Hc Verma II Solutions for Class 12 Science Physics Chapter 38 Electromagnetic Induction are provided here with simple step-by-step explanations. These solutions for Electromagnetic Induction are extremely popular among Class 12 Science students for Physics Electromagnetic Induction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma II Book of Class 12 Science Physics Chapter 38 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma II Solutions. All Hc Verma II Solutions for class Class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 303:

#### Answer:

If the flux through the loop does not vary, there will be no induced emf. Because the magnetic field is non-uniform and does not change with time, there will be no change in the magnetic flux through the loop. Hence, no emf will be induced in the loop.

#### Page No 303:

#### Answer:

When we close the switch, the current takes some time to grow in the circuit. Due to this growth of current, the flux increases; hence, an emf is induced. On the other hand, when we open the switch, there is no path for the current to flow; hence, it suddenly drops to zero. This rate of decrease of current is much greater than the rate of growth of current when the switch is closed. So, when the switch is opened, the induced emf is more.

#### Page No 303:

#### Answer:

When the ends of the coil are not connected, the coil acts as an inductor in which oscillations persist until the current decays slowly. When these ends are connected, the coil forms a close loop; hence, there is inductance across the ends and the coil does not behave like an inductor. Therefore, all oscillations stop at once.

#### Page No 303:

#### Answer:

Consider the above situation in which a magnet is moved towards a conducting circular loop. The north pole of the magnet faces the loop. As the magnet comes closer to the loop, the magnetic field increases; hence, flux through the loop increases. According to Lenz's law, the direction of induced current is such that it opposes the magnetic field that has induced it. Thus, the induced current produces a magnetic field in the direction opposite to the original field; hence, the loop repels the magnet.

On the other hand, when the magnet is going away from the loop, the magnetic field decreases. Hence, flux through the loop decreases. According to Lenz's law, the induced current produces a magnetic field in the opposite direction of the original field; hence, the loop attracts the magnet.

#### Page No 303:

#### Answer:

Consider loops A and B placed coaxially as shown above. Let the direction of the current in loop A be clockwise when the battery is connected to it. According to the right-hand screw rule, the direction of the magnetic field due to this current will be towards left, as seen from the side of B. Due to a sudden flux through loop B, a current will be induced in it. It will only be induced for a moment when the current suddenly jumps from zero to a constant value. After it has attained a constant value, there will be no induced current. Now, according to Lenz's law, the direction of the induced current in loop B will be such that it will oppose the magnetic field due to loop A. Hence, a current will be induced in anti-clockwise direction in loop B. The induced current will flow in loop B as soon as the current grows in loop A and will end when the current through loop A becomes zero. Because the directions of the currents in the loops are opposite, they will repel each other.

#### Page No 303:

#### Answer:

When the battery is suddenly disconnected, a current is induced in loop B due to a sudden change in the flux through it. It is only induced for a moment when the current suddenly falls to zero. There is no induced current after it has fallen to zero. According to Lenz's law, the induced current is such that it increases the decreasing magnetic field. So, if the current in loop A is in clockwise direction, the induced current in loop B will also be in clockwise direction. Hence, the two loops will attract each other.

#### Page No 303:

#### Answer:

The varying magnetic field induces eddy currents on the walls of the copper box. There is a magnetic field due to the induced eddy currents, that is in opposite direction. As copper has good conductivity, thus the magnetic field due to the eddy currents will be strong. The magnetic field induced due to eddy currents in the copper walls cancel the original magnetic field. Thus, magnetic field does not penetrate the enclosure made of copper. The magnetic field inside the box remains zero. This is how a copper box protects the inside material from varying magnetic fields.

#### Page No 303:

#### Answer:

When solid waste is allowed to slide over a permanent magnet, an emf is induced in metallic particles. This is because magnetic flux linked with the particles changes in this case. According to Lenz's law this induced emf opposes its cause i.e. downward motion along the inclined plane of the permanent magnet. On the other hand, non-metallic or insulating particles are free from such effects. As a result, the metallic particles slow down and hence get separated from the waste (or non-metallic particles).

#### Page No 303:

#### Answer:

An aluminium bar falls slowly through a small region containing a magnetic field because of the induced eddy currents (or induced emf) in it. According to Lenz's law this induced eddy current oppose its cause (its motion). Hence, it slows down while falling through a region containing a magnetic field. On the other hand, non-metallic or insulating materials are free from such effects.

#### Page No 303:

#### Answer:

When the circuit is on, eddy currents are produced on the surface of the metallic bob. Due to these eddy currents, thermal energy is generated in it. This thermal energy comes at the cost of the kinetic energy of the bob; hence, oscillations are more quickly damped when the circuit is on compared to when the circuit is off.

#### Page No 304:

#### Answer:

(a) For the largest mutual inductance, the two loops should be placed coaxially. In this case, flux through a loop due to another loop is the largest; hence, mutual inductance is the largest.

(b) For the smallest mutual inductance, the two loops should be placed such that their axes are perpendicular to each other. In this case, flux through a loop due to another loop is the smallest (zero); hence, mutual inductance is the smallest.

#### Page No 304:

#### Answer:

Self-inductance per unit length of solenoid is given as,

$\frac{L}{l}={\mu}_{0}{n}^{2}A$

From the above equation we can see that, self inductance per unit length will depend on the permeability of free space (${\mu}_{0}$), number of turns per unit length (*n*) and area of the cross-section of the solenoid (*A*). All the above factors are constant at the centre and near any end of the solenoid therefore self inductance at both the points will be same.

#### Page No 304:

#### Answer:

In a solenoid energy is stored in the form of magnetic field. If a constant current is flowing from a solenoid then magnetic field inside the solenoid is uniform. Therefore, energy per unit volume (or energy density) in the magnetic field inside the solenoid is constant.

The energy per unit volume in the magnetic field is given as,

$u=\frac{B}{2{\mu}_{0}}$, where *B* is uniform magnetic field inside the solenoid.

Therefore, energy density all points inside a solenoid is same.

#### Page No 304:

#### Answer:

(b) $\frac{1}{8}\mathrm{\omega}B{l}^{2}$

Let us consider a small element *dx* at a distance *x* from the centre of the rod rotating with angular velocity *ω *about its perpendicular bisector. The emf induced in the rod because of this small element is given by

$de=Bvl=B\omega xdx$

The emf induced across the centre and end of the rod is given by

$\int de={\int}_{0}^{l/2}B\omega xdx\phantom{\rule{0ex}{0ex}}\Rightarrow E=B\omega {\left[\frac{{x}^{2}}{2}\right]}_{0}^{l/2}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1}{8}B\omega {l}^{2}$

#### Page No 304:

#### Answer:

(a) zero

Let us consider a small element d*x* at a distance *x* from the centre of the rod rotating with angular velocity *ω *about its perpendicular bisector. The emf induced in the small element of the rod because of its motion is given by

$de=B\omega xdx$

The emf induced between the centre of the rod and one of its end is given by

$\int de={\int}_{0}^{l}B\omega xdx\phantom{\rule{0ex}{0ex}}\Rightarrow e=B\omega {\left[\frac{{x}^{2}}{2}\right]}_{0}^{l/2}\phantom{\rule{0ex}{0ex}}\Rightarrow e=\frac{1}{8}B\omega {l}^{2}$

The emf at both ends is the same. So, the potential difference between the two ends is zero.

#### Page No 304:

#### Answer:

(d) a clockwise current-pulse and then an anticlockwise current-pulse

When the switch is closed, the current will flow in downward direction in part AB of the circuit nearest to the closed loop.

Due to current in wire AB, a magnetic field will be produced in the loop. This magnetic field due to increasing current will be the cause of the induced current in the closed loop. According to Lenz's law, the induced current is such that it opposes the increase in the magnetic field that induces it. So, the induced current will be in clockwise direction opposing the increase in the magnetic field in upward direction.

Similarly, when the circuit is opened, the current will suddenly fall in the circuit, leading to decrease in the magnetic field in the loop. Again, according to Lenz's law, the induced current is such that it opposes the decrease in the magnetic field. So, the induced current will be in anti-clockwise direction, opposing the decrease in the magnetic field in upward direction.

#### Page No 304:

#### Answer:

(c) an anticlockwise current-pulse and then a clockwise current-pulse

According to Lenz's law, the induced current in the loop will be such that it opposes the increase in the magnetic field due to current flow in the circuit. Therefore, the direction of the induced current when the switch is closed is anti-clockwise.

Similarly, when the switch is open, there is a sudden fall in the current, leading to decrease in the magnetic field at the centre of the loop. According to Lenz's law, the induced current in the loop is such that it opposes the decrease in the magnetic field. Therefore, the direction of the induced current when the switch is open is clockwise.

#### Page No 304:

#### Answer:

(b) will move with almost constant speed

As the magnet is moving under gravity, the flux linked with the copper tube will change because of the motion of the magnet. This will produce eddy currents in the body of the copper tube. According to Lenz's law, these induced currents oppose the fall of the magnet. So, the magnet will experience a retarding force. This force will continuously increase with increasing velocity of the magnet till it becomes equal to the force of gravity. After this, the net force on the magnet will become zero. Hence, the magnet will attain a constant speed.

#### Page No 304:

#### Answer:

(c) move away from the solenoid

For the circuit,

$E=-L\frac{di}{dt}$

The current will increase in the solenoid, flowing in clockwise direction in the circuit. Due to this increased current, the flux linked with the copper ring with increase with time, causing an induced current. This induced current will oppose the cause producing it. Hence, the current in the copper ring will be in anticlockwise direction. Now, because the directions of currents in the solenoid and ring are opposite, the ring will be repelled and hence will move away from the solenoid.

#### Page No 304:

#### Answer:

(a) Both A and B are true.

Statement A is true, as an emf can be induced by moving a conductor with some velocity *v* in a magnetic field *B*. It is given by

$e=Bvl$

Statement B is true, as an emf can be induced by changing the magnetic field that causes the change in flux *ϕ* through a conductor or a loop. It is given by

$e=-\frac{d\varphi}{dt}$

#### Page No 304:

#### Answer:

The induced emf across ends A and B is given by

$E=Bvl$

This induced emf will serve as a voltage source for the current to flow across resistor* R*, as shown in the figure. The direction of the current is given by Lenz's law and it is anticlockwise.

$i=\frac{Bvl}{R}$

If the wire is replaced by a semicircular wire, the induced current will remain the same, as it depends on the length of the wire and not on its shape (when *B*, *v* and *R* are kept constant).

#### Page No 304:

#### Answer:

(b) b

The emf developed across the ends of the loop is given by

$e=Bvl$

If R is the resistance of the loop, then the power delivered to the loop is given by

$P=\frac{{e}^{2}}{R}=\frac{{B}^{2}{v}^{2}{l}^{2}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow P\propto {v}^{2}$

This relation is best represented by plot b in the figure.

#### Page No 304:

#### Answer:

Consider loops A and B placed coaxially as above. Let the direction of the current in loop A be clockwise when a battery is connected to it. According to the right-hand screw rule, the direction of the magnetic field due to this current will be towards left. Now, the current through this loop will decrease with time due to increase in resistance with temperature. So, the magnetic field due to this current will also decrease with time. This changing current will induce current in loop B. Now, according to Lenz's law, the direction of the induced current in loop B will be such that it will oppose the decrease in the magnetic field due to loop A. Hence, current will be induced in clockwise direction in loop B. Also, because the direction of the currents in the loops is the same, they will attract each other.

#### Page No 305:

#### Answer:

(c) zero

The magnetic field inside the solenoid is parallel to its axis. If the plane of the loop contains the axis of the solenoid, then the angle between the area vector of the circular loop and the magnetic field is zero. Thus, the flux through the circular loop is given by

$\varphi =BA\mathrm{cos}\theta =BA\mathrm{cos}0\xb0=BA$

Here,

*B *= Magnetic field due to the solenoid

*A* = Area of the circular loop

*θ* = Angle between the magnetic field and the area vector

Now, the induced emf is given by

$e=-\frac{d\varphi}{dt}\phantom{\rule{0ex}{0ex}}\because \phi =BA=\mathrm{constant}\phantom{\rule{0ex}{0ex}}\therefore e=0$

We can see that the induced emf does not depend on the varying current through the solenoid and is zero for constant flux through the loop. Because there is no induced emf, no current is induced in the loop.

#### Page No 305:

#### Answer:

(d) zero

Figure (a) shows the square loop moving in its plane with a uniform velocity *v*.

Figure (b) shows the equivalent circuit.

The induced emf across ends AB and CD is given by

$E=Bvl$

On applying KVL in the equivalent circuit, we get

$E-E+iR=0\phantom{\rule{0ex}{0ex}}\Rightarrow i=0$

No current will be induced in the circuit due to zero potential difference between the closed ends.

#### Page No 305:

#### Answer:

(b) The north pole faces the ring and the magnet moves towards it.

(c) The south pole faces the ring and the magnet moves away from it.

It can be observed that the induced current is in anti-clockwise direction. So, the magnetic field induced in the copper ring is towards the observer.

According to Lenz's law, the current induced in a circuit due to a change in the magnetic flux is in such direction so as to oppose the change in flux.

Two cases are possible:

(1) The magnetic flux is increasing in the direction from the observer to the circular coil.

(2) The magnetic flux is decreasing in the direction from the coil to the observer.

So, from the above mentioned points, the following conclusions can be made:

1. The south pole faces the ring and the magnet moves away from it.

2. The north pole faces the ring and the magnet moves towards it.

#### Page No 305:

#### Answer:

(d) none of these

The potential difference across the two ends is given by

$e=Bvl$

It is non-zero only

(i) if the rod is moving in the direction perpendicular to the magnetic field ($\overrightarrow{v}\perp \overrightarrow{B}$)

(ii) if the velocity of the rod is in the direction perpendicular to the length of the rod

( $\overrightarrow{v}\perp \overrightarrow{l}$)

(iii) if the magnetic field is perpendicular to the length of the rod $\overrightarrow{l}\perp \overrightarrow{B}$

Thus, none of the above conditions is satisfied in the alternatives given.

#### Page No 305:

#### Answer:

(c) it is rotated about a diameter

(d) it is deformed

When translated or rotated about its axis, the magnetic flux through the loop does not change; hence, no emf is induced in the loop.

When rotated about a diameter, the magnetic flux through the loop changes and emf is induced. On deforming the loop, the area of the loop inside the magnetic field changes, thereby changing the magnetic flux. Due to the change in the flux, emf is induced in the loop.

#### Page No 305:

#### Answer:

(a) hold the sheet there if the metal is magnetic

(c) move the sheet away from the pole with uniform velocity if the metal is magnetic

(d) move the sheet away from the pole with uniform velocity if the metal is nonmagnetic

The strong magnetic pole will attract the magnet, so a force is needed to hold the sheet there if the metal is magnetic.

If we move the metal sheet (magnetic or nonmagnetic) away from the pole, eddy currents are induced in the sheet. Because of eddy currents, thermal energy is produced in it. This energy comes at the cost of the kinetic energy of the plate; thus, the plate slows down. So, a force is needed to move the sheet away from the pole with uniform velocity.

#### Page No 305:

#### Answer:

(a) magnetic field at the centre

(b) magnetic flux linked with the solenoid

(c) self-inductance of the solenoid

Iron rod has high permeability. When it is inserted inside a solenoid the magnetic field inside the solenoid increases. As magnetic field increases inside the solenoid thus the magnetic flux also increases. The Self-inductance (*L*) of the coil is directly proportional to the permeability of the material inside the solenoid. As the permeability inside the coil increases. Therefore, the self-inductance will also increase.

#### Page No 305:

#### Answer:

(b) rate of Joule heating if the same current goes through them

(d) time constant if one solenoid is connected to one battery and the other is connected to another battery

Because the solenoids are identical, their self-inductance will be the same.

Resistance of a wire is given by

$R=\rho \frac{l}{A}$

Here,

*l* = Length of the wire

A = Area of cross section of the wire

*ρ* = Resistivity of the wire

Because *ρ* and* l *are the same for both wires, the thick wire will have greater area of cross section and hence less resistance than the thin wire.

$\Rightarrow {R}_{thick}{R}_{thin}$

The time constant for a solenoid is given by

$\tau =\frac{L}{R}$

$\therefore {\tau}_{thick}>{\tau}_{thin}$

Thus, time constants of the solenoids would be different if one solenoid is connected to one battery and the other is connected to another battery.

Also, because the self-inductance of the solenoids is the same and the same current flows through them, the magnetic field energy given by $\frac{1}{2}L{i}^{2}$ will be the same.

Power dissipated as heat is given by

$P={i}^{2}R$

*i* is the same for both solenoids.

$\therefore {P}_{thick}<{P}_{thin}$

Because the resistance of the coils are different, the rate of Joule heating will be different for the coils if the same current goes through them.

#### Page No 305:

#### Answer:

(d) Emf induced in the inductor

At time *t *= 0, the current in the *L-R* circuit is zero. The magnetic field energy is given by

$U=\frac{1}{2}L{i}^{2}$, as the current is zero the magnetic field energy will also be zero. Thus, the power delivered by the battery will also be zero. As, the *LR* circuit is connected to the battery at *t* = 0, at this time the current is on the verge to start growing in the circuit. So, there will be an induced emf in the inductor at the same time to oppose this growing current.

#### Page No 305:

#### Answer:

(b) The end *A* becomes positively charged.

Due to electromagnetic induction, emf *e* is induced across the ends of the rod. This induced emf is given by

$e=Bvl$

The direction of this induced emf is from A to B, that is, A is at the higher potential and B is at the lower potential. This is because the magnetic field exerts a force equal to $qvB$ on each free electron where *q* is $-$1.6 × 10^{-16} C. The force is towards AB by Fleming's left-hand rule; hence, negatively charged electrons move towards the end B and get accumulated near it. So, a negative charge appears at B and a positive charge appears at A.

#### Page No 305:

#### Answer:

(a) $\frac{1}{RC}$

(b) $\frac{R}{L}$

(c) $\frac{1}{\sqrt{LC}}$

The time constant of the *RC* circuit is given by

$\tau =RC$

On taking the reciprocal of the above relation, we get

${f}_{1}=\frac{1}{RC}$ ...(1)

*f*_{1} will have the dimensions of the frequency.

The time constant of the *LR* circuit is given by

$\tau =\frac{L}{R}$

On taking the reciprocal of the above relation, we get

${f}_{2}=\frac{R}{L}$ ...(2)

*f*_{2} will have the dimensions of the frequency.

On multiplying eq. (1) and (2), we get

${f}_{1}{f}_{2}=\frac{1}{LC}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{f}_{1}{f}_{2}}=\frac{1}{\sqrt{LC}}$

Thus, $\sqrt{{f}_{1}{f}_{2}}$ will have the dimensions of the frequency.

#### Page No 305:

#### Answer:

(b) The charge on *C* long after *t* = 0 is *εC*.

(d) The current in *L* long after *t* = *t*_{0} is *ε/R.*

The charge on the capacitor at time ''*t*'' after connecting it with a battery is given by,

$Q=C\epsilon \left[1-{e}^{-t/RC}\right]$

Just after *t* = 0, the charge on the capacitor will be

$Q=C\epsilon \left[1-{e}^{0}\right]=0$

For a long after time, $t\to \infty $

Thus, the charge on the capacitor will be

$Q=C\epsilon \left[1-{e}^{-\infty}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow Q=C\epsilon \left[1-0\right]=C\epsilon $

The current in the inductor at time ''*t*'' after closing the switch is given by

$I=\frac{{V}_{\mathrm{b}}}{R}\left(1-{e}^{-tR/L}\right)$

Just before the time *t*_{0}, current through the inductor is given by

$I=\frac{{V}_{\mathrm{b}}}{R}\left(1-{e}^{-{t}_{0}R/L}\right)$

It is given that the time *t*_{0} is very long.

