chapter 31 faradayÕs law - rmutphysics...faradayÕs law c hap te r o u tlin e 31.1 faradayÕs law...

Faraday’s Law

CHAPTE R OUTL I N E

31.1 Faraday’s Law of Induction

31.2 Motional emf

31.3 Lenz’s Law

31.4 Induced emf and ElectricFields

31.5 Generators and Motors

31.6 Eddy Currents

31.7 Maxwell’s Equations

! In a commercial electric power plant, large generators produce energy that is transferredout of the plant by electrical transmission. These generators use magnetic induction togenerate a potential difference when coils of wire in the generator are rotated in a magneticfield. The source of energy to rotate the coils might be falling water, burning fossil fuels, or anuclear reaction. (Michael Melford/Getty Images)

Chapter 31

967

968

The focus of our studies in electricity and magnetism so far has been the electric fieldsproduced by stationary charges and the magnetic fields produced by moving charges.This chapter explores the effects produced by magnetic fields that vary in time.

Experiments conducted by Michael Faraday in England in 1831 and indepen-dently by Joseph Henry in the United States that same year showed that an emf canbe induced in a circuit by a changing magnetic field. The results of these experi-ments led to a very basic and important law of electromagnetism known as Faraday’slaw of induction. An emf (and therefore a current as well) can be induced in variousprocesses that involve a change in a magnetic flux.

With the treatment of Faraday’s law, we complete our introduction to the funda-mental laws of electromagnetism. These laws can be summarized in a set of four equa-tions called Maxwell’s equations. Together with the Lorentz force law, they represent acomplete theory for describing the interaction of charged objects.

31.1 Faraday’s Law of Induction

To see how an emf can be induced by a changing magnetic field, consider a loop ofwire connected to a sensitive ammeter, as illustrated in Figure 31.1. When a magnet ismoved toward the loop, the galvanometer needle deflects in one direction, arbitrarilyshown to the right in Figure 31.1a. When the magnet is brought to rest and heldstationary relative to the loop (Fig. 31.1b), no deflection is observed. When the magnetis moved away from the loop, the needle deflects in the opposite direction, as shown inFigure 31.1c. Finally, if the magnet is held stationary and the loop is moved eithertoward or away from it, the needle deflects. From these observations, we conclude thatthe loop detects that the magnet is moving relative to it and we relate this detection toa change in magnetic field. Thus, it seems that a relationship exists between currentand changing magnetic field.

These results are quite remarkable in view of the fact that a current is set up eventhough no batteries are present in the circuit! We call such a current an inducedcurrent and say that it is produced by an induced emf.

Now let us describe an experiment conducted by Faraday and illustrated in Figure31.2. A primary coil is connected to a switch and a battery. The coil is wrapped aroundan iron ring, and a current in the coil produces a magnetic field when the switch isclosed. A secondary coil also is wrapped around the ring and is connected to a sensitiveammeter. No battery is present in the secondary circuit, and the secondary coil is notelectrically connected to the primary coil. Any current detected in the secondarycircuit must be induced by some external agent.

Initially, you might guess that no current is ever detected in the secondary circuit.However, something quite amazing happens when the switch in the primary circuit iseither opened or thrown closed. At the instant the switch is closed, the galvanometerneedle deflects in one direction and then returns to zero. At the instant the switch isopened, the needle deflects in the opposite direction and again returns to zero.

Michael FaradayBritish Physicist and Chemist(1791–1867)

Faraday is often regarded as thegreatest experimental scientist ofthe 1800s. His manycontributions to the study ofelectricity include the invention ofthe electric motor, electricgenerator, and transformer, aswell as the discovery ofelectromagnetic induction andthe laws of electrolysis. Greatlyinfluenced by religion, he refusedto work on the development ofpoison gas for the British military.(By kind permission of thePresident and Council of theRoyal Society)

S ECT I O N 31 . 1 • Faraday’s Law of Induction 969

Finally, the galvanometer reads zero when there is either a steady current or nocurrent in the primary circuit. The key to understanding what happens in this experi-ment is to note first that when the switch is closed, the current in the primary circuitproduces a magnetic field that penetrates the secondary circuit. Furthermore, when

Active Figure 31.1 (a) When a magnet is moved toward a loop of wire connected to asensitive ammeter, the ammeter deflects as shown, indicating that a current is inducedin the loop. (b) When the magnet is held stationary, there is no induced current in theloop, even when the magnet is inside the loop. (c) When the magnet is moved awayfrom the loop, the ammeter deflects in the opposite direction, indicating that theinduced current is opposite that shown in part (a). Changing the direction of themagnet’s motion changes the direction of the current induced by that motion.

Ammeter

Ammeter

Ammeter

(b)

(a)

N S

(c)

N S

N S

I

I

At the Active Figures linkat http://www.pse6.com, youcan move the magnet andobserve the current in theammeter.

At the Active Figures linkat http://www.pse6.com, youcan open and close the switchand observe the current in theammeter.

Active Figure 31.2 Faraday’s experiment. When the switch in the primary circuit isclosed, the ammeter in the secondary circuit deflects momentarily. The emf induced inthe secondary circuit is caused by the changing magnetic field through the secondary coil.

Ammeter

Secondarycoil

Primarycoil

Iron

Switch

+–

Battery

970 CHAPTE R 31 • Faraday’s Law

This statement, known as Faraday’s law of induction, can be written

(31.1)

where !B " ∫ B ! dA is the magnetic flux through the circuit. (See Section 30.5.)If the circuit is a coil consisting of N loops all of the same area and if !B is the

magnetic flux through one loop, an emf is induced in every loop. The loops are in series,so their emfs add; thus, the total induced emf in the coil is given by the expression

(31.2)

The negative sign in Equations 31.1 and 31.2 is of important physical significance, asdiscussed in Section 31.3.

Suppose that a loop enclosing an area A lies in a uniform magnetic field B, as inFigure 31.3. The magnetic flux through the loop is equal to BA cos # ; hence, theinduced emf can be expressed as

(31.3)

From this expression, we see that an emf can be induced in the circuit in several ways:

• The magnitude of B can change with time.

• The area enclosed by the loop can change with time.

• The angle # between B and the normal to the loop can change with time.

• Any combination of the above can occur.

$ " %ddt

(BA cos #)

$ " %N d!B

dt

$ " %d!B

dt

the switch is closed, the magnetic field produced by the current in the primary circuitchanges from zero to some value over some finite time, and this changing field inducesa current in the secondary circuit.

As a result of these observations, Faraday concluded that an electric current canbe induced in a circuit (the secondary circuit in our setup) by a changingmagnetic field. The induced current exists for only a short time while the magneticfield through the secondary coil is changing. Once the magnetic field reaches a steadyvalue, the current in the secondary coil disappears. In effect, the secondary circuitbehaves as though a source of emf were connected to it for a short time. It is customaryto say that an induced emf is produced in the secondary circuit by the changingmagnetic field.

The experiments shown in Figures 31.1 and 31.2 have one thing in common: ineach case, an emf is induced in the circuit when the magnetic flux through the circuitchanges with time. In general,

The emf induced in a circuit is directly proportional to the time rate of change ofthe magnetic flux through the circuit.

! PITFALL PREVENTION31.1 Induced emf

Requires a ChangeThe existence of a magnetic fluxthrough an area is not sufficientto create an induced emf. Theremust be a change in the magneticflux in order for an emf to beinduced.

Faraday’s law

B

θ

θ

Loop ofarea A

Figure 31.3 A conducting loopthat encloses an area A in thepresence of a uniform magneticfield B. The angle between B andthe normal to the loop is #.

Quick Quiz 31.1 A circular loop of wire is held in a uniform magnetic field,with the plane of the loop perpendicular to the field lines. Which of the following willnot cause a current to be induced in the loop? (a) crushing the loop; (b) rotating theloop about an axis perpendicular to the field lines; (c) keeping the orientation of theloop fixed and moving it along the field lines; (d) pulling the loop out of the field.

S ECT I O N 31 . 1 • Faraday’s Law of Induction 971

Some Applications of Faraday’s Law

The ground fault interrupter (GFI) is an interesting safety device that protects users ofelectrical appliances against electric shock. Its operation makes use of Faraday’s law. Inthe GFI shown in Figure 31.5, wire 1 leads from the wall outlet to the appliance to beprotected, and wire 2 leads from the appliance back to the wall outlet. An iron ringsurrounds the two wires, and a sensing coil is wrapped around part of the ring. Becausethe currents in the wires are in opposite directions, the net magnetic flux through thesensing coil due to the currents is zero. However, if the return current in wire 2changes, the net magnetic flux through the sensing coil is no longer zero. (This canhappen, for example, if the appliance becomes wet, enabling current to leak toground.) Because household current is alternating (meaning that its direction keepsreversing), the magnetic flux through the sensing coil changes with time, inducing anemf in the coil. This induced emf is used to trigger a circuit breaker, which stops thecurrent before it is able to reach a harmful level.

Another interesting application of Faraday’s law is the production of sound in anelectric guitar (Fig. 31.6). The coil in this case, called the pickup coil, is placed near thevibrating guitar string, which is made of a metal that can be magnetized. A permanent

Quick Quiz 31.2 Figure 31.4 shows a graphical representation of the fieldmagnitude versus time for a magnetic field that passes through a fixed loop and is ori-ented perpendicular to the plane of the loop. The magnitude of the magnetic field atany time is uniform over the area of the loop. Rank the magnitudes of the emf gener-ated in the loop at the five instants indicated, from largest to smallest.

Quick Quiz 31.3 Suppose you would like to steal power for your home fromthe electric company by placing a loop of wire near a transmission cable, so as toinduce an emf in the loop (an illegal procedure). Should you (a) place your loop sothat the transmission cable passes through your loop, or (b) simply place your loopnear the transmission cable?

Figure 31.4 (Quick Quiz 31.2) The time behavior of a magnetic field through a loop.

B

a b c d et

Figure 31.5 Essential components of a groundfault interrupter.

Circuitbreaker

Sensingcoil

Alternatingcurrent

Ironring

1

2


magnet inside the coil magnetizes the portion of the string nearest the coil. When thestring vibrates at some frequency, its magnetized segment produces a changingmagnetic flux through the coil. The changing flux induces an emf in the coil that isfed to an amplifier. The output of the amplifier is sent to the loudspeakers, whichproduce the sound waves we hear.

(b)

Figure 31.6 (a) In an electric guitar, a vibrating magnetized string induces an emf in apickup coil. (b) The pickups (the circles beneath the metallic strings) of this electricguitar detect the vibrations of the strings and send this information through anamplifier and into speakers. (A switch on the guitar allows the musician to select whichset of six pickups is used.)

Char

les D

. Win

ters

(a)

Magnetizedportion of

string

Guitar string

To amplifier

N S

N S

MagnetPickup

coil

Example 31.1 One Way to Induce an emf in a Coil

A coil consists of 200 turns of wire. Each turn is a square ofside 18 cm, and a uniform magnetic field directed perpendic-ular to the plane of the coil is turned on. If the field changeslinearly from 0 to 0.50 T in 0.80 s, what is the magnitude ofthe induced emf in the coil while the field is changing?

Solution The area of one turn of the coil is (0.18 m)2 "0.032 4 m2 . The magnetic flux through the coil at t " 0 iszero because B " 0 at that time. At t " 0.80 s, the magneticflux through one turn is !B " BA " (0.50 T)(0.032 4 m2) "0.016 2 T & m2. Therefore, the magnitude of the induced emfis, from Equation 31.2,

4.1 V" 4.1 T&m2/s "

&$&" N '!B

't" 200

(0.016 2 T&m2 % 0)0.80 s

You should be able to show that 1 T & m2/s " 1 V.

What If? What if you were asked to find the magnitude ofthe induced current in the coil while the field is changing?Can you answer this question?

Answer If the ends of the coil are not connected to acircuit, the answer to this question is easy—the current iszero! (Charges will move within the wire of the coil, but theycannot move into or out of the ends of the coil.) In orderfor a steady current to exist, the ends of the coil must beconnected to an external circuit. Let us assume that the coilis connected to a circuit and that the total resistance of thecoil and the circuit is 2.0 (. Then, the current in the coil is

I "$R

"4.1 V2.0 (

" 2.0 A

Example 31.2 An Exponentially Decaying B Field

A loop of wire enclosing an area A is placed in a regionwhere the magnetic field is perpendicular to the plane ofthe loop. The magnitude of B varies in time according tothe expression B " Bmaxe%at, where a is some constant. Thatis, at t " 0 the field is Bmax, and for t ) 0, the field decreases

exponentially (Fig. 31.7). Find the induced emf in the loopas a function of time.

