ch21 giancoli7e manual

8/15/2019 Ch21 Giancoli7e Manual

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© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21-1

Responses to Questions

1. The advantage of using many turns large num( r) be N = in Faraday’s experiments is that the emf andinduced current are proportional to N , which makes it easier to experimentally measure thosequantities. Many turns in the primary coil will make a larger magnetic flux, and many turns in theoutput 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 enclosedloop area: cos , B BA θ Φ = so the flux is proportional to the magnitude of the magnetic field. Themagnetic flux depends not only on the field itself, but also on the area and on the angle between thefield and the area. Thus, they also have different units (magnetic field tesla T;= = magnetic

2flux 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 andcurrent are induced in the ring due to the changing magnetic flux. As the magnet gets closer tothe 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 isclockwise. In this case, the number of magnetic field lines coming through the loop and pointingtoward you is increasing (remember that magnetic field lines point toward the south pole of themagnet). The induced current in the loop will oppose this change in flux and will attempt tocreate magnetic field lines through the loop that point away from you. A clockwise inducedcurrent 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 andcurrent are not induced in the ring since the magnetic flux through the ring is not changing whilethe magnet is held steady.

(c) A current is induced in the ring when you withdraw the magnet. An emf and current are inducedin 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 ruleagain, the direction of the induced current when you withdraw the south pole from the ring iscounterclockwise. 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 influx and will create more magnetic field lines through the loop that point toward you.A counterclockwise induced current will provide this opposing magnetic field.

E LECTROMAGNETIC INDUCTION AND F ARADAY ’S L AW 21


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21-2 Chapter 21


4. ( a) The magnetic field through the loop due to the current-carrying wire will be into the page. As thewire loop is pulled away, the flux will decrease since the magnetic field is inversely proportionalto 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 magneticfield 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 ofthe field lines are parallel to the face of the loop. Thus there will be no magnetic flux in the loopand 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 piercethe loop from top to bottom. The flux will increase as the magnet gets closer, so a current will beinduced that makes upward-pointing field lines through the loop. The current will becounterclockwise.

6. ( a) Yes. As the battery is connected to the front loop and current starts to flow, it will create anincreasing 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 thesecond loop.

(b) The induced current in the second loop starts to flow as soon as the current in the front loopstarts 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 becomessteady. Once the battery has increased the current in the front loop from zero to its steady-statevalue, the magnetic field it creates is also steady. Since the magnetic flux through the secondloop is no longer changing, the induced current goes to zero.

(d ) The induced current in the second loop is counterclockwise. Since the increasing clockwisecurrent in the front loop is causing an increase in the number of magnetic field lines downthrough 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 counterclockwisecurrent will be induced in the second loop.

(e) Yes. Since both loops carry currents and create magnetic fields while the current in the frontloop is increasing from the battery, each current will “feel” the magnetic field caused by theother 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 secondloop 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 pointingat each other and repel. We can also explain it by saying that the two currents are in oppositedirections, 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 thefirst coil and stop when the current falls to zero in the first coil. The current in the second loop will beclockwise. All of the reasoning is similar to that given for Question 6, except now the current isdecreasing instead of increasing.

8. ( a) The induced current in A R 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 thatopposes this increase in flux, with the field pointing to the right through the center of the coil.A current through A R 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 A R 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 thatopposes 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 A R is to the left as B R in coil B is increased. As B R increases, the

current in coil B decreases, which also decreases the magnetic field coil B produces. As themagnetic field from coil B decreases, the magnetic flux through coil A decreases (there are nowfewer magnetic field lines in coil A pointing to the left). The induced emf in coil A creates acurrent to produce a magnetic field pointing to the left through the center of the coil, opposingthe decrease in flux. A current through A R 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 magneticfield created by the combination. If there is an external magnetic field, it will not be influenced by thesignal current, and then by Newton’s third law, the signal current will not be influenced by an externalmagnetic field either.

10. One advantage of placing the two insulated wires carrying ac close together is that the magnetic fieldcreated by the changing current moving one way in one wire is approximately cancelled out by themagnetic field created by the current moving in the opposite direction in the second wire. Also, sincelarge loops of wire in a circuit can generate a large self-induced back emf, by placing the two wiresclose to each other, or even twisting them about each other, the effective area of the current loop isdecreased 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 tothe rotation speed of the motor). This allows a large current to flow to the refrigerator. The powersource for the house can be treated as an emf with an internal resistance. This large current to therefrigerator motor from the power source reduces the voltage across the power source because of itsinternal resistance. Since the power source voltage has decreased, other items (like lights) will have alower voltage across them and receive less current, so may “dim.” As the motor speeds up to itsnormal operational speed, the back emf increases to its normal level and the current delivered to themotor is now limited to its usual amount. This current is no longer enough to significantly reduce theoutput voltage of the power source, so the other devices then get their normal voltage. Thus, the lightsflicker just when the refrigerator motor first starts up. A heater, on the other hand, draws a largeamount of current (it is a very low-resistance device) at all times. (The heat-producing element is not amotor, so has very little induction associated with it.) The source is then continually delivering a largecurrent, which continually reduces the output voltage of the power source. In an ideal situation, thesource could provide any amount of current to the whole circuit in either situation. In reality, though, thehigher 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. Thiscan be shown using the right-hand rule. The charges in the top metal armature segment are moving inthe direction of the velocity shown with the green arrow (up and to the right), and these movingcharges are in a magnetic field shown with the blue arrows (to the right). The right-hand rule says thatthe charges experience a force into the page producing the induced current. This induced current is alsoin the same magnetic field. Using another right-hand rule, a current-carrying wire, with the currentgoing into the page (as in the upper armature segment) in a magnetic field pointed to the right, willexperience a force in a downward direction. This downward force exerts a counterclockwise torque onthe armature while it is rotation in a clockwise direction. The back emf is opposing the motion of thearmature 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 brakesdo not need to act on ferromagnetic wheels. The external magnetic field of the eddy brake just needs tointeract 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 conductionelectrons (the right-hand rule says moving charges in a magnetic field will experience a magneticforce, making them move and creating an eddy current). This eddy current is also in the brakingmagnetic field. The right-hand rule says these currents will experience a force opposing the originalmotion of the piece of metal and the eddy current brake will begin to slow the wheel. Goodconductors, such as copper and aluminum, have many free conduction electrons and will allow largeeddy 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 ofit in the tube. This flux change induces a current to flow around the tube walls to create an opposingmagnetic field in the tube (Lenz’s law). This induced magnetic field pushes against the falling magnetand reduces its acceleration. The speed of the falling magnet increases until the magnetic force on themagnet is the same size as the gravity force. The opposing magnetic field cannot cause the magnet toactually 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 andfalls at a constant terminal velocity. The weight of the magnet is balanced by the upward force fromthe eddy currents.

