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:

#### Question 1:

A metallic loop is placed in a nonuniform magnetic field. Will an emf be induced in the loop?

#### 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:

#### Question 2:

An inductor is connected to a battery through a switch. Explain why the emf induced in the inductor is much larger when the switch is opened as compared to the emf induced when the switch is closed.

#### 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:

#### Question 3:

The coil of a moving-coil galvanometer keeps on oscillating for a long time if it is deflected and released. If the ends of the coil are connected together, the oscillation stops at once. Explain.

#### 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:

#### Question 4:

A short magnet is moved along the axis of a conducting loop. Show that the loop repels the magnet if the magnet is approaching the loop and attracts the magnet if it is going away from the loop.

#### 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:

#### Question 5:

Two circular loops are placed coaxially but separated by a distance. A battery is suddenly connected to one of the loops establishing a current in it. Will there be a current induced in the other loop? If yes, when does the current start and when does it end? Do the loops attract each other or do they repel?

#### 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:

#### Question 6:

The battery discussed in the previous question is suddenly disconnected. Is a current induced in the other loop? If yes, when does it start and when does it end? Do the loops attract each other or repel?

#### 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:

#### Question 7:

If the magnetic field outside a copper box is suddenly changed, what happens to the magnetic field inside the box? Such low-resistivity metals are used to form enclosures which shield objects inside them against varying magnetic fields.

#### 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:

#### Question 8:

Metallic (nonferromagnetic) and nonmetallic particles in a solid waste may be separated as follows. The waste is allowed to slide down an incline over permanent magnets. The metallic particles slow down as compared to the nonmetallic ones and hence are separated. Discuss the role of eddy currents in the process.

#### 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:

#### Question 9:

A pivoted aluminium bar falls much more slowly through a small region containing a magnetic field than a similar bar of an insulating material. Explain.

#### 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:

#### Question 10:

A metallic bob *A* oscillates through the space between the poles of an electromagnet (figure). The oscillations are more quickly damped when the circuit is on, as compared to the case when the circuit is off. Explain.

Figure

#### 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:

#### Question 11:

Two circular loops are placed with their centres separated by a fixed distance. How would you orient the loops to have (a) the largest mutual inductance (b) the smallest mutual inductance?

#### 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:

#### Question 12:

Consider the self-inductance per unit length of a solenoid at its centre and that near its ends. Which of the two is greater?

#### 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:

#### Question 13:

Consider the energy density in a solenoid at its centre and that near its ends. Which of the two is greater?

#### 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:

#### Question 1:

A rod of length *l* rotates with a small but uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field *B *exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is

(a) zero

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

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

(d) *B*ω*l*^{2}.

#### 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:

#### Question 2:

A rod of length *l* rotates with a uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field *B *exists parallel to the axis of rotation. The potential difference between the two ends of the rod is

(a) zero

(b) $\frac{1}{2}Bl{\mathrm{\omega}}^{2}$

(c) *Bl*ω^{2}

(d) 2*Bl*ω^{2}.

#### 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:

#### Question 3:

Consider the situation shown in figure. If the switch is closed and after some time it is opened again, the closed loop will show

(a) an anticlockwise current-pulse

(b) a clockwise current-pulse

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

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

Figure

#### 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:

#### Question 4:

Solve the previous question if the closed loop is completely enclosed in the circuit containing the switch.

#### 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:

#### Question 5:

A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some time the magnet

(a) will stop in the tube

(b) will move with almost contant speed

(c) will move with an acceleration *g*

(d) will oscillate.

#### 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:

#### Question 6:

Figure shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will

(a) remain stationary

(b) move towards the solenoid

(c) move away from the solenoid

(d) move towards the solenoid or away from it depending on which terminal (positive or negative) of the battery is connected to the left end of the solenoid.

Figure

#### 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:

#### Question 7:

Consider the following statements:

(A) An emf can be induced by moving a conductor in a magnetic field.

(B) An emf can be induced by changing the magnetic field.

(a) Both A and B are true.

(b) A is true but B is false.

(c) B is true but A is false.

(d) Both A and B are false.

#### 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:

#### Question 8:

Consider the situation shown in figure. The wire *AB* is slid on the fixed rails with a constant velocity. If the wire *AB* is replaced by a semicircular wire, the magnitude of the induced current will

(a) increase

(b) remain the same

(c) decrease

(d) increase or decrease depending on whether the semi-circle bulges towards the resistance or away from it.

Figure

#### 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:

#### Question 9:

Figure shows a conducting loop being pulled out of a magnetic field with a speed *v*. Which of the four plots shown in figure (b) may represent the power delivered by the pulling agent as a function of the speed *v*?