Solution Because B is perpendicular to the plane of theloop, the magnetic flux through the loop at time t ) 0 is

S ECT I O N 31 . 2 • Motional emf 973

31.2 Motional emf

In Examples 31.1 and 31.2, we considered cases in which an emf is induced in astationary circuit placed in a magnetic field when the field changes with time. In thissection we describe what is called motional emf, which is the emf induced in aconductor moving through a constant magnetic field.

The straight conductor of length ! shown in Figure 31.9 is moving through auniform magnetic field directed into the page. For simplicity, we assume that theconductor is moving in a direction perpendicular to the field with constant velocityunder the influence of some external agent. The electrons in the conductor experi-ence a force FB " q v " B that is directed along the length !, perpendicular to bothv and B (Eq. 29.1). Under the influence of this force, the electrons move to thelower end of the conductor and accumulate there, leaving a net positive charge atthe upper end. As a result of this charge separation, an electric field E is produced

t

B

Bmax

Figure 31.7 (Example 31.2) Exponential decrease in themagnitude of the magnetic field with time. The induced emfand induced current vary with time in the same way.

Conceptual Example 31.3 Which Bulb Is Shorted Out?

Two bulbs are connected to opposite sides of a circular loopof wire, as shown in Figure 31.8a. A changing magnetic field(confined to the smaller circular area shown in the figure)induces an emf in the loop that causes the two bulbs tolight. When the switch is closed, the resistance-free wiresconnected to the switch short out bulb 2 and it goes out.What happens if the wires containing the closed switchremain connected at points a and b, but the switch and thewires are lifted up and moved to the other side of the field,as in Figure 3.18b? The wire is still connected to bulb 2 as itwas before, so does it continue to stay dark?

Solution When the wire is moved to the other side, eventhough the connections have not changed, bulb 1 goes outand bulb 2 glows. The bulb that is shorted depends onwhich side of the changing field the switch is positioned! InFigure 31.8a, because the branch containing bulb 2 is infi-nitely more resistant than the branch containing theresistance-free switch, we can imagine removing the branchwith the bulb without altering the circuit. Then we have asimple loop containing only bulb 1, which glows.

When the wire is moved, as in Figure 31.8b, there aretwo possible paths for current below points a and b. We canimagine removing the branch with bulb 1, leaving only asingle loop with bulb 2.

× × × × × × ×× × × × × × ×× × × × × × ×× × × × × × ×

× × × × ×

× × × × ×× × × × × × ×× × × × × × ×× × × × × × ×× × × × × × ×

× × × × ×

× × × × ×

Bulb 1 Bulb 1

Bulb 2 Bulb 2

Switch

Switch

a b aa b

(b)(a)

Figure 31.8 (Conceptual Example 31.3) (a) When the wirewith the switch is located as shown, bulb 2 goes out when theswitch is closed. (b) What happens when the switch and thewires are moved to the other side of the magnetic field?

!B " BA cos 0 " ABmaxe%at

Because ABmax and a are constants, the induced emf calcu-lated from Equation 31.1 is

"

This expression indicates that the induced emf decaysexponentially in time. Note that the maximum emf occursat t " 0, where $max " aABmax. The plot of $ versus t issimilar to the B-versus-t curve shown in Figure 31.7.

aABmaxe%at

$ " %d!B

dt" %ABmax

ddt

e%at

inside the conductor. The charges accumulate at both ends until the downwardmagnetic force qvB on charges remaining in the conductor is balanced by theupward electric force qE. At this point, electrons move only with random thermalmotion. The condition for equilibrium requires that

qE " qvB or E " vB

The electric field produced in the conductor is related to the potential differenceacross the ends of the conductor according to the relationship 'V " E! (Eq. 25.6).Thus, for the equilibrium condition,

'V " E! " B !v (31.4)

where the upper end of the conductor in Figure 31.9 is at a higher electric potentialthan the lower end. Thus, a potential difference is maintained between the endsof the conductor as long as the conductor continues to move through the uniform magnetic field. If the direction of the motion is reversed, the polarity of thepotential difference is also reversed.

A more interesting situation occurs when the moving conductor is part of a closedconducting path. This situation is particularly useful for illustrating how a changingmagnetic flux causes an induced current in a closed circuit. Consider a circuit consist-ing of a conducting bar of length ! sliding along two fixed parallel conducting rails, asshown in Figure 31.10a.

For simplicity, we assume that the bar has zero resistance and that the stationarypart of the circuit has a resistance R. A uniform and constant magnetic field B isapplied perpendicular to the plane of the circuit. As the bar is pulled to the right witha velocity v under the influence of an applied force Fapp, free charges in the barexperience a magnetic force directed along the length of the bar. This force sets upan induced current because the charges are free to move in the closed conductingpath. In this case, the rate of change of magnetic flux through the loop and the corre-sponding induced motional emf across the moving bar are proportional to thechange in area of the loop. If the bar is pulled to the right with a constant velocity, thework done by the applied force appears as internal energy in the resistor R . (SeeSection 27.6.)

Because the area enclosed by the circuit at any instant is !x, where x is the positionof the bar, the magnetic flux through that area is

!B " B !x

Using Faraday’s law, and noting that x changes with time at a rate dx/dt " v, we findthat the induced motional emf is

(31.5)

Because the resistance of the circuit is R , the magnitude of the induced current is

(31.6)

The equivalent circuit diagram for this example is shown in Figure 31.10b.

I "&$&

R"

B!vR

$ " %B!v

$ " %d!B

dt" %

ddt

(B!x) " %B! dxdt


Figure 31.9 A straight electrical conductor of length ! movingwith a velocity v through a uniform magnetic field B directedperpendicular to v. Due to the magnetic force on electrons, theends of the conductor become oppositely charged. This establishesan electric field in the conductor. In steady state, the electric andmagnetic forces on an electron in the wire are balanced.

Bin×

×

×

×

×

×

×

×

×

×

×

×

!

++

−−

×

×

×

×

FB

Fe

E

–

v

Active Figure 31.10 (a) A conduct-ing bar sliding with a velocity v alongtwo conducting rails under theaction of an applied force Fapp. Themagnetic force FB opposes themotion, and a counterclockwisecurrent I is induced in the loop.(b) The equivalent circuit diagramfor the setup shown in part (a).

At the Active Figures linkat http://www.pse6.com, youcan adjust the applied force,the magnetic field, and theresistance to see the effects onthe motion of the bar.

(b)

R

!B vε =

I

R FB

(a)

x

Fapp

v

Bin

!

× × ×

× × ×

× × ×

× × ×

× × ×

×I

×

×

×

×

×

×

Motional emf

Let us examine the system using energy considerations. Because no battery is in thecircuit, we might wonder about the origin of the induced current and the energy deliv-ered to the resistor. We can understand the source of this current and energy by notingthat the applied force does work on the conducting bar, thereby moving chargesthrough a magnetic field. Their movement through the field causes the charges tomove along the bar with some average drift velocity, and hence a current is established.The change in energy in the system during some time interval must be equal to thetransfer of energy into the system by work, consistent with the general principle ofconservation of energy described by Equation 7.17.

Let us verify this mathematically. As the bar moves through the uniform magneticfield B, it experiences a magnetic force FB of magnitude I!B (see Section 29.2). Thedirection of this force is opposite the motion of the bar, to the left in Figure 31.10a.Because the bar moves with constant velocity, the applied force must be equal inmagnitude and opposite in direction to the magnetic force, or to the right in Figure31.10a. (If FB acted in the direction of motion, it would cause the bar to accelerate,violating the principle of conservation of energy.) Using Equation 31.6 and the factthat Fapp " I!B, we find that the power delivered by the applied force is

(31.7)

From Equation 27.23, we see that this power input is equal to the rate at which energyis delivered to the resistor, so that Equation 7.17 is confirmed in this situation.

" " Fappv " (I!B)v "B2 !2v2

R"

$2

R

S ECT I O N 31 . 2 • Motional emf 975

Example 31.4 Motional emf Induced in a Rotating Bar

A conducting bar of length ! rotates with a constant angularspeed * about a pivot at one end. A uniform magnetic fieldB is directed perpendicular to the plane of rotation, asshown in Figure 31.11. Find the motional emf inducedbetween the ends of the bar.

Solution Consider a segment of the bar of length dr havinga velocity v. According to Equation 31.5, the magnitude of

the emf induced in this segment is

d$ " Bv dr

Because every segment of the bar is moving perpendicularto B, an emf d$ of the same form is generated across eachsegment. Summing the emfs induced across all segments,which are in series, gives the total emf between the ends

Interactive

Quick Quiz 31.4 As an airplane flies from Los Angeles to Seattle, it passesthrough the Earth’s magnetic field. As a result, a motional emf is developed betweenthe wingtips. Which wingtip is positively charged? (a) the left wing (b) the rightwing.

Quick Quiz 31.5 In Figure 31.10, a given applied force of magnitude Fappresults in a constant speed v and a power input ". Imagine that the force is increasedso that the constant speed of the bar is doubled to 2v. Under these conditions, the newforce and the new power input are (a) 2F and 2" (b) 4F and 2" (c) 2F and 4" (d) 4Fand 4".

Quick Quiz 31.6 You wish to move a rectangular loop of wire into a regionof uniform magnetic field at a given speed so as to induce an emf in the loop. Theplane of the loop remains perpendicular to the magnetic field lines. In which orienta-tion should you hold the loop while you move it into the region of magnetic field inorder to generate the largest emf? (a) with the long dimension of the loop parallel tothe velocity vector (b) with the short dimension of the loop parallel to the velocityvector (c) either way—the emf is the same regardless of orientation.


Example 31.5 Magnetic Force Acting on a Sliding Bar

The conducting bar illustrated in Figure 31.12 moves on twofrictionless parallel rails in the presence of a uniform mag-netic field directed into the page. The bar has mass m andits length is !. The bar is given an initial velocity vi to theright and is released at t " 0.

(A) Using Newton’s laws, find the velocity of the bar as afunction of time.

(B) Show that the same result is reached by using an energyapproach.

Solution (A) Conceptualize this situation as follows. As thebar slides to the right in Figure 31.12, a counterclockwisecurrent is established in the circuit consisting of the bar,the rails, and the resistor. The upward current in the barresults in a magnetic force to the left on the bar as shown inthe figure. As a result, the bar will slow down, so ourmathematical solution should demonstrate this. The text ofpart (A) already categorizes this as a problem in usingNewton’s laws. To analyze the problem, we determine from

Equation 29.3 that the magnetic force is FB " % I!B, wherethe negative sign indicates that the retarding force is tothe left. Because this is the only horizontal force actingon the bar, Newton’s second law applied to motion in the

Interactive

of the bar:

To integrate this expression, note that the linear speed v ofan element is related to the angular speed * through therelationship v " r* (Eq. 10.10). Therefore, because B and *are constants, we find that

What If? Suppose, after reading through this example, youcome up with a brilliant idea. A Ferris wheel has radial metal-lic spokes between the hub and the circular rim. Thesespokes move in the magnetic field of the Earth, so each

12 B*!2$ " B ! v dr " B* !!

0 r dr "

$ " ! Bv dr

spoke acts like the bar in Figure 31.11. You plan to use theemf generated by the rotation of the Ferris wheel to powerthe lightbulbs on the wheel! Will this idea work?

Answer The fact that this is not done in practice suggeststhat others may have thought of this idea and rejected it. Letus estimate the emf that is generated in this situation. Weknow the magnitude of the magnetic field of the Earth fromTable 29.1, B " 0.5 + 10%4 T. A typical spoke on a Ferriswheel might have a length on the order of 10 m. Supposethe period of rotation is on the order of 10 s. This gives anangular speed of

Assuming that the magnetic field lines of the Earth arehorizontal at the location of the Ferris wheel and perpendic-ular to the spokes, the emf generated is

This is a tiny emf, far smaller than that required to operatelightbulbs.

An additional difficulty is related to energy. Assumingyou could find lightbulbs that operate using a potential dif-ference on the order of millivolts, a spoke must be part of acircuit in order to provide a voltage to the bulbs. Conse-quently, the spoke must carry a current. Because thiscurrent-carrying spoke is in a magnetic field, a magneticforce is exerted on the spoke and the direction of the forceis opposite to its direction of motion. As a result, the motorof the Ferris wheel must supply more energy to performwork against this magnetic drag force. The motor must ulti-mately provide the energy that is operating the lightbulbsand you have not gained anything for free!

" 2.5 + 10%3 V " 1 mV

$ " 12 B*!2 " 1

2 (0.5 + 10%4 T)(1 s%1)(10 m)2

* "2,

T"

2,

10 s" 0.63 s%1 " 1 s%1

v

!

×Bin

dr

O

r

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

Figure 31.11 (Example 31.4) A conducting bar rotatingaround a pivot at one end in a uniform magnetic field that isperpendicular to the plane of rotation. A motional emf isinduced across the ends of the bar.

At the Interactive Worked Example link at http://www.pse6.com, you can explore the induced emf for different angularspeeds and field magnitudes.

