KKMishra's Tutorials: H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, Concepts of Physics, Part-II

Electromagnetic Induction

Questions for Short Answer

1. A metallic loop is placed in a non-uniform magnetic field. Will an emf be induced in the loop?

ANSWER: Since there is no change of flux with time, no emf will be induced in the loop.

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 the switch is closed, the current takes some time to grow. In fact, in one time constant, it grows to 63% of the full current. But when the switch is opened, there is no other path for the current to flow and it suddenly drops to zero in fractions of a second. Thus the rate of change of the current in the circuit is much higher when the switch is opened. So the emf induced in the inductor, in this case, is much larger than in the case when the switch is closed.

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 coil is released after deflection, it oscillates due to the mechanical energy stored in the suspended wire. During oscillations, the flux of the magnetic field through the coil produced by the poles of the permanent magnets changes, and an emf is developed between the ends of the coil. But no current flows as the ends are not connected. As soon as we connect the ends of the coil, a current begins to flow due to this induced emf and according to Lenz's law, this induced current will produce flux in the coil that will oppose the movement of the coil. So the oscillation stops at once.

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: When the short magnet is away from the loop on its axis, the magnetic flux through the loop is small compared to the case when it is near to it. So whether we take the magnet closer on the axis of the loop or take it away, there is a change in flux through the loop. This change of flux through the loop will produce an induced current in it. According to Lenz's law, "the direction of the induced current is such that it opposes the change that has induced it". Since the induced current here is due to the movement of the short magnet, it will oppose the movement of the magnet. Thus when the magnet is approaching the loop, it will be repelled by the loop and will be attracted by the loop when going away from it.

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: Initially there is no current in either of the loops and no magnetic fields. When the battery is suddenly connected in one of the loops, a current is established in this loop, and magnetic flux through it is also established. But the current and the magnetic flux so produced take a very small amount of time to reach this value from zero. In this small time, there is a change of magnetic flux through the loop. This flux change is also through the second loop that is placed coaxially. This change of the flux through the other loop will also induce a current in this loop.

The induced current in the second loop remains for a very short time. It starts as soon as the battery is connected in the first loop and a current begins to grow and ends just when the current reaches a constant value in the first loop.

According to Lenz's law, the induced current will oppose the change that has induced it, so the direction of the induced current in the second loop will be such that it will produce a flux that is opposite to the flux produced by the first one. So the loops repel each other in this short time.

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: Originally there is a current established in the first loop. When the battery is suddenly disconnected, the current in the loop becomes zero in a very short time. So there is a change of current from say 'i', to zero. Thus there is a change of magnetic flux through this loop. Since the second loop is coaxially placed parallel to the first one, the magnetic flux through the second loop also changes in that very short time. So this change of flux through the second loop induces a current in it.

This induced current in the second loop starts just when the battery is disconnected and ends when the current in the first loop just becomes zero. The current remains only for this very short interval of time.

According to Lenz's law, this induced current will oppose the change that has induced it. So the induced current will try to keep the magnetic flux stable as it was when the battery was connected. With the disconnection of the battery, the flux starts reducing, so the induced current in the other loop will produce a magnetic flux in the same direction. Thus the direction of the induced current will be the same as it was in the first loop and the loops will attract each other.

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 that shield objects inside them against varying magnetic fields.

ANSWER: When the magnetic field outside the copper box suddenly changes, eddy currents develop on the surface of the copper box because it is a low-resistivity metal. According to Lenz's law, the directions of the eddy currents are such that they oppose the change of magnetic field. Thus the varying magnetic field can not affect the objects inside the box.

8. Metallic (nonferromagnetic) and nonmetallic particles in 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 the particles slide over permanent magnets, due to the movement, the magnetic flux near the particles changes. This flux change induces eddy currents in the metallic particles. According to Lenz's law, these induced eddy currents oppose the movement that is the cause of the induced currents. So the metallic particles slow down and the nonmetallic particles are unaffected.

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

ANSWER: In a magnetic field, when the pivoted aluminum bar falls, the magnetic flux over it changes due to the change of angle between the bar and the magnetic field. See the diagram below.
Diagram for Q-9

Due to the change of flux, eddy currents are induced on the surface of the aluminum bar that opposes the movement (Cause of the induced eddy currents). Thus the aluminum bar falls much more slowly than a similar bar of insulating material that is free from eddy currents.

10. A metallic bob A oscillates through the space between the poles of an electromagnet (figure 38-Q1). The oscillations are more quickly damped when the circuit is on, as compared to the case when the circuit is off. Explain.
The figure for Q-10

ANSWER: When the circuit is off, the electromagnet is demagnetized. There is no magnetic field between the ends of the magnet and the metallic bob dampens due to the general causes of friction at the point of hanging and air resistance. As soon as the circuit is kept on, the ends of the electromagnet get poles of opposite nature and the space between them is filled with the magnetic field. When the bob passes through this space, the magnetic flux over it changes. This change of magnetic flux produces eddy currents on the surface of the metallic bob. These induced eddy currents oppose the movement of the bob that causes them to be induced. So the oscillations are more quickly damped than the case when the circuit is off.

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

ANSWER: When the magnetic flux produced by one loop goes through the boundary of the second loop, an induced current appears in the second loop whenever there is a change of flux.

(a) when both the loops are placed coaxially and parallel to each other, maximum magnetic flux passes through them. Hence the mutual inductance is the largest.

(b) When the loops are placed with their planes and axes perpendicular to each other, minimum common flux passes through them. Hence in this case the mutual inductance is the smallest.

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

ANSWER: Self-inductance of a solenoid is

L =µₒn²πr²l, where l is the length of the solenoid.

Thus the self-inductance per unit length of the coil is,

L/l =µₒn²πr²

Since this expression has been derived taking the magnetic field inside a solenoid as the same everywhere and the R.H.S. of the above expression depends only on n (number of terms per unit length) and the radius of the solenoid, r. So the self-inductance of the coil is the same near the ends and at the center of the coil.

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

ANSWER: The energy density inside a solenoid is,

u =B²/(2µₒ).

In deriving this expression, it is assumed that the magnetic field throughout the volume of the solenoid is uniform and zero outside. Hence the energy density is the same at its center and near its ends.

OBJECTIVE-I

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

(a) zero

(b) ⅛⍵Bl²

(c) ½⍵Bl²

(d) B⍵l².

ANSWER: (b)

EXPLANATION: Small emf developed across a small length dr is

dƐ =vBdr = ωrBdr

→Ɛ = ∫ωrBdr

=ωB∫rdr

=ωB[r²/2]

{between limits r =0 to r =l/2.}

=½ωB*l²/4

=⅛ωBl².

Hence option (b) is correct.

2. A rod of length l rotates with 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) ½Blω²

(c) Blω²

(d) 2Blω².

ANSWER: (a)

EXPLANATION: Each end will be at ⅛ωBl² higher or lower potential than the center depending upon the directions of the magnetic field and the rotation. So both the ends are at equal potential and hence the potential difference between the ends is zero. Option (a) is correct.

3. Consider the situation shown in Figure (38-Q2). 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 for Q-3

ANSWER: (d)

EXPLANATION: When the switch is closed, the current in the left loop increases from zero to a constant value in a very very short time. Hence the magnetic field around the wire changes for a very short time and this changing magnetic field induces a current-pulse in the closed-loop. According to Lenz's law direction of this current pulse will be to oppose the flux through it. Since the direction of flux through the loop due to the nearest wire-current is 'coming out of paper', a clockwise current pulse in the closed loop will produce flux 'going into the paper' through it.

When the switch is opened again, the magnetic flux density through the ring begins to decrease. Hence the induced current pulse in the closed-loop opposes the change and tries to maintain the flux density. In this effort, the induced current in the loop produces a flux in the same direction. Since the direction of the flux before opening the switch was 'coming out of the paper', the same will be the direction of the flux due to the induced current pulse in the closed loop. Such flux can be produced by an anticlockwise current pulse in the loop. Hence option (d) is correct.

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

ANSWER: (c)

EXPLANATION: In this case, the direction of increasing clockwise current in the outer loop will produce increasing magnetic flux inside the loop 'going into the plane of the loop'. To oppose it, the direction of the induced current pulse in the closed loop will be anticlockwise so that the flux inside it is coming out of the plane.

When the switch is opened again, the magnetic flux through the closed loop begins to decrease. The induced current in the loop opposes this decrease in flux, i.e. it tries to maintain the flux by producing the flux in the same direction. So there will be a clockwise induced current pulse in the closed-loop. Option (c) is correct.

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 constant speed

(c) will move with an acceleration g

(d) will oscillate.

ANSWER: (b)

EXPLANATION: After the release from rest, the magnet will fall with acceleration g. This will result in the changing flux linked with the copper tube. It will result in a current in the tube that will resist this change in flux. So a magnetic force will act on the bar magnet opposite to the direction of movement. This opposing force will increase till a velocity is reached when this force is equal to the weight of the magnet. At this stage the net force on the magnet being zero, there will be no acceleration, and the magnet will fall with almost a constant speed. Option (b) is correct.

6. Figure (38-Q3) 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 for Q-6

ANSWER: (c)

EXPLANATION: As the switch is closed, the magnetic flux through the solenoid and hence through the loop will change. The induced current in the loop will be such that it will oppose the increase in the flux through it, and it can be achieved by having the direction of the current opposite to that in the solenoid. This will produce the magnetic flux opposite to that produced by the solenoid so that the net flux is reduced. Since the currents in the solenoid and the loop are opposite, they will repel each other and the loop will move away from the solenoid. Option (c) is correct.

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)

EXPLANATION: When a conductor is moved in a magnetic field, a magnetic force is acted upon the electrons due to the component of the speed perpendicular to the magnetic field. This results in an induced emf in the conductor. Statement A is correct.

Due to a change in the magnetic field, there is a flux change inside a loop placed there. This induces an emf in the loop. Statement B is also correct.

Option (a) is correct.

8. Consider the situation shown in the figure (38-Q4). 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 semicircle bulges towards the resistance or away from it.
Figure for Q-8

ANSWER: (b)

EXPLANATION: The induced emf does not depend on the shape of the wire because all the electrons in the wire AB or semicircular wire move in the same direction. Hence option (b) is correct.

9. Figure (38-Q5a) shows a conducting loop being pulled out of a magnetic field with a speed v. Which of the four plots shown in figure (38-Q5b) may represent the power delivered by the pulling agent as a function of the speed v?
Figure for Q-9

ANSWER: (b)

EXPLANATION: In this loop, the emf will be induced in the wire length 'l' that is perpendicular to the velocity. This emf is

ξ =Blv.

If the resistance of the circuit is R, then the current in the loop,

i =ξ/R =Blv/R

Power P =ξi

→P =BlvBlv/R =(B²l²/R)v²

So, P ∝ v²

This is an equation of a parabola symmetrical to the P-axis and passing through the origin. Since we are interested in positive v, plot b is appropriate. Hence option (b) is correct.

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: (a)

EXPLANATION: When there is a constant current in the loop, a constant magnetic flux is generated by it and the same constant flux passes through the other loop. When due to the temperature variation the resistance increases, the current decreases. So the flux generated by the current begins to decrease. The induced current in the other loop will be such that it will oppose the change of the flux. In order to keep the flux the same, the induced current in the loop will be in the same sense. Since the current in both the loops is in the same sense, they will attract each other. Option (a) is correct.

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)

EXPLANATION: When the current in the solenoid is changed, the magnetic flux along the axis of the solenoid changes. Since the plane of the small conducting loop is parallel to this flux, there is no flux change perpendicular to the plane of the loop. Hence no current is induced in the loop. Option (c) is correct.

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 (38-Q6). The current induced in the loop is

(a) Blv/R clockwise

(b) Blv/R anticlockwise

(c) 2Blv/R anticlockwise

(d) zero.
Figure for Q-12

ANSWER: (d)

EXPLANATION: There will not be any emf induced in the sides parallel to the direction of the velocity. The other two sides will have equal emf induced but there will be no current in the loop because the ends connected of these two sides are equipotential. Hence option (d) is correct.

OBJECTIVE-II

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), (c).

EXPLANATION: The induced current opposes the cause of its occurrence. Since the induced current is anticlockwise when viewed from the magnet, the north pole of the equivalent magnetic dipole of the ring is towards the magnet. It means that the direction of the magnetic flux by the ring is towards the magnet. By this flux, the ring opposes the flux by the magnet.

(1) one condition is that the north pole of the magnet is facing the ring and approaching it and the ring is opposing the increase of flux towards it. Option (b) is correct.

(2) the second condition is that the south pole of the magnet is towards the ring and is going away from the ring resulting in the decrease of flux near the ring. The induced current in the ring produces flux towards the magnet to maintain the original flux. Option (c) is correct.

The other two conditions will induce a clockwise current in the ring.

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 v || l

(b) if v || B

(c) if l || B

(d) none of these.

ANSWER: (d)

EXPLANATION: The potential difference in a rod moving in a magnetic field is due to the magnetic force on the free electrons in the rod, and they concentrate towards one end. If the velocity is along the length any magnetic force on electrons will not be able to concentrate them towards an end. Option (a) is not correct.

If the velocity is along the magnetic field, no magnetic force will be on the electrons. Option (b) is not correct.

If the length is parallel to the magnetic field, even a perpendicular-to-field movement can not produce potential difference because the force on the electrons will be transverse to the length and electrons can not concentrate at one end. So option (c) is also not correct.

Hence none of these options are correct. Option (d) is correct.

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 about a diameter

(d) it is deformed.

ANSWER: (c), (d).

EXPLANATION: emf is induced in a loop when the flux density perpendicular to its plane changes. The flux density does not change if the loop is translated or rotated about its axis. No emf is induced in these two cases. Options (a) and (b) are not correct.

When the loop is rotated or deformed the flux density through it changes, hence emf will be induced. Options (c) and (d) are correct.

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), (c), (d).

EXPLANATION: If the sheet is magnetic, it will be attracted towards the magnetic pole and a force will be needed to hold it. Option (a) is correct.

For a nonmagnetic metal sheet, there will be no attraction, thus no force needed. Option (b) is not correct.

When a magnetic or nonmagnetic metal sheet is moved away from the pole, the magnetic flux near the sheet changes. Thus induced eddy currents develop in the sheet that opposes the movement. Hence a force will be needed to keep it moving with constant velocity. Options (c) and (d) are correct.

