NCERT solutions for Class 12 Physics Chapter 6 - Electromagnetic Induction

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

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 A $1.0m$ 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 $\omega r$.

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) 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 $1.5*10^{-3}$ .

Q6.6 (b) A horizontal straight wire $10m$ long extending from east to west is falling 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 direction of the emf?

Answer:

If we apply the Flemings right-hand rule, we see that the direction of induced emf is from west to east.

Q6.6(c) A horizontal straight wire $10m$ long extending from east to west is falling with a speed of $5.0 m\: s^{-1}$ , at right angles to the horizontal component of the earth’s magnetic field, $0.30\times 10^{-4}\: wb\: m^{-2}$ .

Which end of the wire is at the higher electrical potential?

Answer:

The eastern wire will be at the higher potential end.

Q6.7 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.

Q6.8 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:

$e=M\frac{di}{dt}$

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

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

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

Q1: 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.

Q2: 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.

Q3: 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.

Q4: 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 $2cm^{2}$ 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 $90^{0}$ 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 $7.5mC$ . The combined resistance of the coil and the galvanometer is $0.50\Omega$ . Estimate the field strength of magnet.

Answer:

Given,

Area of search coil :

$A=2cm^2=2*10^{-4}m^2$

The resistance of coil and galvanometer

$R=0.5\Omega$

The number of turns in the coil:

$N=25$

Charge flowing in the coil

$Q=7.5mC=7.5*10^{-3}C$

Now.

Induced emf in the search coil

$e=N\frac{d\phi }{dt}=N\frac{\phi _{final}-\phi _{initial}}{dt}=N\frac{BA-0}{dt}=\frac{NBA}{dt}$

$e=iR=\frac{dQ}{dt}R=\frac{NBA}{dt}$

$B=\frac{RdQ}{NA}=\frac{0.5*7.5*10^{-3}}{25*2*10^{-4}}=0.75T$

Hence magnetic field strength for the magnet is 0.75T.

Q5(a): Figure 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 $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown. Give the polarity and magnitude of the induced emf.

Answer:

Given

Length of the rod

$l=15cm=0.15m$

Speed of the rod

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

Strength of the magnetic field

$B=0.5T$

induced emf in the rod

$e=Bvl=0.5*0.12*0.15=9*10^{-3}V$

Hence 9mV emf is induced and it is induced in a way such that P is positive and Q is negative.

Q5(b) Figure 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 $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown.

(b)Is there an excess charge built up at the ends of the rods when $K$ is open? What if $K$ is closed?

Answer:

Yes, there will be excess charge built up at the end of the rod when the key is open. This is because when we move the conductor in a magnetic field, the positive and negative charge particles will experience the force and move into the corners.

When we close the key these charged particles start moving in the closed loop and continuous current starts flowing.

Q5(c) Figure 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 $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown.

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod $PQ$ even though they do experience magnetic force due to the motion of the rod. Explain.

Answer:

When the key K is open there is excess charge at both ends of the rod. this charged particle creates an electric field between both ends. This electric field exerts electrostatic force in the charged particles which cancel out the force due to magnetic force. That's why net force on a charged particle, in this case, is zero.

Q5 (d) Figure 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 $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown. Give the polarity and magnitude of the induced emf.

(d) What is the retarding force on the rod when K is closed?

Answer:

Induced emf = 9mV (calculated in a part of this question)

The resistance of loop with rod = 9 $m\Omega$

Induced Current

$i = \frac{e}{R}=\frac{9mV}{9m\Omega }=1A$

Now,'

Force on the rod

$F= Bil= 0.5*1*0.15=7.5*10^{-2}$

Hence retarding force when k is closed is $7.5*10^{-2}$.

Q5 (e) Figure 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 $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown.

(e) How much power is required (by an external agent) to keep the rod moving at the same speed $(=12cm\: s^{-1})$ when $K$ is closed? How much power is required when $K$ is open?