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

FBvi

Bin ×

I

!

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

R

×

×

×

×

×

×

Figure 31.12 (Example 31.5) A conducting bar of length ! ontwo fixed conducting rails is given an initial velocity vi to the right.

S ECT I O N 31 . 3 • Lenz’s Law 977

31.3 Lenz’s Law

Faraday’s law (Eq. 31.1) indicates that the induced emf and the change in flux haveopposite algebraic signs. This has a very real physical interpretation that has come tobe known as Lenz’s law1:

The induced current in a loop is in the direction that creates a magnetic field thatopposes the change in magnetic flux through the area enclosed by the loop.

1 Developed by the German physicist Heinrich Lenz (1804–1865).

horizontal direction gives

From Equation 31.6, we know that I " B!v/R, and so wecan write this expression as

Integrating this equation using the initial condition thatv " vi at t " 0, we find that

where the constant - " mR/B 2!2. From this result, we seethat the velocity can be expressed in the exponential form

(1)

To finalize the problem, note that this expression for v indi-cates that the velocity of the bar decreases with time underthe action of the magnetic retarding force, as we expectfrom our conceptualization of the problem.

(B) The text of part (B) immediately categorizes this as aproblem in energy conservation. Consider the sliding bar asone system possessing kinetic energy, which decreases becauseenergy is transferring out of the system by electrical transmis-sion through the rails. The resistor is another system possess-ing internal energy, which rises because energy is transferringinto this system. Because energy is not leaving the combinationof two systems, the rate of energy transfer out of the bar equalsthe rate of energy transfer into the resistor. Thus,

"resistor " % "bar

where the negative sign is necessary because energy isleaving the bar and "bar is a negative number. Substitutingfor the electrical power delivered to the resistor and the

v " vie%t/-

ln # vvi$ " %# B 2!2

mR $ t " %t-

!v

vi

dvv

"%B 2!2

mR !t

0 dt

dvv

" %# B 2!2

mR $ dt

m dvdt

" %B 2!2

R v

Fx " ma " m dvdt

" %I!B

time rate of change of kinetic energy for the bar, we have

Using Equation 31.6 for the current and carrying out thederivative, we find

Rearranging terms gives

To finalize this part of the problem, note that this is thesame expression that we obtained in part (A).

What If? Suppose you wished to increase the distancethrough which the bar moves between the time when it is ini-tially projected and the time when it essentially comes to rest.You can do this by changing one of three variables: vi, R, or B,by a factor of 2 or . Which variable should you change in orderto maximize the distance, and would you double it or halve it?

Answer Increasing vi would make the bar move farther.Increasing R would decrease the current and, therefore, themagnetic force, making the bar move farther. Decreasing Bwould decrease the magnetic force and make the bar movefarther. But which is most effective?

We use Equation (1) to find the distance that the barmoves by integration:

From this expression, we see that doubling vi or R willdouble the distance. But changing B by a factor of causesthe distance to be four times as great!

12

" %vi -(0 % 1) " vi- " vi # mRB 2!2 $

x " !.

0 vie%t/- dt " %vi -e%t/- %.

0

v "dxdt

" vie%t/-

12

dvv

" %# B 2!2

mR $ dt

B 2!2v2

R" %mv

dvdt

I 2R " %ddt

(12 mv2)

At the Interactive Worked Example link at http://www.pse6.com, you can study the motion of the bar after it is released.

Lenz’s law

That is, the induced current tends to keep the original magnetic flux through thecircuit from changing. We shall show that this law is a consequence of the law ofconservation of energy.

To understand Lenz’s law, let us return to the example of a bar moving to theright on two parallel rails in the presence of a uniform magnetic field (the externalmagnetic field, Fig. 31.13a.) As the bar moves to the right, the magnetic fluxthrough the area enclosed by the circuit increases with time because the areaincreases. Lenz’s law states that the induced current must be directed so that themagnetic field it produces opposes the change in the external magnetic flux.Because the magnetic flux due to an external field directed into the page is increas-ing, the induced current, if it is to oppose this change, must produce a fielddirected out of the page. Hence, the induced current must be directed counter-clockwise when the bar moves to the right. (Use the right-hand rule to verify thisdirection.) If the bar is moving to the left, as in Figure 31.13b, the external mag-netic flux through the area enclosed by the loop decreases with time. Because thefield is directed into the page, the direction of the induced current must beclockwise if it is to produce a field that also is directed into the page. In either case,the induced current tends to maintain the original flux through the area enclosedby the current loop.

Let us examine this situation using energy considerations. Suppose that the bar isgiven a slight push to the right. In the preceding analysis, we found that this motionsets up a counterclockwise current in the loop. What happens if we assume that the


Figure 31.13 (a) As theconducting bar slides on the twofixed conducting rails, themagnetic flux due to the externalmagnetic field into the pagethrough the area enclosed by theloop increases in time. By Lenz’slaw, the induced current must becounterclockwise so as to producea counteracting magnetic fielddirected out of the page. (b) Whenthe bar moves to the left, theinduced current must be clockwise.Why?

×

×

×

×

×

FB v

Bin×

I

×

×

×

×

×

×

×

×

×

×

× ×

×

×

×

×

×

×

×

×

×

×

×

×

R

(a)

FBv

×

I

×

×

×

×

×

×

×

×

×

×

× ×

×

×

×

×

×

×

×

×

R

(b)

×

×

×

×

×

×

×

×

×

Figure 31.14 (a) When the magnet is moved toward the stationary conducting loop, a current is induced in the direction shown. The magnetic field lines shown are those due to the bar magnet. (b) This induced current produces its own magnetic field directed to the left that counteracts the increasing external flux. Themagnetic field lines shown are those due to the induced current in the ring. (c) When the magnet is moved away from the stationary conducting loop, a current is induced in the direction shown. The magnetic field lines shown are those due to the bar magnet. (d) This induced current produces a magnetic field directed to theright and so counteracts the decreasing external flux. The magnetic field lines shown are those due to the induced current in the ring.

Exa

mpl

e

N SI

(b)

I

v

(a)

NS

(d)(c)

I

vS N

I

NS

current is clockwise, such that the direction of the magnetic force exerted on the bar isto the right? This force would accelerate the rod and increase its velocity. This, in turn,would cause the area enclosed by the loop to increase more rapidly; this would resultin an increase in the induced current, which would cause an increase in the force,which would produce an increase in the current, and so on. In effect, the system wouldacquire energy with no input of energy. This is clearly inconsistent with all experienceand violates the law of conservation of energy. Thus, we are forced to conclude that thecurrent must be counterclockwise.

Let us consider another situation, one in which a bar magnet moves toward astationary metal loop, as in Figure 31.14a. As the magnet moves to the right toward theloop, the external magnetic flux through the loop increases with time. To counteractthis increase in flux due to a field toward the right, the induced current produces itsown magnetic field to the left, as illustrated in Figure 31.14b; hence, the inducedcurrent is in the direction shown. Knowing that like magnetic poles repel each other,we conclude that the left face of the current loop acts like a north pole and that theright face acts like a south pole.

If the magnet moves to the left, as in Figure 31.14c, its flux through the areaenclosed by the loop decreases in time. Now the induced current in the loop is in thedirection shown in Figure 31.14d because this current direction produces a magneticfield in the same direction as the external field. In this case, the left face of the loop isa south pole and the right face is a north pole.

S ECT I O N 31 . 3 • Lenz’s Law 979

Quick Quiz 31.7 Figure 31.15 shows a magnet being moved in the vicinityof a solenoid connected to a sensitive ammeter. The south pole of the magnet is thepole nearest the solenoid, and the ammeter indicates a clockwise (viewed fromabove) current in the solenoid. Is the person (a) inserting the magnet or (b) pullingit out?

Quick Quiz 31.8 Figure 31.16 shows a circular loop of wire being droppedtoward a wire carrying a current to the left. The direction of the induced current inthe loop of wire is (a) clockwise (b) counterclockwise (c) zero (d) impossible todetermine.

Figure 31.15 (Quick Quiz 31.7)

Rich

ard

Meg

na/F

unda

men

tal P

hoto

grap

hs

Figure 31.16 (Quick Quiz 31.8)I

v


Conceptual Example 31.6 Application of Lenz’s Law

A metal ring is placed near a solenoid, as shown in Figure31.17a. Find the direction of the induced current in the ring

(A) at the instant the switch in the circuit containing thesolenoid is thrown closed,

(B) after the switch has been closed for several seconds, and

(C) at the instant the switch is thrown open.

Solution (A) At the instant the switch is thrown closed, thesituation changes from one in which no magnetic flux existsin the ring to one in which flux exists and the magnetic fieldis to the left as shown in Figure 31.17b. To counteract thischange in the flux, the current induced in the ring must setup a magnetic field directed from left to right in Figure31.17b. This requires a current directed as shown.

(B) After the switch has been closed for several seconds,no change in the magnetic flux through the loop occurs;hence, the induced current in the ring is zero.

(C) Opening the switch changes the situation from one inwhich magnetic flux exists in the ring to one in which there isno magnetic flux. The direction of the induced current is asshown in Figure 31.17c because current in this direction

produces a magnetic field that is directed right to left and socounteracts the decrease in the flux produced by the solenoid.

ε(c)

(a) (b)ε εSwitch

Figure 31.17 (Example 31.6) A current is induced in a metalring near a solenoid when the switch is opened or thrownclosed.

Conceptual Example 31.7 A Loop Moving Through a Magnetic Field

A rectangular metallic loop of dimensions ! and w andresistance R moves with constant speed v to the right, as inFigure 31.18a. The loop passes through a uniform magneticfield B directed into the page and extending a distance 3walong the x axis. Defining x as the position of the right sideof the loop along the x axis, plot as functions of x

(A) the magnetic flux through the area enclosed by the loop,

(B) the induced motional emf, and

(C) the external applied force necessary to counter themagnetic force and keep v constant.

Solution (A) Figure 31.18b shows the flux through thearea enclosed by the loop as a function of x. Before the loopenters the field, the flux is zero. As the loop enters the field,the flux increases linearly with position until the left edge of

the loop is just inside the field. Finally, the flux through theloop decreases linearly to zero as the loop leaves the field.

(B) Before the loop enters the field, no motional emf isinduced in it because no field is present (Fig. 31.18c). As theright side of the loop enters the field, the magnetic fluxdirected into the page increases. Hence, according to Lenz’slaw, the induced current is counterclockwise because it mustproduce its own magnetic field directed out of the page. Themotional emf % B!v (from Eq. 31.5) arises from the magneticforce experienced by charges in the right side of the loop.When the loop is entirely in the field, the change in magneticflux is zero, and hence the motional emf vanishes. Thishappens because, once the left side of the loop enters thefield, the motional emf induced in it cancels the motional emfpresent in the right side of the loop. As the right side of theloop leaves the field, the flux begins to decrease, a clockwise

(d)

0 w 3w 4w x

Fx

B 2!2v

R

ΦB

0 w 3w 4w x

(b)

B!w

(a)

3w

× × × × ×

× × × × ×

× × × × ×

× × × × ×

× × × × ×

vBin

w

!

0 x

(c)

ε

x

B!v

– B!v

Figure 31.18 (Conceptual Example 31.7) (a) Aconducting rectangular loop of width w and length !moving with a velocity v through a uniform magnetic fieldextending a distance 3w. (b) Magnetic flux through thearea enclosed by the loop as a function of loop position.(c) Induced emf as a function of loop position.(d) Applied force required for constant velocity as afunction of loop position.

S ECT I O N 31 . 4 • Induced emf and Electric Fields 981

31.4 Induced emf and Electric Fields

We have seen that a changing magnetic flux induces an emf and a current in aconducting loop. In our study of electricity, we related a current to an electric fieldthat applies electric forces on charged particles. In the same way, we can relate aninduced current in a conducting loop to an electric field by claiming that an electricfield is created in the conductor as a result of the changing magnetic flux.

We also noted in our study of electricity that the existence of an electric field isindependent of the presence of any test charges. This suggests that even in the absenceof a conducting loop, a changing magnetic field would still generate an electric field inempty space.

This induced electric field is nonconservative, unlike the electrostatic field producedby stationary charges. We can illustrate this point by considering a conducting loop ofradius r situated in a uniform magnetic field that is perpendicular to the plane of theloop, as in Figure 31.19. If the magnetic field changes with time, then, according toFaraday’s law (Eq. 31.1), an emf $ " % d!B/dt is induced in the loop. The inductionof a current in the loop implies the presence of an induced electric field E, which mustbe tangent to the loop because this is the direction in which the charges in the wiremove in response to the electric force. The work done by the electric field in moving atest charge q once around the loop is equal to q$. Because the electric force acting onthe charge is qE, the work done by the electric field in moving the charge once aroundthe loop is qE(2,r), where 2,r is the circumference of the loop. These two expressionsfor the work done must be equal; therefore, we see that

Using this result, along with Equation 31.1 and the fact that !B " BA " ,r2B for acircular loop, we find that the induced electric field can be expressed as

(31.8)

If the time variation of the magnetic field is specified, we can easily calculate the in-duced electric field from Equation 31.8.