∴ ${t}_{0}\to \infty $

$I=\frac{\epsilon}{R}\left(1-{e}^{-\infty}\right)=\frac{\epsilon}{R}$

When the switch is opened, the current through the inductor after a long time will become zero.

#### Page No 306:

#### Answer:

(a) The quantity$\int $E.*dl *can also be written as:

$\int $E.*dl* = *V *(*V* = Voltage)

Unit of voltage is J/C.

Voltage can be written as:

$\mathrm{Voltage}=\frac{\mathrm{Energy}}{\mathrm{Charge}}$

Dimensions of energy = [ML^{2}T^{-2}]

Dimensions of charge = [IT]

Thus, the dimensions of voltage can be written as:

[ML^{2}T^{-2}] ×[IT]^{−1} = [ML^{2}I^{−1}T^{−3}]

(b) The quantity *vBl* is the product of quantities *v, B* and *L.*

Dimensions of velocity *v* = [LT^{−1}]

Dimensions of length *l* = [L]

The dimensions of magnetic field *B* can be found using the following formula:

$B=\frac{F}{qv}$

Dimensions of force *F* = [MLT^{−2}^{]}

Dimensions of charge *q* = [IT]

Dimensions of velocity = [LT^{−1}]

The dimensions of a magnetic field can be written as:

MI^{−1}T^{−2}

∴ Dimensions of *vBl* = [LT^{−1}] × [MI^{−1}T^{−2}] × [L]= [ML^{2}I^{−1}T^{−3}]

(c) The quantity $\frac{d\mathrm{\varphi}}{dt}$ is equal to the emf induced; thus, its dimensions are the same as that of the voltage.

Voltage can be written as:

$\mathrm{Voltage}=\frac{\mathrm{Energy}}{\mathrm{Charge}}$

Dimensions of energy = [ML^{2}T^{-2}]

Dimensions of charge = [IT]

The dimensions of voltage can be written as:

[ML^{2}T^{-2}] ×[IT]^{−1} = [ML^{2}I^{−1}T^{−3}]

∴ Dimensions of $\frac{d\mathrm{\varphi}}{dt}$ = [ML^{2}I^{−1}T^{−3}]

#### Page No 306:

#### Answer:

According to the principle of homogeneity of dimensions, the dimensions of each term on both the sides of a correct equation must be the same.

Now,

*ϕ* = *at*^{2} + *bt* + *c*

(a) The dimensions of the quantities *at*^{2}, *bt*, *c *and *ϕ* must be the same.

Thus, the units of the quantities are as follows:

$a=\left(\frac{\mathrm{\varphi}}{{t}^{2}}\right)=\left[\frac{\mathrm{\varphi}/t}{t}\right]=\frac{\mathrm{Volt}}{\mathrm{s}}\phantom{\rule{0ex}{0ex}}b=\left[\frac{\mathrm{\varphi}}{t}\right]=\mathrm{Volt}\phantom{\rule{0ex}{0ex}}c=\left[\mathrm{\varphi}\right]=\mathrm{Weber}$

(b) The emf is written as:

$E=\frac{d\varphi}{dt}$ = 2*at* + *b* = 2 × 0.2 × 2 + 0.4 (∵ *a* = 0.2, *b* = 0.4 and *c* = 0.6)

On substituting *t* = 2 s, we get

*E* = 0.8 + 0.4

= 1.2 V

#### Page No 306:

#### Answer:

Given:

Area of the loop = 2.0 × 10^{−3} m^{2}

The following conclusions can be made from the graph given above:

The magnetic flux at point O is 0.

The magnetic flux at point A is given by

*ϕ*_{2} = *B.A* = 0.01 × 2 × 10^{−3}

= 2 × 10^{−5} [∵ *ϕ*_{1} = 0]

The change in the magnetic flux in 10 ms is given by

$\u2206$*ϕ* = 2 × 10^{−5}

The emf induced is given by

$e=\frac{-\u2206\mathrm{\varphi}}{\u2206t}=-\left(\frac{2\times {10}^{-5}-0}{10\times {10}^{-3}}\right)=-2\mathrm{mV}$

The magnetic flux at point B is given by

*ϕ*_{3} = *B.A* = 0.03 × 2 × 10^{−3}

= 6 × 10^{−5}

The change in the magnetic flux in 10 ms is given by

$\u2206$*ϕ* = 6 × 10^{−5} − 2 × 10^{−5} = 4 × 10^{−5}

The emf induced is given by

$e=-\frac{\u2206\varphi}{\u2206t}=-4\mathrm{mV}$

The magnetic flux at point C is given by

*ϕ*_{4} = *B.A* = 0.01 × 2 × 10^{−3}

= 2 × 10^{−5}

The change in the magnetic flux in 10 ms is given by

$\u2206$ϕ = (2 × 10^{−5} − 6 × 10^{−5} ) = − 4 × 10^{−5}

The emf induced is given by

$e=-\frac{\u2206\mathrm{\varphi}}{\u2206t}=4\mathrm{mV}$

The magnetic flux at point D is given by

*ϕ*_{5}* = B.A* = 0

The change in the magnetic flux in 10 ms is given by

*$\u2206$ϕ =* 0 − 2 × 10^{−5}

The emf induced is given by

$e=\frac{-\u2206\varphi}{\u2206t}=-\frac{(-2)\times {10}^{-5}}{10\times {10}^{-3}}=2\mathrm{mV}$

(b) Emf is not constant in the intervals 10 ms‒20 ms and 20 ms‒30 ms.

#### Page No 306:

#### Answer:

Given:

Magnetic field intensity, *B = *0.50 T

Radius of the loop, *r = *5.0 cm = 5 × 10^{−2} m

∴ Area of the loop, *A* = $\mathrm{\pi}{r}^{2}$

Initial magnetic flux in the loop, *ϕ*_{1} = *B *× *A*

*ϕ*_{1}_{ }= 0.5 × $\mathrm{\pi}$(5 × 10^{−2})^{2} = 125$\mathrm{\pi}$* *× 10^{−5}

As the loop is removed from the magnetic field, magnetic flux (*ϕ*_{2}) = 0.

Induced emf *ε* is given by

$\mathrm{\epsilon}=\frac{{\mathrm{\varphi}}_{1}-{\mathrm{\varphi}}_{2}}{t}\phantom{\rule{0ex}{0ex}}=\frac{125\mathrm{\pi}\times {10}^{-5}}{5\times {10}^{-1}}\phantom{\rule{0ex}{0ex}}=25\mathrm{\pi}\times {10}^{-4}$

= 25 × 3.14 × 10^{−4}

= 78.5 × 10^{−4} V = 7.8 × 10^{−3} V

#### Page No 306:

#### Answer:

Given:

Area of the loop, *A* = 1 mm^{2}

Current through the wire,* i* = 10 A

Separation between the wire and the loop, *d* = 20 cm

Time, *dt* = 0.1 s

The average emf induced in the loop is given by

$e=\frac{\mathrm{d\varphi}}{\mathrm{d}t}\phantom{\rule{0ex}{0ex}}=\frac{BA}{\mathrm{d}t}=\frac{{\mu}_{0}i}{2\pi d}\times \frac{A}{\mathrm{d}t}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 10}{2\mathrm{\pi}\times 2\times {10}^{-1}}\times \frac{{10}^{-6}}{1\times {10}^{-1}}\phantom{\rule{0ex}{0ex}}=1\times {10}^{-10}\mathrm{V}$

#### Page No 306:

#### Answer:

(a) When the coil is removed from the magnetic field:

Initial magnetic flux through the coil, *ϕ*_{1} = *BA*

∴ *ϕ*_{1} = 50 × 0.5 × 0.5 T-m^{2}

= 12.5 T-m^{2}

Now,

Initial magnetic flux through the coil, *ϕ*_{2} = 0

Time taken, *t* = 0.25 s

The average emf induced is given by

$e=-\frac{\u2206\mathrm{\varphi}}{\u2206t}=\frac{{\mathrm{\varphi}}_{1}-{\mathrm{\varphi}}_{2}}{dt}\phantom{\rule{0ex}{0ex}}=\frac{12.5-0}{0.25}=\frac{125\times {10}^{-1}}{25\times {10}^{-2}}=50\mathrm{V}$

(b) When the coil is taken back to its original position:

Initial magnetic flux through the coil, *ϕ*_{1} = 0

Initial magnetic flux through the coil, *ϕ*_{2} = 12.5 T-m^{2}

Time taken, *t* = 0.25 s

The average emf induced is given by

$e=\frac{12.5-0}{0.25}=50\mathrm{V}$

(c) When the coil is moving outside the magnetic field:

Initial magnetic flux, *ϕ*_{1} = 0

Final magnetic flux, *ϕ*_{2} = 0

Because there is no change in the magnetic flux, no emf is induced.

#### Page No 306:

#### Answer:

Given:

Resistance of the coil,* R* = 25 Ω

(a) During the removal the emf induced in the coil,

*e* = 50 V

time taken, *t* = 0.25 s

current in the coil, $i=\frac{e}{\mathrm{R}}=2\mathrm{A}$

Thus, the thermal energy developed is given by

*H = **I*^{2}*RT*

= 4 × 25 × 0.25 = 25 J

(b) During the restoration of the coil,

emf induced in it, *e* = 50 V

time taken, *t* = 0.25 s

current in the coil, $i=\frac{e}{\mathrm{R}}=2\mathrm{A}$

Thus, the thermal energy developed is given by

*H = **i*^{2}*RT = *25 J

(c) We know that energy is a scalar quantity. Also, the net thermal energy is the algebraic sum of the two energies calculated.

∴ Net thermal energy developed

= 25 J + 25 J = 50 J

#### Page No 306:

#### Answer:

Given:

Area of the coil, *A* = 5 cm^{2} = 5 × 10^{−4} m^{2}

The magnetic field at time *t* is given by

*B = **B*_{0} sin ω*t* = 0.2 sin (300*t*)

Angle of the normal of the coil with the magnetic field, *θ* = 60°

(a) The emf induced in the coil is given by

$e=\frac{-d\theta}{dt}=\frac{d}{dt}(BA\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}=\frac{d}{dt}\left[\left({B}_{0}\mathrm{sin}\mathrm{\omega}t\right)\times 5\times {10}^{-4}\times 1/2\right]\phantom{\rule{0ex}{0ex}}={B}_{0}\times \frac{5}{2}\times {10}^{-4}\frac{d}{dt}(\mathrm{sin}\mathrm{\omega}t)\phantom{\rule{0ex}{0ex}}=\frac{{B}_{0}5}{2}{10}^{-4}\omega \left(\mathrm{cos}\mathrm{\omega}t\right)\phantom{\rule{0ex}{0ex}}=\frac{0.2\times 5}{2}\times 300\times {10}^{-4}\times \mathrm{cos}\mathrm{\omega}t\phantom{\rule{0ex}{0ex}}=15\times {10}^{-3}\mathrm{cost}\mathrm{\omega}t$

The induced emf becomes maximum when cos *ωt* becomes maximum, that is, 1.

Thus, the maximum value of the induced emf is given by

${e}_{max}=15\times {10}^{-3}=0.015\mathrm{V}$

(b) The induced emf at *t* = $\left(\frac{\mathrm{\pi}}{900}\right)\mathrm{s}$ is given by

*e* = 15 × 10^{−3} × cos ω*t*

= 15 × 10^{−3} × cos $\left(300\times \frac{\mathrm{\pi}}{900}\right)$

= 15 × 10^{−3} × $\frac{1}{2}$

$=\frac{0.015}{2}=0.0075=7.5\times {10}^{-3}\mathrm{V}$

(c) The induced emf at *t* = $\frac{\mathrm{\pi}}{600}\mathrm{s}$ is given by

*e* = 15 × 10^{−3} × cos $\left(300\times \frac{\mathrm{\pi}}{600}\right)$

= 15 × 10^{−3}^{ }× 0 = 0 V

#### Page No 306:

#### Answer:

It is given that the magnitude of the magnetic field is 0.10 T and it is perpendicular to the area of the loop.

Also,

Area of the loop, *A* = 1 cm^{2} = 10^{−4} m

Time taken to remove the magnet completely, *T* = 2 s

Initial magnetic flux, *ϕ* = $\overrightarrow{B}.\overrightarrow{A}$ = *BA* cos(0) = 10^{−1} × 10^{−4} × 1 = 10^{−}^{5}

Now, the induced emf in the magnetic field is given by

$e=-\frac{\u2206\mathrm{\varphi}}{\u2206t}=\frac{{10}^{-5}-0}{1}={10}^{-5}=10\mathrm{\mu V}$

#### Page No 306:

#### Answer:

Given:

Induced emf, *e* = 20 mV = 20 × 10^{−3} V

Area of the loop, *A* = (2 × 10^{−2})^{2} = 4 × 10^{−4} m^{2}

Time taken to rotate the loop, Δ*t* = 0.2 s

The average induced emf is given by

$e=-\frac{\u2206\varphi}{\u2206t}=\frac{{\varphi}_{i}-{\varphi}_{f}}{t}\phantom{\rule{0ex}{0ex}}\varphi =\overrightarrow{B}.\overrightarrow{A}=BA\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{\varphi}_{i}=B(4\times {10}^{-4})\mathrm{cos}0=B(4\times {10}^{-4})\phantom{\rule{0ex}{0ex}}{\varphi}_{f}=B(4\times {10}^{-4})\mathrm{cos}{180}^{\mathrm{o}}=-B(4\times {10}^{-4})\phantom{\rule{0ex}{0ex}}e=\frac{B(4\times {10}^{-4})-\left[-B(4\times {10}^{-4})\right]}{0.2}\phantom{\rule{0ex}{0ex}}20\times {10}^{-3}=\frac{8B\times {10}^{-4}}{2\times {10}^{-1}}$

$\Rightarrow 20\times {10}^{-3}=4\times B\times {10}^{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{20\times {10}^{-3}}{4\times {10}^{-3}}=5\mathrm{T}$

#### Page No 306:

#### Answer:

The magnetic flux through the coil is given by

*ϕ* = *B.A *= *BA *cos 0° = *BA*

It is given that the loop is withdrawn from the magnetic field.

∴ Final flux = 0

The average induced emf is given by

$e=-\frac{\u2206\mathrm{\varphi}}{\u2206t}=\frac{\mathrm{BA}-0}{t}=\frac{BA}{t}\phantom{\rule{0ex}{0ex}}$

The current in the loop is given by

$i=\frac{e}{R}=\frac{BA}{tR}$

The charge flowing through the area of the cross section of the wire is given by

$q=it=\frac{BA}{R}$

#### Page No 306:

#### Answer:

Given:

Radius of the solenoid, *r* = 2 cm = 2 × 10^{−2} m

Number of turns per centimetre, *n* = 100 = 10000 turns/m

Current flowing through the coil, *i* = 5 A

The magnetic field through the solenoid is given by

*B* = μ_{0}*ni* = 4π × 10^{−7} × 10000 × 5

= 20π × 10^{−3} T

Flux linking with per turn of the second solenoid = *B*π*r*^{2} = *B*π × 10^{−4}

Total flux linking the second coil, *ϕ*_{1} = *Bn*_{2}π*r*^{2}

*∴ **ϕ*_{1}_{ }= 100 × π × 10^{−4} × 20π × 10^{−3}

When the direction of the current is reversed, the total flux linking the second coil is given by

*ϕ*_{2} = −*Bn*_{2}π*r*^{2}

* *= −(100 × π × 10^{−4} × 20π × 10^{−3} )

The change in the flux through the second coil is given by

Δ*ϕ* = *ϕ*_{2} − *ϕ*_{1}

= 2 × (100 × π × 10^{−4}^{ }× 20π × 10^{−3})

Now,

$e=\frac{\u2206\mathrm{\varphi}}{\u2206t}=\frac{4{\mathrm{\pi}}^{2}\times {10}^{-4}}{\u2206t}\phantom{\rule{0ex}{0ex}}$

The current through the solenoid is given by

$I=\frac{e}{R}=\frac{4{\mathrm{\pi}}^{2}\times {10}^{-4}}{\u2206t\times 20}$

The charge flown through the galvanometer is given by

$q=I\u2206t=\frac{4{\mathrm{\pi}}^{2}\times {10}^{-4}}{20\times dt}\times \u2206t\phantom{\rule{0ex}{0ex}}=2\times {10}^{-4}\mathrm{C}$

#### Page No 306:

#### Answer:

(a) The effective length of each side is the length that is perpendicular to the velocity of the corners.

Thus, the effective length of each side is *a *sin *θ*.

Net effective length for four sides = 4 × $\frac{a}{2}$ = 2*a*

∴ Induced emf = *Bvl = *2*Bau*

(b) Current in the frame is given by

*i* $=\frac{e}{R}=\frac{2auB}{R}$

(c) Total charge *q*, which flows through the sides of the frame, is given by

$q=\frac{\u2206\varphi}{R}$

Here,

Δ*Φ* = Change in the flux

*R* = Resistance of the coil

$\therefore q=\frac{\u2206\varphi}{R}$

$=\frac{B({a}^{2}-0)}{R}\phantom{\rule{0ex}{0ex}}=\frac{B{a}^{2}}{R}$

#### Page No 307:

#### Answer:

Given:

Initial flux, *ϕ*_{1} = 0.35 weber

Final flux *ϕ*_{2} = 0.85 weber

∴ Δ*ϕ* = *ϕ*_{2} − *ϕ*_{1}

= (0.85 − 0.35) weber

= 0.5 weber

Also,

Δ*t* = 0.5 s

The magnitude of the induced emf is given by

$e=\frac{\u2206\mathrm{\varphi}}{\u2206t}=\frac{0.5}{0.5}=1\mathrm{V}$

The induced current is anti-clockwise when seen from the side of the magnet.

#### Page No 307:

#### Answer:

When the wire loop is rotated in its own plane in a uniform magnetic field, the magnetic flux through it remains the same. Because there is no change in the magnetic flux, the emf induced in the wire loop is zero.

#### Page No 307:

#### Answer:

Given:

Initial velocity, *u* = 1 cm/s

Magnetic field, *B* = 0.6 T

(a) At *t* = 2 s:

Distance moved by the coil = 2 × 1 cm/s = 2 cm = 2 × 10^{$-$2} m

Area under the magnetic field at* t* = 2s, *A* = 2 × 5 × 10^{$-$4} m^{2}

Initial magnetic flux = 0

Final magnetic flux = *BA *= 0.6 × (10 × 10^{$-$4}) T-m^{2}

Change in the magnetic flux, Δ*ϕ* = 0.6 × (10 × 10^{$-$4}) $-$ 0

Now, induced emf in the coil is

$e=\frac{\u2206\mathrm{\varphi}}{\u2206t}\phantom{\rule{0ex}{0ex}}=\frac{0.6\times (10-0)\times {10}^{-4}}{2}\phantom{\rule{0ex}{0ex}}=3\times {10}^{-4}\mathrm{V}$

(b) At *t* = 10 s:

Distance moved by the coil = 10 × 1 = 10 cm

At this time square loop is completely inside the magnetic field, so there is no change in the flux linked with the coil with time.

Therefore, induced emf in the coil at this time is zero.

(c) At *t* = 22 s:

Distance moved = 22 × 1 = 22 cm

At this time loop is moving out of the field.

Initial magnetic flux = 0.6 × (5 × 5 × 10^{$-$4}) T-m

At this time 2 cm part of the loop is ou t of the field.

Therefore, final magnetic flux = 0.6 × (3 × 5 × 10^{$-$4}) T-m

Change in the magnetic flux, Δ*ϕ* = 0.6 × (3 × 5 × 10^{$-$4}) $-$ 0.6 × (5 × 5 × 10^{$-$4}) = $-$6 × 10^{$-$4} T-m^{2}

Now, induced emf is

$e=\frac{\u2206\mathrm{\varphi}}{\u2206t}\phantom{\rule{0ex}{0ex}}=\frac{-6\times {10}^{-4}}{2}\phantom{\rule{0ex}{0ex}}=-3\times {10}^{-4}\mathrm{V}$

(d) At *t* = 30 s:

At this time loop is completely out of the field, so there is no change in the flux linked with the coil with time.

