CBSE Class 12 Physics Chapter 6 – Electromagnetic Induction Important Question Answer

Very short type question answer – 1 Marks

1. A metallic wire coil is stationary in a non – uniform magnetic field. What is the emf. Induced in the coil?
Ans. NO emf is induced in the coil as there is no change in the magnetic flux linked with the secondary coil.

2. Why does metallic piece becomes very hot when it is surrounded by a coil carrying high frequency (H.F) alternating current?
Ans. When a metallic piece is surrounded by a coil carrying high frequency (H.F) alternating current, it becomes hot because eddy currents are produced which in turn produces joule’s heating effect.

3. An electrical element X when connected to an alternating voltage source has current through it leading the voltage by  radian. Identify X and write expression for its reactance?
Ans. X is a purely capacitive circuit

4. A transformer steps up 220 V to2200 V. What is the transformation ratio?
Ans. 

5. The induced emf is also called back emf. Why?
Ans. it is because induced emf produced in a circuit always opposes the cause which produces it.

Short type question answer – 2 Marks

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  

Short type question answers – 3 Marks

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.

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, 

Long term question answers – 5 Marks

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.