The emf for any closed path can be expressed as the line integral of E & ds over thatpath: $ " &E & ds. In more general cases, E may not be constant, and the path may notbe a circle. Hence, Faraday’s law of induction, $ " % d!B/dt, can be written in thegeneral form

(31.9)

The induced electric field E in Equation 31.9 is a nonconservative field thatis generated by a changing magnetic field. The field E that satisfies Equation 31.9

' E&ds " %d!B

dt

E " %1

2,r d!B

dt" %

r2

dBdt

E "$

2,r

q$ " qE(2,r)

current is induced, and the induced emf is B!v. As soon as theleft side leaves the field, the emf decreases to zero.

(C) The external force that must be applied to the loop tomaintain this motion is plotted in Figure 31.18d. Before theloop enters the field, no magnetic force acts on it; hence, theapplied force must be zero if v is constant. When the right sideof the loop enters the field, the applied force necessary tomaintain constant speed must be equal in magnitude and op-posite in direction to the magnetic force exerted on that side.When the loop is entirely in the field, the flux through theloop is not changing with time. Hence, the net emf induced in

the loop is zero, and the current also is zero. Therefore, no ex-ternal force is needed to maintain the motion. Finally, as theright side leaves the field, the applied force must be equal inmagnitude and opposite in direction to the magnetic forceacting on the left side of the loop.

From this analysis, we conclude that power is suppliedonly when the loop is either entering or leaving the field.Furthermore, this example shows that the motional emfinduced in the loop can be zero even when there is motionthrough the field! A motional emf is induced only when themagnetic flux through the loop changes in time.

Figure 31.19 A conducting loopof radius r in a uniform magneticfield perpendicular to the plane ofthe loop. If B changes in time, anelectric field is induced in adirection tangent to thecircumference of the loop.

E×

E

E E

Bin

× ×

× × ××

× × × ××

× × × ××

× × ××

× × ×

r

Faraday’s law in general form

! PITFALL PREVENTION31.2 Induced Electric

FieldsThe changing magnetic fielddoes not need to be in existenceat the location of the inducedelectric field. In Figure 31.19,even a loop outside the region ofmagnetic field will experiencean induced electric field. Foranother example, consider Figure 31.8. The light bulbs glow(if the switch is open) eventhough the wires are outside theregion of the magnetic field.


31.5 Generators and Motors

Electric generators take in energy by work and transfer it out by electrical transmission.To understand how they operate, let us consider the alternating current (AC) gener-ator. In its simplest form, it consists of a loop of wire rotated by some external meansin a magnetic field (Fig. 31.21a).

cannot possibly be an electrostatic field because if the field were electrostatic, andhence conservative, the line integral of E & ds over a closed loop would be zero(Section 25.1); this would be in contradiction to Equation 31.9.

Quick Quiz 31.9 In a region of space, the magnetic field increases at aconstant rate. This changing magnetic field induces an electric field that (a) increases intime (b) is conservative (c) is in the direction of the magnetic field (d) has a constantmagnitude.

Example 31.8 Electric Field Induced by a Changing Magnetic Field in a Solenoid

A long solenoid of radius R has n turns of wire per unitlength and carries a time-varying current that variessinusoidally as I " Imax cos *t, where Imax is the maximumcurrent and * is the angular frequency of the alternatingcurrent source (Fig. 31.20).

(A) Determine the magnitude of the induced electric fieldoutside the solenoid at a distance r ) R from its long centralaxis.

Solution First let us consider an external point and takethe path for our line integral to be a circle of radius rcentered on the solenoid, as illustrated in Figure 31.20. Bysymmetry we see that the magnitude of E is constant on thispath and that E is tangent to it. The magnetic flux throughthe area enclosed by this path is BA " B,R 2; hence, Equa-tion 31.9 gives

(1)

The magnetic field inside a long solenoid is given byEquation 30.17, B " /0nI. When we substitute the expres-sion I " Imax cos *t into this equation for B and then substi-tute the result into Equation (1), we find that

(2) (for r ) R )E "/0nI max*R2

2r sin *t

" ,R 2 /0nImax * sin *t

E(2,r) " %,R 2 /0 nImax

ddt

(cos *t)

' E&ds " E(2,r) " %,R 2 dBdt

' E&ds " %ddt

(B,R 2) " %,R 2 dBdt

Hence, the amplitude of the electric field outside the sole-noid falls off as 1/r and varies sinusoidally with time.

(B) What is the magnitude of the induced electric fieldinside the solenoid, a distance r from its axis?

Solution For an interior point (r 0 R), the flux through anintegration loop is given by B,r 2. Using the same proce-dure as in part (A), we find that

(3) (for r 0 R )

This shows that the amplitude of the electric field inducedinside the solenoid by the changing magnetic flux throughthe solenoid increases linearly with r and varies sinusoidallywith time.

E "/0nI max*

2 r sin *t

E(2,r) " %,r 2 dB

dt " ,r 2 /0nImax * sin *t

Path ofintegration

R

r

Imax cos tω

Figure 31.20 (Example 31.8) A long solenoid carrying a time-varying current given by I " Imax cos *t. An electric field isinduced both inside and outside the solenoid.

In commercial power plants, the energy required to rotate the loop can be derivedfrom a variety of sources. For example, in a hydroelectric plant, falling water directedagainst the blades of a turbine produces the rotary motion; in a coal-fired plant, theenergy released by burning coal is used to convert water to steam, and this steam isdirected against the turbine blades. As a loop rotates in a magnetic field, the magneticflux through the area enclosed by the loop changes with time; this induces an emf anda current in the loop according to Faraday’s law. The ends of the loop are connectedto slip rings that rotate with the loop. Connections from these slip rings, which act asoutput terminals of the generator, to the external circuit are made by stationarybrushes in contact with the slip rings.

Suppose that, instead of a single turn, the loop has N turns (a more practicalsituation), all of the same area A, and rotates in a magnetic field with a constantangular speed *. If # is the angle between the magnetic field and the normal to theplane of the loop, as in Figure 31.22, then the magnetic flux through the loop atany time t is

!B " BA cos # " BA cos *t

where we have used the relationship # " *t between angular position and angularspeed (see Eq. 10.3). (We have set the clock so that t " 0 when # " 0.) Hence, theinduced emf in the coil is

(31.10)

This result shows that the emf varies sinusoidally with time, as plotted in Figure 31.21b.From Equation 31.10 we see that the maximum emf has the value

$max " NAB* (31.11)

which occurs when *t " 90° or 270°. In other words, $ " $max when the magneticfield is in the plane of the coil and the time rate of change of flux is a maximum. Fur-thermore, the emf is zero when *t " 0 or 180°, that is, when B is perpendicular to theplane of the coil and the time rate of change of flux is zero.

The frequency for commercial generators in the United States and Canada is 60 Hz,whereas in some European countries it is 50 Hz. (Recall that * " 2,f, where f is thefrequency in hertz.)

$ " %N d!B

dt " %NAB

ddt

(cos *t) " NAB* sin *t

S ECT I O N 31 . 5 • Generators and Motors 983

(a)

Slip rings N

Brushes

Externalcircuit

Loop

S

t

ε

(b)

εmax

Externalrotator

Active Figure 31.21 (a) Schematic diagram of an AC generator. An emf is induced ina loop that rotates in a magnetic field. (b) The alternating emf induced in the loopplotted as a function of time.

At the Active Figures linkat http://www.pse6.com, youcan adjust the speed ofrotation and the strength of thefield to see the effects on theemf generated.

Normal

θ

B

Figure 31.22 A loop enclosing anarea A and containing N turns,rotating with constant angularspeed * in a magnetic field. Theemf induced in the loop variessinusoidally in time.


The direct current (DC) generator is illustrated in Figure 31.23a. Such genera-tors are used, for instance, in older cars to charge the storage batteries. The compo-nents are essentially the same as those of the AC generator except that the contacts tothe rotating loop are made using a split ring called a commutator.

In this configuration, the output voltage always has the same polarity and pulsateswith time, as shown in Figure 31.23b. We can understand the reason for this by notingthat the contacts to the split ring reverse their roles every half cycle. At the same time,the polarity of the induced emf reverses; hence, the polarity of the split ring (which isthe same as the polarity of the output voltage) remains the same.

A pulsating DC current is not suitable for most applications. To obtain a moresteady DC current, commercial DC generators use many coils and commutators distrib-uted so that the sinusoidal pulses from the various coils are out of phase. When thesepulses are superimposed, the DC output is almost free of fluctuations.

Motors are devices into which energy is transferred by electrical transmissionwhile energy is transferred out by work. Essentially, a motor is a generator operating

Quick Quiz 31.10 In an AC generator, a coil with N turns of wire spins in amagnetic field. Of the following choices, which will not cause an increase in the emfgenerated in the coil? (a) replacing the coil wire with one of lower resistance (b) spinning the coil faster (c) increasing the magnetic field (d) increasing thenumber of turns of wire on the coil.

Example 31.9 emf Induced in a Generator

An AC generator consists of 8 turns of wire, each of areaA " 0.090 0 m 2 , and the total resistance of the wire is 12.0 (.The loop rotates in a 0.500-T magnetic field at a constantfrequency of 60.0 Hz.

(A) Find the maximum induced emf.

Solution First, note that * " 2,f " 2,(60.0 Hz) " 377 s%1.Thus, Equation 31.11 gives

$max " NAB* " 8(0.090 0 m2)(0.500 T)(377 s%1)

" 136 V

(B) What is the maximum induced current when the outputterminals are connected to a low-resistance conductor?

Solution From Equation 27.8 and the results to part (A),we have

11.3 AImax "$max

R"

136 V12.0 (

"

Commutator

(a)

Brush

N

S

t

ε

(b)

Armature

Active Figure 31.23 (a) Schematic diagram of a DC generator. (b) The magnitude ofthe emf varies in time but the polarity never changes.

At the Active Figures linkat http://www.pse6.com, youcan adjust the speed ofrotation and the strength of thefield to see the effects on theemf generated.

S ECT I O N 31 . 5 • Generators and Motors 985

in reverse. Instead of generating a current by rotating a coil, a current is supplied tothe coil by a battery, and the torque acting on the current-carrying coil causes it torotate.

Useful mechanical work can be done by attaching the rotating coil to some exter-nal device. However, as the coil rotates in a magnetic field, the changing magnetic fluxinduces an emf in the coil; this induced emf always acts to reduce the current in thecoil. If this were not the case, Lenz’s law would be violated. The back emf increases inmagnitude as the rotational speed of the coil increases. (The phrase back emf is used toindicate an emf that tends to reduce the supplied current.) Because the voltage avail-able to supply current equals the difference between the supply voltage and the backemf, the current in the rotating coil is limited by the back emf.

When a motor is turned on, there is initially no back emf; thus, the current is verylarge because it is limited only by the resistance of the coil. As the coil begins to rotate,the induced back emf opposes the applied voltage, and the current in the coil isreduced. If the mechanical load increases, the motor slows down; this causes the backemf to decrease. This reduction in the back emf increases the current in the coil andtherefore also increases the power needed from the external voltage source. For thisreason, the power requirements for starting a motor and for running it are greater forheavy loads than for light ones. If the motor is allowed to run under no mechanicalload, the back emf reduces the current to a value just large enough to overcomeenergy losses due to internal energy and friction. If a very heavy load jams the motor sothat it cannot rotate, the lack of a back emf can lead to dangerously high current inthe motor’s wire. This is a dangerous situation, and is explored in the What If? sectionof Example 31.10.

A current application of motors in automobiles is seen in the development ofhybrid drive systems. In these automobiles, a gasoline engine and an electric motor arecombined to increase the fuel economy of the vehicle and reduce its emissions. Figure31.24 shows the engine compartment of the Toyota Prius, which is one of a smallnumber of hybrids available in the United States. In this automobile, power to thewheels can come from either the gasoline engine or the electric motor. In normaldriving, the electric motor accelerates the vehicle from rest until it is moving at a speedof about 15 mi/h (24 km/h). During this acceleration period, the engine is not running, so that gasoline is not used and there is no emission. When a hybrid vehiclebrakes, the motor acts as a generator and returns some of the kinetic energy of the ve-hicle back to the battery as stored energy. In a normal vehicle, this kinetic energy issimply lost as it is transformed to internal energy in the brakes and roadway.

Figure 31.24 The engine compartment of the Toyota Prius, a hybrid vehicle.

Phot

o by

Bre

nt R

oman

s/w

ww.