Therefore, induced emf in the coil at this time is zero.

#### Page No 307:

#### Answer:

Resistance of the coil, *R* = 45 mΩ = 4.5 × 10^{−3} Ω

The heat produced is found by taking the sum of the individual heats produced.

Thus, the net heat produced is given by

*H* = *H*_{1} + *H*_{2} + *H*_{3}_{ }+ *H*_{4}

(a) Heat developed for the first 5 seconds:

Emf induced, *e* = 3 × 10^{−4} V

Current in the coil, $i=\frac{e}{\mathrm{R}}=\frac{3\times {10}^{-4}}{4.5\times {10}^{-3}}$ = 6.7 × 10^{−2} A

*H*_{1} = (6.7 × 10^{−2})^{2} × 4.5 × 10^{−3} × 5

There is no change in the emf from 5 s to 20 s and from 25 s to 30 s.

Thus, the heat developed for the above mentioned intervals is given by

*H*_{2} = *H*_{4} = 0

Heat developed in interval *t* = 25 s to 30 s:

The current and voltage induced in the coil will be the same as that for the first 5 seconds.

*H*_{3} = (6.7 × 10^{−2})^{2} × 4.5 × 10^{−3}^{ }× 5

Total heat produced:

*H* = *H*_{1}* + **H*_{3}

= 2 × (6.7 × 10^{−2})^{2} × 4.5 × 10^{−2} × 5

= 2 × 10^{−4} J

#### Page No 307:

#### Answer:

The magnetic field lines pass through coil* abcd* only in the part above the cylindrical region.

Radius of the cylindrical region, *r* = 10 cm

Resistance of the coil,* R* = 4 Ω

The rate of change of the magnetic field in the cylindrical region is constant and is given by

$\frac{dB}{dt}=0.010\mathrm{T}/\mathrm{s}$

The change in the magnetic flux is given by

$\frac{d\mathrm{\varphi}}{dt}=\frac{dB}{dt}A\phantom{\rule{0ex}{0ex}}$

The induced emf is given by

$e=\frac{d\mathrm{\varphi}}{dt}=\frac{dB}{dt}\times A=0.01\left(\frac{\mathrm{\pi}\times {r}^{2}}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{0.01\times 3.14\times 0.01}{2}\phantom{\rule{0ex}{0ex}}=\frac{3.14}{2}\times {10}^{-4}=1.57\times {10}^{-4}\mathrm{V}$

The current in the coil is given by

$i=\frac{e}{R}=\frac{1.57\times {10}^{-4}}{4}\phantom{\rule{0ex}{0ex}}=0.39\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=3.9\times {10}^{-5}\mathrm{A}$

#### Page No 307:

#### Answer:

(a) When switch *S*_{1} is closed and switch *S*_{2} is open:

Rate of change of the magnetic field = 20 mT/s = $2\times {10}^{-4}\mathrm{T}/\mathrm{s}$

Net resistance of the coil *adef*, *R* = 4 × 4 = 16 Ω

Area of the coil *adef* = (10^{−2} )^{−2} = 10^{−4} m^{2}

The emf induced is given by

$e=\frac{d\mathrm{\varphi}}{dt}=\mathrm{A}.\frac{d\mathrm{B}}{dt}\phantom{\rule{0ex}{0ex}}$

= 10^{−4} × 2 × 10^{−2}

= 2 × 10^{−6} V

The current through the wire *ad** *is given by

*i*$=\frac{e}{R}=\frac{2\times {10}^{-6}}{16}$

= 1.25 × 10^{−7} A along *ad*

(b) When switch *S*_{2} is closed and switch *S*_{1} is open:

Net resistance of the coil *abcd*, R = 16 Ω

The induced emf is given by

$e=\mathrm{A}\times \frac{d\mathrm{B}}{dt}=2\times {10}^{-6}\mathrm{V}$

The current through wire *ad* is given by

$i=\frac{20\times {10}^{-6}}{16}=1.25\times {10}^{-7}\mathrm{A}\phantom{\rule{0ex}{0ex}}$ along *da*

(c) When both *S*_{1} and *S*_{2} are open, no current is passed, as the circuit is open. Thus, *i* = 0.

(d) When both *S*_{1} and *S*_{2} are closed, the circuit forms a balanced a Wheatstone bridge and no current flows along *ad*. Thus, *i* = 0.

#### Page No 307:

#### Answer:

The magnetic field due to coil 1 at the centre of coil 2 is given by

$B=\frac{{\mathrm{\mu}}_{0}Ni{a}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}$

The flux linked with coil 2 is given by

$\varphi =B.A\text{'}=\frac{{\mathrm{\mu}}_{0}Ni{a}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\mathrm{\pi}a{\text{'}}^{2}$

Now, let *y* be the distance of the sliding contact from its left end.

Given:

$v=\frac{dy}{dt}$

Total resistance of the rheostat = *R*

When the distance of the sliding contact from the left end is *y*, the resistance of the rheostat (*R*') is given by

$R\text{'}=\frac{R}{L}y$

The current in the coil is the function of distance *y* travelled by the sliding contact of the rheostat. It is given by

$i=\frac{\epsilon}{\left({\displaystyle \frac{R}{L}}y+r\right)}$

The magnitude of the emf induced can be calculated as:

$e=\frac{d\mathrm{\varphi}}{dt}=\frac{{\mu}_{0}N{a}^{2}a{\text{'}}^{2}\mathrm{\pi}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\frac{di}{dt}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow e=\frac{{\mathrm{\mu}}_{0}N\mathrm{\pi}{a}^{2}a{\text{'}}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\frac{d}{dt}\frac{\epsilon}{\left({\displaystyle \frac{R}{L}}y+r\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow e=\frac{{\mathrm{\mu}}_{0}N\mathrm{\pi}{a}^{2}a{\text{'}}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\left[\epsilon \frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^{2}}\right]$

(a) For *y* = *L*,

$e=\frac{{\mathrm{\mu}}_{0}N\mathrm{\pi}{a}^{2}\mathit{}a{\mathit{\text{'}}}^{\mathit{2}}\epsilon \mathit{}Rv}{2L({a}^{2}+{x}^{2}{)}^{3/2}(R+r{)}^{2}}$

(b) For *y* = *L*/*2*,

$\frac{R}{L}y=\frac{R}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow e=\frac{{\mu}_{0}N\pi {a}^{2}a{\text{'}}^{2}}{2L({a}^{2}+{x}^{2}{)}^{3/2}}\frac{\epsilon Rv}{{\left({\displaystyle \frac{R}{2}}+r\right)}^{2}}$

#### Page No 307:

#### Answer:

Given:

Number of turns of the coil, *N* = 50

Magnetic field through the circular coil, $\overrightarrow{B}$ = 0.200 T

Radius of the circular coil, *r* = 2.00 cm = 0.02 m

Angle through which the coil is rotated, *θ* = 60°

Time taken to rotate the coil, *t** *= 0.100 s

(a) The emf induced in the coil is given by

$e=-\frac{N\u2206\varphi}{\u2206t}=\frac{N(\overrightarrow{{B}_{f}}.{\overrightarrow{A}}_{f}-\overrightarrow{{B}_{i}}.{\overrightarrow{A}}_{i})}{T}\phantom{\rule{0ex}{0ex}}=\frac{NB.A(\mathrm{cos}{0}^{\mathrm{o}}-\mathrm{cos}60\xb0)}{T}\phantom{\rule{0ex}{0ex}}=\frac{50\times 2\times {10}^{-1}\times \mathrm{\pi}(0.02{)}^{2}}{2\times 0.1}\phantom{\rule{0ex}{0ex}}=5\times 4\times {10}^{-5}\times \mathrm{\pi}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}\times {10}^{-2}\mathrm{V}=6.28\times {10}^{-3}\mathrm{V}$

(b) The current in the coil is given by

$i=\frac{e}{R}=\frac{6.28\times {10}^{-3}}{4}\phantom{\rule{0ex}{0ex}}=1.57\times {10}^{-3}\mathrm{A}$

The net charge passing through the cross section of the wire is given by

$Q=it=1.57\times {10}^{-3}\times {10}^{-1}\phantom{\rule{0ex}{0ex}}=1.57\times {10}^{-4}\mathrm{C}$

#### Page No 307:

#### Answer:

Given:

Number of turns in the coil,* n* = 100 turns

Magnetic field, *B* = 4 × 10^{−4}

Area of the loop, *A* = 25 cm^{2} = 25 × 10^{−4} m^{2}

(a) When the coil is perpendicular to the field:

*ϕ*_{1} = *nBA*

When the coil goes through the half turn:

*ϕ*_{2} = *nBA* cos 180° = −*nBA*

∴ Δ*ϕ* = 2*nBA*

When the coil undergoes 300 revolutions in 1 minute, the angle swept by the coil is

300 × 2π rad/min = 10π rad/s

10π rad is swept in 1 s.

π rad is swept in $\left(\frac{1}{10\mathrm{\pi}}\right)\mathrm{\pi}=\frac{1}{10}\mathrm{s}$

$e=\frac{d\mathrm{\varphi}}{dt}=\frac{2nBA}{dt}\phantom{\rule{0ex}{0ex}}=\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{1/10}\phantom{\rule{0ex}{0ex}}=2\times {10}^{-3}\mathrm{V}$

(b) *ϕ*_{1} *= nBA, **ϕ*_{2} = *nBA** *(θ = 360°)

Δ*ϕ* = 0, thus emf induced will be zero.

(c) The current flowing in the coil is given by

$i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}$

= 0.5 × 10^{−3} = 5 × 10^{−4} A

Hence, the net charge is given by

*Q* = *idt* = 5 × 10^{−4}^{ }× $\frac{1}{10}$

= 5 × 10^{−5} C

#### Page No 307:

#### Answer:

Given:

Radius of the coil, *r* = 10 cm = 0.1 m

Resistance of the coil, *R* = 40 Ω

Number of turns in the coil, *N* = 1000

Angle of rotation, *θ* = 180°

Horizontal component of Earth's magnetic field, *B*_{H} = 3 × 10^{−5} T

Magnetic flux, *ϕ* = *NBA* cos 180°

⇒ *ϕ *= −*NBA*

= −1000 × 3 × 10^{−5} × π × 1 × 1 × 10^{−2}

= 3π × 10^{−4} Wb

*d*ϕ = 2*NBA* = 6π × 10^{−4} Wb

$e=\frac{d\mathrm{\varphi}}{dt}=\frac{6\pi \times {10}^{-4}}{dt}$ V

Thus, the current flowing in the coil and the total charge are:

$i=\frac{e}{R}=\frac{6\mathrm{\pi}\times {10}^{-4}}{40dt}=\frac{4.71\times {10}^{-5}}{dt}\phantom{\rule{0ex}{0ex}}Q=\frac{4.71\times {10}^{-5}\times dt}{dt}\phantom{\rule{0ex}{0ex}}=4.71\times {10}^{-5}\mathrm{C}$

#### Page No 307:

#### Answer:

Given,

Radius of the circular coil, *R *= 5.0 cm

Angular speed of circular coil, $\omega $ = 80 revolutions/minute

Magnetic field acting perpendicular to the axis of rotation, *B* = 0.010 T

The emf induced in the coil $\left(e\right)$ is given by,

$e=\frac{d\varphi}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow e=\frac{dB.A\mathrm{cos}\theta}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow e=-BA\mathrm{sin}\theta \frac{d\theta}{dt}$

$\Rightarrow $*e* = −*BAω*sin*θ*

($\frac{d\theta}{dt}=\omega $ = the rate of change of angle between the arc vector and B)

(a) For maximum emf, sin*θ* = 1

*$\therefore $ e* = *BAω*

*$\Rightarrow $e* = 0.010 × 25 × 10^{−4} × 80 × $\frac{2\mathrm{\pi}\times \mathrm{\pi}}{60}$

*$\Rightarrow $e* = 0.66 × 10^{−3} = 6.66 × 10^{−4} V

(b) The direction of the induced emf changes every instant. Thus, the average emf becomes zero.

(c) The emf induced in the coil is *e* = −*BAω*sin*θ** = *−*BAω*sin *ωt*

The average of the squares of emf induced is given by

${{e}_{\mathrm{av}}}^{2}=\frac{{\int}_{0}^{T}{B}^{2}{A}^{2}{\omega}^{2}{\mathrm{sin}}^{2}\omega t\mathrm{d}t}{{\int}_{0}^{T}\mathrm{d}t}\phantom{\rule{0ex}{0ex}}\Rightarrow {{e}_{\mathrm{av}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega}^{2}{\int}_{0}^{T}{\mathrm{sin}}^{2}\omega t\mathrm{d}t}{{\int}_{0}^{T}\mathrm{d}t}\phantom{\rule{0ex}{0ex}}\Rightarrow {{e}_{\mathrm{av}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega}^{2}{\int}_{0}^{T}\left(1-\mathrm{cos}2\omega t\right)\mathrm{d}t}{2T}\phantom{\rule{0ex}{0ex}}\Rightarrow {{e}_{\mathrm{av}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega}^{2}}{2T}{\left[t-\frac{\mathrm{sin}2\omega t}{2\omega}\right]}_{0}^{T}\phantom{\rule{0ex}{0ex}}\Rightarrow {{e}_{\mathrm{av}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega}^{2}}{2T}\left[T-\frac{\mathrm{sin}4\pi -\mathrm{sin}0}{2\omega}\right]=\frac{{B}^{2}{A}^{2}{\omega}^{2}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {{e}_{\mathrm{av}}}^{2}=\frac{(6.66\times {10}^{-4}{)}^{2}}{2}=22.1778\times {10}^{-8}{\mathrm{V}}^{2}\left[\because BA\omega =6.66\times {10}^{-4}\mathrm{V}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {{e}_{\mathrm{av}}}^{2}=2.2\times {10}^{-7}{\mathrm{V}}^{2}$

#### Page No 308:

#### Answer:

Given:

T = 1 minute

Heat produced in the circuit is calculated using the following relation:

$H=\underset{0}{\overset{T}{\int}}{i}^{2}Rdt$

$\Rightarrow H=\underset{0}{\overset{1\mathrm{min}}{\int}}\frac{{B}^{2}{A}^{2}{\omega}^{2}}{{R}^{2}}\mathrm{sin}\left(\omega t\right)Rdt$

$=\frac{{B}^{2}{A}^{2}{\omega}^{2}}{2R}.\underset{0}{\overset{1\mathrm{min}}{\int}}\left(1-\mathrm{cos}2\mathrm{\omega}t\right)dt\phantom{\rule{0ex}{0ex}}=\frac{{B}^{2}{A}^{2}{\omega}^{2}}{2R}{\left(1-\frac{\mathrm{sin}2\omega t}{2\omega}\right)}_{0}^{1\mathrm{min}}\phantom{\rule{0ex}{0ex}}=\frac{{B}^{2}{A}^{2}{\omega}^{2}}{2R}\left(60-\frac{\mathrm{sin}2\times 80\times 2\mathrm{\pi}/60\times 60}{2\times 80\times 2\mathrm{\pi}/60}\right)\phantom{\rule{0ex}{0ex}}=\frac{60}{2R}\times {\pi}^{2}{r}^{4}\times {B}^{2}\times {\left(80\times \frac{2\pi}{60}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{60}{200}\times 10\times \frac{64}{9}\times 10\times 625\times {10}^{-8}\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=\frac{625\times 6\times 64}{9\times 2}\times {10}^{-11}=1.33\times {10}^{-7}\mathrm{J}$

#### Page No 308:

#### Answer:

Magnetic flux through the wheel (initially):

${\varphi}_{1}=BA=\frac{2\times {10}^{-4}\times \mathrm{\pi}{\left(0.1\right)}^{2}}{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\times {10}^{-6}\mathrm{Wb}$

As the wheel rotates, the wooden (non-metal) part of the wheel comes inside the magnetic field and the iron part of the wheel comes outside the magnetic field. Thus, the magnetic flux through the wheel becomes zero.

i.e. ${\varphi}_{2}=0$

*dt* = 2 s

The average emf induced in the wheel is given by

$e=-\frac{d\mathrm{\varphi}}{dt}\phantom{\rule{0ex}{0ex}}=-\left(\frac{{\varphi}_{2}-{\varphi}_{1}}{dt}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}\times {10}^{-6}}{2}\phantom{\rule{0ex}{0ex}}=1.57\times {10}^{-6}\mathrm{V}$

#### Page No 308:

#### Answer:

Given:

Length of the rod, *l *= 20 cm = 0.2 m

Velocity of the rod, *v* = 10 cm/s = 0.1 m/s

Magnetic field, *B* = 0.10 T

(a) The force on a charged particle moving with velocity *v* in a magnetic field is given by

*$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$
F = qvB* sin

*θ*

Here,

*θ =*90

^{o}

Now,

*F*= (1.6 × 10

^{−19}) × (1 × 10

^{−1}) × (1 × 10

^{−1})

= 1.6 × 10

^{−21}N

(b) The electrostatic force on the charged particle is

*qE.*

Here,

*qE = qvB*

⇒

*E*= (1 × 10

^{−1}) × (1 × 10

^{−1})

= 1 × 10

^{−2}V/m

It is created because of the induced emf.

(c) Motional emf between the ends of the rod,

*e*=

*Bvl*

⇒

⇒

*e*= 0.1 × 0.1 × 0.2

= 2 × 10

^{−3}V

#### Page No 308:

#### Answer:

Given:

Length of the stick, *l *= 1 m

Magnetic field, *B* = 0.2 T

Velocity of the stick, *v* = 2 m/s

Thus, we get

Induced emf,* e* = *Blv* = 0.2 × 1 × 2 = 0.4 V

#### Page No 308:

#### Answer:

Given:

*l* = 10 m

*v* = 3 × 10^{7} m/s

*B* = 3 × 10^{−10} T

Now,

Motional emf = *Bvl*

= (3 × 10^{−10}^{ }) × (3 × 10^{7} ) × (10)

= 9 × 10^{−2}^{ }

= 0.09 V

#### Page No 308:

#### Answer:

Here,

Velocity of the train, *v* = 180 km/h = 50 m/s

Earth's magnetic field, *B* = 0.2 × 10^{−4}^{ }T

Separation between the railings, *l* = 1 m

Induced emf, *e* = *Bvl* = 0.2 × 10^{−4} × 50

= 10^{−3} V

So, the voltmeter will record 1 mV as the reading.

#### Page No 308:

#### Answer:

(a) The emf induced in loop *abc* is zero, as there is no change in the magnetic flux through it.

(b) The emf induced is given by

$e=\left(\overrightarrow{v}\times \overrightarrow{B}\right).\overrightarrow{l}$* *

Emf induced in segment *bc*, *e* = *Bvl* (With positive polarity at point C)

(c) There is no emf induced in segment *bc*, as the velocity is parallel to its length.

(d) The emf induced in segment *ab* is calculated by the following formula:

*e* = B.*v*. (Effective length of *ab*)

The effective length of *ab* is along the direction perpendicular to its velocity.

Emf induced, *e* = B.*v*.(*bc*)

#### Page No 308:

#### Answer:

(a) The emf induced between the ends of the wire is calculated using the following formula:

*e *= *Bv* (Effective length of the wire)

Effective length of the wire = Component of length perpendicular to the velocity

Here, the component of length moving perpendicular to *v* is 2*r*.

∴ Induced emf, *e = **Bv*2*r*

(b) When the velocity is parallel to the diameter of the semicircular wire, the component of its length perpendicular to its velocity is zero.

