13. Flux is a measure of flow of Magnetic Field Lines
through an Area
Case ⅠⅠⅠ
Magnetic Flux фB
14. When there is a change in magnetic flux passing
through an area bounded by closed conducting loop,
an emf known as induced emf is produced in it
Faraday’s Law of ElectroMagnetic Induction
16. Induced emf is proportional to the rate of change of flux
17. The magnitude of induced emf in a closed loop equals
time rate of change of magnetic flux through the loop.
Faraday’s Law of ElectroMagnetic Induction
18. ф = (3t2 + 2t) Wb through a loop.
Find induced emf at t = 1s
A. 2V B. 4V C. 6V D. 8V
19. A coil of circular cross-section having 1000 turns and
4 cm2 face area is placed with its axis parallel to a magnetic
field which decreases by 10–2 Wb m–2 in 0.01 s. The e.m.f.
induced in the coil is: [JEE Main 2014]
A. 400 mV
B. 200 mV
C. 4 mV
D. 0.4 mV
20. At time t = 0 magnetic field of 1000 Gauss is passing
perpendicularly through the area defined by the closed loop
shown in the figure. If the magnetic field reduces linearly
to 500 Gauss, in the next 5 s, then induced EMF in the loop is:
[JEE Main 2020]
A. 56 μV
B. 28 μV
C. 48 μV
D. 36 μV
21. The flux linked with a coil at any instant 't' is given by
ф = 10t2 - 50t +250 Wb. The induced emf at t = 3s is
A. -190 V B. -10 V C. 10 V D. 190 V
[AIEEE 2006]
23. Induced emf/current opposes the cause
by which it is produced.
Negative sign in Faraday’s Law formula
suggests oppose to the cause.
εinduced emf
=
Lenz’s law
24. 1. If flux is increasing,
magnetic field due to induced current will be opposite
to the existing magnetic field.
Direction of Induced current:
Lenz’s law
25. 2. If flux is decreasing,
magnetic field due to induced current is along the
existing magnetic field.
Direction of Induced current:
Lenz’s law
26. 1. If flux is decreasing, B due to induced i is along the existing.
2. If flux is increasing, B due to induced i is opposite to existing.
Direction of Induced current:
Lenz’s law
27. Determine direction of induced emf/current as the conducting
loop enters the space with Magnetic Field
A. Clockwise B. Anticlockwise
36. A small bar magnet is moved through a coil at constant speed from
one end to the other. Which of the following series of observations
will be seen on the galvanometer G attached across the coil ?
Three positions shown describe : (1) the magnet's entry, (2) magnet
is completely inside and (3) magnet's exit.
[JEE Main 2020]
A.
B.
C.
D.
37. Induced emf gives rise to induced current
R : Resistance of loop
Induced Current
R
38. A coil having n turns and resistance R Ω is connected with a
galvanometer of resistance 4R Ω. This combination is moved in
time t second from a magnetic flux ɸ1 weber to ɸ2 weber.
The induced current in the circuit is
A.
B.
C.
D.
39. Induced emf gives rise to induced current
R : Resistance of loop
R
❖ Induced emf does not depends on nature of the coil and its resistance.
❖ Induced current is dependent on resistance of coil (or circuit).
❖ Induced emf exists in open circuit also.
❖ Magnitude of induced emf is directly proportional to the relative speed
of coil magnet system, (e v).
Induced Current
40. In the figure, the magnet is moved towards the coil
with a speed v and induced emf is e. If magnet and coil
recede away from one another each moving with speed
v the induced emf of the coil will be-
N S
v
Coil
Magnet
A. e B. 2e
C. e/2 D. 4e
41. q : Amount of charge
flown through the
loop of Resistance R
when magnetic flux
through it changes
by ΔФ
Amount of Charge
42. A circular coil of n turns and radius r is placed in a uniform
magnetic field B. Initially the plane of the coil is perpendicular
to the field. Now the coil is rotated by 90°. If its resistance is R
then quantity of charge passing through the coil is -
D.
