Mastering Physics Solutions Chapter 23 Magnetic Flux and Faraday’s Law of Induction

Mastering Physics Solutions Chapter 23 Magnetic Flux and Faraday’s Law of Induction

Mastering Physics Solutions

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1CQ
Explain the difference between a magnetic field and a magnetic flux.
Solution:
Magnetic field:
It is the amount of magnetic force experience by a charged particle moving with a velocity at a
given point in space
Magnetic flux:
It is the measure of the amount of magnetic field passing through a given area of any coil

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1P
A 0.055-T magnetic field passes through a circular ring of radius 3.1 cm at an angle of 16° with the normal. Find the magnitude of the magnetic flux through the ring.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2CQ
(Answers to odd-numbered Conceptual Questions can be found in the back of the book)
A metal ring with a break in its perimeter is dropped from a field-free region of space into a region with a magnetic fieId What effect does the magnetic field have on the ring?
Solution:
An induced emf will be developed in a conductor if it is moving in a magnetic field and hence some current flows through the conductor In the case of a broken ring, the magnetic field does
induce an emf between the ends of the broken ring, but the flow of the current will be prevented along the circumference because of the break in the ring.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3CQ
In a common classroom demonstration, a magnet is dropped down a long. vertical copper tube. The magnet moves very slowly as it moves through the tube, taking several seconds to reach the bottom Explain this behavior
Solution:
The eddy current in the copper tube produces a magnetic field that opposes the direction of fall. Due to this repulsion. the magnet falls slowly taking much time to reach the bottom.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3P
A magnetic field is oriented at an angle of 47° to the normal of a rectangular area 5.1 cm by 6.8 cm. If the magnetic flux through this surface has a magnitude of 4.8 × 10−5 T· m2, what is the strength of the magnetic field?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4CQ
Many equal-arm balances have a small metal plate attached to one of the two arms. The plate passes between the poles of a magnet mounted in the base of the balance. Explain the purpose of this arrangement.
Solution:
An electrical current is induced in a piece of metal due to the relative motion of a nearby magnet is known as eddy current
The metal plate moving between the poles of a magnet experiences eddy currents that retard its motion. This helps to damp out oscillations in the balance, resulting in the more accurate readings.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4P
Find the magnitude of the magnetic flux through the floor of a house that measures 22 m by 18 m. Assume that the Earth’s magnetic field at the location of the house has a horizontal component of 2.6 × 10−5 T pointing north, and a downward vertical component of 4.2 × 10−5 T.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5CQ

Solution:
When the switched in closed, the current in the wire coil produces a magnetic field in the iron rod. This increases the magnetic flux through the metal ring and a corresponding induced e.m.f.
The current produced by the induced e.m.f. generates a magnetic field opposite in direction to the field in the rod causing the ring to fly into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5P
The magnetic field produced by an MRI solenoid 2.5 m long and 1.2 m in diameter is 1.7 T. Find the magnitude of the magnetic flux through the core of this solenoid.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6CQ

Solution:
The break prevents a current from circulating around the ring. This in turn will prevent the ring from experiencing a force that would throw it into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6P
At a certain location, the Earth’s magnetic field has a magnitude of 5.9 × 10−5 T and points in a direction that is 72° below the horizontal. Find the magnitude of the magnetic flux through the top of a desk at this location that measures 130 cm by 82 cm.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7CQ

Solution:
The rod initially moves to left due to the downward current. As it moves, the motional emf which it generates will begin to oppose the emf of the battery. When both the emfs are balanced, the current stops flowing in the rod, from this point it moves with constant speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7P
A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core of magnitude 1.28 × 10−4 T · m2. (a) Find the current in this solenoid, (b) How would your answer to part (a) change if the diameter of the solenoid were doubled? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8CQ
A penny is placed on edge in the powerful magnetic field of an MR1 solenoid. If the penny is tipped over, it takes several seconds for it to land on one of its faces. Explain.
Solution:
As the penny begins to tip over; there is a large change in magnetic flux due to magnetic field of the solenoid.
This change in flux produces induced current in the penny which opposes it from falling down. So it takes more seconds to land.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8P
A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1250 turns per meter and a diameter of 6.00 cm, and carries a current of 2.50 A. Find the magnetic flux through the loop when (a) L = 3.00 cm, (b) L = 6.00 cm, and (c) L = 12.0 cm.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9CQ
Recently, NASA tested a power generation system that involves connecting a small satellite to the space shuttle with a conducting wire several miles long. Explain how such a system can generate electrical power.
Solution:
Since the e.m.f. is given as the product of length of the wire, speed of the shuttle and perpendicular component of magnetic field.
The long conducting wire connected to the shuttle moves through a field can generate an induced e.m.f.
With large value of speed and length the induced emf is great enough to provide electrical power.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9P
A 0.45-T magnetic field is perpendicular to a circular loop of wire with 53 turns and a radius of 15 cm. If the magnetic field is reduced to zero in 0.12 s, what is the magnitude of the induced emf?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10CQ
Explain what happens when the angular speed of the coil in an electric generator is increased.
Solution:
When the angular speed of the coil in an electric generator is increased; the magnitude of the induced emf increases because induced emf is directly proportional to angular speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11CQ
The inductor in an RL circuit determines how long it takes for the current to reach a given value, but it has no effect on the final value of the current. Explain.
Solution:
When current reaches a given value in an RL circuit, it stops changing; the back emf in an inductor vanishes.
So the final current in the circuit is determined by the resistor and the e.m.f. of the battery.
The inductor behaves like an ideal wire with zero resistance when the current reaches a given value.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12CQ
When the switch in a circuit containing an inductor is opened, it is common for a spark to jump across the contacts of the switch. Why?
Solution:
An inductor acts to resist any change in the current whether increasing or decreasing.
When the switch in a circuit containing an inductor is opened, the inductor tries to maintain the original current, therefore the continuing current causes a spark to jump the gap.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.13P