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) the magnetic field at the center

(b) magnetic flux linked with the solenoid

(c) self-inductance of the solenoid

(d) rate of Joule heating.

ANSWER: (a), (b), (c).

EXPLANATION: Before the rod is inserted in the solenoid, the magnetic field at the center is µₒni. As the rod is inserted the magnetic field becomes, µni, where µ is the permeability of the iron. Option (a ) correct.

The magnetic flux also changes from "(µₒni)πr²" to "(µni)πr²" after the iron rod is inserted. Option (b) is correct.

Self-inductance of the coil also changes from "µₒn²πr²l" to "µn²πr²l". Hence option (c) is also correct.

The rate of Joule heating depends on the resistance of the coil which does not change. Hence option (d) is not correct.

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

ANSWER: (b), (d).

EXPLANATION: The self-inductance of a solenoid depends only on the geometrical factors that are the same in both solenoids. Hence self-inductance is not different for them. Option (a) is not correct.

The thin wired solenoid will have higher resistance, so Joule heating will be more in it for the same current. Option (b) is correct.

Magnetic field energy in a solenoid =½Li. So it will be the same in both solenoids, not different. Option (c) is not correct.

The time constant =L/R. L is the same for both but not R. Hence time constant is different for them. Option (d) is correct.

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)

EXPLANATION: As the connection is made, the current begins to grow. So a small current is present just after the connection. Magnetic field energy depends on the current. Hence it is also not zero. Power delivered by the battery =i²R, so it is not zero either. Thus options (a), (b), and (c) are not correct.

Emf is induced when there is a change in the current. When the battery is connected, the current just begins, not changes. Hence no emf. Option (d) is correct.

8. A rod AB moves with a uniform velocity v in a uniform magnetic field as shown in figure (38-Q7).

(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.
The figure for Q-8

ANSWER: (b)

EXPLANATION: Since total electrons remain in the rod, there is no question of the rod being electrically charged. Option (a) is not correct.

From the right-hand rule, the magnetic force on a positive charge moving with the given velocity will be towards end A. Since there are free electrons in the rod, not positive charges, the magnetic force will be towards end B on the electrons. The electrons in the rod will concentrate towards end B and end A will be deficient in electrons. So end A will be positively charged. Option (b) is correct. Option (c) is not correct

With end A positively charged and end B negatively, an emf is developed between the ends A and B. But there is no current flowing, it is like a disconnected battery. So no Joule heating because i =0. Option (d) is not correct.

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

(a) 1/RC

(b) R/L

(c) 1/√(LC)

(d) C/L.

ANSWER: (a), (b), (c).

EXPLANATION: Dimensions of L, C, and R are as follows.

L = M L² T⁻² I⁻²

C = M⁻¹ L⁻² T⁴ I²

R = M L² T⁻³ I⁻²

So,

[RC] = [T]

and dimensions of 1/RC = [T⁻¹], which is the dimensions of frequency. Option (a) is correct.

[R/L] = [T⁻¹], which is also the dimensions of frequency. Option (b) is correct.

[LC] = [T²], hence,

[1/√(LC)] = [T⁻¹]. It is also the dimensions of frequency. So option (c) is also correct.

[C/L] = [M⁻² L⁻⁴ T⁶ I⁴], and it is not the dimensions of frequency. Option (d) is not correct.

10. The switches in figure (38-Qa) and (38-8b) are closed at t = 0 and reopened after a long time at t = tₒ.
The figure for Q -10

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

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

(c) The current in L just before t = tₒ is ξ/R.

(d) The current in L long after t = tₒ is ξ/R.

ANSWER: (b), (c).

EXPLANATION: The charge on a capacitor becomes maximum after a long time and it is = ξC. Hence option (a) is not correct but option (b) is correct.

The current in the inductance reaches the maximum before a long time. So just before, t = tₒ, the full current is developed in L and this current is equal to ξ/R. So option (c) is correct. Since at t = tₒ the switch is reopened, after this, the current decreases from ξ/R. So option (d) is not correct.

EXERCISES

1. Calculate the dimensions of

(a) ∫E.dl

(b) vBl, and

(c) dⲪᵦ/dt.

The symbols have their usual meanings.

ANSWER: The dimensions of

(a) ∫E.dl

Electric field E =force per unit charge

Its dimensions =[MLT⁻²]/[IT]

=[MLI⁻¹T⁻³]

Dimensions of dl =[L]

Hence dimensions of

∫E.dl =[MLI⁻¹T⁻³][L]

=[ML²I⁻¹T⁻³].

(b) Dimensions of vBl:-

Since, the magnetic force F =qvBsinθ

→B =F/(qv.sinθ)

So, vBl =vFl/(qv.sinθ)

=Fl/(q.sinθ)

Sine is dimensionless due to a ratio. Hence the dimensions of,

vBl = [MLT⁻²][L]/[IT]

= [ML²I⁻¹T⁻³].

(c) Dimensions of dⲪᵦ/dt:-

It is the magnetic flux per unit of time. The magnetic flux is BAcosθ. So,

BAcosθ =FA.tanθ.(qv)⁻¹

Dimensions of flux

=[MLT⁻²][L²][I⁻¹T⁻¹][L⁻¹T¹]

=[ML²I⁻¹T⁻²]

Hence dimensions of dⲪᵦ/dt

=[ML²I⁻¹T⁻³].

2. The flux of the magnetic field through a closed conducting loop changes with time according to the equation,

Φ =at²+bt+c.

(a) Write the SI units of a, b and c.

(b) If the magnitude of a, b, and c are 0.20, 0.40, and 0.60 respectively, find the induced emf at t = 2 s.

ANSWER: For the given equation to be dimensionally correct, each term should have the same units and dimensions as that of flux. The unit of flux is Weber or volt-sec.

(a) Hence unit of the term at² is also volt-sec. Thus the unit of 'a' is

volt-sec/sec² =volt/s.

Unit of b = Flux/sec =volt-sec/sec

= volt.

The unit of c is the same as the unit of flux, i.e. volt-s (or Weber).

(b) At t =2 s, and magnitudes of a, b, and c as 0.20, 0.40, and 0.60 respectively, the magnitude of induced emf is

= dΦ/dt

= (2at+b)

= (20.202 +0.40)

= (0.80 +0.40)

= 1.20 volt

3. (a) The magnetic field in a region varies as shown in figure (38-E1). Calculate the average induced emf in a conducting loop of area 2.0x10⁻³ m² placed perpendicular to the field in each of the 10 ms intervals shown. (b) In which intervals is the emf not constant? Neglect the behavior near the ends of 10 ms intervals.
Figure for Q-3

ANSWER: (a) Area of the loop, A= 2x10⁻³ m². At t = 0, B =0. So the flux is zero. But at t =10 ms, B =0.01 T. Flux now =BA

=0.01*2x10⁻³ weber

=2x10⁻⁵ weber

Hence the change in flux in time Δt =10 ms,

ΔΦ = 2x10⁻⁵ weber,

Hence the induced emf,

ξ = -ΔΦ/Δt

= -2x10⁻⁵/(10x10⁻³) volt

= -2x10⁻³ volt

= -2.0 mV.

At t =20 m, B =0.03 T

So Φ =BA

=0.03*2x10⁻³ weber

=6.0x10⁻⁵ weber

In this interval of 10 ms, change in flux,

ΔΦ =6.0x10⁻⁵ -2.0x10⁻⁵ weber

=4.0x10⁻⁵ weber

Hence the average induced emf in this interval,

ξ = -ΔΦ/Δt

= -4.0x10⁻⁵/(10*10⁻³) volt

= -4.0x10⁻³ volt

= -4.0 mV

At t =30 ms, again the value of B =0.01 T. Hence flux = 2.0x10⁻⁵ weber. Change in flux ΔΦ =(2.0 -6.0)x10⁻⁵ weber

= -4.0x10⁻⁵ weber.

Hence the induced emf,

ξ = -ΔΦ/Δt

=4.0x10⁻⁵/(10x10⁻³) volt

=4.0x10⁻³ volt

=4.0 mV

At t = 40 ms, B = 0, hence Φ =0.

Change in flux ΔΦ =0 -2.0x10⁻⁵ weber

= -2.0x10⁻⁵ weber.

Hence induced emf

ξ = -ΔΦ/Δt

= 2.0x10⁻⁵/(10x10⁻³) volt

=2.0x10⁻³ volt

=2.0 mV.

(b) Between 10 ms to 20 ms and 20 ms to 30 ms, the magnetic field B does not change linearly. Hence the emf between these two intervals will not be constant.

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 Δt =0.50 s, and ΔΦ =ΔB*A

=0.50π(0.05)² weber

=0.50*7.8x10⁻³ weber

The magnitude of average emf induced.

ξ =ΔΦ/Δt

=0.50*7.8x10⁻³/0.5 V

=7.8x10⁻³ V

5. A conducting circular loop of area 1 mm² is placed coplanar with a long straight wire at a distance of 20 cm from it. The straight wire carries an electric current that changes from 10 A to zero in 0.1 s. Find the average emf induced in the loop in 0.1 s.

ANSWER: The magnetic field produced by the current in the wire will be circular in a plane perpendicular to it. So these magnetic fields will pass perpendicularly to the circular loop. Distance of the loop from the wire,

d =20 cm =0.20 m.

Current in the wire, i =10 A

Hence the magnetic field due to the current near the circular loop,

B = µₒi/(2πd)

= (µₒ/4π)*2i/d

= 10⁻⁷210/0.20 T

= 1x10⁻⁵ T

Area of the loop,

A = 1 mm² = 1x10⁻⁶ m²

Hence flux at the loop,

φ =BA

= 1x10⁻⁵*1x10⁻⁶ weber

=1x10⁻¹¹ weber

In an interval of time Δt =0.1 s, the current becomes zero. Hence the magnetic field and flux will also become zero.

So, Δφ =0 -1x10⁻¹¹ weber

=-1x10⁻¹¹ weber.

The average emf induced in this time interval will be

ξ = -ΔΦ/Δt

= 1x10⁻¹¹/0.1 V

=1x10⁻¹⁰ V.

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 to 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: Numer of turns, N =50. Area of the cross-section of the coil, A = 0.5*0.5 m² =0.25 m². B =1.0 T. So the flux through one turn,

Φ =BA

=1.0*0.25 weber

=0.25 Weber.

In a time interval, Δt =0.25 s the coil is in a place that is without a magnetic field. So the flux is also zero. Hence,

ΔΦ =0 -0.25 = -0.25 weber.

(a) Hence the magnitude of the average emf induced in the coil during its removal,

ξ = N(ΔΦ/Δt) = 50(0.25/0.25) V

= 50 V.

(b) During the restoration, the change in flux ΔΦ =0.25 -0 =0.25 Weber

Time interval Δt = 0.25 s.

Hence the average magnitude of emf induced in the coil during its restoration

ξ = N*(ΔΦ/Δt)

=50*(0.25/0.25) V

=50 V.

(c) During the motion of the coil the initial and final strength of the magnetic field is the same and so is the flux through the coil. Hence the change in flux ΔΦ =0, The time interval

Δt =0.25 +0.25 =0.50 s

Hence the magnitude of the average emf developed

ξ = N*(ΔΦ/Δt)

=50*(0/0.50) V

= Zero.

7. Suppose the resistance of the coil in the previous problem is 25 𝛀. 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: Resistance of the coil, R =25 Ω.

(a) Energy dissipated during its removal,

U = Power*Time

=(ξ²/R)*Δt

= (50²/25)*0.25 J

= 25 J.

(b) Energy dissipated during its restoration is again

U = (ξ²/R)*Δt

= (50²/25)*0.25 J

= 25 J.

(c) The energy dissipated during its motion will be the sum of energy dissipated during removal and restoration because it is a scalar quantity. So,

U = 25 +25 J =50 J.

8. A conducting loop of area 5.0 cm² is placed in a magnetic field which varies sinusoidally with time as B = Bₒsin ωt where Bₒ =0.20 T and ω =300 s⁻¹. The normal to the coil makes an angle of 60° with the field. Find (a) 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: (a) Angle between normal to the coil and the field, ß =60°. Hence emf developed at time t,

ξ =dΦ/dt

=d(BAcosß)/dt

=A.cosß*d(BₒSin ωt)/dt

=A.cos 60°BₒωCos ωt

=(5x10⁻⁴)½0.20300Cos (300t)

= 0.015*Cos(300t)

The maximum value of Cos (300t) will be =1, hence maximum emf developed,

ξₘₐₓ =0.015 V.

(b) emf induced at t =(π/900) s

ξ = 0.015*Cos (300π/900)

=0.015*Cos(π/3)

=0.015*½ V

=7.5x10⁻³ V.

(c) Emf induced at t =(π/600) s

ξ = 0.015*Cos (300π/600) V

=0.015*Cos(π/2) V

=0.015*0 V =zero.

9. Figure (38-E2) shows a conducting square loop placed parallel to the pole faces of a ring magnet. The pole-faces have an area of 1 cm² 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 for Q-9

ANSWER: Area of pole faces, A =1 cm² each =1.0x10⁻⁴ m². The field between the poles, B =0.10 T. Since the plane of the loop is perpendicular to the magnetic field, the flux Φ =BA

→Φ =0.10*1.0x10⁻⁴ weber

=1.0x10⁻⁵ weber

After removing, the flux inside the loop is zero in time Δt =1.0 s. Hence the average emf developed is

ξ =ΔΦ/Δt

=1.0x10⁻⁵/1.0 V

= 10x10⁻⁶ V

= 10 µV.

10. A conducting square loop with having edge 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: Area of the square loop, A =(0.02)² m² = 4x10⁻⁴ m².

Initial flux Φ =BA.cosß

→Φ =B4x10⁻⁴cos0°

= 4x10⁻⁴B weber

Final flux Φ' =B4x10⁻⁴cos180°

→Φ' =-4x10⁻⁴B weber

Change in flux ΔΦ =-Φ' -Φ

→ΔΦ =-8x10⁻⁴B weber

Δt = 0.20 s

Hence the induced emf ξ = -ΔΦ/Δt

→20x10⁻³ = 8x10⁻⁴B/0.20

→B =½*10 T =5 T.

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 that 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: Initial magnetic flux through the coil, φ = BA.

After the withdrawal the magnetic flux =0.

Change in flux, Δφ =BA -0 =BA.

