v kˆ x
y
z
⎧ ⎛ qBt ⎞ ˆ ⎛ qBt ⎞ ˆ ⎫ v = v ⎨cos α ˆi + sin α cos ⎜ ⎟ j − sin α sin ⎜ ⎟k⎬ . ⎝ m ⎠ ⎝ m ⎠ ⎭ ⎩
38. Motion of a charged particle in space containing both electric and magnetic fields : If a region contains electric and magnetic fields E and B respectively and a particle having a charge q is in motion with a velocity v , then the force acting on it is given by F = q(E + v × B)
(i)
known as Lorentz force. Several cases are possible, depending upon the situation and complexity. Here, we intend to discuss two simple but important (from particle point of view) cases viz, (i) E is parallel to B and (ii) E is perpendicular to B . In each case the initial velocity v 0 of particle will be taken normal to both E and B . E is parallel to B : Consider a situation in which a constant electric field Y E = E0 ˆi and a constant magnetic field B = B0 ˆi exist in space pointing along B
u E the X-axis. Let a particle having a charge q and mass m be projected with an X initial velocity u the along Y-axis. Let us analyse its further motion. Z The case is too simple in the sense that the speed of the particle normal to the direction of E , (i.e., the velocity u ) remains constant as no work is done on or against it in the direction. It follows that the particle executes rotatory motion about an axis parallel to the X axis as a helix due to the magnetic force. However, the electric force qE keeps on accelerating
the particle which as a result, circulates the helical path once in the same time, but the increasing speed amounts to the pitch of the helix, getting continuously Y enhanced with time. Thus, the acceleration along X-axis due to electric force is u qE0 F ax = x = . C m m X O ∴ The velocity vx at time t will be Z qE0 t . vx = axt = m The initial velocity u (normal to the field) makes the charged particle rotate about the X-axis, as shown in figure. Applying RHR-1 to the particle at origin (initially), it is obvious that the centripetal force (magnetic force) acts along the negative Z-axis. Hence, the projection the helical path on the Y-Z plane will be as shown in the figure. Thus, the velocity components and position coordinates corresponding to the Y-Z plane will be identical to that discussed in Example, with the only substitution of v = u and α = π/2, so that Y sin α = 1 and cos α = 0. P u mu ⎛ qBt ⎞ y= sin ⎜ ⎟ ⎝ m ⎠ qB θ O
mu ⎡ ⎛ qBt ⎞ ⎤ z= − 1 − cos ⎜ ⎟ ⎢ qB ⎣ ⎝ m ⎠ ⎥⎦
9
C
–Z
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Electromagnetism-I
⎛ qBt ⎞ vy = u cos θ = u cos ⎜ ⎟ ⎝ m ⎠ ⎛ qBt ⎞ and vz = – u sin θ = – u sin ⎜ ⎟. ⎝ m ⎠ 1 Using s = ut + at 2 along X-axis, where E produces a constant acceleration, the displacement of 2 the particle along X-axis is ⎛ 1 qE0 2 ⎞ x = 0+⎜ t ⎟. ⎝2 m ⎠ ∴ The instantaneous position vector of the charged particle can be expressed as : r = xiˆ + yjˆ + zkˆ ⎛ qE0 t 2 ⎞ ˆ mu qBt ⎞ ˆ ⎛ qBt ⎞ ˆ mu ⎛ r =⎜ sin ⎜ ⎟i + ⎟j− ⎜ 1 − cos ⎟k qB qb ⎝ m ⎠ ⎝ 2m ⎠ ⎝ m ⎠
(ii)
qE0 t ˆ ⎛ qBt ⎞ ˆ ⎛ qBt ⎞ ˆ v= i + ucos ⎜ ⎟ j − u sin ⎜ ⎟k . m ⎝ m ⎠ ⎝ m ⎠ E is perpendicular to B : We know venture into the case where E = E0 ˆi and B = B0 ˆj , and a
charged particle is released at the origin O. The electric field keeps on accelerating it along the X-axis, while the magnetic field B (directed along Y-axis) makes the particle rotate in a plane normal to it, (i.e., the X-Z plane). There is no movement along the Y-axis, as neither there is any initial velocity component nor there is any acceleration in that direction. Thus, the overall motion is purely in the X-Z plane. First of all let us express the velocity at any instant as v = v ˆi + v kˆ x
z
where vx and vz are the components of the velocity along X and Z axis respectively. Net force on the particle at time t is F = Fe + Fm = qE + q(v × B) i.e.,
ˆ ˆj] , = q[E0 ˆi + (v x ˆi + v zk)B 0 ˆ F = q(E0 − v zB0 )i + qv xB0kˆ = Fx ˆi + Fzkˆ
so that Fx = q(E0 – vz B0) and Fz = qvxB0. ∴ The acceleration along X and Z axis will be respectively F q a x = x = (E0 − v zB0 ) m m Fx qv zB0 and . az = = m m qB0 da x qB0 ⎛ −dv z ⎞ (a z ) Now, = ⎜ ⎟=− dt m ⎝ dt ⎠ m = − or,
qB0 ⎛ qv xB0 ⎞ ⎜ ⎟ m ⎝ m ⎠
d2 v x dt 2
2
⎛ qB ⎞ + ⎜ 0 ⎟ vx = 0 . ⎝ m ⎠
This is the differential equation of a SHM whose general solution can be expressed as vx = v1 sin (ωt + δ) 10
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Electromagnetism-I 2
⎛ qB ⎞ where ω2 = ⎜ 0 ⎟ and ; v1 and δ are constants which depend upon a particular situation and ⎝ m ⎠ can be obtained by the use of boundary conditions. Initially, at time t = 0, vx = vy = vz = 0. ∴ 0 = v1 sin (δ) or, δ = 0. ∴ vx = v1 sin ωt dv x = ωv1 cos(ωt) = a x . Also, dt qE0 . Again, initially at time t = 0, ax = m qE0 qE0 ∴ or, v1 = . = ωv1 m mω qB0 E ∴ v1 = 0 . But, ω = m B0
∴
E0 sin ωt B0
vx =
Again, from az =
…(i)
dv z q = v x B0 . dt m
Substituting for vx, we have qE0 dv sin ωt = z . az = m dt Integrating between the limits t = 0, vz = 0 and t = t and vz = vz, qE0 …(ii) vz = (1 − cos ωt) mω Integration of expressions corresponding to vx and vz between the limits, t = 0, vx = 0, vz = 0 and t = t, vx = vx, vz = vz, yields E x = 0 (1 − cos ωt) …(iii) B0 ω and
z=
E0 (ωt − sin ωt) B0 ω
…(iv)
qB0 . …(v) m Thus, equation (i) and (ii) combinedly give the velocity at any instant while equations (iii) and (iv) yield the position coordinates on the X-Z plane. The path traveled by the particle is a cycloid. In general, a cycloid is represented by the parametric form : X x = a(θ ± sinθ) and y = a(1 ± cosθ) A D where ‘a’ is known as the radius of the generating circle, and 2πa 2a is the base of the cycloid. In our discussion see figure, –Z O C E E0 2πa 2πa a= . B0 ω
where
ω=
39. The frequency of a charged particle in a uniform magnetic field is known as cyclotron frequency as the particles in a circular accelerator, called cyclotron, move with this frequency. 40. Cyclotron is a device used to accelerate charged particles to high speed for nuclear reaction. It was invented by Lawrence. 11
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41. Force on a current carrying conductor : i
l
i
F
B
B
i
i
42. When a conductor of length l carrying a current i is placed at an angle θ to the direction of the magnetic field of flux density or induction B, the force on the conductor or wire is given by F F = i( l x B) or F = ilB sin θ Its direction can be determined by using Fleming’s left hand rule, whose statement is as follows: Stretch the fore finger, middle finger and thumb of the left hand mutually perpendicular to each other. If the fore finger represents the direction of magnetic field, the middle finger that of current, then the thumb will represent the direction of force on the conductor.