A. zero B. C.
43. ❖ Induced charge in any coil (or circuit) does not depends on
time in which change in flux occurs i.e. it is independent
from rate of change of flux or relative speed of coil–
magnet system.
❖ Induced charge depends on change in flux through the coil
and nature of the coil (or circuit) i.e. resistance.
Amount of Charge
48. A boat is moving due east in a region where the earth's
magnetic field is 5.0 × 10–5 NA–1 m–1 due north and
horizontal. The boat carries a vertical aerial 2m long. If the
speed of the boat is 1.50 ms–1, the magnitude of the induced
emf in the wire of aerial is: [JEE 2011]
A. 0.03 mV B. 0.15 mV C. 0.01 mV D. zero
49. D. None
A. B.
C.
If the rod here has length L & resistance r and it moves at
constant speed v, then current in the Resistance R will be:
R
R
51. A conducting rod AB of length = 1 m is moving at a velocity
v = 4 m/s making an angle 30º with its length. A uniform
magnetic field B = 2T exists in a direction perpendicular to the
plane of motion. Then- A. VA – VB = 8V B. VA – VB = 4V
C. VB – VA = 8V D. VB – VA = 4V
30o
A B
52. l
A
B
Motional EMF of an Arbitrary Shaped Wire
Join initial & final points by Straight Line &
this is the effective length l.
53. v
R
Find the emf induced between the ends of a
semicircular wire of Radius R, moving in a way
that its diameter is perpendicular to velocity
A. BvR B. 2BvR C. 2πBvR D. BvR / 2
54. Where l’ is the component
of length between ends A &
B which is perpendicular to
Join initial & final points by Straight Line &
this is the effective length l.
A
B
l’
Motional EMF of an Arbitrary Shaped Wire
55. v
Find the emf induced between the ends of a
semicircular wire of Radius R, moving in a way
that its diameter is perpendicular to velocity
B. 0.6BvR
A. 0.5BvR C. 0.8BvR D. 1.6BvR
θ=53o
60. A. zero B. 25 volt C. 20 volt D. 15 volt
A uniform magnetic field exists in region given by
A rod of length 5 m is placed along y-axis is moved along x-
axis with constant speed 1 m/sec. Then induced e.m.f. in
the rod will be-
61. If any two out of , or
Motional EMF (General form)
are parallel or anti-parallel
62. There will be no induced emf in a straight conductor
moving in a uniform magnetic field, if-
(a) it is moving parallel to magnetic field
(b)it is moving along its length
(c) it is moving in the magnetic field with its
length parallel to field
Then correct statement (s) is/are -
A. a only B. a,b only C. a,c only D. a,b,c
63. v
R
Find the emf induced in the circular wire of Radius R,
moving with velocity v perpendicular to magnetic field
A. BvR B. 2BvR C. 2πBvR D. None of these
64. l
v
Loop ABCD is moving in a magnetic field perpendicular
to its plane as shown. Determine net emf induced in the loop
b A. B.
C. D. zero
2Bvl
Bvl
2Bvb
65. Motional EMF due to Rotation
Rod AB rotating at angular speed ω about end A
ω
A B
66. ω
dx
A B
Motional EMF due to Rotation
A
Rod AB rotating at angular speed ω about end A
B
x
68. A rod of length L and resistance r rotates about one end.
Its other end touches a conducting ring of negligible
resistance. A resistance R is connected between the centre
and periphery. Find the current in resistance R.
D.
A. B.
C.
R
69. Motional EMF due to Rotation
Rod AB rotating at angular speed ω about end A
ω
dx
A B
A B
x
70. A metallic rod of length ‘l’ is tied to a string of length 2l
and made to rotate with angular speed ω on a horizontal table
with one end of the string fixed. If there is a vertical magnetic
field ‘B’ in the region, the e.m.f. induced across the ends of the
rod is
A.
B.
C.
D.