15. The nonferrous materials are not magnetic, but they are conducting. As they pass by the permanentmagnets, eddy currents will be induced in them. The eddy currents provide a “braking” mechanismwhich will cause the metallic materials to slide more slowly down the incline than the nonmetallicmaterials. The nonmetallic materials will reach the bottom with larger speeds. The nonmetallicmaterials can therefore be separated from the metallic, nonferrous materials by placing bins at differentdistances 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 materialsshould 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 fallthrough the region of magnetic field. The smaller eddy currents then experience a smaller opposingforce 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 themagnetic field, eddy currents are created in the sheet. The magnetic force on these induced currentsopposes 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 throughthe loop, it sets up an induced current in the loop, which opposes the change in flux. This current actslike a magnet that is opposing the physics magnet, repelling it and so reducing its speed. Then, afterthe midpoint of the magnet passes through the loop, the induced current will reverse its direction andattract 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, andthe resulting force opposes the motion of the swinging metal bar. This opposing force acts on the barno matter which direction it is swinging through the magnetic field, thus damping the motion during

both directions of the swing.


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20. To determine the ratio of turns on the two coils of a transformer without taking it apart, apply a knownac input source voltage to one pair of leads and carefully measure the output voltage across the othertwo leads. Then, source output source output/ / ,V V N N = 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 twowires is very small, then those two wires are a pair. If the resistance between two wires is infinite, thenthose 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 betweenhousehold wires and anything that is grounded (other wires, people, etc.) would more easily causeelectrical breakdown of the air, and then much more sparking would occur. Basically, this wouldsupply 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 shortcircuits and accidental electrocutions.

22. When 120 V dc is applied to the transformer, there is no induced back emf that would usually occurwith 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 thecoils, which are usually wound with many loops of very fine, low-resistance wire, and could melt theinsulation 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 fluxfrom one coil also goes through the other coil.

(b) To create the least amount of mutual inductance with two flat circular coils, you would placethem with their faces at right angles. This way, almost none of the magnetic flux from one coilgoes through the other coil.

24. ( a) No. Although the current through an LR circuit is described by 0 /1 ,t

L RV I e R

−⎛ ⎞⎜ ⎟= −⎜ ⎟⎝ ⎠

we can

substitute 0 max .V

I R

= Thus, a given fraction of a maximum possible current, max/ , I I is equal to

/1t

L Re−

− and is independent of the battery emf.

(b) Yes. Since 0 /1 ,t

L RV 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.

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


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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 emfaccording to 1/ 2 .C X fC π = 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 thecapacitor gets very small.

(c) The impedance of a pure inductance varies directly with the frequency of the source emfaccording to 2 . L X fLπ = Thus, as the source frequency gets very small, the impedance of theinductor gets very small, and as the source frequency gets very large, the impedance of theinductor gets very large.

(d ) The impedance of an LRC circuit with a small R near resonance is very sensitive to thefrequency of the source emf. If the frequency is set at resonance exactly, where , L C X X = thenthe LRC circuit’s impedance is very small and equal to R. The impedance increases rapidly asthe 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 thesource frequency is much higher or much lower than the resonance frequency. If the sourcefrequency is much higher than resonance, then the impedance is directly proportional to thefrequency of the source emf. Basically, at extremely high frequencies, the circuit impedance isequal to 2 . fLπ If the source frequency is much lower than resonance (nearly zero), then theimpedance is inversely proportional to the frequency of the source emf. Basically, at extremelylow frequencies, the circuit impedance is equal to 1/2 . fC π

27. To make the impedance of an LRC circuit a minimum, make the resistance very small and make thereactance of the capacitor equal to the reactance of the inductor: , L C X X = or

1 12 .

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 thecircuit also oscillates and is alternately stored in the magnetic field of the inductor and the electric fieldof the capacitor.

29. Yes. The instantaneous voltages across the different elements in the circuit will be different, but thecurrent 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 intothe page. Since the initial magnetic flux is into the page, the current will be induced when themagnetic 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 counterclockwisecurrent. Moving the coil sideways, as in answer ( c), does not change the flux, so no currentwould be induced.

