# Chapter 6 – Electromagnetic Induction Questions and Answers: NCERT Solutions for Class 12 Physics

Class 12 Physics NCERT book solutions for Chapter 6 - Electromagnetic Induction Questions and Answers.

Education Blogs Chapter 6 – Electromagnetic Induction Questions and Answers: NCERT Solutions for Class...

Class 12 Physics NCERT book solutions for Chapter 6 - Electromagnetic Induction Questions and Answers.

Ans: The direction of the induced current in a closed loop could be given by Lenz’s law. The following 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.

Now, by using Lenz’s rule, the direction of the induced current in the given situation is found to be along qrpq.

Ans: On using Lenz’s law, we find the direction of the induced current here to be along prqp.

Ans: Using Lenz’s law, we find the direction of the induced current to be along yzxy.

Ans: Using Lenz’s law, we find the direction of the induced current to be along zyxz.

Ans: Using Lenz’s law, we found the direction of the induced current to be along xryx.

Ans: Here we find that, no current is induced since the field lines are lying in the same plane as that of the closed loop.

a) A wire of irregular shape turning into a circular shape;

The wire is expanding to form a circle, which means that force would be acting outwards on each part of the wire because of the magnetic field (acting in the downwards direction). Now, the direction of induced current should be such that it will produce magnetic field in the upward direction (towards the reader). Therefore, the force on wire will be towards the inward direction, i.e., induced current would be flowing in anticlockwise direction in the loop from cbad.

2.0A2.0A

to

4.0A4.0A

in

0.1s0.1s

, what is the induced emf in the loop while the current is changing?

Number of turns on the solenoid

=15turns/cm=1500turns/m=15turns/cm=1500turns/m

Number of turns per unit length,

n=1500 turnsn=1500 turns

The solenoid has a small loop of area,A=2.0cm2=2×10−4m2A=2.0cm2=2×10−4m2

Current carried by the solenoid changes from

2A to 4A2A to 4A

Now, the change in current in the solenoid,

di=4−2=2Adi=4−2=2A

Change in time,

dt=0.1 sdt=0.1 s

Induced emf in the solenoid could be given by Faraday’s law as:

ε=dϕdtε=dϕdt………………………… (1)

Where, induced flux through the small loop, ϕ=BAϕ=BA…………… (2)

Equation (1) would now reduce to:

ε=ddt(BA)=Aμ0n×(didt)ε=ddt(BA)=Aμ0n×(didt)

Substituting the given values into this equation, we get,

ε=2×10−4×4π×10−7×1500×20.1ε=2×10−4×4π×10−7×1500×20.1

∴ε=7.54×10−6V∴ε=7.54×10−6V

Therefore, the induced voltage in the loop is found to be, ε=7.54×10−6Vε=7.54×10−6V.

a) longer side? For how long does the induced voltage last in this case?

Length of the rectangular wire,

l=8cm=0.08 ml=8cm=0.08 m

Width of the rectangular wire,

b=2cm=0.02mb=2cm=0.02m

Now, the area of the rectangular loop,

A=lb=0.08×0.02=16×10−4m2A=lb=0.08×0.02=16×10−4m2

Magnetic field strength,

B=0.3TB=0.3T

Velocity of the loop,

v=1 cm/s=0.01 m/sv=1 cm/s=0.01 m/s

Emf developed in the loop could be given as:

ε=Blvε=Blv

Substituting the given values,

ε=0.3×0.08×0.01=2.4×10−4Vε=0.3×0.08×0.01=2.4×10−4V

Time taken to travel along the width, t=Distance travelledvelocity=bvt=Distance travelledvelocity=bv

⇒t=0.020.01=2s⇒t=0.020.01=2s

Therefore, the induced voltage is found to be 2.4×10−4V2.4×10−4Vwhich lasts for 2s.