B
i or v
Fleming’s left hand rule
Ex-5. Figure shows a circular arc ab of radius r carrying a current i resting in a uniform magnetic field of strength B. Find the magnitude and direction of the force (magnetic) exerted on the arc. Sol. The length of straight line joining ab is b ⎛θ⎞ B = 2r sin ⎜ ⎟ . r i ⎝2⎠ θ
The required force is Fnet = i(ab × B) or,
⎛θ⎞ Fnet = 2Bir sin ⎜ ⎟ . ⎝2⎠
a
r
(in a direction normal to the line joining ab to the right). 43. Force between two current carrying conductors : If i1 and i2 are the strengths of currents passing through two infinitely long, straight and parallel wires separated by a distance r, the magnetic field induction B, due to the flow of current in the µ i first conductor at a distance r on the second conductor is B1 = 0 1 . 2πr The force exerted by this induction on a length l on the second conductor is F = i2B1l µ ii The force per unit length is F = 0 1 2 . 2πr
i2
i1 r
If the current is in the same direction, there will be attraction and if the current is passing in opposite direction, there will be repulsion. 44. An ampere is that steady current when flowing in each of two long straight parallel wires separated by a distance of one metre apart in vacuum causes each wire exert to each other a force of 2 × 10–7 N per each metre length of wire. 12
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Ex-6. A square loop of side a carries a current i2 with a long straight conductor carrying a current i1
lying in the plane of the loop, with its closer edge, which is parallel to the conductor, distant b. Find the force exerted on the loop.
+∞
Sol. The relevant diagram is shown in figure. The magnetic field due to the
⊗
i1
conductor is normal to the plane of the square loop. The total force on
i2 r
q
the loop can be obtained as the vector sum of the forces on its four
F1
arms, namely pq, qr, rs and sp respectively.
F2 s
p
Let the current i2 flow in the loop in the clockwise sense as shown by arrows. Evidently, the relative configurations of the edges qr and ps with
b
a
⊗
–∞
respect to the long conductor are identical and hence the forces on
them will be equal in magnitude; however, the current directions in them being opposite, the forces cancel each other. Force on side pq is given by F1 = B1i2 a(sin 90°) where B1 is the magnetic field at the site of pq, which is equal to
µ0 i1 . 2π a
µ0i1i2 . (toward left) 2π And, force on the side rs is F2 = B2i2 a sin 90° where B2 is the magnetic field at the location of rs, which is equal to µ0i1 F2 = 2π(a + b) µ0i1i2a ∴ F2 = (toward right) 2π(a + b)
∴ F1 =
∴ The net force (toward left) acting on the loop is µ0i1i2b . Fnet = F1 – F2 = 2π(a + b) 45. Force and torque on current loop or coil in a magnetic field : The total magnetic force on any closed current loop in a uniform magnetic field is zero. 46. Torque or couple on a current carrying rectangular conductor or loop in a uniform magnetic field is given by τ = iAB sinθ where A is
θ
the area of the rectangular loop, B is the magnetic induction and θ R
is the angle between the normal to the plane of the rectangular loop and the field. If there are n turns, then the torque acting on the coil suspended is given by τ = ni AB sin θ or τ = MB sinθ where M is the
Q
i i
Bil
S
i i
Bil
B
P
magnetic dipole moment equal to niA and θ is the angle between the magnetic induction and the normal to the plane of the coil i.e., the direction of the magnetic moment. When the plane of the coil makes an angle θ with the field, then the couple acting on the coil is
τ = niAB Cosθ or τ = MB cos θ. 13
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Electromagnetism-I
Moving coil galvanometer :
47. A moving coil galvanometer consists of a powerful horse shoe magnet with concave poles to produce a uniform radial magnetic field. On a cylindrical soft iron core, a coil is wound and the coil is suspended with a phosphorbronze fibre. The plane of the coil always lies in the direction of the magnetic field because the magnetic field is radial. The whole apparatus is kept inside a brass case provided with a glass window. Cθ i= ampere NAB
θ R
Q
b i
Bil
S
Bil
B
P
Here C is the couple per unit twist on the suspension fibre, ‘i’ is the current passing through the galvanometer coil and θ is the angle of deflection by the needle. C/BAN is a constant (k) is called as galvanometer constant. Deflection is measured more accurately using Lamp and Scale arrangement. The ray reflected onto the scale by the mirror is deflected by 'x'. The distance between the mirror attached to the suspension fibre of the galvanometer and the scale is D, then Tan 2θ ≈ 2θ when 2θ is very small x x C x ; θ= . Then 2θ = ; i= BAN 2D D 2D 48. The current sensitivity of a galvanometer is the deflection in mm produced on a scale kept at a distance of one metre by a constant current of one microampere. 49. The reciprocal of the current sensitivity is called figure of merit and is expressed in µA/mm. 50. A current upto 10–9A can be measured using moving coil galvanometer. 51. Table Galvanometer : It has a rectangular coil of insulated copper supported on two bearings. The poles of the magnet are concave. It has a light aluminium pointer which moves on a scale. The whole arrangement is kept in an ebonite case with a glass window. 52. Tangent Galvanometer : a) It is portable and minimum measurable current is of the order of 10–6 A. b) works on the principle of Tangent law B = BH Tanθ Here B = Magnetic induction due to passage of current in µ i the coil = 0 2r
BH
θ B
i
i
⎛ 2rBH ⎞ ⎟⎟ c) current measured by Tangent galvanometer is i = ⎜⎜ ⎝ µ 0n ⎠
Tanθ = KTanθ r = Radius of coil n = number of turns of coil d) Reading is more accurate when θ = 45° since relative error
di 1 ∝ and it is minimum i sin2θ
for 45° e) Sensitiveness is maximum when θ = 0° since sensitiveness f)
dθ ∝ cos2θ, which is maximum di
for θ = 0°. During experiment, plane of the coil should be along the magnetic meridian [to fulfill the requirement of tangent law] 14
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Electromagnetism-I
