Responses to Questions
1. The advantage of using many turns (N=large number) in Faraday’s experiments is that the emf and induced current are proportional to N, which makes it easier to experimentally measure those quantities. Many turns in the primary coil will make a larger magnetic flux, and many turns in the output coil will produce a larger output voltage and current.
2. Magnetic flux is proportional to the total number of magnetic field lines passing through an enclosed loop area: Φ =B BAcos ,θ so the flux is proportional to the magnitude of the magnetic field. The magnetic flux depends not only on the field itself, but also on the area and on the angle between the field and the area. Thus, they also have different units (magnetic field tesla T;= = magnetic
2
flux T m= ⋅ =Wb .) Another difference is that magnetic field is a vector, but magnetic flux is a scalar. 3. (a) A current is induced in the ring when you move the south pole toward the ring. An emf and
current are induced in the ring due to the changing magnetic flux. As the magnet gets closer to the ring, more magnetic field lines are going through the ring. Using Lenz’s law and the right-hand rule, the direction of the induced current when you bring the south pole toward the ring is clockwise. In this case, the number of magnetic field lines coming through the loop and pointing toward you is increasing (remember that magnetic field lines point toward the south pole of the magnet). The induced current in the loop will oppose this change in flux and will attempt to create magnetic field lines through the loop that point away from you. A clockwise induced current will provide this opposing magnetic field.
(b) A current is not induced in the ring when the magnet is held steady within the ring. An emf and current are not induced in the ring since the magnetic flux through the ring is not changing while the magnet is held steady.
(c) A current is induced in the ring when you withdraw the magnet. An emf and current are induced in the ring due to the changing magnetic flux. As you pull the magnet out of the ring toward you, fewer magnetic field lines are going through the ring. Using Lenz’s law and the right-hand rule again, the direction of the induced current when you withdraw the south pole from the ring is counterclockwise. In this case, the number of magnetic field lines coming through the loop and pointing toward you is decreasing. The induced current in the loop will oppose this change in flux and will create more magnetic field lines through the loop that point toward you. A counterclockwise induced current will provide this opposing magnetic field.
4. (a) The magnetic field through the loop due to the current-carrying wire will be into the page. As the wire loop is pulled away, the flux will decrease since the magnetic field is inversely proportional to the distance from the wire. Current will be induced to increase the inward magnetic field, which means that the induced current will be clockwise.
(b) If the wire loop is stationary but the current in the wire is decreased, then the inward magnetic field through the loop will again be decreasing, so again the induced current will be clockwise. 5. (a) There will be no magnetic field lines “piercing” the loop from top to bottom or vice versa. All of
the field lines are parallel to the face of the loop. Thus there will be no magnetic flux in the loop and no change of flux in the loop, so there will be no induced current.
(b) Now, since the magnet is much thicker than the loop, there will be some field lines that pierce the loop from top to bottom. The flux will increase as the magnet gets closer, so a current will be induced that makes upward-pointing field lines through the loop. The current will be
counterclockwise.
6. (a) Yes. As the battery is connected to the front loop and current starts to flow, it will create an increasing magnetic field that points away from you and down through the two loops. Because the magnetic flux will be increasing in the second loop, an emf and current will be induced in the second loop.
(b) The induced current in the second loop starts to flow as soon as the current in the front loop starts to increase and create a magnetic field (basically, immediately upon the connection of the battery to the front loop).
(c) The current in the second loop stops flowing as soon as the current in the front loop becomes steady. Once the battery has increased the current in the front loop from zero to its steady-state value, the magnetic field it creates is also steady. Since the magnetic flux through the second loop is no longer changing, the induced current goes to zero.
(d) The induced current in the second loop is counterclockwise. Since the increasing clockwise current in the front loop is causing an increase in the number of magnetic field lines down through the second loop, Lenz’s law states that the second loop will attempt to oppose this change in flux. To oppose this change, the right-hand rule indicates that a counterclockwise current will be induced in the second loop.
(e) Yes. Since both loops carry currents and create magnetic fields while the current in the front loop is increasing from the battery, each current will “feel” the magnetic field caused by the other loop.
(f) The force between the two loops will repel each other. The front loop is creating a magnetic field pointed toward the second loop. This changing magnetic field induces a current in the second loop to oppose the increasing magnetic field, and this induced current creates a magnetic field pointing toward the front loop. These two magnetic fields will act like two north poles pointing at each other and repel. We can also explain it by saying that the two currents are in opposite directions, and opposing currents exert a repelling force on each other.
7. Yes, a current will be induced in the second coil. It will start when the battery is disconnected from the first coil and stop when the current falls to zero in the first coil. The current in the second loop will be clockwise. All of the reasoning is similar to that given for Question 6, except now the current is decreasing instead of increasing.
(b) The induced current in RAis to the left as coil B is moved away from coil A. As B recedes from A, the magnetic flux through coil A decreases (there are now fewer magnetic field lines in coil A pointing to the left). The induced emf in coil A creates a current to produce a magnetic field that opposes this decrease in flux, pointing to the left through the center of the coil. A current through
A
R to the left will produce this opposing field.
(c) The induced current in RAis to the left as RB in coil B is increased. As RB increases, the current in coil B decreases, which also decreases the magnetic field coil B produces. As the magnetic field from coil B decreases, the magnetic flux through coil A decreases (there are now fewer magnetic field lines in coil A pointing to the left). The induced emf in coil A creates a current to produce a magnetic field pointing to the left through the center of the coil, opposing the decrease in flux. A current through RAto the left will produce this opposing field. 9. The net current in the wire and shield combination is 0. Thus to the “outside,” there is no magnetic
field created by the combination. If there is an external magnetic field, it will not be influenced by the signal current, and then by Newton’s third law, the signal current will not be influenced by an external magnetic field either.
10. One advantage of placing the two insulated wires carrying ac close together is that the magnetic field created by the changing current moving one way in one wire is approximately cancelled out by the magnetic field created by the current moving in the opposite direction in the second wire. Also, since large loops of wire in a circuit can generate a large self-induced back emf, by placing the two wires close to each other, or even twisting them about each other, the effective area of the current loop is decreased and the induced current is minimized.
13. Eddy current brakes will work on metallic wheels, such as copper and aluminum. Eddy current brakes do not need to act on ferromagnetic wheels. The external magnetic field of the eddy brake just needs to interact with the “free” conduction electrons in the metal wheels in order to have the braking effect. First, the magnetic field creates eddy currents in the moving metal wheel using the free conduction electrons (the right-hand rule says moving charges in a magnetic field will experience a magnetic force, making them move and creating an eddy current). This eddy current is also in the braking magnetic field. The right-hand rule says these currents will experience a force opposing the original motion of the piece of metal and the eddy current brake will begin to slow the wheel. Good
conductors, such as copper and aluminum, have many free conduction electrons and will allow large eddy currents to be created, which in turn will provide good braking results.