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unds

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31.6 Eddy Currents

As we have seen, an emf and a current are induced in a circuit by a changingmagnetic flux. In the same manner, circulating currents called eddy currents areinduced in bulk pieces of metal moving through a magnetic field. This can easily bedemonstrated by allowing a flat copper or aluminum plate attached at the end of arigid bar to swing back and forth through a magnetic field (Fig. 31.25). As the plateenters the field, the changing magnetic flux induces an emf in the plate, which inturn causes the free electrons in the plate to move, producing the swirling eddycurrents. According to Lenz’s law, the direction of the eddy currents is such thatthey create magnetic fields that oppose the change that causes the currents. For thisreason, the eddy currents must produce effective magnetic poles on the plate,which are repelled by the poles of the magnet; this gives rise to a repulsive forcethat opposes the motion of the plate. (If the opposite were true, the plate wouldaccelerate and its energy would increase after each swing, in violation of the law ofconservation of energy.)

As indicated in Figure 31.26a, with B directed into the page, the induced eddycurrent is counterclockwise as the swinging plate enters the field at position 1. This isbecause the flux due to the external magnetic field into the page through the plate isincreasing, and hence by Lenz’s law the induced current must provide its ownmagnetic field out of the page. The opposite is true as the plate leaves the field atposition 2, where the current is clockwise. Because the induced eddy current alwaysproduces a magnetic retarding force FB when the plate enters or leaves the field, theswinging plate eventually comes to rest.

If slots are cut in the plate, as shown in Figure 31.26b, the eddy currents andthe corresponding retarding force are greatly reduced. We can understand this


Example 31.10 The Induced Current in a Motor

Assume that a motor in which the coil has a total resistanceof 10 ( is supplied by a voltage of 120 V. When the motor isrunning at its maximum speed, the back emf is 70 V. Findthe current in the coil

(A) when the motor is turned on and

(B) when it has reached maximum speed.

Solution (A) When the motor is first turned on, the backemf is zero (because the coil is motionless). Thus, thecurrent in the coil is a maximum and equal to

(B) At the maximum speed, the back emf has its maximumvalue. Thus, the effective supply voltage is that of the exter-nal source minus the back emf. Hence, the current isreduced to

What If? Suppose that this motor is in a circular saw. Youare operating the saw and the blade becomes jammed in a

5.0 AI "$ % $back

R"

120 V % 70 V10 (

"50 V10 (

"

12 AI "$R

"120 V10 (

"

piece of wood so that the motor cannot turn. By whatpercentage does the power input to the motor increase whenit is jammed?

Answer You may have everyday experiences with motorsbecoming warm when they are prevented from turning.This is due to the increased power input to the motor. Thehigher rate of energy transfer results in an increase in theinternal energy of the coil, an undesirable effect. When themotor is jammed, the current is that given in part (A). Letus set up the ratio of power input to the motor whenjammed to that when it is not jammed:

where the subscripts (A) and (B) refer to the currents inparts (A) and (B) of the example. Substituting these values,

This represents a 476% increase in the input power! Such ahigh power input can cause the coil to become so hot that itis damaged.

"jammed

"not jammed"

(12 A)2

(5.0 A)2 " 5.76

"jammed

"not jammed"

I (A)2 R

I (B)2 R

"I (A)

2

I (B)2

v

Pivot

S

N

Figure 31.25 Formation of eddycurrents in a conducting platemoving through a magnetic field.As the plate enters or leaves thefield, the changing magnetic fluxinduces an emf, which causes eddycurrents in the plate.

by realizing that the cuts in the plate prevent the formation of any large currentloops.

The braking systems on many subway and rapid-transit cars make use of electro-magnetic induction and eddy currents. An electromagnet attached to the train is posi-tioned near the steel rails. (An electromagnet is essentially a solenoid with an ironcore.) The braking action occurs when a large current is passed through the electro-magnet. The relative motion of the magnet and rails induces eddy currents in the rails,and the direction of these currents produces a drag force on the moving train. Becausethe eddy currents decrease steadily in magnitude as the train slows down, the brakingeffect is quite smooth. As a safety measure, some power tools use eddy currents to stoprapidly spinning blades once the device is turned off.

Eddy currents are often undesirable because they represent a transformation ofmechanical energy to internal energy. To reduce this energy loss, conducting partsare often laminated—that is, they are built up in thin layers separated by a noncon-ducting material such as lacquer or a metal oxide. This layered structure increasesthe resistance of eddy current paths and effectively confines the currents to individ-ual layers. Such a laminated structure is used in transformer cores (see Section33.8) and motors to minimize eddy currents and thereby increase the efficiency ofthese devices.

S ECT I O N 31 . 6 • Eddy Currents 987

Active Figure 31.26 (a) As the conducting plate enters the field (position 1), the eddycurrents are counterclockwise. As the plate leaves the field (position 2), the currentsare clockwise. In either case, the force on the plate is opposite the velocity, andeventually the plate comes to rest. (b) When slots are cut in the conducting plate, theeddy currents are reduced and the plate swings more freely through the magnetic field.

FB

Pivot

×

×

× ×

×

×

×

×

×× ×

1 2

Bin

v

v

FB

(a)

(b)

Choose to let a solid or aslotted plate swing through themagnetic field and observe theeffect at the Active Figures linkat http://www.pse6.com.

Quick Quiz 31.11 In equal-arm balances from the early twentieth century(Fig. 31.27), it is sometimes observed that an aluminum sheet hangs from one of thearms and passes between the poles of a magnet. This causes the oscillations of theequal-arm balance to decay rapidly. In the absence of such magnetic braking, the oscil-lation might continue for a very long time, so that the experimenter would have to waitto take a reading. The oscillations decay because (a) the aluminum sheet is attracted tothe magnet; (b) currents in the aluminum sheet set up a magnetic field that opposesthe oscillations; (c) aluminum is paramagnetic.

31.7 Maxwell’s Equations

We conclude this chapter by presenting four equations that are regarded as the basis ofall electrical and magnetic phenomena. These equations, developed by James ClerkMaxwell, are as fundamental to electromagnetic phenomena as Newton’s laws are tomechanical phenomena. In fact, the theory that Maxwell developed was more far-reaching than even he imagined because it turned out to be in agreement with thespecial theory of relativity, as Einstein showed in 1905.

Maxwell’s equations represent the laws of electricity and magnetism that we havealready discussed, but they have additional important consequences. In Chapter 34 weshall show that these equations predict the existence of electromagnetic waves(traveling patterns of electric and magnetic fields), which travel with a speed

the speed of light. Furthermore, the theory shows thatsuch waves are radiated by accelerating charges.

For simplicity, we present Maxwell’s equations as applied to free space, that is, inthe absence of any dielectric or magnetic material. The four equations are

(31.12)'S

E&dA "q10

c " 1/√/010 " 3.00 + 108 m/s,


Figure 31.27 (Quick Quiz 31.11) In an old-fashioned equal-arm balance, analuminum sheet hangs between the poles of a magnet.

John

W. J

ewet

t, Jr

.

Gauss’s law

(31.13)

(31.14)

(31.15)

Equation 31.12 is Gauss’s law: the total electric flux through any closed surfaceequals the net charge inside that surface divided by #0. This law relates an electricfield to the charge distribution that creates it.

Equation 31.13, which can be considered Gauss’s law in magnetism, states that thenet magnetic flux through a closed surface is zero. That is, the number ofmagnetic field lines that enter a closed volume must equal the number that leave thatvolume. This implies that magnetic field lines cannot begin or end at any point. If theydid, it would mean that isolated magnetic monopoles existed at those points. The factthat isolated magnetic monopoles have not been observed in nature can be taken as aconfirmation of Equation 31.13.

Equation 31.14 is Faraday’s law of induction, which describes the creation of anelectric field by a changing magnetic flux. This law states that the emf, which is theline integral of the electric field around any closed path, equals the rate ofchange of magnetic flux through any surface area bounded by that path. Oneconsequence of Faraday’s law is the current induced in a conducting loop placed in atime-varying magnetic field.

Equation 31.15, usually called the Ampère–Maxwell law, is the generalized form ofAmpère’s law, and describes the creation of a magnetic field by an electric field andelectric currents: the line integral of the magnetic field around any closed path isthe sum of $0 times the net current through that path and #0$0 times the rate ofchange of electric flux through any surface bounded by that path.

Once the electric and magnetic fields are known at some point in space, the forceacting on a particle of charge q can be calculated from the expression

F " qE 2 qv " B (31.16)

This relationship is called the Lorentz force law. (We saw this relationship earlier asEquation 29.16.) Maxwell’s equations, together with this force law, completely describeall classical electromagnetic interactions.

It is interesting to note the symmetry of Maxwell’s equations. Equations 31.12 and31.13 are symmetric, apart from the absence of the term for magnetic monopoles inEquation 31.13. Furthermore, Equations 31.14 and 31.15 are symmetric in that the lineintegrals of E and B around a closed path are related to the rate of change of magneticflux and electric flux, respectively. Maxwell’s equations are of fundamental importancenot only to electromagnetism but to all of science. Heinrich Hertz once wrote, “Onecannot escape the feeling that these mathematical formulas have an independent exis-tence and an intelligence of their own, that they are wiser than we are, wiser even thantheir discoverers, that we get more out of them than we put into them.”

' B &ds " /0I 2 10 /0 d!E

dt

' E&ds " %d!B

dt

'S

B &dA " 0

Summary 989

Faraday’s law of induction states that the emf induced in a circuit is directlyproportional to the time rate of change of magnetic flux through the circuit:

(31.1)

where !B " &B & d A is the magnetic flux.

$ " %d!B

dt

S U M M A RY

Take a practice test forthis chapter by clicking on thePractice Test link athttp://www.pse6.com.

Faraday’s law

Ampère–Maxwell law

The Lorentz force law

Gauss’s law in magnetism

When a conducting bar of length ! moves at a velocity v through a magneticfield B , where B is perpendicular to the bar and to v, the motional emf induced inthe bar is

$ " %B!v (31.5)

Lenz’s law states that the induced current and induced emf in a conductor are insuch a direction as to set up a magnetic field that opposes the change that producedthem.

A general form of Faraday’s law of induction is

(31.9)

where E is the nonconservative electric field that is produced by the changingmagnetic flux.

When used with the Lorentz force law, F " qE 2 q v " B, Maxwell’s equationsdescribe all electromagnetic phenomena:

(31.12)

(31.13)

(31.14)

(31.15)' B &ds " /0I 2 10/0 d!E

dt

' E&ds " %d!B

dt

'S

B &d A " 0

'S

E&d A "q10

$ " ' E&ds " %d!B

dt


1. What is the difference between magnetic flux and magneticfield?

2. A loop of wire is placed in a uniform magnetic field. Forwhat orientation of the loop is the magnetic flux a maxi-mum? For what orientation is the flux zero?

3. As the bar in Figure Q31.3 moves to the right, an electricfield is set up directed downward in the bar. Explain why

E

Bin

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−−−

v

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v

Bout

the electric field would be upward if the bar were movingto the left.

4. As the bar in Figure Q31.3 moves perpendicular to thefield, is an external force required to keep it moving withconstant speed?

The bar in Figure Q31.5 moves on rails to the right with avelocity v, and the uniform, constant magnetic field isdirected out of the page. Why is the induced currentclockwise? If the bar were moving to the left, what wouldbe the direction of the induced current?

5.

Q U E S T I O N S

Figure Q31.5 Questions 5 and 6.Figure Q31.3 Questions 3 and 4.

R

S

N

v

6. Explain why an applied force is necessary to keep the barin Figure Q31.5 moving with a constant speed.

7. Wearing a metal bracelet in a region of strong magneticfield could be hazardous. Explain.

8. When a small magnet is moved toward a solenoid, an emf isinduced in the coil. However, if the magnet is movedaround inside a toroid, no measurable emf is induced.Explain.

How is energy produced in dams that is then transferredout by electrical transmission? (That is, how is the energyof motion of the water converted to energy that is trans-mitted by AC electricity?)

10. Will dropping a magnet down a long copper tube producea current in the walls of the tube? Explain.

11. A piece of aluminum is dropped vertically downwardbetween the poles of an electromagnet. Does the magneticfield affect the velocity of the aluminum?

12. What happens when the rotational speed of a generatorcoil is increased?

13. When the switch in Figure Q31.13a is closed, a current isset up in the coil and the metal ring springs upward (Fig.Q31.13b). Explain this behavior.

9.

14. Assume that the battery in Figure Q31.13a is replacedby an AC source and the switch is held closed. If helddown, the metal ring on top of the solenoid becomeshot. Why?

15. A bar magnet is held above a loop of wire in a horizontalplane, as shown in Figure Q31.15. The south end of themagnet is toward the loop of wire. The magnet is droppedtoward the loop. Find the direction of the current throughthe resistor (a) while the magnet is falling toward the loopand (b) after the magnet has passed through the loop andmoves away from it.

