NCERT Solutions for Class 12 Physics chapter 6 Electromagnetic Induction

 

NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction: In the chapter moving charges and magnetism you have studied that moving charges produces a magnetic field. The solutions of NCERT class 12 physics chapter 6 electromagnetic induction focus on the question based on the concept that the changing magnetic field can produce an emf across an electrical conductor. The equipment like an electric generator and the transformer work on this principle. In chapter 1 you have learnt about electric flux and in NCERT grade 12 chapter 6 electromagnetic induction you will learn about magnetic flux. The questions explained in the CBSE NCERT solutions for class 12 physics chapter 6 electromagnetic induction are based on the depth of understanding the concept. Many questions based on finding the directions of the magnetic field and current are explained in the NCERT solutions for class 12 physics chapter 6 electromagnetic induction which will be useful in exams. NCERT solutions are an important tool to perform better in the exams.

NCERT solutions for class 12 physics chapter 6 electromagnetic induction exercise:

Q 6.1(a) Predict the direction of induced current in the situations described by the following

  

Answer:

To oppose the magnetic field current should flow in anti-clockwise, so the direction of the induced current is qrpq

Q 6.1 (b)  Predict the direction of induced current in the situations described by the following Figs.

           

Answer:

Current in the wire in a way such that it opposes the change in flux through the loop. Here hence current will induce in the direction of p--->r--->q in the first coil and y--->z--->x in the second coil.

Q 6.1 (c) Predict the direction of induced current in the situations described by the following Figs.(c)

Answer:

When we close the key, the current will flow through the first loop and suddenly magnetic flux will flow through it such that magnetic rays will go from right to left of the first loop. Now, to oppose this change currently in the second loop will flow such that magnetic rays go from left to right which is the direction yzxy

Q 6.1 (d) Predict the direction of induced current in the situations described
            by the following Fig. (d)

       

Answer:

hen we increase the resistance of the rheostat, the current will decrease which means flux will decrease so current will be induced to increase the flux through it. Flux will increase if current flows in xyzx.

On the other hand, if we decrease the resistance that will increase the current which means flux will be an increase, so current will induce to reduce the flux. Flux will be reduced if current goes in direction zyxz

Q 6.1 (e)   Predict the direction of induced current in the situations described  by the following Fig(e)

           

Answer:

As we release the tapping key current will induce to increase the flux. Flux will increase when current flows in direction xryx.

Q 6.1 (f)  Predict the direction of induced current in the situations described by the following Fig (f)

            

Answer:

The current will not induce as the magnetic field line are parallel to the plane. In other words, since flux through the loop is constant (zero in fact), there won't be any induction of the current.

Q6.2 (a) 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;

            

Answer:

By turning the wire from irregular shape to circle, we are increasing the area of the loop so flux will increase so current will induce in such a way that reduces the flux through it. By right-hand thumb rule direction of current is adcba.

Q6.2 (b)  Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19 b :

 A circular loop being deformed into a narrow straight wire.

            

Answer:

Here, by changing shape, we are decreasing the area or decreasing the flux, so the current will induce in a manner such that it increases the flux. Since the magnetic field is coming out of the plane, the direction of the current will be adcba.

Q6.3   A long solenoid with 15 turns per cm has a small loop of area 2.0 cm^{2} 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 0.1s, what is the induced emf in the loop while the current is changing?

Answer:

Given in a solenoid,

The number of turn per unit length :

  n = 15turn/cm=1500turn/m 

loop area :

 A=2cm^2=2*10^{-4}m

Current in the solenoid :

initial current = I_{initial}=2

finalcurrent = I_{final}=4

change in current :

 \Delta I= 4-2 = 2

change in time:

\Delta t=0.1s

Now, the induced emf :

e=\frac{d\phi }{dt}=\frac{d(BA)}{dt}=\frac{d(\mu _0nIA)}{dt} = \mu _0nA\frac{dI}{dt}=\mu _0nA\frac{\Delta I}{\Delta t}

 

e=4\pi*10^{-7}*1500*2*10^{-4}*\frac{2}{0.1}=7.54*10^{-6}

hence induced emf in the loop is 7.54*10^{-6}.

Q6.4   A rectangular wire loop of sides 8 cm and 2cm  with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^{-1} 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?