14. As a magnet falls through a metal tube, an increase in the magnetic flux is created in the areas ahead of it in the tube. This flux change induces a current to flow around the tube walls to create an opposing magnetic field in the tube (Lenz’s law). This induced magnetic field pushes against the falling magnet and reduces its acceleration. The speed of the falling magnet increases until the magnetic force on the magnet is the same size as the gravity force. The opposing magnetic field cannot cause the magnet to actually come to a stop, since then the flux would become a constant and the induced current would disappear, as would the opposing magnetic field. Thus, the magnet reaches a state of equilibrium and falls at a constant terminal velocity. The weight of the magnet is balanced by the upward force from the eddy currents.
15. The nonferrous materials are not magnetic, but they are conducting. As they pass by the permanent magnets, eddy currents will be induced in them. The eddy currents provide a “braking” mechanism which will cause the metallic materials to slide more slowly down the incline than the nonmetallic materials. The nonmetallic materials will reach the bottom with larger speeds. The nonmetallic materials can therefore be separated from the metallic, nonferrous materials by placing bins at different distances from the bottom of the incline. The closest bin will catch the metallic materials, since their projectile velocities off the end of the incline will be small. The bin for the nonmetallic materials should be placed farther away to catch the higher-velocity projectiles.
16. The slots in the metal bar prevent the formation of large eddy currents, which would slow the bar’s fall through the region of magnetic field. The smaller eddy currents then experience a smaller opposing force to the motion of the metal bar. Thus, the slotted bar falls more quickly through the magnetic field. 17. This is similar to the situation accompanying Fig. 21–20. As the aluminum sheet is moved through the
magnetic field, eddy currents are created in the sheet. The magnetic force on these induced currents opposes the motion. Thus it requires some force to pull the sheet out.
18. The speed of the magnet in case (b) will be larger than that in case (a). As the bar magnet falls through the loop, it sets up an induced current in the loop, which opposes the change in flux. This current acts like a magnet that is opposing the physics magnet, repelling it and so reducing its speed. Then, after the midpoint of the magnet passes through the loop, the induced current will reverse its direction and attract the falling magnet, again reducing its speed.
20. To determine the ratio of turns on the two coils of a transformer without taking it apart, apply a known ac input source voltage to one pair of leads and carefully measure the output voltage across the other two leads. Then, Vsource/Voutput =Nsource/Noutput, which provides us with the ratio of turns on the two coils. To determine which leads are paired with which, you could use an ohmmeter, since the two source wires are in no way electrically connected to the two output wires. If the resistance between two wires is very small, then those two wires are a pair. If the resistance between two wires is infinite, then those two wires are not a pair.
21. Higher voltages, such as 600 V or 1200 V, would be dangerous if they were used in household wires. Having a larger voltage than the typical 120 V would mean that any accidental contact with a “live” wire would send more current through a person’s body. Such a large potential difference between household wires and anything that is grounded (other wires, people, etc.) would more easily cause electrical breakdown of the air, and then much more sparking would occur. Basically, this would supply each of the charges in the household wires with much more energy than the lower voltages, which would allow them to arc to other conductors. This would increase the possibility of more short circuits and accidental electrocutions.
22. When 120 V dc is applied to the transformer, there is no induced back emf that would usually occur with 120 V ac. This means that the 120 V dc encounters much less resistance than the 120 V ac, resulting in too much current in the primary coils. This large amount of current could overheat the coils, which are usually wound with many loops of very fine, low-resistance wire, and could melt the insulation and burn out or short out the transformer.
23. (a) To create the largest amount of mutual inductance with two flat circular coils of wire, you would place them face-to-face and very close to each other. This way, almost all of the magnetic flux from one coil also goes through the other coil.
(b) To create the least amount of mutual inductance with two flat circular coils, you would place them with their faces at right angles. This way, almost none of the magnetic flux from one coil goes through the other coil.
24. (a) No. Although the current through an LR circuit is described by 0 1 / , t L R V
I e
R −
⎛ ⎞
⎜ ⎟
= ⎜ − ⎟
⎝ ⎠
we can
substitute VR0 =Imax. Thus, a given fraction of a maximum possible current, I I/ max, is equal to
/ 1
t L R e−
− and is independent of the battery emf. (b) Yes. Since 0 1 / ,
t L R V
I e
R −
⎛ ⎞
⎜ ⎟
= −
⎜ ⎟
⎝ ⎠
if a given value of the current is desired, then it is dependent on the value of the battery emf.
25. (a) Yes. (b) Yes.
26. (a) The frequency of the source emf does not affect the impedance of a pure resistance. The impedance of a pure resistance is independent of the source emf frequency.
(b) The impedance of a pure capacitance varies inversely with the frequency of the source emf according to XC =1/2πfC. As the source frequency gets very small, the impedance of the capacitor gets very large, and as the source frequency gets very large, the impedance of the capacitor gets very small.
(c) The impedance of a pure inductance varies directly with the frequency of the source emf according to XL =2π fL. Thus, as the source frequency gets very small, the impedance of the inductor gets very small, and as the source frequency gets very large, the impedance of the inductor gets very large.
(d) The impedance of an LRC circuit with a small R near resonance is very sensitive to the
frequency of the source emf. If the frequency is set at resonance exactly, where XL =XC, then the LRC circuit’s impedance is very small and equal to R. The impedance increases rapidly as the source frequency is either increased or decreased a small amount from resonance.
(e) The impedance of an LRC circuit with a small R very far from resonance depends on whether the source frequency is much higher or much lower than the resonance frequency. If the source frequency is much higher than resonance, then the impedance is directly proportional to the frequency of the source emf. Basically, at extremely high frequencies, the circuit impedance is equal to 2πfL. If the source frequency is much lower than resonance (nearly zero), then the impedance is inversely proportional to the frequency of the source emf. Basically, at extremely low frequencies, the circuit impedance is equal to 1/2π fC.
27. To make the impedance of an LRC circuit a minimum, make the resistance very small and make the reactance of the capacitor equal to the reactance of the inductor: XL =XC, or
1 1
2 .
2 2
fL f
fC LC
π
π π
= → =
28. In an LRC circuit, the current and the voltage in the circuit both oscillate. The energy stored in the circuit also oscillates and is alternately stored in the magnetic field of the inductor and the electric field of the capacitor.
29. Yes. The instantaneous voltages across the different elements in the circuit will be different, but the current through each element in the series circuit is the same.
Responses to MisConceptual Questions
1. (b, d) The right-hand rule shows that a clockwise current will create a flux in the loop that points into the page. Since the initial magnetic flux is into the page, the current will be induced when the magnetic flux decreases. This happens when the size of the coil decreases or the magnetic field becomes tilted. Increasing the magnetic field, as in answer (a), will create a counterclockwise current. Moving the coil sideways, as in answer (c), does not change the flux, so no current would be induced.
3. (d) A current is induced in the loop when the flux through the loop is changing. As the loop passes through line J it enters a region with a magnetic field, so the flux through the loop increases and a current will be induced. When the loop passes line K, the flux remains constant, as there is no change in field, so no current is induced. As the loop passes line L, the magnetic field flux through the loop decreases and a current is again induced.
4. (c,a) The magnetic field near a long straight wire is inversely proportional to the distance from the wire. For C, the loop remains at the same distance from the wire, so the magnetic flux through the wire remains constant and no current is induced in the loop. For D, the magnetic field from the long wire points into the page in the region of the loop. As the loop moves away from the wire, the magnetic flux decreases, so a clockwise current is induced in the loop to oppose the change in flux.