Answer:

Force on the rod

$F=7.5*10^{-2}$

Speed of the rod

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

Power required to keep moving the rod at the same speed

$P=Fv=7.5*10^{-2}*0.12=9*10^{-3}=9mW$

Hence required power is 9mW.

When the key is open, no power is required to keep moving rod at the same speed.

Q5(f) Figure 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 $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown.

(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?

Answer:

Current in the circuit $i$ = 1A

The resistance of the circuit $R$ = 9m $\Omega$

The power which is dissipated as the heat

$P_{heat}=i^2R=1^2*9*10^{-3}=9mW$

Hence 9mW of heat is dissipated.

We are moving the rod which induces the current. The external agent through which we are moving our rod is the source of the power.

Q54 (g) Figure 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 $rod=15cm$ , $B=0.50T$ , resistance of the closed loop containing the $rod=9.0m\Omega$ . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of $12cm\: s^{-1}$ in the direction shown.

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Answer:

If the magnetic field is parallel to the rail then, the motion of the rod will not cut across the magnetic field lines and hence no emf will induce. Hence emf induced is zero in this case.

Q6 An air-cored solenoid with length $30cm$ , area of cross-section $25cm^{2}$ and number of turns $500$ , carries a current of $2.5A$ . The current is suddenly switched off in a brief time of $10^{-3}s$ . 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.

Answer:

Given

Length od the solenoid $l=30cm=0.3m$

Area of the cross-section of the solenoid $A=25cm^2=25*10^{-4}m^2$

Number of turns in the solenoid $N=500$

Current flowing in the solenoid $I=2.5A$

The time interval for which current flows $\Delta t=10^{-3}s$

Now.

Initial flux:

$\phi _{initial}=NBA=N\left ( \frac{\mu _0NI}{l} \right )A=\frac{\mu _0N^2IA}{l}$

$\phi _{initial}=\frac{4\pi*10^{-7}*500^2*2.5*25*10^{-4}}{0.3}=6.55*10^{-3}Wb$

Final flux: since no current is flowing,

$\phi _{final}=0$

Now

Induced emf:

$e=\frac{d\phi}{dt}=\frac{\Delta \phi}{\Delta t}=\frac{\phi_{final}-\phi_{initial}}{\Delta t}=\frac{6.55*10^{-3}-0}{10^{-3}}=6.55V$

Hence 6.55V of average back emf is induced.

Q7(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.

Answer:

Here let's take a small element dy in the loop at y distance from the wire

Area of this element dy :

$dA=a*dy$

The magnetic field at dy (which is y distance away from the wire)

$B=\frac{\mu _0I}{2\pi y}$

The magnetic field associated with this element dy

$d\phi =BdA$

$d\phi=\frac{\mu _0I}{2\pi y}*ady=\frac{\mu _0Ia}{2\pi}\frac{dy}{y}$

$\phi=\int_{x}^{a+x}\frac{\mu _0Ia}{2\pi}\frac{dy}{y}=\frac{\mu _0Ia}{2\pi}[lnx]^{a+x}_a=\frac{\mu _0Ia}{2\pi}ln[\frac{a+x}{x}]$

Now As we know

$\phi=MI$ where M is the mutual inductance

so

$\phi=MI=\frac{\mu _0Ia}{2\pi}ln[\frac{a+x}{x}]$

$M=\frac{\mu _0a}{2\pi}ln[\frac{a+x}{x}]$

Hence mutual inductance between the wire and the loop is:

$\frac{\mu _0a}{2\pi}ln[\frac{a+x}{x}]$

Q7 (b) Now assume that the straight wire carries a current of $50A$ and the loop is moved to the right with a constant velocity, $V=10m/s$ . Calculate the induced emf in the loop at the instant when $X=0.2m$ . Take $a=0.1m$ and assume that the loop has a large resistance.