Solution:
Given that, the figure in the given question shows four different situations in which a metal ring moves to the right with constant speed through a region with a varying magnetic field. The intensity of the color indicates the intensity of the field and in each case the field either increases or decreases at a uniform rate from the left edge of the colored region to the right edge.
For figure (1):–
The magnetic field is coming out of the page. As the ring moves and leaves the magnetic field which is coming out of the page, it will create an EMF that tries to account for this change. Hence the induced emf will be a anti clock wise or counter clockwise in order to create a magnetic field out of the page through the ring.
For figure (2):–

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.14P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.15P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.16P
A single conducting loop of wire has an area of 7.2 × 10−2 m2 and a resistance of 110 Ω. Perpendicular to the plane of the loop is a magnetic field of strength 0.48 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.32 A?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.17P
The area of a 120-turn coil oriented with its plane perpendicular to a 0.20-T magnetic field is 0.050 m2. Find the average induced emf in this coil if the magnetic field reverses its direction in 0.34 s.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.18P
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 4.25 s and the local 0.125-T magnetic field is perpendicular to the plane of the loop.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.19P
A magnetic field increases from 0 to 0.25 T in 1.8 s. How many turns of wire are needed in a circular coil 12 cm in diameter to produce an induced emf of 6.0 V?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.20P

Solution:
(a)
According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. So the induced current is in
clock wise direction.
At location (2) the ring is in the uniform magnetic field. So the induced current is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced current is in
counter clock wise direction.
(b) So the at location (1) the induced current is clock wise direction and, at location (2) it is zero and at location (3) it is in counter clock wise direction.
Hence best explanation is (I).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.21P

Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. The induced magnetic field opposes the ring so it is in upward direction.
At location (2) the ring is in the uniform magnetic field. So the force is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced magnetic field opposes the ring to leave from field .So it is in upward direction.
(b) So the at location (1) the force is in upward direction and, at location (2) it is zero and at location (3) it is in upward direction.
Hence best explanation is (III)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.22P

(a) Is the retarding effect of eddy currents on the solid disk greater than, less than, or equal to the retarding effect on the slotted disk?
(b) Choose the best explanation from among the following:
I. The solid disk experiences a greater retarding force because eddy currents in it flow freely and are not interrupted by the slots.
II. The slotted disk experiences the greater retarding force because the slots allow more magnetic field to penetrate the disk.
III. The disks are the same size and made of the same material· therefore, they experience the same retarding force.
Solution:
(a) The retarding effect is greatest on the solid disk. Because the holes in the slotted disk interrupt the flow of eddy currents, and it is the eddy currents that produce the opposing magnetic field responsible for the retarding effect.
(b) The retarding effect is greatest on the solid disk. So the solid disk experiences a greater retarding force.
So the best explanation is (I)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.23P

Solution:
When the disk swing to the right as far as it can go, it is still in the constant magnetic field. Therefore the change in the magnetic field during its swing is zero. So the induced current in the disk is minimum (zero).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.24P

Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
As the solid disk is swings from right to left where the magnetic the field is in to the page. So the induced current should produce the induced magnetic field which is opposite to the filed, that is filed must directed out of the page. Hence the induced current is in counter clock wise direction.
(b)The induced current should be in counter clock wise direction in order oppose the field by pointing out of the page.
Best explanation is(II)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.25P
A bar magnet with its north pole pointing downward is falling toward the center of a horizontal conducting ring. As viewed from above, is the direction of the induced current in the ring clockwise or counterclockwise? Explain.
Solution:
According to the Lenz’s law, the polarity of the induced EMF is such that it opposes the change in the magnetic flux which is the cause of its production.
When the north pole of a bar magnet is moving downward then the amount of magnetic flux linked with the ring increases. So, the current induced in such a direction that it opposes the increases in flux. This will happen when current is in the counterclockwise direction. Hence, the direction of the induced current is in the Counterclockwise direction.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.26P

Solution:

a) When the loop is above the magnet , the magnetic field increases and is directed
out of the page. According to Lenz law the current in the loop will oppose the
Increasing field by flowing clockwise
b) When the loop is below the magnet, the magnetic field decreases and is directed out of the page. According to Lenz law the current opposes the decreasing magnetic field by flowing in counter clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.27P

Solution:
a) The poles of the loop field line up with the magnet causing repulsive force, resisting to the loop which is moving down towards the magnet. So the tension in the string is less than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction force , resisting to the loop to move down away from magnet So the tension in the string is again less than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.28P

Solution:
a) The poles of the loop field line up with the magnet causing repulsion and resisting to the magnet. So the tension in the string is greater than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction and resisting the loop to move upward away from the magnet. So the tension again greater than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.29P

Solution:
a) Since the current in the wire is constant, the magnetic field does not vary with Time, so the induced current is zero
b) Since the current in the wire is increasing, the magnetic field through the circuit is increasing. Since the magnetic field is directed out of the page, the induced, is the circuit will induce magnetic field into the page. So the current flows is clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.30P

Solution:

If the current in the wire changes its direction, the direction of magnetic field will be into the page. According to Lenz law, the current induced in the circuit will oppose this change by flowing counter clock wise generating a field which is directed out of the page.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1CQ
Explain the difference between a magnetic field and a magnetic flux.
Solution:
Magnetic field:
It is the amount of magnetic force experience by a charged particle moving with a velocity at a
given point in space
Magnetic flux:
It is the measure of the amount of magnetic field passing through a given area of any coil