Let the time of withdrawal = Δt.

emf induced Ɛ =Δφ/Δt

→Ɛ = BA/Δt

Hence the current in the loop,

i = Ɛ/R

→i = BA/(R*Δt)

→i*Δt = BA/R

→The flow of charge through any cross-section of wire = BA/R.

12. A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm has 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: Magnetic field inside the solenoid, B =µₒni.

The magnetic flux through the inner coil,

φ =B(πr²)n =πn'r²B

where n' = the total number of turns in the inner coil.

When the current in the solenoid is reversed, the magnetic field becomes -B.

Hence the magnetic flux through the coil now is

φ' =-πn'r²B

Hence the change in the flux,

Δφ =φ -φ'

=2πn'r²B.

Suppose this change occurs in a time Δt, then the emf developed in the coil,

E =Δφ/Δt

For a resistance R of the coil, the current in the coil,

i' =E/R.

So the charge flown through the galvanometer

Q =i'*Δt

=EΔt/R

=Δφ/R

=2πn'r²B/R

=2πn'r²*µₒni/R

Given values, n =100, r =1 cm =0.01 m, r' =2 cm =0.02 m, n' =100 turns/cm =10⁴ turns/m, i =5 A, R =20 Ω.

Hence Q=2π1000.00014π10⁻⁷10⁴5/20 C

=2π²*10⁻⁵ C

=1.97x10⁻⁴ C

≈2x10⁻⁴ C.

13. Figure (38-E3) shows a metallic square frame of an 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 that flows through a side of the frame by the time the square is deformed into a straight line.
Figure for Q-13

ANSWER: (a) The projection of an edge of the square perpendicular to the velocity

=a.sinθ

(Where θ is the angle between an edge and the direction of velocity.)

This length will be the length effective for the generation of emf.

The emf generated in one edge

=Bu(a.sinθ)

=auBsinθ

emf generated in four edges

Ɛ = 4auBsinθ

(Given that the angle at the corners =60°, hence θ = 30°.)

=4auB*sin30°

=2auB.

(b) Total resistance of the frame = R.

Hence the induced current in the frame at this moment,

i = Ɛ/R =2auB/R.

(c) Initial flux =Ba²

Final flux =B*0 =0 since the frame is now a straight line and its area will be zero. Change in flux in a time interval say Δt,

Δφ =Ba²-0 =Ba²

Emf induced E' =Δφ/Δt

Induced current i =E'/R

→i =Δφ/RΔt

→i*Δt =Δφ/R

Hence the charge flew through one side,

Q =i*Δt

=Δφ/R

=Ba²/R.

14. The north pole of a magnet is brought down along the axis of a horizontal circular coil (figure 38-E4). As a result, the flux through the coil changes from 0.35 weber to 0.85 webers 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 for Q-14

ANSWER: Δφ =0.85 -0.35 Weber

→Δφ =0.50 weber

Δt = 0.50 s

Hence the induced emf

E =Δφ/Δt

=0.50/0.50 V

=1.0 V.

From Lenz's law, the induced current will oppose the cause that has induced it. The cause here is that the downward magnetic flux through the loop is increasing. So the induced current will produce a magnetic flux that is upward. Such flux will be produced by a current in the loop that is anticlockwise when viewed from the side of the magnet.

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 loop is rotated in its own plane with some angular velocity, there is no change in the magnetic flux through the loop. Since there is no change in the magnetic flux through the loop, there will be no emf induced.

16. Figure (38-E5) shows a square loop of side 5 cm being moved towards the 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 (a) at t = 2 s. (b) t = 10 s, (c) t =22 s and (d) t =30 s.
The figure for Q -16

ANSWER: (a) At t = 0, the magnetic flux through the loop =0.

At t = 2 sec, the area of the loop inside the field =2 cm*5 cm =10 cm² =0.001 m².

Flux through the loop at this time =B*A

=0.6*0.001 weber

=6x10⁻⁴ weber.

Change in flux in 2 s =6x10⁻⁴ Weber

So emf induced =6x10⁻⁴/2 V

=3x10⁻⁴ V.

(b) From t =5 s to t =10 s, the loop is fully inside the magnetic field and there is no change in magnetic flux through the loop at t =10 s. So no emf induced i.e. induced emf =zero.

(c) At t =20 s, the loop is fully inside the magnetic field. Magnetic flux through the loop =BA

=0.6*0.0025 weber

=0.0015 weber.

At t =22 s, the area of the loop inside the magnetic field =5 cm*3 cm

=15 cm²

=0.0015 m²

The magnetic flux through the loop at t =22 s,

=0.6*0.0015 weber

=0.0009 Weber.

The change in magnetic flux through the loop in 2 s =0.0015 -0.0009 Weber

=0.0006 weber

The emf induced in the loop at t = 22 s,

=0.0006/2 V

=0.0003 V

=3x10⁻⁴ V.

(d) At t = 25 s, the loop is totally out of the magnetic field. Hence at t = 30 s, there is no change in the magnetic flux through the loop. So the induced emf in the loop =zero.

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: The heat will be produced only when there is an induced current in the loop. The induced current will be from t =0 to t = 5 s and then t =20 s to t =25 s. These are at the entry and exit times. At both of these times, the heat produced will be the same.

Between t = 0 to t = 5 s, change in magnetic flux,

Δφ = 0.6*25/10000 weber

= 0.0015 weber

emf induced, E =Δφ/Δt

→E =0.0015/5 V

=0.0003 V

R = 4.5 mΩ

=0.0045 Ω

Hence the induced current,

i =E/R

=0.0003/0.0045 A

=1/15 A.

The heat produced =i²Rt

= (1/15)²0.00455 J

= 1x10⁻⁴ J

The same will be the heat produced during exit, i.e. t = 20 s to t =25 s.

Hence total heat produced between t =0 to t =30 s will be

=2*1x10⁻⁴ J

=2x10⁻⁴ J.

18. A uniform magnetic field B exists in the cylindrical region of radius 10 cm as shown in figure (38-E6). 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 the 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 38-E6

ANSWER: Area of the magnetic fields inside the loop, A =πr²/2

→A =π(10 cm)²/2 =50π cm²

=0.005π m².

Magnetic flux through the loop, φ =BA

dφ/dt =A.dB/dt

(Since A is constant, and given dB/dt = 0.010 T/s)

Hence emf induced, E =dφ/dt

→E =A.dB/dt

=0.005π*0.010 V

=5πx10⁻⁵ V

R =4 Ω, hence the induced current,

i =E/R

=5πx10⁻⁵/4 A

=3.9x10⁻⁵ A.

19. The magnetic field in the cylindrical region shown in figure (38-E7) 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 S₁ is closed but S₂ is open. (b) S₁ is open but S₂ is closed. (c) both S₁ and S₂ are open and (d) both S₁ and S₂ are closed.
Figure for Q-19

ANSWER: The area of each square loop,

A =(1 cm)² =1 cm² =1x10⁻⁴ m².

Magnetic flux through each square,

φ =BA

Emf induced in a loop, E =dφ/dt

→E =A.dB/dt

=1x10⁻⁴*20x10⁻³ V

=2x10⁻⁶ V.

(a) When S₁ is closed and S₂ is open.

The current will be induced only in the left loop.

The total resistance of one loop, R =4*4 Ω

→ R = 16 Ω

Current induced, i =E/R

→ i =2x10⁻⁶/16 A

= 1.25x10⁻⁷ A.

The magnetic field going into the plane of paper is increasing, hence the induced current in the loop will oppose it. So it will produce a magnetic field that is coming out of the plane of the paper. It will happen when the induced current in the left loop is anticlockwise, So the current in the wire ad will be from a to d.

(b) When S₁ is open but S₂ is closed.

In this case, the current will be induced in the right loop only. Since both the loops are similar, the same magnitude of current will flow in this loop.

So, i = 1.25x10⁻⁷ A.

Also in this right loop, the induced current will be anticlockwise. So the current in the wire ad will be from d to a.

(c) Both S₁ and S₂ are open.

In this case, emf will be induced in each loop but there will be no current as there is no circuit complete. Hence the current in the wire ad is zero.

(d) Both S₁ and S₂ are closed

In both of the loops, equal emf will be induced which will further induce an equal anticlockwise current in them. Since the wire ad is common to both loops, the direction of currents will be equal and opposite in it. This will result in zero current in the wire ad.

20. Figure (38-E8) shows a circular coil of N turns and radius a, connected to a battery of emf Ɛ 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 center of the loop is at a distance x from the center of the coil. In the beginning, the sliding contact of the rheostat is at the left end and then onwards it is moved towards the 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 for Q-20

ANSWER: The magnetic field at the small circular loop,

B =µₒiNa²/{2(a²+x²)³ˊ²}

The magnetic flux at the smaller loop,

φ =BA

=µₒiNa²*πa'²/{2(a²+x²)³ˊ²}

emf induced in this loop,

E =dφ/dt

={½πµₒNa²a'²/(a²+x²)³ˊ²}di/dt

At time t, the resistance of the rheostat,

r' =R-Rvt/L

The total resistance of the circuit =r'+r

=R+r-Rvt/L

The current in the circuit,

i = Ɛ/(r+R-Rvt/L)

di/dt =ƐRv/{L(r+R-Rvt/L)²}

Hence the emf in the small loop,

E ={½πµₒNa²a'²/(a²+x²)³ˊ²}*ƐRv/{L(r+R-Rvt/L)²}

(a) At the time t =0 when the contact begins to slide,

E ={½πµₒNa²a'²/(a²+x²)³ˊ²}*ƐRv/{L(r+R)²}

E =½πµₒNa²a'²ƐRv/{L(a²+x²)³ˊ²(r+R)²}

(b) When the contact slid through half the length, the time

t =½L/v

Now emf,

E ={½πµₒNa²a'²/(a²+x²)³ˊ²}*ƐRv/{L(r+R-R/2)²}

E =½πµₒNa²a'²ƐRv/{L(a²+x²)³ˊ²(r+½R)²}

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°. 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: (a) Initial magnetic flux,

φ =n*BA

=500.20π(0.02)² weber

=0.004π weber

Final magnetic flux through the coil,

φ' =nB*A cos60°

=0.004π*½ weber

=0.004π/2 weber

Change in flux, Δφ =0.004π -0.004π/2 weber

→Δφ =0.004π/2 weber

=0.002π weber

Δt =0.100 s

Average emf induced = Δφ/Δt

=0.002π/0.100 V

=0.0628 V

=6.28x10⁻² V.

(b) R =4.00 Ω

Net charge crossing a cross-section of the wire of the coil =i*Δt

=(E/R)*Δt

=(6.28x10⁻²/4)*0.100 C

=0.00157 C

=1.57x10⁻³ C.

22. A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0x10⁻⁴ T about a diameter that is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm² 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: N =100, B =4.0x10⁻⁴ T, Angular speed =300 revolutions per minute. Area, A = 25 cm², Resistance, R =4.0 Ω.

(a) The angle covered in half a turn,

ß = 180°,

Time to cover half turn,

Δt =½*(60/300) s

=0.10 s

Δφ =NBA(cos0° -cos180°)

=1004.0x10⁻⁴25x10⁻⁴*(1+1) web

=2.0x10⁻⁴ weber

Average emf developed,

E =Δφ/Δt =2.0x10⁻⁴/0.10 V

=2x10⁻³ V.

(b) Since the magnetic flux at the initial and final position in a complete turn will be the same, the change in flux Δφ =0. Hence the average emf developed Δφ/Δt will be zero. In fact, the average emf in each half-turn will be equal and opposite hence the net average emf = zero.

(c) Net charge displaced in part (a) is.

Q = i*Δt

= (E/R)*Δt

= (2x10⁻³/4.0)*0.10 C

= 0.50x10⁻⁴ C

= 5.0x10⁻⁵ C.

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 that flows through the galvanometer. If the horizontal component of the earth's magnetic field is BH = 3.0x10⁻⁵ T.

ANSWER: If the time taken in this rotation =Δt.

Emf developed in this rotation,

E = Δφ/Δt

=NAB(cos0° -cos180°)/Δt

=1000π(0.1)²3x10⁻⁵*2/Δt

=6πx10⁻⁴/Δt

The charge that flows through the galvanometer,

Q =i*Δt

=(E/R)*Δt

=6πx10⁻⁴/40 C

=0.47x10⁻⁴ C

=4.7x10⁻⁵ C.

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: (a) A =area of circular coil,

=π*(0.05)² m²

=2.5πx10⁻³ m².

B =0.010 T

Ѡ =dß/dt =80 rev/minute

=80/60 rev/s

=4/3 rev/s

=4*2π/3 rad/s

=8π/3 rads/s.

emf developed in a small time dt,

E =dφ/dt

→E =d(BACosß)/dt

=-BAsinßdß/dt

=-BAѠ*sinß.

The maximum value of sinß =1

Hence the maximum value of emf induced,

Em = BAѠ

=0.012.5πx10⁻³8π/3 V

=6.6x10⁻⁴ V.

(b) Since the direction of the induced emf changes every half revolution, the average emf induced in each complete revolution will be zero, and hence over a long time, it will also be zero.

(c) Average of the squares of emf induced over a long period, say between 0 to T,

=(1/T)∫E²dt

=(1/T)∫B²A²Ѡ²sin²ß.dt

=(B²A²Ѡ²/2T)∫{1-cos(2Ѡt)}dt

=(B²A²Ѡ²/2T)[t -{sin(2Ѡt)}/2Ѡ]

=(B²A²Ѡ²/2T)[T-{sin(2ѠT)}/2Ѡ]

{Putting the limit t =0 to t =T}

=(B²A²Ѡ²/2T)[T-sin(2ß)2Ѡ]

For a long period let us put ß =2π for a full revolution.

=B²A²Ѡ²/2 ----- (i)

=(0.01)²(2.5πx10⁻³)²(8π/3)²/2 V²

=2.2x10⁻⁷ V².

One interesting thing:-

Maximum emf, Em =BAѠ

Em² =B²A²Ѡ²

Minimum emf, E =0, →E² =0

Average of these two squares,

=(B²A²Ѡ² -0)/2

=B²A²Ѡ²/2

It is the same as (i).