Figure

#### 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:

#### Question 10:

Two circular loops of equal radii are placed coaxially at some separation. The first is cut and a battery is inserted in between to drive a current in it. The current changes slightly because of the variation in resistance with temperature. During this period, the two loops

(a) attract each other

(b) repel each other

(c) do not exert any force on each other

(d) attract or repel each other depending on the sense of the current.

#### 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:

#### Question 11:

A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop is

(a) clockwise

(b) anticlockwise

(c) zero

(d) clockwise or anticlockwise depending on whether the resistance is increased or decreased.

#### 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:

#### Question 12:

A conducting square loop of side *l* and resistance *R* moves in its plane with a uniform velocity *v* perpendicular to one of its sides. A uniform and constant magnetic field *B* exists along the perpendicular to the plane of the loop as shown in figure. The current induced in the loop is

(a) *Blv*/*R* clockwise

(b) *Blv*/*R* anticlockwise

(c) 2*Blv*/*R* anticlockwise

(d) zero.

Figure

#### 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:

#### Question 1:

A bar magnet is moved along the axis of a copper ring placed far away from the magnet. Looking from the side of the magnet, an anticlockwise current is found to be induced in the ring. Which of the following may be true?

(a) The south pole faces the ring and the magnet moves towards it.

(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.

(d) The north pole faces the ring and the magnet moves away from it.

#### 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:

#### Question 2:

A conducting rod is moved with a constant velocity *v* in a magnetic field. A potential difference appears across the two ends

(a) if $\overrightarrow{v}\parallel \overrightarrow{l}$

(b) if $\overrightarrow{v}\parallel \overrightarrow{B}$

(c) if $\overrightarrow{l}\parallel \overrightarrow{B}$

(d) none of these.

#### 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:

#### Question 3:

A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if

(a) it is translated

(b) it is rotated about its axis

(c) it is rotated about a diameter

(d) it is deformed.

#### 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:

#### Question 4:

A metal sheet is placed in front of a strong magnetic pole. A force is needed to

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

(b) hold the sheet there if the metal is nonmagnetic

(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.

Neglect any effect of paramagnetism, diamagnetism and gravity.

#### 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:

#### Question 5:

A constant current *i* is maintained in a solenoid. Which of the following quantities will increase if an iron rod is inserted in the solenoid along its axis?

(a) magnetic field at the centre

(b) magnetic flux linked with the solenoid

(c) self-inductance of the solenoid

(d) rate of Joule heating.

#### 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:

#### Question 6:

Two solenoids have identical geometrical construction but one is made of thick wire and the other of thin wire. Which of the following quantities are different for the two solenoids?

(a) self-inductance

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

(c) magnetic field energy 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.

#### 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:

#### Question 7:

An *LR* circuit with a battery is connected at *t* = 0. Which of the following quantities is not zero just after the connection?

(a) Current in the circuit

(b) Magnetic field energy in the inductor

(c) Power delivered by the battery

(d) Emf induced in the inductor

#### 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:

#### Question 8:

A rod *AB* moves with a uniform velocity *v* in a uniform magnetic field as shown in figure.

(a) The rod becomes electrically charged.

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

(c) The end *B* becomes positively charged.

(d) The rod becomes hot because of Joule heating.

Figure

#### 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:

#### Question 9:

*L*, *C* and *R* represent the physical quantities inductance, capacitance and resistance respectively. Which of the following combinations have dimensions of frequency?

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

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

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

(d) *C*/*L*.

#### 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:

#### Question 10:

The switches in figure (a) and (b) are closed at *t* = 0 and reopened after a long time at *t* = *t*_{0}.

Figure

(a) The charge on *C* just after *t* = 0 is *εC*.

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

(c) The current in *L* just before *t* = *t*_{0} is *ε/R.*

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

#### 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:

#### Question 1:

Calculate the dimensions of (a) $\int \overrightarrow{E}.d\overrightarrow{l,}$ (b) *vBl* and (c) $\frac{d{\mathrm{\Phi}}_{B}}{dt}$. The symbols have their usual meaning.

#### 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:

#### Question 2:

The flux of magnetic field through a closed conducting loop changes with time according to the equation, Φ = *at*^{2} + *bt* + *c*. (a) Write the SI units of *a*, *b* and *c*. (b) If the magnitudes of *a*, *b* and *c* are 0.20, 0.40 and 0.60 respectively, find the induced emf at *t* = 2 s.

#### 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:

#### Question 3:

(a) The magnetic field in a region varies as shown in figure. Calculate the average induced emf in a conducting loop of area 2.0 × 10^{−3} m^{2} placed perpendicular to the field in each of the 10 ms intervals shown. (b) In which intervals is the emf not constant? Neglect the behaviour near the ends of 10 ms intervals.

Figure

#### 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:

#### Question 4:

A conducting circular loop having a radius of 5.0 cm, is placed perpendicular to a magnetic field of 0.50 T. It is removed from the field in 0.50 s. Find the average emf produced in the loop during this time.

