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 3 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
3 Marks Questions
1. How is the mutual inductance of a pair of coils affected when
(1) Separation between the coils is increased.
(2) The number of turns of each coil is increased.
(3) A thin iron sheet is placed between two coils, other factors remaining the same. Explain answer in each case.
Ans. (1) When the Separation between the coils is increased, the flux linked with the
secondary coils decreases, hence mutual induction decreases.
(2) Since m =
so when 
increases, mutual induction increase.
(3) Mutual induction will increase because
(Relative permeability of material)
2. Distinguish between resistances, reactance and impedance of an a.c. circuit?
Ans.
| Resistance | Reactance | Impedance | |
| 1 | Opposition offered by the resistor to the flow of current | Opposition offered by the inductor or capacitor to the flow of current | Opposition offered by the combination of resistor, inductor or capacitor |
| 2 | It is independent of the frequency of the source. | It depends on the frequency of the source | It depends on the frequency of the source |
3. A sinusoidal voltage V = 200 sin 314t is applied to a resistor of 10
resistance. Calculate
(1) rms value of the voltage
(2) rms value of the current
(3) Power dissipated as heat in watt.
Ans. V = 200 sin 314t
V = Vo sin wt
Vo = 200V, w = 314 rad/s.
R = 10
(1) V rms = 
V rms = 
200 = 282.8 V
(2) 
I rms = 28.28 A
(3) Since circuit is purely resistive
P = 7.998 watt
4. Obtain an expression for the self inductance of a long solenoid? Hence define one Henry?
Ans. Consider a long solenoid of area A through which current I is flowing
Let N be the total number of turns in the solenoid
Total flux 
= NBA
Here = B = 
Where n is no. of turns per unit length of the solenoid
N = nl
Equation (1) & (2)
[n = N/
]
One Henry – if current is changing at a rate of 1 A/s in a coil induces an emf. of 1 volt in it then the inductance of the coil is one henry.
5. A conducting rod rotates with angular speed w with one end at the centre and other end at circumference of a circular metallic ring of radius R, about an axis passing through the centre of the coil perpendicular to the plane of the coil A constant magnetic field B parallel to the axis is present everywhere. Show that the emf. between the centre and the metallic ring is 
Ans. Consider a circular loop connect the centre with point P with a resistor.
The potential difference across the resistor = induced emf.
Rate of change of area of loop.
If the resistor QP is rotated with angular velocity w and turns by an angle 
in time t then
Area swept 
6. (a) At a very high frequency of a.c., capacitor behaves as a conductor. Why?
(b) Draw the graph showing the variation of reactance of
(i) A capacitor
(ii) An inductor with a frequency of an a.c. circuit.
Ans. (a) 
For a.c. when 
Thus at a very high frequency of a.c. capacitor behaves as a conductor
(b)
7. Calculate the current drawn by the primary of a transformer which steps down 200 V to 20 V to operate a device of resistance 20
. Assume the efficiency of the transformer to be 80%?
Ans.
= 80% Ep = 200V Es = 20V Z = 20
Ip = 0.125 A
8. An a.c. voltage E = Eo sin wt is applied across an inductance L. obtain the expression for current I?
Ans.E = Eo sin wt (Given)
Emf produced across L = 
Total emf of the circuit = 
Since there is no circuit element across which potential drop may occur
Integrating
(peak vlue of current)
I = Io sin 
9. A series circuit with L = 0.12H, C = 0.48mF and R = 25is connected to a 220V variable frequency supply. At what frequency is the circuit current maximum?
Ans.L = 0.12H, C = 0.48mF = 
R = 25 Ev = 220V
In the circuit when I is maximum, R will be minimum
f = 21 Hz
10. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field 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 
in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Ans. 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
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
=
Time taken to travel along the width. ts
Hence, the induced voltage is 
which lasts for 2 s.
(b) Emf developed, e = Bbv
= 
Time taken to travel along the length,
t
Hence, the induced voltage is
which lasts for 8 s.
11. 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.
Ans. Initial current, 
Final current, 
Change in current,
Time taken for the change, t = 0.1
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
e = L 
s
s
Hence, the self-induction of the coil is 4 H.
12. Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of
. If the cut is joined and the loop has a resistance of 1.6 how much power is dissipated by the loop as heat? What is the source of this power?
Ans. Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
Initial value of the magnetic field, 
Rate of decrease of the magnetic field, 
Emf developed in the loop is given as:
Where,
= Change in flux through the loop area
= AB
Resistance of the loop, R = 1.6
The current induced in the loop is given as:
Power dissipated in the loop in the form of heat is given as:
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.
13. An air-cored solenoid with length 30 cm, area of cross-section 
and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of
. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Ans. Length of the solenoid, l = 30 cm = 0.3 m
Area of cross-section, A = 
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time, t = 
s
Average back emf, 
Where,
= Change in flux
= NAB ……(2)
Where,
B = Magnetic field strength
Where,
= Permeability of free space = 
Using equations (2) and (3) in equation (1), we get
Hence, the average back emf induced in the solenoid is 6.5 V.
14. A line charge 
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. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,
= 0 (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off?
Ans. 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.,
Where
V=linear velocity of the wheel
∴ Angular velocity, 