53. Shunt : a) A low resistance connected in parallel to galvanometer to protect it from large current is known as shunt. ig i G b) i = ig + is i(S ) c) ig = G+S S i – ig i(G) d) is = G+S ig S 1 = = e) is G + S n GS G+S g) If the range of galvanometer is increased to n times, 1/nth of main current passes through galvanometer. Hence sensitiveness decrease by n times. 54. Ammeter : a) It is a device used to measure current in electrical circuits. b) Galvanometer can be converted in to Ammeter by connecting low resistance parallel to it. c) To increase the range by ‘n’ times or to decrease the Sensitiveness by ‘n’ times, shunt to be connected across Galvanometer. ig(G) G S= = i − ig (n − 1) i new range old dividisions / A Here n = = = ig old range new divisions / A f) effective resistance of circuit =
GS G+S e) Resistance of ideal Ammeter is zero and its conductivity is infinity f) Ammeter must be always connected in series to the conductor. g) Among low range and high range Ammeter, low range Ammeter has more resistance. h) As shunt value decreases sensitivity decreases, accuracy increases. 55. Voltmeter : r a) Voltmeter is a device used to measure P.D. across the i – ig i conductor in electric circuits. b) Galvanometer is converted into voltmeter by connecting high ig resistance in series to it. R c) Voltmeter is always connected in parallel to the conductor G [P.D. across which is to be measured] V d) Resistance of voltmeter = G +R = ig
d) Resistance of Ammeter =
e) Here ‘V’ is range of voltmeter (e) resistance of ideal voltmeter is infinity and conductivity is zero. f) Among low range and high range voltmeters, high range voltmeter has more resistance. g) P.D. across the ends of voltmeter is, V = ig + (G + R) V h) Resistance to be connected in series to galvanometer to convert into voltmeter is R = − G . ig i) To increase the range by n times, ig ( G + R ) new rangeV 2 R = ⇒ 1+ n = old rangeV1 ig ( G ) G Hence resistance to be connected in series to galvanometer is R = G(n – 1). j) As series resister value increases sensitivity decreases, accuracy increases. 15
RK
Electromagnetism-I
Assertion & Reason : In each of the following questions, a statement is given and a corresponding statement or reason is given just below it. In the statements, marks the correct answer as 1) If both Assertion and Reason are true and Reason is correct explanation of Assertion. 2) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. 3) If Assertion is true but Reason is false. 4) If both Assertion and Reason are false. 5) If Assertion is false but Reason is true.
1. 2. 3.
[A] [R] [A] [R] [A]
6.
[R] [A] [R] [A] [R] [A]
7.
[R] [A]
4. 5.
8.
9.
[R] [A] [R] [A] [R]
10. [A] [R] 11. [A] [R] 12. [A] [R] 13. [A] [R] 14. [A] [R] 15. [A] [R] 16. [A] [R]
: Magnetic field interacts with a moving charge and not with stationary charge. : A moving charge produces a magnetic field. : A linear solenoid carrying current is equivalent to bar magnet. : The magnetic field lines of both are same. : Force experienced by moving charge will be maximum if direction of velocity of charge is perpendicular to applied magnetic field. : Force on moving charge is independent of direction of applied magnetic field. : Torque on coil is maximum, when coil is suspended in radial magnetic field. : Torque depends upon the magnitude of the applied magnetic field. : The coil is wound over the metallic frame in moving coil galvanometer. : The metallic frame help in making steady deflection without any oscillation. : When two long parallel wires, hanging freely are connected in series to a battery, they come closer to each other. : Wires carrying current in opposite direction repel each other. : When the observation point lies along the length of current element, magnetic field is zero. : Magnetic field close to current element is zero. : A moving coil galvanometer has a permanent magnet with cylindrical pole faces. : The radial magnetic field produced keeps the constant of the galvanometer to be constant throughout the deflection range. : A moving coil galvanometer has a permanent magnet with cylindrical pole faces. : The radial magnetic field produced makes the coil to experience a constant torque in any deflected position. : A moving coil galvanometer has a permanent magnet with cylindrical pole faces. : The radial magnetic field produced makes the calibration of the scale to be uniform throughout the scale. : A soft iron cylinder is arrange in the space of the coil suspended between the pole pieces of the magnet. : To increase the sensitivity of the galvanometer. : The resistance of a voltmeter is high. : The voltmeter is connected always in parallel. : The resistance of an ideal voltmeter is infinite. : When connected the voltmeter should not draw any current from main circuit thus keeping the current in that circuit constant. : The resistance of potentiometer during the measurement of potential differences, can be considered as infinite. : It does not draw any current from secondary circuit. : The resistance of an ammeter is very low. : The ammeter is always connected in series. : The ideal ammeter has zero resistance. : The ammeter is always connected in series. As such it should not alter the current in the circuit. 16
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Electromagnetism-I