∴ Induced emf, *e* = 0

#### Page No 308:

#### Answer:

Given:

Length of the rod, *l* = 10 cm = 0.1 m

Angle between the velocity and length of the rod, *θ* = 60°

Magnetic field, *B* = 1 T

Velocity of the rod, *v *= 20 cm/s = 0.2 m/s

The motional emf induced in the rod is given by

$e=\left(\overrightarrow{v}\times \overrightarrow{B}\right).\overrightarrow{l}$

∴ *e* = B*vl* sin 60°

We take the component of the length vector that is perpendicular to the velocity vector.

∴ *e* = $1\times 0.2\times 0.1\times \sqrt{\frac{3}{2}}$

= 17.32 × 10^{−3}^{ }V

#### Page No 308:

#### Answer:

(a) The maximum value of the emf is between the end points of the diameter perpendicular to the velocity.

The value of the maximum emf is given by

*E*_{max} = *vB*(2*R*)

(b) The minimum value of emf is between the end points of the diameter parallel to the velocity of the ring.

The minimum value of emf is given by

*E*_{min} = 0

#### Page No 308:

#### Answer:

Because the force exerted by the magnetic field on the rod is given by *F*_{magnetic} = *ilB*, the direction of this force is opposite to that of the motion of the rod.

Now, let the external force on it be *F*.

Because the velocity is constant, the net force acting on the wire must be zero.

Thus, *F = **F*_{magnetic} = *ilB* is acting in the direction of the velocity.

#### Page No 308:

#### Answer:

The induced emf is given by

*e* = *Bvl*

Total resistance, *R *= *r* × Total length of the wire

Because the length of the movable wire is *l* and the distance travelled by the movable wire in time *t *is *vt*, the total length of the loop is 2 (*l* +* vt*).

∴ *e* = *i* × 2*r** *(*l* +* vt*)

*Bvl* = 2*ri* (*l* + *vt*)

$\Rightarrow i=\frac{Bvl}{2r(l+vt)}$

#### Page No 308:

#### Answer:

Emf induced in the circuit, *e* = *Bvl*

Current in the circuit, $i=\frac{e}{R}=\frac{Bvl}{2r(l+vt)}$

(a) Force *F* needed to keep the sliding wire moving with a constant velocity *v *will be equal in magnitude to the magnetic force on it. The direction of force *F* will be along the direction of motion of the sliding wire.

Thus, the magnitude of force *F* is given by

$F=ilB=\frac{Bvl}{2r(l+vt)}\times lB\phantom{\rule{0ex}{0ex}}=\frac{{B}^{2}{l}^{2}v}{2r(l+vt)}$

(b) The magnitude of force *F* at *t* = 0 is given by

${F}_{0}=ilB=lB\left(\frac{lBv}{2rl}\right)\phantom{\rule{0ex}{0ex}}=\frac{l{B}^{2}v}{2r}...\left(1\right)$

Let at time *t* = *T, *the value of the force be* **F*_{0}/2.

Now,

$\frac{{F}_{0}}{2}=\frac{{l}^{2}{B}^{2}v}{2r(l+vT)}$

On substituting the value of *F*_{0}_{ }from (1), we get

$\frac{l{B}^{2}v}{4r}=\frac{{l}^{2}{B}^{2}v}{2r(l+vT)}\phantom{\rule{0ex}{0ex}}\Rightarrow 2l=l+vT\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{l}{v}$

#### Page No 309:

#### Answer:

(a) When wire PQ is moving with a speed *v*, the emf induced across it is given by

*e = Blv *

Total resistance of the circuit = *r* +* R*

∴ Current in the circuit, *i* = $\frac{Blv}{r+R}$

(b) Force acting on the wire at the given instant, *F *= *ilB*

On substituting the value of *i *from above, we get

$F=\frac{\left(Blv\right)\left(lB\right)}{(R+r)}=\frac{{B}^{2}{l}^{2}v}{R+r}$

Acceleration of the wire is given by

*a*$=\frac{{B}^{2}{l}^{2}v}{m(R+r)}$

(c) Velocity can be expressed as:

*v* = *v*_{0} + *at* = ${v}_{0}-\frac{{B}^{2}{l}^{2}v}{m(R+r)}t$ (As force is opposite to velocity)

Velocity as the function of *x* is given by

$v={v}_{0}-\frac{{B}^{2}{l}^{2}x}{m(R+r)}$

$\left(d\right)a=v\frac{dv}{dx}=\frac{{B}^{2}{l}^{2}v}{m(R+r)}\phantom{\rule{0ex}{0ex}}dx=\frac{m(R+r)}{{B}^{2}{l}^{2}}dv$

On integrating both sides, we get

$x=\frac{m(R+r){v}_{0}}{{B}^{2}{l}^{2}}$

#### Page No 309:

#### Answer:

Given:

Total resistance of the frame,* R* = 2.0 Ω

Magnetic field,* B* = 0.020 T

Dimensions of the frame:

Length, *l* = 32 cm = 0.32 m

Breadth, *b* = 8 cm = 0.08 m

(a) Let the velocity of the frame be *v.*

The emf induced in the rectangular frame is given by

*e = Blv*

Current in the coil, $i=\frac{Blv}{R}$

The magnetic force on the rectangular frame is given by

*F* =* **ilB* = 3.2 × 10^{−5}^{ }N

On putting the value of* i*, we get

$\frac{{B}^{2}{l}^{2}v}{R}=3.2\times {10}^{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{(0.020{)}^{2}\times (0.08{)}^{2}\times v}{2}=3.2\times {10}^{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{3.2\times {10}^{-5}}{6.4\times {10}^{-3}\times 4\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=25\mathrm{m}/\mathrm{s}$

(b) Emf induced in the loop, *e = vBl
⇒ e =* 25 × 0.02 × 0.08

= 4 × 10

^{−2}V

(c) Resistance per unit length is given by

*r*$=\frac{2}{0.8}$

Ratio of the resistance of part, $\frac{ad}{cb}=\frac{2\times 0.72}{0.8}=1.8\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

${V}_{\mathrm{ab}}=iR=\frac{Blv}{2}\times 1.8\phantom{\rule{0ex}{0ex}}=\frac{0.2\times 0.08\times 25\times 1.8}{2}\phantom{\rule{0ex}{0ex}}=0.036\mathrm{V}=3.6\times {10}^{-2}\mathrm{V}$

(d) Resistance of

*cd:*

${R}_{\mathrm{cd}}=\frac{2\times 0.8}{0.8}=0.2\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}V=i{R}_{\mathrm{cd}}=\frac{2\times 0.08\times 25\times 0.2}{2}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-3}\mathrm{V}$

${R}_{\mathrm{cd}}=\frac{2\times 0.8}{0.8}=0.2\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}V=i{R}_{\mathrm{cd}}=\frac{2\times 0.08\times 25\times 0.2}{2}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-3}\mathrm{V}$

#### Page No 309:

#### Answer:

Given:

Separation between the parallel arms, *l* = 20 cm = 20 × 10^{−2} m

Velocity of the sliding wire, *v* = 20 cm/s = 20 × 10^{−2} m/s

Horizontal component of the earth's magnetic field, *B*_{H} = 3 × 10^{−5} T

Current through the wire, *i* = 2 µA = 2 × 10^{−6} A

Resistance* *of the wire*, R* = 0.2 Ω

Let the vertical component of the earth's magnetic field be *B*_{v} and the angle of the dip be *δ.*

Now,

$i=\frac{{B}_{v}lv}{R}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {B}_{\mathrm{v}}=\frac{iR}{lv}$_{}

$=\frac{2\times {10}^{-5}\times 2\times {10}^{-1}}{20\times {10}^{-2}\times 20\times {10}^{-2}}=\frac{2\times 2\times {10}^{-7}}{2\times 2\times {10}^{-2}}$

$=1\times {10}^{-5}\mathrm{T}$

We know, ^{}

$\mathrm{tan}\delta =\frac{{B}_{\mathrm{v}}}{{B}_{\mathrm{H}}}=\frac{1\times {10}^{-5}}{3\times {10}^{-5}}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \delta ={\mathrm{tan}}^{-1}\left(\frac{1}{3}\right)$

#### Page No 309:

#### Answer:

Component of weight along its motion, *F*' = *mg*sin*θ*

The emf induced in the rod due to its motion is given by

*e* = *Bl'v'*

Here,

*l*' = Component of the length of the rod perpendicular to the magnetic field

*v*' = Component of the velocity of the rod perpendicular to the magnetic field

$i=\frac{B\times l\times v\mathrm{cos\theta}}{R}$

$\left|\overrightarrow{F}\right|=i\left|\overrightarrow{l}\times \overrightarrow{B}\right|=ilB\mathrm{sin}(90-\theta )\phantom{\rule{0ex}{0ex}}F=ilB=\frac{Blv\mathrm{cos}\theta}{R}\times l\times B\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}F=\frac{{B}^{2}{l}^{2}v{\mathrm{cos}}^{2}\theta}{R}\phantom{\rule{0ex}{0ex}}$

The direction of force *F* is opposite to *F.*'

Because the rod is moving with a constant velocity, the net force on it is zero.

Thus,

*F $-$ F*' = 0

*F = F*'

or

$\frac{{B}^{2}{l}^{2}v{\mathrm{cos}}^{2}\mathrm{\theta}}{R}=mg\mathrm{sin}\theta $

$\therefore B=\sqrt{\frac{Rmg\mathrm{sin}\theta}{{l}^{2}v{\mathrm{cos}}^{2}\theta}}$

#### Page No 309:

#### Answer:

(a) When both wires move in same direction:

The sliding wires constitute two parallel sources of emf.

The net emf is given by

*e* = B*lv*

*⇒ e* = (1 × 4 × 10^{−2}^{ }) × 5 × (10^{−2})

= 20 × 10^{−4} V

The resistance of the sliding wires is 2 Ω.

∴ Net resistance = $\frac{2\times 2}{2+2}$ + 19 = 20 Ω

Net current through 19 Ω = $\frac{2\times {10}^{-4}}{20}$ = 0.1 mA

(b) When both wires move in opposite directions with the same speed, the direction of the emf induced in both of them is opposite. Thus, the net emf is zero.

∴ Net current through 19 Ω = 0

#### Page No 309:

#### Answer:

(a) When the wires move in the same direction, their polarity remains the same. The circuit remains incomplete. Therefore, no current flows in the circuit.

(b) When the wires move in opposite directions, their polarities are reversed. Thus, current flows in the circuit.

${V}_{{\mathrm{P}}_{2}{\mathrm{Q}}_{2}}=Blv$

= 1 × 0.04 × 0.05

= 2 × 10^{−3} V

*R =* 2 Ω

Current in the circuit is given by

$i=\frac{2\times {10}^{-3}}{2}$

= 1 × 10^{−3} A = 1 mA

#### Page No 309:

#### Answer:

Given:

Magnetic field, *B* = 1 T

Velocity of the sliding wire, *v *= 5 × 10^{−2} m/s

Resistance of the connected resistor, *R* = 10 Ω

(a) When the switch is thrown to the middle rail:

Length of the sliding wire = 2 × 10^{−2} m

Induced emf, *E = Bvl*

= 1 × (5 × 10^{−2}) × (2 × 10^{−2}) V

= 10 × 10^{−4} = 10^{−3} V

Current in the 10 Ω resistor is given by

$i=\frac{E}{R}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-3}}{10}={10}^{-4}=0.1\mathrm{mA}$

(b) When the switch is thrown to the bottom rail:

The length of the sliding wire becomes 4 × 10^{−2} m.

The induced emf is given by

*E = Bvl*'

= 1 × (5 × 10^{−2}) × (4 × 10^{−2})

= 20 × 10^{−4} V

Now,

Current, *i* = $\frac{20\times {10}^{-4}}{10}$ A

= 2 × 10^{−4} A = 0.2 mA

#### Page No 309:

#### Answer:

Current passing through the circuit initially = *i*

Initial emf = *ir*

Emf induced due to motion of *ab*, *e* = *Blv*

Net emf, *e*_{net}= *ir* − *Blv*

Net resistance = 2*r*

Thus, the current passing through the circuit is $\frac{ir-Blv}{2r}$.

#### Page No 309:

#### Answer:

Because current *i* passes through the sliding wire, the magnetic force on the wire (*F*) is *il*B.

Now,

Acceleration of the sliding wire, *a* = $\frac{ilB}{m}$

Velocity of the sliding wire, *v* = *u + at*

∵ *u = *0

∴ *v* = $\frac{ilBt}{m}$

#### Page No 309:

#### Answer:

Let us consider the above free body diagram.

As the net force on the wire is zero, *ilB = mg.*

When the wire is replaced by a wire of double mass, we have

Now, let *a' *be the acceleration of the wire in downward direction and *t* be the time taken by the wire to fall*.*

Net force on the wire = 2*mg* − *ilB = **F*_{net}

On applying Newton's second law, we get

2*mg* − *ilB =* 2 *ma*' ...(1)

$\Rightarrow a\text{'}=\frac{2mg-ilB}{2m}\phantom{\rule{0ex}{0ex}}s=ut+\frac{1}{2}a\text{'}{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{1}{2}\times \frac{2mg-ilB}{2m}\times {t}^{2}[\because s=l]\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{\frac{4ml}{2mg-ilB}}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{\frac{4ml}{2mg-mg}}[\mathrm{From}(1\left)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{\frac{2l}{g}}$

#### Page No 310:

#### Answer:

Given:

Width of rectangular frame = *d*

Mass of rectangular frame = *m*

Resistance of the coil = *R*

(a) As the frame attains the speed *v*

Emf developed in side AB = *Bdv *(When it attains a speed *v*)

Current = $\frac{\mathrm{B}dv}{\mathrm{R}}$

The magnitude of the force on the current carrying conductor moving with speed *v* in direction perpendicular to the magnetic field as well as to its length is given by

$F=ilB$

Therefore, Force *F*_{B} = $\frac{\mathrm{B}{d}^{2}v}{\mathrm{R}}$

As the force is in direction opposite to that of the motion of the frame .

Therefore,Net force is given by

${F}_{\mathrm{net}}\mathit{}\mathit{=}\mathit{}F\mathit{-}{F}_{B}\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{net}}\mathit{=}F-\frac{\mathrm{B}{d}^{2}{v}^{2}}{\mathrm{R}}=\frac{\mathrm{RF}-\mathrm{B}{d}^{2}v}{\mathrm{R}}$

Applying Newton's second law

$\frac{\mathrm{RF}-\mathrm{B}{d}^{2}{v}^{2}}{\mathrm{R}}=ma$

Net acceleration is given by *a*= $\frac{RF-B{d}^{2}v}{mR}$

(b)

Velocity of the frame becomes constant when its acceleration becomes 0.

Let the velocity of the frame be *v*_{0}

$\frac{\mathit{F}}{\mathit{m}}\mathit{-}\frac{{\mathit{B}}^{\mathit{2}}{\mathit{d}}^{\mathit{2}}{\mathit{v}}_{\mathit{0}}}{\mathit{m}\mathit{R}}\mathit{=}\mathit{}\mathit{0}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{\mathit{F}}{\mathit{m}}\mathit{=}\frac{{\mathit{B}}^{{}_{\mathit{2}}}{\mathit{d}}^{\mathit{2}}{\mathit{v}}_{\mathit{0}}}{\mathit{m}\mathit{R}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{v}_{\mathit{0}}\mathit{=}\frac{\mathit{F}\mathit{R}}{{\mathit{B}}^{\mathit{2}}{\mathit{d}}^{\mathit{2}}}$

As the speed thus calculated depends on *F, R, B* and *d* all of them are constant, therefore the velocity is also constant.

Hence, proved that the frame moves with a constant velocity till the whole frame enters.

(c)

Let the velocity at time *t* be *v*.

The acceleration is given by

$a=\frac{\mathrm{d}v}{\mathrm{d}t}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{RF-{\mathrm{d}}^{2}{\mathrm{B}}^{2}{v}^{2}}{m\mathrm{R}}=\frac{\mathrm{d}v}{\mathrm{d}t}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}v}{\mathrm{RF}-{d}^{2}{\mathrm{B}}^{2}{v}^{2}}=\frac{\mathrm{d}t}{m\mathrm{R}}\phantom{\rule{0ex}{0ex}}\mathrm{Integrating}\phantom{\rule{0ex}{0ex}}\Rightarrow \underset{0}{\overset{v}{\int}}\frac{\mathrm{d}v}{\mathrm{RF}-{d}^{2}{\mathrm{B}}^{2}{v}^{2}}=\underset{0}{\overset{t}{\int}}\frac{\mathrm{d}t}{m\mathrm{R}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left[\mathrm{ln}(RF-{d}^{2}{\mathrm{B}}^{2}v)\right]}_{0}^{v}=-{d}^{2}{B}^{2}{\left[\frac{t}{Rm}\right]}_{0}^{t}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ln}(RF-{d}^{2}{\mathrm{B}}^{2}v)-\mathrm{ln}\left(RF\right)=-\frac{{d}^{2}{B}^{2}t}{Rm}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{d}^{2}{\mathrm{B}}^{2}v}{\mathrm{RF}}=1-{e}^{-\frac{{d}^{2}{\mathrm{B}}^{2}t}{\mathrm{R}m}}\phantom{\rule{0ex}{0ex}}v=\frac{FR}{{l}^{2}{\mathrm{B}}^{2}}\left(1-{e}^{-\frac{{\mathrm{B}}^{2}{d}^{2}{v}_{0}t}{\mathrm{R}{v}_{0}m}}\right)\phantom{\rule{0ex}{0ex}}v={v}_{0}(1-{e}^{-Ft/{v}_{0}m})\left[\because F=\frac{{B}^{2}{d}^{2}{v}_{0}}{R}\right]$

#### Page No 310:

#### Answer:

According to Fleming's left hand rule the force in the wire ab will be in the upward direction.

Moreover, a moving wire ab is equivalent to a battery of emf* vBl* as shown in the figure.

At the given instant, the net emf across the wire (*e*) is *E − **Bvl.*

(a) The current through the wire is given by

$i=\frac{E-Bvl}{r}$

The direction of the current is from *b* to *a*.

(b) The force acting on the wire at the given instant is given by

$F=ilB=\left(\frac{E-Bvl}{r}\right)$ towards right

(c) The velocity of the wire attains a value such that it satisfies* E* *= Bvl.*

The net force on the wire becomes zero. Thus, the wire moves with a constant velocity *v*.

∴ $v=\frac{E}{Bl}$

#### Page No 310:

#### Answer:

(a) When the speed of the wire is *v*, the emf developed in the loop is, *e* = *Blv.*

(b)

Magnitude of the induced current in the wire, *I* = $\frac{\mathit{B}\mathit{l}\mathit{v}}{\mathit{R}}$

As wire is moving the magnetic flux passing through the loop is increasing with time. Therefore, the direction of the current should be as such to oppose the change in magnetic flux. Therefore in order to induce the current in anticlockwise direction the current flows from *b* to *a*

(c)

Due to motion of the wire in the magnetic field there is a force in upward direction (perpendicular to the wire).

The magnitude of the force on the wire carrying current *i* is given by $F=ilB$

The net force on the wire = $mg-F=mg-ilB$

Downward acceleration of the wire due to current, *a*$=\frac{mg-F}{m}$

$a=\frac{mg-ilB}{m}\phantom{\rule{0ex}{0ex}}a=g-\frac{{B}^{2}{l}^{2}v}{\mathrm{R}m}\left[\because i=\frac{Blv}{R}\right]$

(d) Let the wire start moving with a constant velocity.