2. ( c) A common misconception is that a moving loop would experience a change in flux. However, ifthe loop is moving through a constant field without rotation, then the flux through the loop willremain constant and no current will be induced.


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3. ( d ) A current is induced in the loop when the flux through the loop is changing. As the loop passesthrough line J it enters a region with a magnetic field, so the flux through the loop increases anda current will be induced. When the loop passes line K, the flux remains constant, as there is nochange in field, so no current is induced. As the loop passes line L, the magnetic field fluxthrough 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 fromthe wire. For C, the loop remains at the same distance from the wire, so the magnetic fluxthrough the wire remains constant and no current is induced in the loop. For D, the magneticfield from the long wire points into the page in the region of the loop. As the loop moves awayfrom the wire, the magnetic flux decreases, so a clockwise current is induced in the loop tooppose the change in flux.

5. ( c) The induced current creates a magnetic field that opposes the motion of the magnet. The result isthat the magnet has less kinetic energy than it would have had if the loop were not present. Thechange 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 isnot a conductor, no current will be induced. The emf is produced regardless of whether there is aconducting 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 andtherefore the magnetic flux in the loop. Rotating the loop changes the angle between the loopand wire, which will change the flux. Since the magnetic field strength decreases with distancefrom the wire, moving the loop away from the wire, either with or without rotation, will decreasethe flux in the wire. If the loop is moved parallel to the wire, then the flux through the loop willnot 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 thesecond will also change, inducing a current in the second coil.

9. ( b) A generator converts mechanical energy into electric energy. The generator’s magnetic fieldremains unchanged as the generator operates, so energy is not being pulled from the magneticfield. Resistance in the coils removes electric energy from the system in the form of heat—itdoes not provide the energy. In order for the generator to work, an external force is necessary torotate 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 thecoil. Since the generator’s voltage output is proportional to the rate of change of the flux throughthe coil, increasing the maximum flux results in a greater voltage output. When the magneticfield 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.

11. ( d ) A common misconception is that a transformer only changes voltage. However, power isconserved across a transformer, where power is the product of the voltage and current. When atransformer increases the voltage, it must also proportionately decrease the current.


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21-8 Chapter 21


12. ( d ) If the charger unit had a battery, then it could run the laptop without being plugged in. A motorconverts electrical energy into mechanical energy, but the laptop charger output is electricalenergy, not mechanical energy. A generator converts mechanical energy into electric energy, butthe input to the laptop is electrical, not mechanical. The cables to and from the charger unit can beconsidered 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 andvoltage) is constant across the transformer. Therefore, if the voltage increases, then the currentmust decrease across the transformer. If the current increases, then the voltage must decreaseacross the transformer. A transformer works due to the induced voltage created by the changingflux. 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-uptransformer with twice as many loops in the secondary coil as in the primary. It is true that an accurrent 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 thesecondary coil.

15. ( b) Many people may not realize that generators (rotating coils in a magnetic field) are the heart ofmost 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 tooquickly, so they slow down the swipe. The credit card has a magnetic stripe with informationencoded in the magnetic field. Swiping the card more rapidly increases the induced emf, due tothe 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 accircuit, 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 thecurrent and the voltage source, so ( c) is not correct. The current and voltage across a resistor arealways 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

∆Φ − −= = = ≈

∆

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 befrom right to left.

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 theinduced current is counterclockwise.


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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 inducedcurrent 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 loopchanges from 0° to 90°. The magnitude of the average induced emf is given by Eq. 21–2a.

2cos (0.0925 m) 1.5 T(cos 90 cos 0 )0.20 V

0.20 s B ABt t

θ π ∆Φ ∆ ° − °= − = − = − =∆ ∆

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 V0.16 s

B A Bt t

π −∆Φ ∆ . − . − .= − = − = − = ×∆ ∆

7. ( a) When the plane of the loop is perpendicular to the field lines, the flux is given by the maximumof Eq. 21–1.

2 2 2 2(0 50 T) (0 080 m) 1.005 10 Wb 1.0 10 Wb B BA B r π π

− −

Φ = = = . . = × ≈ × (b) The angle is 90 42 48 .θ = ° − ° = °

(c) Use Eq. 21–1.

2 2 3cos (0.50 T) (0.080 m) cos 48 6.7 10 Wb B BA B r θ π π −Φ = = = ° = ×

8. ( a) As the resistance is increased, the current in the outer loop will decrease. Thus the flux throughthe inner loop, which is out of the page, will decrease. To oppose this decrease, the inducedcurrent in the inner loop will produce a flux out of the page, so the direction of the inducedcurrent 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 aflux 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 emffrom Eq. 21–3.

2(0.800 T)(0.120 m)(0.180 m/s) 1.73 10 V B υ −= = = ×

(b) Because the upward flux is increasing, the induced flux will be into the page, so the inducedcurrent is clockwise. Thus the induced emf in the rod is down, which means that the electric fieldwill be down. The electric field is the induced voltage per unit length.

21.73 10 V0.144 V/m, down

0.120 m E

−×= = =

10. ( a) The magnetic flux through the loop is into the page and decreasing, because the area isdecreasing. 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 5 0 s

3.717 10 V 3.7 10 V

B B A

t t π

− −

∆∆Φ −= = =

∆ ∆ .