5. (c) The induced current creates a magnetic field that opposes the motion of the magnet. The result is that the magnet has less kinetic energy than it would have had if the loop were not present. The change in gravitational potential energy provides the energy for the current.
6. (c) Since the flux through the loop is increasing, an emf will be produced. However, since plastic is not a conductor, no current will be induced. The emf is produced regardless of whether there is a conducting path for current or not.
7. (b) A current will be induced in the loop whenever the magnetic flux through the loop is changed. Increasing the current in the wire will increase the magnetic field produced by the wire and therefore the magnetic flux in the loop. Rotating the loop changes the angle between the loop and wire, which will change the flux. Since the magnetic field strength decreases with distance from the wire, moving the loop away from the wire, either with or without rotation, will decrease the flux in the wire. If the loop is moved parallel to the wire, then the flux through the loop will not change, so no current will be created in the loop.
8. (c) When a steady current flows in the first coil, it creates a constant magnetic field and therefore a constant magnetic flux through the second coil. Since the flux is not changing, a current will not be induced in the second coil. If the current in the first coil changes, then the flux through the second will also change, inducing a current in the second coil.
9. (b) A generator converts mechanical energy into electric energy. The generator’s magnetic field remains unchanged as the generator operates, so energy is not being pulled from the magnetic field. Resistance in the coils removes electric energy from the system in the form of heat—it does not provide the energy. In order for the generator to work, an external force is necessary to rotate the generator’s axle. This external force does work, which is converted to electric energy. 10. (c) Increasing the rotation frequency increases the rate at which the flux changes; therefore, answer (a)
will increase the output voltage. Answers (b), (d), and (e) all increase the maximum flux in the coil. Since the generator’s voltage output is proportional to the rate of change of the flux through the coil, increasing the maximum flux results in a greater voltage output. When the magnetic field is parallel to the generator’s axis, little to no flux passes through the coil. With less flux passing through the coil, there will be a lower output voltage.
12. (d) If the charger unit had a battery, then it could run the laptop without being plugged in. A motor converts electrical energy into mechanical energy, but the laptop charger output is electrical energy, not mechanical energy. A generator converts mechanical energy into electric energy, but the input to the laptop is electrical, not mechanical. The cables to and from the charger unit can be considered transmission lines, but they are not the important component inside the charger unit. A transformer can convert a high-voltage input into a low-voltage output without power loss. This is a significant function of the charger unit.
13. (a) In a transformer with no lost flux, the power across the transformer (product of current and voltage) is constant across the transformer. Therefore, if the voltage increases, then the current must decrease across the transformer. If the current increases, then the voltage must decrease across the transformer. A transformer works due to the induced voltage created by the changing flux. A dc circuit does not have a changing flux, so a transformer does not work with dc current. 14. (e) It may appear that (b) is the correct answer if the problem is interpreted as an ac step-up
transformer with twice as many loops in the secondary coil as in the primary. It is true that an ac current would double the voltage and cut the current in half; however, this is a dc current. A dc current does not produce a changing flux, so no current or voltage will be induced in the secondary coil.
15. (b) Many people may not realize that generators (rotating coils in a magnetic field) are the heart of most electric power plants that produce the alternating current in wall outlets.
16. (b) A common misconception by users of credit card readers is that they are swiping their cards too quickly, so they slow down the swipe. The credit card has a magnetic stripe with information encoded in the magnetic field. Swiping the card more rapidly increases the induced emf, due to the greater rate of change of magnetic flux.
17. (b) In a series circuit (whether ac or dc), the current is the same at every point in the circuit. In an ac circuit, the current is out of phase with the voltage across an inductor and the voltage across a capacitor, so (a) is incorrect. In nonresonant ac circuits, there is a phase difference between the current and the voltage source, so (c) is not correct. The current and voltage across a resistor are always in phase, so the resistor does not change the phase in a circuit, so (d) is incorrect.
Solutions to Problems
1. The magnitude of the average induced emf is given by Eq. 21–2b. 38 Wb ( 58 Wb)
2 564.7 V 560 V
0.34 s B
N t
ΔΦ − −
= = = ≈
Δ Ᏹ
A direction for the emf cannot be specified without more information.
2. As the magnet is pushed into the coil, the magnetic flux increases to the right. To oppose this increase, flux produced by the induced current must be to the left, so the induced current in the resistor will be from right to left.
4. As the solenoid is pulled away from the loop, the magnetic flux to the right through the loop decreases. To oppose this decrease, the flux produced by the induced current must be to the right, so the induced current is counterclockwise as viewed from the right end of the solenoid.
5. The flux changes because the loop rotates. The angle between the field and the normal to the loop changes from 0° to 90°. The magnitude of the average induced emf is given by Eq. 21–2a.
2
cos (0.0925 m) 1.5 T( cos 90 cos 0 ) 0.20 V 0.20 s
B AB
t t
θ π
ΔΦ Δ ° − °
= − = − = − =
Δ Δ
Ᏹ
6. We choose up as the positive direction. The average induced emf is given by Eq. 21–2a. 2
2 (0 054 m) ( 0 25 T 0 48 T)
4.2 10 V 0.16 s
B A B
t t
π −
ΔΦ Δ . − . − .
= − = − = − = ×
Δ Δ
Ᏹ
7. (a) When the plane of the loop is perpendicular to the field lines, the flux is given by the maximum of Eq. 21–1.
2 (0 50 T) (0 080 m)2 1.005 10 2Wb 1.0 10 2Wb
B BA B rπ π − −
Φ = = = . . = × ≈ ×
(b) The angle is θ = ° − ° =90 42 48 .°
(c) Use Eq. 21–1.
2 2 3
cos (0.50 T) (0.080 m) cos 48 6.7 10 Wb
B BA θ B rπ π −
Φ = = = ° = ×
8. (a) As the resistance is increased, the current in the outer loop will decrease. Thus the flux through the inner loop, which is out of the page, will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise.
(b) If the small loop is placed to the left, then the flux through the small loop will be into the page and will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux into the page, so the direction of the induced current will be clockwise.
9. (a) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from Eq. 21–3.
2 (0.800 T)(0.120 m)(0.180 m/s) 1.73 10 V
B υ −
= A = = ×
Ᏹ
(b) Because the upward flux is increasing, the induced flux will be into the page, so the induced current is clockwise. Thus the induced emf in the rod is down, which means that the electric field will be down. The electric field is the induced voltage per unit length.
2 1.73 10 V
0.144 V/m, down 0.120 m
E= = × − =
A Ᏹ
10. (a) The magnetic flux through the loop is into the page and decreasing, because the area is
decreasing. To oppose this decrease, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise.
(b) The average induced emf is given by Eq. 21–2a.
2 2
avg
2 2
(0.65 T) [(0.100 m) (0.030 m) ] 0 50 s
3.717 10 V 3.7 10 V
B B A
t t
π
− −
Δ
ΔΦ −
= = =
Δ Δ .