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1P
A 0.055-T magnetic field passes through a circular ring of radius 3.1 cm at an angle of 16° with the normal. Find the magnitude of the magnetic flux through the ring.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2CQ
(Answers to odd-numbered Conceptual Questions can be found in the back of the book)
A metal ring with a break in its perimeter is dropped from a field-free region of space into a region with a magnetic fieId What effect does the magnetic field have on the ring?
Solution:
An induced emf will be developed in a conductor if it is moving in a magnetic field and hence some current flows through the conductor In the case of a broken ring, the magnetic field does
induce an emf between the ends of the broken ring, but the flow of the current will be prevented along the circumference because of the break in the ring.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3CQ
In a common classroom demonstration, a magnet is dropped down a long. vertical copper tube. The magnet moves very slowly as it moves through the tube, taking several seconds to reach the bottom Explain this behavior
Solution:
The eddy current in the copper tube produces a magnetic field that opposes the direction of fall. Due to this repulsion. the magnet falls slowly taking much time to reach the bottom.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3P
A magnetic field is oriented at an angle of 47° to the normal of a rectangular area 5.1 cm by 6.8 cm. If the magnetic flux through this surface has a magnitude of 4.8 × 10−5 T· m2, what is the strength of the magnetic field?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4CQ
Many equal-arm balances have a small metal plate attached to one of the two arms. The plate passes between the poles of a magnet mounted in the base of the balance. Explain the purpose of this arrangement.
Solution:
An electrical current is induced in a piece of metal due to the relative motion of a nearby magnet is known as eddy current
The metal plate moving between the poles of a magnet experiences eddy currents that retard its motion. This helps to damp out oscillations in the balance, resulting in the more accurate readings.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4P
Find the magnitude of the magnetic flux through the floor of a house that measures 22 m by 18 m. Assume that the Earth’s magnetic field at the location of the house has a horizontal component of 2.6 × 10−5 T pointing north, and a downward vertical component of 4.2 × 10−5 T.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5CQ

Solution:
When the switched in closed, the current in the wire coil produces a magnetic field in the iron rod. This increases the magnetic flux through the metal ring and a corresponding induced e.m.f.
The current produced by the induced e.m.f. generates a magnetic field opposite in direction to the field in the rod causing the ring to fly into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5P
The magnetic field produced by an MRI solenoid 2.5 m long and 1.2 m in diameter is 1.7 T. Find the magnitude of the magnetic flux through the core of this solenoid.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6CQ

Solution:
The break prevents a current from circulating around the ring. This in turn will prevent the ring from experiencing a force that would throw it into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6P
At a certain location, the Earth’s magnetic field has a magnitude of 5.9 × 10−5 T and points in a direction that is 72° below the horizontal. Find the magnitude of the magnetic flux through the top of a desk at this location that measures 130 cm by 82 cm.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7CQ

Solution:
The rod initially moves to left due to the downward current. As it moves, the motional emf which it generates will begin to oppose the emf of the battery. When both the emfs are balanced, the current stops flowing in the rod, from this point it moves with constant speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7P
A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core of magnitude 1.28 × 10−4 T · m2. (a) Find the current in this solenoid, (b) How would your answer to part (a) change if the diameter of the solenoid were doubled? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8CQ
A penny is placed on edge in the powerful magnetic field of an MR1 solenoid. If the penny is tipped over, it takes several seconds for it to land on one of its faces. Explain.
Solution:
As the penny begins to tip over; there is a large change in magnetic flux due to magnetic field of the solenoid.
This change in flux produces induced current in the penny which opposes it from falling down. So it takes more seconds to land.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8P
A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1250 turns per meter and a diameter of 6.00 cm, and carries a current of 2.50 A. Find the magnetic flux through the loop when (a) L = 3.00 cm, (b) L = 6.00 cm, and (c) L = 12.0 cm.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9CQ
Recently, NASA tested a power generation system that involves connecting a small satellite to the space shuttle with a conducting wire several miles long. Explain how such a system can generate electrical power.
Solution:
Since the e.m.f. is given as the product of length of the wire, speed of the shuttle and perpendicular component of magnetic field.
The long conducting wire connected to the shuttle moves through a field can generate an induced e.m.f.
With large value of speed and length the induced emf is great enough to provide electrical power.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9P
A 0.45-T magnetic field is perpendicular to a circular loop of wire with 53 turns and a radius of 15 cm. If the magnetic field is reduced to zero in 0.12 s, what is the magnitude of the induced emf?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10CQ
Explain what happens when the angular speed of the coil in an electric generator is increased.
Solution:
When the angular speed of the coil in an electric generator is increased; the magnitude of the induced emf increases because induced emf is directly proportional to angular speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11CQ
The inductor in an RL circuit determines how long it takes for the current to reach a given value, but it has no effect on the final value of the current. Explain.
Solution:
When current reaches a given value in an RL circuit, it stops changing; the back emf in an inductor vanishes.
So the final current in the circuit is determined by the resistor and the e.m.f. of the battery.
The inductor behaves like an ideal wire with zero resistance when the current reaches a given value.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12CQ
When the switch in a circuit containing an inductor is opened, it is common for a spark to jump across the contacts of the switch. Why?
Solution:
An inductor acts to resist any change in the current whether increasing or decreasing.
When the switch in a circuit containing an inductor is opened, the inductor tries to maintain the original current, therefore the continuing current causes a spark to jump the gap.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.13P