Answer:

Given,

Current in the straight wire

$I$ = 50A

Speed of the Loop which is moving in the right direction

$V=10m/s$

Length of the square loop

$a=0.1m$

Distance from the wire to the left side of the square

$X=0.2m$

Now,

Induced emf in the loop :

$E=B_xav=\frac{\mu _0I}{2\pi x}av=\left ( \frac{4\pi*10^{-7}*50}{2\pi*0.2} \right )*0.1*10=5*10^{-5}V$

Hence emf induced is $5*10^{-5}V$ .

Q8: A line charge $\lambda$ 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 . A uniform magnetic field extends over a circular region within the rim. It is given by, $B=\: -B_{0}K\: \: \: \: (r\leq a;\: a< R) =0\: \: \: \: \: \: \: \: \: \: (otherwise)$
What is the angular velocity of the wheel after the field is suddenly switched off?

Answer:

Given,

The radius of the wheel =R

The mass of the wheel = M

Line charge per unit length when the total charge is Q

$\lambda=\frac{Q}{2\pi r}$

Magnetic field :

$B=\: -B_{0}K\: \: \: \: (r\leq a;\: a< R) =0$

Magnetic force is balanced by centrifugal force when $v$ is the speed of the wheel that is

$BQv=\frac{mv^2}{r}$

$B2\pi r\lambda=\frac{Mv}{r}$

$v=\frac{B2\pi \lambda r^2}{M}$

Angular velocity of the wheel

$w=\frac{v}{r}=\frac{B2\pi \lambda r^2}{MR}$

when $(r\leq a;\: a< R)$

$w=-\frac{B2\pi \lambda a^2}{MR}$

It is the angular velocity of the wheel when the field is suddenly shut off.

Class 12 physics NCERT Chapter 6: Higher Order Thinking Skills (HOTS) Questions

Q.1 A conducting circular loop is placed in a uniform magnetic field $0.4 \mathrm{~T}$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $1 \mathrm{~mm} \mathrm{~s}^{-1}$. The induced emf in the loop when the radius is $5 \mathrm{~cm}$ is:
a) $2 \pi \times 10^{-5} \mathrm{~V}$
b) $4 \pi \times 10^{-5} \mathrm{~V}$
c) $\pi \times 10^{-3} \mathrm{~V}$
d) $5 \pi \times 10^{-4} \mathrm{~V}$

Answer:

$e=\frac{d \phi}{d t}=\frac{d}{d t}(B A)$

$=B \frac{d}{d t}\left(\pi r^2\right)=\pi B \times 2 r\left(\frac{d r}{d t}\right)$

$\begin{aligned} & =\pi \times 0.4 \times 2 \times 5 \times 10^{-2} \times 10^{-3} \\ e & =4 \pi \times 10^{-5} \mathrm{~V}\end{aligned}$

Hence, the correct option is (2).

Q. 2 At time t=0 magnetic field of 1500 gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnets field reduces linearly to 1000 gauss in the next 10 sec. Then induced FMF in the loop is ____ $\mu \mathrm{v}$.

Answer:

$\begin{aligned} & \text { Area of loop }=\text { area of the rectangle } \text { - area of } 2 \text { triangle } \\ & =32 \times 8-2 \times \frac{1}{2} \times 8 \times 4 \\ & =256-32 \\ & =224 \mathrm{~cm}^2\end{aligned}$

Magnetise flux passing through an area is given by

$\oint=\int B \cdot d A=\int B d A \cos \theta$

$\phi_{\text {initial }}=1500 \times 224 \mathrm{~cm}^2$

$\begin{aligned} & =336000 \text { gauss cm } & \\ & =3.36 \times 10^5 \times 10^{-8} \\ & =3.36 \times 10^{-3}\end{aligned}$

$\phi_{\text {final }}=1000 \times 224 \mathrm{~cm}^2=224000$ gauss $\mathrm{cm}^2=2.24 \times 10^{-3}$ since the film changes linearly.

$\begin{aligned} \varepsilon=\frac{d \phi}{d t}=\frac{\left|\phi_{\text {final }}-\phi_{\text {initial }}\right|}{\Delta t} & =\frac{1.12 \times 10^{-3}}{10}=1.12 \times 10^{-4} = 112 \mu \mathrm{v} \\ & \end{aligned}$

Hence, the answer is $112$.