#### 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:

#### Question 5:

A conducting circular loop of area 1 mm^{2} is placed coplanarly with a long, straight wire at a distance of 20 cm from it. The straight wire carries an electric current which changes from 10 A to zero in 0.1 s. Find the average emf induced in the loop in 0.1 s.

#### 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:

#### Question 6:

A square-shaped copper coil has edges of length 50 cm and contains 50 turns. It is placed perpendicular to a 1.0 T magnetic field. It is removed from the magnetic field in 0.25 s and restored in its original place in the next 0.25 s. Find the magnitude of the average emf induced in the loop during (a) its removal, (b) its restoration and (c) its motion.

#### 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:

#### Question 7:

Suppose the resistance of the coil in the previous problem is 25Ω. Assume that the coil moves with uniform velocity during its removal and restoration. Find the thermal energy developed in the coil during (a) its removal, (b) its restoration and (c) its motion.

#### 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:

#### Question 8:

A conducting loop of area 5.0 cm^{2} is placed in a magnetic field which varies sinusoidally with time as *B* = *B*_{0} sin ω*t* where *B*_{0} = 0.20 T and ω = 300 s^{−1}. The normal to the coil makes an angle of 60° with the field. Find (a) the maximum emf induced in the coil, (b) the emf induced at τ = (π/900)s and (c) the emf induced at *t* = (π/600) s.

#### 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:

#### Question 9:

Figure shows a conducting square loop placed parallel to the pole-faces of a ring magnet. The pole-faces have an area of 1 cm^{2} each and the field between the poles is 0.10 T. The wires making the loop are all outside the magnetic field. If the magnet is removed in 1.0 s, what is the average emf induced in the loop?

Figure

#### 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:

#### Question 10:

A conducting square loop having edges of length 2.0 cm is rotated through 180° about a diagonal in 0.20 s. A magnetic field *B* exists in the region which is perpendicular to the loop in its initial position. If the average induced emf during the rotation is 20 mV, find the magnitude of the magnetic field.

#### 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:

#### Question 11:

A conducting loop of face-area *A* and resistance *R* is placed perpendicular to a magnetic field *B*. The loop is withdrawn completely from the field. Find the charge which flows through any cross-section of the wire in the process. Note that it is independent of the shape of the loop as well as the way it is withdrawn.

#### 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:

#### Question 12:

radiuA long solenoid ofs 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.

#### 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:

#### Question 13:

Figure shows a metallic square frame of edge *a* in a vertical plane. A uniform magnetic field *B* exists in the space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They start pulling the corners at *t* = 0 and displace the corners at a uniform speed *u*. (a) Find the induced emf in the frame at the instant when the angles at these corners reduce to 60°. (b) Find the induced current in the frame at this instant if the total resistance of the frame is *R*. (c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.

Figure

#### 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:

#### Question 14:

The north pole of a magnet is brought down along the axis of a horizontal circular coil (figure). As a result, the flux through the coil changes from 0.35 weber to 0.85 weber in an interval of half a second. Find the average emf induced during this period. Is the induced current clockwise or anticlockwise as you look into the coil from the side of the magnet ?

Figure

#### 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:

#### Question 15:

A wire-loop confined in a plane is rotated in its own plane with some angular velocity. A uniform magnetic field exists in the region. Find the emf induced in the loop.

#### 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:

#### Question 16:

Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edge enters the 20 cm wide magnetic field at *t* = 0. Find the emf induced in the loop at (a) *t* = 2 s, (b) *t* = 10 s, (c) *t* = 22 s and (d) *t* = 30 s.

Figure

#### 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:

#### Question 17:

Find the total heat produced in the loop of the previous problem during the interval 0 to 30 s if the resistance of the loop is 4.5 mΩ.

#### 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:

#### Question 18:

A uniform magnetic field *B* exists in a cylindrical region of radius 10 cm as shown in figure. A uniform wire of length 80 cm and resistance 4.0 Ω is bent into a square frame and is placed with one side along a diameter of the cylindrical region. If the magnetic field increases at a constant rate of 0.010 T/s, find the current induced in the frame.

Figure

#### 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:

#### Question 19:

The magnetic field in the cylindrical region shown in figure increases at a constant rate of 20.0 mT/s. Each side of the square loop *abcd* and *defa* has a length of 1.00 cm and a resistance of 4.00 Ω. Find the current (magnitude and sense) in the wire *ad* if (a) the switch *S*_{1}_{ }is closed but *S*_{2} is open, (b) *S*_{1} is open but *S*_{2} is closed, (c) both *S*_{1} and *S*_{2} are open and (d) both *S*_{1} and *S*_{2} are closed.