ε=Blvε=Blv

Substituting the given values,

ε=0.3×0.02×0.01=0.6×10−4Vε=0.3×0.02×0.01=0.6×10−4V

Time taken to travel along the width, t=Distance travelledvelocity=lvt=Distance travelledvelocity=lv

⇒t=0.080.01=8s⇒t=0.080.01=8s

Therefore, the induced voltage is found to be 0.6×10−4V0.6×10−4Vwhich lasts for 8s.

1.0m1.0m

long metallic rod is rotated with an angular frequency of 400rads−1400rads−1about 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

0.5T0.5T

parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Length of the rod,

l=1ml=1m

Angular frequency, ω=400rad/sω=400rad/s

Magnetic field strength,

B=0.5TB=0.5T

One end of the rod has zero linear velocity, while the other end has a linear velocity of lωlω.

Average linear velocity of the rod, ν=lω+02=lω2ν=lω+02=lω2

Emf developed between the centre and the ring,

ε=Blv=Bl(lω2)=(Bl2ω2)ε=Blv=Bl(lω2)=(Bl2ω2)

On substituting the given values,

∴ε=0.5×(1)2×4002=100V∴ε=0.5×(1)2×4002=100V

Therefore, the emf developed between the centre and the ring is

100V100V

8.0cm8.0cm

and 20 turns is rotated about its vertical diameter with an angular speed of 50rads−150rads−1in a uniform horizontal magnetic field of magnitude 3.0×10−2T3.0×10−2T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance

10Ω10Ω

, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Max induced emf

=0.603V=0.603V

Average induced emf

=0V=0V

Max current in the coil

=0.0603A=0.0603A

Average power loss

=0.018W =0.018W

(Power comes from the external rotor)

Radius of the circular coil,

r=8cm=0.08mr=8cm=0.08m

Area of the coil, A=πr2=π×(0.08)2m2A=πr2=π×(0.08)2m2

Number of turns on the coil,

N=20N=20

Angular speed, ω=50rad/sω=50rad/s

Magnetic field strength, B=3×10−2TB=3×10−2T

Resistance of the loop,

R=10 ΩR=10 Ω

Maximum induced emf could be given as:

ε=NωAB=20×50×π×(0.08)2×3×10−2ε=NωAB=20×50×π×(0.08)2×3×10−2

∴ε=0.603V∴ε=0.603V

The maximum emf induced in the coil is found to be

0.603V0.603V

Over a full cycle, the average emf induced in the coil is found to be zero.

Maximum current is given as:

I=εRI=εR

⇒I=0.60310⇒I=0.60310

⇒I=0.0603A⇒I=0.0603A

Average power loss due to joule heating:

∴P=eI2=0.603×0.06032=0.018W∴P=eI2=0.603×0.06032=0.018W

We know that the current induced in the coil would produce a torque opposing the rotation of the coil. Since 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.

10m10m

long extending from east to west is falling with a speed of 5.0ms−15.0ms−1, at right angles to the horizontal component of the earth’s magnetic field, 0.30×10−4Wbm−20.30×10−4Wbm−2.

a) What is the instantaneous value of the emf induced in the wire?

Length of the wire,

l=10ml=10m

Falling speed of the wire,

v=5.0m/sv=5.0m/s

Magnetic field strength, B=0.3×10−4Wbm−2B=0.3×10−4Wbm−2

Emf induced in the wire is thus found to be,

ε=Blvε=Blv

⇒ε=0.3×10−4×5×10⇒ε=0.3×10−4×5×10

∴ε=1.5×10−3V∴ε=1.5×10−3V

Hence, the emf induced in the wire is ε=1.5×10−3Vε=1.5×10−3V.

5.0A to 0.0A5.0A to 0.0A

in

0.1s0.1s

. If an average emf of

200V200V

is induced, give an estimate of the self-inductance of the circuit.