16. Find the direction of the current in the resistor in FigureQ31.16 (a) at the instant the switch is closed, (b) after theswitch has been closed for several minutes, and (c) at theinstant the switch is opened.

17. Quick Quiz 31.4 describes the emf induced between thewingtips of an airplane by its motion in the Earth’smagnetic field. Can this emf be used to power a light inthe passenger compartment? Explain your answer.

18. Do Maxwell’s equations allow for the existence of magneticmonopoles? Explain.

19. Induction welding has many important industrial applica-tions. One example is the manufacture of airtight tubes,represented in Figure Q31.19. A sheet of metal is rolledinto a cylinder and forced between compression rollers tobring its edges into contact. The tube then enters a coilcarrying a time-varying current. The seam is welded wheninduced currents around the tube raise its temperature.Typically, a sinusoidal current with a frequency of 10 kHzis used. (a) What causes a current in the tube? (b) Why is ahigh frequency like 10 kHz chosen, rather than the 120 Hzcommonly used for power transmission? (c) Why do theinduced currents raise the temperature mainly of theseam, rather than all of the metal of the tube? (d) Why is itnecessary to bring the edges of the sheet together with thecompression rollers before the seam can be welded?

Questions 991

(a)

Iron coreMetal ring

S

(b)

R

S

ε

Figure Q31.15

Figure Q31.16

Figure Q31.19

Figure Q31.13 Questions 13 and 14.

Cour

tesy

of C

entra

l Scie

ntifi

c Com

pany


produced by the solenoid is half as strong as at the centerof the solenoid. Assume the solenoid produces negligiblefield outside its cross-sectional area. The current in thesolenoid is increasing at a rate of 270 A/s. (a) What isthe induced current in the ring? At the center of the ring,what are (b) the magnitude and (c) the direction of themagnetic field produced by the induced current in thering?

8. An aluminum ring of radius r1 and resistance R is placedaround the top of a long air-core solenoid with n turns permeter and smaller radius r 2 as shown in Figure P31.7.Assume that the axial component of the field produced bythe solenoid over the area of the end of the solenoid ishalf as strong as at the center of the solenoid. Assume thatthe solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing ata rate of 'I/'t. (a) What is the induced current in thering? (b) At the center of the ring, what is the magneticfield produced by the induced current in the ring?(c) What is the direction of this field?

9. (a) A loop of wire in the shape of a rectangle of width wand length L and a long, straight wire carrying a current Ilie on a tabletop as shown in Figure P31.9. (a) Determinethe magnetic flux through the loop due to the current I.(b) Suppose the current is changing with time accordingto I " a 2 bt, where a and b are constants. Determine theemf that is induced in the loop if b " 10.0 A/s, h "1.00 cm, w " 10.0 cm, and L " 100 cm. What is the direc-tion of the induced current in the rectangle?

Section 31.1 Faraday’s Law of InductionSection 31.3 Lenz’s Law1. A 50-turn rectangular coil of dimensions 5.00 cm +

10.0 cm is allowed to fall from a position where B " 0 to anew position where B " 0.500 T and the magnetic field isdirected perpendicular to the plane of the coil. Calculatethe magnitude of the average emf that is induced in thecoil if the displacement occurs in 0.250 s.

2. A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magneticfield that increases uniformly in magnitude from 0.500 Tto 2.50 T in 1.00 s. What is the resulting induced current ifthe loop has a resistance of 2.00 (?

3. A 25-turn circular coil of wire has diameter 1.00 m. It isplaced with its axis along the direction of the Earth’smagnetic field of 50.0 /T, and then in 0.200 s it is flipped180°. An average emf of what magnitude is generated inthe coil?

4. A rectangular loop of area A is placed in a region wherethe magnetic field is perpendicular to the plane of theloop. The magnitude of the field is allowed to vary intime according to B " Bmaxe%t/-, where Bmax and - areconstants. The field has the constant value Bmax for t 0 0.(a) Use Faraday’s law to show that the emf induced in theloop is given by

(b) Obtain a numerical value for $ at t " 4.00 s whenA " 0.160 m2, Bmax " 0.350 T, and - " 2.00 s. (c) For thevalues of A, Bmax, and - given in (b), what is the maximumvalue of $?

A strong electromagnet produces a uniform mag-netic field of 1.60 T over a cross-sectional area of0.200 m2. We place a coil having 200 turns and a totalresistance of 20.0 ( around the electromagnet. We thensmoothly reduce the current in the electromagnet until itreaches zero in 20.0 ms. What is the current induced inthe coil?

6. A magnetic field of 0.200 T exists within a solenoid of500 turns and a diameter of 10.0 cm. How rapidly (that is,within what period of time) must the field be reduced tozero, if the average induced emf within the coil during thistime interval is to be 10.0 kV?

An aluminum ring of radius 5.00 cm and resistance3.00 + 10%4 ( is placed on top of a long air-core sole-noid with 1 000 turns per meter and radius 3.00 cm, asshown in Figure P31.7. Over the area of the end of thesolenoid, assume that the axial component of the field

7.

5.

$ "AB max

- e%t/-

1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide

= coached solution with hints available at http://www.pse.com = computer useful in solving problem

= paired numerical and symbolic problems

P R O B L E M S

5.00 cm

3.00 cm

I

I

Figure P31.7 Problems 7 and 8.

Problems 993

10. A coil of 15 turns and radius 10.0 cm surrounds a longsolenoid of radius 2.00 cm and 1.00 + 103 turns/meter(Fig. P31.10). The current in the solenoid changes asI " (5.00 A) sin(120t). Find the induced emf in the 15-turn coil as a function of time.

11. Find the current through section PQ of length a " 65.0 cmin Figure P31.11. The circuit is located in a magnetic fieldwhose magnitude varies with time according to the expres-sion B " (1.00 + 10%3 T/s)t. Assume the resistance perlength of the wire is 0.100 (/m.

12. A 30-turn circular coil of radius 4.00 cm and resistance1.00 ( is placed in a magnetic field directed perpen-dicular to the plane of the coil. The magnitude of themagnetic field varies in time according to the expressionB " 0.010 0t 2 0.040 0t 2, where t is in seconds andB is in tesla. Calculate the induced emf in the coil att " 5.00 s.

A long solenoid has n " 400 turns per meter and carries acurrent given by I " (30.0 A)(1 % e%1.60t ). Inside the sole-noid and coaxial with it is a coil that has a radius of6.00 cm and consists of a total of N " 250 turns of finewire (Fig. P31.13). What emf is induced in the coil by thechanging current?

13.

14. An instrument based on induced emf has been used tomeasure projectile speeds up to 6 km/s. A small magnet isimbedded in the projectile, as shown in Figure P31.14.The projectile passes through two coils separated by adistance d. As the projectile passes through each coil apulse of emf is induced in the coil. The time intervalbetween pulses can be measured accurately with an oscillo-scope, and thus the speed can be determined. (a) Sketch agraph of 'V versus t for the arrangement shown. Considera current that flows counterclockwise as viewed from thestarting point of the projectile as positive. On your graph,indicate which pulse is from coil 1 and which is from coil2. (b) If the pulse separation is 2.40 ms and d " 1.50 m,what is the projectile speed?

A coil formed by wrapping 50 turns of wire in the shape ofa square is positioned in a magnetic field so that thenormal to the plane of the coil makes an angle of 30.0°with the direction of the field. When the magnetic field isincreased uniformly from 200 /T to 600 /T in 0.400 s, anemf of magnitude 80.0 mV is induced in the coil. What isthe total length of the wire?

16. When a wire carries an AC current with a known frequency,you can use a Rogowski coil to determine the amplitude I maxof the current without disconnecting the wire to shunt the

15.

I

w

h

L


I

15-turn coil

R

Figure P31.10

a

aQ2a

P

×

×

×

×

×

×

×

×

B

×

×

×

×

×

×

×

×

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×

×

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×

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×

×

×

×

×

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×

×

×

×

×

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×

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×

×

×

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×

×

×

×

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×

Figure P31.11

N turns

n turns/m I

× × × × × × × × × × × × × × × × × × × × × ×

R

Figure P31.13

V1 V2

d

vS N

Figure P31.14

ε I(t )

Figure P31.16


current in a meter. The Rogowski coil, shown in FigureP31.16, simply clips around the wire. It consists of atoroidal conductor wrapped around a circular return cord.The toroid has n turns per unit length and a cross-sectionalarea A. The current to be measured is given by I(t) "I max sin *t. (a) Show that the amplitude of the emfinduced in the Rogowski coil is $max " /0nA* Imax. (b) Explain why the wire carrying the unknown currentneed not be at the center of the Rogowski coil, and why thecoil will not respond to nearby currents that it does notenclose.

17. A toroid having a rectangular cross section (a " 2.00 cmby b " 3.00 cm) and inner radius R " 4.00 cm consistsof 500 turns of wire that carries a sinusoidal currentI " I max sin *t, with I max " 50.0 A and a frequencyf " */2, " 60.0 Hz. A coil that consists of 20 turns ofwire links with the toroid, as in Figure P31.17. Determinethe emf induced in the coil as a function of time.

18. A piece of insulated wire is shaped into a figure 8, as inFigure P31.18. The radius of the upper circle is 5.00 cmand that of the lower circle is 9.00 cm. The wire has a uni-form resistance per unit length of 3.00 (/m. A uniformmagnetic field is applied perpendicular to the plane of thetwo circles, in the direction shown. The magnetic field isincreasing at a constant rate of 2.00 T/s. Find the magni-tude and direction of the induced current in the wire.

Section 31.2 Motional emf

Section 31.3 Lenz’s Law

19. An automobile has a vertical radio antenna 1.20 m long.The automobile travels at 65.0 km/h on a horizontal road

where the Earth’s magnetic field is 50.0 /T directedtoward the north and downward at an angle of 65.0° belowthe horizontal. (a) Specify the direction that the automo-bile should move in order to generate the maximummotional emf in the antenna, with the top of the antennapositive relative to the bottom. (b) Calculate the magni-tude of this induced emf.

20. Consider the arrangement shown in Figure P31.20. Assumethat R " 6.00 (, ! " 1.20 m, and a uniform 2.50-T mag-netic field is directed into the page. At what speed shouldthe bar be moved to produce a current of 0.500 A in theresistor?

Figure P31.20 shows a top view of a bar that can slidewithout friction. The resistor is 6.00 ( and a 2.50-T mag-netic field is directed perpendicularly downward, intothe paper. Let ! " 1.20 m. (a) Calculate the appliedforce required to move the bar to the right at a constantspeed of 2.00 m/s. (b) At what rate is energy deliveredto the resistor?

22. A conducting rod of length ! moves on two horizontal,frictionless rails, as shown in Figure P31.20. If a constantforce of 1.00 N moves the bar at 2.00 m/s through a mag-netic field B that is directed into the page, (a) what is thecurrent through the 8.00-( resistor R ? (b) What is the rateat which energy is delivered to the resistor? (c) What is themechanical power delivered by the force Fapp?

23. Very large magnetic fields can be produced using a proce-dure called flux compression. A metallic cylindrical tube ofradius R is placed coaxially in a long solenoid of somewhatlarger radius. The space between the tube and the sole-noid is filled with a highly explosive material. When theexplosive is set off, it collapses the tube to a cylinder ofradius r 0 R . If the collapse happens very rapidly, inducedcurrent in the tube maintains the magnetic flux nearlyconstant inside the tube. If the initial magnetic field in thesolenoid is 2.50 T, and R/r " 12.0, what maximum valueof magnetic field can be achieved?

24. The homopolar generator, also called the Faraday disk, is alow-voltage, high-current electric generator. It consists ofa rotating conducting disk with one stationary brush (asliding electrical contact) at its axle and another at apoint on its circumference, as shown in Figure P31.24.A magnetic field is applied perpendicular to the plane ofthe disk. Assume the field is 0.900 T, the angular speed is3 200 rev/min, and the radius of the disk is 0.400 m. Findthe emf generated between the brushes. When supercon-ducting coils are used to produce a large magnetic field, ahomopolar generator can have a power output of several

21.

N ′ = 20a

b

R

N = 500

Figure P31.17

× ×× × × ×

× ×× × × ×

× ×× × × ×

× ×× × × ×

× ×× × × ×

× ×× × × ×

× ×× × × ×

× ×× × × ×

Figure P31.18

Problem 71 in Chapter 29 can be assigned with this section.

Fapp!R

Figure P31.20 Problems 20, 21, and 22.

Problems 995

megawatts. Such a generator is useful, for example, inpurifying metals by electrolysis. If a voltage is applied tothe output terminals of the generator, it runs in reverse asa homopolar motor capable of providing great torque, usefulin ship propulsion.