Figure

#### 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:

#### Question 20:

Figure shows a circular coil of *N *turns and radius *a*, connected to a battery of emf *ε* through a rheostat. The rheostat has a total length *L* and resistance *R*. the resistance of the coil is *r*. A small circular loop of radius *a*' and resistance* r*' is placed coaxially with the coil. The centre of the loop is at a distance *x* from the centre of the coil. In the beginning, the sliding contact of the rheostat is at the left end and then onwards it is moved towards right at a constant speed *v*. Find the emf induced in the small circular loop at the instant (a) the contact begins to slide and (b) it has slid through half the length of the rheostat.

Figure

#### 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:

#### Question 21:

A circular coil of radius 2.00 cm has 50 turns. A uniform magnetic field *B *= 0.200 T exists in the space in a direction parallel to the axis of the loop. The coil is now rotated about a diameter through an angle of 60.0°. The operation takes 0.100 s. (a) Find the average emf induced in the coil. (b) If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 Ω, calculate the net charge crossing a cross-section of the wire of the coil.

#### 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:

#### Question 22:

A closed coil having 100 turns is rotated in a uniform magnetic field *B* = 4.0 × 10^{−4} T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm^{2} and its resistance is 4.0 Ω. Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a).

#### 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:

#### Question 23:

A coil of radius 10 cm and resistance 40 Ω has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180°. Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is *B _{H}* = 3.0 × 10

^{−5}T.

#### 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:

#### Question 24:

A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field *B* = 0.010 T exists in a direction perpendicular to the axis of rotation. Find (a) the maximum emf induced, (b) the average emf induced in the coil over a long period and (c) the average of the squares of emf induced over a long period.

#### 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:

#### Question 25:

Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Ω. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.

#### 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:

#### Question 26:

Figure

Figure shows a circular wheel of radius 10.0 cm whose upper half, shown dark in the figure, is made of iron and the lower half of wood. The two junctions are joined by an iron rod. A uniform magnetic field *B* of magnitude 2.00 × 10^{−4} T exists in the space above the central line as suggested by the figure. The wheel is set into pure rolling on the horizontal surface. If it takes 2.00 seconds for the iron part to come down and the wooden part to go up, find the average emf induced during this period.

#### 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:

#### Question 27:

A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 cm s^{−1} perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion. (a) Find the average magnetic force on the free electrons of the rod. (b) For what electric field inside the rod, the electric force on a free elctron will balance the magnetic force? How is this electric field created? (c) Find the motional emf between the ends of the rod.

#### 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:

#### Question 28:

A metallic metre stick moves with a velocity of 2 m s^{−1} in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.2 T. Find the emf induced between the ends of the stick.

#### 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:

#### Question 29:

A 10 m wide spacecraft moves through the interstellar space at a speed 3 × 10^{7} m s^{−1}. A magnetic field *B* = 3 × 10^{−10} T exists in the space in a direction perpendicular to the plane of motion. Treating the spacecraft as a conductor, calculate the emf induced across its width.

#### 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:

#### Question 30:

The two rails of a railway track, insulated from each other and from the ground, are connected to a millivoltmeter. What will be the reading of the millivoltmeter when a train travels on the track at a speed of 180 km h^{−1}? The vertical component of earth's magnetic field is 0.2 × 10^{−4} T and the rails are separated by 1 m.

#### 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:

#### Question 31:

A right-angled triangle *abc*, made from a metallic wire, moves at a uniform speed *v* in its plane as shown in figure. A uniform magnetic field *B* exists in the perpendicular direction. Find the emf induced (a) in the loop *abc*, (b) in the segment *bc*, (c) in the segment *ac* and (d) in the segment *ab*.

Figure

#### 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:

#### Question 32:

A copper wire bent in the shape of a semicircle of radius *r* translates in its plane with a constant velocity *v*. A uniform magnetic field *B* exists in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if (a) the velocity is perpendicular to the diameter joining free ends, (b) the velocity is parallel to this diameter.

#### 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:

#### Question 33:

A wire of length 10 cm translates in a direction making an angle of 60° with its length. The plane of motion is perpendicular to a uniform magnetic field of 1.0 T that exists in the space. Find the emf induced between the ends of the rod if the speed of translation is 20 cm s^{−1}.

#### 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:

#### Question 34:

A circular copper-ring of radius *r* translates in its plane with a constant velocity *v*. A uniform magnetic field *B* exists in the space in a direction perpendicular to the plane of the ring. Consider different pairs of diametrically opposite points on the ring. (a) Between which pair of points is the emf maximum? What is the value of this maximum emf? (b) Between which pair of points is the emf minimum? What is the value of this minimum emf ?

#### 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:

#### Question 35:

Figure shows a wire sliding on two parallel, conducting rails placed at a separation *l*. A magnetic field *B* exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity *v *?