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: Heat produced in a time interval dt =i²Rdt

Heat produced over a time T,

H =∫i²Rdt

=∫(E/R)²Rdt

=(1/R)∫E²dt

=(1/R)∫{BAѠsin(Ѡt)}²dt

=(B²A²Ѡ²/R)½∫(1-cos2Ѡt)dt

=(B²A²Ѡ²/2R)[t -(sin2Ѡt)/2Ѡ]

=(B²A²Ѡ²/2R)[T -(sin2ѠT)/2Ѡ]

(After putting the limits)

T =60 s, 2ѠT =2608π/3 =160*2π

Hence sin2ѠT =0.
Now,

H =B²A²Ѡ²T/2R
=0.01²(2.5πx10⁻³)²(8π/3)²60/200 J
=1.3x10⁻⁷ J.

26. Figure (38-E9) 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 B of magnitude 2.00x10⁻⁴ 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.
The figure for Q-26

ANSWER: In 2 seconds the iron part comes down, at this moment there is no magnetic field through the iron part. Hence the magnetic flux through the iron part, φ = 0.

Initial magnetic flux, φ' =BA

→φ' =2x10⁻⁴*½π(0.10)²

=3.14x10⁻⁶ weber

Δt =2 s.

Hence the average emf developed =Δφ/Δt

=(3.14x10⁻⁶ -0)/2 V

=1.57x10⁻⁶ V.

27. A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 cm/s 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 electron will balance the magnetic force? How is the electric field created? (c) Find the motional emf between the ends of the rod.

ANSWER: (a) Average magnetic force on a free electron of the rod, F =evB,

→F =1.6x10⁻¹⁹0.100.10 N

=1.6x10⁻²¹ N.

(b) When this magnetic force is balanced by the electric force due to the electric field inside the rod, the electric force

F =1.6x10⁻²¹ N.

If the electric field inside the rod =E, then,

eE =F

→E =F/e

=1.6x10⁻²¹/1.6x10⁻¹⁹ V/m

=1.0x10⁻² V/m.

This electric field is created due to the accumulation of opposite charges at the ends of the rod.

(c) The motional emf between the ends of the rod,

V =vBl

=0.100.100.20 volts

=2x10⁻³ Volts.

28. A metallic meter stick moves with a velocity of 2 m/s in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.20 T. Find the emf induced between the ends of the stick.

ANSWER: The length of the meter stick,

l =1 m.

Velocity v =2 m/s,

Magnetic field, B=0.20 T

The emf induced between the ends of the stick =Bvl

=0.2021 V

=0.4 V.

29. A 10 m wide spacecraft moves through interstellar space at a speed of 3x10⁷ m/s. A magnetic field B =3x10⁻¹⁰ 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: B =3x10⁻¹⁰ T, v =3x10⁷ m/s

Width, l =10 m

The emf induced across its width =Bvl

=3x10⁻¹⁰3x10⁷10 V

=9x10⁻² V =0.09 V.

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? The vertical component of the earth's magnetic field is 0.2x10⁻⁴ T and the rails are separated by 1 m.

ANSWER: The emf will be induced across the width of the train and since the separated railway tracks are in contact with the train through the wheels, the emf will exist between the railway track.

emf =Bvl

=0.2x10⁻⁴(180x10³/3600)1 V

=1.0x10⁻³ V

=1.0 mV.

31. A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in figure (38-E10). 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 ac and (d) in the segment ab.
Figure for Q-31

ANSWER: (a) Since the loop abc moves in its plane and the uniform magnetic field is perpendicular to its plane, the magnetic flux through it remains constant. Since the emf induced in a loop is proportional to the rate of change of flux through it, the emf induced in this loop is zero.

(b) The emf induced in the segment bc is

=vB(bc)

From the right-hand rule, the positive end will be at c.

(c) Since the segment ac is moving along its length, its exposed portion perpendicular to velocity is zero. Hence the induced emf in the segment ac,

=vB(ac)*cos90° =0.

Hence the emf induced is zero.

(d) In the segment ab, the emf induced

=vB{(ab)*cos(∠abc)}

=vB(bc).

From the right-hand rule, the positive end will be at a.

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 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) Since the velocity is perpendicular to the diameter joining free ends, the projection of the semicircular wire perpendicular to the velocity =2r.

The emf induced between the ends,

=vB*2r

=2rvB.

(b) As is clear from the figure each segment AB and BC has the same projected length r perpendicular to the velocity. Hence each segment will have equal emf induced =vBr.
Diagram for Q-32

Also, the ends A and C will be at the same potential, either positive or negative depending upon the sense of the magnetic field. Since the potentials of A and C are the same, the emf between the ends A and C is zero.

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 space. Find the emf induced between the ends of the rod if the speed of translation is 20 cm/s.

ANSWER: B =1.0 T, l =0.10 m, v =0.20 m/s. The angle between the length and the velocity =60°. Hence the emf induced between the ends,

E = vBl*sin60°

=0.21.00.1*(√3/2) V

=0.017 V

=17x10⁻³ V.

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 emf between a pair of diametrically opposite points is,

=vB2rsinß, where ß is the angle between the diameter and the velocity. Since sinß is maximum (= 1) for ß =90°, the maximum emf =2vBr, between the ends of the diameter perpendicular to the velocity.

(b) Minimum emf will be for ß =0° for which sinß =0. Value of this emf =vB2r0 =0 (zero). Since ß =0°, this pair of points are located at the ends of the diameter that is parallel to the velocity.

35. Figure (38-E11) 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 for Q-35

ANSWER: The emf developed at the ends of the wire, E =vBL. Since the conducting wires on which it slides are not connected, there is no current in the moving wire, i =0.

The force on a current-carrying conductor in a magnetic field,

F =ilB =zero. If the wire is moving with a uniform velocity, no force is needed.

36. Figure (38-E12) 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 the 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 for Q-36

ANSWER: At time t, let the distance of the moving wire from the left = x. In the next small time dt, distance moved =dx.

dφ = Bldx,

emf induced, E =dφ/dt

→E =Bldx/dt

=Blv

At this time, the length of the wire in the circuit,

=2l +2x =2l+2vt =2(l+vt)

Given the resistance of wires = r per unit length. Hence the resistance in the circuit,

R = 2r(l +vt)

Current in the circuit,

i =E/R

=Blv/{2r(l+vt)}

The equivalent circuit diagram can be drawn as below:-
Diagram for Q-36

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 Fₒ, find the time at which the force needed will be Fₒ/2.

ANSWER: (a) Force on a current-carrying conductor in a magnetic field is,

F = ilB,

The equal and opposite force will be needed to move the wire at uniform velocity.

Here the force required to move the wire at uniform velocity at time t is given as,

F =(lB)*Blv/{2r(l+vt)}

=B²l²v/{2r(l+vt)}.

(b) Initial force at time t = 0,

Fₒ =B²l²v/(2rl)

If at time t = t' the force becomes Fₒ/2, then

B²l²v/{2r(l+vt')} =½B²l²v/(2rl)

→1/(l+vt') = 1/2l

→l+vt' =2l

→vt' =l

→t' = l/v.

38. Consider the situation shown in figure (38-E13). 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 the right with a speed vₒ. 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 function of x and (d) the maximum distance the wire will move.
Figure for Q-38

ANSWER: (a) Initially, the emf induced in the wire PQ = vₒBl.

Resistance of the circuit =R+r

Hence the current induced in the circuit,

i =vₒBl/(R+r).

So, when the speed of the wire is v, the current induced in the loop is,

=vBl/(R+r).

(b) Due to the current of the wire in the magnetic field, an opposing force will act on the wire,

F =ilB

=vBl*Bl/(R+r)

=vB²l²/(R+r)

(Towards left)

Acceleration of the wire, a =F/m

=vB²l²/{m(R+r)}.

Directed towards left.

(c) The velocity of the wire after time t,

v =vₒ -at

=vₒ -vB²l²t/{m(R+r)}

=vₒ -B²l²(vt)/{m(R+r)}

=vₒ -B²l²x/{m(R+r)}.

(d) For the maximum distance that wire will move (xₘₐₓ), v = 0.

→B²l²xₘₐₓ/{m(R+r)} =vₒ

→xₘₐₓ =mvₒ(R+r)/B²l².

39. A rectangular frame of wire abcd has dimensions 32 cm x 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 3.2x10⁻⁵ N (figure 38-E14). It is found that the frame moves at a constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between points a and b, and (d) the potential difference between points c and d.
Figure for Q-39

ANSWER: (a) Suppose the current in the loop is = i. Only the part 'ab' of the loop is in the magnetic field and it is perpendicular to the motion. The emf will be induced only in this part. This emf,

E =vBl, where v is the constant speed.

Current in the loop,

i = E/R

=vBl/R

Since the loop is moving with a constant speed, the magnitude of the magnetic force on the loop will be equal to the applied force but opposite in direction. So, magnetic force,

F =3.2x10⁻⁵ N.

But the force on a current-carrying conductor in a magnetic field,

F =ilB =vBl*lB/R =vB²l²/R

→3.2x10⁻⁵ =v(0.02)²(0.08)²*/2

→v =23.2x10⁻⁵/{(0.02)²(0.08)²}

=25 m/s.

(b) The emf induced in the loop is the same as the emf induced in the part ab. It is calculated in above part (a) as,

E = vBl

=250.020.08 V

=0.04 V

=4x10⁻² V.

(c) Total resistance of the loop =2 Ω.

Resistance of the part adcb, (excluding part ab) =(72/80)*2 Ω.

→R' = 1.8 Ω.

Current in the loop, i =vBl/R

→i =250.020.08/2 A

→i =0.02 A

Hence the potential difference between points a and b due to this current =iR'

=0.02*1.8 V

=0.036 V

=3.6x10⁻² V.

(d) The potential difference between points c and d =iR"

=(0.02 A){(8/80)2 Ω}

=0.02*0.2 V

=0.004 V

=4.0x10⁻³ V.

40. Figure (38-E15) 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. If the horizontal component of the earth's magnetic field is 3.0x10⁻⁵ T, calculate the dip at the place.
Figure for Q-40

ANSWER: Speed v =20 cm/s =0.20 m/s.

l = 20 cm =0.20 m

R =0.20 Ω

BH =3.0x10⁻⁵ T

i =2.0x10⁻⁶ A

Since the wire moves horizontally, the emf induced in it is due to the earth's vertical component of the magnetic field.

emf induced, E =Bvvl

→E =Bv0.200.20 V

=Bv*4x10⁻² V

Current induced =E/R

=Bv*4x10⁻²/0.20 A

Equating it with the given current,

0.20Bv = 2x10⁻⁶

→Bv =1.0x10⁻⁵ T

Hence the dip at the place, θ is given as,

tan θ =Bv/BH

=1.0x10⁻⁵/3.0x10⁻⁵

=1/3

→θ = tan⁻¹(1/3).

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 (38-E16). 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=√(mgRsinθ)/vl²cos²θ.
The figure for Q-41

ANSWER: emf induced in the wire ab,

E =vBl.cosθ

{we will take the component of B that is perpendicular to the plane of movement of wire ab}

Current in the wire ab,

i =E/R =vBl.cosθ/R

The magnetic force in the wire,

F =ilB.cosθ

=vBl.cosθ*(lB.cosθ)/R

=vB²l²cos²θ/R

The direction of this force will be opposite to the movement.

Weight of the wire, W =mg

Component of the weight along the plane =mg.sinθ

Since the wire is moving with a uniform speed, the net force on the wire is zero. i.e.

vB²l²cos²θ/R -mgsinθ =0

→B² = mgR.sinθ/vl²cos²θ

→B =√{(mgR.sinθ)/(vl²cos²θ)}

42. Consider the situation shown in figure (38-E17). The wire P₁Q₂ and P₂Q₂ are made to slide on the rails with the same speed of 5 cm/s. Find the electric current in the 19 Ω resistor if (a) both the wires move towards the right and (b) if P₁Q₁ moves towards the left but P₂Q₂ moves towards the right.
The figure for Q-42

ANSWER: (a) When both the wires move towards the right with the same speed, the emf induced in each wire will be the same and in the same direction. The emf induced E =vBl,

→E =0.051.00.04 V

= 0.002 V.

It will be as if two cells are joined in parallel, hence the emf will remain the same = 0.002 V.

The equivalent resistance of the moving wires that are in parallel,

1/r =1/2 +1/2 = 1

→r = 1 Ω.

The total resistance of the circuit,

R =1 Ω +19 Ω =20 Ω.

Hence the current through the 19 Ω resistor,

i =E/R =0.002/20 A

= 0.1x10⁻³ A

= 0.1 mA.

(b) When both the wires move parallelly with equal and opposite speeds in a uniform magnetic field, the EMFs induced in each wire are equal but opposite in direction. So the net emf available for the 19 Ω resistor is zero.

43. Suppose the 19 Ω resistor of the previous problem is disconnected. Find the current through P₂Q₂ in the two situations (a) and (b) of that problem.

ANSWER: (a) When the 19 Ω resistor is disconnected, the points P₁ and P₂ have the same potential. Similarly, the points Q₁ and Q₂ have equal potentials. It is like two cells connected in parallel but there is no circuit completed. So the current through the wire P₂Q₂ will be zero.

(b) In the second case, when both wires slide in opposite directions, the EMFs induced in the wires are equal and opposite. This situation is as if two cells are connected in series. The emf in the circuit will be the sum of both EMFs, i.e.

E =0.002 +0.002 V =0.004 V

The circuit completes through both the wires and the resistance of the circuit

R =2 Ω +2 Ω =4 Ω

So the current in the circuit and hence in the wire P₂Q₂,

i =E/R

=0.004/4 A

=0.001 A

=1.0x10⁻³ A

=1.0 mA.

44. Consider the situation shown in the figure (38-E18). The wire PQ has negligible resistance and is made to slide on the three rails with a constant speed of 5 cm/s. Find the current in the 10 Ω resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.
The figure for Q-44

ANSWER: (a) When the switch s is thrown to the middle rail, the emf induced in the bottom half of the wire PQ is not contributing to the circuit that is completed through the top and middle rail. The emf induced in the upper half of the wire and contributing to the circuit,

E =vBl

=0.051.00.02 V

=0.001 V

Resistance in the circuit, R =10 Ω

Hence the current in the 10 Ω resistor,

=E/R

=0.001/10 A

=0.0001 A

=0.1 mA.

(b) When the switch is thrown to the bottom wire, the emf induced in the whole wire PR is contributing to the circuit that is through the top and bottom rails. The emf induced,

E =vBl

=0.051.00.04 V

=0.002 V.