Initial current, I1=5.0AI1=5.0A

Final current, I2=0.0AI2=0.0A

Change in current, dI=I1−I2=5AdI=I1−I2=5A

Time taken for the change,

t=0.1st=0.1s

Average emf, ε=200Vε=200V

For self-inductance (L) of the coil, we have the relation for average emf that could be given as:

ε=Ldidtε=Ldidt

⇒L=ε(didt)⇒L=ε(didt)

Substituting the given values we get,

∴L=200(50.1)=4H∴L=200(50.1)=4H

Therefore, we found the self induction in the coil to be 4H.

1.5H1.5H

. If the current in one coil changes from

0 to 20A in 0.5s0 to 20A in 0.5s

, what is the change of flux linkage with the other coil?

Mutual inductance of a pair of coils, μ=1.5Hμ=1.5H

Initial current, I1=0AI1=0A

Final current, I2=20AI2=20A

Change in current, dI=I2−I1=20−0=20AdI=I2−I1=20−0=20A

Time taken for the change,

t=0.5st=0.5s

Induced emf, ε=dϕdtε=dϕdt…………………. (1)

Where, dϕdϕis the change in the flux linkage with the coil.

Emf is related with mutual inductance could be given as:

ε=μdIdtε=μdIdt…………………….. (2)

Equating equations (1) and (2), we get,

dϕdt=μdIdtdϕdt=μdIdt

⇒dϕ=1.5×(20)⇒dϕ=1.5×(20)

∴dϕ=30Wb∴dϕ=30Wb

Hence, we found the change in the flux linkage to be 30Wb.

1800km/h1800km/h

. What is the voltage difference developed between the ends of the wing having a span of

25m,25m,

if the Earth’s magnetic field at the location has a magnitude of 5×10−4T5×10−4Tand the dip angle is 30∘30∘.

v=1800 km/h=500 m/sv=1800 km/h=500 m/s

Wingspan of jet plane,

l=25ml=25m

Earth’s magnetic field strength, B=5.0×10−4TB=5.0×10−4T

Angle of dip, δ=30∘δ=30∘

Vertical component of Earth’s magnetic field could be given by,

BV=BsinδBV=Bsinδ

⇒BV=5×10−4sin30∘=2.5×10−4T⇒BV=5×10−4sin30∘=2.5×10−4T

Voltage difference between the ends of the wing can be calculated as,

ε=BV×l×vε=BV×l×v

Substituting the given values,

⇒ε=2.5×10−4×25×500⇒ε=2.5×10−4×25×500

∴ε=3.125V∴ε=3.125V

Hence, the voltage difference developed between the ends of the wings is

3.125V3.125V

0.3T0.3T

at the rate of

0.02Ts−10.02Ts−1

. If the cut is joined and the loop has a resistance of

1.6Ω1.6Ω

how much power is dissipated by the loop as heat? What is the source of this power?

Sides of the rectangular loop are 8cm and 2cm. Hence, area of the rectangular wire loop would be,

A = length × widthA = length × width

Initial value of the magnetic field,

B′=0.3TB′=0.3T

Rate of decrease of the magnetic field, dBdt=0.02T/sdBdt=0.02T/s

Emf developed in the loop is given as:

ε=dϕdtε=dϕdt

Where,

ε=d(AB)dt=AdBdtε=d(AB)dt=AdBdt

⇒ε=16×10−4×0.02=0.32×10−4V⇒ε=16×10−4×0.02=0.32×10−4V

Resistance of the loop, R=1.6ΩR=1.6Ω

The current induced in the loop could be given as:

i=εRi=εR

Substituting the given values,

⇒i=0.32×10−41.6=2×10−5A⇒i=0.32×10−41.6=2×10−5A

Power dissipated in the loop in the form of heat could be given as:

P=i2RP=i2R

⇒P=(2×10−5)2×1.6⇒P=(2×10−5)2×1.6

∴P=6.4×10−10W∴P=6.4×10−10W

The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

12cm12cm

with its sides parallel to X and Y axes is moved with a velocity of

18cm18cm

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 10−3Tcm−110−3Tcm−1 along the negative x-direction (that is it increases by 10−3Tcm−110−3Tcm−1as one moves in the negative x-direction), and it is decreasing in time at the rate of 10−3Ts−110−3Ts−1. Determine the direction and magnitude of the induced current in the loop if its resistance is