Figure

#### 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:

#### Question 36:

Figure shows a long U-shaped wire of width *l *placed in a perpendicular magnetic field *B*. A wire of length *l* is slid on the U-shaped wire with a constant velocity *v *towards right. The resistance of all the wires is *r *per unit length. At *t* = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, showing the induced emf as a battery. Calculate the current in the circuit.

Figure

#### 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:

#### Question 37:

Consider the situation of the previous problem. (a) Calculate the force needed to keep the sliding wire moving with a constant velocity *v*. (b) If the force needed just after *t* = 0 is *F*_{0}, find the time at which the force needed will be *F*_{0}/2.

#### 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:

#### Question 38:

Consider the situation shown in figure. The wire *PQ* has mass *m*, resistance *r* and can slide on the smooth, horizontal parallel rails separated by a distance *l*. The resistance of the rails is negligible. A uniform magnetic field *B* exists in the rectangular region and a resistance *R* connects the rails outside the field region. At *t* = 0, the wire *PQ* is pushed towards right with a speed *v*_{0}. Find (a) the current in the loop at an instant when the speed of the wire *PQ* is *v*, (b) the acceleration of the wire at this instant, (c) the velocity *v* as a functions of *x* and (d) the maximum distance the wire will move.

Figure

#### 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:

#### Question 39:

A rectangular frame of wire *abcd *has dimensions 32 cm × 8.0 cm and a total resistance of 2.0 Ω. It is pulled out of a magnetic field *B* = 0.020 T by applying a force of 3.2 × 10^{−5} N (figure). It is found that the frame moves with constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between the points *a* and *b* and (d) the potential difference between the points *c *and *d*.

Figure

#### 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:

#### Question 40:

Figure shows a metallic wire of resistance 0.20 Ω sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 µA passes through the wire when it is slid at a rate of 20 cm s^{−1}. If the horizontal component of the earth's magnetic field is 3.0 × 10^{−5} T, calculate the dip at the place.

Figure

#### 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:

#### Question 41:

A wire *ab* of length *l*, mass *m* and resistance *R* slides on a smooth, thick pair of metallic rails joined at the bottom as shown in figure. The plane of the rails makes an angle θ with the horizontal. A vertical magnetic field *B* exists in the region. If the wire slides on the rails at a constant speed *v*, show that $B=\sqrt{\frac{mgR\mathrm{sin\theta}}{v{l}^{2}{\mathrm{cos}}^{2}\mathrm{\theta}}}$

Figure

#### 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:

#### Question 42:

Consider the situation shown in figure. The wires *P*_{1}*Q*_{1} and *P*_{2}*Q*_{2} are made to slide on the rails with the same speed 5 cm s^{−1}. Find the electric current in the 19 Ω resistor if (a) both the wires move towards right and (b) if *P*_{1}*Q*_{1} moves towards left but *P*_{2}*Q*_{2} moves towards right.

Figure

#### 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:

#### Question 43:

Suppose the 19 Ω resistor of the previous problem is disconnected. Find the current through *P*_{2}*Q*_{2} in the two situations (a) and (b) of that problem.

#### 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:

#### Question 44:

Consider the situation shown in figure. The wire *PQ* has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm s^{−1}. Find the current in the 10 Ω resistor when the switch *S* is thrown to (a) the middle rail (b) the bottom rail.

Figure

#### 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:

#### Question 45:

The current generator *I _{g}*' shown in figure, sends a constant current

*i*through the circuit. The wire

*cd*is fixed and

*ab*is made to slide on the smooth, thick rails with a constant velocity

*v*towards right. Each of these wires has resistance

*r*. Find the current through the wire

*cd*.

Figure

#### 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:

#### Question 46:

The current generator *I _{g}*' shown in figure, sends a constant current

*i*through the circuit. The wire

*ab*has a length

*l*and mass

*m*and can slide on the smooth, horizontal rails connected to

*I*. The entire system lies in a vertical magnetic field

_{g}*B*. Find the velocity of the wire as a function of time.

Figure

#### 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:

#### Question 47:

The system containing the rails and the wire of the previous problem is kept vertically in a uniform horizontal magnetic field *B* that is perpendicular to the plane of the rails (figure). It is found that the wire stays in equilibrium. If the wire *ab* is replaced by another wire of double its mass, how long will it take in falling through a distance equal to its length?

Figure

#### 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:

#### Question 48:

The rectangular wire-frame, shown in figure, has a width *d*, mass *m*, resistance *R* and a large length. A uniform magnetic field *B* exists to the left of the frame. A constant force *F* starts pushing the frame into the magnetic field at *t* = 0. (a) Find the acceleration of the frame when its speed has increased to *v*. (b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find this velocity *v*_{0}. (c) Show that the velocity at time *t* is given by

v = v_{0}(1 − e^{−Ft}^{/mv0}).

