ELECTROMAGNETIC INDUCTION AND FARADAY’S LAW

21Responses to Questions

1. The advantage of using many turns 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: 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 magnetic

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.

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21-1

21-2   Chapter 21

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.

8. (a) The induced current in is to the right as coil B is moved toward coil A. As B approaches A, the magnetic flux through coil A increases (there are now more 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 increase in flux, with the field pointing to the right through the center of the coil. A current through to the right will produce this opposing field.

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Electromagnetic Induction and Faraday’s Law   21-3

(b) The induced current in is 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

to the left will produce this opposing field.

(c) The induced current in is to the left as in coil B is increased. As 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 to 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.

11. When the motor first starts up, there is only a small back emf in the circuit (back emf is proportional to the rotation speed of the motor). This allows a large current to flow to the refrigerator. The power source for the house can be treated as an emf with an internal resistance. This large current to the refrigerator motor from the power source reduces the voltage across the power source because of its internal resistance. Since the power source voltage has decreased, other items (like lights) will have a lower voltage across them and receive less current, so may “dim.” As the motor speeds up to its normal operational speed, the back emf increases to its normal level and the current delivered to the motor is now limited to its usual amount. This current is no longer enough to significantly reduce the output voltage of the power source, so the other devices then get their normal voltage. Thus, the lights flicker just when the refrigerator motor first starts up. A heater, on the other hand, draws a large amount of current (it is a very low-resistance device) at all times. (The heat-producing element is not a motor, so has very little induction associated with it.) The source is then continually delivering a large current, which continually reduces the output voltage of the power source. In an ideal situation, the source could provide any amount of current to the whole circuit in either situation. In reality, though, the higher current in the wires causes bigger losses of energy along the way to the devices, so the lights dim.

12. Figure 21–17 shows that the induced current in the upper armature segment points into the page. This can be shown using the right-hand rule. The charges in the top metal armature segment are moving in the direction of the velocity shown with the green arrow (up and to the right), and these moving charges are in a magnetic field shown with the blue arrows (to the right). The right-hand rule says that the charges experience a force into the page producing the induced current. This induced current is also in the same magnetic field. Using another right-hand rule, a current-carrying wire, with the current going into the page (as in the upper armature segment) in a magnetic field pointed to the right, will experience a force in a downward direction. This downward force exerts a counterclockwise torque on the armature while it is rotation in a clockwise direction. The back emf is opposing the motion of the armature during its operation.

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21-4   Chapter 21

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.

19. As the metal bar enters (or leaves) the magnetic field during the swinging motion, areas of the metal bar experience a change in magnetic flux. This changing flux induces eddy currents with the “free” conduction electrons in the metal bar. These eddy currents are then acted on by the magnetic field, and the resulting force opposes the motion of the swinging metal bar. This opposing force acts on the bar no matter which direction it is swinging through the magnetic field, thus damping the motion during both directions of the swing.

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Electromagnetic Induction and Faraday’s Law   21-5

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, 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 we can

substitute Thus, a given fraction of a maximum possible current, is equal to

and is independent of the battery emf.

(b) Yes. Since 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.

The rms voltages across either an inductor or a capacitor of an LRC circuit can be greater than the rms source voltage because the different voltages are out of phase with each other. At any given instant, the voltage across either the inductor or the capacitor could be negative, for example, thus allowing for a very large positive voltage on the other device. (The rms voltages, however, are always positive by definition.)

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21-6   Chapter 21

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 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 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 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 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

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: or

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.

Solutions to Problems

1. The magnitude of the average induced emf is given by Eq. 21–2b.

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.

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Electromagnetic Induction and Faraday’s Law   21-7

3. As the coil is pushed into the field, the magnetic flux through the coil increases into the page. To oppose this increase, the flux produced by the induced current must be out of the page, so the induced current is counterclockwise.

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21-8   Chapter 21

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.

6. We choose up as the positive direction. The average induced emf is given by Eq. 21–2a.

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.

(b) The angle is

(c) Use Eq. 21–1.

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.

(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.

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.

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Electromagnetic Induction and Faraday’s Law   21-9

(c) We find the average induced current from Ohm’s law.

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, and the changing current in the solenoid causes the field to change.

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.

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, Because the inward flux is decreasing, the induced flux will be into the page, so the induced current is clockwise, given by 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

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.

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21-10   Chapter 21

(b) The power required is the external force times the speed of the rod.

16. From the derivation in Problem 15a, we have an expression relating the external force to the magnetic field.

17. (a) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from Eq. 21–3.

(b) Find the induced current from Ohm’s law.

(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.

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.

  

(b) The rate at which thermal energy is produced in the wire is the power dissipated in the wire.

19. The charge that passes a given point is the current times the elapsed time, The current will be

the emf divided by the resistance, The resistance is given by Eq. 18–3, and the

emf is given by Eq. 21–2b. Combine these equations to find the charge during the operation.

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Electromagnetic Induction and Faraday’s Law   21-11

20. From Eq. 21–5, the induced voltage is proportional to the angular speed. Thus their quotient is a constant.

21. (a) The rms voltage is found from the peak induced emf. The peak induced emf is calculated from Eq. 21–5.

(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.

(b) The area can be found from Eq. 21–5.

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.

So the generator’s rms emf must be 14.61 volts. We find the frequency of the generator from Eq. 21–5.

24. When the motor is running at full speed, the back emf opposes the applied emf, to give the net voltage across the motor.

 (2 significant figures)

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21-12   Chapter 21

25. From Eq. 21–5, the induced voltage (back emf) is proportional to the angular speed. Thus their quotient is a constant.

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.

27. We find the number of turns in the secondary from Eq. 21–6.

28. Because this is a step-down transformer. Use Eq. 21–6 to find the voltage ratio and Eq. 21–7 to find the current ratio.

29. Use Eqs. 21–6 and 21–7 to relate the voltage and current ratios.

30. We find the ratio of the number of turns from Eq. 21–6.

(2 significant figures)

If the transformer is connected backward, then the role of the turns will be reversed:

31. (a) Use Eqs. 21–6 and 21–7 to relate the voltage and current ratios.

(b) Because this is a step-down transformer.

32. (a) We assume 100% efficiency and find the input voltage from

Since this is a step-up transformer.

(b)

33. Use Eqs. 21–6 and 21–7 to relate the voltage and current ratios.

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Electromagnetic Induction and Faraday’s Law   21-13

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.

(b) The power loss in the lines is given by

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.

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.

The power lost is given by Eq. 18–6a.

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.

(b) At the house, the voltage is changed from 7200 V to 240 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.

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21-14   Chapter 21

Thus there must be 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 Using the step-down transformer efficiency, we calculate the current in the transmission lines and the loss in the transmission lines.

The power to be delivered is 2.0 MW. The power that must be delivered to the step-down transformer

is The power that must be present at the start of the transmission must be

to compensate for the transmission line loss. The power that must enter the transmission lines from the 99% efficient step-up transformer is

So the power saved is as follows.

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.

The transmission lines must have a diameter greater than or equal to 3.3 cm.

39. Find the induced emf from Eq. 21–9.

40. Because the current is increasing, the emf is negative. We find the self-inductance from Eq. 21–9.

41. Use the relationship for the inductance of a solenoid, as given in Example 21–13.

42. Use the relationship for the inductance of a solenoid, as given in Example 21–13.

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Electromagnetic Induction and Faraday’s Law   21-15

43. (a) Use the relationship for the inductance of a solenoid, as given in Example 21–13.

(b) Apply the same equation again, solving for the number of turns but using the permeability of iron.

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.

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.

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.

The resistance is calculated from the resistivity, area, and length of the wire.

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 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.

As in Eq. 21–8a, the mutual inductance is the proportinality constant in the above relationship.

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ab

increasingI

induced%

R L

21-16   Chapter 21

47. The magnetic energy in the field is derived from Eq. 21–10.

48. The initial energy stored in the inductor is found from an equation in Section 21–11.

The final current is found from the final energy, which is 5 times the initial energy.

The current increases at a constant rate.

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.

50. (a) We set I equal to 75% of the maximum value in the equation and solve for the time constant.

(b) The resistance can be calculated from the time constant,

51. The potential difference across the resistor is proportional to the current, so we set the current in the equation equal to 0.025I0 and solve for the time.

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

so the entire battery voltage appears across the inductor. Apply Eq. 21–9 to find the initial rate of change of the current.

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Electromagnetic Induction and Faraday’s Law   21-17

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.

Find the time to reach the maximum current if the rate of current change remained at

53. For an LR circuit, we have Solve for

(a)

(b)

(c)

54. The reactance of a capacitor is given by Eq. 21–12b,

(a)

(b)

55. We find the frequency from Eq. 21–11b for the reactance of an inductor.

56. We find the frequency from Eq. 21–12b for the reactance of a capacitor.

57. We find the reactance from Eq. 21–11b and the current from Ohm’s law.

58. We find the reactance from Ohm’s law and the inductance by Eq. 21–11b.

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21-18   Chapter 21

59. (a) We find the reactance from Eq. 21–12b.

(b) We find the peak value of the current from Ohm’s law.

60. We find the impedance from Eq. 21–14.

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)

The inductor has essentially no effect on the impedance at this relatively low frequency.

(b)

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)

(b)

63. Use Eq. 21–15, with no capacitive reactance.

64. The total impedance is given by Eq. 21–15.

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Electromagnetic Induction and Faraday’s Law   21-19

The phase angle is given by Eq. 21–16a.

The voltage is lagging the current, or the current is leading the voltage.

The rms current is given by Eq. 21–14.

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.

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.

(b) The phase angle is given by Eq. 21–6a with no capacitive reactance.

The current is lagging the source voltage.

(c) The power dissipated is given by

(d) The rms voltage reading is the rms current times the resistance or reactance of the element.

Note that, because the maximum voltages occur at different times, the two readings do not add up to the applied voltage of 120 V.

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21-20   Chapter 21

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.

(b) The phase angle is given by Eq. 21–16a with no inductive reactance.

The current is leading the source voltage.

(c) The rms voltage reading is the rms current times the resistance or reactance of the element.

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.

(b) Repeat the calculation with a frequency of 6.0 kHz.

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.

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Electromagnetic Induction and Faraday’s Law   21-21

70. (a) The resonant frequency is given by Eq. 21–19.

(b) The inductance can be found from the resonant frequency and the capacitance.

71. (a) We find the capacitance from the resonant frequency, Eq. 21–19.

(b) At resonance the impedance is the resistance, so the current is given by Ohm’s law.

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.

73. (a) We calculate the inductance from the resonance frequency.

(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.

(c) The maximum energy in the inductor is equal to the initial energy in the capacitor.

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21-22   Chapter 21

74. (a) The clockwise current in the left-hand loop produces a magnetic field which is into the page within the loop and out of the page outside the loop. Thus the right-hand loop is in a magnetic field that is directed out of the page. Before the current in the left-hand loop reaches its steady state, there will be an induced current in the right-hand loop that will produce a magnetic field into the page to oppose the increase of the field from the left-hand loop. Thus the induced current will be clockwise.

(b) After a long time, the current in the left-hand loop is constant, so there will be no induced current in the right-hand coil.

(c) If the second loop is pulled to the right, then the magnetic field out of the page from the left-hand loop through the second loop will decrease. During the motion, there will be an induced current in the right-hand loop that will produce a magnetic field out of the page to oppose the decrease of the field from the left-hand loop. Thus the induced current will be counterclockwise.

(d) Since equilibrium has been reached (“a long time”), there is a steady current in the left-hand loop and steady magnetic flux passing through the right-hand loop. There is no emf being induced in the right-hand loop, so closing a switch in the right-hand loop has no effect.

75. The electrical energy is dissipated because there is current flowing in a resistor. The power dissipation by a resistor is given by so the energy dissipated is The current is created by the induced emf caused by the changing B field. The emf is calculated by Eq. 21–2a.

76. (a) Because this is a step-down transformer.(b) Assuming 100% efficiency, the power in both the primary and secondary is 45 W. Find the

current in the secondary from the relationship

(c)

(d) Find the resistance of the bulb from Ohm’s law. The bulb is in the secondary circuit.

77. The coil should have a diameter about equal to the diameter of a standard flashlight D-cell so that it will be simple to hold and use. This would give the coil a radius of about 1.5 cm. As the magnet passes through the coil, the field changes direction, so the change in flux for each pass is twice the maximum flux. Let us assume that the magnet is shaken with a frequency of about two shakes per second, so the magnet passes through the coil four times per second. We obtain the number of turns in the coil using Eq. 21–2b.

The answer will vary somewhat based on the approximations used. For example, a flashlight with a smaller diameter, for AA batteries perhaps, would require more turns.

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Electromagnetic Induction and Faraday’s Law   21-23

78. (a) Use Ohm’s law to see that the current is

(b) Find the back emf from the normal operating current.

(c) Use Eq. 18–6a.

(d) Use Eq. 18–6a.

79. Because the transformers are assumed to be perfect, the power loss is due to resistive heating in the transmission lines. Since the town requires 55 MW, the power at the generating plant must be

Thus the power lost in the transmission is 0.838 MW. This can be used to

determine the current in the (two) transmission lines.

To produce 55.838 MW of power at 273.6 A, the following voltage is required.

The voltage has 2 significant figures.

80. (a) From the efficiency of the transformer, we have Use this to calculate the current in the primary.

(b) The voltage in both the primary and secondary is proportional to the number of turns in the respective coil. The secondary voltage is calculated from the secondary power and resistance since

81. (a) The voltage drop across the lines is due to the resistance.

(b) The power input is given by

(c) The power loss in the lines is due to the current in the resistive wires.

(d) The power output is given by

This could also be found by subtracting the power lost from the input power.

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21-24   Chapter 21

82. A side view of the rail and bar is shown in the figure. From the discussion in Section 21–3, the emf in the bar is produced by the components of the magnetic field, the length of the bar, and the velocity of the bar, which are all mutually perpendicular. The magnetic field and the length of the bar are already perpendicular. The component of the velocity of the bar that is perpendicular to the magnetic field is so the induced emf is given by the following.

This produces a current in the wire, which can be found by Ohm’s law. That current is pointing into the page on the diagram.

Because the current is perpendicular to the magnetic field, the force on the wire from the magnetic field can be calculated from Eq. 20–2 and will be horizontal, as shown in the diagram.

For the wire to slide down at a steady speed, the net force along the rail must be zero. Write Newton’s second law for forces along the rail, with up the rail being positive.

83. We find the current in the transmission lines from the power transmitted to the user and then find the power loss in the lines.

84. The induced current in the coil is the induced emf divided by the resistance. The induced emf is found from the changing flux by Eq. 21–2b. The magnetic field of the solenoid, which causes the flux, is given by Eq. 20–8. For the area used in Eq. 21–2b, the cross-sectional area of the solenoid (not the coil) must be used, because all of the magnetic flux is inside the solenoid.

As the current in the solenoid increases, a magnetic field from right to left is created in the solenoid and the loop. The induced current will flow in such a direction as to oppose that field, so must flow from left to right through the resistor.

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B

NF

mg

vBF

Electromagnetic Induction and Faraday’s Law   21-25

85. Putting an inductor in series with the device will protect it from sudden surges in current. The growth of current in an LR circuit is given in Section 21–12.

The maximum current is 55 mA, and the current is to have a value of 7.5 mA after a time of 120 microseconds. Use this data to solve for the inductance.

Put an inductor of value 98 mH in series with the device.

86. The emf is related to the flux change by Eq. 21–2b. The flux change is caused by the changing magnetic field.

87. We find the peak emf from Eq. 21–5.

88. (a) The electric field energy density is given by Eq. 17–11, and the magnetic field energy density is given by Eq. 21–10.

The energy density in the magnetic field is 3.6 billion times greater than the energy density in the electric field.

(b) Set the two densities equal and solve for the magnitude of the electric field.

89. If there is no current in the secondary, then there will be no induced emf from the mutual inductance. Therefore, we set the ratio of the voltage to current equal to the inductive reactance (Eq. 21–11b) and solve for the inductance.

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21-26   Chapter 21

90. For the current and voltage to be in phase, the net reactance of the capacitor and inductor must be zero, which means that the circuit is at resonance. Thus Eq. 21–19 applies.

91. The inductance of the solenoid is given by The (constant) length of the

wire is given by so since we also know that The fact that the wire is tightly wound gives Find the ratio of the two inductances.

92. (a) From the text of the problem, the Q factor is the ratio of the voltage across the capacitor or inductor to the voltage across the resistor, at resonance. The resonant frequency is given by Eq. 21–19.

(b) Find the inductance from the resonant frequency and the resistance from the Q factor.

Solutions to Search and Learn Problems

1. (a) Kitchens, bathrooms, outdoor areas, and near swimming pools are especially dangerous for touching ground because of the abundance of water in those areas. As shown in Section 19–7, high current through the body can stop the heart. Wet skin has a much smaller resistance than dry skin, so for a given voltage it is more dangerous to ground an object with wet skin than with dry skin.

(b) The GFCI works using Ampere’s law. Both wires from the power source pass through the center of an iron ring. If the ac current in the circuit passes through both wires (one going in and one going out), then the net current through the ring is exactly zero and there is no magnetic field induced in the ring. If even a small amount of the electric current is shorted, so the exact same current no longer flows through both of the wires, then an ac-induced magnetic field is created in the ring. The changing flux induces a current in the small wire wrapped around the ring, which shuts off the power to the GFCI. This is similar to fuses and circuit breakers, which shut off the power if too much current is flowing through the wires. GFCIs are much more sensitive and they act more quickly than fuses and circuit breakers. This is why electrical codes now require them in kitchens and bathrooms, even though homes already have fuses or circuit breakers. As with a circuit breaker, a GFCI can be reset and used again after it has been triggered.

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Electromagnetic Induction and Faraday’s Law   21-27

2. (a) Use Eq. 21–2a to calculate the emf induced in the ring, where the flux is the magnetic field multiplied by the area of the ring. Then using Eq. 18–6b, calculate the average power dissipated in the ring as it is moved away. The thermal energy is the average power times the time.

(b) The temperature change is calculated from the thermal energy using Eq. 14–2.

3. (a) Since the coils are directly connected to the wheels, the torque provided by the motor (Eq. 20–10) balances the torque caused by the frictional force.

(b) To maintain this speed, the power loss due to the friction must equal the net power provided by the coils. The power provided by the coils is the current through the coils multiplied by the back emf.

(c) The power dissipated in the coils is the difference between the power produced by the coils and the net power provided to the wheels.

(d) We divide the net power by the total power to determine the percent used to drive the car.

4. The power dissipated (wasted) in a transmission line is proportional to the square of the current passing through the lines, or To minimize the ratio of the power dissipated to the power delivered, it is desirable to have the smallest current possible, which requires a large transmission voltage. However, it is also desirable, for safety reasons, for the voltage at the consumer end to be low. The use of transformers in ac circuits makes it possible to have high voltage in the transmission lines while having low voltage for the consumers.

5. The sinusoidal varying current in the power line creates a sinusoidal varying magnetic field encircling the power line, with an amplitude given by Eq. 20–6. Using Eq. 21–1, approximate the flux by using the average distance from the power line to the coil. Take the change in flux to be the maximum flux through the coil, as suggested in the problem hint. Set the maximum emf equal to 170 V and note that the flux changes from the maximum to zero in a quarter cycle. Set the area of the coil equal to the product of the length and the height of the rectangle, and solve for the length.

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21-28   Chapter 21

This is stealing and unethical because the current in the rectangle creates a back emf in the initial wire, which results in a power loss to the electric company just as if the wire had been directly connected to the line.

6. The average induced emf is given by Eq. 21–2b. Because the coil orientation changes by the change in flux is the opposite of twice the initial flux. The average current is the induced emf divided by the resistance, and the charge that flows in a given time is the current times the elapsed time.

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