An artist s impression of the Skerries SeaGen Array, a tidal energy generator under emf can be induced in a circuit by a

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1 chpter 31 Frdy s Lw 31.1 Frdy s Lw of nduction 31.2 Motionl emf 31.3 Lenz s Lw 31.4 nduced emf nd Electric Fields 31.5 Genertors nd Motors 31.6 Eddy Currents o fr, our studies in electricity nd mgnetism he focused on the electric fields produced by sttionry chrges nd the mgnetic fields produced by moing chrges. This chpter explores the effects produced by mgnetic fields tht ry in time. Experiments conducted by Michel Frdy in Englnd in 1831 nd independently by Joseph Henry in the United ttes tht sme yer showed tht n An rtist s impression of the kerries egen Arry, tidl energy genertor under emf cn be induced in circuit by deelopment ner the islnd of Anglesey, North Wles. When it is brought on line, possibly in 2012, it will offer 10.5 MW of power from genertors turned by tidl chnging mgnetic field. The results of strems. The imge shows the underwter bldes tht re drien by the tidl currents. these experiments led to ery bsic The second blde system hs been rised from the wter for sericing. We will study genertors in this chpter. (Mrine Current Turbines TM Ltd.) nd importnt lw of electromgnetism known s Frdy s lw of induction. An emf (nd therefore current s well) cn be induced in rious processes tht inole chnge in mgnetic flux Frdy s Lw of nduction To see how n emf cn be induced by chnging mgnetic field, consider the experimentl results obtined when loop of wire is connected to sensitie mmeter s illustrted in Actie Figure 31.1 (pge 894). When mgnet is moed towrd the loop, the reding on the mmeter chnges from zero in one direction, 893

2 894 CHAPTE 31 Frdy s Lw By kind permission of the President nd Council of the oyl ociety Michel Frdy British Physicist nd Chemist ( ) Frdy is often regrded s the gretest experimentl scientist of the 1800s. His mny contributions to the study of electricity include the inention of the electric motor, electric genertor, nd trnsformer s well s the discoery of electromgnetic induction nd the lws of electrolysis. Gretly influenced by religion, he refused to work on the deelopment of poison gs for the British militry. rbitrrily shown s negtie in Actie Figure When the mgnet is brought to rest nd held sttionry reltie to the loop (Actie Fig. 31.1b), reding of zero is obsered. When the mgnet is moed wy from the loop, the reding on the mmeter chnges in the opposite direction s shown in Actie Figure 31.1c. Finlly, when the mgnet is held sttionry nd the loop is moed either towrd or wy from it, the reding chnges from zero. From these obsertions, we conclude tht the loop detects tht the mgnet is moing reltie to it nd we relte this detection to chnge in mgnetic field. Therefore, it seems tht reltionship exists between current nd chnging mgnetic field. These results re quite remrkble becuse current is set up een though no btteries re present in the circuit! We cll such current n induced current nd sy tht it is produced by n induced emf. Now let s describe n experiment conducted by Frdy nd illustrted in Actie Figure A primry coil is wrpped round n iron ring nd connected to switch nd bttery. A current in the coil produces mgnetic field when the switch is closed. A secondry coil lso is wrpped round the ring nd is connected to sensitie mmeter. No bttery is present in the secondry circuit, nd the secondry coil is not electriclly connected to the primry coil. Any current detected in the secondry circuit must be induced by some externl gent. nitilly, you might guess tht no current is eer detected in the secondry circuit. omething quite mzing hppens when the switch in the primry circuit is either opened or thrown closed, howeer. At the instnt the switch is closed, the mmeter reding chnges from zero in one direction nd then returns to zero. At the instnt the switch is opened, the mmeter chnges in the opposite direction nd gin returns to zero. Finlly, the mmeter reds zero when there is either stedy current or no current in the primry circuit. To understnd wht hppens in this experiment, note tht when the switch is closed, the current in the primry circuit produces mgnetic field tht penetrtes the secondry circuit. Furthermore, when the switch is thrown closed, the mgnetic field produced by the current in the primry circuit chnges from zero to some lue oer some finite time, nd this chnging field induces current in the secondry circuit. As result of these obsertions, Frdy concluded tht n electric current cn be induced in loop by chnging mgnetic field. The induced current exists only while the mgnetic field through the loop is chnging. Once the mgnetic field reches stedy lue, the current in the loop disppers. n effect, the loop When mgnet is moed towrd loop of wire connected to sensitie mmeter, the mmeter shows tht current is induced in the loop. When the mgnet is held sttionry, there is no induced current in the loop, een when the mgnet is inside the loop. When the mgnet is moed wy from the loop, the mmeter shows tht the induced current is opposite tht shown in prt. ACTVE FGUE 31.1 A simple experiment showing tht current is induced in loop when mgnet is moed towrd or wy from the loop. N b N c N

