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 5 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
5 Marks Questions
1. (a) State the condition under which the phenomenon of resonance occurs in a series LCR circuit. Plot a graph showing the variation of current with frequency of a.c. sources in a series LCR circuit.
(b) Show that in a series LCR circuit connected to an a.c. source exhibits resonance at its natural frequency equal to 
Ans. (a) In a series LCR circuit
Resonance occurs when 
The variation of current with frequency of a.c. source in series LCR circuit
(b) Electrical resonance takes place in a series LCR circuit when circuit allows maximum alternating current for which
Impedance Z = 
For electrical resonance
Where w is the natural frequency of the circuit.
2. In a step up transformer, transformation ratio is 100. The primary voltage is 200 V and input is 1000 watt. The number of turns in primary is 100. Calculate
(1) Number of turns in the secondary
(2) Current in the primary
(3) The voltage across the secondary
(4) Current in the secondary
(5) Write the formula for transformation ratio?
Ans. (1) k = 100, Ep = 200V
EpIp = 1000 W, Np = 100
NS = 10,000
(2) EpIp = 1000W
Ip = 5A
(3) 
Es = 20000 Volt
(4) 
Is = 0.05 A
(5) For step up Trans former k > 1 
3. Drive an expression for the average power consumed in a.c. series LCR circuit. Hence define power factor?
Ans. For an a.c. series circuit
E = Eo sin wt
And I = Io sin (wt +
)
Where is the phase angle by which current leads the emf.
Now using dw = EIdt
dw = (Eo sin wt) (Io sin (wt +) )dt
dw = EoIo sin wt (sin wt cos + cos wt sin)dt
dw = EoIo (
wt cos + sin wt cos wt sin)dt
Integrating within limits t = o to t = T
Hence average power consumed in a.c circuit is given by
Pav = EvIv cos —–(1)
Power factor – In the above expression
Cos is termed as power factor
When cos = 1 = 
It means circuit is purely resistive and Pav = EvIv
When cos = 0 =
It means circuit is purely capacitive or inductive.
Pav = 0
Hence
4. Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).
(a)
(b)
(c)
(d)
(e)
(f)
Ans. 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 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 field lines are lying in the plane of the closed loop.
5. A long solenoid with 15 turns per cm has a small loop of area 2.0 
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?
Ans. 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 = 
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:
……(i)
Where,
= Induced flux through the small loop
= BA …(ii)
B = Magnetic field
= 
= Permeability of free space
Hence, equation (i) reduces to:
Hence, the induced voltage in the loop is 
6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 
in a uniform horizontal magnetic field of magnitude 
. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 
, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Ans. 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, 
Number of turns on the coil, N = 20
Angular speed, 
= 50 rad/s
Magnetic field strength, B = 
Resistance of the loop, R = 10 
Maximum induced emf is given as:
= 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:
Average power loss due to joule heating:
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 square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 
in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 
along the negative x-direction (that is it increases by 
as one moves in the negative x-direction), and it is decreasing in time at the rate of
. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 .
Ans. Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, 
Velocity of the loop, v = 8 cm/s = 0.08 m/s
Gradient of the magnetic field along negative x-direction,
And, rate of decrease of the magnetic field,
Resistance of the loop, 
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
Rate of change of the flux due to explicit time variation in field B is given as:
Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:
∴Induced current, 
s
s
Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.
8. It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 
with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 
turn to bring its plane parallel to the field direction). The total charge flown 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 . Estimate the field strength of magnet.
Ans. Area of the small flat search coil, A = 
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 
Total resistance of the coil and galvanometer, R = 0.50
Induced current in the coil,
Induced emf is given as:
Where,
= Charge in flux
Combining equations (1) and (2), we get
Initial flux through the coil, 
= BA
Where,
B = Magnetic field strength
Final flux through the coil, 
Integrating equation (3) on both sides, we have
But total charge, 
Hence, the field strength of the magnet is 0.75 T.
9. 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 m . Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 
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 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 (=) 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?
Ans. Length of the rod, l = 15 cm = 0.15 m
Magnetic field strength, B = 0.50 T
Resistance of the closed loop, R = 9 =
(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
= 
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:
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
= 
= 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.
10. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.
(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.
Ans.
(a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).
Magnetic flux associated with element
Where,
dA = Area of element dy = a dy
B = Magnetic field at distance y
s
I = Current in the wire
= Permeability of free space =
y tends from x to a+x.s
For mutual inductance M, the flux is given as:
(b) Emf induced in the loop, e = B’av
Given,
I = 50 A
x = 0.2 m
a = 0.1 m
v = 10 m/s