= × ≈ ×


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21-10 Chapter 21


(c) We find the average induced current from Ohm’s law.2

23.717 10 V 1.5 10 A2.5

I R

−−×= = = ×

Ω

11. ( a) Because the current is constant, there will be no change in flux, so the induced current will bezero.

(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 thedirection 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 thedirection 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 thedirection 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 shortcoil, 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, 0 solenoid solenoid/ , B IN µ = and the changing current in the solenoid causes the field to change.

0 solenoidshort short

solenoidshort short 0 short solenoid short

solenoid7 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 −

−

⎛ ⎞∆ ⎜ ⎟

∆ ∆⎝ ⎠= = =∆ ∆ ∆

× ⋅= = ×

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 tothe horizontal component of the Earth’s magnetic field. This occurs when the car is traveling in theeast or west direction. We calculate the magnitude of the emf using Eq. 21–3, where B is thehorizontal 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 − −= = × ° = × =

14. As the loop is pulled from the field, the flux through the loop decreases, causing an induced emf whosemagnitude is given by Eq. 21–3, . B υ = Because the inward flux is decreasing, the induced fluxwill be into the page, so the induced current is clockwise, given by / . I R= Because this current inthe left-hand side of the loop is in a downward magnetic field, there will be a magnetic force to theleft. To keep the rod moving, there must be an equal external force to the right, given by . F I B=

2 2 2 2(0.550 T) (0.350 m) (3.10 m/s)

0.499 N0.230

B B

F I B B B R R Rυ = = = = = =

Ω

15. ( a) There is a conventional current flowing in the rod, from top to bottom, equal to the induced emfdivided by the resistance of the bar. That current is in a magnetic field and, by the right-handrule, will have a force on it (and the bar) to the left in Fig. 21–11a. The external force must equalthat 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 = = = = =


11/29



(b) The power required is the external force times the speed of the rod.2 2 2 B

P F R

υ υ = =

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

2 2external

external 2 2(0.350 N)(0.25 )

1.208 T 1.2 T(0.200 m) (1.50 m/s)

F R B F B

Rυ

υ

Ω= → = = = ≈

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

(0.35 T)(0.300 m)(1.6 m/s) 0.168 V 0.17 V B υ = = = ≈

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

3 30.168 V 7.149 10 A 7.1 10 A21.0 2.5

I R

− −= = = × ≈ ×Ω + Ω

(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 equalexternal 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 B − − −= = × = × ≈ ×

18. ( a) There is an emf induced in the coil since the flux through the coil changes. The current in thecoil is the induced emf divided by the resistance of the coil. The resistance of the coil is foundfrom Eq. 18–3.

coilwire

B NA R

t A ρ ∆= =

∆

coilcoil wire

wire2 3 3

830[ (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 Bt I R t

A ρ ρ

π π

π

− −

−

∆

∆∆= = =

∆

× ×= = ≈

× Ω ⋅

(b) The rate at which thermal energy is produced in the wire is the power dissipated in the wire.8

2 2 23 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 ⋅= = =

×

= × ≈ ×

19. The charge that passes a given point is the current times the elapsed time, .Q I t = ∆

The current will bethe emf divided by the resistance, . I

R=

The resistance is given by Eq. 18–3,

wire, R

A ρ = 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 A A Bt R I

t t A R t A

ρ ρ ρ

∆

∆ ∆∆Φ ∆= = = = = =∆ ∆ ∆


12/29

21-12 Chapter 21


2 2 2loop 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 BQ I t

r

π π π

ρ ρ π ρ

π −

−

∆ ∆ ∆= ∆ = = =

×= =

× Ω ⋅

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

1 2 22 1

1 2 1

2500 rpm(12.7 V) 28.86 V 29 V

1100 rpmω

ω ω ω = → = = = ≈

21. ( a) The rms voltage is found from the peak induced emf. The peak induced emf is calculated fromEq. 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 AV

ω

ω π π

= →

= = =

= ≈

(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 rms2 2(70.0 A) 99.0 A I I = = =

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

peak rms

2 2 2rms

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 −

= = →

= = = ≈ ×

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 4 212

rmsrms

circuit bulb wire bulb wire

(1.68 10 m)[85(4)(0.060 m)]1.254

[ (5.9 10 m)]

25.0 W2.083 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 =

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

2

214.61 V

2 2

(14.61 V) 2 (14.61 V) 216.54 Hz 17 Hz 17 cycles/s

2 2 (85)(0.65 T)(0.060 m)

fNBA

f NBA

π

π π

= = →

= = = ≈ =

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

applied back back applied 120 V (8.20 A)(3.65 ) 90 V IR IR− = → = − = − Ω =

(2 significant figures)


13/29



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

1 2 2 2 22 1 1 1

1 2 1 1 1

2 2300 rpm(72 V) 92 V

2 1800 rpm f f f f

ω π ω ω ω π

= → = = = = =

26. The back emf is proportional to the rotation speed (Eq. 21–5). Thus if the motor is running at halfspeed, then the back emf is half the original value, or 54 V. Find the new current from writing a loopequation for the motor circuit, from Fig. 21–19.

back back

120 V 54 V0 13 A

5.0 IR I

R

− −− − = → = = =

Ω

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

S S SS P

P P P

13,500 V(148 turns) 17,100 turns

117 VV N V

N N V N V

= → = = ≈

28. Because S P , N N < 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 turnsor 0.33 3.0

360 turns 3 120 turnsV 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 S SP P P

P P P S P S P S

25 V; 0.21

120 VV N I V I N 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 V50

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 SS

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 SP P PS P

P P P S P S S

0.35 A; (120 V) 6.2 V

6.8 AV N I V N I I

V V V N I N V I I

= = → = → = = =

(b) Because S P ,V V < 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 W3.8 V

25 A P

P I V V I

= → = = =

Since P S ,V V < this is a step-up transformer.