4.50mΩ4.50mΩ

Side of the square loop,

s=12cm=0.12ms=12cm=0.12m

Area of the square loop, A=0.12×0.12=0.0144m2A=0.12×0.12=0.0144m2

Velocity of the loop,

v=8cm/s=0.08m/sv=8cm/s=0.08m/s

Gradient of the magnetic field along negative x-direction,

dBdx=10−3Tcm−1=10−1Tm−1dBdx=10−3Tcm−1=10−1Tm−1

And, rate of decrease of the magnetic field,

dBdt=10−3Ts−1dBdt=10−3Ts−1

Resistance of the loop,

R=4.5mΩ=4.5×10−3ΩR=4.5mΩ=4.5×10−3Ω

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

dϕdt=A×dBdx×vdϕdt=A×dBdx×v

⇒dϕdt=144×10−4m2×10−1×0.08⇒dϕdt=144×10−4m2×10−1×0.08

⇒dϕdt=11.52×10−5Tm2s−1⇒dϕdt=11.52×10−5Tm2s−1

Rate of change of the flux due to explicit time variation in field B is given as:

dϕ′dt=A×dBdxdϕ′dt=A×dBdx

⇒dϕ′dt=144×10−4×10−3=1.44×10−5Tm2s−1⇒dϕ′dt=144×10−4×10−3=1.44×10−5Tm2s−1

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

e=1.44×10−5+11.52×10−5e=1.44×10−5+11.52×10−5

=12.96×10−5V=12.96×10−5V

∴Induced current, i=eR∴Induced current, i=eR

⇒i=12.96×10−54.5×10−3⇒i=12.96×10−54.5×10−3

∴i=2.88×10−2A∴i=2.88×10−2A

Therefore, the direction of the induced current is such that there is an increase in the flux through the loop along the positive z-direction.

2cm22cm2

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

90∘90∘

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

7.5mC7.5mC

. The combined resistance of the coil and the galvanometer is

0.50Ω0.50Ω

. Estimate the field strength of magnet.

Area of the small flat search coil,

A=2cm2=2×10−4m2A=2cm2=2×10−4m2

Number of turns on the coil,

N=25N=25

Total charge flowing in the coil,

Q=7.5mC=7.5×10−3CQ=7.5mC=7.5×10−3C

Total resistance of the coil and galvanometer,

R=0.50ΩR=0.50Ω

Induced current in the coil,

I=Induced emf(ε)RI=Induced emf(ε)R

……………………………………………. (1)

Induced

emfemf

is given as:

ε=−Ndϕdtε=−Ndϕdt

………………………………………………… (2)

Where,

dϕ=Induced fluxdϕ=Induced flux

Combining equations (1) and (2), we get

I=−NdϕdtRI=−NdϕdtR

Idt=−NRdϕIdt=−NRdϕ

……………………………………………. (3)

Initial flux through the coil,

ϕi=BAϕi=BA

Where,

B=Magnetic field strengthB=Magnetic field strength

Final flux through the coil,

ϕf=0ϕf=0

Integrating equation (3) on both sides, we have

∫Idt=−NR∫ϕiϕfdϕ∫Idt=−NR∫ϕiϕfdϕ

But total charge could be given as,

Q=∫IdtQ=∫Idt

⇒Q=−NR(ϕf−ϕi)=−NR(−ϕi)=+NϕiR⇒Q=−NR(ϕf−ϕi)=−NR(−ϕi)=+NϕiR

Q=NBARQ=NBAR

⇒B=QRNA⇒B=QRNA

Substituting the given values, we get,

⇒B=7.5×10−3×0.525×2×10−4⇒B=7.5×10−3×0.525×2×10−4

∴B=0.75T∴B=0.75T

Therefore, the field strength of the magnet is found to be

0.75T0.75T

=15cm=15cm

,

B=0.50TB=0.50T

, resistance of the closed loop containing the rod

=9.0mΩ=9.0mΩ

. Assume the field to be uniform.