Figure

#### 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:

#### Question 49:

Figure shows a smooth pair of thick metallic rails connected across a battery of emf *ε* having a negligible internal resistance. A wire *ab* of length *l* and resistance *r* can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field *B*. At an instant *t*, the wire is given a small velocity *v* towards right. (a) Find the current in it at this instant. What is the direction of the current? (b) What is the force acting on the wire at this instant? (c) Show that after some time the wire *ab* will slide with a constant velocity. Find this velocity.

Figure

#### 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:

#### Question 50:

A conducting wire *ab* of length *l*, resistance *r* and mass *m* starts sliding at *t* = 0 down a smooth, vertical, thick pair of connected rails as shown in figure. A uniform magnetic field* B* exists in the space in a direction perpendicular to the plane of the rails. (a) Write the induced emf in the loop at an instant *t* when the speed of the wire is *v*. (b) What would be the magnitude and direction of the induced current in the wire? (c) Find the downward acceleration of the wire at this instant. (d) After sufficient time, the wire starts moving with a constant velocity. Find this velocity *v _{m}*. (e) Find the velocity of the wire as a function of time. (f) Find the displacement of the wire as a function of time. (g) Show that the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached.

Figure

#### 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:

#### Question 51:

A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speed of 100 revolutions per minute. If the length of each spoke is 30.0 cm and the horizontal component of the earth's magnetic field is 2.0 × 10^{−5} T, find the emf induced between the axis and the outer end of a spoke. Neglect centripetal force acting on the free electrons of the spoke.

#### 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:

#### Question 52:

A conducting disc of radius *r* rotates with a small but constant angular velocity *ω* about its axis. A uniform magnetic field *B* exists parallel to the axis of rotation. Find the motional emf between the centre and the periphery of the disc.

#### 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:

#### Question 53:

Figure shows a conducting disc rotating about its axis in a perpendicular magnetic field *B*. A resistor of resistance *R* is connected between the centre and the rim. Calculate the current in the resistor. Does it enter the disc or leave it at the centre? The radius of the disc is 5.0 cm, angular speed ω = 10 rad/s, *B* = 0.40 T and *R* = 10 Ω.

Figure

#### 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:

#### Question 54:

The magnetic field in a region is given by $\overrightarrow{B}=\overrightarrow{k}\frac{{B}_{0}}{L}y$ where *L* is a fixed length. A conducting rod of length *L* lies along the *Y-*axis between the origin and the point (0, *L*, 0). If the rod moves with a velocity *v* = *v*_{0} $\overrightarrow{i}$, find the emf induced between the ends of the rod.

#### 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:

#### Question 55:

Figure shows a straight, long wire carrying a current *i* and a rod of length *l* coplanar with the wire and perpendicular to it. The rod moves with a constant velocity *v* in a direction parallel to the wire. The distance of the wire from the centre of the rod is *x*. Find the motional emf induced in the rod.

Figure

#### 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:

#### Question 56:

Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair of thick metallic rails laid parallel to the wire. At one end the rails are connected by resistor of resistance *R*. (a) What force is needed to keep the rod sliding at a constant speed *v*? (b) In this situation what is the current in the resistance *R*? (c) Find the rate of heat developed in the resistor. (d) Find the power delivered by the external agent exerting the force on the rod.

#### 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:

#### Question 57:

Figure shows a square frame of wire having a total resistance *r* placed coplanarly with a long, straight wire. The wire carries a current *i* given by *i *= *i*_{0}_{ }sin ω*t*. Find (a) the flux of the magnetic field through the square frame, (b) the emf induced in the frame and (c) the heat developed in the frame in the time interval 0 to $\frac{20\mathrm{\pi}}{\mathrm{\omega}}$.

Figure

#### 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:

#### Question 58:

A rectangular metallic loop of length *l *and width *b* is placed coplanarly with a long wire carrying a current *i* (figure). The loop is moved perpendicular to the wire with a speed *v* in the plane containing the wire and the loop. Calculate the emf induced in the loop when the rear end of the loop is at a distance *a* from the wire. solve by using Faraday's law for the flux through the loop and also by replacing different segments with equivalent batteries.

Figure

#### 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:

#### Question 59:

Figure shows a conducting circular loop of radius *a* placed in a uniform, perpendicular magnetic field *B*. A thick metal rod *OA* is pivoted at the centre *O*. The other end of the rod touches the loop at *A*. The centre *O* and a fixed point *C* on the loop are connected by a wire *OC* of resistance *R*. A force is applied at the middle point of the rod *OA* perpendicularly, so that the rod rotates clockwise at a uniform angular velocity ω. Find the force.

Figure

#### 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:

#### Question 60:

Consider the situation shown in the figure of the previous problem. Suppose the wire connecting *O* and *C* has zero resistance but the circular loop has a resistance *R* uniformly distributed along its length. The rod *OA* is made to rotate with a uniform angular speed ω as shown in the figure. Find the current in the rod when ∠ *AOC* = 90°.

