NCERT Solutions for Electromagnetic Induction Class 12 Physics
Book Solutions1
Answer
The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures 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 prqp.
(c) The direction of the induced current is along vzxy.
(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 field lines are lying in the plane of the closed loop.
2
Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
Answer
According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.
(a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.
(b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along a’d’c’b’
3
Answer
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as:
Where, Ф = Induced flux through the small loop `
= BA ... (ii)
B = Magnetic field
= μ0ni… (iii)
μ0 = Permeability of free space = 4π×10−7 H/m
Hence, equation (i) reduces to:
= 7.54 × 10-6V
Hence, the induced voltage in the loop is 7.54 × 10-6V.
4
Answer
Length of the rectangular wire, l = 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 × 0.02
= 16 × 10−4 m2
Magnetic field 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 × 0.08 × 0.01 = 2.4 × 10−4 V
Time taken to travel along the width,
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 × 10−4 V
Time taken to travel along the length,
Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s.
5
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis 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 field 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 = 1 m
Angular frequency, ω = 400 rad/s
Magnetic field strength, B = 0.5 T
One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.
Average linear velocity of the rod,
Hence, the emf developed between the centre and the ring is 100 V.
6
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 = πr2 = π × (0.08)2 m2
Number of turns on the coil, N = 20
Angular speed, ω = 50 rad/s
Magnetic field strength, B = 3 × 10−2 T
Resistance of the loop, R = 10 Ω
Maximum induced emf is given as:
e = Nω AB
= 20 × 50 × π × (0.08)2 × 3 × 10−2
= 0.603 V
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:
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.
7
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s−1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10−4 Wb m−2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Answer
Length of the wire, l = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 0.3 × 10−4 Wb m−2
(a) Emf induced in the wire,
e = Blv
= 0.3 × 10-4 × 5 × 10
= 1.5 × 10-3 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.
8
Answer
Initial current, I1 = 5.0 A
Final current, I2 = 0.0 A
Change in current,
dI = I1 – I2 = 5A
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:
Hence, the self-induction of the coil is 4 H.
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 flux linkage with the other coil?
Answer
Mutual inductance of a pair of coils, µ = 1.5 H
Initial current, I1 = 0 A
Final current I2 = 20 A
Change in current, dI = I1 – I2 = 20A
Time taken for the change, t = 0.5 s
Induced emf,
Where
Emf is related with mutual inductance as:
Equating equations (1) and (2), we get
dϕ = 1.5 × (20)
= 30 Wb
Hence, the change in the flux linkage is 30 Wb.
10
Answer
Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 × 10−4 T
Angle of dip, δ = 30°
Vertical component of Earth’s magnetic field,
BV = B sinδ
= 5 × 10−4 sin 30°
= 2.5 × 10−4 T
Voltage difference between the ends of the wing can be calculated as:
e = (BV) × l × v
= 2.5 × 10−4 × 25 × 500
= 3.125 V
Hence, the voltage difference developed between the ends of the wings is
3.125 V.
11
Answer
Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length × width
= 8 × 2 = 16 cm2
= 16 × 10−4 m2
B’ = 0.3T
Rate of decrease of the magnetic field,
Where, dϕ = Change in flux through the loop area = AB
= 16 × 10-4 × 0.02 = 0.32 × 10-4 V
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as:
Power dissipated in the loop in the form of heat is given as:
P = i2R
= (2 × 10-5)2 × 1.6
= 6.4 × 10-10 W
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.
12
Answer
Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12 × 0.12 = 0.0144 m2
Velocity of the loop, v = 8 cm/s = 0.08 m/s
Gradient of the magnetic field along negative x-direction,
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.
13
Answer
Area of the small flat search coil, A = 2 cm2 = 2 × 10−4 m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 × 10−3 C
Total resistance of the coil and galvanometer, R = 0.50 Ω
Induced current in the coil,
Combining equations (1) and (2), we get
Initial flux through the coil, ϕ = BA
Where, B = Magnetic field strength
Final flux through the coil, ϕ = 0
Integrating equation (3) on both sides, we have
Hence, the field strength of the magnet is 0.75 T.
14
Figure 6.20 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 field 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 field 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 do experience 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 field is parallel to the rails instead of being perpendicular?
Answer
Length of the rod, l = 15 cm = 0.15 m
Magnetic field strength, B = 0.50 T
Resistance of the closed loop, R = 9 mΩ = 9 × 10−3 Ω
(a) Induced emf = 9 mV; polarity of the induced emf is such that end P shows positive while end Q 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
= 9 × 10−3 v
= 9 mV
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 flow 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 flow 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 = IBl
Where, I = Current flowing through the rod
(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
= 75 × 10-3 × 0.12
= 9 mW
When key K is open, no power is expended.
(f) 9 mW; power is provided by an external agent.
Power dissipated as heat = I2 R
= (1)2 × 9 × 10−3
= 9 mW
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 field lines.
15
Answer
Length of the solenoid, l = 30 cm = 0.3 m
Area of cross-section, A = 25 cm2 = 25 × 10−4 m2
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time, t = 10−3 s
Average back emf,
Hence, the average back emf induced in the solenoid is 6.5 V.
16
Answer
As the magnetic field will be variable with distance from long straight wire, so the flux through square loop can be calculate by integration.
17
Answer
Line charge per unit length
Where, r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field,
At distance r, the magnetic force is balanced by the centripetal force i.e.