25. Review problem. A flexible metallic wire with linear den-sity 3.00 + 10%3 kg/m is stretched between two fixedclamps 64.0 cm apart and held under tension 267 N. Amagnet is placed near the wire as shown in Figure P31.25.Assume that the magnet produces a uniform field of4.50 mT over a 2.00-cm length at the center of the wire,and a negligible field elsewhere. The wire is set vibratingat its fundamental (lowest) frequency. The section of thewire in the magnetic field moves with a uniform amplitudeof 1.50 cm. Find (a) the frequency and (b) the amplitudeof the electromotive force induced between the ends ofthe wire.

26. The square loop in Figure P31.26 is made of wires withtotal series resistance 10.0 (. It is placed in a uniform0.100-T magnetic field directed perpendicularly into theplane of the paper. The loop, which is hinged at eachcorner, is pulled as shown until the separation betweenpoints A and B is 3.00 m. If this process takes 0.100 s,what is the average current generated in the loop? Whatis the direction of the current?

A helicopter (Figure P31.27) has blades of length 3.00 m,extending out from a central hub and rotating at2.00 rev/s. If the vertical component of the Earth’s mag-netic field is 50.0 /T, what is the emf induced between theblade tip and the center hub?

27.

28. Use Lenz’s law to answer the following questions concern-ing the direction of induced currents. (a) What is thedirection of the induced current in resistor R in FigureP31.28a when the bar magnet is moved to the left?(b) What is the direction of the current induced in theresistor R immediately after the switch S in Figure P31.28b

B

Figure P31.24

V

N

Figure P31.25

3.00 m

3.00 m

3.00 m

3.00 m

A

BFigure P31.26

Figure P31.27

Ross

Har

rison

Kot

y/Ge

tty Im

ages

v

R

S N

R

S

(a) (b)

I

(c)

R

(d)

v

++

––

ε

Figure P31.28


is closed? (c) What is the direction of the induced currentin R when the current I in Figure P31.28c decreasesrapidly to zero? (d) A copper bar is moved to the rightwhile its axis is maintained in a direction perpendicular toa magnetic field, as shown in Figure P31.28d. If the top ofthe bar becomes positive relative to the bottom, what is thedirection of the magnetic field?

A rectangular coil with resistance R has N turns, each oflength ! and width w as shown in Figure P31.29. The coilmoves into a uniform magnetic field B with constant velocityv. What are the magnitude and direction of the totalmagnetic force on the coil (a) as it enters the magnetic field,(b) as it moves within the field, and (c) as it leaves the field?

29.

31. Two parallel rails with negligible resistance are 10.0 cm apartand are connected by a 5.00-( resistor. The circuit alsocontains two metal rods having resistances of 10.0 ( and15.0 ( sliding along the rails (Fig. P31.31). The rods arepulled away from the resistor at constant speeds of 4.00 m/sand 2.00 m/s, respectively. A uniform magnetic field ofmagnitude 0.010 0 T is applied perpendicular to the planeof the rails. Determine the current in the 5.00 -( resistor.

Section 31.4 Induced emf and Electric Fields32. For the situation shown in Figure P31.32, the magnetic

field changes with time according to the expressionB " (2.00t 3 % 4.00t 2 2 0.800)T, and r2 " 2R " 5.00 cm.(a) Calculate the magnitude and direction of the forceexerted on an electron located at point P2 when t " 2.00 s.(b) At what time is this force equal to zero?

A magnetic field directed into the page changes with timeaccording to B " (0.030 0t 2 2 1.40)T, where t is inseconds. The field has a circular cross section of radiusR " 2.50 cm (Fig. P31.32). What are the magnitude anddirection of the electric field at point P1 when t " 3.00 sand r1 " 0.020 0 m ?

34. A long solenoid with 1 000 turns per meter andradius 2.00 cm carries an oscillating current given byI " (5.00 A) sin(100,t). What is the electric field inducedat a radius r " 1.00 cm from the axis of the solenoid?What is the direction of this electric field when the currentis increasing counterclockwise in the coil?

Section 31.5 Generators and Motors

33.

A coil of area 0.100 m2 is rotating at 60.0 rev/s withthe axis of rotation perpendicular to a 0.200-T magneticfield. (a) If the coil has 1 000 turns, what is the maximumemf generated in it? (b) What is the orientation of the coilwith respect to the magnetic field when the maximuminduced voltage occurs?

36. In a 250-turn automobile alternator, the magnetic flux ineach turn is !B " (2.50 + 10%4 Wb) cos(*t), where * isthe angular speed of the alternator. The alternator isgeared to rotate three times for each engine revolution.When the engine is running at an angular speed of1 000 rev/min, determine (a) the induced emf in thealternator as a function of time and (b) the maximum emfin the alternator.

37. A long solenoid, with its axis along the x axis, consists of200 turns per meter of wire that carries a steady current of15.0 A. A coil is formed by wrapping 30 turns of thin wirearound a circular frame that has a radius of 8.00 cm. Thecoil is placed inside the solenoid and mounted on anaxis that is a diameter of the coil and coincides with they axis. The coil is then rotated with an angular speedof 4.00, rad/s. (The plane of the coil is in the yz plane

35.

w

v

Bin

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×!

Figure P31.29

a

R

b

Motion towardthe loop

S

N

Figure P31.30

××××××

××××××

××××××

××××××

××××××

××××××

××××××

××××××

××××××

××××××

××××××

10.0 Ω 15.0 Ω

5.00 Ω 2.00 m/sB4.00 m/s

Figure P31.31

Bin

× × × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × ×

× × × × ×

× × × × × ×

× × × × ×

r2P2

P1

R

r1


30. In Figure P31.30, the bar magnet is moved toward theloop. Is Va % Vb positive, negative, or zero? Explain.

Problems 28 and 62 in Chapter 29 can be assigned withthis section.

Problems 997

at t " 0.) Determine the emf generated in the coil as afunction of time.

38. A bar magnet is spun at constant angular speed * aroundan axis as shown in Figure P31.38. A stationary flat rectan-gular conducting loop surrounds the magnet, and att " 0, the magnet is oriented as shown. Make a qualitativegraph of the induced current in the loop as a function oftime, plotting counterclockwise currents as positive andclockwise currents as negative.

39. A motor in normal operation carries a direct current of0.850 A when connected to a 120-V power supply. Theresistance of the motor windings is 11.8 (. While innormal operation, (a) what is the back emf generated bythe motor? (b) At what rate is internal energy produced inthe windings? (c) What If? Suppose that a malfunctionstops the motor shaft from rotating. At what rate will inter-nal energy be produced in the windings in this case? (Mostmotors have a thermal switch that will turn off the motorto prevent overheating when this occurs.)

40. A semicircular conductor of radius R " 0.250 m is rotatedabout the axis AC at a constant rate of 120 rev/min (Fig.P31.40). A uniform magnetic field in all of the lower halfof the figure is directed out of the plane of rotation andhas a magnitude of 1.30 T. (a) Calculate the maximumvalue of the emf induced in the conductor. (b) What is thevalue of the average induced emf for each complete rota-tion? (c) What If? How would the answers to (a) and (b)change if B were allowed to extend a distance R above theaxis of rotation? Sketch the emf versus time (d) when thefield is as drawn in Figure P31.40 and (e) when the field isextended as described in (c).

41. The rotating loop in an AC generator is a square 10.0 cmon a side. It is rotated at 60.0 Hz in a uniform field of0.800 T. Calculate (a) the flux through the loop as afunction of time, (b) the emf induced in the loop, (c) the

current induced in the loop for a loop resistance of1.00 (, (d) the power delivered to the loop, and (e) thetorque that must be exerted to rotate the loop.

Section 31.6 Eddy Currents42. Figure P31.42 represents an electromagnetic brake that

uses eddy currents. An electromagnet hangs from arailroad car near one rail. To stop the car, a large currentis sent through the coils of the electromagnet. The movingelectromagnet induces eddy currents in the rails, whosefields oppose the change in the field of the electromagnet.The magnetic fields of the eddy currents exert force onthe current in the electromagnet, thereby slowing the car.The direction of the car’s motion and the direction of thecurrent in the electromagnet are shown correctly in thepicture. Determine which of the eddy currents shown onthe rails is correct. Explain your answer.

A conducting rectangular loop of mass M, resistanceR, and dimensions w by ! falls from rest into a magneticfield B as shown in Figure P31.43. During the time interval

43.

S

N

ω

Figure P31.38

A R

Bout

C

Figure P31.40

I

N

Sv

N

S

Figure P31.42

v

!

Bout

w

Figure P31.43

before the top edge of the loop reaches the field, the loopapproaches a terminal speed vT. (a) Show that

(b) Why is vT proportional to R? (c) Why is it inversely propor-tional to B2?

Section 31.7 Maxwell’s Equations44. An electron moves through a uniform electric field

E " (2.50 i 2 5.00 j) V/m and a uniform magnetic fieldB " (0.400k)T. Determine the acceleration of theelectron when it has a velocity v " 10.0 i m/s.

A proton moves through a uniform electric field given byE " 50.0 j V/m and a uniform magnetic field B "(0.200 i 2 0.300 j 2 0.400k)T. Determine the accelerationof the proton when it has a velocity v " 200 i m/s.

Additional Problems46. A steel guitar string vibrates (Figure 31.6). The compo-

nent of magnetic field perpendicular to the area of apickup coil nearby is given by

B " 50.0 mT 2 (3.20 mT) sin(2, 523 t/s)

The circular pickup coil has 30 turns and radius 2.70 mm.Find the emf induced in the coil as a function of time.

47. Figure P31.47 is a graph of the induced emf versus timefor a coil of N turns rotating with angular speed * in a uni-form magnetic field directed perpendicular to the axis ofrotation of the coil. What If ? Copy this sketch (on a largerscale), and on the same set of axes show the graph of emfversus t (a) if the number of turns in the coil is doubled;(b) if instead the angular speed is doubled; and (c) if theangular speed is doubled while the number of turns in thecoil is halved.

45.

vT "MgRB2w2

48. A technician wearing a brass bracelet enclosing area0.005 00 m2 places her hand in a solenoid whose magneticfield is 5.00 T directed perpendicular to the plane of thebracelet. The electrical resistance around the circumfer-ence of the bracelet is 0.020 0 (. An unexpected powerfailure causes the field to drop to 1.50 T in a time of20.0 ms. Find (a) the current induced in the bracelet and(b) the power delivered to the bracelet. Note: As this

problem implies, you should not wear any metal objectswhen working in regions of strong magnetic fields.

49. Two infinitely long solenoids (seen in cross section) passthrough a circuit as shown in Figure P31.49. The magni-tude of B inside each is the same and is increasing at therate of 100 T/s. What is the current in each resistor?

50. A conducting rod of length ! " 35.0 cm is free to slide ontwo parallel conducting bars as shown in Figure P31.50.Two resistors R1 " 2.00 ( and R 2 " 5.00 ( are connectedacross the ends of the bars to form a loop. A constantmagnetic field B " 2.50 T is directed perpendicularly intothe page. An external agent pulls the rod to the left with aconstant speed of v " 8.00 m/s. Find (a) the currents inboth resistors, (b) the total power delivered to the resis-tance of the circuit, and (c) the magnitude of the appliedforce that is needed to move the rod with this constantvelocity.

51. Suppose you wrap wire onto the core from a roll of cello-phane tape to make a coil. Describe how you can use a barmagnet to produce an induced voltage in the coil. What isthe order of magnitude of the emf you generate? State thequantities you take as data and their values.

52. A bar of mass m, length d, and resistance R slides withoutfriction in a horizontal plane, moving on parallel rails asshown in Figure P31.52. A battery that maintains aconstant emf $ is connected between the rails, and aconstant magnetic field B is directed perpendicularly tothe plane of the page. Assuming the bar starts from rest,show that at time t it moves with a speed

v "$Bd

(1 % e%B 2d 2t/mR)


(mV)

1 2t(ms)

ε

10

5

–5

–10

3

Figure P31.47

0.500 m 0.500 m

0.500 m6.00 Ω5.00 Ω

r2 = 0.150 m

BoutBin

r1 = 0.100 m

3.00 Ω

× ××××

××

Figure P31.49

××××××

××××××

××××××

××××××

××××××

××××××

5.00 Ω

B××××××

××××××

2.00 Ω v

Figure P31.50

d B (out of page) ε

Figure P31.52

53. Review problem. A particle with a mass of 2.00 + 10%16 kgand a charge of 30.0 nC starts from rest, is accelerated by astrong electric field, and is fired from a small source insidea region of uniform constant magnetic field 0.600 T. Thevelocity of the particle is perpendicular to the field. Thecircular orbit of the particle encloses a magnetic flux of15.0 /Wb. (a) Calculate the speed of the particle. (b) Cal-culate the potential difference through which the particleaccelerated inside the source.