Answer:

Given:

Length of rectangular loop :

l=8cm=0.08m

Width of the rectangular loop:

b=2cm=0.02m

Area of the rectangular loop:

A=l*b=(0.08)(0.02)m^2=16*10^{-4}m^2

Strength of the magnetic field

B=0.3T

The velocity of the loop :

v=1cm/s=0.01m/s

Now,

a) Induced emf in long side wire of rectangle:

e=Blv=0.3*0.08*0.01=2.4*10^{-4}V

this emf will be induced till the loop gets out of the magnetic field, so 

time for which emf will induce :

t=\frac{distnce }{velocity}=\frac{b}{v}=\frac{2*10^{-2}}{0.01}=2s

Hence a 2.4*10^{-4}V emf will be induced for 2 seconds.

b) Induced emf when we move along the width of the rectangle:

e=Bbv=0.3*0.02*0.01=6*10^{-5}V

time for which emf will induce :

t=\frac{distnce }{velocity}=\frac{l}{v}=\frac{8*10^{-2}}{0.01}=8s

Hence a 6*10^{-5}V emf will induce for 8 seconds.

Q6.5 1.0m  long metallic rod is rotated with an angular frequency of  400 rad\: s^{-1} 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.

Answer:

Given 

length of metallic rod :

l=1m

Angular frequency of rotation :

\omega = 400s^{-1}

Magnetic field (which is uniform)

B= 0.5T

Velocity: here velocity at each point of the rod is different. one end of the rod is having zero velocity and another end is having velocity \omega r. and hence we take the average velocity of the rod so,

Average velocity =\frac{0+\omega l}{2}=\frac{\omega l }{2}

Now,

Induce emf 

e=Blv=Bl\frac{wl}{2}=\frac{Bl^2\omega}{2}

e=\frac{0.5*1^2*400}{2}=100V

Hence emf developed is 100V.

Q6.6  A circular coil of radius 8.0 cm and 20 turns  is rotated about its vertical diameter with an angular speed of 50 \: rads\: s^{-1}  in a uniform horizontal magnetic field of magnitude 3.0\times 10^{-2} T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10\Omega , calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Answer:

Given

The radius of the circular loop r=8cm=0.08m

Number of turns N = 20

Flux through each turn 

\phi = B.A=BAcos\theta =BAcos\omega t=B\pi r^2cos\omega t

Flux through N turn

\phi =NB\pi r^2cos\omega t

Induce emf:

e=\frac{d\phi }{dt}=\frac{d(NB\pi r^2cos\omega t)}{dt}=NBr^2\pi \omega sin\omega

Now,

maximum induced emf (when sin function will be maximum)

 e_{max}=NB\pi r^2\omega=20*50*\pi *(0.08)^2*3*10^{-2}=0.603V

 

Average induced emf

e_{average}=0  as the average value of sin function is zero,

 

Maximum current when resistance R of the loop is 10\Omega.

I_{max}=\frac{e_{max}}{R}=\frac{0.603}{10}=0.0603A

 

Power loss :

P_{loss}= \frac{1}{2}E_0I_0=\frac{1}{2}(0.603)(0.0603)=0.018W

Here, power is getting lost as emf is induced and emf is inducing because we are MOVING the conductor in the magnetic field. Hence external force through which we are rotating is the source of this power.

Q6.7 (a) A horizontal straight wire 10m  long extending from east to west isfalling with a speed of 5.0 m\: s^{-1}, at right angles to the horizontal component of the earth’s magnetic field, 0.30\times 10^{-4}\: wb\: m^{-2}.

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

Answer:

Given

Length of the wire l=10m

Speed of the wire v=5m/s

The magnetic field of the earth B=0.3*10^{-4}Wbm^{-2}

Now,

The instantaneous value of induced emf :

e=Blv=0.3*10^{-4}*10*5=1.5*10^{-3}

Hence instantaneous emf induce is 1.5*10^{-3}.

Q6.8  Current in a circuit falls from 5.0A to 0.0A  in 0.1s . If an average emf  of 200V  induced, give an estimate of the self-inductance of the circuit.

Answer:

Given

Initial current I_{initial}=5A

Final current I_{final}=0A

Change in time I_{final}=\Delta t=0.1s

Average emf e=200V

Now,

As we know, in an inductor

e=L\frac{di}{dt}=L\frac{\Delta I}{\Delta t}=L\frac{I_{final}-I_{initial}}{\Delta t}

L= \frac{e\Delta t}{I_{final}-I_{initial}}=\frac{200*0.1}{5-0}=4H

Hence self-inductance of the circuit is 4H.

Q6.9  A pair of adjacent coils has a mutual inductance of 1.5H. If the current in one coil changes from 0 to 20A in 0.5s, what is the change of flux linkage with the other coil?