#### 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:

#### Question 61:

Consider a variation of the previous problem (figure). Suppose the circular loop lies in a vertical plane. The rod has a mass *m*. The rod and the loop have negligible resistances but the wire connecting *O* and *C *has a resistance *R*. The rod is made to rotate with a uniform angular velocity ω in the clockwise direction by applying a force at the midpoint of *OA* in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle θ with the vertical.

#### 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:

#### Question 62:

Figure shows a situation similar to the previous problem. All parameters are the same except that a battery of emf *ε* and a variable resistance *R* are connected between *O* and *C*. Neglect the resistance of the connecting wires. Let *θ* be the angle made by the rod from the horizontal position (show in the figure), measured in the clockwise direction. During the part of the motion 0 < *θ* < π/4 the only forces acting on the rod are gravity and the forces exerted by the magnetic field and the pivot. However, during the part of the motion, the resistance *R* is varied in such a way that the rod continues to rotate with a constant angular velocity ω. Find the value of *R* in terms of the given quantities.

Figure

#### 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:

#### Question 63:

A wire of mass *m* and length *l* can slide freely on a pair of smooth, vertical rails (figure). A magnetic field *B* exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance *C*. Find the acceleration of the wire neglecting any electric resistance.

#### 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:

#### Question 64:

A uniform magnetic field *B* exists in a cylindrical region, shown dotted in figure. The magnetic field increases at a constant rate $\frac{dB}{dt}$. Consider a circle of radius *r* coaxial with the cylindrical region. (a) Find the magnitude of the electric field *E* at a point on the circumference of the circle. (b) Consider a point *P* on the side of the square circumscribing the circle. Show that the component of the induced electric field at *P* along *ba* is the same as the magnitude found in part (a)

Figure

#### 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:

#### Question 65:

The current in an ideal, long solenoid is varied at a uniform rate of 0.01 As^{−1}. The solenoid has 2000 turns/m and its radius is 6.0 cm. (a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds. (b) Find the electric field induced at a point on the circumference of the circle. (c) Find the electric field induced at a point outside the solenoid at a distance 8.0 cm from its axis.

#### 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:

#### Question 66:

An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction to the same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor.

#### 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:

#### Question 67:

A magnetic flux of 8 × 10^{−4} weber is linked with each turn of a 200-turn coil when there is an electric current of 4 A in it. Calculate the self-inductance of the coil.

#### 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:

#### Question 68:

The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of 0.8 A s^{−1}. Find the emf induced in it.

#### 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:

#### Question 69:

Find the value of *t*/*τ* for which the current in an *LR* circuit builds up to (a) 90%, (b) 99% and (c) 99.9% of the steady-state value.

#### 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:

#### Question 70:

An inductor-coil carries a steady-state current of 2.0 A when connected across an ideal battery of emf 4.0 V. If its inductance is 1.0 H, find the time constant of the circuit.

#### 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:

#### Question 71:

A coil having inductance 2.0 H and resistance 20 Ω is connected to a battery of emf 4.0 V. Find (a) the current at the instant 0.20 s after the connection is made and (b) the magnetic field energy at this instant.

#### 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:

#### Question 72:

A coil of resistance 40 Ω is connected across a 4.0 V battery. 0.10 s after the battery is connected, the current in the coil is 63 mA. Find the inductance of the coil.

#### 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:

#### Question 73:

An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 Ω resistor and a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the circuit is switched on.

#### 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:

#### Question 74:

The time constant of an *LR* circuit is 40 ms. The circuit is connected at *t* = 0 and the steady-state current is found to be 2.0 A. Find the current at (a) *t* = 10 ms (b) *t* = 20 ms, (c) *t* = 100 ms and (d) *t* = 1 s.

#### 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:

#### Question 75:

An L-R circuit has *L* = 1.0 H and *R* = 20 Ω. It is connected across an emf of 2.0 V at *t* = 0. Find *di*/*dt* at (a) *t* = 100 ms, (b) *t* = 200 ms and (c) *t* = 1.0 s.

#### 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:

#### Question 76:

What are the values of the self-induced emf in the circuit of the previous problem at the times indicated therein?

#### 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:

#### Question 77:

An inductor-coil of inductance 20 mH having resistance 10 Ω is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at (a) *t* = 0, (b) *t* = 10 ms and (c) *t* = 1.0 s.

#### 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:

#### Question 78:

An *LR* circuit contains an inductor of 500 mH, a resistor of 25.0 Ω and an emf of 5.00 V in series. Find the potential difference across the resistor at *t* = (a) 20.0 ms, (b) 100 ms and (c) 1.00 s.