54. An induction furnace uses electromagnetic induction toproduce eddy currents in a conductor, thereby raising theconductor’s temperature. Commercial units operate at fre-quencies ranging from 60 Hz to about 1 MHz and deliverpowers from a few watts to several megawatts. Inductionheating can be used for welding in a vacuum enclosure, toavoid oxidation and contamination of the metal. At highfrequencies, induced currents occur only near the surfaceof the conductor—this is the “skin effect.” By creating aninduced current for a short time at an appropriately highfrequency, one can heat a sample down to a controlleddepth. For example, the surface of a farm tiller can betempered to make it hard and brittle for effective cuttingwhile keeping the interior metal soft and ductile to resistbreakage.

To explore induction heating, consider a flat conduct-ing disk of radius R , thickness b, and resistivity 3. A sinu-soidal magnetic field Bmax cos *t is applied perpendicularto the disk. Assume that the frequency is so low that theskin effect is not important. Assume the eddy currentsoccur in circles concentric with the disk. (a) Calculate theaverage power delivered to the disk. (b) What If? By whatfactor does the power change when the amplitude of thefield doubles? (c) When the frequency doubles? (d) Whenthe radius of the disk doubles?

The plane of a square loop of wire with edge lengtha " 0.200 m is perpendicular to the Earth’s magnetic fieldat a point where B " 15.0 /T, as shown in Figure P31.55.The total resistance of the loop and the wires connecting itto a sensitive ammeter is 0.500 (. If the loop is suddenlycollapsed by horizontal forces as shown, what total chargepasses through the ammeter?

55.

56. Magnetic field values are often determined by using adevice known as a search coil. This technique depends onthe measurement of the total charge passing through acoil in a time interval during which the magnetic flux

linking the windings changes either because of the motionof the coil or because of a change in the value of B.(a) Show that as the flux through the coil changes from!1 to !2, the charge transferred through the coil will begiven by Q " N(!2 % !1)/R, where R is the resistance ofthe coil and a sensitive ammeter connected across it and Nis the number of turns. (b) As a specific example, calculateB when a 100-turn coil of resistance 200 ( and cross-sectional area 40.0 cm2 produces the following results.A total charge of 5.00 + 10%4 C passes through the coilwhen it is rotated in a uniform field from a position wherethe plane of the coil is perpendicular to the field to a posi-tion where the coil’s plane is parallel to the field.

57. In Figure P31.57, the rolling axle, 1.50 m long, is pushedalong horizontal rails at a constant speed v " 3.00 m/s. Aresistor R " 0.400 ( is connected to the rails at points a andb, directly opposite each other. (The wheels make good elec-trical contact with the rails, and so the axle, rails, and R forma closed-loop circuit. The only significant resistance in thecircuit is R.) A uniform magnetic field B " 0.080 0 T is verti-cally downward. (a) Find the induced current I in the resis-tor. (b) What horizontal force F is required to keep the axlerolling at constant speed? (c) Which end of the resistor, a orb, is at the higher electric potential? (d) What If ? After theaxle rolls past the resistor, does the current in R reversedirection? Explain your answer.

58. A conducting rod moves with a constant velocity v in adirection perpendicular to a long, straight wire carrying acurrent I as shown in Figure P31.58. Show that the magni-tude of the emf generated between the ends of the rod is

In this case, note that the emf decreases with increasing r,as you might expect.

&$&"/0vI!

2,r

Problems 999

A

aF F

a

Figure P31.55

B

v

R

a

b

Figure P31.57

r

I

!v

Figure P31.58

59. A circular loop of wire of radius r is in a uniform magneticfield, with the plane of the loop perpendicular to thedirection of the field (Fig. P31.59). The magnetic fieldvaries with time according to B(t) " a 2 bt, where a and bare constants. (a) Calculate the magnetic flux through theloop at t " 0. (b) Calculate the emf induced in the loop.(c) If the resistance of the loop is R , what is the inducedcurrent? (d) At what rate is energy being delivered to theresistance of the loop?

60. In Figure P31.60, a uniform magnetic field decreases at aconstant rate dB/dt " % K, where K is a positive constant.A circular loop of wire of radius a containing a resistanceR and a capacitance C is placed with its plane normal tothe field. (a) Find the charge Q on the capacitor when it isfully charged. (b) Which plate is at the higher potential?(c) Discuss the force that causes the separation of charges.

A rectangular coil of 60 turns, dimensions 0.100 m by0.200 m and total resistance 10.0 (, rotates with angularspeed 30.0 rad/s about the y axis in a region where a 1.00-Tmagnetic field is directed along the x axis. The rotation isinitiated so that the plane of the coil is perpendicular tothe direction of B at t " 0. Calculate (a) the maximuminduced emf in the coil, (b) the maximum rate of changeof magnetic flux through the coil, (c) the induced emf att " 0.050 0 s, and (d) the torque exerted by the magneticfield on the coil at the instant when the emf is a maximum.

62. A small circular washer of radius 0.500 cm is held directlybelow a long, straight wire carrying a current of 10.0 A.The washer is located 0.500 m above the top of a table(Fig. P31.62). (a) If the washer is dropped from rest, whatis the magnitude of the average induced emf in the washerfrom the time it is released to the moment it hits the table-top? Assume that the magnetic field is nearly constant overthe area of the washer, and equal to the magnetic field atthe center of the washer. (b) What is the direction of theinduced current in the washer?

61.

63. A conducting rod of length ! moves with velocity v parallelto a long wire carrying a steady current I. The axis of therod is maintained perpendicular to the wire with the nearend a distance r away, as shown in Figure P31.63. Showthat the magnitude of the emf induced in the rod is

&$&"/0Iv2,

ln #1 2!

r $

64. A rectangular loop of dimensions ! and w moves with aconstant velocity v away from a long wire that carries acurrent I in the plane of the loop (Fig. P31.64). The totalresistance of the loop is R. Derive an expression that givesthe current in the loop at the instant the near side is adistance r from the wire.

The magnetic flux through a metal ring varies with time taccording to !B " 3(at3 % bt2) T & m2 , with a " 2.00 s%3

and b " 6.00 s%2. The resistance of the ring is 3.00 (.Determine the maximum current induced in the ringduring the interval from t " 0 to t " 2.00 s.

66. Review problem. The bar of mass m in Figure P31.66 ispulled horizontally across parallel rails by a massless stringthat passes over an ideal pulley and is attached to a

65.


B

Figure P31.59

R C

Bin

× × × ×

× × × ×

× × × ×

× × × ×Figure P31.60

h

I

Figure P31.62

vIR

r w

!

Figure P31.64

r

v

I

!

Figure P31.63

suspended object of mass M. The uniform magnetic fieldhas a magnitude B, and the distance between the rails is !.The rails are connected at one end by a load resistor R.Derive an expression that gives the horizontal speed of thebar as a function of time, assuming that the suspendedobject is released with the bar at rest at t " 0. Assume nofriction between rails and bar.

67. A solenoid wound with 2 000 turns/m is supplied withcurrent that varies in time according to I "(4A) sin(120,t), where t is in seconds. A small coaxialcircular coil of 40 turns and radius r " 5.00 cm is locatedinside the solenoid near its center. (a) Derive an expres-sion that describes the manner in which the emf in thesmall coil varies in time. (b) At what average rate is energydelivered to the small coil if the windings have a total resis-tance of 8.00 (?

68. Figure P31.68 shows a stationary conductor whose shape issimilar to the letter e. The radius of its circular portion isa " 50.0 cm. It is placed in a constant magnetic field of0.500 T directed out of the page. A straight conductingrod, 50.0 cm long, is pivoted about point O and rotateswith a constant angular speed of 2.00 rad/s. (a) Deter-mine the induced emf in the loop POQ. Note that the areaof the loop is #a2/2. (b) If all of the conducting materialhas a resistance per length of 5.00 (/m, what is theinduced current in the loop POQ at the instant 0.250 safter point P passes point Q?

69. A betatron accelerates electrons to energies in the MeVrange by means of electromagnetic induction. Electrons ina vacuum chamber are held in a circular orbit by a

magnetic field perpendicular to the orbital plane. Themagnetic field is gradually increased to induce an electricfield around the orbit. (a) Show that the electric field is inthe correct direction to make the electrons speed up.(b) Assume that the radius of the orbit remains constant.Show that the average magnetic field over the areaenclosed by the orbit must be twice as large as themagnetic field at the circumference of the circle.

70. A wire 30.0 cm long is held parallel to and 80.0 cm above along wire carrying 200 A and resting on the floor (Fig.P31.70). The 30.0-cm wire is released and falls, remainingparallel with the current-carrying wire as it falls. Assumethat the falling wire accelerates at 9.80 m/s2 and derive anequation for the emf induced in it. Express your result as afunction of the time t after the wire is dropped. What isthe induced emf 0.300 s after the wire is released?

A long, straight wire carries a current that is given byI " Imax sin(*t 2 4) and lies in the plane of a rectangularcoil of N turns of wire, as shown in Figure P31.9. The quanti-ties Imax, *, and 4 are all constants. Determine the emfinduced in the coil by the magnetic field created by thecurrent in the straight wire. Assume Imax " 50.0 A,* " 200, s%1, N " 100, h " w " 5.00 cm, and L " 20.0 cm.

72. A dime is suspended from a thread and hung between thepoles of a strong horseshoe magnet as shown in FigureP31.72. The dime rotates at constant angular speed *

71.

Problems 1001

R

M

m

B

!

g

Figure P31.66

The movingstraight rod

P

QO

Bout

θ

a

Figure P31.68

30.0 cm

80.0 cm

I = 200 AFigure P31.70

ω

N

S

Figure P31.72

about a vertical axis. Letting # represent the anglebetween the direction of B and the normal to the face ofthe dime, sketch a graph of the torque due to inducedcurrents as a function of # for 0 0 # 0 2,.

Answers to Quick Quizzes31.1 (c). In all cases except this one, there is a change in the

magnetic flux through the loop.

31.2 c, d " e, b, a. The magnitude of the emf is proportionalto the rate of change of the magnetic flux. For the situa-tion described, the rate of change of magnetic flux isproportional to the rate of change of the magneticfield. This rate of change is the slope of the graph inFigure 31.4. The magnitude of the slope is largest at c.Points d and e are on a straight line, so the slope is thesame at each point. Point b represents a point of rela-tively small slope, while a is at a point of zero slopebecause the curve is horizontal at that point.

31.3 (b). The magnetic field lines around the transmissioncable will be circular, centered on the cable. If you placeyour loop around the cable, there are no field linespassing through the loop, so no emf is induced. Theloop must be placed next to the cable, with the plane ofthe loop parallel to the cable to maximize the fluxthrough its area.

31.4 (a). The Earth’s magnetic field has a downward compo-nent in the northern hemisphere. As the plane fliesnorth, the right-hand rule illustrated in Figure 29.4indicates that positive charge experiences a forcedirected toward the west. Thus, the left wingtipbecomes positively charged and the right wingtip nega-tively charged.

31.5 (c). The force on the wire is of magnitude Fapp " FB "I!B, with I given by Equation 31.6. Thus, the force isproportional to the speed and the force doubles.Because " " Fappv, the doubling of the force and thespeed results in the power being four times as large.

31.6 (b). According to Equation 31.5, because B and v arefixed, the emf depends only on the length of the wiremoving in the magnetic field. Thus, you want the longdimension moving through the magnetic field lines sothat it is perpendicular to the velocity vector. In this case,the short dimension is parallel to the velocity vector.

31.7 (a). Because the current induced in the solenoid isclockwise when viewed from above, the magnetic fieldlines produced by this current point downward inFigure 31.15. Thus, the upper end of the solenoid actsas a south pole. For this situation to be consistent withLenz’s law, the south pole of the bar magnet must beapproaching the solenoid.

31.8 (b). At the position of the loop, the magnetic field linesdue to the wire point into the page. The loop is enteringa region of stronger magnetic field as it drops toward thewire, so the flux is increasing. The induced current mustset up a magnetic field that opposes this increase. To dothis, it creates a magnetic field directed out of the page.By the right-hand rule for current loops, this requires acounterclockwise current in the loop.

31.9 (d). The constant rate of change of B will result in aconstant rate of change of the magnetic flux. Accordingto Equation 31.9, if d !B/dt is constant, E is constant inmagnitude.

31.10 (a). While reducing the resistance may increase thecurrent that the generator provides to a load, it does notalter the emf. Equation 31.11 shows that the emf dependson *, B, and N, so all other choices increase the emf.

31.11 (b). When the aluminum sheet moves between thepoles of the magnet, eddy currents are established inthe aluminum. According to Lenz’s law, these currentsare in a direction so as to oppose the original change,which is the movement of the aluminum sheet in themagnetic field. The same principle is used in commonlaboratory triple-beam balances. See if you can find themagnet and the aluminum sheet the next time you usea triple-beam balance.


chapter 31 faradayÕs law - rmutphysics...faradayÕs law c hap te r o u tlin e 31.1 faradayÕs law...

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