(Image will be uploaded soon)

Suppose K is open and the rod is moved with a speed of

12cm⋅s−112cm⋅s−1

in the direction shown. Give the polarity and magnitude of the induced

emfemf

Length of the rod,

l=15cm=0.15ml=15cm=0.15m

Magnetic field strength,

B=0.50TB=0.50T

Resistance of the closed loop,

R=9mΩ=9×10−3ΩR=9mΩ=9×10−3Ω

Induced

emf=9mVemf=9mV

Here, polarity of the induced

emfemf

is such that end P shows positive while end Q shows negative ends.

Speed of the rod,

v=12cm/s=0.12m/sv=12cm/s=0.12m/s

We know that the induced

emfemf

could be given as:

ε=Bvlε=Bvl

Substituting the given values, we get,

ε=0.5×0.12×0.15ε=0.5×0.12×0.15

⇒ε=9×10−3v⇒ε=9×10−3v

∴ε=9mV∴ε=9mV

Therefore, the magnitude of the induced emf is found to be

ε=9mVε=9mV

an =d the polarity of the induced emf is such that end P shows positive while end Q shows negative.

F=IBlF=IBl

Where,

I=Current flowing through the rodI=Current flowing through the rod

Substituting the given values, we get,

I=eR=9×10−39×10−3=1AI=eR=9×10−39×10−3=1A

⇒F=1×0.5×0.15⇒F=1×0.5×0.15

∴F=75×10−3N∴F=75×10−3N

Therefore, we found the retarding force on the rod when the key K is closed to be,

F=75×10−3NF=75×10−3N

(=12cm⋅s−1)(=12cm⋅s−1)

when K is closed? How much power is required when K is open?

Speed of the rod,

v=12cm/s=0.12m/sv=12cm/s=0.12m/s

Now, power could be given as:

P=FvP=Fv

Substituting the given values, we get,

⇒P=75×10−3×0.12⇒P=75×10−3×0.12

⇒P=9×10−3W⇒P=9×10−3W

∴P=9mW∴P=9mW

Therefore, we found the power that is required (by an external agent) to keep the rod moving at the same speed

(=12cm⋅s−1)(=12cm⋅s−1)

when K is closed to be

P=9mWP=9mW

and when key K is open, no power is expended.

Power dissipated as heat , P=I2RPower dissipated as heat , P=I2R

⇒P=(1)2×9×10−3⇒P=(1)2×9×10−3

∴P=9mW∴P=9mW

The power dissipated as heat in the closed circuit is found to be

P=9mWP=9mW

and the source of this power is found to be an external agent.

emfemf

would be induced in the coil because the motion of the rod does not cut across the field lines.

30cm30cm

, area of cross-section

25cm225cm2

and number of turns

500500

, carries a current of

2.5A2.5A

. The current is suddenly switched off in a brief time of

10−3s10−3s

. How much is the average back

emfemf

induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Length of the solenoid,

l=30cm=0.3ml=30cm=0.3m

Area of cross-section,

A=25cm2=25×10−4m2A=25cm2=25×10−4m2

Number of turns on the solenoid,

N=500N=500

Current in the solenoid,

I=2.5AI=2.5A

Current flows for time,

t=10−3st=10−3s

Average back

emfemf

e=dϕdte=dϕdt

……………………………………. (1)

Where,

dϕ=Change in flux=NABdϕ=Change in flux=NAB

…………………………………………… (2)

Where,

B=Magnetic field strength=μ0NIlB=Magnetic field strength=μ0NIl

………………………… (3)

Where,

μ0=Permeability of free space=4π×10−7TmA−1μ0=Permeability of free space=4π×10−7TmA−1

Substituting equations (2) and (3) in equation (1), we get,

ε=μ0N2IAltε=μ0N2IAlt

⇒ε=4π×10−7×(500)2×2.5×25×10−40.3×10−3⇒ε=4π×10−7×(500)2×2.5×25×10−40.3×10−3

∴ε=6.5V∴ε=6.5V

Hence, the average back

emfemf

induced in the solenoid is found to be

6.5V6.5V

a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in below figure.