17. [A] : The magnetic field ‘B’ due to a conductor of radius ‘r’ carrying a current ‘i’, at a distance of ‘x’ from the axis of the wire (x BQ. [R] : Two infinitely long thin insulated straight wires carrying currents i and i lie along x and y axes respectively. The locus of the points where the magnetic field is zero is straight line. 22. [A] : Two very long straight wires are connected in series with a battery and are arranged parallel to each other. Then the mutual force between the wires is repulsive. [R] : If the two wires are connected in parallel with the battery and one arranged parallel to each other, the mutual force between wires is attractive. 23. [A] : Moving electric charge produce both electric and magnetic field. [R] : A current carrying conductor produces a magnetic field in the surrounding space. 24. [A] : If a straight conductor carries current normally outwards from the plane of paper, lines of induction are concentric circles in counter clockwise direction. [R] : A Straight conductor carrying current produces a magnetic field, which is radially symmetric. 25. [A] : An infinite long conductor carries a current i. The work done to move a unit north pole round it once is µoi. [R] : A straight cylindrical pipe carries current i, the field inside is zero. 26. [A] : For a given length l of a wire carrying a current i, the number of circular turns which produce the maximum magnetic moment is one. [R] : A current carrying circular coil behaves as a magnetic shell. 27. [A] : When a charged particle moves at right angles to the magnetic field, the momentum of the particle will remain constant. [R] : When a charged particle is moving in magnetic field, work done by the magnetic field on the charged particle is zero. 28. [A] : When a charged particle enters perpendicular to the direction of a uniform magnetic field in the time period of rotation depends on nature the particle. [R] : When the particle enters the magnetic field at angle θ (≠0, ≠1800, ≠900), the path of the particle will be helical. B 29. [A] : The force experienced by a semi circular wire radius r when it is i carrying a current i and is placed in a uniform magnetic field of induction B as shown in fig(a) is zero. (a) [R] : The force experienced by a semi circular wire of radius r when it is carrying a current i and is placed in a uniform magnetic field of B i induction B as shown in fig(b) is 2B ir. (b) 17
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Electromagnetism-I
30. [A] : If an electron is not deflected in passing through a certain region of space, then there is no electric and magnetic fields. [R] : When a charged particle moves parallel to electric field its kinetic energy increases. 31. [A] : When a current carrying coil is placed in a uniform magnetic field, the net force on it always is zero. [R] : The current carrying coil acts like a dipole. 32. [A] : In moving coil Galvanometer, phosphorous bronze wire is used as suspension wire. [R] : Phosphorous bronze has high young’s modulus , less rigidity modules. 33. [A] : For more current sensitivity, of Moving coil Galvanometer, B, A, N should be more and c should be small. [R] : In moving coil Galvanometer, voltage sensitivity also varies as that of current sensitivity. 34. [A] : In Moving coil Galvanometer, concave shape magnetic poles makes current as linear function of deflection(θ). [R] : In case of redial magnetic field normal to the plane of coil always makes 90° with magnetic field. 35. [A] : In Moving coil Galvanometer external magnetic fields have no effect on the deflection. [R] : In tangent galvanometer external magnetic fields may influence the deflection. 36. [A] : Shunt increases the range of galvanometer. [R] : Shunt increases the life of galvanometer. 37. [A] : Ammeter must be always connected in series. [R] : Conductance of ideal ammeter is infinity. 38. [A] : Voltmeter must be always connected in parallel. [R] : Conductance of ideal voltmeter is infinity. 39. [A] : The resistance of ammeter is less than that of galvanometer. [R] : In ammeter zero division is at one end of the scale, in Galvanometer zero division is at the middle of the scale. 40. [A] : A beam of charged particles is passed through a magnetic field. The work done on the beam by the field is dependent on the deflection of the beam. [R] : The time rate of the work done by a magnetic field B on a charged particle of moving in a helical path is zero. 41. [A] : On increasing length of the, Potentiometer wire, its sensitivity increases. [R] : On increasing length of the Potentiometer wire, potential gradient decreases. 42. To convert galvanometer into an ammeter, a shunt is used. [A] : The accuracy of the meter increases. [R] : The range of the meter increases. 43. To convert a galvanometer into a voltmeter, a large resistance is connected in series with the galvanometer. Due to use of large resistance. [A] : Accuracy of the voltmeter increases. [R] : Sensitivity decreases. 44. [A] : Range and least count of ammeter and voltmeter are linearly related with each other. [R] : Least count and sensitivity of an ammeter are related inversely with each other. 45. A current i is passing through a circular coil. [A] : To induce magnetic induction of constant magnitude (at centre of the coil), the current required in it is directly proportional to its radius. [R] : Induction of induced magnetic field at centre is inversely proportional to radius of the coil. 46. [A] : In Meter bridge, meter bridge wire is replaced by another wire having twice the cross sectional area, the accuracy increases. [R] : Meter bridge works on the principle of wheat stone bridge. 18
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Electromagnetism-I
47. [A] : If a moving charged particle traces a helical path in a uniform magnetic field, axis of the helix is parallel to the magnetic field. [R] : If in a region a uniform magnetic field and a uniform electric field both exist, a charged particle moving in this region cannot trace circular path. 48. [A] : In tangent Galvanometer plane of the current carrying coil must be in magnetic meridian. [R] : When plane of the coil is in magnetic meridian, B and B H are perpendicular to each other. 49. [A] : [R] : 50. [A] : [R] : 51. [A] : [R] : 52. [A] : [R] : 53. [A] : [R] : 54. [A] : [R] : 55. [A] : [R] : 56. [A] : [R] : 57. [A] : [R] : 58. [A] : [R] : 59. [A] : [R] : 60. [A] : [R] : 61. [A] : [R] : 62. [A] : [R] : 63. [A] : [R] : 64. [A] : [R] :