Resistance R =10 Ω.

Hence the current in the 10 Ω resistor,

i = E/R

= 0.002/10 A

= 0.0002 A

= 0.2 mA.

45. The current generator I₉, shown in figure (38-E19), 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 the right. Each of these wires has resistance r. Find the current through the wire cd.
The figure for Q-45

ANSWER: Let the current in the wire cd = i'. The current in the wire ab = i -i'. The potential difference across cd = i'r. The same potential difference will be across the wire ab. Since the wire ab is moving with a velocity v, the induced emf =Bvl. Its polarity will be opposite to the current i-i'.
Diagram for Q-45

So the net potential difference across ab =(i-i')r-Blv. Equating the potential difference across both wires we have

i'r =(i-i')r-Blv

→2i'r =ir -Blv

→i' =(ir -Blv)/2r.

Watch this solution in the video below↓

46. The current generator I₉, shown in figure (38-E20), 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 I₉. The entire system lies in a vertical magnetic field B. Find the velocity of the wire as a function of time.
The figure for Q-46

ANSWER: The current through the wire ab = i. The force on the wire in the magnetic field B,

F = ilB, the direction of this force from Flemming's left-hand rule is towards the right.

Mass of the wire =m,

Acceleration of the wire, a =F/m

→a = ilB/m

If the initial velocity u =0, at t =0 then from, v =u+at, the velocity at time t is,

v =0 +(ilB/m)t

→v =ilBt/m.

In the same direction as the force i.e. towards the right away from the generator.

47. The system containing the rails and 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 38-E21). 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?
The figure for Q-47

ANSWER: The direction of the magnetic field is coming out of the plane of the figure, hence the force on the wire will be upwards. If the current = i, then due to the equilibrium,

ilB =mg,

When the mass is doubled, net downward force,

F = 2mg -ilB =2mg -mg =mg

Downward acceleration,

a =mg/2m =g/2

Let time t is taken to travel a distance of l, then from

s =ut +½at²

→l = 0 +½(g/2)t²

→t² =4l/g

→t = 2√(l/g).

48. The rectangular wire-frame, shown in figure (38-E22), 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 the velocity vₒ. (c) Show that the velocity at time t is given by

v =vₒ(1 -e^-Ft/mvₒ).
The figure for Q-48

ANSWER: (a) When the speed is v, the emf induced in the width d of the frame,

E =Bvd

The current due to this emf

i =E/R =Bvd/R

Force on the width d, =iBd =B²d²v/R

It will be opposite to the movement. Hence the net force on the loop

=F -B²d²v/R

Acceleration of the loop,

a = Force/mass

=(F -B²d²v/R)/m

=(RF -B²d²v)/mR.

(b) Due to the acceleration speed increases and so does the opposing force. A time comes when the external and opposing forces become equal in magnitude and the net force is zero. At this time there is no acceleration and the frame moves with a constant acquired speed, say vₒ.

For this condition,

F -B²d²vₒ/R =0,

→B²d²vₒ =FR

→vₒ =RF/B²d². All the entities on RHS are constant, so vₒ is constant. The velocity will remain constant till the whole frame enters the magnetic field. After that same emf will be induced in the opposite width d and net emf will be zero. So no current in the circuit. The frame will move with acceleration under the force F.

(c) As in (a) above, acceleration at time t is given as,

a =dv/dt = (RF-B²d²v)/mR

→dv/(RF-B²d²v) =dt/mR

Integrating both sides we get,

(-1/B²d²) [ln(RF-B²d²v)] =(1/mR)[t]

Putting limits from 0 to v for speed and 0 to t for the time we get,

(-1/B²d²){ln(RF-B²d²v)/RF}=t/mR

→ln{(RF-B²d²v)/RF}=-B²d²t/mR

→ln{1 -v*B²d²/RF} =-B²d²Ft/RFm

→ln{1 -v/vₒ} =-Ft/mvₒ

→1 -v/vₒ =e^-Ft/mvₒ

→v/vₒ =1 -e^-Ft/mvₒ

→v =vₒ(1 -e^-Ft/mvₒ).

Proved.

49. Figure (38-E23) shows a smooth pair of thick metallic rails connected across a battery of emf Ɛ 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 the 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.
The figure for Q-49

ANSWER: (a) Induced emf in the moving wire =Bvl.

Its direction will be positive at b and negative at a. So the net emf in the circuit

=Ɛ -Bvl

So the current in the wire,

i = (Ɛ -Blv)/r.

Since the velocity given is small, the direction of the current will be from b to a.

(b) The force acting on the wire ab,

F =ilB

= {(Ɛ -Blv)/r}*lB

= lB(Ɛ -Blv)/r.

Towards the right.

(c) Under the force the wire will accelerate towards the right and due to this the opposing emf Blv will also increase. A time will come when the net emf in the circuit becomes zero and hence the current. Suppose the velocity at this instant is vₒ, then

Ɛ =Blvₒ

→vₒ = Ɛ/Bl.

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(38-E24). 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 the velocity vₘ. (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 a steady-state is reached.
The figure for Q-50

ANSWER: (a) Induced emf in the loop is due to the movement of wire ab in the magnetic field, and it is equal to,

E =Bvl.

(b) Magnitude of the current in the wire

i =E/r =Bvl/r.

The positive charge in the moving wire will accumulate at the end b and the negative at a. So the current will be from b to a.

(c) Upward force on the wire due to the induced emf (opposing the movement)

=ilB

=(Bvl/r)*lB

=vB²l²/r

Downward force

=weight

=mg

Net downward force,

F = mg -vB²l²/r

Downward acceleration of the wire at this instant,

a = F/m

=(mg -vB²l²/r)/m

=g -vB²l²/mr.

(d) Due to the acceleration v increases and a time comes when the opposing force vB²l²/r becomes equal to the weight mg. Now the net force on the wire is zero and it moves with a constant velocity vₘ. In this situation,

vₘB²l²/r =mg

→vₘ =mgr/B²l².

(e) Acceleration of the wire,

dv/dt =a =g -vB²l²/mr

→dv/dt =g -vg(B²l²/mgr)

→dv/dt =g -gv/vₘ =(gvₘ -gv)/vₘ

→dv/(gvₘ -gv) =dt/vₘ

Integrating both sides we get,

(-1/g)[ln(gvₘ -gv)] =[t]/vₘ

Putting the limits for velocity from 0 to v and for time 0 to t, we have

{ln(gvₘ -gv) -ln(gvₘ)} =-gt/vₘ

→ln{(vₘ -v)/vₘ} =-gt/vₘ

→(vₘ -v)/vₘ = e^-gt/vₘ

→vₘ -v = vₘe^-gt/vₘ

→v = vₘ -vₘe^-gt/vₘ

= vₘ(1 -e^-gt/vₘ).

(f) v =dx/dt

→dx = vdt

→X =∫dx =∫vdt

=∫vₘ(1 -e^-gt/vₘ)dt

=∫(vₘ -vₘe^-gt/vₘ)dt

=[vₘt + vₘ(vₘ/g)e^-gt/vₘ)]

Put limits of t from 0 to t.

=vₘt +(vₘ²/g)e^-gt/vₘ)-vₘ²/g
=vₘt -(vₘ²/g)(1-e^-gt/vₘ).

(g) Rate of heat developed is the power consumed by the wire resistance,
P =Vi
=BlvBlv/r
=B²l²v²/r
When the steady-state is achieved,
v =vₘ
So, P =B²l²vₘ²/r
=(B²l²/r)(mgr/B²l²)²
=m²g²r/B²l²

Rate of decrease of gravitational potential energy,
dU/dt =d(mgx)dt =mg.dx/dt =mgvₘ
=mg(mgr/B²l²)
=m²g²r/B²l².
Thus P =dU/dt.

Hence proved.

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 stroke is 30.0 cm and the horizontal component of the earth's magnetic field is 2.0x10⁻⁵ T, Find the emf induced between the axis and the outer end of the spoke. Neglect centripetal force acting on the free electrons of the spoke.

ANSWER: Consider a small length dx at a distance x from the center. Its velocity will be v =ωx. emf induced in this small length due to the earth's horizontal magnetic field, dE =Bv*dx

=Bωxdx,

Hence E =∫dE =Bω∫xdx

=Bω[x²/2]

Taking x from 0 to l, we have,

E =½Bωl²

Given, l =30 cm =0.30 m

ω =100*2π/60 rad/s =10π/3 rad/s,

B =2.0x10⁻⁵ T, hence

E =½2.0x10⁻⁵(10π/3)*(0.30)² V

=9.4x10⁻⁶ V.

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

ANSWER: When a conducting rod or flat surface of length l, moves perpendicular to the magnetic field with a velocity v, the emf induced is =Bvl, whatever its width may be. So if we consider a dx wide metallic ring on the moving disc at distance x from the center, the emf induced between the inner circle of radius x and outer circle of radius x+dx is, dE =Bvdx

→dE =Bωx.dx

Now, E =∫dE = ∫Bωx.dx

=Bω[x²/2]

=½Bωr².

{Putting the value of x from 0 to r}

53. Figure (38-E25) shows a conducting disc rotating about its axis in a perpendicular magnetic field B. A resistor of resistance R is connected between the center and the rim. Calculate the current in the resistor. Does it enter the disc or leave it at the center? The radius of the disc is 5.0 cm, angular speed ⍵ = 10 rads/s, B =0.40 T and R =10 Ω.
The figure for Q-51

ANSWER: As in the previous problem, emf induced between the center and the rim, E =½Bωr².

r =5 cm =0.05 m, ω =10 rad/s, B =0.4 T, hence,

E =½0.410*(0.05)² V

=0.005 V

Current in the 10 Ω resistor, i =E/R

→i =0.005/10 A =0.5x10⁻³ A =0.5 mA.

From Flaming's right-hand rule the positive charge will concentrate at the center and the negative at the rim. When the resistor is connected as in the figure, the current will leave the disc at the center.

54. The magnetic field in a region is given by B =kBₒy/L 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 =vₒi, find the emf induced between the ends of the rod.

ANSWER: Consider a very small element dy on the rod at y distance from the origin. The small emf generated in this element,

dE =Bxv*dy

=k(Bₒy/L)xvₒi*dy

=j(Bₒvₒ/L)ydy

Unit vector j shows that the direction of the emf is along Y-axis. Hence the magnitude of full emf developed in the rod is,

E =∫dE =(Bₒvₒ/L)∫ydy

=(Bₒvₒ/L)[y²/2]

Putting the limit for y between 0 to L, we get

E =(Bₒvₒ/L)*(L²/2)

=½BₒvₒL.

55. Figure (38-E26) 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 center of the rod is x. Find the motional emf induced in the rod.
The figure for Q-55

ANSWER: Magnetic field due to the current carrying wire at a distance r from the wire

B =µₒi/2πr.

Consider a small element dr on the rod. The small emf induced in it,

dE =Bv*dr

=(µₒi/2πr)vdr

=(µₒiv/2π)(1/r)dr

Hence the emf induced in the rod,

E =∫dE =(µₒiv/2π)∫(1/r)dr

=(µₒiv/2π)[ln r]

Putting the limits for the rod length, r =(x-l/2) to (x+l/2), we get

E =(µₒiv/2π)*ln{(x+l/2)/(x-l/2)}

=(µₒiv/2π)*ln {(2x+l)/(2x-l)}

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 a 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 resistor 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) In this case there will be a current in the rod,

i' = E/R

A current-carrying rod will feel a force in the perpendicular magnetic field. This force

F =i'lB =ElB/R

Here i and l are fixed for uniform v, but B is varying. Let the force on a small length element dr is dF, then

dF =(µₒiv/2πR)ln{(2x+l)/(2x-l)}*(µₒi/2πr)dr

=(µₒi/2π)²(v/R)ln{(2x+l)/2x-l)}(dr/r)

Force on the whole rod,

F =∫dF

=(µₒi/2π)²(v/R)ln{(2x+l)/(2x-l)}[ln r]

Putting the same limit for r =(x-l/2) to (x+l/2), we get

F =(µₒi/2π)²(v/R)[ln{(2x+l)/(2x-l)]².

=(v/R)[(µₒi/2π) ln{(2x+l)/(2x-l)}]²

This much force is needed to keep the rod moving on the rails with a constant velocity v.

(b) current in the rod is

i' =E/R

=(µₒiv/2πR) ln{(2x+l)/(2x-l)}

The expression for E is derived in the previous problem.

(c) The rate of heat developed in the resistor,

H =i'²R

=[(µₒiv/2πR) ln{(2x+l)/(2x-l)}]²*R

=(1/R)[(µₒiv/2π) ln{(2x+l)/(2x-l)}]²

(d) Power delivered by the external agent exerting force on the rod,

P = F*v

=(v/R)[(µₒi/2π) ln{(2x+l)/(2x-l)}]²v

=(v²/R)[(µₒi/2π) ln{(2x+l)/(2x-l)}]²

=(1/R)[(µₒiv/2π) ln{(2x+l)/(2x-l)}]²

= H

So the power required to maintain v is the same as the rate of heat generated. The answer is the same as in (c)

The expression of force F is taken from the part (a).