Solution:
Given that, the figure in the given question shows four different situations in which a metal ring moves to the right with constant speed through a region with a varying magnetic field. The intensity of the color indicates the intensity of the field and in each case the field either increases or decreases at a uniform rate from the left edge of the colored region to the right edge.
For figure (1):–
The magnetic field is coming out of the page. As the ring moves and leaves the magnetic field which is coming out of the page, it will create an EMF that tries to account for this change. Hence the induced emf will be a anti clock wise or counter clockwise in order to create a magnetic field out of the page through the ring.
For figure (2):–

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.14P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.15P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.16P
A single conducting loop of wire has an area of 7.2 × 10−2 m2 and a resistance of 110 Ω. Perpendicular to the plane of the loop is a magnetic field of strength 0.48 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.32 A?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.17P
The area of a 120-turn coil oriented with its plane perpendicular to a 0.20-T magnetic field is 0.050 m2. Find the average induced emf in this coil if the magnetic field reverses its direction in 0.34 s.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.18P
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 4.25 s and the local 0.125-T magnetic field is perpendicular to the plane of the loop.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.19P
A magnetic field increases from 0 to 0.25 T in 1.8 s. How many turns of wire are needed in a circular coil 12 cm in diameter to produce an induced emf of 6.0 V?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.20P

Solution:
(a)
According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. So the induced current is in
clock wise direction.
At location (2) the ring is in the uniform magnetic field. So the induced current is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced current is in
counter clock wise direction.
(b) So the at location (1) the induced current is clock wise direction and, at location (2) it is zero and at location (3) it is in counter clock wise direction.
Hence best explanation is (I).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.21P

Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. The induced magnetic field opposes the ring so it is in upward direction.
At location (2) the ring is in the uniform magnetic field. So the force is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced magnetic field opposes the ring to leave from field .So it is in upward direction.
(b) So the at location (1) the force is in upward direction and, at location (2) it is zero and at location (3) it is in upward direction.
Hence best explanation is (III)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.22P

(a) Is the retarding effect of eddy currents on the solid disk greater than, less than, or equal to the retarding effect on the slotted disk?
(b) Choose the best explanation from among the following:
I. The solid disk experiences a greater retarding force because eddy currents in it flow freely and are not interrupted by the slots.
II. The slotted disk experiences the greater retarding force because the slots allow more magnetic field to penetrate the disk.
III. The disks are the same size and made of the same material· therefore, they experience the same retarding force.
Solution:
(a) The retarding effect is greatest on the solid disk. Because the holes in the slotted disk interrupt the flow of eddy currents, and it is the eddy currents that produce the opposing magnetic field responsible for the retarding effect.
(b) The retarding effect is greatest on the solid disk. So the solid disk experiences a greater retarding force.
So the best explanation is (I)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.23P

Solution:
When the disk swing to the right as far as it can go, it is still in the constant magnetic field. Therefore the change in the magnetic field during its swing is zero. So the induced current in the disk is minimum (zero).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.24P

Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
As the solid disk is swings from right to left where the magnetic the field is in to the page. So the induced current should produce the induced magnetic field which is opposite to the filed, that is filed must directed out of the page. Hence the induced current is in counter clock wise direction.
(b)The induced current should be in counter clock wise direction in order oppose the field by pointing out of the page.
Best explanation is(II)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.25P
A bar magnet with its north pole pointing downward is falling toward the center of a horizontal conducting ring. As viewed from above, is the direction of the induced current in the ring clockwise or counterclockwise? Explain.
Solution:
According to the Lenz’s law, the polarity of the induced EMF is such that it opposes the change in the magnetic flux which is the cause of its production.
When the north pole of a bar magnet is moving downward then the amount of magnetic flux linked with the ring increases. So, the current induced in such a direction that it opposes the increases in flux. This will happen when current is in the counterclockwise direction. Hence, the direction of the induced current is in the Counterclockwise direction.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.26P

Solution:

a) When the loop is above the magnet , the magnetic field increases and is directed
out of the page. According to Lenz law the current in the loop will oppose the
Increasing field by flowing clockwise
b) When the loop is below the magnet, the magnetic field decreases and is directed out of the page. According to Lenz law the current opposes the decreasing magnetic field by flowing in counter clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.27P

Solution:
a) The poles of the loop field line up with the magnet causing repulsive force, resisting to the loop which is moving down towards the magnet. So the tension in the string is less than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction force , resisting to the loop to move down away from magnet So the tension in the string is again less than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.28P

Solution:
a) The poles of the loop field line up with the magnet causing repulsion and resisting to the magnet. So the tension in the string is greater than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction and resisting the loop to move upward away from the magnet. So the tension again greater than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.29P

Solution:
a) Since the current in the wire is constant, the magnetic field does not vary with Time, so the induced current is zero
b) Since the current in the wire is increasing, the magnetic field through the circuit is increasing. Since the magnetic field is directed out of the page, the induced, is the circuit will induce magnetic field into the page. So the current flows is clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.30P

Solution:

If the current in the wire changes its direction, the direction of magnetic field will be into the page. According to Lenz law, the current induced in the circuit will oppose this change by flowing counter clock wise generating a field which is directed out of the page.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.31P
A long, straight, current-carrying wire passes through the center of a circular coil. The wire is perpendicular to the plane of the coil, (a) If the current in the wire is constant, is the induced emf in the coil zero or nonzero? Explain, (b) If the current in the wire increases, is the induced emf in the coil zero or nonzero? Explain, (c) Does your answer to part (b) change if the wire no longer passes through the center of the coil but is still perpendicular to its plane? Explain.
Solution:
a) The magnetic field is parallel to the plane of the loop so the induced emf is zero
b) Though the current increases , the magnetic field is still parallel to the plane of the loop so the induced emf is zero
c)
Since the magnetic field is still parallel to the plane of the loop. So the answer part (b) does not change