(b) SP

12 V3.2

3.8 VV V

= =

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

S S SS P

P P P

1240 turns(120 V) 450.9 V 450 V

330 turnsV N N

V V V N N

= → = = = ≈


14/29

21-14 Chapter 21


S SPP 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.6town

town rms rms rms 3rms

output

63town

output rms 3rms

35 10 W778 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 + × = ≈

×

(b) The power loss in the lines is given by 2loss rms . P I R= 2 2

loss loss rms2 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 fromEq. 18–9a.

3

3100 10 W

8.333 A12 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 isstepped down by a factor of 20.

S P PS 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

lost3

9 92 5

50 10 W (0.4167 A) (5 10 /m)( )

50 10 W5.76 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 V20 (2 significant figures)

12,000 V N V N V

= = =

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

2S SP P

240 V3.3 107200 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.

64out

out out line lineout

2 4 2 7lost line line

2.0 10 W1.667 10 A

120 V

(1.667 10 A) 2(0.100 ) 5.556 10 W

P P V I I

V

P I R

×= → = = = ×

= = × Ω = ×


15/29



Thus there must be 6 7 72.0 10 W 5.556 10 W 5.756 10 W× + × ≈ × 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 41.667 10 A.× Using the step-down transformer efficiency, we calculate thecurrent in the transmission lines and the loss in the transmission lines.

out line out out line lineend

43out out

lineline

2 3 2 5lost 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

is2 0 M W

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 5873 MW. + . = . to compensate for the transmission line loss. The power that

must enter the transmission lines from the 99% efficient step-up transformer is2.5873 MW

2.6134 MW.0.99

= So the power saved is as follows.

7 6 75.756 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 powerloss 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. Weassume there are two lines to have a complete circuit.

2 22

loss 2

6 8 3

2 3 2

0 02540.025

4 4(925 10 W)(2.65 10 m)2(185 10 m)0.03256 m 3.3 c m

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 ⋅ ×= = = ≈

×

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

Lt

∆ −= − = − =

∆

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

∆ ∆= − → = − = − − =

∆ ∆ − −

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 H0.60 m

N A L

µ π π − −× ⋅ ×

= = =

42. Use the relationship for the inductance of a solenoid, as given in Example 21–13.2

07 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 −

= → = = ≈× ⋅


16/29

21-16 Chapter 21


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

2 7 2 20 (4 10 T m/A)(2600) (0.0125 m) 0.01479 H 0.015 H

(0.282 m) N A

L µ π π

−× ⋅= = = ≈

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

= → = = ≈× ⋅

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 thecoil, given by Ohm’s law, and an induced emf due to theinductance of the coil, given by Eq. 21–9. Since the current isincreasing, the inductance will create a potential difference tooppose the increasing current, so there is a drop in the potential dueto the inductance. The potential difference across the coil is the sum of the two potential drops.

ab (3.00 A)(2.25 ) (0.112 H)(3.80 A/s) 7.18 V I V IR L t ∆= + = Ω + =∆

45. The self-inductance of an air-filled solenoid was determined in Example 21–13. We solve this equationfor the length of the tube, using the diameter of the wire as the length per turn.

220 0

0 2

2 3 2

2 7 20

(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

−

−

= = =

×= = = ≈

× ⋅

The length of the wire needed ( L) is equal to the number of turns (the length of the solenoid divided bythe diameter of the wire) multiplied by the circumference of the turn.

346.16 m

(0.12 m) 21,490 m 21 km0.81 10 m

L Dd

π π −= = = ≈×

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

×

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 1 11 0 . N I

B µ = When the current in the solenoid changes,

the magnetic field of the solenoid changes, and thus the flux through the coil changes, inducing an emfin the coil.

1 12 0

2 1 2 10

( ) N I

N A N A B N N A I

t t t

µ µ

⎛ ⎞∆ ⎜ ⎟∆ ∆⎝ ⎠= − = − = −

∆ ∆ ∆

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

1 2 1 1 20 0

( ) N N A I N N A M

t µ µ

∆= − → =

∆

ab

increasing I

− +

induced

R L


17/29



47. The magnetic energy in the field is derived from Eq. 21–10.2

12

02 2 2

2 21 1 12 2 2 7

0 0

Energy storedVolume

(0.72 T)Energy (Volume) (0.010 m) (0.36 m) 23 J(4 10 T m/A)

Bu

B Br 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 51 1

2 2 (45.0 10 H)(50.0 10 A) 5.63 10 JU LI − − −= = × × = ×

The final current is found from the final energy, which is 5 times the initial energy.2 21 1

f 0 f 0 f 02 25 5( ) 5U U LI LI I I = → = → =

The current increases at a constant rate.

f 0

3f 0 0 0 0

0.115 A/s

5 (50.0 10 A)( 5 1) ( 5 1) 0.537 s0.115 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 −

−∆= = →

∆ ∆

− − ×∆ = = = − = − =

49. The magnetic energy in the field is derived from Eq. 21–10. The volume of a relatively thin sphericalshell, like the first 10 km above the Earth’s surface, is the surface area of the sphere times its thickness.