M.C.G can be used for the measurement of currents even in mines. In case of M.C.G galvanometer constant does not depend on earth’s magnetic field. A linear solenoid carrying current is equivalent to a bar magnet. The magnetic field lines of both are same. Free electrons always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it. The average velocity of free electron is zero. Torque on the coil is maximum, when coil is suspended in a radial magnetic field. Torque depends upon the magnitude of the applied magnetic field. The coil is wound over the metallic frame in moving coil galvanometer. The metallic frame help in making steady deflection without any oscillation. Out of galvanometer, ammeter and voltmeter, resistance of ammeter is lowest and resistance of voltmeter is highest. An ammeter is connected in series and a voltmeter is connected in parallel, in a circuit. For a point on the axis of a circular coil carrying current, magnetic field is maximum at the centre of the coil. Magnetic field is inversely proportional to the distance of point from the circular coil. A circular loop carrying current lies in XY plane with its centre at origin having a magnetic flux in negative Z-axis. Magnetic flux direction is independent of the direction of current in conductor. The energy of charged particle moving in a uniform magnetic field does not change. Work done by the magnetic field on the charge is zero. The electron passing through crossed magnetic and electric field is always deflected from its path. If velocity of electron is equal to ratio of electric and magnetic field applied than electron beam remains undeflected. A small coil carrying current, in equilibrium, is perpendicular to the direction of the uniform magnetic field. Torque is maximum when plane of coil and direction of the magnetic fields is parallel to each other. Earth’s magnetic field does not affect the working of a moving coil galvanometer. The earth’s magnetic field is quite weak as compared to magnetic field produced in the moving coil galvanometer. Soft iron core is used in a moving coil galvanometer. Soft iron is ferromagnetic in nature. In ammeter, current in shunt is always greater than current in galvanometer. Value of shunt resistance is negative if current in shunt is less than current in galvanometer. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field. Forces acting on each pair of the opposite sides of the coil are equal and opposite. If an electron and proton enter in an electric field with equal energy, then path of electron more curved than that of proton. Electron has a tendency to form large curve due to small mass. 19
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Electromagnetism-I
65. [A] : For a given charged particle moving in a given magnetic field, the radius of circular path is directly proportional to the momentum of particle. [R] : The effect of magnetic field on a charge particle, change only its path from linear to circular. 66. [A] : If a proton and an α-particle enter a uniform magnetic field perpendicularly, with the same speed, the time period of revolution of α - particle is double that of proton. [R] : In a magnetic field, the time period of revolution of a charged particle is directly proportional to the mass of the particle and is inversely proportional to charge of particle. 67. [A] : If an electron while coming vertically from outerspace enter the earth’s magnetic field, it is deflected towards west. [R]: Electron has negative charge. 68. [A] : The workdone by magnetic field on a moving charge is zero. [R] : In magnetic field force is perpendicular to the velocity. 69. [A] : An electron and proton enters a magnetic field with equal velocities, then the force experienced by proton will be more than electron. [R] : The mass of proton is 1837 times more than the mass of electron. 70. [A] : A charged particle moving in a uniform magnetic field penetrates a layer of lead and there by loses half of its kinetic energy. The radius of curvature of its path is now reduced to half of its initial value. [R] : Kinetic energy is inversely proportional to radius of curvature. 71. [A] : Tangent Galvanometer can not be used in mines. [R] : In case of Tangent Galvanometer reduction factor depends on earth’s magnetic field. 72. Match the following : List – I a) In MCG b) in tangent galvanometer
List – II e) i α θ f) i α tan θ µo i dl sin θ 4π r2 µ i h) B= o 4π r 3) a-f, b-e, c-h, d-g
c) Biot Savart’s law
g) B=
d) Ampere’s law 1) a-e, b-f, c-g, d-h 2) a-h, b-g, c-f, d-e 73. Match the following : List – I a) infinite resistance b) zero resistance c) Lorentz’s magnetic force d) at centre of current
4) a-g, b-h, c-e, d-f
List – II e) ideal voltmeter f) ideal ammeter g) q(VxB)
µ i h) B= o 2 r 3) a-f, b-e, c-h, d-g
carrying circular loop 1) a-e, b-f, c-g, d-h 2) a-h, b-g, c-f, d-e 74. Match the following. List – I a) right hand thumb rule b) Biot-Savart law c) Fleming’s left hand rule d) Fleming’s right hand rule The correct code is 1) a-g, b-e, c-h, d-f 2) a-f, b-g, c-h, d-e
4) a-g, b-h, c-e, d-f
List – II e) magnitude of magnetic field f) direction of induced current g) direction of magnetic field g) direction of force due to magnetic fields
3) a-g, b-e, c-h, d-f 20
4) a-e, b-g, c-f, d-h
RK
Electromagnetism-I
75. The magnetic inductions due to a circular coil carrying current at its centre, at distances on its axis x = R and x = 2R are represented by BC, BR, B2R. What are the ascending order of these values. 2) B2RBR, BC 3) BC, B2R, BR 4) BR, B2R, BC 1) BC, BR, B2R 76. The resistance of moving coil galvanometer is ‘G’ and that of an ammeter is RA and that of voltmeter is RV arrange them in decreasing order 2) RV > G > RA 3) G > RV > RA 4) RA > G > RV 1) G > RA > RV
KEY : 1)
1
2)
1
3)
3
4)
2
5)
1
6)
5
7)
2
8)
1
9)
1
10) 1
11) 1
12) 2
13) 1
14) 1
15) 2
16) 1
17) 1
18) 1
19) 1
20) 2
21) 2
22) 2
23) 2
24) 1
25) 2
26) 2
27) 5
28) 2
29) 2
30) 4
31) 1
32) 1
33) 2
34) 1
35) 2
36) 2
37) 2
38) 3
39) 2
40) 5
41) 1
42) 2
43) 2
44) 2
45) 2
46) 5
47) 2
48) 1
49) 1
50) 1
51) 1
52) 2
53) 1
54) 2
55) 1
56) 3
57) 1
58) 5
59) 2
60) 1
61) 1
62) 1
63) 1
64) 4
65) 2
66) 1
67) 1
68) 1
69) 5
70) 4
71) 1
72) 1
73) 1
74) 1
75) 2
76) 2
MULTI CORRECT ANSWERS : 1.
2.
3.
4.
5. 6.