#### 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:

#### Question 79:

An inductor-coil of resistance 10 Ω and inductance 120 mH is connected across a battery of emf 6 V and internal resistance 2 Ω. Find the charge which flows through the inductor in (a) 10 ms, (b) 20 ms and (c) 100 ms after the connections are made.

#### 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:

#### Question 80:

An inductor-coil of inductance 17 mH is constructed from a copper wire of length 100 m and cross-sectional area 1 mm^{2}. Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper = 1.7 × 10^{−8} Ω-m.

#### 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:

#### Question 81:

An *LR* circuit having a time constant of 50 ms is connected with an ideal battery of emf *ε*. find the time elapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches half its maximum value and (c) the magnetic field energy stored in the circuit reaches half its maximum value.

#### 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:

#### Question 82:

A coil having an inductance *L* and a resistance *R* is connected to a battery of emf *ε*. Find the time taken for the magnetic energy stored in the circuit to change from one fourth of the steady-state value to half of the steady-state value.

#### 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:

#### Question 83:

A solenoid having inductance 4.0 H and resistance 10 Ω is connected to a 4.0 V battery at *t* = 0. Find (a) the time constant, (b) the time elapsed before the current reaches 0.63 of its steady-state value, (c) the power delivered by the battery at this instant and (d) the power dissipated in Joule heating at this instant.

#### 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:

#### Question 84:

The magnetic field at a point inside a 2.0 mH inductor-coil becomes 0.80 of its maximum value in 20 µs when the inductor is joined to a battery. Find the resistance of the circuit.

#### 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:

#### Question 85:

An *LR *circuit with emf *ε* is connected at *t* = 0. (a) Find the charge *Q* which flows through the battery during 0 to *t*. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during this period. (d) Find the magnetic field energy stored in the circuit at time *t*. (e) Verify that the results in the three parts above are consistent with energy conservation.

#### 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:

#### Question 86:

An inductor of inductance 2.00 H is joined in series with a resistor of resistance 200 Ω and a battery of emf 2.00 V. At *t* = 10 ms, find (a) the current in the circuit, (b) the power delivered by the battery, (c) the power dissipated in heating the resistor and (d) the rate at which energy is being stored in magnetic field.

#### 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:

#### Question 87:

Two coils *A* and *B* have inductances 1.0 H and 2.0 H respectively. The resistance of each coil is 10 Ω. Each coil is connected to an ideal battery of emf 2.0 V at *t* = 0. Let *i _{A}* and

*i*be the currents in the two circuit at time

_{B}*t*. Find the ratio

*i*/

_{A}*i*at (a)

_{B}*t*= 100 ms, (b)

*t*= 200 ms and (c)

*t*= 1 s.

#### 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:

#### Question 88:

The current in a discharging *LR* circuit without the battery drops from 2.0 A to 1.0 A in 0.10 s. (a) Find the time constant of the circuit. (b) If the inductance of the circuit 4.0 H, what is its resistance?

#### 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:

#### Question 89:

A constant current exists in an inductor-coil connected to a battery. The coil is short-circuited and the battery is removed. Show that the charge flown through the coil after the short-circuiting is the same as that which flows in one time constant before the short-circuiting.

#### 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:

#### Question 90:

Consider the circuit shown in figure. (a) Find the current through the battery a long time after the switch *S* is closed. (b) Suppose the switch is again opened at *t* = 0. What is the time constant of the discharging circuit? (c) Find the current through the inductor after one time constant.

Figure

#### 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:

#### Question 91:

A current of 1.0 A is established in a tightly wound solenoid of radius 2 cm having 1000 turns/metre. Find the magnetic energy stored in each metre of the solenoid.

#### 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:

#### Question 92:

Consider a small cube of volume 1 mm^{3} at the centre of a circular loop of radius 10 cm carrying a current of 4 A. Find the magnetic energy stored inside the cube.

#### 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:

#### Question 93:

A long wire carries a current of 4.00 A. Find the energy stored in the magnetic field inside a volume of 1.00 mm^{3} at a distance of 10.0 cm from the wire.

#### 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:

#### Question 94:

The mutual inductance between two coils is 2.5 H. If the current in one coil is changed at the rate of 1 As^{−1}, what will be the emf induced in the other coil?

#### 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:

#### Question 95:

Find the mutual inductance between the straight wire and the square loop of figure.

#### 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:

#### Question 96:

Find the mutual inductance between the circular coil and the loop shown in figure.

#### 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:

#### Question 97:

A solenoid of length 20 cm, area of cross-section 4.0 cm^{2} and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-sectional area 8.0 cm^{2} and length 10 cm. Find the mutual inductance between the solenoids.

#### 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:

#### Question 98:

The current in a long solenoid of radius *R* and having *n* turns per unit length is given by *i* = *i*_{0} sin *ωt*. A coil having *N* turns is wound around it near the centre. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid ant the coil.

#### 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