Q.3 The configuration of a long straight wire and a wire loop with a = 15 cm and b = 20 cm is depicted in Fig. The wire's current fluctuates in accordance with the relationship i = 6t2 – 12 t, where t is measured in seconds and I is in amperes. What is the emf (in 10-7 volts) in the square loop at t = 6 s?

Answer:

The field (due to the current in the long straight wire) through the part of the rectangle above the wire is out of the page (by the right-hand rule) and below the wire, it is into the page. Thus, a strip below the wire (where the strip borders the long wire and reaches a distance of b - an away from it) has exactly the equal but opposite flux that cancels the contribution from the section above the wire, since the height of the part above the wire is b - a.

$ \phi_B=\int BdA=\int_{b-a}^a\left\{\frac{\mu_0 i}{2 \pi r}\right\}(b d r)=\frac{\mu_0 i b}{2 \pi} \ln \left\{\frac{a}{b-a}\right\}$

As per Faraday law:

$\quad \varepsilon=-\frac{d \phi_B}{d t}=\frac{-d}{d t}\left[\frac{\mu_0 i b}{2 \pi} \ln \left(\frac{a}{b-a}\right)\right]$
$ \Rightarrow \varepsilon=\frac{-\mu_0 b}{2 \pi} \ln \left(\frac{a}{b-a}\right) \frac{d i}{d t} $
$ \Rightarrow \varepsilon=\frac{-\mu_0 b}{2 \pi} \ln \left(\frac{a}{b-a}\right) \frac{d}{d t}\left(6 t^2-12 t\right)$
$\\ \Rightarrow \varepsilon=\frac{\mu_0 b(12 t-12)}{2 \pi} \ln \left(\frac{a}{b-a}\right)$

$\text{As, }a=0.15 \mathrm{~m}, b=0.2 \mathrm{~m}, t=6 \mathrm{~s}$
So,
$\varepsilon=\frac{4 \pi \times 10^{-7} \times 0.2(12 \times 6-12)}{2 \pi} \ln \left\{\frac{0.15}{0.20 - 0.15}\right\} =24 \times 10^{-7} \times =26.36 \times 10^{-7} \mathrm{~V}$

Hence, the answer is (26.36)

Q. 4 The self-inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross-section of the coil corresponding to current 2 mA is:

Answer:

Here, $\mathrm{N}=400, \mathrm{~L}=10 \mathrm{mH}=10 \times 10^{-3} \mathrm{H}, $

$\mathrm{I}=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A}$

Total magnetic flux linked with the coil,

$
\phi=\mathrm{NLI}=400 \times\left(10 \times 10^{-3}\right) \times 2 \times 10^{-3}=8 \times 10^{-3} \mathrm{~Wb}
$

Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn

$
=\frac{\phi}{\mathrm{N}}=\frac{8 \times 10^{-3}}{200}=4 \times 10^{-5} \mathrm{~Wb}
$

Q.5 Two solenoids of an equal number of turns have their lengths and radii in the same ratio 1: 2. The ratio of their self-inductances will be:
Answer:

Self-inductance of solenoids, $\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}=\frac{\mu_0 N^2 \pi r^2}{l}$
Where $l$ is the length of the solenoids, N is the total number of turns of the solenoid and A is the area of the cross-section of the solenoid.

$
\begin{aligned}
& \therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\left(\frac{\mathrm{N}_1}{\mathrm{~N}_2}\right)^2\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2\left(\frac{l_2}{l_1}\right) \\
& \text { Here, } \mathrm{N}_1=\mathrm{N}_2, \frac{l_1}{l_2}=\frac{1}{2}, \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{2} \\
& \therefore \quad \frac{\mathrm{~L}_1}{\mathrm{~L}_2}=\left(\frac{1}{2}\right)^2\left(\frac{2}{1}\right)=\frac{1}{2}
\end{aligned}
$