Choose the correct statements : a) The work of Oersted demonstrated that magnetic effect could be produced by moving electric charge. b) The work of Faraday and Henry demonstrated that magnetic effects could be produced by moving electric charge. c) The work of Oersted demonstrated that currents could be produced by moving magnets d) The work of Faraday and Henry demonstrated that currents could be produced by moving magnets. Choose the correct statements : a) The electric force on a charge does not depend on the speed of the charge b) The electric force on a charge depends on the speed of the charge. c) The magnetic force on a charge does not depend on the speed of the charge d) The magnetic force on a charge depends on the speed of the charge. F = qvBsinθ is the equation of the Lorentz force F on a charge q moving with a velocity v in a magnetic field B. Which of the following pairs of quantities are always perpendicular to each other? a) F – v b) F – B c) v – B d) v – q In which of the following situations, the magnetic field will not exert a force ? a) On a static charge, in the same direction as the direction of the magnetic field b) On a static charge, in a direction perpendicular to the direction of the magnetic field c) On a moving charge, in the same direction as the direction of the magnetic field d) On a moving charge, in a direction perpendicular to the direction of the magnetic field Which of the particles passing normally through a magnetic field will not be affected ? a) Electrons b) Neutrons c) Protons d) Neutrinos A proton is moving in a uniform magnetic field. The initial direction of proton makes an angle θ with the magnetic field. Which of the following pairs representing θ and the path of the proton are correct ? d) 0°i2. When the currents are in the same direction, the magnetic field at a point midway between the wires is 10 µ T. If the direction of i2 is reversed, the field becomes 30 µ T. The ratio i1/i2 is 1) 4 2) 3 3) 2 4) 1 93. Consider a long, straight wire of cross-sectional area A carrying a current i. Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction i and separated from the wire by a distance r. The opposite to the current with a speed v= nAe magnetic field seen by the observer is very nearly µ i µ i 2µ o i 2) zero 3) o 4) 1) o 2πr πr πr 94. A long, straight wire carries a current along the Z-axis. One can find two points in the X-Y plane such that 1) the magnetic fields are equal 2) the directions of the magnetic fields are the same 3) the magnitudes of the magnetic fields are equal 4) the field at one point is opposite to that at the other point 95. A long, straight wire of radius R carries a current distributed uniformly over its cross-section. The magnitude of the magnetic field is 1) maximum at the axis of the wire 2) minimum at the axis of the wire 3) maximum at the surface of the wire 4) minimum at the surface of the wire 96. A hollow tube is carrying an electric current along its length distributed uniform over its surface. The magnetic field 1) increases linearly from the axis to the surface 2) is constant inside the tube 3) is zero at the axis 4) is zero just outside the tube 97. In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero 1) outside the cable 2) inside the inner conductor 3) inside the outer conductor 4) in between the two conductors 98. A steady electric current is flowing through a cylindrical conductor 1) the electric field at the axis of the conductor is zero 2) the magnetic field at the axis of the conductor is zero 3) the electric field in the vicinity of the conductor is zero 4) the magnetic field in the vicinity of the conductor is zero 99. The magnetic effect of electric current was discovered by 1) Fleming 2) Faraday 3) Ampere 4) Oersted 100. A proton moving with a constant velocity passes through a region of space without any change in the velocity. If E and B represent the electric and magnetic fields respectively, this region of space may have 3) E ≠ 0, B=0 4) E ≠ 0, B ≠ 0 1) E=0, B=0 2) E=0, B ≠ 0 32
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Electromagnetism-I
101. An electron is injected into a region of uniform magnetic flux density, with components of velocity parallel to and normal to the flux. The path of the electrons is 1) a helix 2) straight line 3) parabola 4) ellipse 102. An electron having a charge e moves with a velocity v in X-direction. An electric field acts on it in Y-direction. The force on the electron acts in the 1) positive direction of Y-axis 2) negative direction of Y-axis 3) positive direction of Z-axis 4) negative direction of Z-axis 103. An electron moving with a speed u along the positive x-axis at y = 0 enters a region of uniform magnetic field B = −B 0 k which exits to the right of y-axis the electron exits from the region after some time with the speed v at ordinate y. Then 1) v > u, y < 0 2) v = u, y > 0 3) v > u, y > 0 4) v = u, y < 0 104. The energy in a current carrying coil is stored in the form of 1) electric field 2) magnetic field 3) dielectric strength 4) heat 105. Magnetic field is not associated with 1) a charge in uniform motion 2) an accelerated charge 3) a decelerated charge 4) a stationary charge 106. There is a magnetic field acting in a plane perpendicular to this sheet of paper downwards into the paper. Particles in vacuum move in the plane of the paper x x from left to right. The path indicated could be travelled by a x x x 1) cation 2) neutron x x 3) proton 4) electron 107. The time rate of the work done by a magnetic field B on a charged particle q moving in a helical path is 1) qB 2) aB/v 3) qB2 4) zero 108. A current is passed through a straight wire. The magnetic field established around it has its lines of forces 1) circular and endless 2) oval in shape and endless 3) straight 4) radially outward 109. If a copper rod carries a direct current, the magnetic field associated with the current will be 1) only inside the rod 2) only outside the rod 3) both inside and outside the rod 4) neither inside nor outside the rod 110. If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be 1) only inside the pipe 2) only outside the pipe 3) neither inside nor outside the pipe 4) both inside and outside the pipe 111. A live cable is hidden in a wall. Its position can be located with the help of a 1) wattmeter 2) moving coil galvanometer 3) magnetic needle 4) hot wire ammeter 112. A current i flows along infinitely long straight thin conductor. Then the magnetic induction at an point on the conductor is µ 2i 2i 1) ∞ 2) zero 3) 0 . 4) r 4π r 113. A current i ampere flows along an infinitely long straight thin walled tube. Then the magnetic induction at any point inside the tube is µ 2i 2i 1) ∞ 2) zero 3) 0 . 4) r 4π r 33