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.32P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.33P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.34P

Solution:
Current carrying conductor produces magnetic field around the conductor The direction of magnetic force is given by magnetic-field right hand rule According to right hand rule, point your right hand thumb in the direction of current and curl your ringers around the wire The direction of ringers gives the direction of magnetic fieId

As shown in the above figure, the magnetic field due to current carrying conductor at ring A is out of the page and increasing. According to Faraday’s law, this increasing magnetic field creates the induced emf in the ring A.
According to Lenz’s law, the direction of magnetic field created by induced emf must oppose the original magnetic field. So this magnetic field due to induced emf must be into the page.
From right hand rule, to create the magnetic field that directed into the page, the induced current must directed clock wise.
Magnetic field in upper half of the ring B directed out of the page and lower half of the ring directed into the page. Hence the net magnetic field is equal to zero. So the induced emf in this loop is equal to zero.
As shown in the above figure, the magnetic field due to current carrying conductor at ring B is into the page and increasing. According to Faraday’s law, this increasing magnetic field creates the induced emf in the ring B.According to Lenz’s law, the direction of magnetic field created by induced emf must oppose the original magnetic field. So this magnetic field due to induced emf must be out of the page.
From right hand rule, to create the magnetic field that directed out the page, the induced current must directed counter clock wise.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.35P
A conducting rod slides on two wires in a region with a magnetic field. The two wires arc not connected. Is a force required to keep the rod moving with constant speed? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.36P
A metal rod 0.76 m long moves with a speed Of 2.0 m/s perpendicular to a magnetic field. If the induced emf between the ends of the rod is 0.45 V, what is the strength of the magnetic field?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.37P
A Boeing KC-135A airplane has a wingspan of 39.9 m and flies at constant altitude in a northerly direction with a speed of 850 km/h. If the vertical component of the Earth’s magnetic field is 5.0 × 10−6 T, and its horizontal component is 1.4 × 10−6 T, what is the induced emf between the wing tips?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.38P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.39P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.40P
(a) Find the current that flows in the circuit shown in Example. (b) What speed must the rod have if the current in the circuit is to be 1.0 A?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.41P
Suppose the mechanical power delivered to the rod in Example is 8.9 W. Find (a) the current in the circuit and (b) the speed of the rod.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.42P
The maximum induced em f in a generator rotating at 210 rpm is 45 V. Flow fast must the rotor of the generator rotate if it is to generate a maximum induced emf of 55 V?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.43P
A rectangular coil 25 cm by 35 cm has 120 turns. This coil produces a maximum emf of 65 V when it rotates with an angular speed of 190 rad/s in a magnetic field of strength B. Find the value of B.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.44P
A 1.6-m wire is wound into a coil with a radius of 3.2 cm. If this coil is rotated at 85 rpm in a 0.075-T magnetic field, what is its maximum emf?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.45P
A circular coil with a diameter of 22.0 cm and 155 turns rotates about a vertical axis with an angular speed of 1250 rpm. The only magnetic field in this system is that of the Earth. At the location of the coil, the horizontal component of the magnetic field is 3.80 × 10-5 T, and the vertical component is 2.85 × 10−5 T. (a) Which component of the magnetic field is important when calculating the induced emf in this coil? Explain, (b) Find the maximum emf induced in the coil.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.46P
A generator is designed to produce a maximum emf of 170 V while rotating with an angular speed of 3600 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.050 T, how many turns of wire are needed?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.47P
Find the induced em f when the current ill a 45.0-mH inductor increases from 0 to 515 mA in 16.5 ms.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.48P
How many turns should a solenoid of cross-sectional area 0.035 m2 and length 0.22 m have if its inductance is to be 45 mH?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.49P
The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoid? (b) What is the induced emf in the solenoid if its current drops from 3.2 A to 0 in 55 ms?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.50P
Determine the inductance of a solenoid with 640 turns in a length of 25 cm. The circular cross section of the solenoid has a radius of 4.3 cm.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.51P
A solenoid with a cross-sectional area of 1.81 × 10−3 m2 is 0.750 m long and has 455 turns per meter. Find the induced emf in this solenoid if the current in it is increased from 0 to 2.00 A in 45.5 ms.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.52P
A solenoid has N turns of area A distributed uniformly along its length, ℓ. When the current in this solenoid increases at the rate of 2.0 A/s, an induced emf of 75 mV is observed, (a) What is the inductance of this solenoid? (b) Suppose the spacing between coils is doubled. The result is a solenoid that is twice as long but with the same area and number of turns. Will the induced emf in this new solenoid be greater than, less than, or equal to 75 mV when the current changes at the rate of 2.0 A/s? Explain, (c) Calculate the induccd emf for part (b).
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.53P
How long does it take for the current in an RL circuit with R = 130 Ω and L = 68 mH to reach half its final value?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.54P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.55P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.56P
The current in an RL circuit increases to 95% of its filial value 2.24 s after the switch is closed, (a) What is the time constant for this circuit? (b) If the inductance in the circuit is 0.275 H, what is the resistance?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.57P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.58P
The number of turns per meter in a solenoid of fixed length is doubled. At the same time, the current in the solenoid is halved. Does the energy stored in the inductor increase, decrease, or stay the same? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.59P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.60P
A solenoid is 1.5 m long and has 470 turns per meter. What is the cross-sectional area of this solenoid if it stores 0.31 J of energy when it carries a current of 12 A?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.61P
In the Alcator fusion experiment at MIT, a magnetic field of 50.0 T is produced, (a) What is the magnetic energy density in this field? (b) Find the magnitude of the electric field that would have the same energy density found in part (a).
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.62P

Solution:


b) From the above equation, the energy stored in an inductor is inversely proportional to the square of the equivalent resistance. So the value of R should be less to store more energy in the inductor.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.63P

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.64P
Consider the circuit shown in Figure, which contains a 6.0-V battery, a 37-mH inductor, and four 55-Ω resistors, (a) Is more energy stored in the inductor just elfter the switch is closed or long after the switch is closed? Explain, (b) Calculate the energy stored in the inductor one characteristic time interval after the switch is closed, (c) Calculate the energy stored in the inductor long after the switch is closed.