(Image will be uploaded soon)

in the loop at a distance y from the long straight wire (as shown in the given figure).

(Image will be uploaded soon)

Magnetic flux associated with element

dy,dϕ=BdAdy,dϕ=BdA

Where,

dA=Area of element dy=a dydA=Area of element dy=a dy

B=magnetic field at distance y =μ0I2πyB=magnetic field at distance y =μ0I2πy

I=Current in the wireI=Current in the wire

μ0=Permeability of free space=4π×10−7H/mμ0=Permeability of free space=4π×10−7H/m

Carrying out the substitutions accordingly, we get,

⇒dϕ=μ0Ia2πdyy⇒dϕ=μ0Ia2πdyy

⇒ϕ=μ0Ia2π∫dyy⇒ϕ=μ0Ia2π∫dyy

Now, the limit of y will be from x to

a+xa+x

, on applying the limits we get,

⇒ϕ=μ0Ia2x∫xa+xdyy⇒ϕ=μ0Ia2x∫xa+xdyy

⇒ϕ=μ0Ia2π[logey]a+xx⇒ϕ=μ0Ia2π[logey]xa+x

⇒ϕ=μ0Ia2πloge(a+xx)⇒ϕ=μ0Ia2πloge(a+xx)

For mutual inductance M, the flux could be given as:

ϕ=MIϕ=MI

⇒MI=μ0Ia2πloge(ax+1)⇒MI=μ0Ia2πloge(ax+1)

∴M=μ0a2πloge(ax+1)∴M=μ0a2πloge(ax+1)

Therefore, the expression for the mutual inductance between the given long straight wire and the square loop of side a is found to be,

M=μ0a2πloge(ax+1)M=μ0a2πloge(ax+1)

50A50A

and the loop is moved to the right with a constant velocity,

v=10m/sv=10m/s

. Calculate the induced

emfemf

in the loop at the instant when

x=0.2mx=0.2m

. Take

a=0.1ma=0.1m

and assume that the loop has a large resistance.

EmfEmf

induced in the loop,

ε=B′av=(μ0I2πx)avε=B′av=(μ0I2πx)av

We are given the following,

I=50AI=50A

x=0.2mx=0.2m

a=0.1ma=0.1m

v=10m/sv=10m/s

On substituting the given values into the equation, we get,

ε=4π×10−7×50×0.1×102π×0.2ε=4π×10−7×50×0.1×102π×0.2

∴ε=5×10−5V∴ε=5×10−5V

Therefore, induced

emfemf

in the loop at the given instant is found to be,

ε=5×10−5Vε=5×10−5V

λλ

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 (figure). A uniform magnetic field extends over a circular region within the rim. It is given by,

A=−B0k(r≤a;a<R)A=−B0k(r≤a;a<R)

A=0 (otherwise)A=0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

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=λ=Total chargeLength=Q2πr=λ=Total chargeLength=Q2πr

Where,

r = Distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field,

B⃗ =−B0kB→=−B0k

At distance r, the magnetic force would be balanced by the centripetal force i.e.,

BQv=Mv2rBQv=Mv2r

Where,

v=linear velocity of the wheelv=linear velocity of the wheel

⇒B2πrλ=Mvr⇒B2πrλ=Mvr

⇒v=B2πλr2M⇒v=B2πλr2M

∴Angular velocity,ω=vR=B2πλr2M∴Angular velocity,ω=vR=B2πλr2M

r≤ar≤a

, and

a<Ra<R

we would get:

ω=2πB0a2λMRkω=2πB0a2λMRk

Therefore, we found the angular velocity of the wheel after the field is suddenly switched off to be given as,

ω=2πB0a2λMRkω=2πB0a2λMRk

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