= × ≈ ×
(c) We find the average induced current from Ohm’s law. 2
2 3.717 10 V
1.5 10 A 2.5
I R
−
− ×
= = = ×
Ω Ᏹ
11. (a) Because the current is constant, there will be no change in flux, so the induced current will be zero.
(b) The decreasing current in the wire will cause a decreasing field into the page through the loop. To oppose this decrease, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise.
(c) The decreasing current in the wire will cause a decreasing field out of the page through the loop. To oppose this decrease, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise.
(d) The increasing current in the wire will cause an increasing field out of the page through the loop. To oppose this increase, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise.
12. The emf induced in the short coil is given by Eq. 21–2b, where N is the number of loops in the short coil, and the flux change is measured over the area of the short coil. The magnetic flux comes from the B field created by the solenoid. The field in a solenoid is given by Eq. 20–8, B=μ0INsolenoid/Asolenoid, and the changing current in the solenoid causes the field to change.
0 solenoid short short
solenoid
short short 0 short solenoid short
solenoid
7 2
4 (4 10 T m/A)(14)(600) (0.0125 m) (5.0 A)
1.7 10 V (0.25 m) (0.60 s)
IN
N A
N A B N N A I
t t t
μ
μ
π − π −
⎛ ⎞
Δ⎜ ⎟
Δ ⎝ ⎠ Δ
= = =
Δ Δ Δ
× ⋅
= = ×
A
A Ᏹ
The induced emf will oppose the emf that is being used to create the 5.0-A current.
13. Since the antenna is vertical, the maximum emf will occur when the car is traveling perpendicular to the horizontal component of the Earth’s magnetic field. This occurs when the car is traveling in the east or west direction. We calculate the magnitude of the emf using Eq. 21–3, where B is the horizontal component of the Earth’s magnetic field.
5 4
[(5.0 10 T) cos 38 ](0.55 m)(30.0 m/s) 6.501 10 V 0.65 mV x
B υ − −
= A = × ° = × =
Ᏹ
14. As the loop is pulled from the field, the flux through the loop decreases, causing an induced emf whose magnitude is given by Eq. 21–3, Ᏹ=BAυ. Because the inward flux is decreasing, the induced flux will be into the page, so the induced current is clockwise, given by I =Ᏹ/ .R Because this current in the left-hand side of the loop is in a downward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, given by F=I BA .
2 2 (0.550 T) (0.350 m) (3.10 m/s)2 2
0.499 N 0.230
B B
F I B B B
R R R
υ
= = = = = =
Ω
A A
A ᏱA A
15. (a) There is a conventional current flowing in the rod, from top to bottom, equal to the induced emf divided by the resistance of the bar. That current is in a magnetic field and, by the right-hand rule, will have a force on it (and the bar) to the left in Fig. 21–11a. The external force must equal that magnetic force in order for the bar to move with a constant speed.
2 2
external magentic B B
F F I B B B
R R R
υ υ
(b) The power required is the external force times the speed of the rod. 2 2 2
B P F
R
υ υ
= = A
16. From the derivation in Problem 15a, we have an expression relating the external force to the magnetic field.
2 2
external
external 2 (0.350 N)(0.25 )2 1.208 T 1.2 T
(0.200 m) (1.50 m/s)
F R
B
F B
R
υ
υ
Ω
= A → = = = ≈
A
17. (a) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from Eq. 21–3.
(0.35 T)(0.300 m)(1.6 m/s) 0.168 V 0.17 V B υ
= A = = ≈
Ᏹ
(b) Find the induced current from Ohm’s law.
3 3
0.168 V
7.149 10 A 7.1 10 A 21.0 2.5
I R
− −
= = = × ≈ ×
Ω + Ω
Ᏹ
(c) The induced current in the rod will be down. Because this current is in an upward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, given by Eq. 20–2.
3 4 4
(7.149 10 A)(0.300 m)(0.35 T) 7.506 10 N 7.5 10 N
F=I BA = × − = × − ≈ × −
18. (a) There is an emf induced in the coil since the flux through the coil changes. The current in the coil is the induced emf divided by the resistance of the coil. The resistance of the coil is found from Eq. 18–3.
coil
wire B
NA R
t A
ρ Δ
= =
Δ
A Ᏹ
coil
coil wire wire
2 3 3
8
30[ (0.110 m) ][ (1.3 10 m)](8.65 10 T/s)
0.1504 A 0.15 A (1.68 10 m)30(2 )(0.110 m)
B
NA NA A B
t I
R t
A
ρ ρ
π π
π
− −
−
Δ
Δ Δ
= = =
Δ
× ×
= = ≈
× Ω ⋅
A A
Ᏹ
(b) The rate at which thermal energy is produced in the wire is the power dissipated in the wire. 8
2 2 2
3 2
wire
3 3
(1.68 10 m)30(2 )(0.110 m) (0.1504 A)
(1.3 10 m) 1.484 10 W 1.5 10 W
P I R I A
ρ π
π
−
−
− −
× Ω⋅
= = =
×
= × ≈ ×
A
19. The charge that passes a given point is the current times the elapsed time, Q I t= Δ. The current will be the emf divided by the resistance, I .
R
= Ᏹ The resistance is given by Eq. 18–3,
wire , R
A
ρ
= A and the emf is given by Eq. 21–2b. Combine these equations to find the charge during the operation.
loop
loop loop wire
wire
wire
; ;
B
A B
A B t A A B
R I
t t A R t
A
ρ
ρ ρ
Δ
Δ Δ
ΔΦ Δ
= = = = = =
Δ Δ Δ
A
A A
2 2 2 loop wire loop wire loop wire
loop
3 2
8
(2 ) 2
(0.066 m) (1.125 10 m) (0.670 T)
5.23 C
2(1.68 10 m)
A A B r r B r r B
Q I t
r
π π π
ρ ρ π ρ
π −
−
Δ Δ Δ
= Δ = = =
×
= =
× Ω⋅ A
20. From Eq. 21–5, the induced voltage is proportional to the angular speed. Thus their quotient is a constant.
1 2 2
2 1
1 2 1
2500 rpm
(12.7 V) 28.86 V 29 V 1100 rpm
ω
ωᏱ = ωᏱ → Ᏹ =Ᏹ ω = = ≈
21. (a) The rms voltage is found from the peak induced emf. The peak induced emf is calculated from Eq. 21–5.
peak
2 peak
rms (550)(0.55 T)(2 rad/rev)(120 rev/s) (0.040 m)
2 2 2
810.7 V 810 V NB A
NB A V
ω
ω π π
= →
= = =
= ≈
Ᏹ
Ᏹ
(b) To double the output voltage, you must double the rotation frequency to 240 rev/s. 22. (a) The peak current is found from the rms current.
peak 2 rms 2(70.0 A) 99.0 A
I = I = =
(b) The area can be found from Eq. 21–5.
peak rms
2 2 2
rms 2
2 2 (150 V)
0.01394 m 1.4 10 m (950)(0.030 T)(85 rev/s)(2 rad/rev)
NB A V
V A
NB
ω
ω π −
= = →
= = = ≈ ×
Ᏹ
23. We assume that the voltage given is an rms value and that the power is an average power, as in Eq. 18–9a. We calculate the resistance of the wire on the armature using Eq. 18–3.