57. Figure (38-E27) shows a square frame of wire having a total resistance r placed coplanar with a long, straight wire. The wire carries a current 'i' given by i = iₒsin⍵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 20π/⍵.
The figure for Q-57

ANSWER: (a) Consider a small area strip dA parallel to the wire on the loop. dA is at a distance r from the wire and has a width dr as shown in the diagram below.
Diagram for Q57

Magnetic field due to the long current carrying wire at distance r from it,

B = µₒi/2πr

Magnetic flux on the small strip with area dA,

dφ =B.dA

=(µₒi/2πr)(adr)

=(µₒia/2π)(dr/r)

φ =∫dφ

=(µₒia/2π)∫(dr/r)

=(µₒia/2π)*[ln r]

Putting the limit of r from b to (a+b), we get

φ =(µₒia/2π)*ln {(a+b)/b)}

=(µₒia/2π)*ln {(1 +a/b)}

(b) The emf induced in the frame

E =dφ/dt

=(µₒa/2π)ln(1+a/b)di/dt

=(µₒa/2π)ln(1+a/b)d(iₒsinωt)/dt

=(µₒa/2π)ln(1+a/b)iₒωcosωt

=(µₒaiₒω.cos ωt/2π) ln(1+a/b)

(c) Current in the loop at time t

i' =E/r

Heat developed in a time interval dt,

dH =i'²r dt

=(E²/r) dt

=(1/r){(µₒaiₒω/2π) ln(1+a/b)} cos²(ωt)dt

Hence the heat developed in the frame in time t,

H =∫dH

=(1/r){(µₒaiₒω/2π)ln(1+a/b)}² ∫cos²(ωt)dt

=(1/r){(µₒaiₒω/2π)ln(1+a/b)}² ∫{½(1+cos2ωt)}dt

=(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² ∫(1+cos2ωt)dt

=(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² [t+(1/2ω)sin2ωt)]

Putting the value of t from 0 to 20π/ω, we get

H =(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² [20π/ω+(1/2ω)sin40π)]

=(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² (20π/ω)

=(10π/ωr){(µₒaiₒω/2π)ln(1+a/b)}²

=(10πµₒ²a²iₒ²ω²/4π²ωr) {ln(1+a/b)}²

=(5µₒ²a²iₒ²ω/2πr) ln²(1 +a/b)

58. A rectangular metallic loop of length l and width b is placed coplanar with a long wire carrying current i (Figure 38-E28). 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.
The figure for Q-58

ANSWER: (a) consider a dr wide strip on the loop parallel to the current carrying a long wire at a distance r from the wire. Area of the strip, dA =bdr

The magnetic field at distance r from the wire B =µₒi/2πr

Magnetic flux at the strip

dφ =B.dA

=(µₒi/2π)(bdr/r)

=(µₒib/2π)(dr/r)

Hence total flux on the loop

φ=∫dφ

=(µₒib/2π) [ln r]

=(µₒib/2π) ln{(x+l)/x}

=(µₒib/2π) ln(1+l/x)

for r = x to x+l

i.e. when the rear end of the loop is at a distance x from the wire.

From Faraday's law emf induced in the loop = -dφ/dt

=-(µₒib/2π){1/(1+l/x)}(-l/x²)dx/dt

= µₒiblv/{2πx(l+x)}

{since v =dx/dt}

When the rear end of the loop is at a distance x =a from the wire, emf induced in the loop

=µₒiblv/{2πa(l+a)}.

Let us solve this problem by considering different segments of the loop. The emf induced in the two sides of the loop having length l is zero because the projection of length on a plane perpendicular to the velocity is zero.

Now the emf induced in a side of loop having length b and parallel to the wire is

=Bvb

When this loop segment is at a distance x from the wire, the emf induced,

=(µₒi/2πx)vb

=µₒivb/2πx

So when the rear end of the loop is at a distance 'a' from the wire, the emf induced in the rear segment of the width b = µₒivb/2πa

and in the farther segment, emf

=µₒivb/2π(l+a)

Since these two emf will have the same direction in the rods, they will oppose each other in the circuit of the loop. Hence the net emf in the loop

=µₒivb/2πa -µₒivb/2π(l+a)

=(µₒivb/2π){1/a -1/(l+a)}

=µₒiblv/{2πa(l+a)}.

It is the same as derived by Faraday's law above.

59. Figure (38-E29) 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 center O. The other end of the rod touches the loop at A. The center O and 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 ⍵. Find the force.
The figure for Q-59

ANSWER: Let us find the emf induced in the rod for a uniform angular velocity ω. We can either calculate it by calculus as in earlier problems or by the average speed of the rod. Since the tangential speed of the rod is zero at the center and ωa at the loop, the average speed of the rod =½ωa. Hence the emf induced in the rod

=B(½ωa)a

=½Bωa².

Current through the rod, i =(½Bωa²)/R

Hence the force on the rod due to this current,

=iBa

=(½Bωa²)Ba/R

=½B²a³ω/R.

This is the resisting force on the rod due to the magnetic field and opposite to the direction of motion. Hence to maintain a uniform angular velocity net force should be zero on the rod. So the applied force also should have the same magnitude. Thus

F =½B²a³ω/R.

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 ⍵ as shown in the figure. Find the current in the rod when ∠AOC =90°.

ANSWER: In this case, there will be two resistances connected in parallel in the circuit. Since the resistance R is uniformly distributed in the loop, one resistance will be for a quarter of the loop

r' =R/4, and the other resistance will be for the rest of the loop which is three-quarters. So,

r" =3R/4.

If the equivalent resistance of these two parallel resistances is r, then

1/r =1/r' +1/r" =4/R +4/3R

→1/r =(4/R)(1 +1/3) =(4/R)*4/3

→1/r =16/3R

→r =3R/16.

So the current in the rod

=(induced emf)/Resistance

=(½Bωa²)/r

=(½Bωa²)/(3R/16)

=8Bωa²/3R.

61. Consider a variation of the previous problem (figure 38-E29). 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 the O and C has a resistance R. The rod is made to rotate with a uniform angular velocity ⍵ 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: As we have calculated the emf induced in the rod is,

E = ½B⍵a²

Current in the rod,

i = E/R

=½B⍵a²/R

Force on the rod due to the current in the magnetic field

F' =iBl

=(½B⍵a²/R)Ba

=½B²⍵a³/R

When the rod makes an angle θ with the vertical, the component of weight in the direction of the applied force F,

W' = mg Cos(90° -θ)

=mg Sin θ

Diagram for Q-61

The force by the magnetic field due to current, F' will be opposing the motion. Since the rod is made to move with a uniform angular velocity ⍵, Equal and opposite force should be on the rod. Thus,

W' +F =F'

mg Sinθ +F = ½B²⍵a³/R

→F = ½B²⍵a³/R -mg Sinθ

This much magnitude of force should be applied.

62. Figure (38-E30) shows a situation similar to the previous problem. All parameters are the same except that a battery of emf Ɛ and a variable resistance R are connected between O and C. The connecting wires have zero resistance. No external force is applied to the rod (except gravity, forces by the magnetic field, and by the pivot). In what way should the resistance R be changed so that the rod may rotate with uniform angular velocity in the clockwise direction? Express your answer in terms of the given quantities and the angle θ made by the rod OA with the horizontal.
Figure for Q-62

ANSWER: Suppose the rod makes an angle θ with the horizontal and it moves with a uniform angular velocity ⍵. Due to the clockwise movement, the induced emf E will have a positive end at A and negative at O. So in the circuit, the induced emf E and the battery emf Ɛ are connected in series. So the total emf in the circuit =E+Ɛ.

Current in the rod,

i =(E+Ɛ)/R

Force due to this current in the magnetic field will be opposing the movement and the weight component perpendicular to the rod will be balancing it to keep the motion uniform. So,

mg Cosθ =ilB

→mg Cosθ =(E+Ɛ)aB/R

→R =(E+Ɛ)aB/mgCosθ

=(½B⍵a² +Ɛ)aB/mgCosθ

=aB(2Ɛ +⍵a²B)/(2mg Cosθ).

63. A wire of mass m and length l can slide freely on a pair of smooth, vertical rails (figure 38-E31). 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.
Figure for Q-63

ANSWER: Suppose at any instant the velocity of the wire is v. The emf induced in the wire,

E =Bvl

this will be the potential difference across the capacitor. If the charge on the capacitor is Q, then from

Q =CV

we have, Q =C*Bvl =CBvl

Current at any instant in the circuit,

i =dQ/dt

=d(CBvl)/dt

=CBl*dv/t

=CBla,

where a is the acceleration of the wire.

Force on the wire due to this current in the magnetic field,

F =iBl

=(CBla)Bl

=CB²l²a

This force will be opposing the motion, in the upward direction. There will be the weight of the wire acting downwards,

W =mg

The net force on the wire =W-F

So the acceleration of the wire,

a =(W-F)/m

→a =(mg -CB²l²a)/m

→ma +CB²l²a =mg

→a =mg/(m +CB²l²).

64. A uniform magnetic field B exists in a cylindrical region, shown dotted in figure (38-E32). The magnetic field increases at a constant rate of dB/dt. Consider a circle of radius r coaxial with a 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 for Q-64

ANSWER: (a) The work done by the induced electric field E in moving a unit charge completely around a circuit is the induced emf in the circuit. So the induced emf in the given circular path,

Ɛ = ∮E.dl

Since the directions of E and dl are the same on the circular path,

Ɛ = ∮E.dl

=E*2πr,

{Due to symmetry E will be the same everywhere on the circle}

→Ɛ =2πrE

Now consider the magnetic flux through the circular area.

φ =B*A =πr²B at any instant.

Induced emf due to the changing magnetic flux is given as

Ɛ =dφ/dt

=πr²*dB/dt

Equating the two expressions we get,

2πrE =πr²dB/dt

→E =½r.dB/dt

It will be the magnitude of the electric field at a point on the given circle.

(b) Join the point P to the center of the circle O. Let OP = r'. Now draw a circle with center O and radius r'. It will pass through the point P. Let OP make an angle θ with the vertical. The direction of the electric field E' at P will be tangential at that point as shown in the diagram below.
Diagram for Q-64

The magnitude of E' as from (a) above will be,

E' =½r'dB/dt.

The angle between E' and the side of the square will also be equal to θ. Hence the component of E' along ba,

= E'*Cosθ

=½r'Cosθ.dB/dt

=½rdB/dt.

{Because r'Cosθ =r}

It is the same as derived in (a).

65. The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A/s. 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 the 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: Magnetic field inside a long solenoid,

B =µₒni

(a) Consider a circle of radius r inside the solenoid with its axis coinciding with the axis of the solenoid. Flux through this circle,

φ =BA =µₒni*πr²

=µₒπnr²i

Rate of change of flux through this circle,

dφ/dt =µₒπnr²di/dt

Here, n =2000 turns/m, r =0.01 m, di/dt =0.01 A/s

→dφ/dt=(4πx10⁻⁷)π20001x10⁻⁴*0.01 weber/s

=7.9x10⁻⁹ Weber/s

Hence the change in magnetic flux in 2 s,

=(7.9x10⁻⁹ Weber/s)*(2 s)

=1.6x10⁻⁸ Weber.

(b) Let the electric field induced at a point on the circle =E. Then,

E*2πr =dφ/dt {=emf induced}

→E =(µₒπnr²di/dt)/(2πr)

=½µₒnrdi/dt

=½(4πx10⁻⁷)20000.01*0.01 V/m

=4πx10⁻⁸ V/m

=1.2x10⁻⁷ V/m.

(c) Consider a circle of radius 8.0 cm with its axis coinciding with the axis of the solenoid. Flux changing through this circle is the same as the flux changing through the solenoid. Hence,

dφ/dt =µₒπnr²di/dt

=(4πx10⁻⁷)π2000(0.06)²0.01 W/s

=0.288π²x10⁻⁷ weber/s

If E' is the magnitude of the electric field at a point on the 8.0 cm radius circle outside the solenoid, then

E'2π0.08 =dφ/dt

→E' =0.288π²x10⁻⁷/(0.16π) V/m

=1.8πx10⁻⁷ V/m

=5.6x10⁻⁷ V/m.

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: Induced emf Ɛ =L*di/dt

Ɛ =20 V, di/dt ={2.5 -(-2.5)}/0.1 A/s

→di/dt =50 A/s

Hence, 20 =L*50

→L =20/50 H

=0.40 H.

67. A magnetic flux of 8x10⁻⁴ 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: The magnetic flux linked with each turn,

φ =8x10⁻⁴ Weber,

Number of turns, N =200,

i =4 A. If the sef-inductance of the coil is L, then

Nφ =Li

→L =Nφ/i

=200*8x10⁻⁴/4 H

=4x10⁻² H.

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. Find the emf induced in it.

ANSWER: B =µₒni, A =πr², di/dt =0.8 A/s, r =2 cm =0.02 m.

n =240/0.12 turns/m =2000 turns/m.

Magnetic field inside the solenoid,

B=µₒni.

Magnetic flux through each turn of the solenoid,

φ' =BA =µₒπnr²i

Magnetic flux through N turns of the solenoid,

φ =µₒπnNr²i

emf induced =dφ/dt

=µₒπnNr²di/dt

=(4πx10⁻⁷)π2000240(0.02)²*0.8 V

=6.0x10⁻⁴ V.

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: If iₒ is the steady-state current, then current at time t is,

i =iₒ(1 -e^-t/𝞃)

→1 -e^-t/𝞃 =i/iₒ

→e^-t/𝞃 = 1 -i/iₒ

→t/𝞃 = -ln(1 -i/iₒ)

(a) For the current builds up to 90%,

i/iₒ =0.90

So, t/𝞃 =-ln(1-0.90) =-ln (0.1)

= 2.3

(b) For the current builds up to 99%,

i/iₒ =0.99

So, t/𝞃 =-ln(1-0.99) =-ln(0.01)

=4.6

(c) For the current builds up to 99.9%,

i/iₒ =0.999

So, t/𝞃 =-ln(1-0.999) =-ln(0.001)

=6.9

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: iₒ =2.0 A, Ɛ =4.0 V, Hence the resistance of the coil,

R =Ɛ/iₒ =4.0/2.0 Ω =2.0 Ω

Given L =1.0 H

So the time constant,

𝝉 =L/R =1.0/2.0 s =0.5 s.

71. A coil having an inductance of 2.0 H and resistance of 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 L =2.0 H, R =20 Ω.

Hence 𝝉 =L/R =2.0/20 s =0.1 s, and

Steady-state current iₒ =4.0/20 A =0.20 A.

(a) So, from i =iₒ(1 -e^-t/𝝉)

Current after 0.20 s,

=0.20*(1-e^-0.20/0.10) A

=0.20*(1-e⁻²) A

=0.20*0.86 A

=0.17 A.

(b) The magnetic field energy at this instant

U =½Li²

=½2.0(0.17)² J

≈0.03 J.

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: R =40 Ω, Ɛ =4.0 V. t =0.10 s, i =63 mA =0.063 A.

iₒ =Ɛ/R =4.0/40 A =0.10 A

Hence, i = iₒ(1 -e^-t/𝝉)

→0.063 =0.10*(1-e^-0.10/𝝉)

→e^-0.10/𝝉 =1 -0.63 =0.37

→-0.10/𝝉 =ln 0.37 =-0.99

→𝝉 =0.10

→L/R =0.10

→L =0.10R =0.1040 H

→L =4.0 H.

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 current is switched on.