Answer:

Given

Mutual inductance between two coils:

M = 1.5H

Currents in a coil: 

I_{initial}=0

I_{final}=20

Change in current:

 di=20-0=20

The time taken for the change 

dt=0.5s

The relation between emf and mutual inductance:

e=M\frac{di}{dt}

e= \frac{d\phi }{dt}=M\frac{di}{dt}

d\phi =Mdid\phi =Mdi=1.5*20=30Wb

Hence, the change in flux in the coil is 30Wb.

Q6.10     A jet plane is travelling towards west at a speed of 1800km/h. What is the voltage difference developed between the ends of the wing having a span of 25m , if the Earth’s magnetic field at the location has a  magnitude of 5\times 10^{-4}T and the dip angle is 30^{0}.

Answer:

Given

Speed of the plane:

v=1800kmh^{-1}=\frac{1800*1000}{60*60}=500m/s

Earth's magnetic field at that location:

B = 5 *10^{-4}T

The angle of dip that is angle made with horizontal by earth magnetic field:

 \delta = 30 ^0

Length of the wings 

l=25m

Now, Since the only the vertical component of the magnetic field will cut the wings of plane perpendicularly, only those will help in inducing emf.

The vertical component of the earth's magnetic field :

B_{vertical}=Bsin\delta =5*10^{-4}sin30=5*10^{-4}*0.5=2.5*10^{-4}

So now, Induce emf :

e=B_{vertical}lv=2.5*10^{-4}*25*500=3.125V

Hence voltage difference developed between the ends of the wing is  3.125V.

NCERT solutions for class 12 physics chapter 6 electromagnetic induction additional exercise:

Q6.11 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 0.3T  at the rate of   0.02\: T s^{-1}. If the cut is joined and the loop has a resistance of 1.6\Omega , how much power is dissipated by the loop as heat? What is the source of this power?

Answer:

Given,

Area of the rectangular loop which is held still:

A = l*b=(0.08)(0.02)m^2=16*10^{-4}

The resistance of the loop:

R=1.6\Omega

The initial value of the magnetic field :

B_{initial}= 0.3T

Rate of decreasing of this magnetic field:

\frac{dB}{dt}=0.02T/s

Induced emf in the loop :

e=\frac{d\phi }{dt}=\frac{d(BA)}{dt}=A\frac{dB}{dt}=16*10^{-4}*0.02=0.32*10^{-4}V

Induced Current :

I_{induced}=\frac{e}{R}=\frac{0.32*10^{-4}}{1.6}=2*10^{-5}A

The power dissipated in the loop:

P=I_{induced}^2R=(2*10^{-5})^2*1.6=6.4*10^{-10}W

The external force which is responsible for changing the magnetic field is the actual source of this power.

Q6.12 A square loop of side 12 cm  with its sides parallel to X and Y  axes is moved with a velocity of 8\: cm\: s^{-1} 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^{-3}Tcm^{-1}  along the negative x-direction (that is it increases by  10^{-3}Tcm^{-1}as one moves in the negative x-direction), and it is decreasing in time at the rate of 10^{-3}Ts^{-1}. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50m\Omega.

Answer:

Given,

Side of the square loop

l=12cm=0.12m

Area of the loop:

A=0.12*0.12m^2=144*10^{-4}m^2

The resistance of the loop:

R=4.5m\Omega = 4.5*10^{-3}\Omega

The velocity of the loop in the positive x-direction

v=8cm/s=0.08m/s

The gradient of the magnetic field in the negative x-direction

\frac{dB}{dx}=10^{-3}T/cm=10^{-1}T/m

Rate of decrease of magnetic field intensity

\frac{dB}{dt}=10^{-3}T/s

Now, Here emf is being induced by means of both changing magnetic field with time and changing with space. So let us find out emf induced by both changing of space and time, individually.

Induced emf due to field changing with time:

e_{withtime}=\frac{d\phi }{dt}=A\frac{dB}{dt}=144*10^{-4}*10^{-3}=1.44*10^{-5}Tm^2/s

Induced emf due to field changing with space:

e_{withspace}=\frac{d\phi }{dt}=\frac{d(BA)}{dt}=A\frac{dB}{dx}\frac{dx}{dt}=A\frac{dB}{dx}v

e_{withspace}=144*10^{4}*10^{-1}*0.08=11.52*10^{-5}Tm^2/s

Now, Total induced emf :

e_{total}=e_{withtime}+e_{withspace}=1.44*10^{-5}+11.52*10^{-5}=12.96*10^{-5}V

Total induced current :

I=\frac{e}{R}=\frac{12.96*10^{-5}}{4.5*10^{-3}}=2.88*10^{-2}A

Since the flux is decreasing, the induced current will try to increase the flux through the loop along the positive z-direction.