Now,

Let the speed of the wire be *v*_{m}

As speed is constant, acceleration, *a* = 0

$a=g-\frac{{B}^{2}{l}^{2}{v}_{\mathrm{m}}}{Rm}=0$

$\frac{{B}^{2}{l}^{2}{v}_{\mathrm{m}}}{Rm}=g\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{\mathrm{m}}=\frac{gRm}{{B}^{2}{l}^{2}}$

(e)

The acceleration of the wire can be expressed as time rate of change of velocity

$\frac{dv}{dt}=a$

$\therefore \frac{dv}{dt}=\left(\frac{mg-{B}^{2}{l}^{2}v/R}{m}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dv}{{\displaystyle \frac{mg-{B}^{2}{l}^{2}v/R}{m}}}=dt\phantom{\rule{0ex}{0ex}}\underset{0}{\overset{v}{\int}}\frac{mdv}{mg-{\displaystyle \frac{{B}^{2}{l}^{2}v}{R}}}=\underset{0}{\overset{t}{\int dt}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{m}{{\displaystyle -\frac{{B}^{2}{l}^{2}}{R}}}{\left[\mathrm{log}\left(mg-\frac{{B}^{2}{l}^{2}v}{R}\right)\right]}_{0}^{v}=t\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-mR}{{B}^{2}{l}^{2}}\left[\mathrm{log}\left(mg-\frac{{\mathrm{B}}^{2}{l}^{2}v}{\mathrm{R}}\right)-\mathrm{log}\left(mg\right)\right]=t\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}\left[\frac{mg-{\displaystyle \frac{{B}^{2}{l}^{2}v}{R}}}{mg}\right]=\frac{-t{B}^{2}{l}^{2}}{mR}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}\left[1-\frac{{\mathrm{B}}^{2}{l}^{2}v}{\mathrm{R}mg}\right]=\frac{-t{B}^{2}{l}^{2}}{m\mathrm{R}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1-\frac{{B}^{2}{l}^{2}v}{Rmg}={e}^{-\frac{t{B}^{2}{l}^{2}}{mR}}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(1-{e}^{-\frac{t{B}^{2}{l}^{2}}{mR}}\right)=\frac{{B}^{2}{l}^{2}v}{Rmg}\phantom{\rule{0ex}{0ex}}v=\frac{Rmg}{{B}^{2}{l}^{2}}\left(1-{e}^{-\frac{{\mathrm{B}}^{2}{l}^{2}t}{m\mathrm{R}}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow v={v}_{\mathrm{m}}\left(1-{e}^{-\frac{gt}{{v}_{\mathrm{m}}}}\right)\left[\because {v}_{m}=\frac{Rmg}{{B}^{2}{l}^{2}}\right]$

(f)

The velocity can be expressed as time rate of change of position

*x* is the position of the wire at instant *t*.

∵ $\frac{dx}{dt}=v$

Thus, the displacement of the wire can be expressed as:

$s={\int}_{0}^{t}dx={\int}_{0}^{t}v\mathrm{d}t$

$\therefore s={x}_{t}-{x}_{0}={v}_{m}\underset{0}{\overset{t}{\int}}\left(1-{e}^{-\frac{gt}{{v}_{m}}}\right).dt\phantom{\rule{0ex}{0ex}}s={v}_{m}.{\left(t+\frac{{v}_{m}}{g}.{e}^{-\frac{gt}{{v}_{m}}}\right)}_{0}^{t}\phantom{\rule{0ex}{0ex}}s=\left({v}_{m}t+\frac{{{v}_{m}}^{2}}{g}{e}^{-\frac{gt}{{v}_{m}}}\right)-\frac{{{v}_{m}}^{2}}{g}\phantom{\rule{0ex}{0ex}}s={v}_{m}t-\frac{{{v}_{m}}^{2}}{g}\left(1-{e}^{-\frac{gt}{{v}_{m}}}\right)$

(g)

Rate of development of heat in the wire is given by *P = V × i*

$V=Blv\phantom{\rule{0ex}{0ex}}i=\frac{Blv}{R}$

Therefore, the rate of development of heat in the wire is given by

$P=Blv\times \frac{Blv}{R}=\frac{{B}^{2}{l}^{2}{v}^{2}}{R}\phantom{\rule{0ex}{0ex}}P=\frac{{B}^{2}{l}^{2}{{v}_{m}}^{2}{\left(1-{e}^{-\frac{gt}{{v}_{m}}}\right)}^{2}}{R}\left[\because v={v}_{m}\left(1-{e}^{-\frac{gt}{{v}_{m}}}\right)\right]$

Rate of decrease in potential energy is given by

$\frac{\mathrm{d}U}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(mgx\right)=mg.\frac{dx}{dt}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}U}{\mathrm{d}t}=mgv\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}U}{\mathrm{d}t}=mg.{v}_{m}\left(1-{e}^{-\frac{gt}{{v}_{m}}}\right)\left[\because v={v}_{m}\left(1-{e}^{-\frac{gt}{{v}_{m}}}\right)\right]$

After the steady state, *i*.*e*., *t* → ∞,

$\frac{dU}{dt}=mg{v}_{\mathrm{m}}\phantom{\rule{0ex}{0ex}}P=\frac{{l}^{2}{B}^{2}}{R}{{v}_{m}}^{2}\phantom{\rule{0ex}{0ex}}P=\frac{{l}^{2}{B}^{2}}{R}\times {v}_{m}\times \frac{mg\mathrm{R}}{{l}^{2}{B}^{2}}\left[\because {v}_{m}=\frac{mg\mathrm{R}}{{l}^{2}{B}^{2}}\right]\phantom{\rule{0ex}{0ex}}P=mg{v}_{m}$

Thus, after the steady state, $P=\frac{\mathrm{d}U}{\mathrm{d}t}$

#### Page No 310:

#### Answer:

Given:

Length of the spoke of the bicycle's wheel, *l* = 0.3 m

Earth's magnetic field, $\overrightarrow{B}$ = 2.0 × 10^{−5} T

Length of each spoke = 30.0 cm = 0.3 m

Angular speed of the wheel, $\omega =\frac{100}{60}\times 2\mathrm{\pi}=\frac{10}{3}\mathrm{\pi}\mathrm{rad}/\mathrm{s}$

Linear speed of the spoke, $\mathrm{v}=\frac{l}{2}\times \mathrm{\omega}=\frac{0.3}{2}\times \frac{10}{3}\mathrm{\pi}$

Now,

Emf induced in the spoke of the wheel, *e* = *Blv*

$\Rightarrow e=(2.0\times {10}^{-5})\times (0.3)\times \left(\frac{0.3}{2}\times \frac{10}{3}\times \mathrm{\pi}\right)$

= 3$\pi $ × 10^{−6} V

= 3 × 3.14 × 10^{−6}V

= 9.42 × 10^{−6} V

#### Page No 310:

#### Answer:

The angular velocity of the disc is *ω. *Also, the magnetic field of magnitude *B *is perpendicular to the disc.

Let us take a circular element of thickness *da* at a distance *a* from the centre.

Linear speed of the element at *a* from the centre, *v = ωa*

Now,

$de=Blv\phantom{\rule{0ex}{0ex}}de=B\times da\times a\mathrm{\omega}\phantom{\rule{0ex}{0ex}}\Rightarrow e={\int}_{0}^{r}\left(B\omega a\right)da\phantom{\rule{0ex}{0ex}}e=\frac{1}{2}B\omega {r}^{2}$

#### Page No 310:

#### Answer:

Given:

Magnetic field perpendicular to the disc, *B* = 0.40 T

Angular speed, *ω* = 10 rad/s

Resistance, *R* = 10 Ω

Radius of the disc, *r* = 5 cm = 0.5 m

Let us consider a rod of length 0.05 m fixed at the centre of the disc and rotating with the same ω.

Now,

$v=\frac{l}{2}\times \omega =\frac{0.05}{2}\times 10\phantom{\rule{0ex}{0ex}}e=Blv\phantom{\rule{0ex}{0ex}}=0.40\times 0.05\times \frac{0.05}{2}\times 10\phantom{\rule{0ex}{0ex}}=5\times {10}^{-3}\mathrm{V}\phantom{\rule{0ex}{0ex}}i=\frac{e}{R}\phantom{\rule{0ex}{0ex}}=\frac{5\times {10}^{-3}}{10}=0.5\mathrm{mA}$

As the disc is rotating in the anti-clockwise direction, the emf induced in the disc is such that the centre is at the higher potential and the periphery is at the lower potential. Thus, the current leaves from the centre.

#### Page No 310:

#### Answer:

Magnetic field in the given region, $\overrightarrow{B}=\frac{{B}_{0}}{L}y\hat{k}$

Length of the rod on the *y*-axis = *L*

Velocity of the rod, *v* = *v*_{0}$\hat{i}$

We will consider a small element of length* **dy* on the rod.

Now,

Emf induced in the element:

d*e* = *Bv**dy*

$\Rightarrow de=\frac{{B}_{0}}{L}y\times {v}_{0}\times dy\phantom{\rule{0ex}{0ex}}=\frac{{B}_{0}{v}_{0}}{L}y\mathrm{d}y\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}e=\frac{{B}_{0}{v}_{0}}{L}\underset{0}{\overset{\mathrm{L}}{\int}}ydy\phantom{\rule{0ex}{0ex}}=\frac{{B}_{0}{v}_{0}}{L}{\left[\frac{{y}^{2}}{2}\right]}_{0}^{L}\phantom{\rule{0ex}{0ex}}=\frac{{B}_{0}{v}_{0}}{L}\frac{{L}^{2}}{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{B}_{0}{v}_{0}L$

#### Page No 310:

#### Answer:

Here, the magnetic field $\overrightarrow{B}$ due to the long wire varies along the length of the rod. We will consider a small element of the rod of length *da* at a distance *a* from the wire. The magnetic field at a distance *a* is given by

$\overrightarrow{B}=\frac{{\mathrm{\mu}}_{0}i}{2\pi a}$

Now,

Induced emf in the rod:

$de=Bvda\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}i}{2\pi a}\times v\times da$

Integrating from $x-\frac{l}{2}$ to $x+\frac{l}{2}$, we get

$e=\underset{x-\frac{l}{2}}{\overset{x+\frac{l}{2}}{\int}}de\phantom{\rule{0ex}{0ex}}=\underset{x-\frac{l}{2}}{\overset{x+\frac{l}{2}}{\int \frac{{\mu}_{0}i}{2\pi a}}}vda\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}iv}{2\mathrm{\pi}}\left[\mathrm{ln}\left(x+\frac{l}{2}\right)-\mathrm{ln}\left(x-\frac{l}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}iv}{2\mathrm{\pi}}\mathrm{ln}\left[\frac{x+{\displaystyle \frac{l}{2}}}{x-{\displaystyle \frac{l}{2}}}\right]$

#### Page No 311:

#### Answer:

(a) Here, the magnetic field $\overrightarrow{B}$ due to the long wire varies along the length of the rod.

We will consider a small element of the rod of length *da* at a distance *a* from the wire. The magnetic field at a distance *a* is given by

$\overrightarrow{B}=\frac{{\mathrm{\mu}}_{0}i}{2\pi a}$

Now,

Induced emf in the rod:

$de=Bvda\phantom{\rule{0ex}{0ex}}de=\frac{{\mathrm{\mu}}_{0}i}{2\pi a}\times v\times da$

Integrating $x-\frac{l}{2}$ and $x+\frac{l}{2}$, we get

$e=\underset{x-\frac{l}{2}}{\overset{x+\frac{l}{2}}{\int}}de\phantom{\rule{0ex}{0ex}}=\underset{x-\frac{l}{2}}{\overset{x+\frac{l}{2}}{\int \frac{{\mu}_{0}i}{2\pi a}}}vda\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}iv}{2\mathrm{\pi}}\left[\mathrm{ln}\left(x+\frac{l}{2}\right)-\mathrm{ln}\left(x-\frac{l}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}iv}{2\mathrm{\pi}}\mathrm{ln}\left[\frac{x+{\displaystyle \frac{l}{2}}}{x-{\displaystyle \frac{l}{2}}}\right]$

Emf induced in the rod due to the current-carrying wire:

$e=\frac{{\mathrm{\mu}}_{0}iv}{2\mathrm{\pi}}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)$

Now, let the current produced in the circuit containing the rod and the resistance be *i'*.

$i\text{'}=\frac{e}{R}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}iv}{2\pi R}\mathrm{ln}\left(\frac{2x+1}{2x-1}\right)$

Force on the element:

*dF** = **i'Bl*

$\Rightarrow \mathrm{d}F=\frac{{\mathrm{\mu}}_{0}iv}{2\pi R}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\times \left(\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}a}\right)\times da\phantom{\rule{0ex}{0ex}}={\left(\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}}\right)}^{2}\frac{v}{R}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\frac{dx}{a}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}F={\left(\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}}\right)}^{2}\frac{v}{R}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\underset{x-l/2}{\overset{x+l/2}{\int}}\frac{da}{a}\phantom{\rule{0ex}{0ex}}={\left(\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}}\right)}^{2}\frac{v}{R}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\phantom{\rule{0ex}{0ex}}=\frac{v}{R}{\left[\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\right]}^{2}$

(b) Current, *i*' = $\frac{e}{R}=\frac{{\mathrm{\mu}}_{0}iv}{2\pi R}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)$

(c) The rate of heat, that is, power, developed is given by

*w* = *i*^{2} *R*

$w={\left[\frac{{\mathrm{\mu}}_{0}iv}{2\pi R}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\right]}^{2}R\phantom{\rule{0ex}{0ex}}=\frac{1}{R}{\left[\frac{{\mathrm{\mu}}_{0}iv}{2\mathrm{\pi}}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\right]}^{2}$

(d) Power delivered by the external agency is the same as the rate of heat developed.

Here,

*p *= *i*^{2}*R*

$=\frac{1}{R}{\left[\frac{{\mathrm{\mu}}_{0}iv}{2\mathrm{\pi}}\mathrm{ln}\left(\frac{2x+l}{2x-l}\right)\right]}^{2}$

#### Page No 311:

#### Answer:

Let us consider an element of the loop of length d*x* at a distance *x* from the wire.

(a) Area of the element of loop A = *a*d*x*

Magnetic field at a distance* x* from the wire, $B=\frac{{\mu}_{0}i}{2\pi x}$

The magnetic flux of the element is given by

$d\varphi =\frac{{\mu}_{0}i\times a\mathrm{d}x}{2\pi x}$

The total flux through the frame is given by

$\varphi =\int \mathrm{d}\varphi \phantom{\rule{0ex}{0ex}}={\int}_{b}^{a+b}\frac{{\mu}_{0}iadx}{2\pi x}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}ia}{2\mathrm{\pi}}\mathrm{ln}\left[1+\frac{a}{b}\right]\phantom{\rule{0ex}{0ex}}$

(b) The emf induced in the frame is given by

$e=\frac{d\varphi}{dt}\phantom{\rule{0ex}{0ex}}=\frac{d}{dt}\frac{{\mu}_{0}ia}{2\pi}\mathrm{ln}\left[1+\frac{a}{b}\right]\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}a}{2\pi}\mathrm{ln}\left[1+\frac{a}{b}\right]\frac{d}{dt}({i}_{0}\mathrm{sin}\omega t)\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}a{i}_{0}\omega \mathrm{cos}\omega t}{2\pi}\mathrm{ln}\left[1+\frac{a}{b}\right]$

(c) The current through the frame is given by

$i=\frac{e}{r}\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}a{i}_{0}\omega \mathrm{cos}\omega t}{2\pi r}\mathrm{ln}\left[1+\frac{a}{b}\right]$

The heat developed in the frame in the given time interval can be calculated as:

$H={i}^{2}rt\phantom{\rule{0ex}{0ex}}={\left[\frac{{\mu}_{0}a{i}_{0}\omega \mathrm{cos}\omega t}{2\pi r}\mathrm{ln}\left(1+\frac{a}{b}\right)\right]}^{2}\times r\times t\phantom{\rule{0ex}{0ex}}\mathrm{Using}t=\frac{20\mathrm{\pi}}{\omega},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}H=\frac{{\mathrm{\mu}}_{0}^{2}\times {i}^{2}\times {\omega}^{2}}{4\mathrm{\pi}\times {r}^{\mathit{2}}}{\mathrm{ln}}^{2}\left[1+\frac{a}{b}\right]\times r\times \frac{20\mathrm{\pi}}{\omega}\phantom{\rule{0ex}{0ex}}=\frac{5{{\mathrm{\mu}}_{0}}^{2}{a}^{2}{{i}_{0}}^{2}\omega}{2\mathrm{\pi}r}{\mathrm{ln}}^{2}\left[1+\frac{a}{b}\right]$

#### Page No 311:

#### Answer:

Consider an element of the loop of length d*x* at a distance *x* from the current-carrying wire.

The magnetic field at a distance *x* from the the current-carrying wire is given by

*B* = $\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}x}$

Area of the loop = *b*d*x*

Magnetic flux through the loop element:

$\mathrm{d}\varphi =\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}x}b\mathrm{d}x$

The magnetic flux through the loop is calculated by integrating the above expression.

Thus, we have

$\varphi =\underset{a}{\overset{a+l}{\int}}\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}x}bdx\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}i}{2\pi}b\underset{a}{\overset{a+l}{\int}}\left(\frac{dx}{x}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}i}{2\pi x}\mathrm{ln}\left(\frac{a+l}{a}\right)\phantom{\rule{0ex}{0ex}}$

The emf can be calculated as:

$e=-\frac{d\varphi}{dt}=\frac{d}{dt}\left[\frac{{\mu}_{0}ib}{2\pi}\mathrm{log}\left(\frac{a+l}{a}\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{{\mu}_{0}ib}{2\pi}\frac{a}{a+l}\left(\frac{va-(a+l)v}{{a}^{2}}\right)\left(\because \frac{da}{dt}=v\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}ib}{2\pi}\frac{a}{a+l}\frac{vl}{{a}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}ibvl}{2\pi (a+l)a}$

Calculation of the emf using the emf method:

The emf. induced in AB and CD due to their motion in the magnetic field are opposite to each other.

The magnetic field at AB is given by

${B}_{\mathrm{AB}}=\frac{{\mu}_{0}i}{2\pi a}$

Now,

Length = *b*

Velocity = *v*

The emf induced in AB is given by

${e}_{\mathrm{AB}}=\frac{{\mu}_{0}ivb}{2\pi a}$

The magnetic field at CD is given by

${\mathrm{B}}_{\mathrm{CD}}=\frac{{\mathrm{\mu}}_{0}i}{2\pi (a+l)}$

The emf induced in side CD is given by

${e}_{\mathrm{CD}}=\frac{{\mathrm{\mu}}_{0}ibv}{2\mathrm{\pi}(a+l)}$

The net emf induced is given by

${e}_{\mathrm{net}}=\frac{{\mathrm{\mu}}_{0}ibv}{2\mathrm{\pi}a}-\frac{{\mathrm{\mu}}_{0}ibv}{2\mathrm{\pi}(a+l)}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}ibl(a+l)-\mathrm{\mu}ibva}{2\mathrm{\pi}a(a+l)}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}ibvl}{2\mathrm{\pi}a(a+l)}$

#### Page No 311:

#### Answer:

Calculation of the emf induced in the rotating rod:

It is given that the angular velocity of the disc is *ω* and the magnetic field perpendicular to the disc is having magnitude *B*.

Let us take an element of the rod of thickness d*r* at a distance *r* from the centre.

Now,

Linear speed of the element at *r* from the centre, *v = ωr*

$de=Blv\phantom{\rule{0ex}{0ex}}de=B\times dr\times \omega r\phantom{\rule{0ex}{0ex}}\Rightarrow e={\int}_{0}^{a}\left(B\omega r\right)dr\phantom{\rule{0ex}{0ex}}=\frac{1}{2}B\omega {a}^{2}$

Because it is connected to resistance *R*, the current in the circuit containing the rod, wire and circular loop is given by

$i=\frac{B{a}^{2}\omega}{2R}$

The direction of the current is from point A to point O in the rod.

The magnitude of the force that is applied on the rod is given by

$F=ilB\phantom{\rule{0ex}{0ex}}=\frac{B{a}^{2}\omega}{2R}\times a\times B\phantom{\rule{0ex}{0ex}}=\frac{{B}^{2}{a}^{2}\omega}{2R}$

#### Page No 311:

#### Answer:

Calculation of the emf induced in the rotating rod:

It is given that the angular velocity of the disc is *ω* and the magnetic field perpendicular to the disc is having magnitude *B*.

Let us take an element of the rod of thickness d*r* at a distance *r* from the centre.

Now,

Linear speed of the element at *r* from the centre, *v = ωr*

$de=Blv\phantom{\rule{0ex}{0ex}}de=B\times dr\times \omega r\phantom{\rule{0ex}{0ex}}\Rightarrow e={\int}_{0}^{a}\left(B\omega r\right)dr\phantom{\rule{0ex}{0ex}}=\frac{1}{2}B\omega {a}^{2}$

As ∠*AOC* = 90°, the minor and major segments of AC are in parallel with the rod.

The resistances of the segments are $\frac{R}{4}$ and $\frac{3R}{4}$.

The equivalent resistance is given by

$R\text{'}=\frac{{\displaystyle \frac{R}{4}}\times {\displaystyle \frac{3R}{4}}}{R}=\frac{3R}{16}$

The motional emf induced in the rod rotating in the clockwise direction is given by

$e=\frac{1}{2}B\omega {a}^{2}$

The current through the rod is given by

$i=\frac{e}{R\text{'}}\phantom{\rule{0ex}{0ex}}=\frac{B{a}^{2}\omega}{2R\text{'}}\phantom{\rule{0ex}{0ex}}=\frac{B{a}^{2}\omega}{2\times 3R/16}\phantom{\rule{0ex}{0ex}}=\frac{B{a}^{2}\omega \times 16}{2\times 3R}=\frac{8B{a}^{2}\omega}{3R}$

#### Page No 311:

#### Answer:

When the circular loop is in the vertical plane, it tends to rotate in the clockwise direction because of its weight.

Let the force applied be *F* and its direction be perpendicular to the rod.

The component of *mg* along *F* is *mg* sin* θ*.

The magnetic force is in perpendicular and opposite direction to *mg* sin *θ*.

Now,

Current in the rod:

$i=\frac{B{a}^{2}\omega}{2R}$

The force on the rod is given by

${F}_{\mathrm{B}}=iBl=\frac{{B}^{2}{a}^{2}\omega}{2R}$

Net force = *F*− $\frac{{B}^{2}{a}^{2}\omega}{2R}$ + *mg* sin *θ*

The net force passes through the centre of mass of the rod.

Net torque on the rod about the centre O:

*$\tau =\left(F-\frac{{B}^{2}{a}^{3}\omega}{2R}+mg\mathrm{sin}\theta \right)\frac{\mathrm{OA}}{2}$*

Because the rod rotates with a constant angular velocity, the net torque on it is zero.

i.e. $\tau =0$

*$\left(F-\frac{{B}^{2}{a}^{3}\omega}{2R}+mg\mathrm{sin}\theta \right)\frac{\mathrm{OA}}{2}=0$*

*∴ $F=\frac{{B}^{2}{a}^{3}\omega}{2R}-mg\mathrm{sin}\theta $*

#### Page No 311:

#### Answer:

It is given that the rod is rotated with angular speed in clockwise direction.

The emf induced in the rod (*e*) is $\frac{\mathit{B}\mathit{\omega}{\mathit{a}}^{\mathit{2}}}{\mathit{2}}$, with O at the lower potential and A at the higher potential.

The equivalent circuit can be drawn as:

$i=\frac{e+E}{R}=\frac{{\displaystyle \frac{1}{2}}B\omega {a}^{2}+E}{R}\phantom{\rule{0ex}{0ex}}=\frac{B\omega {a}^{2}+2E}{2R}$

Because the rod rotates with uniform angular velocity, the net torque about point O is zero.

Now,

Net force on the rod, *F*_{net} = *mg* cos θ $-$ *ilB*

Net torque, *τ* = (*mg* cos θ $-$ *ilB*).(*r*/2)* = *0

∴ *mg* cos θ = *ilB*

$R=\frac{(B\omega {a}^{2}+2E)}{2\mathrm{R}}(a\mathit{\times}B)\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{(B\omega {a}^{2}+2E)aB}{2mg\mathrm{cos}\theta}$

#### Page No 311:

#### Answer:

Let the velocity of the rod at an instant be *v* and the charge on the capacitor be *q.*

The emf induced in the rod is given by

*e* = *Bl*v

The potential difference across the terminals of the capacitor and the ends of the rod must be the same, as they are in parallel.

∴ $\frac{q}{C}=Blv$

And,

*q* = *C* × *Blv* = *CBlv*

Current in the circuit:

*i* = $\frac{dq}{dt}=\frac{d\left(CBlv\right)}{dt}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow i=CBl\frac{dv}{dt}=CBla$ (*a* = acceleration of the rod)

The force on the rod due to the magnetic field and its weight are in opposite directions.

∴ *mg* − *il*B = *ma*

⇒ *m*g − *cBla* × *lB** *= *ma*

⇒ *ma* + *cB*^{2}*l*^{2}*a* = *mg*

⇒ *a*(*m* + *cB*^{2}*l*^{2}) = *mg*

$\Rightarrow a=\frac{mg}{m+c{B}^{2}{l}^{2}}$

#### Page No 311:

#### Answer:

(a) The emf induced in the circle is given by

$e=\frac{d\varphi}{dt}=\frac{d(B.A)}{dt}\phantom{\rule{0ex}{0ex}}=A\frac{dB}{dt}$

The emf induced can also be expressed in terms of the electric field as:

*E*.*dl* = *e*

For the circular loop, $A=\pi {r}^{2}$

$\Rightarrow E2\mathrm{\pi}r=\mathrm{\pi}{r}^{2}\frac{dB}{dt}$

Thus, the electric field can be written as:

$E=\frac{\mathrm{\pi}{r}^{2}}{2\mathrm{\pi}r}\frac{\mathrm{d}B}{\mathrm{d}t}=\frac{r}{2}\frac{\mathrm{d}B}{\mathrm{d}t}$

(b) When the square is considered:

*E*.*dl* = *e*

For the square loop, $A={\left(2r\right)}^{2}$

$\Rightarrow E\times 2r\times 4=\frac{dB}{dt}(2r{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{dB}{dt}\frac{4{r}^{2}}{8r}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{r}{2}\frac{dB}{dt}$

The electric field at the given point has the value same as that in the above case.

#### Page No 312:

#### Answer:

Given:

Rate of change of current in the solenoid,$\frac{di}{dt}$ = 0.01 A/s for 2s

$\therefore \frac{di}{dt}$ = 0.02 A/s

*n* = 2000 turns/m

R = 6.0 cm = 0.06 m

*r* = 1 cm = 0.01 m

(a) *ϕ = BA*

Area of the circle, *A* = π × 1 × 10^{−4}

$\u2206i=\frac{\mathrm{d}i}{\mathrm{d}t}\times \u2206t$ = (0.01 A/s) × 2 = 0.02 A/s

$\Rightarrow \u2206\varphi ={\mathrm{\mu}}_{0}nA\u2206i$

Now,

Δ*ϕ** *= (4$\mathrm{\pi}$ × 10^{−7}) × (2 × 10^{3}) × ($\mathrm{\pi}$ × 10^{−4}) × (2 × 10^{−2})

= 16π^{2} × 10^{−10} Wb

= 157.91 × 10^{−10} Wb

= 1.6 × 10^{−5} Wb

∴$\frac{d\mathrm{\varphi}}{dt}$ for 1s = 0.785 Wb

(b) The emf induced due to the change in the magnetic flux is given by

$e=\frac{\mathrm{d\varphi}}{\mathrm{d}t}\phantom{\rule{0ex}{0ex}}\int Edl=\frac{\mathrm{d\varphi}}{\mathrm{d}t}\phantom{\rule{0ex}{0ex}}\Rightarrow E\times 2\mathrm{\pi}r\mathit{=}\frac{\mathrm{d}\varphi}{\mathrm{d}t}$

The electric field induced at the point on the circumference of the circle is given by

$E=\frac{0.785\times {10}^{-8}}{2\mathrm{\pi}\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=1.2\times {10}^{-7}\mathrm{V}/\mathrm{m}$

(c) For the point located outside the solenoid,

$\frac{\mathrm{d}\varphi}{\mathrm{d}t}={\mathrm{\mu}}_{0}n\frac{\mathrm{d}i}{\mathrm{d}t}\mathrm{A}\phantom{\rule{0ex}{0ex}}=(4\mathrm{\pi}\times {10}^{-7})\times \left(2000\right)\times (0.01)\times (\mathrm{\pi}\times (0.06{)}^{2})\phantom{\rule{0ex}{0ex}}E.dl=\frac{\mathrm{d}\varphi}{\mathrm{d}t}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{\mathrm{d}\varphi /\mathrm{d}t}{2\pi r}\phantom{\rule{0ex}{0ex}}$

The electric field induced at a point outside the solenoid at a distance of 8.0 cm from the axis is given by

$E=\frac{(4\mathrm{\pi}\times {10}^{-7})\times \left(2000\right)\times (0.01\times \mathrm{\pi}\times (0.06{)}^{2})}{(\mathrm{\pi}\times 8\times {10}^{-2})}\times \frac{d\mathrm{B}}{dt}\phantom{\rule{0ex}{0ex}}=5.64\times {10}^{-7}\mathrm{V}/\mathrm{m}$

#### Page No 312:

#### Answer:

Let the self-inductance of the inductor be *L*.

Average emf induced in the inductor, *V* = 20 V

Change in current, *di *= *i*_{2} −* **i*_{1} = 2.5 − (−2.5) = 5 A

Time taken for the change, *dt* = 0.1 s

The voltage induced in the inductor is given by

$V=L\frac{di}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow 20=L\left(\frac{5}{0.1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 20=L\times 50\phantom{\rule{0ex}{0ex}}\Rightarrow L=\frac{20}{50}=\frac{4}{10}=0.4\mathrm{H}$

#### Page No 312:

#### Answer:

Given:

Magnetic flux linked with each turn, *ϕ* = 8 × 10^{−4} Wb

Number of turns, *n* = 200

Current, *i* = 4 A

Self-inductance is calculated as:

$L=\left(\frac{n\varphi}{i}\right)$

= $\left(\frac{200}{4}\right)\times \left(8\right)\times \left({10}^{-4}\right)$

= 4 × 10^{−2} H

#### Page No 312:

#### Answer:

Given:

Number of turns, *N = *240

Radius of the solenoid, *r* = 2 cm

Length of the solenoid, *l* = 12 cm

The emf induced in the solenoid is given by

$e=L\frac{\mathrm{d}i}{\mathrm{d}t}$

The self-inductance of the solenoid is given by

$L=\frac{{\mu}_{0}{N}^{2}A}{l}\phantom{\rule{0ex}{0ex}}L=\frac{4\mathrm{\pi}\times {10}^{-7}\times {240}^{2}\times \mathrm{\pi}\times (2\times {10}^{-2}{)}^{2}}{12\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Thus, the emf induced in the solenoid is given by

$e=\frac{4\mathrm{\pi}\times {10}^{-7}\times {240}^{2}\times \mathrm{\pi}\times (2\times {10}^{-2}{)}^{2}}{12\times {10}^{-2}}\times 0.8\phantom{\rule{0ex}{0ex}}=60577.3824\times {10}^{-8}=6\times {10}^{-4}\mathrm{V}$

#### Page No 312:

#### Answer:

Current *i* in the* LR* circuit at time *t* is given by

*i* = *i*_{0}(1 −* **e*^{−t}^{/τ})

Here,

*i*_{0}_{ }= Steady-state value of the current

(a) When the value of the current reaches 90% of the steady-state value:

$i=\frac{90}{100}\times {i}_{0}$

$\frac{90}{100}{i}_{0}$ = i_{o}(1 − *e*^{−t}^{/τ})

⇒ 0.9 = 1 − *e*^{−t}^{/τ}

⇒ *e*^{−t}^{/τ} = 0.1

On taking natural logarithm (ln) of both sides, we get

ln (*e*^{−t}^{/τ}) = ln 0.1

$-\frac{t}{\tau}=-2.3\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{t}{\tau}=2.3$

(b) When the value of the current reaches 99% of the steady-state value:

$\frac{99}{100}{i}_{0}$ = *i*_{0}(1 − *e*^{−t}^{/τ})

*e*^{−t}^{/τ} = 0.01

On taking natural logarithm (ln) of both sides, we get

ln *e*^{−t}^{/τ} = ln 0.01

⇒ - $\frac{t}{\tau}$ = − 4.6

⇒ $\frac{t}{\tau}$ = 4.6

(c) When the value of the current reaches 99.9% of the steady-state value:

$\frac{99.9}{100}{i}_{0}$ = *i*_{0}(1 − *e*^{−t}^{/τ})

⇒ *e*^{−t}^{/τ} = 0.001

On taking natural logarithm (ln) of both sides, we ge

ln *e*^{−t}^{/τ} = ln 0.001

⇒ ^{-}$\frac{t}{\tau}$ = − 6.9

⇒ $\frac{t}{\tau}$ = 6.9

#### Page No 312:

#### Answer:

We know that time constant is the ratio of the self-inductance (*L*) of the coil to the resistance (*R*) of the circuit.

Given:

Current in the circuit, *i* = 2 A

Emf of the battery, *E* = 4 V

Self-inductance of the coil, *L* = 1 H

Now,

Resistance of the coil:

$R=\frac{E}{i}=\frac{4}{2}=2\mathrm{\Omega}$

Time constant:

$\tau =\frac{L}{R}=\frac{1}{2}=0.5\mathrm{s}$

#### Page No 312:

#### Answer:

Given:

Self-inductance of the coil, *L* = 2.0 H

Resistance in the coil, *R* = 20 Ω

Emf of the battery, *e *= 4.0 V

The steady-state current is given by

${i}_{0}=\frac{e}{R}=\frac{4}{20}$ A

The time-constant is given by

$\tau =\frac{L}{R}=\frac{2}{20}=0.1$ s

(a) Current at an instant 0.20 s after the connection is made:

*i* =* **i*_{0}(1 − *e ^{−t}^{/τ}*)

= $\frac{4}{20}$(1 −

*e*

^{−0.2/0.1})

= $\frac{1}{5}$(1 −

*e*

^{−2})

= 0.17 A

(b) Magnetic field energy at the given instant:

$\frac{1}{2}L{i}^{2}$ = $\frac{1}{2}$ × 2(0.17)

^{2}

= 0.0289 = 0.03 J

#### Page No 312:

#### Answer:

Given:

Resistance, *R* = 40 Ω

Emf of the battery, *E* = 4 V

Now,

The steady-state current in the *LR* circuit is given by

${i}_{0}=\frac{4}{40}=0.1\mathrm{A}$

At time, *t* = 0.1 s, the value of current *i* is 63 mA = 0.063 A

The current at time *t* is given by

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

⇒ 0.063 = 0.1(1 − *e*^{−tR}* ^{/L}*)

⇒ 63 =100(1 −

*e*

^{−4}

^{/L})

⇒ 63 = 100(1 −

*e*

^{−4}

^{/}

*)*

^{L}⇒ 1 − 0.63 =

*e*

^{−4}

^{/L}

⇒

*e*

^{−4}

*= 0.37*

^{/L}⇒ $-\frac{4}{L}$ = ln (0.37)

⇒

*L*= $\frac{-4}{-0.994}$

= 4.024 H

= 4 H

#### Page No 312:

#### Answer:

Given:

Self-inductance, *L* = 5.0 H

Resistance, *R* = 100 Ω

Emf of the battery = 2.0 V

At time,* t* = 20 ms (after switching on the circuit)

*t *= 20 ms = 20 × 10^{−3} s = 2 × 10^{−2} s

The steady-state current in the circuit is given by

${i}_{0}=\frac{2}{100}$

The time constant is given by

$\tau =\frac{L}{R}=\frac{5}{100}$ s

The current at time *t* is given by

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

$i=\frac{2}{100}\left(1-{e}^{\left(\frac{-2\times {10}^{-2}\times 100}{5}\right)}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{100}(1-{e}^{-2/5})\phantom{\rule{0ex}{0ex}}=\frac{2}{100}(1-0.670)$

= 0.00659 = 0.0066

Now,

*V = iR* = 0.0066 × 100

= 0.66 V

#### Page No 312:

#### Answer:

Given:

Time constant of the given *LR* circuit, *τ* = 40 ms

Steady-state current in the circuit, *i*_{0} = 2 A

(a) Current at time *t* = 10 ms:

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

= 2(1 − *e*^{−10}^{/40})

= 2(1 − *e*^{−1}^{/4})

= 2(1 − 0.7788)

= 0.4422 A

= 0.44 A

(b) Current at time *t* = 20 ms:

*i* = *i*_{0}(1 − *e*^{−t}* ^{/τ}*)

= 2(1 −

*e*

^{−20}

^{/40})

= 2(1 −

*e*

^{−1}

^{/2})

= 2(1 − 0.606)

= 0.788 A

= 0.79 A

(c) Current at

*t*= 100 ms:

*i*=

*i*

_{0}(1 −

*e*

^{−t}

^{/τ})

= 2(1 −

*e*

^{−100}

^{/40})

= 2(1 −

*e*

^{−10}

^{/4})

= 2(1 −

*e*

^{−5}

^{/2})

= 2(1−0.082)

=1.835 A

= 1.8 A

(d) Current at

*t*= 1 s:

*i*=

*i*

_{0}(1 −

*e*

^{−t}

^{/τ})

= 2(1 −

*e*

^{−1000}

^{/40})

= 2(1 −

*e*

^{−100}

^{/4})

= 2(1 −

*e*

^{−25})

= 2 × 1 A

= 2 A

#### Page No 312:

#### Answer:

Given:

Inductance, *L* = 1.0 H

Resistance in the circuit,* R* = 20 Ω

Emf of the battery = 2.0 V

Now,

Time constant:

$\tau =\frac{L}{R}=\frac{1}{20}=0.05\mathrm{s}$

Steady-state current:

${i}_{0}=\frac{e}{R}=\frac{2}{20}=0.1\mathrm{A}$

Current at time *t*:

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

or

*i* = *i*_{0} − *i*_{0}(*e*^{−}^{t}^{/τ})

On differentiating both sides with respect to *t*, we get

$\frac{di}{dt}=-({i}_{0}\times \left(\frac{-1}{\tau}\right){e}^{-t/\tau})\phantom{\rule{0ex}{0ex}}=\frac{{i}_{0}}{\tau}{e}^{-t/\tau}$

(a) At time *t* = 100 ms,

$\frac{di}{dt}=\frac{0.1}{0.05}\times {e}^{-0.1/0.05}=0.27\mathrm{A}/\mathrm{s}$

(b) At time *t* = 200 ms,

$\frac{di}{dt}=\frac{0.1}{0.05}\times {e}^{-0.2/0.05}\phantom{\rule{0ex}{0ex}}=0.0366\mathrm{A}/\mathrm{s}$

(c) At time *t* = 1 s,

$\frac{di}{dt}=\frac{0.1}{0.05}\times {e}^{-1/0.05}\phantom{\rule{0ex}{0ex}}=41\times {10}^{-9}\mathrm{A}/\mathrm{s}$

#### Page No 312:

#### Answer:

Given:

Self-inductance, *L* = 1 H

For an inductor of self-inductance *L*, the emf induced is given by

*e = L*$\frac{di}{dt}$

(a) At* t** *= 100 ms,

$\frac{di}{dt}$ = 0.27 A/s

∴ Induced emf, *e *= *L*$\frac{di}{dt}$ = 1 × 0.27 = 0.27 V

(b) At *t* = 200 ms,

$\frac{di}{dt}=0.036$ A/s

∴ Induced emf = *L*$\frac{di}{dt}$ = 1 × 0.036 = 0.036 V

(c) At *t* = 1 s,

$\frac{di}{dt}$ = 4.1 × 10^{−9} A/s

∴ Induced emf = $L\frac{di}{dt}$ = 4.1 × 10^{−9} V

#### Page No 312:

#### Answer:

Given:

Self-inductance, *L* = 20 mH

Emf of the battery, *e* = 5.0 V

Resistance, *R* = 10 Ω

Now,

Time constant of the coil:

$\tau =\frac{L}{R}=\frac{20\times {10}^{-3}}{10}$ = 2 × 10^{−3} s

Steady-state current:

${i}_{0}=\frac{e}{R}=\frac{5}{10}=0.5$

The current in the *LR* circuit at time *t *is given by

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

⇒ *i* = *i*_{0} − *i*_{0}*e*^{−t}^{/τ}

On differentiating both sides, we get

$\frac{di}{dt}=\frac{{i}_{0}}{\tau}{e}^{-t/\tau}$

The rate of change of the induced emf is given by

$R\frac{di}{dt}=R\frac{{i}_{0}}{\tau}\times {e}^{-t/\tau}$

(a) At time *t* = 0 s, the rate of change of the induced emf is given by

$R\frac{di}{dt}=R\frac{{i}_{0}}{\tau}\phantom{\rule{0ex}{0ex}}=10\times \frac{0.5}{2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{3}\mathrm{V}/\mathrm{s}$

(b) At time *t* = 10 ms, the rate of change of the induced emf is given by

$R\frac{di}{dt}=R\frac{{i}_{0}}{\tau}\times {e}^{-t/\tau}$

Now,

For *t* = 10 ms = 10 × 10^{−3} s = 10^{−2} s,

$\frac{dE}{dt}=10\times \frac{5}{10}\times \frac{1}{2\times {10}^{-3}}\times {{e}^{-0.01/(2}}^{\times {10}^{-3})}\phantom{\rule{0ex}{0ex}}$

= 16.844 = 17 V/s

(c) At time *t* = 1 s, the rate of change of the induced emf is given by

$\frac{dE}{dt}=\frac{Rdi}{dt}=R\frac{{i}_{0}}{\tau}\times {e}^{-t/\tau}$

$=10\times \frac{5\times {10}^{-1}}{2\times {10}^{-3}}\times {{e}^{-1/(2}}^{\times {10}^{-3})}$

= 0.00 V/s

#### Page No 312:

#### Answer:

Given:

Inductance of the inductor, *L* = 500 mH

Resistance of the resistor connected, *R* = 25 Ω

Emf of the battery, *E* = 5 V

For the given circuit, the potential difference across the resistance is given by

*V = iR*

The current in the *LR* circuit at time *t* is given by

*i* = *i*_{0} (1 − *e*^{−tR}* ^{/L}*)

∴ Potential difference across the resistance at time

*t*,

*V =*(

*i*

_{0}(1 −

*e*

^{−tR}

*)*

^{/L}*R*

(a) For

*t*= 20 ms,

*i*=

*i*

_{0}(1 −

*e*

^{−tR}

*)*

^{/L}$=\frac{E}{R}(1-{e}^{-tR/L})\phantom{\rule{0ex}{0ex}}=\frac{5}{25}(1-{e}^{-(2\times {10}^{-3}\times 25)/(500\times {10}^{-3})}\phantom{\rule{0ex}{0ex}}=\frac{1}{5}(1-{e}^{-1})=\frac{1}{5}(1-0.3678)\phantom{\rule{0ex}{0ex}}=\frac{0.632}{5}=0.1264\mathrm{A}$

Potential difference:

*V*=

*iR*= (0.1264) × (25)

= 3.1606 V = 3.16 V

(b) For

*t*= 100 ms,

*i*=

*i*

_{0}(1 −

*e*

^{−tR}

^{/L})

$=\frac{5}{25}\left(1-{e}^{(-100\times {10}^{-3})\times (25/500\times {10}^{-3})}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}(1-{e}^{-50})\phantom{\rule{0ex}{0ex}}=\frac{1}{5}(1-0.0067)\phantom{\rule{0ex}{0ex}}=\frac{0.9932}{5}=0.19864\mathrm{A}$

Potential difference:

*V*=

*iR*

= (0.19864) × (25) = 4.9665 = 4.97 V

(c) For

*t*= 1 s,

$i=\frac{5}{25}\left(1-{e}^{-1\times 25/500\times {10}^{-3}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}(1-{e}^{-50})\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\times 1=\frac{1}{5}\mathrm{A}\phantom{\rule{0ex}{0ex}}$

Potential difference:

*V*=

*iR*

$=\left(\frac{1}{5}\times 25\right)=5\mathrm{V}$

#### Page No 312:

#### Answer:

Given:

Inductance, *L* = 120 mH = 0.120 H

Resistance, *R* = 10 Ω

Emf of the battery, *E* = 6 V

Internal resistance of the battery, *r* = 2 Ω

The current at any instant in the *LR* circuit is given by

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

Charge *dQ* flown in time *dt* is given by

*dQ = idt = i _{0}*(1

*− e*)

^{−t}^{/τ}*dt*

*Q*= ∫

*dQ*

$=\underset{0}{\overset{t}{\int}}{i}_{0}=\underset{0}{\overset{t}{\int}}{i}_{0}(1-{e}^{-t/\tau})dt\phantom{\rule{0ex}{0ex}}={i}_{0}\left[\underset{0}{\overset{t}{\int}}dt-\underset{0}{\overset{t}{\int}}{e}^{-t/\tau}dt\right]\phantom{\rule{0ex}{0ex}}={i}_{0}\left[t-\left\{\left(-\tau \right){{\left|{e}^{-t/\tau}\right|}^{t}}_{0}\right\}\right]\phantom{\rule{0ex}{0ex}}={i}_{0}\left[t+\tau \left\{{e}^{-t/\tau}-1\right\}\right]$

The steady-state current and the time constant for the given circuit are as follows:

${i}_{0}=\frac{E}{{R}_{\mathrm{total}}}=\frac{6}{10+2}=\frac{6}{12}=0.5\mathrm{A}\phantom{\rule{0ex}{0ex}}\tau =\frac{L}{R}=\frac{120}{12}=0.01\mathrm{s}$

Now,

(a) At time

*t*= 0.01 s,

*Q*= 0.5 [0.01 + 0.01(

*e*

^{−0.1/0.01}− 0.01)]

= 0.00108 = 1.8 × 10

^{−3}= 1.8 mΩ

(b) At

*t*= 20 ms = 2 × 10

^{−2}s = 0.02 s,

*Q*= 0.5 [0.02 + 0.01(

*e*

^{−0.02/0.01}− 0.01)]

= 0.005676 = 5.7 × 10

^{−3}C

= 5.7 mC

(c) At

*t*= 100 ms = 0.1 s,

*Q*= 0.5 [0.1 + 0.1 (

*e*

^{−0.1/0.01}− 0.01)]

= 0.045 C = 45 mc

#### Page No 312:

#### Answer:

Given:

Inductance, *L* = 17 mH

Length of the wire, *l* = 100 m

Cross-sectional area of the wire, *A* = 1 mm^{2} = 1 × 10^{−6} m^{2}

Resistivity of copper, ρ = 1.7 × 10^{−8} Ω-m

Now,

$R=\frac{\mathrm{\rho}l}{A}\phantom{\rule{0ex}{0ex}}=\frac{1.7\times {10}^{-8}\times 100}{1\times {10}^{-6}}=1.7\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

The time constant of the L-R circuit is given by

$\tau =\frac{L}{R}=\frac{17\times {10}^{-3}}{1.7}\phantom{\rule{0ex}{0ex}}={10}^{-2}\mathrm{s}=10\mathrm{ms}$

#### Page No 312:

#### Answer:

Given:

Time constant of the* LR* circuit = 50 ms

Emf of the battery = *ε*

The time constant of the *LR *circuit is given by

$\tau =\frac{L}{R}=50\mathrm{ms}=0.05\mathrm{s}\phantom{\rule{0ex}{0ex}}$

Let the current reach half of its maximum value in time *t*.

Now,

$\frac{{i}_{0}}{2}={i}_{0}(1-{e}^{-t/0.05})\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=1-{e}^{-t/0.05}\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{-t/0.03}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}$

On taking natural logarithm (ln) on both sides, we get

$\mathrm{ln}{e}^{-t/0.05}=\mathrm{ln}\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{t}{0.05}=\mathrm{ln}\left(1\right)-\mathrm{ln}\left(2\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{t}{0.05}=0-0.6931\phantom{\rule{0ex}{0ex}}\Rightarrow t=0.05\times 0.6931\phantom{\rule{0ex}{0ex}}=0.03465\mathrm{s}\phantom{\rule{0ex}{0ex}}=35\mathrm{ms}$

(b) Let *t* be the time at which the power dissipated is half its maximum value.

Maximum power = $\frac{{E}^{2}}{R}$

$\therefore \frac{{E}^{2}}{2R}=\frac{{E}^{2}}{R}(1-{e}^{tR/L}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1-{e}^{-R/L}=\frac{1}{\sqrt{2}}=0.707\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{-t\mathrm{R}/\mathrm{L}}=0.293\phantom{\rule{0ex}{0ex}}\Rightarrow t=50\times 1.2275\mathrm{ms}\phantom{\rule{0ex}{0ex}}=61.2\mathrm{ms}$

(c) Current in the coil at the steady state, *i* = $\frac{\epsilon}{R}$

Magnetic field energy stored at the steady state, $U=\frac{1}{2}L{i}^{2}$ or *U*$=\frac{{\epsilon}^{2}}{2{R}^{2}}L$

Half of the value of the steady-state energy = $\frac{1}{4}L\frac{{\epsilon}^{2}}{{R}^{2}}$

Now,

$\frac{1}{4}L\frac{{\epsilon}^{2}}{{R}^{2}}=\frac{1}{2}L\frac{{\epsilon}^{2}}{{R}^{2}}(1-{e}^{-tR/L}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{-tR/L}=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{2-\sqrt{2}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\tau \left[\mathrm{ln}\left(\frac{1}{2-\sqrt{2}}\right)+\mathrm{ln}2\right]\phantom{\rule{0ex}{0ex}}=0.05\left[\mathrm{ln}\left(\frac{1}{2-\sqrt{2}}\right)+\mathrm{ln}2\right]\phantom{\rule{0ex}{0ex}}=0.061\mathrm{s}\phantom{\rule{0ex}{0ex}}=61\mathrm{ms}$

#### Page No 312:

#### Answer:

Given:

Emf of the battery = *ε*

Inductance of the inductor = *L*

Resistance = *R*

Maximum current in the coil = $\frac{\epsilon}{R}$

At the steady state, current in the coil, *i* = $\frac{\epsilon}{R}$.

The magnetic field energy stored at the steady state is given by

$U=\frac{1}{2}L{i}^{2}$

or

*U*$=\frac{{\epsilon}^{2}}{2{R}^{2}}L$

One-fourth of the steady-state value of the magnetic energy is given by

$U\text{'}=\frac{1}{8}L\frac{{E}^{2}}{{R}^{2}}$

Half of the value of the steady-state energy = $\frac{1}{4}L\frac{{E}^{2}}{{R}^{2}}$

Let the magnetic energy reach one-fourth of its steady-state value in time *t*_{1}_{ }and let it reach half of its value in time *t*_{2}.

Now,

$\frac{1}{8}L\frac{{E}^{2}}{{R}^{2}}=\frac{1}{2}L\frac{{E}^{2}}{{R}^{2}}(1-{e}^{-{t}_{1}R/L}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1-{e}^{-{t}_{1}R/L}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}_{1}\frac{R}{L}=\mathrm{ln}2\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}\frac{1}{4}L\frac{{E}^{2}}{{R}^{2}}=\frac{1}{2}L\frac{{E}^{2}}{{R}^{2}}(1-{e}^{-{t}_{2}R/L}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{-{t}_{2}R/L}=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{2-\sqrt{2}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}_{1}=\tau \mathrm{ln}\left(\frac{1}{2-\sqrt{2}}\right)+\mathrm{ln}2$

Thus, the time taken by the magnetic energy stored in the circuit to change from one-fourth of its steady-state value to half of its steady-state value is given by

${t}_{2}-{t}_{1}=\tau \mathrm{ln}\frac{1}{2-\sqrt{2}}$

#### Page No 312:

#### Answer:

Given:

Inductance*, L* = 4.0 H

Resistance, *R* = 10 Ω

Emf of the battery, *E *= 4 V

(a) Time constant

$\tau =\frac{L}{R}=\frac{4}{10}=0.4\mathrm{s}$

(b) As the current reaches 0.63 of its steady-state value,* i* = 0.63 *i*_{0}*.*

Now,

0.63 *i*_{0} = *i*_{0}(1 − *e*^{−t}^{/τ})

⇒ *e*^{−t}^{/τ} = 1 − 0.063 = 0.37

⇒ ln *e*^{−t}^{/τ} = ln 0.37

⇒ $-\frac{t}{\tau}=-0.9942$

⇒ *t* = 0.942 × 0.4

= 0.3977 = 0.4 s

(c) The current in the* LR* circuit at an instant is given by

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

$=\frac{4}{10}(1-{e}^{-0.4/0.4})$

= 0.4 × 0.6321

= 0.2528 A

Power delivered*, P = Vi*

⇒* P* = 4 × 0.2528

= 1.01 = 1 W

(d) Power dissipated in Joule heating, *P' *= *i*^{2}*R** *

⇒* P' *= (0.2258)^{2} × 10

= 0.639 = 0.64 W

#### Page No 312:

#### Answer:

Given:

Inductance of the inductor, *L* = 2.0 mH

Let the resistance in the circuit be *R* and the steady state value of the current be *i*_{0}.

At time *t* , current* i* in the* LR* circuit is given by

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

Here,

*$\tau =\frac{L}{R}$ = *Time constant

On multiplying both sides by µ_{0}*n**, *we get

*n* = Number of turns per unit length of the coil

µ_{0}*ni* = µ_{0}*ni*_{0}(1 − *e*^{−}^{t}^{/τ})

⇒ *B* = *B*_{0}(*1* − *e*^{−tR}^{/L})

⇒ 0.8 *B*_{0} = *B*_{0}$\left(1-{e}^{\frac{-20\times {10}^{-6}\mathrm{R}}{2\times {10}^{-3}}}\right)$

⇒ 0.8 = (1 − *e*^{−R}^{/100})

⇒ *e*^{−R}^{/100} = 0.2

⇒ ln (*e*^{−}^{R}^{/100}) = ln (0.2)

⇒ $-\frac{R}{100}$ = −1.693

⇒ *R* = 169.3 Ω

#### Page No 312:

#### Answer:

(a) Let the current in the *LR* circuit be *i*.

Let the charge flowing through the coil in the infinitesimal time *dt *be* dq*.

Now,

$i=\frac{dq}{dt}$

∴ *dq = idt*

The current in the* LR* circuit after *t* seconds after connecting the battery is given by

*i* = *i*_{0} (1 − *e*^{−t}* ^{/τ}*)

Here,

*i*

_{0}

_{ }= Steady state current

*τ =*Time constant = $\frac{L}{R}$

*dq*=

*i*

_{0}(1 −

*e*

^{−tR}

^{/L})

*dt*

On integrating both sides, we get

$Q=\underset{0}{\overset{t}{\int}}dq$

$={i}_{0}\left[\underset{0}{\overset{t}{\int}}dt-\underset{0}{\overset{t}{\int}}{e}^{-tR/L}dt\right]\phantom{\rule{0ex}{0ex}}={i}_{0}\left[t-\left(-\frac{L}{R}\right)({e}^{-tR/L}-1)\right]\phantom{\rule{0ex}{0ex}}={i}_{0}\left[t-\frac{L}{R}(1-{e}^{-tR/L})\right]$

Thus, the charge flowing in the coil in time

*t*is given by

$Q=\frac{\epsilon}{R}\left[t-\frac{L}{R}(1-{e}^{-tR/L})\right]$

$Where{i}_{o}=\frac{\epsilon}{R}$

(b) The work done by the battery is given by

*W*=

*εQ*

From the above expression for the charge in the LR circuit, we have

$W=\frac{{\epsilon}^{2}}{R}\left[t-\frac{L}{R}(1-{e}^{-tR/L})\right]$

(c) The heat developed in time

*t*can be calculated as follows:

$H=\underset{0}{\overset{t}{\int}}{i}^{2}Rdt\phantom{\rule{0ex}{0ex}}H=\frac{{\epsilon}^{2}}{{R}^{2}}R\underset{0}{\overset{t}{\int}}(1-{e}^{-tR/L}{)}^{2}dt\phantom{\rule{0ex}{0ex}}H=\frac{{\epsilon}^{2}}{R}\underset{0}{\overset{t}{\int}}\left[(1+{e}^{-2tR/L})-2{e}^{-tR/L}\right]dt\phantom{\rule{0ex}{0ex}}H=\frac{{\epsilon}^{2}}{R}{\left(t-\frac{L}{2R}{e}^{-2tR/L}+\frac{L}{R}2{e}^{-tR/L}\right)}_{0}^{t}\phantom{\rule{0ex}{0ex}}H=\frac{{\epsilon}^{2}}{R}\left(t-\frac{L}{2R}{e}^{-2tR/L}+\frac{L}{R}2{e}^{-tR/L}\right)-\left(-\frac{L}{2R}+\frac{2L}{R}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\epsilon}^{2}}{R}\left\{t-\frac{L}{2R}({x}^{2}-4x+3)\right\}(x={e}^{-tR/L})\phantom{\rule{0ex}{0ex}}$

(d) The magnetic energy stored in the circuit is given by

$U=\frac{1}{2}L{i}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow U=\frac{1}{2}L\frac{{\epsilon}^{2}}{{R}^{2}}(1-{e}^{-tR/L}{)}^{2}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow U=\frac{L{\epsilon}^{2}}{2{R}^{2}}(1-x{)}^{2}$

(e)

Taking the sum of total energy stored in the magnetic field and the heat developed in time

*t*

$E=\frac{L{\epsilon}^{2}}{2{R}^{2}}(1-x{)}^{2}+\frac{{\epsilon}^{2}}{R}\left\{t-\frac{L}{2R}({x}^{2}-4x+3)\right\}\phantom{\rule{0ex}{0ex}}E=\frac{L{\epsilon}^{2}}{2{R}^{2}}(1+{x}^{2}-2x)+\frac{{\epsilon}^{2}}{R}t-\frac{L{\epsilon}^{2}}{2{R}^{2}}({x}^{2}-4x+3)\phantom{\rule{0ex}{0ex}}E=\frac{L{\epsilon}^{2}}{2{R}^{2}}(2x-2)+\frac{{\epsilon}^{2}}{R}t\phantom{\rule{0ex}{0ex}}E=\frac{L{\epsilon}^{2}}{{R}^{2}}(x-1)+\frac{{\epsilon}^{2}}{R}t\phantom{\rule{0ex}{0ex}}E=\frac{{\epsilon}^{2}}{R}\left\{t-\frac{L}{R}(1-x)\right\}=\frac{{\epsilon}^{2}}{R}\left\{t-\frac{L}{R}(1-{e}^{-tR/L})\right\}$

The above expression is equal to the energy drawn from the battery. Therefore, the conservation of energy holds good.

#### Page No 312:

#### Answer:

Given:

Inductance of the inductor, *L* = 2 H

Resistance of the resistor connected to the inductor, *R* = 200 Ω,

Emf of the battery connected, *E* = 2 V

(a) The current in the *LR* circuit after *t* seconds after connecting the battery is given by

*i* = *i*_{0}(1 − *e*^{−t}^{/τ})

Here,

*i*_{0}_{ }= Steady state value of current

*i*_{0}_{ }= $\frac{E}{R}=\frac{2}{200}\mathrm{A}$

At time *t* = 10 ms, the current is given by

*i *$=\frac{2}{200}(1-{e}^{-10\times {10}^{-3}\times 200/2})$

*i* = 0.01(1 − *e*^{−1})

*i* = 0.01(1 − 0.3678)

*i* = 0.01 × 0.632 = 6.3 mA

(b) The power delivered by the battery is given by

*P = Vi*

*P = **Ei*_{0}(1 − *e*^{−t}^{/τ})

$P=\frac{{E}^{2}}{R}(1-{e}^{-t/\tau})\phantom{\rule{0ex}{0ex}}P=\frac{2\times 2}{200}(1-{e}^{10\times {10}^{-3}\times 200/2})$

*P *= 0.02(1 − *e*^{−1})

*P* = 0.01264 = 12.6 mW

(c) The power dissipated in the resistor is given by

*P*_{1} = *i*^{2}*R*

*P*_{1} = [*i*_{0}(1 − *e*^{−t}^{/τ})]^{2}* R*

*P*_{1}_{ }= (6.3 mA)^{2} × 200

*P*_{1}_{ }= 6.3 × 6.3 × 200 × 10^{−5}

*P*_{1}_{ }= 79.38 × 10^{−4}

*P*_{1}_{ }= 7.938 × 10^{−3} = 8 mW

(d) The rate at which the energy is stored in the magnetic field can be calculated as:

*W *= $\frac{1}{2}L{i}^{2}$

*W *$=\frac{L}{2}{{i}_{0}}^{2}(1-{e}^{-t/\tau}{)}^{2}$

*W *= 2 × 10^{−2}(0.225)

*W *= 0.455 × 10^{−2}

*W* = 4.6 × 10^{−3}

*W *= 4.6 mW

#### Page No 312:

#### Answer:

Given:

Inductance of the coil A,* L*_{A} = 1.0 H

Inductance of the coil B, *L*_{B} = 2.0 H

Resistance in each coil, *R* = 10 Ω

The current in the LR circuit after *t* seconds after connecting the battery is given by

*i* = *i*_{0} (1 − *e*^{−t}* ^{/τ}*)

Here,

*i*

_{0}

_{ }= Steady state current

*τ =*Time constant = $\frac{L}{R}$

(a) At

*t*= 0.1 s, time constants of the coils A and B are

*τ*

_{A}

_{ }and

*τ*

_{B}, respectively.

Now,

${\tau}_{\mathrm{A}}=\frac{1}{10}=0.1\mathrm{s}\phantom{\rule{0ex}{0ex}}{\mathrm{\tau}}_{\mathrm{B}}=\frac{2}{10}=0.2\mathrm{s}$

Currents in the coils can be calculated as follows:

${i}_{\mathrm{A}}={i}_{0}(1-{e}^{-t/\tau}),\phantom{\rule{0ex}{0ex}}=\frac{2}{10}\left(1-{e}^{\frac{0.1\times 10}{1}}\right)=0.2(1-{e}^{-1})\phantom{\rule{0ex}{0ex}}=0.126424111\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{B}}={i}_{0}(1-{e}^{-t/\tau})\phantom{\rule{0ex}{0ex}}=\frac{2}{10}(1-{e}^{0.1\times 10/2})\phantom{\rule{0ex}{0ex}}=0.2(1-{e}^{-1/2})=0.078693$

$\therefore \frac{{i}_{\mathrm{A}}}{{i}_{\mathrm{B}}}=\frac{0.126411}{0.78693}=1.6$

(b) At

*t*= 200 ms = 0.2 s,

*i*

_{A}= 0.2 (1 −

*e*

^{−0.2 × 10.1})

*i*

_{A}

_{ }= 0.2 × 0.864664716

*i*

_{A}

_{ }= 0.1729329943

*i*

_{B}= 0.2 (1 −

*e*

^{−0.2 × 10.2})

*i*

_{B}= 0.2 × 0.632120 = 0.126424111

$\therefore \frac{{i}_{\mathrm{A}}}{{i}_{\mathrm{B}}}=\frac{0.172932343}{0.126424111}=1.36=1.4$

(c) At time

*t*= 1 s,

*i*

_{A}= 0.2 (1 −

*e*

^{−1}

^{ × 10.1})

= 0.2 − 0.9999549

= 0.19999092

*i*

_{B}= 0.2 (1 −

*e*

^{−1}

^{ × 10.2})

= 0.2 × 0.99326 = 0.19865241

$\therefore \frac{{i}_{\mathrm{A}}}{{i}_{\mathrm{B}}}=\frac{0.19999092}{0.19999092}\approx 1.0$

#### Page No 312:

#### Answer:

The current in the discharging LR circuit after *t* seconds is given by

*i* = *i*_{0} *e*^{−t}^{/τ}

Here,

*i*_{0}_{ }= Steady state current = 2 A

Now, let the time constant be *τ.*

In time *t *= 0.10 s, the current drops to 1 A.

$i={i}_{0}\left(1-{e}^{-t/\tau}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 1=2\left(1-{e}^{-t/\tau}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=1-{e}^{-t/\tau}\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{-t/\tau}=1-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{-t/\tau}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ln}\left({e}^{-t/\tau}\right)=\mathrm{ln}\left(\frac{1}{2}\right)=\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{0.1}{\tau}=-0.693\phantom{\rule{0ex}{0ex}}$

The time constant is given by

$\tau =\frac{0.1}{0.693}=0.144=0.14$ s

(b) Given:

Inductance in the circuit, *L* = 4 H

Let the resistance in the circuit be *R*.

The time constant is given by

$\tau =\frac{L}{R}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{the}\mathrm{above}\mathrm{relation},\mathrm{we}\mathrm{have}\phantom{\rule{0ex}{0ex}}0.14=\frac{4}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{4}{0.14}\phantom{\rule{0ex}{0ex}}\Rightarrow R=28.5728\mathrm{\Omega}$

#### Page No 313:

#### Answer:

Consider an inductance *L*, a resitance *R *and a source of emf $\xi $ are connected in series.

Time constant of this *LR* circuit is, $\tau =\frac{L}{R}$

Let a constant current *i*_{0} ($=\frac{\xi}{R}$) is maitened in the circuit before removal of the battery.

Charge flown in one time constant before the short-circuiting is,

${Q}_{\tau}={i}_{0}\tau $ ...(i)

Discahrge equation for *LR* circuit after short circuiting is given as,

$i={i}_{0}{e}^{-\frac{t}{\tau}}$

Change flown from the inductor in small time *dt* after the short circuiting is given as,

$dQ=idt$

Chrage flown from the inductor after short circuting can be found by interating the above eqation within the proper limits of time,

$Q={\int}_{0}^{\infty}idt\phantom{\rule{0ex}{0ex}}\Rightarrow Q={\int}_{0}^{\infty}{i}_{0}{e}^{-\frac{t}{\tau}}dt\phantom{\rule{0ex}{0ex}}\Rightarrow Q={\left[-\tau {i}_{0}{e}^{-\frac{t}{\tau}}\right]}_{0}^{\infty}\phantom{\rule{0ex}{0ex}}\Rightarrow Q=-\tau {i}_{0}\left[0-1\right]\phantom{\rule{0ex}{0ex}}\Rightarrow Q=\tau {i}_{0}...\left(\mathrm{ii}\right)$

Hence, proved.

#### Page No 313:

#### Answer:

(a) Because the switch is closed, the battery gets connected across the* L‒R* circuit.

The current in the* L‒R *circuit after *t* seconds after connecting the battery is given by

*i* = *i*_{0} (1 − *e*^{−t}* ^{/τ}*)

Here,

*i*

_{0}

_{ }= Steady state current

*τ =*Time constant = $\frac{L}{R}$

After a long time,

*t$\to \infty $.*

Now,

Current in the inductor,

*i*=

*i*

_{0}(1 −

*e*

^{0}) = 0

Thus, the effect of inductance vanishes.

$i=\frac{\epsilon}{{R}_{net}}\phantom{\rule{0ex}{0ex}}i=\frac{\epsilon}{{\displaystyle \frac{{R}_{1}\times {R}_{2}}{{R}_{1}+{R}_{2}}}}=\frac{\epsilon ({R}_{1}+{R}_{2})}{{R}_{1}{R}_{2}}$

(b) When the switch is opened, the resistance are in series.

The time constant is given by

$\tau =\frac{L}{{R}_{net}}=\frac{L}{{R}_{1}+{R}_{2}}$

(c) The inductor will discharge through resistors

*R*

_{1}and

*R*

_{2}.

The current through the inductor after one time constant is given by

*t = τ*

∴ Current,

*i*=

*i*

_{0}

*e*

^{−τ}

^{/τ}Here,

*i*

_{0}= $\frac{\epsilon}{{R}_{1}+{R}_{2}}$

∴

*i*= $\frac{\epsilon}{{R}_{1}+{R}_{2}}\times \frac{1}{e}$

#### Page No 313:

#### Answer:

Given:

Current through the solenoid, *i* = 1.0 A

Radius of the coil, *r* = 2 cm

Number of turns per metre, *n* = 1000

The magnetic energy density is given by $\frac{{B}^{2}}{2{\mu}_{0}}$.

Volume of the solenoid,* V* = $\mathrm{\pi}{r}^{2}l$

For $l=1\mathrm{m}$, $V=\mathrm{\pi}{r}^{2}$.

Thus, the magnetic energy stored in volume *V* is given by

*U* = $\frac{{B}^{2}\mathrm{\pi}{r}^{2}}{2{\mu}_{0}}\phantom{\rule{0ex}{0ex}}$

The magnetic field is given by

*B* = *μ*_{0}*ni*

= (4π × 10^{$-$7}) × (1000) × 1

= 4π × 10^{$-$4} T

$U=\frac{(4\mathrm{\pi}\times {10}^{-4}{)}^{2}\times 4\mathrm{\pi}\times {10}^{-4}}{2\times (4\mathrm{\pi}\times {10}^{-7})}\phantom{\rule{0ex}{0ex}}=8{\mathrm{\pi}}^{2}\times {10}^{-5}\phantom{\rule{0ex}{0ex}}=78.956\times {10}^{-5}\phantom{\rule{0ex}{0ex}}=7.9\times {10}^{-4}\mathrm{J}$

#### Page No 313:

#### Answer:

Given:

Current in the loop, *i *= 4 A

Radius of the loop, *r* = 10 cm = 0.1 m

Volume of the cube, *V* = 1 mm^{3} = $1\times {10}^{-9}\mathrm{m}$

Magnetic field intensity at the centre of the circular loop:

$B=\frac{{\mu}_{0}i}{2r}\phantom{\rule{0ex}{0ex}}=\frac{(4\pi \times {10}^{-7})\times 4}{2\times 0.1}\phantom{\rule{0ex}{0ex}}=8\mathrm{\pi}\times {10}^{-6}\mathrm{T}$

Magnetic energy density = $\frac{{B}^{2}}{2{\mu}_{0}}$

Total energy stored in volume *V*:

*U *= $\frac{{B}^{2}V}{2{\mu}_{0}}$

$=\frac{(8\mathrm{\pi}\times {10}^{-6}{)}^{2}\times (1\times {10}^{-9})}{(4\mathrm{\pi}\times {10}^{-7})\times 2}\phantom{\rule{0ex}{0ex}}=8\mathrm{\pi}\times {10}^{-14}\mathrm{J}$

#### Page No 313:

#### Answer:

Current flowing through the wire, *i** *= 4.00 A

Volume of the region, *V* = 1 mm^{3}

Distance of the region from the wire, *d* = 10 cm = 0.1 m

Magnetic field due to the current-carrying straight wire, *B* = $\frac{{\mu}_{0}i}{2\pi r}$

The magnetic energy stored is given by

$U=\frac{{B}^{2}V}{2{\mu}_{0}}=\frac{{{\mu}_{0}}^{2}{i}^{2}}{4{\pi}^{2}{r}^{2}}\times \frac{1}{2{\mu}_{0}}\times V\phantom{\rule{0ex}{0ex}}U=\frac{{\mu}_{0}{i}^{2}}{4{\pi}^{2}{r}^{2}}\times \frac{1}{2}\times V\phantom{\rule{0ex}{0ex}}U=\frac{(4\mathrm{\pi}\times {10}^{-7})\times (4{)}^{2}\times (1\times {10}^{-9})}{(4{\pi}^{2}\times {10}^{-2})\times 2}\phantom{\rule{0ex}{0ex}}U=2.55\times {10}^{-14}\mathrm{J}$

#### Page No 313:

#### Answer:

Given:

Mutual inductance between the coils, *M* = 2.5 H

Rate of change of current in one coil, $\frac{di}{dt}$ = 1 As^{−1}

The flux in the coil due do another coil carrying current *i* is given by

*ϕ = Mi*

The emf induced in the second coil due to change in the current in the first coil is given by

$e=\frac{d\varphi}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow e=\frac{d\left(Mi\right)}{dt}=M\frac{di}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow e=2.5\times 1=2.5\mathrm{V}$

#### Page No 313:

#### Answer:

The flux through the square frame is given by

$\varphi =Mi$

Let us first calculate the flux through the square frame.

Let us now consider an element of loop of length *dx* at a distance *x* from the wire.

Now,

Area of the element of loop, *A* = *a*d*x*

Magnetic field at a distance* x* from the wire, $B=\frac{{\mu}_{0}i}{2\pi x}$

The magnetic flux of the element is given by

$d\varphi =\frac{{\mu}_{0}i\times a\mathrm{d}x}{2\pi x}$

The total flux through the frame is given by

$\varphi =\int \mathrm{d}\varphi \phantom{\rule{0ex}{0ex}}={\int}_{b}^{a+b}\frac{{\mu}_{0}iadx}{2\pi x}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}ia}{2\mathrm{\pi}}\mathrm{ln}\left[1+\frac{a}{b}\right]\phantom{\rule{0ex}{0ex}}$

Also,

$\varphi =Mi$

Thus, the mutual inductance is calculated as

$Mi=\frac{{\mathrm{\mu}}_{0}ia}{2\mathrm{\pi}}\mathrm{ln}\left[1+\frac{a}{b}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{{\mathrm{\mu}}_{0}a}{2\mathrm{\pi}}\mathrm{ln}\left[1+\frac{a}{b}\right]$

#### Page No 313:

#### Answer:

The magnetic field due to coil 1 at the centre of coil 2 is given by

$B=\frac{{\mathrm{\mu}}_{0}Ni{a}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}$

The flux linked with coil 2 is given by

$\varphi =B.A\text{'}=\frac{{\mathrm{\mu}}_{0}Ni{a}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\mathrm{\pi}a{\text{'}}^{2}$

Now, let *y* be the distance of the sliding contact from its left end.

Given:

$v=\frac{dy}{dt}$

Total resistance of the rheostat = *R*

When the distance of the sliding contact from the left end is *y*, the resistance of the rheostat is given by

$r\text{'}=\frac{R}{L}y$

The current in the coil is the function of distance *y* travelled by the sliding contact of the rheostat. It is given by

$i=\frac{E}{\left({\displaystyle \frac{R}{L}}y+r\right)}$

The magnitude of the emf induced can be calculated as:

$e=\frac{d\mathrm{\varphi}}{dt}=\frac{{\mu}_{0}N{a}^{2}a{\text{'}}^{2}\mathrm{\pi}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\frac{di}{dt}\phantom{\rule{0ex}{0ex}}$

$e=\frac{{\mathrm{\mu}}_{0}N\mathrm{\pi}{a}^{2}a{\text{'}}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\frac{d}{dt}\frac{E}{\left({\displaystyle \frac{R}{L}}y+r\right)}\phantom{\rule{0ex}{0ex}}e=\frac{{\mathrm{\mu}}_{0}N\mathrm{\pi}{a}^{2}a{\text{'}}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^{2}}\right]$

emf induced, $e=\frac{{\mathrm{\mu}}_{0}N\mathrm{\pi}{a}^{2}a{\text{'}}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^{2}}\right]$

The emf induced in the coil can also be given as:

$\frac{\mathrm{d}i}{\mathrm{d}t}=\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^{2}}\right]$

$e=M\frac{\mathrm{d}i}{\mathrm{d}t},\frac{\mathrm{d}i}{\mathrm{d}t}=\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}M=\frac{e}{{\displaystyle \frac{\mathrm{d}i}{\mathrm{d}t}}}=\frac{N{\mu}_{0}\mathrm{\pi}{a}^{2}a{\text{'}}^{2}}{2({a}^{2}+{x}^{2}{)}^{3/2}}$

#### Page No 313:

#### Answer:

Mutual inductance

M = μ_{0}N_{1}N_{2}πr_{1}^{2}*l*

= 4π × 10^{−7} × 4 × 10^{3} × 2 × 10^{3} × 4 × 10^{−4} × 10 × 10^{−2}

= 0.04 × 10^{−2} H

Given that,

For solenoid-1

Area of cross section, *a*_{1} = 4 cm^{2} = 4 × 10^{−4}^{ }m^{2}

Length of the solenoid, *l*_{1} = 20 cm = 0.20 m

Number of turns per unit length , *n*_{1} = 4000/0.2 m = 20000 turns/m

For solenoid-2

Area of cross section, a_{2} = 8 cm^{2}^{ }= 4 × 10^{−4}^{ }m^{2}

Length of the solenoid, *l*_{2} = 10 cm = 0.1 m

Number of turns per unit length,* n*_{2} = 2000/0.1 m = 20000 turns/m

It is given that the solenoid-1 is placed inside the solenoid-2

Let the current through the solenoid-2 be *i.*

The magnetic field due to current in solenoid-2

*B* = μ_{0}*n*_{2}*i* = $\left(4\pi \times {10}^{-7}\right)\times \left(20000\right)\times i$

Now,

Flux through the coil-1 is given by

*ϕ* = *n*_{1}*l*_{1}*.B.a*_{1} = *n*_{1}*l*_{1}(μ_{0}*n*_{2}*i*) × *a*_{1}

$\varphi =2000\times 20000\times 4\mathrm{\pi}\times {10}^{-7}\times i\times 4\times {10}^{-4}$

If the current flowing in the coil-2 changes, then emf is induced in the coil-1

Thus, emf induced in the coil-1 due to change in the current in coil-2 is given by

#### Page No 313:

#### Answer:

Given:

Radius of the long solenoid = *R*

Number of turns per unit length of the long solenoid = *n*

Current in the long solenoid, *i* = *i*_{0} sin *ωt*

Number of turns in the small solenoid = *N*

Radius of the small solenoid = *R*

The magnetic field inside the long solenoid is given by

*B* = *μ*_{0}*ni*

Flux produced in the small solenoid because of the long solenoid,* ϕ* = (μ_{0}*ni*) × (*NπR*^{2})

(a) The emf developed in the small solenoid is given by

*e *= $\frac{d\varphi}{dt}=\frac{d}{dt}\left({\mu}_{0}niN\pi {R}^{2}\right)\phantom{\rule{0ex}{0ex}}$

*e *=* μ _{0}nN *

*πR*

^{2}$\frac{di}{dt}$

Substituting

*i*=

*i*

_{0}sin

*ωt*, we get

*e*=

*μ*

_{0}

*nNπR*

^{2}

*i*

_{0}ω cos ω

*t*

(b) Let the mutual inductance of the coils be

*m*.

Flux

*ϕ*linked with the second coil is given by

*ϕ*= (μ

_{0}

*ni*) × (

*NπR*

^{2})

The flux can also be written as

*ϕ*=

*mi*

∴ (μ

_{0}

*ni*) × (

*NπR*

^{2}) =

*mi*

And,

*m*= πμ

_{0}

*n*

*NR*

^{2}

View NCERT Solutions for all chapters of Class 12