ANSWER: Steady-state current,

iₒ =2.0/100 A =0.02 A,

t =20 ms =0.02 s

𝝉 =L/R =5.0/100 s =0.05 s

Hence, i =iₒ(1-e^-t/𝝉)

→i =0.02*(1-e^-0.02/0.05)

=0.02*(1-e^-0.4)

=0.0066 A

Hence the potential difference across the resistor,

=iR

=0.0066*100 V

=0.66 V.

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: 𝝉 =40 ms =0.040 s

iₒ =2.0 A.

From i =iₒ(1 -e^-t/𝝉)

we have,

(a) At t =10 ms,

t/𝝉 =10/40 =0.25

Hence the current at t =10 ms,

i =2.0*(1 -e^-0.25) A

=0.44 A.

(b) At t = 20 ms,

t/𝝉 =20/40 =0.5

Hence the current at t =20 ms,

i =2.0*(1-e^-0.5) A

=0.79 A.

(c) At t = 100 ms,

t/𝝉 =100/40 =2.5

Hence the current at t =100 ms,

i =2.0*(1 -e^-2.5) A

=1.8 A.

(d) At t =1 s,

t/𝝉 =1/0.04 =25

Hence the current at t =1 s,

i =2.0*(1 -e⁻²⁵) A

=2.0 A.

75. An LR 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: Steady-state current,

iₒ =V/R =2.0/20 A =0.10 A

𝝉 =L/R =1.0/20 s =0.05 s

Hence the current at time t is,

i =iₒ(1 -e^-t/𝝉)

So, di/dt =iₒ{-e^-t/𝝉*(-1/𝝉)}

→di/dt =iₒe^-t/𝝉/𝝉

=0.10 e^-t/0.05/0.05 A/s

=2 e^-20t A/s

(a) at t =100 ms =0.10 s,

di/dt = 2e^-200.10 A/s

= 2*e^-2

= 0.27 A/s

(b) at t = 200 ms =0.20 s

di/dt =2e^-200.20 A/s

=2*e^-4 A/s

=0.036 A/s

(c) at t = 1.0 s

di/dt =2e^-201.0 A/s

= 2/e²⁰ A/s

= 4.1x10⁻⁹ A/s.

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

ANSWER: The magnitude of self-induced emf is given by,

Ɛ = L*di/dt

For the previous problem

di/dt =2 e^-20t A/s

(a) At t =100 ms, the calculated di/dt =0.27 A/s

Hence the value of self-induced emf at t =100 ms,

Ɛ =L*di/dt

=1.0*0.27 V

=0.27 V.

(b) At t = 200 ms, the calculated value of di/dt =0.036 A/s.

Hence the value of self-induced emf, Ɛ

Ɛ =L*di/dt

=1.0*0.036 V

=0.036 V.

(c) At t = 1.0 s, the calculated value of di/dt =4.1x10⁻⁹ A/s

Hence the value of self-induced emf,

Ɛ =L*di/dt

=1.0*4.1x10⁻⁹ V

=4.1x10⁻⁹ V.

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: L =20 mH =0.02 H,

R =10 Ω, 𝝉 =L/R =0.02/10 =0.002 s

Steady-state current,

iₒ =E/R =5.0/10 =0.5 A

At any instant t/𝝉 = t/0.002 =500t

Hence the current at any instant,

i =iₒ(1 -e^-t/𝝉)

=0.5*(1 -e⁻⁵⁰⁰^t)

Hence di/dt = 0.5{-e^-500t(-500)}

→di/dt =250e^-500t

So the self-induced emf at time t is,

Ɛ =-Ldi/dt

=-0.02*250 e^-500t

=-5 e^-500t

The rate of change of the induced emf at time t,

dƐ/dt =-5*(-500) e^-500t

=2500 e^-500t

(a) at t =0,

dƐ/dt =2500*e⁰ V/s

=2500 V/s.

=2.5x10³ V/s.

(b) At t =10 ms =0.01 s

dƐ/dt =2500e^-5000.01 V/s

=2500*e⁻⁵ V/s

≈17 V/s

(c) At t =1.0 s,

dƐ/dt =2500*e⁻⁵⁰⁰ V/s

≈ 0 V/s.

78. An LR circuit contains an inductor of 500 mH, a resistor of 25.0 Ω, and an emf of 5.0 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: L =500 mH =0.50 H,

R =25 Ω, 𝝉 =L/R =0.5/25 s =0.02 s

Steady-state current,

iₒ =5.0/25 A =0.20 A

If i is the current at any time t, then

The potential difference across the resistor =iR

=R*iₒ(1 -e^-t/𝝉)

=250.20(1-e^-50t)

=5(1 -e^-50t)

(a) At t = 20.0 ms =0.02 s

P.D. =5*(1-e⁻¹) V =3.16 V.

(b) At t =100 ms =0.10 s

P.D. =5*(1-e⁻⁵) V =4.97 V

(c) At t =1.00 s

P.D. =5*(1-e⁻⁵⁰) V

=5.00 V.

79. An inductor coil of resistance 10 Ω and inductance 120 mH is connected across a battery of emf 6.0 V and internal resistance 2 Ω. Find the charge that flows through the inductor in (a) 10 ms, (b) 20 ms, and (c) 100 ms after the connections are made.

ANSWER: R =10 Ω, r =2 Ω,

L =120 mH =0.12 H

𝝉 =L/(R+r) =0.12/12 s =0.01 s

Ɛ =6.0 V

Hence steady-state current

iₒ =6.0/(10+2) A =0.5 A.

The current at any time t is,

i =iₒ(1 -e^-t/𝝉)

The flow of charge in a small time dt is,

dQ =idt

→dQ =iₒ(1 -e^-t/𝝉)dt

Hence Q =∫dQ

=∫iₒ(1 -e^-t/𝝉)dt

=iₒ[t -e^-t/𝝉/(-1/𝝉)]

=iₒ[t +𝝉e^-t/𝝉]

Keeping the limits of integration between t =0 to t =t we get,

Q =iₒ{t+𝝉e^-t/𝝉 -𝝉}

=0.5{t+0.01 e^-100t -0.01}

=5x10⁻³{100t+e^-100t -1}

(a) Up till t=10 ms =0.01 s,

100t =100*0.01 s =1.0 s

Charge flows through the inductor,

=5x10⁻³{1+e⁻¹-1} C

=5x10⁻³/e C

=1.8x10⁻³ C

=1.8 mC.

(b) For t =20 ms =0.02 s

100t =100*0.02 s =2 s

Charge flows through the inductor in 20 ms,

Q =5x10⁻³{2 +e⁻² -1} C

=5x10⁻³*1.14 C

=5.7x10⁻³ C

=5.7 mC.

(c) For t =100 ms =0.10 s

100t =100*0.10 s =10 s

Hence the flow of charge through the inductor in 100 ms,

Q =5x10⁻³{10+e⁻¹⁰-1} C

=5x10⁻³{9+e⁻¹⁰} C

=5x10⁻³*9.00 C

=45x10⁻³ C

=45 mC.

80. An inductor coil of the inductance of 17 mH is constructed from copper wire of a length of 100 m and a cross-sectional area of 1 mm². Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper =1.7x 10⁻⁸ Ω-m.

ANSWER: Inductance, L =17 mH =0.017 H

Length of the wire, l =100 m,

The cross-sectional area of the wire, A =1 mm² = 1x10⁻⁶ m².

Given resistivity,

⍴ =1.7x10⁻⁸ Ω-m

The relation between the resistance of a wire R and resistivity is,

⍴ =R*(A/l)

→R =⍴l/A

=1.7x10⁻⁸*100/1x10⁻⁶ Ω

=1.7 Ω

Hence the time constant of the circuit,

𝜏 =L/R

=0.017/1.7 s

=0.01 s

=10 ms.

81. An LR circuit having a time constant of 50 ms is connected with an ideal battery of emf Ɛ. 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 L/R =50 ms =0.05 s,

(a) Since i =iₒ(1 -e^-tR/L)

For i =iₒ/2

iₒ/2 =iₒ(1 -e^-t/0.05)

→1 -e^-20t = 0.5

→e^-20t =0.5

→-20t =ln 0.5 =-0.693

→t = 0.035 s =35 ms.

(b) The power dissipated in heat at an instant is,

P = i²R

=iₒ²R(1 -e^-tR/L)²

=iₒ²R(1 -e^-20t)²,

while the maximum power dissipated in heat is,

Pₒ = iₒ²R

The given condition is,

P = Pₒ/2. So we have,

iₒ²R(1 -e^-20t)² =iₒ²R/2

→(1 -e^-20t)² = 1/2

→ 1-e^-20t =土1/√2 ------(i)

→e^-20t = 1 土1/√2 =1.707 or 0.292

→-20t = ln 1.707 or ln 0.292

→-20t = 0.53 or -1.23

Since for a positive value of t, the value of -20t ≠ 0.53, hence we have

-20t = -1.23

→t =0.061 s =61 ms.

(c) The magnetic field energy stored in the circuit at any time t is given as,

U =½Li²,

The maximum value of this stored energy is when the current is maximum i.e. = iₒ. So the maximum magnetic field energy stored in the circuit,

Uₒ =½Liₒ²

The given condition is,

U = ½Uₒ

→½Li² =½*½Liₒ²

→i² =½iₒ²

→iₒ²(1 -e^-20t)² =½iₒ²

→1 -e^-20t = ±1/√2

It is the same equation (i) in the (b) part of the solution above. Hence after we solve this equation we again get t = 61 ms.

82. A coil having an inductance L and a resistance R is connected to a battery of emf Ɛ. 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: At steady-state the magnetic energy stored in the circuit,

Uₒ =½Liₒ²

If t is the time to reach the magnetic field enrgy value one fourth the steady-state value and the current in the circuit at this time is i, then

¼Uₒ =½Li²

→¼*½Liₒ² =½Liₒ²(1 -e^-tR/L)²

→1 -e^-tR/L =±(1/2)

→e^-tR/L = 3/2 or 1/2

→-tR/L =ln (3/2) or ln (1/2)

(ln 3/2) will have positive value hence we can take only (ln 1/2) on the RHS. So,

-tR/L =ln 1/2 =ln 1 -ln 2 =-ln 2

→t =𝜏 ln 2

Now let t' be the time to reach the magnetic energy value half the steady-state and the current is i'. Then

U' =½Li'²

But U' =½Uₒ

→½Liₒ²(1 -e^t'R/L)² =½*½Liₒ²

→1 -e^t'R/L =±1/√2

→e^-t'R/L =1±(1/√2)

→e^-t'R/L =(√2+1)/√2 or (√2-1)/√2

→-t'R/L =ln(√2+1)/√2 or ln(√2-1)/√2

Only the later value on the RHS is negative hence,

-t'/𝜏 =ln (√2-1)/√2

→t' =𝜏 ln √2/(√2-1)

Hence the required time is =t' -t

=𝜏 ln √2/(√2-1) - 𝜏 ln 2

=𝜏 ln √2/2(√2-1)

=𝜏 ln {1/(2 -√2).

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: (a) Given, L = 4.0 H,

R =10 Ω

Hence the time-constant

𝜏 =L/R =4.0/10 s =0.40 s.

(b) We know that in one time constant the current reaches 0.63 of its steady-state value.

Hence the required time, t = 𝜏 =0.40 s.

Otherwise, we can solve it by

063 iₒ =iₒ(1-e^-t/𝜏)

→e^-t/𝜏 =1 -0.63 =0.37

→-t/0.40 =ln 0.37 = -1

→t =0.40 s

(c) In steady-state, iₒ =Ɛ/R

→iₒ =4.0/10 A =0.40 A

The power delivered by the battery at the instant when i =0.63 iₒ is

P =Ɛi

=0.63Ɛiₒ

=0.634.00.40 W

=1.0 W.

(d) The power dissipated in joule heating at this time is,

P' =i²R

=(0.63iₒ)²R

=(0.630.40)²10 W

=0.64 W.

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: Let the resistance of the coil =R and the maximum value of the current in it is iₒ. Hence the maximum value of the magnetic field inside the coil,

Bₒ =µₒniₒ

Let the current in the coil be 'i' when the value of the magnetic field inside it reaches 0.80Bₒ. So,

0.80Bₒ =µₒni

→0.80µₒniₒ =µₒni

→0.80iₒ =iₒ(1-e^-tR/L)

→0.80 =1 -e^{-20x10⁻⁶R/0.002}

→e^-0.01R =1 -0.80 =0.20

→-0.01R =ln 0.20 =-1.61

→R =1.61/0.01 Ω =161 Ω

85. An LR circuit with emf Ɛ is connected at t =0. (a) Find the charge Q which flows through the battery from 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 consistence with energy conservation.

ANSWER: (a) Current at time t after the connection,

i =iₒ(1 -e^-tR/L)

In a small time interval dt, the small amount of charge flow,

dQ =i dt

→dQ =iₒ(1 -e^-tR/L)dt

Hence total charge flow

Q =∫dQ

=iₒ∫(1-e^-tR/L)dt

=iₒ[t +(L/R)e^-tR/L]

For the limit of time from 0 to t,

Q =iₒ{t+(L/R)e^-tR/L-L/R}

=(Ɛ/R){t -(L/R)(1 -e^-tR/L)}

=(Ɛ/R){t -(L/R)(1 -x)},

where x = e^-tR/L.

(b) W =Work-done by the battery during this period = Total charge flow in this period X emf

=(Ɛ/R){t -(L/R)(1-x)}*Ɛ

=(Ɛ²/R){t -(L/R)(1-x)},

Where x =e^-tR/L

(c) Heat developed during this period

H = ∫i²Rdt

=iₒ²R∫(1 -e^-tR/L)²dt

=iₒ²R∫(1-2e^-tR/L +e^-2tR/L)dt

=iₒ²R[t +(2L/R)e^-tR/L -(L/2R)e^-2tR/L]

Putting the limit of time between 0 to t we get,

H =(Ɛ²/R²)R{t+(2L/R)x-(L/2R)x²-(2L/R) +(L/2R)}

=(Ɛ²/R){t -(L/2R)(x²-1+4-4x)}

=(Ɛ²/R){t -(L/2R)(3-4x+x²)}.

(d) Magnetic field energy stored in the circuit at time t,

U =½Li²

=½Liₒ²(1-e^-tR/L)²

=½L(Ɛ²/R²)(1-e^-tR/L)² +e^-2tR/L)

=(LƐ²/2R²)(1 -x)².