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.65P
You would like to store 9.9 J of energy in the magnetic field of a solenoid. The solenoid has 580 circular turns of diameter 7.2 cm distributed uniformly along its 28-cm length, (a) How much current is needed? (b) What is the magnitude of the magnetic field inside the solenoid? (c) What is the energy density (energy/volume) inside the solenoid?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.66P
Transformer 1 has a primary voltage Vp, and a secondary voltage Vs. Transformer 2 has twice the number of turns on both its primary and secondary coils compared with transformer 1. If the primary voltage on transformer 2 is 2Vp, what is its secondary voltage? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.67P
Transformer 1 has a primary current Ip and a secondary current Is. Transformer 2 has twice as many turns on its primary coil as transformer 1, and both transformers have the same number of turns on the secondary coil. If the primary current on transformer 2 is 3Ip, what is its secondary current? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.68P
The electric motor in a toy train requires a voltage of 3.0 V. Find the ratio of turns on the primary coil to turns on the secondary coil in a transformer that will step the 110-V household voltage down to 3.0 V.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.69P
A disk drive plugged into a 120-V outlet operates on a voltage of 9.0 V. The transformer that powers the disk drive has 125 turns on its primary coil, (a) Should the number of turns on the secondary coil be greater than or less than 125? Explain, (b) Find the number of turns on the secondary coil.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.70P
A transformer with a turns ratio (secondary/primary) of 1:18 is used to step down the voltage from a 120-V wall socket to be used in a battery recharging unit. What is the voltage supplied to the recharger?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.71P
A neon sign that requires a voltage of 11,000 V is plugged into a 120-V wall outlet. What turns ratio (secondary/primary) must a transformer have to power the sign?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.72P
A step-down transformer produces a voltage of 6.0 V across the secondary coil when the voltage across the primary coil is 120 V. What voltage appears across the primary coil of this transformer if 120 V is applied to the secondary coil?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.73P
A step-up transformer has 25 turns on the primary coil and 750 turns on the secondary coil. If this transformer is to produce an output of 4800 V with a 12-mA current, what input current and voltage are needed?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.74GP
An airplane flies level to the ground toward the north pole, (a) Is the induced emf from wing tip to wing tip when the plane is a t the equator greater them, less than, or equal to the wing-tip-to-wing-tip emf when it is at the latitude of New York? (b) Choose the best explanation from among the following:
I. The induced emf is the same because the strength of the Earth’s magnetic field is the same at the equator and at New York.
II. The induced emf is greater at New York because the vertical component of the Earth’s magnetic field is greater there than at the equator.
III. The induced emf is less at New York because at the equator the plane is flying parallel to the magnetic field lines.
Solution:
(a) At equator the plane is moving in the magnetic field direction so the induced emf is low.
When plane flying at the it moving perpendicular to the field as it is in latitude so the induced emf is high.
(Since the rate of change of magnetic flux for the air plane’s wings is the product of downward component of the magnetic field, wingspan and speed of plane.)
(b) At latitude of the vertical component of earth magnetic field is high and it is low at equator, so the induced emf is high at latitude of .
Hence best explanation is .(II)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.75GP
You hold a circular loop of wire at the north magnetic pole of the Earth. Consider the magnetic flux through this loop due to the Earth’s magnetic field. Is the flux when the normal to the loop points horizontally greater than, less than, or equal to the flux when the normal points vertically downward? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.76GP
You hold a circular loop of wire at the equator. Consider the magnetic flux through this loop due to the Earth’s magnetic field. Is the flux when the normal to the loop points north greater than, less than, or equal to the flux when the normal points vertically upward? Explain.
Solution:
The flux through the loop is the greatest when the normal to the loop points the direction of the field.
We know that at the equator the field points to the north. When the normal to the loop points north, the direction of magnetic field and the normal are parallel. So the flux is maximum. But when the normal to the loop points vertically up, then the direction of magnetic field and the normal are perpendicular. Then the magnetic flux becomes zero. So flux when the normal point north is greater than when the normal points vertical upwards.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.77GP
The inductor shown in Figure is connected to an electric circuit with a changing current. At the moment in question, the inductor has an induced emf with the indicated direction. Is the current in the circuit at this time increasing and to the right, increasing and to the left, decreasing and to the right, or decreasing and to the left?