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Electromagnetism-I
114. An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current of i ampere and the radius of the circular loop is r metre. Then the magnetic induction at the centre of the circular art is i µ 2i µ 2i 1) zero 2) infinite 3) 0 . ( π + 1) tesla 4) 0 . ( π − 1) tesla 4π r 4π r 115. In the given figure, the straight parts of the wire very long. The magnetic i induction at O is µ i µ i µ i µ i r 2) 0 + 0 out of the page 1) 0 + 0 out of the page o 4r 2πr 2r 4πr µ i µ i µ i µ i 3) 0 + 0 into the page 4) 0 + 0 into the page 4r 4πr 4r 2πr 116. The variation of magnetic field B due to a long straight current-carrying wire with distance r from the wire is given by 1) B 2) B 3) B 4) B
r r r r 117. A current i ampere flows in the loop having a circular arc of radius r metre subtending an angle θ as shown in the figure. The magnetic field at the centre of 0 the circle is µ iθ µ i sin θ θ 1) 0 2) 0 4πr i 4πr 2 o r 2µ 0 i sin θ µ 0 i sin θ 4) 3) 2r 2r 118. In the figure, there are two semi-circles of radii r1 and r2 in which a current i is flowing. The magnetic induction at the centre O will be µ i µ i 1) 0 (r1 + r2 ) 2) 0 (r1 − r2 ) r1 4 4 o µ i⎛r +r ⎞ µ i⎛r −r ⎞ 3) 0 ⎜⎜ 1 2 ⎟⎟ 4) 0 ⎜⎜ 1 2 ⎟⎟ r2 4 ⎝ r1r2 ⎠ 4 ⎝ r1r2 ⎠ 119. A straight conductor carrying a direct current i ampere is split into circular loop as shown in the figure. The magnetic induction at the centre of the circular loop of radius R metre is µ i 2πi 1) zero 2) 0 . tesla i 4π R µ 2i µ i 3) 0 . tesla 4) 0 tesla 4R 4π R 120. A battery of e.m.f ε volt is connected across a coil of uniform wire as shown. The radius of the coils is ‘a’ metres and the total resistance of the wire is R o o ohm. The magnetic field at the centre of O of the coil is 90 µ0ε µ0ε 2µ 0 ε 1) zero 2) 3) 4) Ra Ra 2Ra 121. A long thin wire is bent as shown in the figure. The radius of the semicircular part is r meters. If a current of i ampere flows through the wire, then the magnetic induction at O in tesla is µ i µ i 1 1 1) 0 (1 + ) out of the page 2) 0 (1 + ) into the page π 4r π r µ i µ i 1 1 3) 0 (1 + ) out of the page 4) 0 (1 + ) into the page 2r 2π 2r 2π 34
RK
Electromagnetism-I
122. In a moving coil galvanometer, we use a radial magnetic field so that the galvanometer scale is 1) logarithmic 2) exponential 3) linear 4) nonlinear 123. The current that must flow through the coil of a galvanometer so as to produce a deflection of one division on its scale is called 1) charge sensitivity of the galvanometer 2) current sensitivity of the galvanometer 3) micro-volt sensitivity 4) reduction factor of galvanometer 124. The sensitivity of a tangent galvanometer will increase when 1) the number of turns in the coil is decreased 2) the number of turns in the coil is increased 3) the radius of the coil is increased 4) it is independent of the radius of the coil 125. A voltmeter always gives a less value of the actual potential difference because 1) some energy is lost in moving the needle 2) internal resistance of a cell plays its role 3) it absorbs some energy 4) both (1) and (2) are true 126. The resistance of an ideal ammeter should be 1) zero 2) moderate 3) very high 127. If magnetic field at O is equal to zero the value of θ is 1) π – 2 2) 2π – 2 3) 2π 4) π/2
4) infinitey
θ O
128. Two coils in two moving coil galvanometers have there are as in the ratio of 2 : 3 and number of turns in the ratio 4 : 5. These two coils carry the same current and are situated in the same field. The defections produced by these two coils will be in the ratio of 1) 8 : 15
2) 15 : 8
3) 15 :4
4) 2 2 : 15
129. A conducting rod of length l and mass m is moving down a smooth inclined plane of inclination θ with constant velocity v. A current i is flowing in the conductor in a direction perpendicular to paper inwards. A vertically upward magnetic field B exists in space. B Then magnitude of magnetic field B is mg mg sin θ 2) tan θ 1) υ il il Side View mg cos θ mg θ 4) 3) il il sin θ 130. Magnetic field at the centre of a circular loop of area A is B. Then magnetic moment of the loop will be 1)
BA 2 µ0π
2)
BA A µ0
3)
BA A µ0
4)
2BA µ0
A π
131. A uniform magnetic field B = B 0 ˆj exists in space. A particle of mass m and charge q is projected
towards negative x-axis with speed v from a point (d, 0, 0). The maximum value of v for which the particle does not hit the y-z plane is 2Bq Bqd Bq Bqd 1) 2) 3) 4) dm m 2dm 2m 132. A nonconducting disc of radius R is rotating about an axis passing through its centre and perpendicular to its plane with an angular velocity ω. Charge q is uniformly distributed over its surface. The magnetic moment of the disc is 1 1 1 1) qωR 2 2) qωR 3) qωR 4) qωR 2 4 2 2 35
RK
Electromagnetism-I
133. A conducting rod of mass m and length l is placed over a smooth horizontal surface. A uniform magnetic field B is acting perpendicular to the rod. Charge q is suddenly passing through the rod and it acquires an initial velocity v on the surface, then q is equal to 2mv Bl mv Blv 2) 3) 4) 1) Bl 2mv Bl 2m 134. A wire of length l is bent in the form of a circular coil of some turns. A current i flows through the coil. The coil is placed in a uniform magnetic field B. The maximum torque on the coil can be iBl 2 iBl 2 iBl 2 2iBl 2 2) 3) 4) 4π π 2π π 135. Dimensions of magnetic flux are 1) MLT–3A2 2) ML2T–2A–1 3) ML–2T2A 4) ML–2TA–1 136. The magnetic field at the centre of an equilateral triangular loop of side 2L and carrying a current is
1)
9µ 0 i 3 3µ 0 i 2 3µ 0 i 3µ 0 i 2) 3) 4) 4πL 4πL πL 4πL 137. Ratio of magnetic field at the centre of a current carrying coil of radius R and at a distance of 3R on its axis is
1)
1) 10 10
2) 20 10
3) 2 10
4)
10
138. Magnetic moment of an electron in nth orbit of hydrogen atom is neh neh meh meh 2) 3) 4) 1) πm 4πm 2πm 4πm 139. The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is x. if the current and radius both are doubled the new ratio will become 1) 2x 2) 4x 3) x/4 4) x/8 140. Magnetic field at the centre of a circular coil of radius R and carrying a current i is a)
µ 0i 2R
b)
i
c)
2
2c ε 0 R
µ 0i 2πR
d)
ic 2 2ε 0R
(c = speed of light) 1) a and b are correct 2) b and c are correct 3) c and d are correct 4) ‘a’ only correct 141. A charged particle is projected in a plane perpendicular to uniform magnetic field. The areal velocity (area swept per unit time) of the particle is a) directly proportional to kinetic energy of particle b) directly proportional to momentum of the particle c) inversely proportional to magnetic field strength d) inversely proportional to charge on particle 1) a, c and d are correct 2) a, b and c are correct 3) ‘a’ only correct 4) All are correct 142. H+, He+ and O+2 all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H+, He+ and O+2 are 1 amu, 4 amu and 16 amu respectively. Then b) O+2 will be deflected most a) H+ will be deflected most d) all will be deflected equally c) He+ and O+2 will be deflected equally 1) a and b are correct 2) b and c are correct 3) a and c are correct 4) ‘c’ only correct 143. A wire of length ‘ ’ carries a current i0 along x-axis. A magnetic field exists which is given by B = B0( ˆi + ˆj + kˆ )T. The magnitude of magnetic force acting on the wire is 1) B 0i0
2) 2 B 0i0
3) 3 B 0i0 36
4) zero
RK
Electromagnetism-I
144. Protons having kinetic energy E emerge from an accelerator as a narrow beam. The beam is bent misses a plane target kept at a distance in front of the accelerator. The magnetic field is (m → mass of a proton, q → change of proton). 1)
2mE q
2)
2 2mE q
3)
4 2mE q
4)
2mE 2q
145. A particle having mass m and charge q is released from rest at the origin in a region in which electric field and magnetic field are given by B = −B o ˆj and E = E 0kˆ . If s is the distance travelled, then the speed of the particle is 1)
qE 0 s m
2)
2qE 0 s m
3)
qE 0 s 2m
4)
2qB 0 s m
146. A solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed w. If ‘µ’ is magnetic moment and ‘L’ is angular momentum, then µ/L = 1) q/m 2) q/4m 3) q/2m 4) m/2q 147. If n is the number of moving charges(q) per unit volume and vd is the drift velocity then current density is nqv d nq 2) nqvd 3) 2nqvd 4) 1) vd 2 148. The figures of merit of two galvanometers with resistance of 100 Ω and 50Ω are respectively 10–8 amp per division and 2 × 10–5 amp per division respectively. The voltage sensitivity is (div/volt) 1) more for first 2) more for second 3) equal for both 4) data is incomplete 149. A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 m in a plane perpendicular to magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 m in the same plane with the same B. 1) 25 keV 2) 50 keV 3) 200 keV 4) 100 keV 150. A particle of charge ‘q’ and mass ‘m’ enters in the uniform magnetic field ‘B’ at an angle θ with velocity v. The forward distance covered during one complete rotation is 2πv cos θ 2πmv sin θ v cos θxm 2πmv cos θ 1) 2) 3) 4) qB qB qB qB 151. An electron move with a constant speed v along a circle of radius r. The magnetic moment of the circulating election is ev evr mvr 2) 3) 4) zero 1) 2πr 2 2e 152. Which of the following particles will describe the smallest circle where projected with the same velocity perpendicular to a magnetic field ? 1) electron 2) proton 3) He+ 4) Li+ 153. Which of the following particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field ? 4) Li+ 1) electron 2) proton 3) He+ 154. A charged particle is projected in a plane parallel to a uniform magnetic field. The area a bounded by the path described by the particle is proportional to 1) the velocity 2) the momentum 3) the kinetic energy 4) none of these 155. Which of the following graph represents best the variation of magnetic field along the axis of circular coil. 1) 2) 3) 4) ↑ ↑ B B ↑ B B
x→
x→
x→ 37
x→
RK
Electromagnetism-I
156. A current of 15 A is directed along the positive x-axis and perpendicular to a magnetic field. The conductor experiences a magnetic force per unit length of 0.12 Nm–1 in the negative y-direction. The magnitude and direction of the magnetic field in the region through which the current passes is 2) 4 × 10–3 T in the +y direction 1) 8 × 10–3 T in the –z direction –3 3) 4 × 10 in the –x direction 4) 8 × 10–3 T in the + z direction 157. Two long straight conductors carry currents of 5 A each. They are 5A 5A separated by a distance of 10 cm. The points P1 and P2 are at 10 P P1 2 cm and 20 cm respectively. Then the magnetic induction fields. O 20 cm 10 cm a) at the point O, midway between the wires is zero 10 cm b) at the point P1, 0.5×10–5 T out of the page –5 c) at the point O, midway between the wires is 4 × 10 T, into the page d) at the point P2, zero 1) a and b are true 2) b and c are true 3) c and d are true 4) b and d are true 158. A rectangular current loop is located near a long straight wire that carriers a current of 12A. The current in the loop is 25A.The net force acting on the 12A loop is 25 A 0.5 m 1) 2.4 × 10–4N, attractive –4 2) 1.8 × 10 N, attractive 3) 2.4 × 10–4N, repulsive 0.1 m 0.15 m 4) 1.2 × 10–4N, repulsive 159. An electrical meter of internal resistance 20 Ω gives a full scale deflection when 1 mA current flows through it. The maximum current, that can be measured by using three resistors of resistance 12 Ω each, in mA is 1) 10 2) 8 3) 6 4) 4
KEY : 1)
3
2)
1
3)
3
4)
1
5)
4
6)
1
7)
1
8)
2
9)
4
10) 2
11) 3
12) 3
13) 2
14) 4
15) 3
16) 3
17) 3
18) 3
19) 3
20) 3
21) 3
22) 3
23) 1
24) 4
25) 2
26) 3
27) 1
28) 3
29) 2
30) 3
31) 4
32) 1
33) 3
34) 1
35) 1
36) 1
37) 2
38) 2
39) 2
40) 2
41) 2
42) 2
43) 1
44) 2
45) 4
46) 3
47) 2
48) 1
49) 4
50) 4
51) 3
52) 3
53) 4
54) 2
55) 2
56) 3
57) 1
58) 1
59) 2
60) 1
61) 2
62) 1
63) 2
64) 1
65) 3
66) 4
67) 2
68) 1
69) 2
70) 1
71) 3
72) 2
73) 2
74) 3
75) 2
76) 4
77) 3
78) 4
79) 3
80) 2
81) 3
82) 3
83) 3
84) 1
85) 4
86) 2
87) 4
88) 3
89) 3
90) 3
91) 2
92) 3
93) 1
95) 2,3
96) 2,3
97) 1
98) 2,3
99) 4
101) 1
102) 2
103) 4
94) 2,3,4 104) 2
105) 4
106) 4
107) 4
108) 1
109) 3
100) 1,2,4 110) 2
111) 3
112) 2
113) 2
114) 4
115) 1
116) 1
117) 1
118) 3
119) 1
120) 1
121) 1
122) 3
123) 2
124) 2
125) 2
126) 1
127) 2
128) 1
129) 2
130) 4
131) 2
132) 1
133) 3
134) 1
135) 3
136) 1
137) 1
138) 2
139) 4
140) 1
141) 1
142) 3
143) 2
144) 1
145) 2
146) 3
147) 2
148) 1
149) 4
150) 4
151) 2
152) 1
153) 4
154) 3
155) 3
156) 4
157) 2
158) 2
159) 3
38
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