(e) From the law of conservation of energy, the total work done by the battery should be equal to the sum of heat developed during this period and the magnetic field energy stored in the circuit. i.e.

W =H +U

Let us verify it.

H+U =(Ɛ²/R){t-(L/2R)(3-4x+x²)} +(LƐ²/2R²)(1-x)²}

=(Ɛ²/R){t-(L/2R)(3-4x+x²)+(L/2R)(1-2x+x²)}

=(Ɛ²/R){t-(L/2R)(3-4x+x²-1+2x-x²)}

=(Ɛ²/R){t-(L/2R)(2-2x)}

=(Ɛ²/R){t-(L/R)(1-x)}

= W.

Hence verified.

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 the magnetic field.

ANSWER: (a) L =2.00 H, R =200 Ω,

𝜏 =L/R =2.00/200 s =0.01 s

Ɛ =2.0 V, So iₒ =Ɛ/R =2.0/200 A =0.01 A

t =10 ms =0.010 s, t/𝜏 = 1.

Hence the current at time t is,

i =iₒ(1-e^-t/𝜏)

=0.01(1 -e⁻¹) A

=0.0063 A

=6.3 mA.

(b) The power delivered by the battery at t =10 ms is

=Ɛi

=(2.00 V)*(6.3 mA)

=12.6 mW.

(c) The power dissipated in heating the resistor,

=i²R

=(6.3x10⁻³)²*200 W

≈80x10⁻⁴ W

=8.0 mW.

(d) The energy stored in the magnetic field at time t is

U =½Li²

=½Liₒ²(1-e^-t/𝜏)²

Hence the rate of energy storage,

dU/dt=½Liₒ²2(1-e^-t/𝜏)(-e^-t/𝜏)(-1/𝜏)

=(Liₒ²/𝜏)(1-e⁻¹)*e⁻¹

{Since t/𝜏 =1}

=(2*0.01²/0.01)(0.63)(0.37) W/s

=0.0046 W/s

=4.6 mW/s.

87. Two coils A and B have inductances of 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_B be the currents in the two circuits at time t. Find the ratio i_A/i_B at (a) t = 100 ms, (b) t =200 ms and (c) t =1 s.

ANSWER: For the coil A

𝜏 =L/H =1.0/10 s =0.10 s

For coil B, 𝜏' =2.0/10 s =0.20 s

Steady-state current for each coil,

iₒ =2.0/10 A =0.20 A

Current at any time t in the coil A,

i_A =iₒ(1 -e^-t/𝜏) =iₒ(1 -e^-10t)

and in the coil B,

i_B =iₒ(1 -e^-t/𝜏') =iₒ(1 -e^-5t)

Hence at any time t, the ratio of the currents in coils

i_A/i_B =(1 -e^-10t)/(1 -e^-5t)

(a) At t =100 ms =0.10 s,

10t =100.10 =1 and 5t =50.10 =0.50

Hence the ratio,

i_A/i_B =(1-e⁻¹)/(1-e⁻⁰῾⁵)

=0.63 A/0.39 A

=1.6

(b) At t =200 ms =0.20 s

10t =100.20 =2 and 5t =50.20 =1.0

Hence the ratio

i_A/i_B =(1-e⁻²)/(1-e⁻¹)

=0.86 A/0.63 A

≈1.4

(c) At t =1 s, 10t =10 and 5t =5

Hence the ratio

i_A/i_B =(1-e⁻¹º)/(1-e⁻⁵)

=0.999 A/0.993 A

≈ 1.

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 is 4.0 H, what is its resistance?

ANSWER: For a discharging LR circuit, the current at time t is given as,

i =iₒe⁻^t/𝜏, where tau is the time constant.

(a) Given iₒ =2.0 A,

at t =0.10 s, i =1.0 A. So

1.0 =2.0 e^-0.10/𝜏

→e^-1/10𝜏 =0.5

→-1/10𝜏 =ln 0.5 = -0.69

→6.9𝜏 =1

→𝜏 =1/6.9 =0.14 s.

(b) Given L =4.0 H. So

𝜏 =L/R

→0.14 =4.0/R

→R =4.0/0.14 Ω

=28.6 Ω

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 flew through the coil after the short-circuiting is the same as that which flows in one time constant before short-circuiting.

ANSWER: Charge flown after short-circuiting

Q =∫dQ

=∫idt

=iₒ∫e^-t/𝜏 dt

=iₒ[e^-t/𝜏]*(-𝜏)

=-iₒ𝜏[e^-t/𝜏]

For the total charge flown through the coil after short-circuiting, we take the limit of time from t =0 to t =∞.

So, Q =-iₒ𝜏{0 -1} =iₒ𝜏

Before short-circuiting, a steady current iₒ is flowing through the coil. Hence in one time constant t =𝜏, the charge flown through the coil

Q' =iₒ*t

=iₒ*𝜏

=iₒ𝜏

=Q.

Proved.

90. Consider the circuit shown in figure (38-E33). (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? (b) Find the current through the inductor after one time constant.
Diagram for Q-90

ANSWER: (a) A long time after the switch is closed, the steady-state is reached. In this case, the current iₒ through the battery is the same as the current through the equivalent resistance R. For the parallel connection

1/R =1/R₁ +1/R₂

→R =R₁R₂/(R₁+R₂)

Hence,

iₒ =Ɛ/R

=Ɛ(R₁+R₂)/R₁R₂

(b) When the switch is again opened, the discharging circuit completes through the upper loop leaving the battery branch. In this circuit, both resistors are in series. So total resistance of the circuit, R =R₁+R₂. Inductance in the circuit is L. Hence the time constant of the discharging circuit,

𝜏 =L/R

=L/(R₁+R₂)

(c) Suppose the steady-state current through the inductor is i'.

So, i'R₁ =Ɛ

→i' =Ɛ/R₁

It will be the initial current at time t =0 for the discharging circuit. Hence after one time constant, i.e. t =𝜏, the current through the inductor is

i =i' e^-t/𝜏

=(Ɛ/R₁)e⁻¹

=Ɛ/(R₁e).

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

ANSWER: Number of turns of the solenoid, n =1000 turns/meter. Current, i =1.0 A, radius of solenoid, r =2 cm =0.02 m.

Hence the magnetic field inside it is,

B =µₒni

=4πx10⁻⁷10001.0 T

=4πx10⁻⁴ T

Energy density inside the solenoid,

u =B²/2µₒ

=16π²x10⁻⁸/(2*4πx10⁻⁷)

=0.2π J/m³

The volume of each meter of the coil,

v =πr²l

=π(0.02)²1 m³

=4πx10⁻⁴ m³

Hence the magnetic energy stored in each meter of the solenoid,

=u*v

=0.2π*4πx10⁻⁴ J

=7.9x10⁻⁴ J.

92. Consider a small cube of volume 1 mm³ at the center of a circular loop of radius 10 cm carrying a current of 4 A. Find the magnetic energy stored inside the cube.

ANSWER: In comparison to the sizes of the loop and the cube at the center, the magnetic field at the cube will be fairly uniform. The magnetic field at the center,

B =µₒi/2a

Hence the energy density at the center,

u =B²/2µₒ

=µₒ²i²/(4a²*2µₒ)

=µₒi²/(8a²)

=4πx10⁻⁷4²/(80.10²) J/m³

=8πx10⁻⁵ J/m³

The volume of the cube,

v =1 mm³ =1x10⁻⁹ m³

Hence the magnetic energy stored in the cube,

U =u*v

=8πx10⁻⁵*1x10⁻⁹ J

=8πx10⁻¹⁴ J.

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 mm³ at a distance of 10.0 cm from the wire.

ANSWER: The magnetic field at a distance of 10.0 cm from the long current-carrying wire,

B =µₒi/(2πd)

Magnetic energy density at this place,

u =B²/2µₒ

=µₒ²i²/(4π²d²*2µₒ)

=µₒi²/(8π²d²)

=4πx10⁻⁷4²/(8π²0.1²) J/m³

=2.55x10⁻⁵ J/m³

Hence the magnetic energy stored in a volume of 1.00 mm³ is,

U =2.55x10⁻⁵*1x10⁻⁹ J

=2.55x10⁻¹⁴ J.

94. The mutual inductance between two coils is 2.5 H. If the current in one coil is changed at the rate of 1 A/s what will be the emf induced in the other coil?

ANSWER: Given, M =2.5 H.

Rate of change of current in the coil,

di/dt =1 A/s.

emf induced in the other coil,

Ɛ =-M*di/dt

=-2.5*1 V

=-2.5 V.

So the emf induced in the other coil will be 2.5 V such that it will oppose the change of flux in the first coil.

95. Find the mutual inductance between the straight wire and the square loop of the figure (38-E27).
The figure for Q-95

ANSWER: We have calculated the emf induced in the coil in the solution of Q-57(b) of this chapter as,

Ɛ =(µₒaiₒ⍵.cos⍵t/2π) ln(1+a/b)

(Link - Exercises, Q51 to 60)

If M is the mutual inductance between them, then it's magnitude

Ɛ =M*di/dt

(µₒaiₒ⍵.cos⍵t/2π) ln(1+a/b) =M.d(iₒsin⍵t)/dt

→(µₒaiₒ⍵.cos⍵t/2π) ln(1+a/b) =M.iₒ⍵cos⍵t

→(µₒa/2π)*ln (1 +a/b) =M

→M = (µₒa/2π)*ln (1 +a/b)

96. Find the mutual inductance between the circular coil and the loop shown in figure (38-E8).
The figure for Q-96

ANSWER: As we have solved in Q-20 of this exercise, emf-induced

Ɛ ={½πµₒNa²a'²/(a²+x²)^3/2}*di/dt

(Link - Exercises, Q11 to Q20)

If M is the mutual inductance between the coils, then

Ɛ =M*di/dt

Comparing the above two expressions,

M =½πµₒNa²a'²/(a²+x²)^3/2.

97. A solenoid of a length of 20 cm, area of cross-section of 4.0 cm², and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-sectional area of 8.0 cm² and length of 10 cm. Find the mutual inductance between the solenoids.

ANSWER: Given, N =4000 turns. Length, l =20 cm =0.20 m, hence,

n =N/l =4000/0.20 =20000 turns/m.

Area of cross-section, A =4.0 cm² =4.0x10⁻⁴ m².

So the magnetic field inside this solenoid,

B =µₒni.

The number of turns in this solenoid =2000

So the magnetic flux through the outer solenoid

φ =BAN

=µₒni4.0x10⁻⁴2000 weber

=4πx10⁻⁷200000.8i weber

=2.0x10⁻²i weber

Hence the emf induced in the outer coil,

Ɛ =dφ/dt

→Ɛ =2.0x10⁻²*di/dt

But, Ɛ = M*di/dt

Hence the mutual inductance,

M =2.0x10⁻² H.

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

Sunday, February 25, 2024

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, Concepts of Physics, Part-II

1. A metallic loop is placed in a non-uniform magnetic field. Will an emf be induced in the loop?

ANSWER: Since there is no change of flux with time, no emf will be induced in the loop.

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.

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.

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.

The induced current in the second loop remains for a very short time. It starts as soon as the battery is connected in the first loop and a current begins to grow and ends just when the current reaches a constant value in the first loop.

According to Lenz's law, the induced current will oppose the change that has induced it, so the direction of the induced current in the second loop will be such that it will produce a flux that is opposite to the flux produced by the first one. So the loops repel each other in this short time.

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?

This induced current in the second loop starts just when the battery is disconnected and ends when the current in the first loop just becomes zero. The current remains only for this very short interval of time.

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 that shield objects inside them against varying magnetic fields.

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

ANSWER: In a magnetic field, when the pivoted aluminum bar falls, the magnetic flux over it changes due to the change of angle between the bar and the magnetic field. See the diagram below. Diagram for Q-9

Due to the change of flux, eddy currents are induced on the surface of the aluminum bar that opposes the movement (Cause of the induced eddy currents). Thus the aluminum bar falls much more slowly than a similar bar of insulating material that is free from eddy currents.

10. A metallic bob A oscillates through the space between the poles of an electromagnet (figure 38-Q1). The oscillations are more quickly damped when the circuit is on, as compared to the case when the circuit is off. Explain. The figure for Q-10

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

ANSWER: When the magnetic flux produced by one loop goes through the boundary of the second loop, an induced current appears in the second loop whenever there is a change of flux.

(a) when both the loops are placed coaxially and parallel to each other, maximum magnetic flux passes through them. Hence the mutual inductance is the largest.

(b) When the loops are placed with their planes and axes perpendicular to each other, minimum common flux passes through them. Hence in this case the mutual inductance is the smallest.

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

ANSWER: Self-inductance of a solenoid is

L =µₒn²πr²l, where l is the length of the solenoid.

Thus the self-inductance per unit length of the coil is,

L/l =µₒn²πr²

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

ANSWER: The energy density inside a solenoid is,

u =B²/(2µₒ).

In deriving this expression, it is assumed that the magnetic field throughout the volume of the solenoid is uniform and zero outside. Hence the energy density is the same at its center and near its ends.

OBJECTIVE-I

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

(a) zero

(b) ⅛⍵Bl²

(c) ½⍵Bl²

(d) B⍵l².

ANSWER: (b)

EXPLANATION: Small emf developed across a small length dr is

dƐ =vBdr = ωrBdr

→Ɛ = ∫ωrBdr

=ωB∫rdr

=ωB[r²/2]

{between limits r =0 to r =l/2.}

=½ωB*l²/4

=⅛ωBl².

Hence option (b) is correct.

2. A rod of length l rotates with 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) ½Blω²

(c) Blω²

(d) 2Blω².

ANSWER: (a)

EXPLANATION: Each end will be at ⅛ωBl² higher or lower potential than the center depending upon the directions of the magnetic field and the rotation. So both the ends are at equal potential and hence the potential difference between the ends is zero. Option (a) is correct.

3. Consider the situation shown in Figure (38-Q2). 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.

ANSWER: In a magnetic field, when the pivoted aluminum bar falls, the magnetic flux over it changes due to the change of angle between the bar and the magnetic field. See the diagram below.
Diagram for Q-9

10. A metallic bob A oscillates through the space between the poles of an electromagnet (figure 38-Q1). The oscillations are more quickly damped when the circuit is on, as compared to the case when the circuit is off. Explain.
The figure for Q-10