Solution:

When the inductor has an inducei emf with the indicated direction, then the current in the inductor increases in the direction opposite to the increase in current of the circuit. Therefore. the current in the circuit can be either increasing to the right or decreasing to the left because the current in the inductor is directed towards left, that is. it is leaving the positive terminal

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.78GP
The Voyager I spacecraft moves through interstellar space with a speed of 8.0 × 103 m/s. The magnetic field in this region of space has a magnitude of 2.0 × 10-10 T. Assuming that the 5.0-m-long antenna on the spacecraft is at right angles to the magnetic field, find the induced emf between its ends.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.79GP
The coils used to measure the movements of a blowfly, as described in Section 23-5, have a diameter of 2.0 mm. In addition, the fly is immersed in a magnetic field of magnitude 0.15 mT. Find the maximum magnetic flux experienced by one of these coils.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.80GP
Computerized jaw tracking, or electrognathography (EGN), is air important tool for diagnosing and treating temporomandibular disorders (TMDs) that affect a person’s ability to bite effectively. The first step in applying EGN is to attach a small permanent magnet to the patient’s gum below the lower incisors. Then, as the jaw undergoes a biting motion, the resulting change in magnetic flux is picked up by wire coils placed on either side of the mouth, as shown in Figure. Suppose this person’s jaw moves to her right and that the north pole of the permanent magnet also points to her right. From her point of view, is the induced current in the coil to (a) her right and (b) her left clockwise or counterclockwise? Explain.

Solution:
Given that, computerized jaw tracking or electrognathograph (EGN) is an important tool for diagnosing and treating temporomandibular disorders (TMDs) that affect a person’s ability to bite effectively.
The first step in applying EGN is to attach a small permanent magnet to the patients gum below the lower incisors. Then, as the jaw undergoes a biting motion, the resulting change in magnetic flux is picked up by wire coils place don either side of the mouth, as shown in figure given in the question.
Suppose this person’s jaw moves to her right and that the north pole of the permanent magnet also points to her right.
(a)
Here, the magnetic field lines leave the north pole and enter the south pole, there is a change in the magnetic flux, specifically and increase to the right. This change by decreasing the magnetic flux on her left and increasing on her right will be accounted by the coils. Thus, from her point of view, the current in the coil on her right will turn counter clock wise, forming a magnetic field that opposes the field created by the magnet in her teeth
(b)
To her left, the current in the coil will turn anti-clock wise forming a magnetic field that accounts for the increase in the magnetic flux on her left side.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.81GP
A rectangular loop of wire 24 cm by 72 cm is bent into em L shape, as shown in Figure. The magnetic field in the vicinity of the loop has a magnitude of 0.035 Tand points in a direction 25° below the y axis. The magnetic field has no x component. Find the magnitude of the magnetic flux through the loop.

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.82GP
A circular loop with a radius of 3.7 cm lies in the x-y plane. The magnetic field in this region of space is uniform and given by (a) What is the magnitude of the magnetic flux through this loop? (b) Suppose we now increase the x component of , leaving the other components unchanged. Does the magnitude of the magnetic flux increase, decrease, or stay the same? Explain, (c) Suppose, instead, that we increase the z component of , leaving the other components unchanged. Does the magnitude of the magnetic flux increase, decrease, or stay the same? Explain.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.83GP
Consider a rectangular loop of wire 5.8 cm by 8.2 cm in a uniform magnetic field of magnitude 1.3 T. The loop is rotated from a position of zero magnetic flux to a position of maximum flux in 21 ms. What is the average induced emf in the loop?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.84GP
A car with a vertical radio antenna 85 cm long drives due east at 25 m/s. The Earth’s magnetic field at this location has a magnitude of 5.9 × 10−5 T and points northward, 72° below the horizontal, (a) Ts the top or the bottom of the antenna at the higher potential? Explain, (b) Find the induced emf between the ends of the antenna.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.85GP
The rectangular coils in a 325-turn generator are 11 cm by 17 cm. What is the maximum emf produced by this generator when it rotates with an angular speed of 525 rpm in a magnetic field of 0.45 T?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.86GP
A cubical box 22 cm 011 a side is placed in a uniform 0.35-T magnetic field. Find the net magnetic flux through the box.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.87GP
Transcranial magnetic stimulation (IMS) is a non invasive method for studying brain function, and possibly for treatment as well. In this technique, a conducting loop is held near a person’s head, as shown in Figure. When the current in the loop is changed rapidly, the magnetic field it croates can change at the rate of 3.00 X 104 T/s. This rapidly changing magnetic field induces an electric current in a restricted region of the brain that can cause a finger to twitch, bright spots to appear in the visual field (magnetophosphenes), or a feeling of complete happiness to overwhelm a person. If the magnetic field changes at the previously mentioned rate over an area of 1.13 × 10 2 m2, what is the induced emf?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.88GP
A magnetic field with the time dependence shown in Figure is at right angles to a 155-tum circular coil with a diameter of 3.75 cm. What is the induced emf in the coil at (a) t = 2.50 ms, (b) t = 7.50 ms, (c) t = 15.0 ms, and (d) t = 25.0 ms?

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.89GP
You would like to construct a 50.0-mH inductor by wrapping insulated copper wire (diameter = 0.0332 cm) onto a tube with a circular cross section of radius 2.67 cm. What length of wire is required if it is wrapped onto the tube in a single, close- packed layer?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.90GP
The time constant of an RL circuit with L = 25 mH is twice the time constant of an RC circuit with C = 45 μ F. Both circuits have the same resistance R. Find (a) the value of R and (b) the time constant of the RL circuit.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.91GP
A 6.0-V battery is connected in series with a 29-mH inductor a 110-Ω resistor, and an open switch, (a) How long after the switch is closed will the current in the circuit be equal to 12 mA? (b) How much energy is stored in the inductor when the current reaches its maximum value?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.92GP
A 9.0-V battery is connected in series with a 31-mH inductor, a 180-Ω. resistor, and an open switch, (a) What is the current in the circuit 0.120 ms after the switch is closed? (b) How much energy is stored in the inductor at this time?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.93GP
Suppose the fly described in Problem 79 turns through an angle of 90° in 37 ms. If the magnetic flux through one of the coils on the insect goes from a maximum to zero during this maneuver, and the coil has 85 turns of wire, find the magnitude of the induced emf.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.94GP
A conducting rod of mass m is in contact with two vertical conducting rails separated by a distance L, as shown in Figure. The entire system is immersed in a magnetic field of magnitude B pointing out of the page. Assuming the rod slides without friction, (a) describe the motion of the rod after it is released from rest, (b) What is the direction of the induced current (clockwise or countcrclockwise) in the circuit? (c) Find the speed of the rod after it has fallen for a long time.