8
wire 1 4 2
2 rms
rms
circuit bulb wire bulb wire
(1.68 10 m)[85(4)(0.060 m)] 1.254 [ (5.9 10 m)]
25.0 W
2.083 A 12.0 V
12 V (2.083 A)(1.254 ) 14.61 V R
A P I
V
V V V V IR
ρ
π
−
−
× Ω⋅
= = = Ω
×
= = =
= + = + = + Ω =
A
So the generator’s rms emf must be 14.61 volts. We find the frequency of the generator from Eq. 21–5.
2
2 14.61 V
2 2
(14.61 V) 2 (14.61 V) 2 16.54 Hz 17 Hz 17 cycles/s 2 2 (85)(0.65 T)(0.060 m)
fNBA
f
NBA
π
π π
= = →
= = = ≈ =
Ᏹ
24. When the motor is running at full speed, the back emf opposes the applied emf, to give the net voltage across the motor.
applied− back =IR→ back = applied−IR=120 V (8.20 A)(3.65 )− Ω = 90 V
Ᏹ Ᏹ Ᏹ Ᏹ
25. From Eq. 21–5, the induced voltage (back emf) is proportional to the angular speed. Thus their quotient is a constant.
1 2 2 2 2
2 1 1 1
1 2 1 1 1
2 (72 V)2300 rpm 92 V
2 1800 rpm
f f
f f
ω π
ω = ω → = ω = π = = =
Ᏹ Ᏹ Ᏹ Ᏹ Ᏹ Ᏹ
26. The back emf is proportional to the rotation speed (Eq. 21–5). Thus if the motor is running at half speed, then the back emf is half the original value, or 54 V. Find the new current from writing a loop equation for the motor circuit, from Fig. 21–19.
back
back 0 120 V 54 V 13 A
5.0
IR I
R
− −
− − = → = = =
Ω Ᏹ Ᏹ
Ᏹ Ᏹ
27. We find the number of turns in the secondary from Eq. 21–6.
S S S
S P
P P P
13,500 V
(148 turns) 17,100 turns 117 V
V N V
N N
V = N → = V = ≈
28. Because NS<NP, this is a step-down transformer. Use Eq. 21–6 to find the voltage ratio and Eq. 21–7 to find the current ratio.
S S S P
P P P S
120 turns 1 360 turns
or 0.33 3.0
360 turns 3 120 turns
V N I N
V = N = = I = N = =
29. Use Eqs. 21–6 and 21–7 to relate the voltage and current ratios.
S S S P S P S P
P P P S P S P S
25 V
; 0.21
120 V
V N I N V I I V
V = N I = N → V = I → I =V = = 30. We find the ratio of the number of turns from Eq. 21–6.
S S
P P
12, 000 V 50 240 V
N V
N =V = = (2 significant figures)
If the transformer is connected backward, then the role of the turns will be reversed:
S S S
S
P P
1 1
(240 V) 4.8 V
50 240 V 50
N V V
V
N =V → = → = =
31. (a) Use Eqs. 21–6 and 21–7 to relate the voltage and current ratios.
S S S P S P P
S P
P P P S P S S
0.35 A
; (120 V) 6.2 V
6.8 A
V N I N V I I
V V
V = N I = N → V = I → = I = =
(b) Because VS<VP, this is a step-down transformer.
32. (a) We assume 100% efficiency and find the input voltage from P=IV.
P P P
P
95 W 3.8 V 25 A P
P I V V
I
= → = = =
Since VP<VS, this is a step-up transformer. (b) S
P
12 V 3.2 3.8 V V
V = =
33. Use Eqs. 21–6 and 21–7 to relate the voltage and current ratios.
S S S
S P
P P P
1240 turns
(120 V) 450.9 V 450 V
330 turns
V N N
V V
S P S
P S
P S P
1240 turns
(15.0 A) 56 A
330 turns
I N N
I I
I = N → = N = =
34. (a) The current in the transmission lines can be found from Eq. 18–9a, and then the emf at the end of the lines can be calculated from Kirchhoff’s loop rule.
6 town
town rms rms rms 3
rms output
6
3 town
output rms 3
rms
35 10 W
778 A 45 10 V
0
35 10 W
(4.6 ) 45 10 V 48,578 V 49 kV(rms) 45 10 V
P
P V I I
V IR V
P
IR V R V
V
×
= → = = =
×
− + = →
×
= + = + = Ω + × = ≈
× Ᏹ
Ᏹ
(b) The power loss in the lines is given by Ploss =Irms2 R.
2 2
loss loss rms
2 6 2
total town loss town rms
(778 A) (4.6 ) Fraction wasted
35 10 W (778 A) (4.6 ) 0.074 7.4%
P P I R
P P P P I R
Ω
= = = =
+ + × + Ω
= =
35. At the power plant, 100 kW of power is delivered at 12,000 V. We first find the initial current from Eq. 18–9a.
3 3 100 10 W
8.333 A 12 10 V
P
P VI I
V
×
= → = = =
×
Then we have a step-up transformer, which steps up the voltage by a factor of 20, so the current is stepped down by a factor of 20.
S P P
S P
P S S
1
(8.333 A) 0.4167 A 20
I N N
I I
I = N → = N = =
The power lost is given by Eq. 18–6a.
2 3 2 5
lost
3
9 9
2 5
50 10 W (0.4167 A) (5 10 /m)( )
50 10 W 5.76 10 m 6 10 m
(0.4167 A) (5 10 /m)
P I R x
x
−
−
= = → × = × Ω →
×
= = × ≈ ×
× Ω
Note that we are not taking into account any power losses that occur after the step-down transformers. 36. (a) At the power plant, the voltage is changed from 12,000 V to 240,000 V.
S S
P P
240,000 V
20 (2 significant figures) 12,000 V
N V
N =V = =
(b) At the house, the voltage is changed from 7200 V to 240 V. 2
S S
P P
240 V 3.3 10 7200 V
N V
N V
−
= = = ×
37. Without the transformers, we find the delivered current, which is the current in the transmission lines, from the delivered power and the power lost in the transmission lines.
6
4 out
out out line line out
2 4 2 7
lost line line
2.0 10 W
1.667 10 A 120 V
(1.667 10 A) 2(0.100 ) 5.556 10 W P
P V I I
V
P I R
×
= → = = = ×
Thus there must be 2.0 10 W 5.556 10 W 5.756 10 W× 6 + × 7 ≈ × 7 of power generated at the start of the process.
With the transformers, to deliver the same power at 120 V, the delivered current from the step-down transformer must still be 1.667 10 A.× 4 Using the step-down transformer efficiency, we calculate the current in the transmission lines and the loss in the transmission lines.
out line out out line line
end
4
3 out out
line
line
2 3 2 5
lost line line
0.99 0.99
(120 V)(1.66 10 A)
1.684 10 A 0.99 (0.99)(1200 V)
(1.684 10 A) 2(0.100 ) 5.671 10 W 0.5671 MW
P P V I V I
V I I
V
P I R
= → = →
×
= = = ×
= = × Ω = × =
The power to be delivered is 2.0 MW. The power that must be delivered to the step-down transformer is 2 0 MW 2.0202 MW.
0.99
. = The power that must be present at the start of the transmission must be
2 0202 MW 0 5671M 2 5873MW. + . = . to compensate for the transmission line loss. The power that must enter the transmission lines from the 99% efficient step-up transformer is
2.5873 MW 2.6134 MW.
0.99 = So the power saved is as follows.
7 6 7
5.756 10 W 2.6134 10 W 5.495 10 W× − × = × ≈ 55 MW
38. We set the power loss equal to 2.5% of the total power. Then, using Eq. 18–6a, we write the power loss in terms of the current (equal to the power divided by the voltage drop) and the resistance. Then, using Eq. 18–3, we calculate the cross-sectional area of each wire and the minimum wire diameter. We assume there are two lines to have a complete circuit.
2 2
2
loss 2
6 8 3
2 3 2
0 025
4 0.025
4 4(925 10 W)(2.65 10 m)2(185 10 m)
0.03256 m 3.3 cm
0.025 0.025 (660 10 V)
P P d
P P I R A
V A V
P d
V
ρ ρ π
ρ
π π
− ⎛ ⎞ ⎛ ⎞
= . = =⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎟⎠ → = = →
× × Ω⋅ ×
= = = ≈
×
A A
A
The transmission lines must have a diameter greater than or equal to 3.3 cm. 39. Find the induced emf from Eq. 21–9.
(10.0 A 25.0 A)
(0.16 H) 6.9 V
0.35 s I
L t
Δ −
= − = − =
Δ Ᏹ
40. Because the current is increasing, the emf is negative. We find the self-inductance from Eq. 21–9. 0.0140 s
( 2.50 V) 0.593 H
[0.0310 A ( 0.0280 A)]
I t
L L
t I
Δ Δ
= − → = − = − − =
Δ Δ − −
Ᏹ Ᏹ
41. Use the relationship for the inductance of a solenoid, as given in Example 21–13.
2 7 2 2 2
0 (4 10 T m/A)(8500) (1.45 10 m) 0.10 H 0.60 m
N A
L=μ = π× − ⋅ π × − =
A
42. Use the relationship for the inductance of a solenoid, as given in Example 21–13. 2
0
7 2
0
(0.13 H)(0.300 m) 3400 turns (4 10 T m/A) (0.029 m)
N A L
L N
A
μ
μ π − π
= → = = ≈
× ⋅
43. (a) Use the relationship for the inductance of a solenoid, as given in Example 21–13.
2 7 2 2
0 (4 10 T m/A)(2600) (0.0125 m) 0.01479 H 0.015 H (0.282 m)
N A
L=μ = π× − ⋅ π = ≈
A
(b) Apply the same equation again, solving for the number of turns but using the permeability of iron.
2
7 2
(0.01479 H)(0.282 m) 75 turns (1200)(4 10 T m/A) (0.0125 m)
N A L
L N
A
μ
μ π − π
= → = = ≈
× ⋅
A A
44. We draw the coil as two elements in series, a pure resistance and a pure inductance. There is a voltage drop due to the resistance of the coil, given by Ohm’s law, and an induced emf due to the
inductance of the coil, given by Eq. 21–9. Since the current is increasing, the inductance will create a potential difference to oppose the increasing current, so there is a drop in the potential due
to the inductance. The potential difference across the coil is the sum of the two potential drops. ab I (3.00 A)(2.25 ) (0.112 H)(3.80 A/s) 7.18 V
V IR L t
Δ
= + = Ω + =
Δ
45. The self-inductance of an air-filled solenoid was determined in Example 21–13. We solve this equation for the length of the tube, using the diameter of the wire as the length per turn.
2
2
0 0
0 2
2 3 2
2 7 2
0
(1.0 H)(0.81 10 m)
46.16 m 46 m (4 10 T m/A) (0.060 m)
N A A
L n A
d Ld
r
μ μ μ
μ π π π
− −
= = =
×
= = = ≈
× ⋅
A A A
A
The length of the wire needed (L) is equal to the number of turns (the length of the solenoid divided by the diameter of the wire) multiplied by the circumference of the turn.
3 46.16 m
(0.12 m) 21,490 m 21 km 0.81 10 m
L D
dπ − π
= = = ≈
×
A
The resistance is calculated from the resistivity, area, and length of the wire. 8
3 2
(1.68 10 m)(21,490 m)
0.70 k (0.405 10 m)
R A
ρ
π
− −
× Ω⋅
= = = Ω
×
A
46. We assume that the solenoid and the coil have the same cross-sectional area. The magnetic field of the solenoid (which passes through the coil) is B1=μ0N I1 1.
A When the current in the solenoid changes, the magnetic field of the solenoid changes, and thus the flux through the coil changes, inducing an emf in the coil.
1 1
2 0
2 1 2 1
0 ( )
N I N A
N A B N N A I
t t t
μ
μ
⎛ ⎞
Δ⎜ ⎟
Δ ⎝ ⎠ Δ
= − = − = −
Δ Δ Δ
A
A Ᏹ
As in Eq. 21–8a, the mutual inductance is the proportinality constant in the above relationship.
1 2 1 1 2
0N N A ( )I M 0N N A t
μ Δ μ
= − → =
Δ
A A
Ᏹ
a b
increasing
I
− +
induced
Ᏹ
47. The magnetic energy in the field is derived from Eq. 21–10. 2
1 2
0
2 2 2
2 2
1 1 1
2 2 2 7
0 0
Energy stored Volume
(0.72 T)
Energy (Volume) (0.010 m) (0.36 m) 23 J
(4 10 T m/A) B
u
B B r L
μ
π π
μ μ π −
= = →
= = = =
× ⋅
48. The initial energy stored in the inductor is found from an equation in Section 21–11.
2 3 3 2 5
1 1
2 2(45.0 10 H)(50.0 10 A) 5.63 10 J
U= LI = × − × − = × −
The final current is found from the final energy, which is 5 times the initial energy.
2 2
1 1
f 5 0 2 f 5(2 0) f 5 0
U = U → LI = LI → I = I
The current increases at a constant rate.
f 0
3
f 0 0 0 0
0.115 A/s
5 (50.0 10 A)
( 5 1) ( 5 1) 0.537 s
0.115 A/s 0.115 A/s 0.115 A/s 0.115 A/s I I
I
t t
I I I I I
t −
−
Δ = = →
Δ Δ
− − ×
Δ = = = − = − =
49. The magnetic energy in the field is derived from Eq. 21–10. The volume of a relatively thin spherical shell, like the first 10 km above the Earth’s surface, is the surface area of the sphere times its thickness.
2 1 2
0
2 4 2
2 6 2 4 15
1 1
Earth
2 2 7
0
(0.50 10 T)
(volume) 4 4 (6.38 10 m) (10 m) 5 10 J
(4 10 T m/A) B
u
B
E u R h
μ
π π
μ π
− −
= →
×
= = = × ≈ ×
× ⋅
50. (a) We set I equal to 75% of the maximum value in the equation I =I0(1−e−t/τ)and solve for the time constant.
/
0 0 (2.56 ms)
0.75 (1 ) 1.847 ms 1.8 ms
ln(0.25) ln(0.25 ms)
t t
I= I =I −e− τ → τ= − = − = ≈
(b) The resistance can be calculated from the time constant, τ =L R/ . 31.0 mH
17 1.847 ms L
R
τ
= = = Ω
51. The potential difference across the resistor is proportional to the current, so we set the current in the equation I=Imaxe2t/τ equal to 0.025I0 and solve for the time.
/
0 0
0.025 t ln(0.025) 3.7
I= I =I e− τ → t= −τ ≈ τ
52. When the switch is initially closed, the inductor prevents current from flowing, so the initial current is 0, as shown in Fig. 21–37. If the current is 0, then there is no voltage drop across the resistor (since
R ),
V =IR so the entire battery voltage appears across the inductor. Apply Eq. 21–9 to find the initial rate of change of the current.
L I I V
V V L
t t L
Δ Δ
= = → =
The maximum value of the current is reached after a long time, when there is no voltage across the inductor, so the entire battery voltage appears across the resistor. Apply Ohm’s law.
max max V
V I R I
R
= → =
Find the time to reach the maximum current if the rate of current change remained at I V. t L Δ = Δ max 0 I(elapsed time) elapsed time ( max 0) t V 0 L L
I I I I
t I R V R
Δ Δ ⎛ ⎞
= +Δ → = − Δ =⎜ − ⎟ =
⎝ ⎠
53. For an LR circuit, we have I=Imax(1−e−t/τ). Solve for t.
/ /
max
max max
(1 t ) t 1 I ln 1 I
I I e e t
I I
τ τ τ
− − ⎛ ⎞
= − → = − → = − ⎜ − ⎟
⎝ ⎠
(a) max
max
0.90 ln 1 I ln(1 0.9) 2.3
I I t
I
τ ⎛ ⎞ τ τ
= → = − ⎜ − ⎟= − − =
⎝ ⎠
(b) max
max
0.990 ln 1 I ln(1 0.99) 4.6
I I t
I
τ ⎛ ⎞ τ τ
= → = − ⎜ − ⎟= − − =
⎝ ⎠
(c) max
max
0.999 ln 1 I ln(1 0.999) 6.9
I I t
I
τ ⎛ ⎞ τ τ
= → = − ⎜ − ⎟= − − =
⎝ ⎠
54. The reactance of a capacitor is given by Eq. 21–12b, 1 . 2 C X
fC
π
=
(a) 1 1 6 428
2 2 (60.0 Hz)(6.20 10 F) C
X
fC
π π −
= = = Ω
×
(b) 1 6 1 6 2.57 10 2
2 2 (1.00 10 Hz)(6.20 10 F) C
X
fC
π π
− −
= = = × Ω
× ×
55. We find the frequency from Eq. 21–11b for the reactance of an inductor. 660
2 3283 Hz 3300 Hz
2 2 (0 0320 H) L
L X
X fL f
L
π
π π
Ω
= → = = = ≈
.
56. We find the frequency from Eq. 21–12b for the reactance of a capacitor.
3 6
1 1 1
10.9 Hz
2 2 2 (6.10 10 )(2.40 10 F)
C
C
X f
fC X C
π π π −
= → = = =
× Ω ×
57. We find the reactance from Eq. 21–11b and the current from Ohm’s law. 3
2
2 2 (10.0 10 Hz)(0.26 H) 16,336 16 k 240 V
1.47 10 A 16,336
L
L
L
X fL
V
V IX I
X
π π
−
= = × = Ω ≈ Ω
= → = = = ×
Ω
58. We find the reactance from Ohm’s law and the inductance by Eq. 21–11b.
2 240 V
2 5.22 10 H
2 2 2 (60.0 Hz)12.2 A
L L
L L
V
V IX X
I
X V
X fL L
f fI
π
π π π −
= → =
59. (a) We find the reactance from Eq. 21–12b.
8
1 1 7368 7400
2 2 (720 Hz)(3.0 10 F) C
X
fC
π π −
= = = Ω ≈ Ω
×
(b) We find the peak value of the current from Ohm’s law. 3
rms
peak 2 rms 2 22.0 10 V7368 0.38 A C
V
I I
X
×
= = = =
Ω
60. We find the impedance from Eq. 21–14. rms
3 rms
120 V
1700 2000 70 10 A
V Z
I −
= = = Ω ≈ Ω
×
61. The impedance of the circuit is given by Eq. 21–15 without a capacitive reactance. The reactance of the inductor is given by Eq. 21–11b.
(a) Z = R2+XL2 = R2+4π2 2 2f L = (36 10× 3Ω +)2 4π2(50 Hz) (55 102 × −3H)2 = 3.6 10× 4Ω The inductor has essentially no effect on the impedance at this relatively low frequency. (b) Z = R2+XL2 = R2+4π2 2 2f L = (36 10× 3Ω +)2 4π2(3.0 10 Hz) (55 10× 4 2 × −3H)2
4 37, 463 3.7 10
= Ω ≈ × Ω
62. The impedance of the circuit is given by Eq. 21–15 without an inductive reactance. The reactance of the capacitor is given by Eq. 21–12b.
(a) 2 2 2 2 2 21 (3.5 103 )2 2 21 6 2
4 4 (60 Hz) (3.0 10 F)
C
Z R X R
f C
π π −
= + = + = × Ω +
×
3609 3600
= Ω ≈ Ω
(b) 2 2 2 2 2 21 (3.5 103 )2 2 12 6 2
4 4 (60,000 Hz) (3.0 10 F)
C
Z R X R
f C
π π −
= + = + = × Ω +
×
3500
= Ω
63. Use Eq. 21–15, with no capacitive reactance.
2 2 2 2 (235 )2 (115 )2 205
L L
Z = R +X → R= Z −X = Ω − Ω = Ω
64. The total impedance is given by Eq. 21–15.
2
2 2 2
2
3 2 4 2
4 9
1
( ) 2
2
1 (8.70 10 ) 2 (1.00 10 Hz)(2.80 10 H)
2 (1.00 10 Hz)(6.25 10 F) 8735.5 8.74 k
L C
Z R X X R fL
fC
π π
π
π
−
−
⎛ ⎞
= + − = +⎜ − ⎟
⎝ ⎠
⎡ ⎤
= × Ω +⎢ × × − ⎥
× ×
⎢ ⎥
⎣ ⎦
The phase angle is given by Eq. 21–16a.
1 1
4 2
4 9
1 1
1 2
2
tan tan
1 2 (1.00 10 Hz)(2.80 10 H)
2 (1.00 10 Hz)(6.25 10 F) tan
787
tan 5.17
8700
L C
fL
X X fC
R R
R
π π φ
π
π
− −
−
− −
−
− −
= =
× × −
× ×
=
− Ω
= = − °
Ω
The voltage is lagging the current, or the current is leading the voltage. The rms current is given by Eq. 21–14.
2 rms
rms 725 V 8.30 10 A
Z 8735.5 V
I = = = × −
Ω
65. We use the rms voltage across the resistor to determine the rms current through the circuit. Then, using the rms current and the rms voltage across the capacitor in Eq. 21–13b, we determine the frequency.
, rms rms
rms , rms
rms rms
6
, rms , rms
2
(3.0 V) 272.1 Hz 270 Hz
2 2 2 (1.0 10 C)(650 )(2.7 V)
R
C
R,
C C
V I
I V
R fC
V I
f
CV CRV
π
π π π −
= =
= = = = ≈
× Ω
Since the voltages in the resistor and capacitor are not in phase, the rms voltage across the power source will not be the sum of the rms voltages across the resistor and capacitor.
66. (a) The rms current is the rms voltage divided by the impedance. The impedance is given by Eq. 21–15 with no capacitive reactance.
2 2 2 2
rms rms
rms
2 2 2 2 3 2 2 2 2
2 (2 )
120 V
4 (2.80 10 ) 4 (60.0 Hz) (0.35 H) 120 V
0.04281 A 4.3 10 A 2803
L
Z R X R fL
V V
I
Z R f L
π
π π
−
= + = +
= = =
+ × Ω +
= = ≈ ×
Ω
(b) The phase angle is given by Eq. 21–6a with no capacitive reactance.
1 1 1
3 2 2 (60.0 Hz)(0.35 H)
tan tan tan 2.7
2.80 10 L
X fL
R R
π π
φ= − = − = − = °
× Ω
The current is lagging the source voltage.
(c) The power dissipated is given by P I= rms2 R=(4.281 10× −2A) (2.80 102 × 3Ω =) 5.1 W . (d) The rms voltage reading is the rms current times the resistance or reactance of the element.
2 3
rms rms
2
rms rms rms
(4.281 10 A)(2.8 10 ) 119.87 V 120 V
2 (4.281 10 A)2 (60.0 Hz)(0.35 H) 5.649 V 5.6 V R
L L
V I R
V I X I πfL π
−
−
= = × × Ω = ≈
= = = × = ≈
67. (a) The rms current is the rms voltage divided by the impedance. The impedance is given by Eq. 21–15 with no inductive reactance.
2 2 2
2
rms rms
rms
2 3 2
2 2 2 2 2 6 2
2 2
1 (2 )
120 V
1 (6.60 10 ) 1
4 4 (60.0 Hz) (1.80 10 F)
120 V
1.774 10 A 1.77 10 A 6763
C
Z R X R
fC
V V
I
Z R
f L
π
π π −
− −
= + = +
= = =
+ × Ω +
×
= = × ≈ ×
Ω
(b) The phase angle is given by Eq. 21–16a with no inductive reactance.
6
1 1 1
3 1 1
2 (60.0 Hz)(1.80 10 F) 2
tan tan tan 12.6
6.60 10 C
X fC
R R
π π
φ= − − = − − = − − × − = − °
× Ω
The current is leading the source voltage.
(c) The rms voltage reading is the rms current times the resistance or reactance of the element.
2 3
rms rms
2
rms rms rms 6
(1.774 10 A)(6.60 10 ) 117 V
1 (1 774 10 A) 1 26.1 V
2 2 (60.0 Hz)(1.80 10 F)
R
C C
V I R
V I X I
fC
π π
−
−
−
= = × × Ω =
= = = . × =
×
Note that, because the maximum voltages occur at different times, the two readings do not add up to the applied voltage of 120 V.
68. (a) The impedance of the circuit is given by Eq. 21–15. Divide the source voltage by the impedance to determine the magnitude of the current in the circuit. Finally, multiply the current by the resistance to determine the voltage drop across the resistor.
in in
2 2
2 6 2
1/(2 )
(130 mV)(520 ) 29.77 mV 30 mV (520 ) 1/[2 (60 Hz)(1.2 10 F)]
R V V R
V IR R
Z R π fC
π −
= = =
+
Ω
= = ≈
Ω + ×
(b) Repeat the calculation with a frequency of 6.0 kHz.
2 6 2
(130 mV)(520 ) 129.88 mV 130 mV (520 ) 1/[2 (6000 Hz)(1.2 10 F)]
R V
π −
Ω
= = ≈
Ω + ×
Thus the capacitor allows the higher frequency to pass but attenuates the lower frequency. 69. The resonant frequency is found from Eq. 21–19. The resistance does not influence the resonant
frequency.
5 5
0 1 1 1 6 1 12 3.627 10 Hz 3.6 10 Hz
2 2 (55.0 10 H)(3500 10 F) f
LC
π π − −
= = = × ≈ ×
70. (a) The resonant frequency is given by Eq. 21–19. 580 kHz 580 kHz 1600 kHz 1600 kHz 2 2
1600 kHz 580 kHz
1 1
2
1 1 580 kHz
2 1600 kHz 1 1
2
580 580
(2800 pF) 367.9 pF 370 pF
1600 1600 LC f f LC f LC C C π π π = → = = → ⎛ ⎞ ⎛ ⎞ =⎜ ⎟ =⎜ ⎟ = ≈ ⎝ ⎠ ⎝ ⎠
(b) The inductance can be found from the resonant frequency and the capacitance.
5
2 2 2 5 2 12
1 1 1 1
2.7 10 H
2 4 4 (5.8 10 Hz) (2800 10 F)
f L
LC f C
π π π
− −
= → = = = ×
× ×
71. (a) We find the capacitance from the resonant frequency, Eq. 21–19.
7
0 2 2 2 3 2
0
1 1 1 1
1.3 10 F
2 4 4 (14.8 10 H)(3600 Hz)
f C
LC Lf
π π π
− −
→
= = = ≈ ×
×
(b) At resonance the impedance is the resistance, so the current is given by Ohm’s law. peak
peak 4.10150 V 37 A V
I
R
= = =
Ω
72. Since the circuit is in resonance, we use Eq. 21–19 for the resonant frequency to determine the necessary inductance. We set this inductance equal to the solenoid inductance calculated in Example 21–13, with the area equal to the area of a circle of radius r, the number of turns equal to the length of the wire divided by the circumference of a turn, and the length of the solenoid equal to the diameter of the wire multiplied by the number of turns. We solve the resulting equation for the number of turns.
0 0 2 2 0 2 0
0 2 2
solenoid
2 2 3 2 7 7 2
0 wire
3
1 1 2
2 4
(18.0 10 Hz) (2.6 10 F)(4 10 T m/A)(12.0 m) 1.1 10 m
43.45 44 loops
wire r
N A r
f L
Nd
LC f C
f C N d μ π μ π π π
π μ π − π −
− → → ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = = = = × × × ⋅ = = × = ≈ A A A
73. (a) We calculate the inductance from the resonance frequency.
0 0
2 2 2 3 2 9
1 2
1 1 0.03189 H 0.032 H
4 4 (19 10 Hz) (2.2 10 F) f
LC L
f C
π
π π −
→
=
= = = ≈
× ×
(b) We set the initial energy in the electric field equal to the maximum energy in the magnetic field and solve for the maximum current.
2 9 2
2 2 0
1 1
0 max max
2 2
(2.2 10 F)(120 V)
0.03152A 0.032 A (0.03189 H)
CV
CV LI I
L
−
→ ×
= = = = ≈
(c) The maximum energy in the inductor is equal to the initial energy in the capacitor.
2 9 2
1 1
,max 2 0 2(2.2 10 F)(120 V) 16 J L