[JEE Main 2013]
71. A rod of length rotates with a small but uniform angular
velocity about its perpendicular bisector. A uniform magnetic
field B exists parallel to the axis of rotation. The potential
difference between the two ends of the rod is-
A. zero B. ½ ωBl2 C. ωBl2 D. 2ωBl2
72. EMF Induced in Rotating Coil
ω
Potential Difference between centre
and any point on the circumference.
Rotating Ring
73. EMF Induced in Rotating Disc
ω
Potential Difference between centre
and any point on the circumference.
Rotating Disc
74. A circular copper disc of radius 25 cm is rotating about its own axis
with a constant angular velocity of 130 rad/s. If a magnetic field of 5 x
10-3 Tesla is applied at right angles to the disc, then the induced
potential difference between the centre and the rim of the disc will
approximately be -
A. 20 x 10-3 V
B. 20 x 10-6 V
C. 20 x 10-9 V
D. Zero
75. A wire ring of radius R is in pure rolling on a surface.
Find the EMF induced across the top and bottom points
of the ring at any instant.
A. zero
B. ½ BωR2
C. BωR2
D. 2BωR2
B
R
A ω
𝛎
76. Two metallic rings of radius R are rolling on a metallic rod.
A magnetic field of magnitude B is applied in the region.
The magnitude of potential difference between points A and C
on the two rings (as shown), will be -
A. 0
B. 4 BωR2
C. 8 BωR2
D. 2 BωR2
R
C ω
A
ω
81. The magnetic field is decreasing. What is the direction of
induced electric field ?
A. B.
C. D.
82. The magnetic field is decreasing. What is the direction of
induced electric field ?
A. B.
C. D.
83. A long solenoid has a circular cross-section of radius R.
The magnetic field through it is increasing at a steady rate
K= dB/dt. Compute the variation of the electric field as a
function of the distance r from the axis of the solenoid if
r<R
r
R
A. Kr B. Kr / 2 C. Kr / 4 D. None
84. A long solenoid has a circular cross-section of radius R.
The magnetic field through it is increasing at a steady rate
K= dB/dt. Compute the variation of the electric field as a
function of the distance r from the axis of the solenoid is r > R
r
R
A. B. C. D.
85. Solenoid of radius R has a time varying magnetic field that
increases continuously at . Which of these
represents variation of Einduced with distance r from axis of
solenoid Ein
r = R
Ein
r = R r
Ein
r = R r
Ein
r = R r
A.
C.
B.
D.
r
86. Properties of Induced Electric Field Lines :
1. Induced electric field lines are not created by source
charges.
2. These are created by changing magnetic field.
3. These lines form closed loops without any source charge
present in loop.
4. The electric field vector is tangent to the electric
field line at each point.
5. is non-electrostatic & non-conservative. It has no
potential associated with it.
87. Self Inductance As ‘i’ changes, ‘Φ’ changes and emf is induced
this emf is called ‘Self-induced emf’
as emf is induced because of ‘i’ of loop itself.
i
88. Self-inductance: Property of a coil by which
it opposes any change in the magnitude of current
flowing through it by inducing an emf in itself.
Self Inductance
i
L: Self-inductance
Units: weber/ampere or Henry (H)
L: depends on geometry of loop
Total flux through the loop per unit current
89. Self Inductance
Corresponding induced emf
L units : Volt-sec/Ampere
Self-inductance: Property of a coil by which
it opposes any change in the magnitude of
current flowing through it by inducing an emf
in itself.
90. When current in a coil changes from 5 A to 2 A in 0.1 s,
average voltage of 50 V is produced. The self-inductance
of the coil is :
A. 6 H B. 0.67 H
C. 3 H D. 1.67 H
[JEE Main 2015]
91. Direction of Induced emf
(a) I is increasing
(b) I is decreasing
e = L dl/dt
e = L dl/dt
I
I
I
I
92. I
If I is increasing which of the following diagram shows
the equivalent emf induced in the inductor
A.
B.
I
I
93. A. 2 V
B. 4 V
C. 6 V
D. 8 V
If current is increases with rate of 2 Amp/sec.
Then find VA - VB. [JEE Main Sept - 2020]
A B
L = 2 H
i = 1 amp
R = 2
94. Self inductance of a Long Solenoid
n: no. of turns per unit length
l: length of solenoid
r: radius of each turn
N: total no. of turns N = nl
I
95. The total number of turns and cross-section area in a
solenoid is fixed. However, its length l is varied by adjusting
the separation between windings. The inductance of solenoid
will be proportional to:
A. l
B. l2
C. 1/l2
D. 1/l
[JEE Main 2019]
96. A thin copper wire of length 100 metres is wound as a
solenoid of length l and radius r. Its self inductance is found
to be L. Now if the same length of wire is wound as a
solenoid of length l but of radius r/2, then its self inductance
will be
A. 4 L B. 2 L C. L D. L/2
97. This question has Statement 1 and Statement 2. Of the four choices given
after the Statements, choose the one that best describes the two
Statements.
Statement 1: Self inductance of a long solenoid of length L, total
number of turns N and radius r is less than (πμON2r2)/L
Statement 2: The magnetic induction in the solenoid in Statement 1
carrying current I is (μONI/L) in the middle of the solenoid but
becomes less as we move towards its ends.
A. Statement I is true, Statement 2 is false.
B. Statement I is true, Statement 2 is true, Statement 2 is the
correct explanation of Statement I.
C. Statement I is false, Statement 2 is true.
D. Statement I is true, Statement 2 is true, Statement 2 is not
the correct explanation of Statement I .
[JEE Main 2012]
98. Primary Secondary
When the current passing through a coil (primary) changes ,
the magnetic flux through neighbouring coil (secondary) changes,
hence an emf is produced in the secondary coil.
This phenomenon is called Mutual Induction.
Фsecondary ∝ iprimary
100. M: Mutual inductance
depends on geometry only
Units : weber/amp or Henry(H)
Mutual Inductance
Emf induced in
secondary coil
101. The coefficient of mutual inductance of the two coils is 5 H. The
current through the primary coil is reduced to zero value from 3A
in 1 millisecond. The induced emf in the secondary coils is
A. 0 B. 1.67 KV
C. 15 KV D. 600 V
102. Mutual Inductance of
1. Coil having N1 turns (radius r1) surrounded
by another coil having N2 turns (radius) r2
[r2 >> r1]
N1
N2
i
1
2
103. 2. A Solenoid S1 (n1, r1, l1) placed inside
another Solenoid S2 (n2, r2, l2) (l2>>l1)
Mutual Inductance of
104. There are two long coaxial solenoids of same length 1. The
inner and outer coils have radii r1 and r2 and number of turns
per unit length n1 and n2, respectively. The ratio of mutual
inductance to the self-inductance of the inner-coil is :
[JEE Main 2019]
A. n1/n2
B. n2r1/n1r2
C. n2r2
2/n1r1
2
D. n2/n1
105. Two coaxial solenoids are made by winding thin insulated wire
over a pipe of cross-sectional area A = 10cm2 and length = 20cm.
If one of the solenoid has 300 turns and the other has 400 turns,
their mutual inductance is (μ0 = 4π × 10 –7 Tm A–1)
[2008]
A. 2.4π × 10–5 H B. 4.8π × 10–4 H
C. 4.8π × 10–5 H D. 2.4π × 10–4 H
107. On passing a current through Inductor
Self-induced emf in a coil opposes
the change in the current that has induced it.
If current is increasing,
induced emf will be
opposite to direction of current.
If current is decreasing
induced emf will be
in the same direction as the current.
I
I
1.
2. I
Inductor
108. The network shown in figure is part of a complete circuit.
If at a certain instant the current (I) is 5 A, and decreasing
at a rate of 103 A/s, then VB – VA =
A 1Ω B
5mH
15V
I A. 15 V B. 10 V
C. 5 V D. 20 V
[IIT-1997]
109. Thus ½ Li2 is the energy stored in
inductor when current through it is i
Energy stored in an Inductor
I
110. Energy Density in Magnetic field
For a long Solenoid (r, l, n)
Neglecting end effects
Assuming uniform field throughout solenoid
I
112. Combination of Inductors
Inductor in Series
L2
L1
(a) L = L1 + L2
(b) L = L1 + L2 ± 2M
(If mutual inductance is also considered)
(Neglecting mutual induction)
113. Relation between Self Inductance and Mutual Inductance
Leq = L1 + L2 + 2M
M
L1
L2
Leq = L1 + L2 - 2M
(Current in coils in
opposite direction)
L2
L1
M (Current in coils in
same direction)
114. M = Mutual inductance of two inductors L1 and L2
K = Coefficient of coupling
For a tight (perfect) coupling K = 1, otherwise K < 1
L2
L1
Relation between Self Inductance and Mutual Inductance
115. Two coils of self inductances L1 and L2 are tightly wrapped
one over the other. The maximum mutual inductance M of
the combination will be
A. L1 + L2 B. L1L2
C. D.
116. Two identical solenoid coils, each of self inductance L are
connected in series. Their turns are in the same sense, and
the distance between them is such that the coefficient of
coupling is half. Then the equivalent inductance of the
combination is
A. 3L/2 B. 2L C. 3L D. 5L/2
117. Growth of Current in LR Circuit
At t = 0 the Key K is closed
R
L
ε t = 0
118. L
Growth of Current in LR Circuit
B
A
At t = ∞
at steady state
we can replace all inductors
of the circuit by a straight
conducting wire
A B
119. The growth of current in an L-R circuit in time t = L/R
is equal to about
A. 37% of maximum B. 63% of maximum
C. 57% of maximum D. 67% of maximum
120. Time Constant (τ)
τ : time after which
growth process of current
completes by 63%
Growth of Current in LR Circuit
For small ‘τ’ current rises faster
i: current in the R
i
t
i0
t = τ
0.63i0
121. q: Charge on the Capacitor
q
t
q0
t = τ
0.63q0
Time Constant (τ)
Time after which
growth process
completes by 63%
RC Charging Circuit
Recalling
122. [AIEEE 2007]
An ideal coil of 10H is connected in series with a resistance of
5Ω and a battery of 5V. 2 second after the connection is
made, the current flowing in ampere in the circuit is
A. (1 – e–1) B. (1 – e)
C. e D. e-1
123. i0 = ε/R
in one time period at t = τ
at t = 0
at t ⟶ ∞
i
t
Decay of Current in LR Circuit
R
L
124. i
t
i0
t = τ
0.37i0
Time after which
decay process
completes by 63%
Time Constant (τ)
Decay of Current in LR Circuit
R
L
125. q = Q0cos ωt
q
t
Q0
T/2
T
Q0 : initial charge
on Capacitor
LC Oscillations A charged capacitor is connected to an inductor
and switch is closed at t = 0.
C
L
+ -
ε
+ -
i
126. i
ω
T/2
i = i0sin ωt
T
i0 = Q0ω
A charged capacitor is connected to an inductor
and switch is closed at t = 0.
LC Oscillations
128. ● In Ideal situation when R = 0, UL + UC is constant.
● If there is some resistance, there is a continuous loss of
energy. Amplitude of charge or current decays with time.
● During oscillations, at all times VC = VL
voltage across capacitor at any instant = emf induced in the inductor.
● Energy stored in capacitor or inductor
oscillates with frequency 2𝜈
LC Oscillations
129. Changing Magnetic field generates Induced Electric Field which
generates emf that generates currents on the surface of metal
130. Changing Magnetic field generates Induced Electric Field which
generates emf that generates currents on the surface of metal
131. Changing Magnetic field generates Induced Electric Field which
generates emf that generates currents on the surface of metal
132. A. normal to the paper, inwards
B. normal to the paper, outwards
C. from east to west
D. from north to south
A metal sheet is placed in a variable magnetic field which is
increasing from zero to maximum. Induced current flows in
the directions as shown in figure. The direction of magnetic
field will be -
133. plates moves through region of magnetic field,
magnetic flux through an area bounded by a
random loop on metal plate changes.
Hence a current is induced.
Induced current appears on surface along
variety of paths.
System itself looks for loops on surface.
Induced current ⇒ Thermal energy ⇒ Damping of K.E.
⇓
Electromagnetic
damping
Eddy Current
cut slots
in metal
⇐
134. 2m
3m
Magnetic field in the region decreases from 1 T to 0.5 T
in 2 sec. Find Average emf produced in the loop
A. 1/2 V B. 2 V C. 3/2 V D. 3 V
Ans: C
135. Given B = 10-2 T and r = 5 cm. If radius r is increasing
at a rate of 1 cm/s. Find induced emf
A. 𝜋 x 10-4 V B. 2𝜋 x 10-5 V
C. 𝜋 x 10-5 V D. 3𝜋 x 10-5 V
Ans: C
136. Determine direction of induced emf/current using Lenz’s
Law, when ends of the wire are moved in the fashion shown
A. Clockwise B. Anticlockwise
Ans: B
137. Two coils P and Q are lying a little distance apart coaxially.
If a current I is suddenly set up in the coil P then the
direction of current induced in coil Q will be-
A. clockwise
B. towards north
C. towards south
D. anticlockwise
Ans: A
138. A coil has 200 turns and area of 70 cm2. The magnetic
field perpendicular to the plane of the coil is 0.3 Wb/m2
and take 0.1 sec to rotate through 180°. The value of the
induced e.m.f. will be -
A. 8.4 V
B. 84 V
C. 42 V
D. 4.2 V
Ans: A
139. In the figure shown a square loop PQRS of side ‘a’ and resistance ‘r’ is
placed in near an infinitely long wire carrying a constant current I. The
sides PQ and RS are parallel to the wire. The wire and the loop are in
the same plane. The loop is rotated by 180° about an axis parallel to
the long wire and passing through the mid points of the side QR and
PS. The total amount of charge which passes through any point of the
loop during rotation is
A.
B.
C.
D. Cannot be found because
time of rotation not given
Ans: B
140. 5.5 × 10–4 magnetic flux lines are passing through a coil of
resistance 10 ohm and number of turns 1000. If the number of
flux lines reduces to 5 × 10–5 in 0.1 sec. The electromotive
force and the current induced in the coil will be respectively-
A. 5 V, 0.5 A B. 5 × 10–4 V, 5 × 10–4A
C. 50 V, 5 A D. none of the above
Ans: A
141. The north pole of a magnet is brought away from a coil,
then the direction of induced current will be -
A. In the clockwise direction
B. In the anticlockwise direction
C. Initially in the clockwise and then
anticlockwise direction
D. Initially in the anti clockwise and
then clockwise
Ans: B
142. Consider the situation shown in figure. If the current I in the
long straight wire XY is increased at a steady rate then the
induced emf' s in loops A and B will be-
A. clockwise in A, anticlockwise in B
B. anticlockwise in A, clockwise in B
C. clockwise in both A and B
D. anticlockwise in both A and B
X
Y
A B
i
Ans: A
143. Consider the situation shown in figure. If the switch is closed
and after some time it is opened again, the closed loop
will show-
S
A. an anticlockwise current-pulse
B. a clockwise current-pulse
C. an anticlockwise current-pulse and then
a clockwise current-pulse
D. a clockwise current-pulse and then an
anticlockwise current-pulse
Ans: D
144. Find the direction of induced emf or current in the
conducting rod AB that moves on the wire PQRS
P
B
A
Q
R S
A. Clockwise B. Anticlockwise
Ans: B
145. The variation of induced emf (e) with time (t) in a coil if a short
magnet is moved along its axis with a constant velocity is best
represented as -
[AIEEE 2004]
A. B.
C. D.
Ans: B
146. v
i
x
Rod AB is moving away from an infinite wire carrying
constant i current as shown. Determine emf induced in the
rod at a moment when it is at distance x from wire:
A. B.
C. D. None of these
Ans: A
147. i
v
l
a
Rod AB of length l moves in such a way that its length
remains perpendicular to long straight current carrying wire.
Determine emf induced in the Rod:
A.
B.
C.
D. None of these
Ans: A
148. i
l
x
v
Loop ABCD is moving away from an infinite wire carrying
constant i current as shown. Determine net emf induced
in the loop at a moment when its one end is at distance
x from wire:
b
A.
B.
C.
D. zero
Ans: B
149. R
V
Calculate the magnitude and direction of Force
acting on wire AB that slide on rail PQST as shown:
P Q
S
T
A
B
A. B. C. D.
Ans: B
150. An aeroplane, with its wings spread 10 m, is flying at a speed of
180 km/h in a horizontal direction. The total intensity of earth's
field at that part is 2.5 x 10-4 Wb/m2 and the angle of dip is 60°.
The emf induced between the tips of the plane wings will be
A. 88.37 mV
B. 62.50 mV
C. 54.125 mV
D. 108.25 mV
[JEE Main 2021]
Ans: D
151. A metal rod moves at a constant velocity in a direction
perpendicular to its length. A constant, uniform magnetic
field exists in space in a direction perpendicular to the rod as
well as its velocity. Select the correct statement(s) from the
following
A. The entire rod is at the same electric potential
B. There is an electric field in the rod
C. The electric potential is highest at the centre of
the rod and decreases towards its ends
D. The electric potential is lowest at the centre of
the rod and increases towards its ends
[1998 JEE]
Ans: B
152. A rod of length 10 cm made up of conducting and non-conducting
material (shaded part is non-conducting). The rod is rotated with
constant angular velocity 10 rad/sec about point O, in constant
magnetic field of 2 tesla as shown in the figure. The induced emf
between the point A and B of rod will be-
A. 0.029 volt
B. 0.1 volt
C. 0.051 volt
D. 0.064 volt
2cm
3cm
O
A
B
ω
Ans: C
153. An equilateral triangular conducting frame is rotated with
angular velocity ω in a uniform magnetic field B as shown.
Side of triangle is l. Choose the correct option(s).
A. Va - Vc = 0
B. Va - Vc = (Bωl2/2)
C. Va - Vb = (Bωl2/2)
D. Vc - Vb = (-Bωl2/2)
Ans: A, C
154. A conducting rod of length l is hinged at point O. It is free
to rotate in a vertical plane. The rod is released from the
position shown. The potential difference between the
two ends of the rod is proportional to-
A. l3/2
B. l2
C. sinθ
D. (sinθ)1/2
Ans: A, D
155. R
V
Calculate the magnitude and direction of Force
acting on wire AB that slide on rail PQST as shown:
P Q
S
T
A
B
A. B. C. D.
Ans: B
156. Conducting Rod of negligible resistance moves with
initial speed vo. If no force is applied on the rod to
maintain this speed, what will be the speed of rod after
a time
D. None
A. B. vo/2
C.
R v
l
Ans: C
157. A conducting wire of length l, resistance R and mass m starts
sliding at t = 0 down a smooth, vertical, thick pair of connected
rails as shown in figure. A uniform magnetic field B exists in the
space in a direction perpendicular to the plane of the rails.
The speed of wire when it moves uniformly is:
a b
A. B.
C. D.
Ans: A
158. A conducting wire of length l, resistance R and mass m starts
sliding at t = 0 down a smooth, vertical, thick pair of connected
rails as shown in figure. A uniform magnetic field B exists in the
space in a direction perpendicular to the plane of the rails.
The velocity of wire as function of time is: (vm being terminal
velocity)
A.
B.
C.
D.
a b
Ans: B
159. i
l
x
v
Loop ABCD is moving away from an infinite wire carrying
constant i current as shown. Determine net emf induced
in the loop at a moment when its one end is at distance
x from wire:
b
A.
B.
C.
D. zero
Ans: B
160. A conducting rod of resistance r moves uniformly with a
constant speed v in magnetic field B normal to the plane of
its motion. What force shall be applied on the rod in order
to keep it moving uniformly
B.
C.
A.
R v
l D. Zero
Ans: C