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.95GP
A single-turn rectangular loop of width W and length L moves parallel to its length with a speed v. The loop moves from a region with a magnetic field perpendicular to the plane of the loop to a region where the magnetic field is zero, as shown in Figure. Find the rate of change in the magnetic flux through the loop (a) before it enters the region of zero field, (b) just after it enters the region of zero field, and (c) once it is fully within the region of zero field, (d) For each of the cases considered in parts (a), (b), and (c), state whether the induced current in the loop is clockwise, counterclockwise, or zero. Explain in each case.

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.96GP
The switch in the circuit shown in Figure is open initially, (a) Find the current in the circuit a long time after the switch is closed, (b) Describe the behavior of the lightbulb from the time the switch is closed until the current reaches the value found in part (a), (c) Now, suppose the switch is opened after having been closed for a long time, if the inductor is large, it is observed that the light flashes brightly and then burns out. Explain this behavior, (d) Find the voltage across the lightbulb just before and just after the switch is opened.

Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.97GP
An electric field E and a magnetic field B have the same energy density, (a) Express the ratio E/B in terms of the fundamental constants ε0 and μ0. (b) Evaluate E/B numerically, and compare your result with the speed of light.
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.98PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.

A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
Suppose the downward vertical component of the magnetic field increases as a car drives over a loop detector. As viewed from above, is the induced current in the loop clockwise, counterclockwise, or zero?
Solution:
As the downward vertical component of magnetic field is increased due to the car, according to Lenz law the induced current opposes the increase in the magnetic field.
So from right hand rule (RHR) the current direction must be in counter clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.99PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.

FIGURE
A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
A car drives onto a loop detector and increases the downward component of the magnetic field within the loop from 1.2 × 10−5, T to 2.6 × 10−5 T in 0.38 s. What is the induced emf in the detector if it is circular, has a radius of 0.67 m, and consists of four loops of wire?
A. 0.66 × 10−4 V
B. 1.5 × 10−4 V
C. 2.1 × 10−4 V
D. 6.2 × 10−4 V
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.100PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.

FIGURE
A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
A truck drives onto a loop detector and increases the downward component of the magnetic field within the loop from 1.2 × 10−5 T to the larger value B in 0.38 s. The detector is circular, has a radius of 0.67 m, and consists of three loops of wire. What is B, given that the induced emf is 8.1 × 10−4 V?
A. 3.6 × 10−5 T
B. 7.3 × 10−5 T
C. 8.5×10−5T
D. 24 × 10−5T
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.101PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.

FIGURE
A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
Suppose a motorcycle increases the downward component of the magnetic field within a loop only from 1.2 × 10-5T to 1.9 × 10-5 T. The detector is square, is 0.75 m on a side, and has four loops of wire. Over what period of time must the magnetic field increase if it is to induce an emf of 1.4 × 10-4 V?
A. 0.028 s
B. 0.11 s
C. 0.35 s
D. 0.60 s
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.102IP
Suppose the ring is initially to the left of the field region, where there is no field, and is moving to the right. When the ring is partway into the field region, (a) is the induced current in the ring clockwise, counterclockwise, or zero, and (b) is the magnetic force exerted on the ring to the right, to the left, or zero? Explain.
Solution:
a) According to right handed rule ; the motion of the ring is towards the right
So, the current will be clock wise
b)
The magnetic forced exerted on the ring is to left the because v, B & F are mutually perpendicular to each other.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.103IP
Suppose the ring is completely inside the field region initially and is moving to the right, (a) Is the induced current in the ring clockwise, counterclockwise, or zero, and (b) is the magnetic force on the ring to the right, to the left, or zero? Explain. The ring now begins to emerge from the field region, still moving to the right, (c) Is the induced current in the ring clockwise, counterclockwise, or zero, and (d) is the magnetic force on the ring to the right, to the left, or zero? Explain.
Solution:
Given that, suppose the ring is completely inside the field region initially and is moving to the right
a) The induced current in the ring is zero actually since if the ring is completely inside the field region and moves to the right, then there is no change in the magnetic field and hence there is no change in the magnetic flux.
b) There is no force on the ring since there is no induced emf., there is no current is produced
c) If the ring is emerging from the field moving to the right, then the ring experiences a decrease in the magnetic field and hence produces or change in the magnetic flux. The ring will compensate for this change by generating a counter clockwise current to form a magnetic field out of the page.
d) As the ring leaves the magnetic field, a force will be generated to the left on the left side of the ring, while the right side of the ring will have zero force since it has dropped out of the field. Hence the net force generated by the ring will be to the left, counter acting the force moving the ring to the right.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.104IP
(a) What external force is required to give the rod a speed of 3.49 m/s, every tiling else remaining the same? (b) What is the current in the circuit in this case?
Solution:

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.105IP
Suppose the direction of the magnetic field is reversed. Everything else in the system remains the same, (a) Is the magnetic force exerted on the rod to the right, to the left, or zero? Explain, (b) Is the direction of the induced current clockwise, counterclockwise, or zero? Explain. (c) Suppose we now adjust the strength of the magnetic field until the speed of the rod is 2.49 m/s, keeping the force equal to 1.60 N. What is the new magnitude of the magnetic field?
Solution: