CBSE Class 12 Physics Chapter-6 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 6 – Electromagnetic Induction 2 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-6 – Electromagnetic Induction
CBSE Class 12 Physics Important Questions Chapter 6 – Electromagnetic Induction
2 Marks Questions
1. IF the rate of change of current of 2A/s induces an emf of 1OmV in a solenoid. What is the self-inductance of the solenoid?
Ans.
2. A circular copper disc. 10 cm in radius rotates at a speed of 2
rad/s about an axis through its centre and perpendicular to the disc. A uniform magnetic field of 0.2T acts perpendicular to the disc.
1) Calculate the potential difference developed between the axis of the disc and the rim.
2) What is the induced current if the resistant of the disc is 2
?
Ans. (1) Radius = 10cm, B = 0.2T w = 2 rad/s
I = 0.0314 A
3. An ideal inductor consumes no electric power in a.c. circuit. Explain?
Ans. P = E rms I rms cos 
But for an ideal inductor 
P=0
4. Capacitor blocks d.c. why?
Ans. The capacitive reactance
For d.c. 
= 0
Since capacitor offers infinite resistance to the flow of d.c. so d.c. cannot pass through the capacitor.
5. Why is the emf zero, when maximum number of magnetic lines of force pass through the coil?
Ans. The magnetic flux will be maximum in the vertical position of the coil. But as the coil rotates 
Hence produced emf 
6. An inductor L of reactance 
is connected in series with a bulb B to an a.c. source as shown in the figure.
Briefly explain how does the brightness of the bulb change when
(a) Number of turns of the inductor is reduced.
(b) A capacitor of reactance 
is included in series in the same circuit.
Ans. (a) Since Z = 
When number of turns of the inductor gets reduced and Z decreases and in turn current increases
Hence the bulb will grow more brightly
(b) When capacitor is included in the circuit
But 
(given)
Z = R (minimum)
Hence brightness of the bulb will become maximum.
7. 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 field at the location has a magnitude of 
and the dip angle is 
Ans. 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 = 
Angle of dip, 
s
Vertical component of Earth’s magnetic field,
BV = B sin
= 
= 
Voltage difference between the ends of the wing can be calculated as:
=
= 3.125 V
Hence, the voltage difference developed between the ends of the wings is
3.125 V.
8. 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?
Ans. Mutual inductance of a pair of coils, µ = 1.5 H
Initial current, 
= 0 A
Final current 
= 20 A
Change in current, 
Time taken for the change, t = 0.5 s
Induced emf, 
Where 
is the change in the flux linkages with the coil.
Emf is related with mutual inductance as:
Equating equations (1) and (2), we get
Hence, the change in the flux linkage is 30 Wb.
9. A horizontal straight wire 10 m long extending from east to west is falling with a speed of
, at right angles to the horizontal component of the earth’s magnetic field, 
Wb
.
(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?
Ans. Length of the wire, l = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 
(a) Emf induced in the wire,
e = Blv
(b) Using Fleming’s right hand rule, its 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.
10. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad 
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.
Ans. Length of the rod, l = 1 m
Angular frequency,
= 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
.
Average linear velocity of the rod,
Emf developed between the centre and the ring,
Hence, the emf developed between the centre and the ring is 100 V.
11. 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.
Ans. 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 