212

02 4 2

2 6 2 4 151 1Earth2 2 7

0

(0.50 10 T)(volume) 4 4 (6.38 10 m) (10 m) 5 10 J

(4 10 T m/A)

Bu

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 /0

(1 )t I I e 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, / . L Rτ =

31.0 mH17

1.847 ms L

Rτ

= = = Ω

51. The potential difference across the resistor is proportional to the current, so we set the current in the

equation /maxt I I e τ =

2 equal to 0.025 I 0 and solve for the time.

/0 00.025 ln(0.025) 3.7

t

I I I e t τ

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 initialrate of change of the current.

L I I V

V V Lt t L

∆ ∆= = → =

∆ ∆


18/29

21-18 Chapter 21


The maximum value of the current is reached after a long time, when there is no voltage across theinductor, 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

∆ =∆

max 0 max 0(elapsed time) elapsed time ( ) 0 I t V L L

I I I I t I R V R

∆ ∆ ⎛ ⎞= + → = − = − =⎜ ⎟

∆ ∆ ⎝ ⎠

53. For an LR circuit, we have /max (1 ).t I I e τ

−= − Solve for .t

/ /max

max max(1 ) 1 ln 1t t

I I I I e e t

I I τ τ τ

− − ⎛ ⎞= − → = − → = − − ⎜ ⎟⎝ ⎠

(a) maxmax

0.90 ln 1 ln(1 0.9) 2.3 I

I I t I

τ τ τ ⎛ ⎞

= → = − − = − − =⎜ ⎟⎝ ⎠

(b) maxmax

0.990 ln 1 ln(1 0.99) 4.6 I I I t I

τ τ τ ⎛ ⎞= → = − − = − − =⎜ ⎟⎝ ⎠

(c) maxmax

0.999 ln 1 ln(1 0.999) 6.9 I

I I t I

τ τ τ ⎛ ⎞

= → = − − = − − =⎜ ⎟⎝ ⎠

54. The reactance of a capacitor is given by Eq. 21–12b,1

.2C

X fC π

=

(a)6

1 1428

2 2 (60.0 Hz)(6.20 10 F)C X fC π π

−= = = Ω

×

(b) 2

6 6

1 12.57 10

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 Hz2 2 (0 0320 H)

L L

X X fL f

Lπ

π π

Ω= → = = = ≈

.

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

3 61 1 1

10.9 Hz2 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 V1.47 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.

2240 V2 5.22 10 H2 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

−

= → =

= → = = = = ×


19/29



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

81 1

7368 74002 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 rms

2.0 10 V2 2 2 0.38 A

7368C

V I I

X

×= = = =

Ω

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

rms3

rms

120 V1700 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 ofthe inductor is given by Eq. 21–11b.

(a)2 2 2 2 2 2 3 2 2 2 3 2 4

4 (36 10 ) 4 (50 Hz) (55 10 H) 3.6 10 L Z R X R f Lπ π −= + = + = × Ω + × = × Ω

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

(b) 2 2 2 2 2 2 3 2 2 4 2 3 24 (36 10 ) 4 (3.0 10 Hz) (55 10 H) L Z R X R f Lπ π −= + = + = × Ω + × ×

437,463 3.7 10= Ω ≈ × Ω

62. The impedance of the circuit is given by Eq. 21–15 without an inductive reactance. The reactance ofthe capacitor is given by Eq. 21–12b.

(a) 2 2 2 3 22 2 2 2 2 6 2

1 1(3.5 10 )

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 3 22 2 2 2 2 6 2

1 1(3.5 10 )

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 2 2(235 ) (115 ) 205 L L Z R X R Z X = + → = − = Ω − Ω = Ω

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

22 2 2

23 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 + × × −⎢ ⎥

× ×⎢ ⎥⎣ ⎦= Ω ≈ Ω


20/29

21-20 Chapter 21


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

1 1

4 24 9

1

1

12

2tan tan

12 (1.00 10 Hz)(2.80 10 H)2 (1.00 10 Hz)(6.25 10 F)

tan

787tan 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. 21–14.2rms

rms725 V

8.30 10 AZ 8735.5

V I

−= = = ×Ω

65. We use the rms voltage across the resistor to determine the rms current through the circuit. Then, usingthe rms current and the rms voltage across the capacitor in Eq. 21–13b, we determine the frequency.

, rms rmsrms , rms

rmsrms6

, rms , rms

2

(3.0 V)272.1 Hz 270 Hz

2 2 2 (1.0 10 C)(650 )(2.7 V)

RC

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 powersource 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. 21–15with no capacitive reactance.

2 2 2 2

rms rmsrms 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 13

2 2 (60.0 Hz)(0.35 H)tan tan tan 2.7

2.80 10 L X fL

R Rπ π

φ − − −= = = = °

× Ω

The current is lagging the source voltage.

(c) The power dissipated is given by2 2 2 3rms (4.281 10 A) (2.80 10 ) 5.1 W . P I R

−= = × × Ω =

(d ) The rms voltage reading is the rms current times the resistance or reactance of the element.2 3

rms rms

2rms 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 fLπ π

−

−

= = × × Ω = ≈

= = = × = ≈

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


21/29



67. ( a) The rms current is the rms voltage divided by the impedance. The impedance is given by Eq. 21–15with no inductive reactance.

