Answers to Even-Numbered Conceptual Questions
2. The magnetic field will have little apparent effect, because the break in the ring will prevent a current from flowing around its circumference. What the magnetic field will do, however, is produce a nonzero emf between the two sides of the break.
4. 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 more accurate readings.
6. Nothing. In this case, the break prevents a current from circulating around the ring. This, in turn, prevents the ring from experiencing a magnetic force that would propel it into the air.
8. As the penny begins to tip over, there is a large change in the magnetic flux through its surface, due to the great intensity of the MRI magnetic field. This change in magnetic flux generates an induced current in the penny that opposes its motion. As a result, the penny falls over slowly, as if it were immersed in molasses. 10. When the angular speed of the coil in an electric generator is increased, the rate at which the magnetic flux
changes increases as well. As a result, the magnitude of the induced emf produced by the generator increases. Of course, the frequency of the induced emf increases as well.
12. When the switch is opened in a circuit with an inductor, the inductor tries to maintain the original current. (In general, an inductor acts to resist any change in the current—whether the current is increasing or decreasing.) Therefore, the continuing current may cause a spark to jump the gap.
Solutions to Problems and Conceptual Exercises
1. Picture the Problem: The image shows a ring of radius 3.1 cm oriented at an angle of θ = 16º from a B = 0.055 T magnetic field.
Strategy: Solve equation 23-1 for the flux. Solution: Calculate the flux:
(
) (
)
2 4 cos 0.055 T 0.031 m cos16 1.6 10 Wb BA θ π − Φ = = ° Φ = ×Insight: The maximum flux through this coil, 1.66×10−4 Wb, occurs when the angle θ is zero.
2. Picture the Problem: The image shows a box immersed in a vertical magnetic field.
Strategy: Use equation 23-1 to calculate the flux through each side. Solution: 1. The sides of the box are parallel to the field, so the magnetic flux through the sides is zero .
2. Calculate the flux through the bottom: Φ =BAcosθ=
(
0.0250 T 0.325 m 0.120 m cos 0)(
)(
)
° = 9.75 10 Wb .× −4 Insight: The height of the box is not important in this problem.3. Picture the Problem: The image shows a rectangular loop oriented 42 degrees from a magnetic field.
Strategy: Solve equation 23-1 for the magnetic field. Solution: Calculate
the magnetic field:
(
)(
)
5 2 4.8 10 Tm 0.020 T cos 0.051 m 0.068 m cos 47 B A θ − Φ × = = = °
Insight: The minimum magnetic field that would produce this flux would occur when the rectangle is parallel to the magnetic field.
4. Picture the Problem: A house has a floor of dimensions 22 m by 18 m. The local magnetic field due to Earth has a horizontal component 2.6×10-5 T and a downward vertical component 4.2×10-5 T.
Strategy: The horizontal component of the magnetic field is parallel to the floor, so it does not contribute to the flux. Use equation 23-1 to calculate the flux using the vertical component.
Solution: Calculate the magnetic flux: BAcosθ B A
(
4.2 10 T 22 m 18 m−5)
(
)(
)
1.7 10 Wb−2 ⊥Φ = = = × = ×
Insight: The flux through the vertical walls of the house is determined by the horizontal component of the magnetic field instead of the vertical component.
5. Picture the Problem: A solenoid of diameter 1.2 m produces a magnetic field of 1.7 T.
Strategy: Multiply the magnetic field by the cross-sectional area of the solenoid to calculate the magnetic flux. Solution: Calculate the magnetic flux:
(
)
2 1.2 m cos 1.7 T cos 0 1.9 Wb 2 BA θ π ⎛ ⎞ Φ = = ⎜ ⎟ ° = ⎝ ⎠
Insight: Note that the length of the solenoid does not affect the flux through the solenoid.
6. Picture the Problem: A magnetic field of magnitude 5.9×10−5 T is directed 72° below the horizontal and passes
through a horizontal region 130 cm by 82 cm.
Strategy: Use equation 23-1 to calculate the flux, where the angle from the vertical is θ = 90°−72°.
Solution: Calculate the flux: Φ =BAcosθ=
(
5.9 10 T 1.30 m 0.82 m cos 90× −5)
(
)(
) (
° − ° =72)
6.0 10 Wb× −5 Insight: Increasing the angle from the horizontal increases the flux through the desk top. For example, if the angle were increased to 80° from the horizontal the total flux would increase to 6.2×10-5 Wb.7. Picture the Problem: When a current flows through a solenoid of diameter 15.0 cm and 375 turns per meter, a flux of 1.28×10−4 Wb is generated in the core of the solenoid.
Strategy: Divide the flux by the cross-sectional area of the solenoid to calculate the magnetic field. Then use equation 22-12 to calculate the current in the solenoid.
Solution: 1. (a) Calculate the
magnetic field inside the solenoid:
(
)
4 2 3 2 1.28 10 T m 5.64 10 T 0.0850 m B A π − − Φ × ⋅ = = = ×
2. Calculate the current:
(
)(
)
0 3 7 1 0 5.64 10 T 11.7 A 4 10 T m/A 385 m B nI B I n μ μ π − − − = × = = = × ⋅3. (b) Because the current is inversely proportional to the square of the diameter of the solenoid, doubling the diameter would multiply the answer to part (a) by a factor of one-fourth. Therefore, the current would be cut to a fourth. Insight: If the diameter were 34.0 cm, the current would become 2.91 A, which is one-fourth that found in part (a).
8. Picture the Problem: A square loop of side A is inside of a long solenoid and is oriented perpendicular to the axis of the solenoid.
Strategy: Use equation 22-12 to calculate the magnetic field of the solenoid. Then use equation 23-1, with θ=0 to calculate the flux through the rectangle. Because the magnetic field is confined within the solenoid, the area will be the smaller of the cross-sectional area of the solenoid or the area of the rectangle.
Solution: 1. Calculate the magnetic
field inside the solenoid: B=μ0nI =
(
4π×10 T m/A 1250 m−7 ⋅)(
-1)
(
2.50 A)
=3.926 mT 2. (a) Find the flux when A=3.00 cm(
A<d)
: Φ =BA=(
3.926 mT 0.0300 m)(
)
2= 3.53 10 Wb × −6 3. (b) Repeat for A=6.00 cm(
A=d)
: Φ =BA=(
3.926 mT) (
π 0.0300 m)
2 = 1.11 10 Wb × −5 4. (c) Repeat for A=12.0 cm(
A>d)
: Φ is the same as that found in part (b), 1.11 10 Wb× −5 Insight: When the length is less than d 2 , the square will completely fit within the solenoid and the flux is2. B
Φ = A When the length is greater than or equal to d, the solenoid will completely fit within the square and the flux
is
(
1 2)
4 .
B πd
Φ =
9. Picture the Problem: A coil of radius 15 cm and 53 turns is oriented perpendicular to a magnetic field. The magnetic field changes from 0.45 T to zero in 0.12 s.
Strategy: Use equation 23-3 to calculate the induced emf, with the flux given by equation 23-1.
Solution: Calculate the emf:
(
) (
)
2 0.45 T 0.15 m 0 53 14 V 0.12 s BA N N t t π
ε
ΔΦ − = = = = Δ ΔInsight: Note that the emf is inversely proportional to the time it takes for the magnetic field to change. In this case, if it dropped to zero in only 0.060 seconds (half the time), the average emf would be twice as large, or 28 V.
10. Picture the Problem: The magnetic flux through a coil oscillates in time as indicated by the graph at right.
Strategy: The magnitude of the flux is greatest when the flux is at a maximum or a minimum on the graph. The magnitude of the emf is greatest when the flux has the greatest positive or negative slope. Solution: 1. (a) The magnetic flux has its greatest magnitude at t = 0 s, 0.2 s, 0.4 s, and 0.6 s .
2. (b) The magnitude of the induced emf is greatest at t = 0.1 s, 0.3 s, and 0.5 s .
Insight: Note that the magnitude of the induced emf is zero when the magnitude of the flux is a maximum and the magnitude of the induced emf is a maximum when the flux is zero.
11. Picture the Problem: The figure shows the flux through a single loop coil as a function of time.
Strategy: Use equation 23-3 to calculate the emf at the times
t=0.05 s, 0.15 s, and 0.50 s. Use the graph to find the change in flux. Solution: 1. (a) Calculate
the emf at t=0.05 s: 10 Wb 0 0.1 kV 0.1 s N t
ε
= − ΔΦ= − − = − Δ 2. (b) Calculate the emf at t =0.15 s:ε
= 0 3. (c) Calculate the emf at t=0.50 s:ε
= −−5 Wb 10 Wb0.6 s 0.2 s−− = 0.04 kV12. Picture the Problem: A wire loop is placed in a magnetic field that is perpendicular to its plane. The field varies with time as shown at right. Strategy: Faraday’s Law states that the magnitude of the induced emf is proportional to the rate of change of the magnetic flux. In this case, the magnetic flux is proportional to the field magnitude because the area of the loop and its orientation remain constant. The rate of change of the flux is therefore determined by the rate of change of the field, which in turn is represented by the slope of the plot at the right. Use the slope of the plot to determine the ranking of the magnitude of the induced emf.
Solution: The graph region with the steepest slope (E) corresponds to the greatest induced emf, and the regions with zero slope (D and F) correspond to zero emf. Using similar reasoning we arrive at the ranking:
D = F < A < B < C < E
Insight: Note that in regions D and F the field is nonzero (and so is the flux) but the induced emf is zero because it depends on the rate of change of the flux, not the magnitude of the flux.
13. Picture the Problem: The figure at the right 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. The direction of the field in each region is indicated.
Strategy: Lenz’s Law states that an induced current always flows in a direction that opposes the change that caused it. In each of the four cases the magnetic flux through the ring changes because of its motion at constant speed to the right. Use the Right-Hand Rule for the magnetic field due to a current in order to determine in which direction the current should flow in order to oppose the change in flux.
Solution: 1. Figure 1 indicates the flux out of the page and through the ring decreases as the ring moves to the right and into a weaker field. The induced current in the ring will oppose this change and attempt to increase the out-of-the-page flux through the ring. The Right-Hand Rule indicates the current will flow counterclockwise in order to do this. 2. Figure 2 indicates the flux out of the page and through the top half of the ring decreases as the ring moves to the right and into a weaker field. However, the out-of-the-page flux through the bottom half of the ring increases as the ring moves to the right and into a stronger field. The net flux through the ring stays approximately constant and there will be no change of flux, so the induced emf in the ring is zero.
3. Figure 3 indicates the flux into the page and through the top half of the ring decreases as the ring moves to the right and into a weaker field. Meanwhile, the out-of-the-page flux through the bottom half of the ring also decreases as the ring moves to the right and into a weaker field. The net flux through the ring stays approximately constant (the flux is approximately zero) and there will be no change of flux, so the induced emf in the ring is zero.
4. Figure 4 indicates the flux into the page and through the top half of the ring decreases as the ring moves to the right and into a weaker field. Meanwhile, the out-of-the-page flux through the bottom half of the ring increases as the ring moves to the right and into a stronger field. The net flux through the ring increases in the out-of-the-page direction. The induced current in the ring will oppose this change and attempt to increase the into-the-page flux through the ring. The Right-Hand Rule indicates the current will flow clockwise in order to do this.
Insight: If the ring in figure 4 were to move toward the top of the page, the bottom half of the ring would leave the out-of-the-page field region and enter the into-the-page region. This would increase the into-the-page flux through the ring, and the induced current would oppose this change and flow in the counterclockwise direction.
14. Picture the Problem: The image shows the emf through a single loop as a function of time.
Strategy: Use equation 23-3 to calculate the emf at the times t=0.25 s and t=0.55 s
Solution: 1. (a) The flux at t = 0.25 s is about 8 Wb. This is greater than the flux at t = 0.55 s, which is about −3 Wb.
2. (b) The two emf values are the same, because at those two times the flux is changing at the same rate.
3. (c) Calculate the slope of the flux between 0.2 s and 0.6 s:
5 Wb 10 Wb 37.5 Wb/s 0.6 s 0.2 s t ΔΦ − − = = − Δ −
4. Calculate the induced emf : N 1 37.5 Wb/s
(
)
0.04 kVt
ε
ΔΦ= − = − − =
Δ
Insight: Note that the emf is zero for times 0.1 s < t < 0.2 s and t > 0.6 s. The voltage is not determined by the magnitude of the flux but by the slope of the flux vs. time graph. For these two time periods the slope is zero.
15. Picture the Problem: The magnetic flux through a coil oscillates in time as indicated by the graph at the right.
Strategy: Use equation 23-4 to estimate the magnitude of the emf from the graph.
Solution: 1. (a) The magnitude of the induced emf is greater near t = 0.5 s, because Φ is changing more rapidly there. (In fact, it is not changing at all at the instant t = 0.4 s.)
2. (b) The induced emf will have maximum magnitude where Φ is changing most rapidly: 0.1 s, 0.3 s, 0.5 s, ... 3. (c) Estimate the emf at t = 0.3 s: 1
(
2 Wb) (
2 Wb)
0.06 kV0.333 s 0.267 s N t
ε
= − ΔΦ≈ − ⎡ − − ⎤= − ⎢ ⎥ Δ ⎣ − ⎦4. Estimate the emf at t = 0.4 s:
ε
= 0 because 0 t ΔΦ=Δ
5. Estimate the emf at t = 0.5 s: 1
(
2 Wb) (
2 Wb)
0.06 kV 0.533 s 0.467 sN t
ε
= − ΔΦ≈ − ⎡⎢ − − ⎤⎥=Δ ⎣ − ⎦
Insight: The magnitude of the emf is the same at t = 0.3 s and t = 0.5 s, but the polarity is opposite. 16. Picture the Problem: The image shows a single loop of area 7.4×10−2 m2 and
resistance 110 Ω. The loop is perpendicular to a magnetic field.
Strategy: Solve Ohm’s Law (equation 21-2) for the necessary emf. Then insert the emf into equation 23-4 to calculate the rate of change in the magnetic field.
Solution: 1. Calculate the emf :
ε
= IR =(
0.32 A 110)(
Ω =)
35 V 2. Solve equation 23-4 for thechange in magnetic field:
(
2 2)
2 35 V 4.9 10 T/s 1 7.2 10 m A B N N t t B t ΝΑε
ε
− ΔΦ Δ = = Δ Δ Δ = = = × Δ ×17. Picture the Problem: A coil with 120 loops is oriented perpendicular to a changing magnetic field. Strategy: Solve equation 23-4 for the emf, where the magnetic flux is given by equation 23-1.
Solution: Calculate the emf :
(
)
(
)
(
)
2 2 0.20 T 0.050 m 2 120 7.1 V 0.34 s BA BA BA N N N t t t
ε
= ΔΦ = − − = = = Δ Δ ΔInsight: This emf is double the value the coil would experience if the magnetic field simply dropped to zero in the same time period.
18. Picture the Problem: A square loop of wire is immersed in a magnetic field. The square is rapidly changed into a circle with the same perimeter.
Strategy: Use equation 23-1 to calculate the flux through the circle and the square. Then insert the fluxes into equation 23-4 to calculate the induced emf. Solution: 1. Calculate
the radius of the circle:
1.22 m 2 0.194 m 2 C πr r π = ⇒ = =
2. Calculate the flux
through the circle:
(
) (
)
2 2 circle 0.125 T 0.194 m 0.0148 Wb BA B rπ π Φ = = = = 3. Find the side length
of the square: 1.22 m 4 0.305 m 4 4 P P= s ⇒ s= = =
4. Calculate the flux
through the square:
(
)(
)
2 2
square BA Bs 0.125 T 0.305 m 0.0116 Wb
Φ = = = =
5. Calculate the emf: circle square 10.0148 0.0116 Wb 7.5 10 V4 4.25 s
N
t
ε
= Φ − Φ = − = × −Δ
Insight: If the field points out of the page, the induced current will be clockwise because the out-of-the-page flux through the coil is increasing. Note that there are only two significant figures in the answer because of the rules for subtraction.
19. Picture the Problem: A number of circular loops of wire are oriented perpendicular to a changing magnetic field. Strategy: Solve equation 23-4 for the number of coils, with the flux given by equation 23-1. The radius of the loops is
(
)
1 1
2 2 0.12 m 0.060 m.
r= d= =
Solution: Find the
number of coils:
(
) (
)
3 2 1.8 s 6.0 V 3.8 10 0.25 T 0.060 m t t N N t BA πε
= ΔΦ ⇒ =ε
Δ =ε
Δ = = × Δ ΔΦInsight: Note that the number of coils is proportional to the desired emf. To obtain an emf of 3.0 V, only half as many (1.9×103) coils would be needed.
20. Picture the Problem: A metal ring is dropped into a localized region of constant magnetic field, as indicated in the figure at the right. The magnetic field is zero above and below the indicated region.
Strategy: Use Lenz’s Law to determine the direction that the induced current must flow in order to oppose the change in the magnetic flux through the ring.
Solution: 1. (a) At location 1 the magnetic flux through the ring is increasing in the out-of-the-page direction. The induced current will flow clockwise in order to produce into-the-page flux to oppose this change. At location 2 the flux through the ring is not changing so that the induced current is zero. At location 3 the out-of-the-page flux through the ring is decreasing, so the induced current will flow counterclockwise to oppose this change by producing out-of-the-page flux. In summary: Location 1, clockwise; location 2, zero; location 3, counterclockwise.
2. The best explanation is I. Clockwise at 1 to oppose the field; zero at 2 because the field is uniform; counterclockwise at 3 to try to maintain the field. Statement II has the current directions reversed, and statement III does not properly recognize the rate of change of the magnetic flux through the loop.
21. Picture the Problem: A metal ring is dropped into a localized region of constant magnetic field, as indicated in the figure at the right. The magnetic field is zero above and below the indicated region.
Strategy: Use Lenz’s Law to determine the direction that the induced current must flow in order to oppose the change in the magnetic flux through the ring. Then use the Right-Hand Rule for magnetic forces to determine the force on the ring.
Solution: 1. (a) From the previous problem we know that at location 1 the induced current will flow clockwise. There is no magnetic force on the top of the ring where the field is zero, but at the bottom of the ring the force on a current flowing from right to left will be in the upward direction. The forces on the sides of the ring cancel, so the net force on the ring is upward. At location 2 no current flows (see previous problem) so there is no magnetic force. At location 3 the current is counterclockwise (see previous problem) and there is no force on the bottom of the ring because it is in the field-free region. At the top of the ring the magnetic force on a current flowing from right-to-left is in the upward direction. In summary: Location 1, upward; location 2, zero; location 3, upward.
2. The best explanation is III. Upward at 1 to oppose entering the field; zero at 2 because the field is uniform; upward at 3 to oppose leaving the field. Statement I has the force direction at location 3 reversed and statement II does not recognize that the induced current (and therefore the magnetic force) is zero at location 2.
Insight: The action of the magnetic force will be to reduce the downward acceleration of the ring both when it enters and when it leaves the field region. At all other times the acceleration of the ring is 9.81 m/s2 downward.
22. Picture the Problem: The figure at the right shows two metal disks of the same size and material oscillating in and out of a region with a magnetic field. One disk is solid; the other has a series of slots.
Strategy: Note that the action of the slots will be to suppress the induced currents in the disk that result from the changing magnetic field through the disk. The disk with slots will therefore experience a much smaller magnetic force than will the solid disk.
Solution: 1. (a) The slots in the second disk will tend to suppress the clockwise and counterclockwise currents that are induced by the change in the magnetic field through the disk. The diminished currents through the disk will result in a diminished magnetic force on the disk, so we conclude that the retarding effect of eddy currents on the solid disk is greater than the retarding effect on the slotted disk.
2. The best explanation is I. The solid disk experiences a greater retarding force because eddy currents in it flow freely and are not interrupted by the slots. Statement II does not recognize that the field penetration through the disk is not affected by the slots, and statement III ignores the importance of the eddy currents in explaining the motion of the disks. Insight: In this common classroom demonstration the motion of the two disks are dramatically different. The motion of the solid disk is quickly damped to zero while the slotted disk continues to swing nearly unimpeded.
23. Picture the Problem: The figure at the right shows two metal disks of the same size and material oscillating in and out of a region with a magnetic field. One disk is solid; the other has a series of slots.
Strategy: Consider the solid disk as it swings into and out of the magnetic field. Note that whenever the magnetic flux through the disk changes, a circulating eddy current is induced in the disk that opposes the change in the flux.
Solution: When the solid disk has swung to the right as far as it can go, it is fully immersed in the magnetic field. At that point the magnetic flux through the disk is not changing. According to Faraday’s Law, there is no induced current if the magnetic flux is not changing, so we conclude that the induced current in the disk is at a minimum at that location.
24. Picture the Problem: The figure at the right shows two metal disks of the same size and material oscillating in and out of a region with a magnetic field. One disk is solid; the other has a series of slots.
Strategy: Consider the solid disk as it swings into the magnetic field from left to right. Note that whenever the magnetic flux through the disk changes, a circulating eddy current is induced in the disk that opposes the change in the flux.
Solution: 1. (a) As the disk enters the field, which points into the page, a current
will be induced that produces a field pointing out of the page. Therefore, the induced current circulates in the counterclockwise direction.
2. The best explanation is II. The induced current is counterclockwise to generate a field within the disk that points out of the page. Statement I does not agree with Lenz’s Law, and statement III is false.
Insight: If the disk swings from right to left, exiting the field region, a current will flow that attempts to sustain the into-the-page magnetic flux through the disk. This current will thus flow in the clockwise direction.
25. Picture the Problem: The image shows a bar magnet, with north pole pointing downward, falling toward the center of a coil.
Strategy: Use Lenz’s Law to determine the direction of induced current.
Solution: The magnetic flux due to the bar magnet increases in the downward direction as viewed from above. According to Lenz’s Law, the current in the loop will flow such that it opposes the change in flux. Therefore, the current in the loop flows counterclockwise as viewed from above. Insight: If the bar magnet were moving upward, the downward flux would be decreasing so the induced current would be clockwise.
26. Picture the Problem: The image shows a loop of wire dropping between the poles of a magnet.
Strategy: Use Lenz’s Law to determine the direction of the induced current. Solution: 1. (a) When the loop is above the magnet, the magnetic field is increasing and directed out of the page. The current in the loop will oppose the increasing field by flowing clockwise.
2. (b) When the loop is below the magnet, the magnetic field is decreasing and is directed out of the page. The current in the loop will oppose the decreasing field by flowing counterclockwise.
Insight: When the loop is directly between the two poles the flux is a maximum, and therefore momentarily not changing. At this point the induced current is zero. 27. Picture the Problem: The image shows a coil lowered at constant speed between
the two poles of a magnet.
Strategy: From Lenz’s Law note that the current in the loop opposes the change in the flux through the loop. The result is an upward (retarding) force on the loop as it is lowered (see section 23-4).
Solution: 1. (a) The loop resists moving downward toward the magnet. (The poles of the loop’s field line up with the magnet poles north-to-north and south-to-south, causing repulsion.) So the string tension is less than the loop’s weight.
2. (b) The loop resists moving downward away from the magnet. (The poles of the loop’s field line up with the magnet poles north-to-south and south-to-north, causing attraction.) So the string tension is again less than the loop’s weight.
Insight: If the string were used to pull the ring upward through the magnet the tension would be greater when the loop is both below and above the magnet.
28. Picture the Problem: The image shows a coil raised at constant speed between the two poles of a magnet.
Strategy: From Lenz’s Law note that the current in the loop opposes the change in the flux through the loop. The result is a downward (retarding) force on the loop as it is raised (see section 23-4).
Solution: 1. (a) The loop resists moving upward toward the magnet. (The poles of the loop’s field line up with the magnet poles north-to-north and south-to-south, causing repulsion.) So the string tension is greater than the loop’s weight. 2. (b) The loop resists moving upward away from the magnet. (The poles of the loop’s field line up with the magnet poles north-to-south and south-to-north, causing attraction.) So the string tension is again greater than the loop’s weight.
Insight: Because the induced current increases with the speed of the ring, the faster the ring is pulled upward, the greater the tension in the string.
29. Picture the Problem: The image shows a loop with resistance R to the left of an upward current.
Strategy: Use Lenz’s Law to determine the direction of the induced current in the loop.
Solution: 1. (a) Because the current in the wire is constant, the magnetic field through the circuit does not vary with time, so the induced current is zero.
2. (b) The magnetic field through the circuit is increasing because the current in the wire is increasing. And, since the magnetic field is directed out of the page, the induced current in the circuit will induce a magnetic field into the page. So, the current in the circuit flows clockwise.
Insight: If the current were decreasing, the outward magnetic field would decrease, inducing a counterclockwise current.
30. Picture the Problem: The image shows a loop with resistance R to the left of a current that switches from upward to downward.
Strategy: Use Lenz’s Law to determine the direction of the induced current in the loop.
Solution: When the current in the wire changes direction, the direction of the magnetic field is reversed, changing its direction from out of the page to into the
page (it is shown in its final condition). According to Lenz’s Law, the current induced in the circuit will oppose this change by flowing counterclockwise, generating a field that is directed out of the page.
Insight: If the current were to switch back to the upward direction, the induced current would flow clockwise. 31. Picture the Problem: The image shows a long, straight current passing through the
center of a wire coil.
Strategy: Use Faraday’s Law (equation 23-4) to determine the emf.
Solution: 1. (a) The induced emf is zero because the magnetic field is parallel to the plane of the loop.
2. (b) If the current in the wire increases, the induced emf is still zero because the field due to the wire is still parallel to the plane of the loop, and there is no time-varying flux through the loop to generate an emf.
3. (c) If the wire no longer passes through the center of the coil but is still perpendicular to its plane, the answer to part (b) does not change because the magnetic field is still parallel to the plane of the loop.
32. Picture the Problem: The image shows a circuit containing a resistor and capacitor. A magnetic field initially pointing into the page reverses to point out of the page.
Strategy: Use Lenz’s Law to determine the direction of current flow. As the current flows onto the capacitor plate it becomes positively charged. Solution: Because the field changes from in to out of the page, the induced current in the circuit will flow clockwise to generate a field directed into the page. Therefore, the top plate will become positively charged.
Insight: As the capacitor plate becomes charged it opposes the induced current. The emf around the loop is still the same as given by equation 23-3, but the voltage across the capacitor (V =Q C) must be subtracted to calculate the current through the resistor.
33. Picture the Problem: The image shows a circuit containing a resistor and capacitor. A magnetic field pointing into the page increases in magnitude. Strategy: Use Lenz’s Law to determine the direction of current flow. As the current flows onto the capacitor plate it becomes positively charged. Solution: If the magnetic field increases into the page, the current in the circuit will flow counterclockwise to generate a field to oppose the increase. So, the bottom plate will become positively charged.
Insight: As the capacitor plate becomes charged it opposes the induced current. The emf around the loop is still the same as given by equation 23-3, but the voltage across the capacitor (V =Q C) must be subtracted out to calculate the current through the resistor.
34. Picture the Problem: The image shows three rings near a wire with a current that is increasing with time.
Strategy: The increasing current in the wire generates an increasing magnetic field that is directed out of the page above the wire and into the page below. Use Lenz’s Law to determine the direction of the induced current in each loop.
Solution: 1. Loop A: The induced emf is clockwise in order to oppose the change in the magnetic flux, which is increasing out of the page.
2. Loop B: Although the field through the loop is varying with time, the net flux through the plane of the loop is zero, so the induced emf is zero.
3 Loop C: The induced emf is counterclockwise in order to oppose the change in the magnetic flux, which is increasing into the page.
Insight: If the current were decreasing, the direction of the induced emf in rings A and C would be reversed. Ring B would still have zero emf.
35. Picture the Problem: A conducting rod slides on two wires in a region with a magnetic field. The two wires are not connected.
Strategy: Consider Faraday’s Law and the magnetic force on a current when answering the conceptual question. Solution: No. The fact that the two wires are not connected means that no current can flow through the rod. As a result, the magnetic field exerts zero force on the rod. If the system is frictionless, no force will be required to keep the rod moving with a constant speed.
Insight: If the wires were connected, as in Figure 23-11, Faraday’s Law predicts an emf will drive a current that flows through the conducting rod in the direction specified by Lenz’s Law, and a magnetic force will be experienced by the rod.
36. Picture the Problem: The image shows a 0.76 m rod moving at 2.0 m/s
perpendicular to a magnetic field. An emf of 0.45 V is induced between the ends of the rod.
Strategy: Solve equation 23-5 for the magnetic field. Solution: Solve equation 23-5 for B:
(
2.0m/s 0.76 m0.45 V)(
)
0.30 T vB B vε
ε
= = = = A AInsight: Rotating the rod by an angle θ decreases the effective length of the rod to cosθ
′ =
A A . If the rod were rotated by 25° the magnetic field would have to be increased to 0.34 T to produce the same emf.
37. Picture the Problem: The image shows an airplane with wingspan 39.9 m flying northward at 850 km/h through a magnetic field with vertical component of Bv = 5.0×10−6 T.
Strategy: Solve equation 23-5 for the induced emf.
Solution: Calculate the induced emf :
(
)(
)
km 1 h 850 5.0 T 39.9 m h 3600 s 47 mV vB με
= ⎛ ⎞⎛ ⎞ = ⎜⎝ ⎟⎜⎠⎝ ⎟⎠ = AInsight: Because the plane is flying horizontally northward, the horizontal component of Earth’s magnetic field does not contribute to the emf between the wing tips. If this were a rocket accelerating upward, then the horizontal component of the magnetic field would determine the emf.
38. Picture the Problem: The image shows a frictionless rod sliding across two rails separated by 0.45 m. A magnetic field of 0.750 T points out of the page.
Strategy: Calculate the emf using Ohm’s Law (equation 21-2). Insert the emf into equation 23-3 and solve for the speed of the rod.
Solution: 1. (a) Calculate
ε
:(
0.155 A 12.5)(
)
1.94 V I Rε
ε
= = Ω = 2. Solve for v:(
0.750 T 0.450 m1.94 V)(
)
5.75 m/s v B ε = = = A3. (b) No, the answer to part (a) would not change because the equation for v is independent of the direction of motion of the bar.
Insight: Lenz’s Law says that as the bar moves to the right, the current flows upward in the resistor. If the bar moved at the same speed to the left, the same current would flow downward through the resistor.
39. Picture the Problem: The image shows a frictionless rod sliding across two rails separated by 0.45 m. A magnetic field of 0.750 T points out of the page.
Strategy: The external force must be equal and opposite the magnetic force to maintain the rod’s constant speed. Use equation 22-4 to calculate the magnetic force. Use equation 21-5 to calculate the power dissipated by the resistor and equation 7-13 to calculate the power generated by the external force. Note that the speed of the rod is 5.75 m/s as found in problem 38.
Solution: 1. (a) Calculate the magnetic force: F =I BA=
(
0.155 A 0.750 T 0.450 m)(
)(
)
= 52.3 mN 2. (b) Calculate the power dissipated: P I R= 2 =(
0.155 A) (
2 12.5 Ω =)
0.300 W3. (c) Calculate the power generated
by the external force: P F v= =
(
52.3 mN 5.75 m/s)(
)
= 0.300 WInsight: The kinetic energy of the rod is constant because it moves at constant speed. The power generated by the external force is completely dissipated by the resistor. The apparatus can be considered a generator because it converts mechanical energy to electrical energy.
40. Picture the Problem: The image shows a frictionless rod moving at constant speed along two rails connected to a 12-Ω resistor. A magnetic field of 2.0 T points out of the page. When the rod moves at 3.1 m/s the light bulb consumes 5.0 W of power. Strategy: Solve equation 21-5 for the current through the circuit. Find the induced emf using Ohm’s Law (equation 21-2). Then solve equation 23-5 for the speed of the rod when a current of 1.0 A passes through the circuit.
Solution: 1. (a) Solve equation 21-5 for I: 5.0 W 0.65 A 12 P I R = = = Ω
2. (b) Calculate the new speed to produce 1.0 A:
(
1.0 A 12)(
(
)
)
4.8 m/s 2.0 T 1.25 m IR v BL BL ε Ω = = = =Insight: The new circuit in part (b) has a higher current and it will dissipate more power than the original circuit. 41. Picture the Problem: The image shows a light bulb of resistance
12 Ω that consumes 7.9 W of power. The moving rod is 1.25 m long, and a magnetic field of 2.0 T is directed out of the page. Strategy: Solve equation 21-5 for the current through the circuit. Find the induced emf using Ohm’s Law (equation 21-2). Then solve equation 23-5 for the speed of the rod.
Solution: 1. (a) Calculate the current: 8.9 W 0.86 A 12 P I R = = = Ω 2. (b) Calculate the
speed of the rod:
(
)(
)
(
)(
)
0.86 A 12 4.1 m/s 2.0 T 1.25 m I R v BL B L ε Ω = = = =Insight: Because this circuit dissipates more power than the circuit of Example 23-3, this circuit must have a higher current and the rod must move with a higher speed.
42. Picture the Problem: The emf produced by a rotating coil is proportional to the speed of the coil rotation. Strategy: Use equation 23-11 to determine the maximum emf as a function of the rotation speed. Place all of the constants on the right side of the equation and develop a ratio for the maximum emf and rotation speed. Solve the resulting ratio for the unknown rotation speed.
Solution: 1. Create a ratio using equation 23-11: max max1 max 2
1 2 NBAω NBA ω ω
ε
ε
ε
= ⇒ = =2. Solve for the final rotation speed: 2 max 2 1
(
)
max155 V
210 rpm 260 rpm 45 V
ω =
ε
ε
ω = =Insight: The emf is linearly proportional to the rotation speed. Doubling the rotation speed will double the emf. 43. Picture the Problem: A rectangular coil 25 cm by 35 cm of 120 turns rotates at a 190 rad/s in a magnetic field. The
rotation creates an emf of 65 volts across the coil. Strategy: Solve equation 23-11 for the magnetic field. Solution: Calculate the magnetic field:
( )(
)(
)(
)
max 65 V 33 mT 120 0.25 m 0.35 m 190 rad/s B NAωε
= = =Insight: If the rotation speed were to decrease to 95 rad/s, the magnetic field would need to increase to 66 mT to maintain the same emf.
44. Picture the Problem: A 3.2-cm-radius coil made from a 1.6-m-long wire creates an emf by rotating at 95 rpm in a 0.070-T magnetic field.
Strategy: Divide the length of the wire by the circumference of the coil to determine the number of loops. Then solve equation 23-11 for the maximum emf.
Solution: 1. Calculate
the number of loops:
(
)
1.6 m 8 2 2 0.032 m L L N C πr π = = = =
2. Calculate
ε
max: max 8 0.075 T(
) (
0.032 m) (
2 85 rev/min)
2 rad 1 min 17 mVrev 60 s
N B Aω π π
ε
= = ⎛ ⎞⎛ ⎞=⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
Insight: The emf is proportional to the loop radius because the number of turns is inversely proportional to the radius of the loop and the emf is proportional to the square of the radius. Creating a single loop from the coil would produce the largest emf.
45. Picture the Problem: A coil with 155 turns of diameter 22.0 cm rotates at 1250 rpm in Earth’s magnetic field. Strategy: Use equation 23-11 to calculate the maximum emf.
Solution: 1. (a) Only the horizontal component of the magnetic field is important. The vertical component is parallel to the plane of the circular coil at all times. Thus, it does not contribute to the flux through the coil.
2. (b) Calculate
ε
max:(
5)
(
)
2max
rev 2 rad 1 min
155 3.80 10 T 0.11 m 1250 29.3 mV min 1 rev 60 s N B Aω π π
ε
= = × − ⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠Insight: If the coil rotated about a horizontal axis, the maximum emf would be 22.0 mV.
46. Picture the Problem: A set of coils with area 0.016 m2 rotates at 3600 rpm in a 0.050-T magnetic field. The coils
produce an emf of 170 V.
Strategy: Solve equation 23-11 for the number of turns N. Solution: Calculate
the number of turns: max
(
)
(
2)
(
)
170 V rev 60 s 560 turns 2 rad min 0.050 T 0.016 m 3600 rev/min N B Aω π
ε
⎛ ⎞⎛ ⎞ = = ⎜ ⎟⎜ ⎟= ⎝ ⎠⎝ ⎠Insight: Note that the magnetic field and number of turns are inversely proportional. To produce the same emf at the same frequency with half the number of coils it would be necessary to double the magnetic field.
47. Picture the Problem: The current through a 45.0-mH inductor increases from zero to 515 mA in 16.5 ms. Strategy: Solve equation 23-12 for the emf.
Solution: Calculate the emf:
(
45.0 10 H3)
515mA 1.40 V16.5ms I L t
ε
= − Δ = − × − = − ΔInsight: The sign of the emf shows that the induced current opposes the increasing current in the circuit.
48. Picture the Problem: A solenoid is 0.22 m long and has a 0.035 m2 cross-sectional area. Its inductance is 45 mH.
Strategy: Solve equation 23-14 for the number of turns. Solution: Calculate the number of turns:
(
)
(
)
2 0 3 7 2 0 0.22 m 45 10 H 470 turns 4 10 T m/A 0.035 m N L A L N A μ μ π − − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ × = = = × ⋅ A AInsight: The number of turns is proportional to the square root of the inductance. To double the inductance to 90 mH, the number of turns would increase by a factor of 2 to 670 turns.
49. Picture the Problem: A 7.3-mH solenoid has 450 turns and a length of 24 cm.
Strategy: Solve equation 23-14 for the cross-sectional area of the solenoid. Use equation 23-12 to calculate the emf across the solenoid as the current drops to zero.
Solution: 1. (a) Solve for the cross-sectional area:
(
)
(
)
(
)
(
)
3 3 2 2 7 2 0 0.24 m 7.3 10 H 6.9 10 m 4 10 T m/A 450 L A N μ π − − − × = = = × × ⋅ A2. (b) Calculate the emf: 7.3 mH 3.2 A 0.42 V
55 ms I L t
ε
= − Δ = − ⎛− ⎞= ⎜ ⎟ Δ ⎝ ⎠Insight: Because the current is decreasing, the solenoid produces a positive emf in order to oppose the decrease in current.
50. Picture the Problem: A solenoid has 640 turns, is 24 cm long, and has a radius of 4.3 cm. Strategy: Solve equation 23-14 for the inductance.
Solution: Calculate the inductance:
(
)
(
)
2 2 2 7 0 640 4 10 T m/A 0.043 m 12 mH 0.25 m N L=μ A= π× − ⋅ π = A
Insight: To double the inductance, you could increase either the radius or the number of turns by 2, or you could cut the length in half while keeping the number of turns the same.
51. Picture the Problem: A solenoid has 455 turns per meter, a length of 0.75 m, and a cross-sectional area of 1.81×10−3 m2. The current through this solenoid increases from zero to 2.00 A in 45.5 ms.
Strategy: Solve equation 23-14 for the inductance. Then insert the inductance into equation 23-12 to calculate the emf. Solution: 1. Calculate L: L=μ0n A2A =
(
4π×10 T m/A 455 /m−7 ⋅)
(
) (
2 0.750 m 1.81 10 m)
(
× −3 2)
=353.2 Hμ 2. Calculate the emf:(
3.532 10 H4)
2.00 A3 15.5 mV45.5 10 s I L t
ε
− − Δ = − = − × = − Δ ×Insight: Because the current is increasing through the solenoid, the emf across the solenoid is negative in order to oppose the change in current.
52. Picture the Problem: An induced emf of 75 mV is created across an inductor when the current increases at 2.0 A/s. Strategy: Solve equation 23-12 for the inductance. Then use equation 23-14 to calculate the inductance when the length has doubled (A′ =2A) but all other parameters have remained the same. Use the new inductance in equation 23-12 to calculate the emf.
Solution: 1. (a) Solve equation 23-12 for L: 75 mV 38 mH
2.0 A/s L I t
ε
= = = Δ Δ2. (b) Because the induced emf is directly proportional to the inductance, and since the inductance of the solenoid is inversely proportional to its length, the induced emf is inversely proportional to the length of the solenoid. So, if the length of the solenoid is doubled, the induced emf will be less than 75 mV.
3. (c) Calculate the new inductance:
2 2 0 0 37.5 mH 18.75 mH 2 2 2 N N L L=μ ⎛⎜ ⎞⎟A=μ ⎛⎜ ⎞⎟A= = = ′ ⎝ A ⎠ ⎝ A⎠
4. Use the inductance to calculate the emf:
ε
2 =L2Δ Δ =I t(
18.75 mH 2.0 A/s)(
)
= 38 mVInsight: Because the inductance is inversely proportional to the solenoid length, doubling the length cuts the inductance in half, and thus cuts the induced emf in half also.
53. Picture the Problem: An RL circuit consists of a 130 Ω resistor and a 63 mH inductor. Strategy: Solve equation 23-16 for the time, where the current is equal to half the final current. Solution: 1. Set I=12Ifinal in equation 23-16:
(
/)
final 1 1 2 2 tR L I I e R R
ε ε
− = = = −2. Solve for the time t:
3 4 1 2 68 10 H ln 2 ln 2 3.6 10 s 130 t R L L e t R − − = ⇒ = = × = × − Ω
Insight: After a time of 2t = 7.2×10−4 s, the current will equal ¾ of its final value. For each time step of 3.6×10−4 s, the current will increase by one-half the remaining difference between the maximum current and the actual current.
54. Picture the Problem: The four electrical circuits shown in the figure at the right have identical batteries, resistors, and inductors.
Strategy: A long time after the switch is closed each inductor will act like an ideal wire. Use this principle and the rules of adding resistors in series and in parallel to determine the ranking of the currents supplied by the batteries. Solution: Let each battery have emf
ε
and each resistor have resistance R. For circuit A the inductor will act like an ideal wire when the switch has been closed for a long time, and the current supplied by the battery will beε
R.For circuits B and C the resistors in parallel with the inductor are shorted out, and make no contribution to the current in the circuit. The current supplied by the battery in those circuits is alsoε
R. For circuit D the equivalent resistance of the two resistors in parallel is R 2, so the current supplied by thebattery after a long time, when the inductor acts as an ideal wire, is 2
ε
R. The ranking of the currents, therefore, is A = B = C < D.Insight: While inductors act like ideal wires in DC circuits, capacitors act like open switches (see Chapter 21). If each inductor were replaced by a capacitor, the currents in each circuit would be A, zero; B,
ε
2 ;R C, 2 3 ;ε
R and D, zero.55. Picture the Problem: A circuit is constructed of four 55-Ω resistors and a 37-mH inductor connected to a 6.0-V battery.
Strategy: Use the rules of series and parallel circuits (equations 21-7 and 21-10) to calculate the effective resistance. Insert the effective resistance into equation 23-15 to calculate the characteristic time constant and into equation 23-16 to calculate the current.
Solution: 1. (a) Calculate the effective resistance:
(
)
1 eq 1 1 2 5 5 55 91.7 2 3 3 3 R R R R R R R − ⎛ ⎞ = +⎜ + ⎟ = + = = Ω = Ω ⎝ ⎠2. Insert the resistance into equation 23-15:
3 4 eq 37 10 H 4.0 10 s 0.40 ms 91.7 L R τ = = × − = × − = Ω
3. (b) Calculate I after two time constants:
(
2)
(
2)
eq 6.0 V 1 1 57 mA 91.7 I e e R τ τε
− − = − = − = Ω4. (c) Calculate the current after a long time:
(
)
(
)
eq 6.0 V 1 1 0 65 mA 91.7 I e Rε
−∞ = − = − = ΩInsight: After two time constants the current is 86% of its maximum value. After five time constants the current is 99.3% of its maximum value, which is indistinguishable from the maximum value when rounded to two digits.
56. Picture the Problem: The image shows the current as a function of time for the RL circuit. After 2.24 s the current reaches 95% of its maximum value.
Strategy: Solve equation 23-16 for the time constant, given that I=0.95
ε
R at t=2.24 s. Then insert the inductance into equation 23-15 to calculate the resistance.Solution: 1. (a) Set the current equal
to 95% its maximum value: 0.95Ifinal Ifinal 1 e(2.24 s)
τ
−
⎡ ⎤
= ⎣ − ⎦
2. Solve for the time constant: 2.24 s 0.75 s ln 0.05
τ = − =
3. (b) Calculate the resistance: 0.275 H 0.37 0.748 s
L R
τ
= = = Ω
Insight: If the resistance of this circuit were decreased to 0.25 Ω, the current would reach only 87% of its maximum in 2.24 s. However, since the maximum current would be 50% greater than when the resistance is 0.37 Ω, at 2.24 s more current will be flowing when the resistance is 0.25 Ω than when it is 0.37 Ω.
57. Picture the Problem: The image shows an RL circuit with a 5.5-Ω resistor. When the switch is closed, the current rises from zero to 0.32 A in 0.15 s. Strategy: Solve equation 23-16 for the inductance. Then insert the inductance into equation 23-16 and solve for the time at which the current is 0.50 A. Finally, divide the voltage by the resistance to calculate the maximum current.
Solution: 1. (a) Solve for the inductance:
(
)
(
)
(
)(
(
) (
)
)
1 0.15 s 5.5 3.8 H ln 1 ln 1 5.5 0.32 A / 9.0 V tR L I e R tR L RI εε
− = − Ω − = = − = − ⎡⎣ − Ω ⎤⎦2. (b) Solve for the time when I = 0.50 A: ln 1 3.79 Hln 1
(
5.5 0.50)(
A)
0.25 s 5.5 9.0 V L RI t Rε
Ω ⎡ ⎤ ⎛ ⎞ = − ⎜ − ⎟= − ⎢ − ⎥= Ω ⎝ ⎠ ⎣ ⎦3. (c) Calculate the maximum current: max
9.0 V 1.6 A 5.5 I R
ε
= = = Ω Insight: The time constant for this circuit is 0.69 s.58. Picture the Problem: 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.
Strategy: Use equations 23-14 and 23-19 to determine the effect of the described changes on the energy stored in the solenoid.
Solution: The energy stored in the inductor remains the same. Doubling the number of turns per length quadruples the inductance of the solenoid, as we can see from equation 23–14. The energy stored in an inductor, however, depends on both the inductance of the inductor and the current it carries, as we see in equation 23–19. In fact, the energy stored in an inductor depends on the square of the current. Therefore, halving the current reduces the stored energy by a factor of four. We conclude that the energy stored in the inductor will stay the same.
Insight: We can also arrive at the solution using a ratio:
(
)
(
)
2 2 2 2 1 2 1 1 0 new new 2new new old
new 2 new new old 2
2 2 2
1 1
old 2 old old 2 0 old old old old old old 2 1 n A I L I I U n I n U L I n A I n I n I μ μ ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ =⎜ ⎟ = ⎝ ⎠ ⎝ ⎠ A A
59. Picture the Problem: The image shows an RL circuit with a 5.5-Ω resistor, a 6.1-mH inductor, and 9.0-V battery.
Strategy: Use equation 23-19 to calculate the maximum energy stored in the inductor, where the current is given by the battery emf divided by the resistance (Ohm’s Law).
Solution: Calculate the stored energy:
(
)
2 2 2 3 1 1 1 9.0 V 6.1 10 H 8.2 mJ 2 2 2 5.5 U LI L R
ε
− ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟ = × ⎜ ⎟ = Ω ⎝ ⎠ ⎝ ⎠Insight: Increasing the inductance will increase the stored energy, but will not affect the maximum current in the circuit. However, it would increase the time required for the current in the circuit to reach its maximum value.
60. Picture the Problem: A 1.5-m-long solenoid with 470 turns stores 0.31 J of energy when it carries 12 A of current. Strategy: Use equation 23-19 to calculate the inductance of the solenoid. Then solve equation 23-14 for the area. Solution: 1. Calculate the inductance:
(
)
(
)
2 2 2 1 2 2 0.31 J 2 0.0043 H 12 A U LI U L I = = = =2. Calculate the area:
(
)(
)
(
)
2 0 2 2 7 1 2 0 0.0043 H 0.010 m 4 10 T m/A 470 m 1.5 m L n A L A n μ μ π − − = = = = × ⋅ A AInsight: Note that the energy stored in the solenoid is linearly proportional to both the cross-sectional area and to the length of the solenoid. Because the area multiplied by the length is the volume, the energy stored is linearly
61. Picture the Problem: A magnetic field of 50.0 T is generated in the Alcator fusion experiment at MIT.
Strategy: Calculate the energy density of the magnetic field using equation 23-20. Set this energy equal to the electric field energy density (equation 20-19) and solve for the electric field.
Solution: 1. (a) Calculate the energy density:
(
)
(
)
2 2 8 3 B 7 0 50.0 T 9.95 10 J/m 2 2 4 10 T m/A B u μ π − = = = × × ⋅2. (b) Set uE =uBand solve equation 20-19 for E:
(
)
2 0 E B 8 3 10 B 12 2 2 0 2 2 9.95 10 J/m 2 1.50 10 V/m 8.85 10 C /N m E u u u E ε ε − = = × = = = × × ⋅Insight: Note that the electric field is equal to the magnetic field multiplied by the speed of light (3.00×108 m/s). We will discover in Chapter 25 that this is not just a coincidence!
62. Picture the Problem: A circuit is constructed with a 62.0-mH inductor in series with two parallel resistors and a 12-V battery. A long time after the switch is closed the inductor stores 0.110 J of energy.
Strategy: Solve equation 23-19 for the current through the battery. Then use Ohm’s Law (equation 21-2) to calculate the current through the 7.50-Ω resistor and subtract that current from the total to calculate the current through the unknown resistor. Divide the voltage across the battery by the current to calculate the resistance.
Solution: 1. (a) Calculate the current through the inductor:
(
)
2 1 2 2 0.110 J 2 1.884 A 62.0 mH U LI U I L = = = =2. Calculate the current through the 7.50-Ω resistor: 12 V 1.60 A 7.50 V I R = = = Ω
3. Calculate the current through the unknown resistor: IR =IL−I7.5=1.884 A 1.600 A= 0.284 A− 4. Divide the voltage by the current: 12 V 42
0.284 A
V R
I
= = = Ω
5. (b) The energy stored in the inductor is proportional to the square of the total current. Decreasing the resistance will increase the current, thereby increasing the energy stored. Therefore, in order to store more energy in the inductor the value of R should be less than the value found in part (a).
63. Picture the Problem: A circuit is constructed with a 62.0-mH inductor in series with two parallel resistors and a 12-V battery. In this problem the right-hand resistor has the value R = 14 Ω.
Strategy: Use equation 21-10 to calculate the effective resistance. Then insert the effective resistance into equation 23-16 to calculate the current at times t=τ and t=2 .τ Substitute these currents into equation 23-19 in order to calculate the stored energy.
Solution: 1. (a) Calculate the effective resistance:
1 eq 1 1 4.88 7.5 14 R − ⎛ ⎞ =⎜ + ⎟ = Ω Ω Ω ⎝ ⎠
2. Calculate the current at t= τ:
(
/)
(
1)
eq 12 V 1 1 1.554 A 4.88 I e e R τ τ
ε
− − = − = − = Ω3. Calculate the stored energy: 1 2 1
(
)(
)
22 2 62.0 mH 1.554 A 0.075 J
U = LI = =
4. (b) Calculate the current at t=2 :τ
(
2 /)
(
2)
eq 12 V 1 1 2.126 A 4.88 I e e R τ τε
− − = − = − = Ω5. Calculate the stored energy: 1 2 1
(
)(
)
22 2 62.0 mH 2.126 A 0.14 J
U = LI = =
6. (c) Because τ =L R/ eq and Req increase when R increases, increasing R causes the characteristic timeτ to decrease. Insight: In this problem the characteristic time is 12.7 ms. Increasing R to 21 Ω would increase the effective resistance to 5.53 Ω and decrease the time constant to 11.2 ms.
64. Picture the Problem: A circuit is constructed with four 55-Ω resistors and a 37-mH inductor connected to a 6.0-V battery.
Strategy: Use the rules of series and parallel circuits (equations 21-7 and 21-10) to calculate the effective resistance. Use equation 23-16 to calculate the current after one time constant and equation 23-19 to calculate the stored energy. Repeat the above for t = ∞.
Solution: 1. (a) The current through the inductor starts at zero and rises after the switch is closed. Because energy stored in the inductor is proportional to the square of the current, more energy is stored long after the switch is closed.
2. (b) Calculate the effective resistance:
(
)
1 eq 1 1 2 5 5 55 91.7 2 3 3 3 R R R R R R R − ⎛ ⎞ = +⎜ + ⎟ = + = = Ω = Ω ⎝ ⎠
3. Calculate the current after one time constant:
(
)
(
1)
eq 6.0 V 1 1 0.04138 A 91.7 I e e R τ τε
− − = − = − = Ω4. Calculate the stored energy: 1 2 1
(
)(
)
21 2 2 37 mH 0.04138 A 32 J
U = LI = = μ
5. (c) Calculate the current a long
time after the switch is closed: eq
(
)
(
)
6.0 V 1 1 0 0.06545 A 91.7 I e R τ
ε
−∞ = − = − = Ω6. Calculate the stored energy: 1 2 1
(
)(
)
22 2 37 mH 0.06545 A 79 J
U∞ = LI = = μ
Insight: Decreasing the resistance will increase the total stored energy, but will not change the fraction of the maximum energy stored after one time step.
65. Picture the Problem: A 28-cm-long solenoid has 580 turns of wire and a diameter of 7.2 cm. The solenoid stores 9.9 J of energy in its magnetic field.
Strategy: Use equation 23-14 to calculate the inductance of the solenoid, and then solve equation 23-19 for the required current. Use equation 22-12 to calculate the magnetic field inside the solenoid and equation 23-20 to calculate the energy density.
Solution: 1. (a) Calculate the inductance:
(
)
( )
(
)
2 2 2 7 1 0 2 580 4 10 T m/A 0.072 m 0.0061 H 0.28 m N L=μ ⎛ ⎞A= π× − ⋅ π × = ⎜ ⎟ ⎝ A ⎠2. Solve for the current: 2 2 9.9 J
(
)
57 A 0.0061 HU I
L
= = =
3. (b) Calculate the magnetic field: 0
(
7)
(
)
5804 10 T m/A 57 A 0.148 T 0.15 T
0.28 m
B=μnI= π× − ⋅ ⎛⎜ ⎞⎟ = =
⎝ ⎠
4. (c) Calculate the energy density:
(
)
(
)
2 2 3 3 B 7 0 0.148 T 8700 J/m 8.7 kJ/m 2 2 4 10 T m/A B u μ π − = = = = × ⋅Insight: The energy density can also be calculated by dividing the stored energy by the volume of the solenoid:
(
) (
)
3 2 2 9.9 J 8.7 kJ/m 0.036 m 0.28 m U u r π π = = = A .66. Picture the Problem: 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.Strategy: Use the characteristics of transformers to answer the conceptual question.
Solution: The two transformers have the same ratio of the number of turns (Np / Ns) in their primary and secondary
circuits. Therefore, the ratio of their primary and secondary voltages must be the same as well. It follows that the secondary voltage in transformer 2 is 2Vs; that is, V Vp s =2Vp 2 .Vs
Insight: When manufacturing a transformer, increasing the number of turns improves the magnitude of the magnetic flux through each coil, which can be helpful for amplifying small voltages or currents. However, a large number of turns also increases the electrical resistance of the device. A wide variety of transformers are available so that a user can select one with optimal characteristics for the particular application.
67. Picture the Problem: Transformer 1 has a primary current Ip and a secondary currentIs. 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.
Strategy: Use the characteristics of transformers to answer the conceptual question.
Solution: The secondary current in transformer 2 is 6Is. The factor of 6 is the result of a factor of 3 in the primary current, multiplied by a factor of 2 due to the doubling of the number of turns in the primary coil.
Insight: The result can also be obtained by calculating a ratio, remembering that the number of turns in the secondary coil of each transformer is the same (that is, Ns, 2=Ns):
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)
(
)
p, 2 p, 2 s, 2 p, 2 p, 2 p p s, 2 s p p s p p p p 3 2 6 I N N I N I N I I = I N N = I N = I N =68. Picture the Problem: A transformer converts a primary voltage of Vp =110 V to a secondary voltage of Vs =3.0 V.
Strategy: Solve equation 23-22 for the ratio primary turns to secondary turns. Solution: Calculate the ratio of turns: p p
s s 110 V 37 3.0 V N V N =V = =
Insight: If the secondary coil has 75 turns, the primary coil would need 2800 turns. 69. Picture the Problem: A transformer converts a primary voltage of Vp =120 V
to a secondary voltage of Vs =9.0 V. It has 125 turns in the primary coil. Strategy: Solve equation 23-22 for the number of turns in the secondary coil. Solution: 1. (a) The voltage needs to be stepped down. To step down a primary voltage to a secondary voltage, the number of turns on the secondary coil must be less than the number of turns on the primary coil. So, the number of turns on the secondary coil must be less than 125.
2. (b) Calculate the number of turns in the secondary coil: s p s p 9.0 V 125 turns 9.4 turns 120 V V N N V ⎛ ⎞ ⎛ ⎞ = ⎜⎜ ⎟⎟= ⎜ ⎟= ⎝ ⎠ ⎝ ⎠
Insight: To create a coil with an integer number of turns you should decrease the number of primary turns to 120. 70. Picture the Problem: A transformer that has a turns ratio of 1:13 is used to step down a 120-V voltage source.
Strategy: Solve equation 23-22 for the voltage in the secondary coil.
Solution: Calculate the secondary voltage: s s p p 1 120 V 6.7 V 18 N V V N ⎛ ⎞ ⎛ ⎞ = ⎜⎜ ⎟⎟= ⎜ ⎟= ⎝ ⎠ ⎝ ⎠
Insight: If the transformer were connected backwards, the secondary voltage would become 120 V × 18 = 2160 V. 71. Picture the Problem: A transformer converts a primary voltage Vp =120 V to a secondary voltage Vs =11,000 V.
Strategy: Solve equation 23-22 for the ratio primary turns to secondary turns. Solution: Calculate the ratio of turns: s s
p p
11,000 V 92 120 V
N V
N =V = =
Insight: If the transformer were connected backwards, it would provide only a voltage of 1.3 V to the neon sign. 72. Picture the Problem: A transformer converts a primary voltage Vp =120 V to a secondary voltage Vs =6.0 V.
Strategy: Use equation 23-22 to calculate the ratio of the primary turns to secondary turns when the primary is 120 V and the secondary is 6.0 V. Then solve the equation for the primary voltage if the secondary voltage is 120 V. Solution: 1. Calculate the turn ratio: p p
s s
120 V 20 6.0 V
N V
N =V = =
2. Solve for the primary voltage: p s p
(
)( )
s 120 V 20 2.4 kV N V V N ⎛ ⎞ ′ = ⎜ ⎟= = ⎝ ⎠
Insight: Care must be taken to ensure that the transformer is properly installed. A reversed transformer will easily damage the circuit and possibly start a fire!
