NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 6
Chapter Name : Electromagnetic Induction

Q6.1 Predict the direction of induced current in the situations described by the following Figs.

Answer. The direction of the induced current in a closed loop is given by Lenz's law. The given
pairs of gures show the direction of the induced current when the North pole of a bar magnet is
moved towards and away from a closed loop respectively.

Using Lenz's rule, the direction of the induced current in the given situations can be predicted as
follows:
(a) The direction of the induced current is along qrpq.
(b) The direction of the induced current is along prap.
(c) The direction of the induced current is along yzxy.
(d) The direction of the induced current is along zyxz.
(e) The direction of the induced current is along xryx.
(f) No current is induced since the eld lines are lying in the plane of the closed loop.

Page : 229 , Block Name : Exercise

Q6.2 Use Lenz’s law to determine the direction of induced current in the situations described by
Fig.
(a) A wire of irregular shape turning into a circular shape.
(b) A circular loop being deformed into a narrow straight wire.

Answer. (a) As the loop changes from irregular to circular shape, it's area increases. Hence the
magnetic ux linked with it increases. According to Lenz's law, the induced current should produce
magnetic ux in the opposite direction of original ux. For this induced ow in the anticlockwise
direction. It means the direction of current will be along adcba.
(b) As the circular loop is being deformed into a narrow straight wire, its area decreases.The
magnetic eld linked with its also decreases. By Lenz's law, the induced current should produce a
ux in the direction of original ux. For this, the induced current should ow in the anticlockwise
direction. It means the direction of current will be along a’d’c’b’.

Page : 230 , Block Name : Exercise

Q6.3 A long solenoid with 15 turns per cm has a small loop of area placed inside the solenoid
normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1
s, what is the induced emf in the loop while the current is changing?

Answer. Number of turns on the solenoid = 15 turns/cm = 1500 tums/m
Number of turns per unit length, n=1500 turns
The solenoid has a small loop of area, A = 2.0cm = 2 × 10 m
2 −4 2

Current carried by the solenoid changes from 2 A to 4 A.
Change in current in the solenoid, di = 4 - 2 = 2 A
Change in time, dt=0.1 s Induced emf in the solenoid is given by Faraday's law as
dϕ
e =
dt
....(i)
Where,
ϕ = Induced ux through the small loop
= BA …..(ii)
B - Magnetic eld
= μ0 ni

μ0 = Permeability of free space
−7
= 4π × 10 H/m

Hence, equation (1) reduces to:
d
e = (BA)
dt

di
= Aμ0 n × ( )
dt

2
−4 −7
= 2 × 10 × 4π × 10 × 1500 ×
0.1
−6
= 7.54 × 10 V

Hence, the induced voltage in the loop is 7.54 × 10 .
−6
V

Page : 230 , Block Name : Exercise

Q6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of
uniform magnetic eld of magnitude 0.3 T directed normal to the loop. What is the emf developed
across the cut if the velocity of the loop is 1 cm s in a direction normal to the (a) longer side, (b)
−1

shorter side of the loop? For how long does the induced voltage last in each case?

Answer. Length of the rectangular wire, I = 8 cm = 0.08 m
Width of the rectangular wire, b = 2 cm = 0.02 m
Hence, area of the rectangular loop,
A = lb
= 0.08 x 0.02
= = 16 × 10 m −4 2

Magnetic eld strength, B = 0.3 T
Velocity of the loop, V = 1 cm/s = 0.01 m/s
(a) Emf developed in the loop is given as:
e = Blv
= 0.3 x 0.08 x 0.01
=2.4 × 10 V
−4

Distance travelled b
t = =
Velocity v

0.02
= = 2s
0.01

Hence, the induced voltage is 2.4 × 10 −4
V which lasts for 2 s.
(b) Emf developed, e = Bbv
=0.3 × 0.02 × 0.01 = 0.6 × 10 V −4

Distance travelled b
t = =
Velocity v

0.08
= = 8s
0.01

Hence, the induced voltage is 0.6 × 10 −4
v which lasts for 8 s.

Page : 230 , Block Name : Exercise

Q6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s about an axis
−1

normal to the rod passing through its one end. The other end of the rod is in contact with a
circular metallic ring. A constant and uniform magnetic eld of 0.5 T parallel to the axis exists
everywhere. Calculate the emf developed between the centre and the ring.

Answer.
Length of the rod, l = 1m

Angular frequency, ω = 400rad/s

Magnetic field strength, B = 0.5T

One end of the rod has zero linear velocity, while the other end has a linear velocity of l ω

lω+0 lω
Average linear velocity of the rod, v = =
2 2

Emf developed between the centre and the ring,

2
lω Bl ω
e = Blv = Bl ( ) =
2 2

2
0.5 × (1) × 400
= = 100V
2

Hence, the emf developed between the centre and the ring is 100V.

Page : 230 , Block Name : Exercise

Q6.6 A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an
angular speed of 50 rad s in a uniform horizontal magnetic eld of magnitude 3.0 × 10 T.
−1 −2

Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of
resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power
loss due to Joule heating. Where does this power come from?

Answer. Max induced emf = 0.603 V
Average induced emf = 0 V
Max current in the coil = 0.0603 A
Average power loss = 0.018 W
(Power comes from the external rotor)
Radius of the circular coil, r = 8 cm = 0.08 m
Area of the coil, A = Πr = Π × (0.08) m
2 2 2

Number of turns on the coil, N = 20
Angular speed, ω = = 50 rad/s
Magnetic eld strength, B = 3 × 10 −2
T

Resistance of the loop, R = 10 Ω
Maximum induced emf is given as:
e = N ωAB

2 −2
= 20 × 50 × π × (0.08) × 3 × 10

= 0.603V

The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero.
Maximum current is given as:
e
I =
R
0.603
= = 0.0603A
10

Average power loss due to joule heating:

eI
P =
2
0.603 × 0.0603
= = 0.018W
2

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an
external agent. It must supply a torque to counter this torque in order to keep the coil rotating
uniformly. Hence, dissipated power comes from the external rotor.

Page : 230 , Block Name : Exercise

Q6.7 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0
ms
−1
, at right angles to the horizontal component of the earth’s magnetic eld,
0.30 × 10
−4
wbm . (a) What is the instantaneous value of the emf induced in the wire? (b) What
−2

is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?

Answer. Length of the wire, I = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic eld strength,B = 0.3 × 10 −4
Wbm
−2

(a) Emf induced in the wire,
e = BI v

−4
= 0.3 × 10 × 5 × 10

−3
= 1.5 × 10 V

(b) Using Fleming's right hand rule, it can be inferred that the direction of the induced emf is from
West to East.
(c) The eastern end of the wire is at a higher potential.

Page : 230 , Block Name : Exercise

Q6.8 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give
an estimate of the self-inductance of the circuit.

Answer. Initial current, I = 5.0A 1

Final current, I = 0.0A 2

Change in current, dI = I − I = 5A 1 2

Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
di
e = L
dt
e
L =
di
( )
dt

200
= = 4H
5

0.1

Hence, the self induction of the coil is 4 H.

Page : 230 , Block Name : Exercise

Q6.9 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes
from 0 to 20 A in 0.5 s, what is the change of ux linkage with the other coil?

Answer. Mutual inductance of a pair of coils, μ = 1.5 H
Initial current, I = 0A 1

Final current I = 20A
2

Change in current, dI = I − I = 20 − 0 = 20A
2 1

Time taken for the change, t = 0.5 S
dϕ
Induced emf, e = dt

Where dϕ is the change in the ux linkage with the coil.
Emf is related with mutual inductance as:
dI
e = μ
dt

Equating equations (1) and (2), we get
dϕ dl
= μ
dt dt

dϕ = 1.5 × (20)

= 30Wb

Hence, the change in the ux linkage is 30 Wb.

Page : 230 , Block Name : Exercise

Q6.10 A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference
developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic eld at the
location has a magnitude of5 × 10 T and the dip angle is 30°,
−4

Answer. Speed of the jet plane, V = 1800 km/h = 500 m/s
Wingspan Of jet plane, I = 25 m
Earth's magnetic eld strength,
B = 5.0 x 10-4 T
Angle of dip, δ = 30
∘

Vertical component of Earth's magnetic eld,
BV = B sin δ

−4 ∘
= 5 × 10 sin 30

−4
= 2.5 × 10 T

Voltage difference between the ends of the wing can be calculated as:
e = (Bv ) × I × v

−4
= 2.5 × 10 × 25 × 500

= 3.125V

Hence, the voltage difference developed between the ends of the wings is 3.125 V.

Page : 230 , Block Name : Exercise

Q6.11 Suppose the loop in Exercise 4 is stationary but the current feeding the electromagnet that
produces the magnetic eld is gradually reduced so that the eld decreases from its initial value of
0.3 T at the rate of 0.02Ts If the cut is joined and the loop has a resistance of 1.6 Ω, how much
−1

power is dissipated by the loop as heat? What is the source of this power?

Answer. Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length x width
= 8 x 2 = 16 cm
= = 16 × 10 m
−4 2

Initial value of the magnetic eld, B = 0.3T ′

Rate of decrease of the magnetic eld,
dB
= 0.02T/s
dt

Emf developed in the loop is given as:
dϕ
e =
dt

Where,
dϕ = Change in ux through the loop area
= AB
d(AB) AdB
∴ e = =
dt dt
−4 −4
= 16 × 10 × 0.02 = 0.32 × 10 V

Resistance of the loop,R = 1.6Ω
The current induced in the loop is given as:
e
i =
R
−4
0.32 × 10
−5
= = 2 × 10 A
1.6

Power dissipated in the loop in the form of heat is given as:
2
P = i R
2
−5
= (2 × 10 ) × 1.6

−10
= 6.4 × 10 W

The source of this heat loss is an external agent, which is responsible for changing the magnetic
eld with time.

Page : 231 , Block Name : Additional Exercise

Q6.12 A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of
8 cm s in the positive x-direction in an environment containing a magnetic eld in the positive
−1

z-direction. The eld is neither uniform in space nor constant in time. It has a gradient of
10
−3
Tcm along the negative x-direction (that is it increases by 10 Tcm as one moves in
−1 −3 −1

the negative x-direction), and it is decreasing in time at the rate of 10 Ts Determine the
−3 −1

direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

Answer. Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12 x 0.12 = 0.0144 m 2

Velocity of the loop, v = 8 cm/s = 0.08 m/s
Gradient of the magnetic eld along negative x-direction,
dB −3 −1 −1 −1
= 10 Tcm = 10 Tm
dx

And, rate of decrease of the magnetic eld,
dB −3 −1
= 10 Ts
dt

Resistance of the loop. R = 4.5mΩ = 4.5 × 10 Ω −3

Rate of change of the magnetic ux due to the motion of the loop in a non-uniform magnetic eld
is given as:
dϕ dB
= A × × v
dt dx
−4 2 −1
= 144 × 10 m × 10 × 0.08
−5 2 −1
= 11.52 × 10 Tm s

Rate of change of the ux due to explicit time variation in eld B is given as:
′
dϕ dB
= A ×
dt dt
−4 −3
= 144 × 10 × 10
−5 2 −1
= 1.44 × 10 Tm s

Since the rate of change of the ux is the induced emf, the total induced emf in the loop can be
calculated as:
−5 −5
e = 1.44 × 10 + 11.52 × 10
−5
= 12.96 × 10 V

Induced current,i =
e

R
−5
12.96×10
=
−3
4.5×10

−2
i = 2.88 × 10 A

Hence, the direction of the induced current is such that there is an increase in the ux through the
loop along positive z-direction.

Page : 231 , Block Name : Additional Exercise

Q6.13 It is desired to measure the magnitude of eld between the poles of a powerful loudspeaker
magnet. A small at search coil of area 2cm with 25 closely wound turns, is positioned normal to
2

the eld direction, and then quickly snatched out of the eld region. Equivalently, one can give it a
quick 90° turn to bring its plane parallel to the eld direction). The total charge own in the coil
(measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the
coil and the galvanometer is 0.50 Ω. Estimate the eld strength of magnet.

Answer. Area of the small at search coil, A = 2cm = 2 × 10 2 −4
m
2

Number of turns on the coil, N = 25
Total charge owing in the coil, Q = 7.5 mc = 7.5 x 10 C −3

Total resistance of the coil and galvanometer, R = 0.50Ω

Induced current in the coil,
Induced emf (e)
I =
R
.........(i)
Induced emf is given as:
dϕ
e = −N
dt
..........(ii)
Where,
dϕ – Charge in ux
Combining equations (1) and (2), we get
dϕ
N
dt
I = −
R

N
I dt = − dϕ
R

Initial ux through the coil,ϕ = BA i

Where,
B = Magnetic eld strength
Final ux through the coil,ϕ = 0 f

Integrating equation (3) on both sides, we have
ϕ
−N ϕ
∫ I dt = ∫ dϕ
R 0

But total charge, Q = ∫ J dt
−N −N Nϕ
∴ Q = (ϕf − ϕi ) = (−ϕi ) = +
R R R

N BA
Q =
R

QR
∴ B =
NA
−3
7.5 × 10 × 0.5
= = 0.75T
−4
25 × 2 × 10

Hence, the eld strength of the magnet is 0.75 T.

Page : 231 , Block Name : Additional Exercise

Q6.14 Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the
poles of a permanent magnet. The rails, the rod, and the magnetic eld are in three mutual
perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the
rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the eld
to be uniform. (a) Suppose K is open and the rod is moved with a speed of 12 cm s–1 in the
direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ
even though they doexperience magnetic force due to the motion of the rod. Explain. (d) What is

the retarding force on the rod when K is closed? (e) How much power is required (by an external
agent) to keep the rod moving at the same speed (=12 cm s–1) when K is closed? How much power
is required when K is open? (f ) How much power is dissipated as heat in the closed circuit? What
is the source of this power? (g) What is the induced emf in the moving rod if the magnetic eld is
parallel to the rails instead of being perpendicular?

Answer. Length of the rod, I = 15 cm = 0.15 m
Magnetic eld strength, B = 0.50 T
Resistance of the closed loop, R = 9mΩ = 9 × 10 Ω −3

(a) Induced emf = 9 mV; polarity of the induced emf is such that end P shows positive while end
shows negative ends. Speed of the rod, v = 12 cm/s = 0.12 m/s
Induced emf is given as:
e = Bvl

= 0.5 × 0.12 × 0.15

−3
= 9 × 10 v

= 9mV

The polarity of the induced emf is such that end P shows positive while end Q shows negative
ends.
(b) Yes; when key K is closed, excess charge is maintained by the continuous ow of current. When
key K is open, there is excess charge built up at both ends of the rods. When key K is closed, excess
charge is maintained by the continuous ow of current.
(c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite
nature at both ends of the rod. There is no net force on the electrons in rod PQ when key K is open
and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force
set-up due to the excess charge of opposite nature at both ends of the rods.
(d) Retarding force exerted on the rod, F = IBI
Where, I = Current owing through the rod
−3
e 9×10
= = = 1A
−3
R 9×10

∴ F = 1 × 0.5 × 0.15

−3
= 75 × 10 N

(e) 9 mW; no power is expended when key K is open
Speed of the rod, v = 12 cm/s = 0.12 m/s
Hence, power is given as:
P = Fv
−3
= 75 × 10 × 0.12
−3
= 9 × 10 W

= 9mW

When key K is open, no power is expended.
(f) 9 mW; power is provided by an external agent.
Power dissipated as heat = I R 2

2 −3
= (1) × 9 × 10

= 9mW

The source of this power is an external agent.

(g) Zero In this case, no emf is induced in the coil because the motion of the rod does not cut
across the eld lines.

Page : 231 , Block Name : Additional Exercise

Q6.15 An air-cored solenoid with length 30 cm, area of cross-section 25cm and number of turns
2

500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10 . How
−3

much is the average back emf induced across the ends of the open switch in the circuit? Ignore the
variation in magnetic eld near the ends of the solenoid.

Answer. Length of the solenoid, I = 30 cm = 0.3 m
Area of cross-section, A = 25cm = 25 × 10 m 2 −4 2

Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current ows for time, t = 10-35
dϕ
Average back emf, e = dt

Where,
dϕ- Change in ux
= NAB... (2)
Where,
B= Magnetic eld strength
NI
= μ0
l

Where,
μ = Permeability of free space = 4n × 10
−7 −1
0 TmA

Using equation (2) and (3) in equation (1) we get
2
μ0 N IA
e =
lt
−7 2 −4
4π × 10 × (500) × 2.5 × 25 × 10
= = 6.5V
−3
0.3 × 10

Hence the average back emf induced in the solenoid is 6.5 V.

Page : 232 , Block Name : Additional Exercise

Q6.16 (a) Obtain an expression for the mutual inductance between a long straight wire and a
square loop of side a as shown in Fig. . (b) Now assume that the straight wire carries a current of 50
A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced
emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large
resistance.

Answer. (a) Take a small element dy in the loop at a distance y from the long straight wire (as
shown in the given gure).
src=https://cdn./ncert/65d7baf8-29e6-11e9-a6cd-02f21f5619c4/c025543c-2ace-11e9-
bc12-02f21f5619c4.jpg>
Magnetic ux associated with element dv, dϕ = BdA
Where,
dA = Area of element dy = a dy
B = Magnetic eld at distance y
μ0 I
=
2πy

I = Current in the wire
μ = Permeability of free space =4n × 10
−7 −1
0 TmA
μ0 la dy
∴ dϕ =
2π y

μ0 I a dy
ϕ = ∫
2π y

y tends from x to a + x
θ+x
μ0 I a dy
∴ ϕ = ∫
2π y
x

μ0 I a a+x
= [log y]
e x
2π

μ0 I a a + x
= loge ( )
2π x

For mutual inductance M, the ux is given as
ϕ = MI

μ0 I a a
∴ M = log ( + 1)
2π e x

μ0 a a
M = log ( + 1)
2π e x
μ I
0
=( )
(b) Emf induced in the loop,e = B av ′ 2πx
αv

Given,
I = 50 A
x = 0.2 m
a = 0.1 m
V = 10 m/s
−7
4π×10 ×50×0.1×10
e =
2π×0.2

−5
e = 5 × 10 V

Page : 232 , Block Name : Additional Exercise

Q6.17 A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and
radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its
axis (Fig.). A uniform magnetic eld extends over a circular region within the rim. It is given by,
B = – B0 k (r ≤ a; a ≤ R )
= 0 (otherwise)
What is the angular velocity of the wheel after the eld is suddenly switched off?

Total charge Q
Answer. Line charge per unit length = λ = Length
=
2πr

Where,
r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic eld, B → = −B k ^
0

At distance r , the magnetic force is balanced by the centripetal force i.e.,
2
Mv
BQv =
r

Where,
v= linear velocity of the wheel
Mv
∴ B2πrλ =
r

2
B2πλr
v =
M

Angular velocity,ω =
2
v B2πλr
=
R MR

For r ≤ and a ≤ R we get:
2
2k0 a λ
ω = − ^
k
MR

Page : 232 , Block Name : Additional Exercise