2 2 22

rms rmsrms

2 3 22 2 2 2 2 6 2

2 2

1

(2 )

120V

1 1(6.60 10 )

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.

61 1 1

3

112 (60.0 Hz)(1.80 10 F)2

tan tan tan 12.66.60 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 3rms rms

2rms rms rms 6

(1.774 10 A)(6.60 10 ) 117 V

1 1(1 774 10 A) 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 addup 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 impedanceto determine the magnitude of the current in the circuit. Finally, multiply the current by theresistance to determine the voltage drop across the resistor.

in in2 2

2 6 2

1/(2 )

(130 mV)(520 )29.77 mV 30 mV

(520 ) 1/[2 (60 Hz)(1.2 10 F)]

RV V R

V IR R Z R fC π

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)] RV

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 resonantfrequency.

5 50 6 12

1 1 1 13.627 10 Hz 3.6 10 Hz

2 2 (55.0 10 H)(3500 10 F) f

LC π π − −= = = × ≈ ×

× ×


22/29

21-22 Chapter 21


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

580 kHz580 kHz

1600 kHz

1600 kHz

2 2

1600 kHz 580 kHz

1 121 1 580 kHz

2 1600 kHz 1 12

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.

52 2 2 5 2 12

1 1 1 12.7 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.

70 2 2 2 3 2

0

1 1 1 11.3 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

150 V37 A

4.10

V I

R= = =

Ω

72. Since the circuit is in resonance, we use Eq. 21–19 for the resonant frequency to determine thenecessary inductance. We set this inductance equal to the solenoid inductance calculated in Example21–13, with the area equal to the area of a circle of radius r , the number of turns equal to the length ofthe wire divided by the circumference of a turn, and the length of the solenoid equal to the diameter ofthe wire multiplied by the number of turns. We solve the resulting equation for the number of turns.

0

0

22

020

0 2 2 solenoid

2 2 3 2 7 7 20 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 − −

−

→ →

⎛ ⎞⎜ ⎟⎝ ⎠= = = =

× × × ⋅= =

×

= ≈

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

0

0

2 2 2 3 2 9

1

21 1

0.03189 H 0.032 H4 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 fieldand solve for the maximum current.

2 9 22 2 01 1

0 max max2 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 21 1

,max 02 2 (2.2 10 F)(120 V) 16 J LU CV µ −= = × =


23/29



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

(b) After a long time, the current in the left-hand loop is constant, so there will be no inducedcurrent 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-handloop through the second loop will decrease. During the motion, there will be an induced currentin the right-hand loop that will produce a magnetic field out of the page to oppose the decreaseof 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 loopand steady magnetic flux passing through the right-hand loop. There is no emf being induced inthe 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 2 , P I R= so the energy dissipated is 2 . E P t I R t = ∆ = ∆ The current is created by the induced emf caused by the changing B field. The emf is calculated by Eq. 21–2a.

2 2 2 2 2 2 22

2 2

3

( ) ( ) [(0.240 m) ] [(0 0.665 T)]( ) (6.10 ) (0.0400 s)( )

6.01 10 J

B A B A B I t t R R t

A B A B E P t I R t R t

R t R t −

∆Φ ∆ ∆= − = − = = −

∆ ∆ ∆

∆ ∆ −= ∆ = ∆ = ∆ = =

∆ Ω∆

= ×

76. ( a) Because S P ,V V < this is a step-down transformer.

(b) Assuming 100% efficiency, the power in both the primary and secondary is 45 W. Find thecurrent in the secondary from the relationship . P IV =

SS S S S

S

45 W3.8 A

12 V P

P I V I V

→= = = =

(c) PP P P PP

45 W0.38 A

120 V P

P I V I V

→= = = =

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

SS S

S

12 V3.2

3.75 AV

V I R R I

→= = = = Ω

77. The coil should have a diameter about equal to the diameter of a standard flashlight D-cell so that itwill 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 maximumflux. Let us assume that the magnet is shaken with a frequency of about two shakes per second, so themagnet passes through the coil four times per second. We obtain the number of turns in the coil usingEq. 21–2b.

20

(3 V)(0.25 s)10,610 turns 10,000 turns

/ 2 2(0.05 T) (0.015 m)

t t N

t B A π

∆ ∆= = = = = ≈

∆Φ ∆ ∆Φ

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


24/29

21-24 Chapter 21


78. ( a) Use Ohm’s law to see that the current is 120 V/3.0 40 A (2 significant figures).Ω =

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

applied back back applied 120 V (2.0 A)(3.0 ) 114 V IR IR→− = = − = − Ω =

(c) Use Eq. 18–6a.2 2(2.0 A) (3.0 ) 12 W P I R= = Ω =

(d ) Use Eq. 18–6a.2 2(40 A) (3.0 ) 4800 W P I R= = Ω =

79. Because the transformers are assumed to be perfect, the power loss is due to resistive heating in thetransmission lines. Since the town requires 55 MW, the power at the generating plant must be55 MW

55.838 MW.0.985

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

determine the current in the (two) transmission lines.

6

20.838 10 W

273.6 A2(56 km)0.10 /km P

P I R I R→

×= = = =Ω

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

6555.838 10 W 2.041 10 V 200 kV

273.6 A P

V I

×= = = × ≈

The voltage has 2 significant figures.

80. ( a) From the efficiency of the transformer, we have S P0.88 . P P = Use this to calculate the current inthe primary.

SS P P P P

P

75 W0.88 0.88 0.7748 A 0.77 A

0.88 0.88(110 V) P

P P I V I V

→= = = = = ≈

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

since 2 / . P V R=

P P P

S S S S

110 V8.2

(75 W)(2.4 )

N V V N V P R

= = = =Ω

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

out in 42, 000 V (740 A )(2)(0.95 ) 40,594 V 41 kVV V IR= − = − Ω = ≈

(b) The power input is given by in in . P IV =

7 7in in (740 A)(42,000 V) 3.108 10 W 3.1 10 W P IV = = = × ≈ ×

(c) The power loss in the lines is due to the current in the resistive wires.2 2 6 6

loss (740 A) 2(0.95 ) 1.040 10 W 1.0 10 W P I R= = Ω = × ≈ ×

(d ) The power output is given by out out . P IV =

7 7out out (740 A)(40,594 V) 3.004 10 W 3.0 10 W P IV = = = × ≈ ×

This could also be found by subtracting the power lost from the input power.7 6 7 7

out in loss 3.108 10 W 1.040 10 W 3.004 10 W 3.0 10 W P P P = − = × − × = × ≈ ×


25/29



82. A side view of the rail and bar is shown in the figure. From thediscussion in Section 21–3, the emf in the bar is produced by thecomponents of the magnetic field, the length of the bar, and the velocityof the bar, which are all mutually perpendicular. The magnetic field andthe length of the bar are already perpendicular. The component of thevelocity of the bar that is perpendicular to the magnetic field is cos ,υ θ so the induced emf is given by the following.

cos B υ θ =

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

cos B I

R Rυ θ = =

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

2 2cos cos B

B B F I B B R R

υ θ υ θ = = =

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

2 2 2

net

2

2 2 2 2 2 2

coscos sin 0 sin

sin (0.60 )(0.040 kg)(9.80 m/s ) sin 6.01.199 m/s 1.2 m/s

cos (0.45 T) (0.32 m) cos 6.0

B B

F F mg mg R

Rmg

B

υ θ θ θ θ

θ υ

θ

→ →= − = =

Ω °= = = ≈

°

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.

222T T T L

T L L L L L L 2 P P P R P I V I P I R RV V V

→ ⎛ ⎞= = = = =⎜ ⎟⎝ ⎠

84. The induced current in the coil is the induced emf divided by the resistance. The induced emf is foundfrom the changing flux by Eq. 21–2b. The magnetic field of the solenoid, which causes the flux, isgiven 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.

ind sol sol solcoil coil sol sol 0

sol

sol solcoil sol 0

sol coil sol 0 sol sol

sol

2 7

ind

2

=

190 turns 0.045 m 4 10 230 turns 2.0 A 5.823

12 0.01 m 0.10 s

5.8 10

( ) ( ) ( T m/A) ( )10 A

( )

B N I I N N A B

R t t

N I N A

t N A N I I

R R t

µ

µ µ

π π −

−

∆∆Φ= = =

∆ ∆

∆

∆ ∆= =

∆

×= =

Ω

×

⋅

×

≈ 2 A−

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

θ

B

NF

mg

vBF


26/29

21-26 Chapter 21


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

( ) ( )/ /max1 1tR L tR LV I e I e R− −= − = −

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

( )/ /maxmax

4

max

1 1

(1.2 10 s)(120 )0.0982 H 98 mH in series

7.5 mAln 1ln 1

55 mA

tR L tR L I I I e e I

tR L

I I

− −

−

→ →= − = −

× Ω= − = − = ≈

⎛ ⎞ ⎛ ⎞−−⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

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.

2 2120 V

280 T/s(35) (6.25 10 m)

B A B B N N t t t NA π

−→

∆Φ ∆ ∆= = = = =

∆ ∆ ∆ ×

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

2 2 peak (125)(0.200 T)(2 rad/rev)(120 rev/s)(6.60 10 m) 82 V NB Aω π

−= = × =

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

2 12 2 2 41 102 2

4 3 4 3

2 26 3 61 1

2 2 70

6 39

4 3

3

(8.85 10 C /N m )(1.0 10 V/m)

4.425 10 J/m 4.4 10 J/m

(2.0 T)1.592 10 J/m 1.6 10

(4 10 T m/A)

1.592 10 J/m3.6 10

4.425 10 J/m

J/m

E

B

E

B

u E

B

uu

u

ε

µ π

−

− −

−

−

=

=

= = × ⋅ ×

= × ≈ ×

= = × ≈ ×× ⋅

×= ×

×

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

(b) Set the two densities equal and solve for the magnitude of the electric field.2

21 102 2

0

8

12 2 2 70 0

2.0 T 6.0 10 V/m(8.85 10 C /N m )(4 10 T m/A)

E B B

u E u

B E

ε µ

ε µ π − −

→= = =

= = = ×

× ⋅ × ⋅

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) andsolve for the inductance.

rms rms

rms rms

220 V2 93 mH

2 2 (60 Hz)(6.3 A) LV V

X fL L I f I

π π π

→= = = = =


27/29



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.

70 2 2 2 2

0

1 1 1 11.1 10 F

2 4 4 (0.13 H)(1360 Hz) f C

LC Lf π π π

−→= = = = ×

91. The inductance of the solenoid is given by2 2 2

0 0 .4

N A N d L

µ µ π = =

The (constant) length of the

wire is given by wire sol , N d π = so since sol 2 sol 12 ,d d = we also know that

ch21 giancoli7e manual

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