3 31.1 Frdy s Lw of nduction 895 The emf induced in the secondry circuit is cused by the chnging mgnetic field through the secondry coil. ACTVE FGUE 31.2 Frdy s experiment. When the switch in the primry circuit is closed, the mmeter reding in the secondry circuit chnges momentrily. Bttery ron Primry coil econdry coil behes s though source of emf were connected to it for short time. t is customry to sy tht n induced emf is produced in the loop by the chnging mgnetic field. The experiments shown in Actie Figures 31.1 nd 31.2 he one thing in common: in ech cse, n emf is induced in loop when the mgnetic flux through the loop chnges with time. n generl, this emf is directly proportionl to the time rte of chnge of the mgnetic flux through the loop. This sttement cn be written mthemticlly s Frdy s lw of induction: Pitfll Preention 31.1 nduced emf equires Chnge The existence of mgnetic flux through n re is not sufficient to crete n induced emf. The mgnetic flux must chnge to induce n emf. e 52 df B (31.1) Frdy s lw of induction where F B 5 r B? da is the mgnetic flux through the loop. (ee ection 30.5.) f coil consists of N loops with the sme re nd F B is the mgnetic flux through one loop, n emf is induced in eery loop. The loops re in series, so their emfs dd; therefore, the totl induced emf in the coil is gien by e 52N df B (31.2) The negtie sign in Equtions 31.1 nd 31.2 is of importnt physicl significnce nd will be discussed in ection uppose loop enclosing n re A lies in uniform mgnetic field B s in Figure The mgnetic flux through the loop is equl to BA cos u; hence, the induced emf cn be expressed s e 52 d 1BA cos u 2 (31.3) From this expression, we see tht n emf cn be induced in the circuit in seerl wys: The mgnitude of B cn chnge with time. The re enclosed by the loop cn chnge with time. The ngle u between B nd the norml to the loop cn chnge with time. Any combintion of the boe cn occur. Quick Quiz 31.1 A circulr loop of wire is held in uniform mgnetic field, with the plne of the loop perpendiculr to the field lines. Which of the Norml to loop B u u Loop of re A Figure 31.3 A conducting loop tht encloses n re A in the presence of uniform mgnetic field B. The ngle between B nd the norml to the loop is u.

4 896 CHAPTE 31 Frdy s Lw Alternting current ron ring Circuit breker ensing coil Figure 31.4 Essentil components of ground fult circuit interrupter. 1 2 following will not cuse current to be induced in the loop? () crushing the loop (b) rotting the loop bout n xis perpendiculr to the field lines (c) keeping the orienttion of the loop fixed nd moing it long the field lines (d) pulling the loop out of the field ome Applictions of Frdy s Lw The ground fult circuit interrupter (GFC) is n interesting sfety deice tht protects users of electricl pplinces ginst electric shock. ts opertion mkes use of Frdy s lw. n the GFC shown in Figure 31.4, wire 1 leds from the wll outlet to the pplince to be protected nd wire 2 leds from the pplince bck to the wll outlet. An iron ring surrounds the two wires, nd sensing coil is wrpped round prt of the ring. Becuse the currents in the wires re in opposite directions nd of equl mgnitude, there is no mgnetic field surrounding the wires nd the net mgnetic flux through the sensing coil is zero. f the return current in wire 2 chnges so tht the two currents re not equl, howeer, circulr mgnetic field lines exist round the pir of wires. (Tht cn hppen if, for exmple, the pplince becomes wet, enbling current to lek to ground.) Therefore, the net mgnetic flux through the sensing coil is no longer zero. Becuse household current is lternting (mening tht its direction keeps reersing), the mgnetic flux through the sensing coil chnges with time, inducing n emf in the coil. This induced emf is used to trigger circuit breker, which stops the current before it is ble to rech hrmful leel. Another interesting ppliction of Frdy s lw is the production of sound in n electric guitr. The coil in this cse, clled the pickup coil, is plced ner the ibrting guitr string, which is mde of metl tht cn be mgnetized. A permnent mgnet inside the coil mgnetizes the portion of the string nerest the coil (Fig. 31.5). When the string ibrtes t some frequency, its mgnetized segment produces chnging mgnetic flux through the coil. The chnging flux induces n emf in the coil tht is fed to n mplifier. The output of the mplifier is sent to the loudspekers, which produce the sound wes we her. Mgnetized portion of string N Guitr string N Pickup coil To mplifier Mgnet. Cengge Lerning/Chrles D. Winters b Figure 31.5 () n n electric guitr, ibrting mgnetized string induces n emf in pickup coil. (b) The pickups (the circles beneth the metllic strings) of this electric guitr detect the ibrtions of the strings nd send this informtion through n mplifier nd into spekers. (A switch on the guitr llows the musicin to select which set of six pickups is used.) Exmple 31.1 nducing n emf in Coil A coil consists of 200 turns of wire. Ech turn is squre of side d 5 18 cm, nd uniform mgnetic field directed perpendiculr to the plne of the coil is turned on. f the field chnges linerly from 0 to 0.50 T in 0.80 s, wht is the mgnitude of the induced emf in the coil while the field is chnging?

5 31.1 Frdy s Lw of nduction cont. OLUTON Conceptulize From the description in the problem, imgine mgnetic field lines pssing through the coil. Becuse the mgnetic field is chnging in mgnitude, n emf is induced in the coil. Ctegorize We will elute the emf using Frdy s lw from this section, so we ctegorize this exmple s substitution problem. Elute Eqution 31.2 for the sitution described here, noting tht the mgnetic field chnges linerly with time: 0 e 0 5 N DF B Dt 5 N D1BA2 Dt 5 NA DB Dt 5 Nd B 2 f 2 B i Dt T ubstitute numericl lues: 0 e m V 0.80 s WHAT F? Wht if you were sked to find the mgnitude of the induced current in the coil while the field is chnging? Cn you nswer tht question? Answer f the ends of the coil re not connected to circuit, the nswer to this question is esy: the current is zero! (Chrges moe within the wire of the coil, but they cnnot moe into or out of the ends of the coil.) For stedy current to exist, the ends of the coil must be connected to n externl circuit. Let s ssume the coil is connected to circuit nd the totl resistnce of the coil nd the circuit is 2.0 V. Then, the mgnitude of the induced current in the coil is 5 0 e V 2.0 V A Exmple 31.2 An Exponentilly Decying Mgnetic Field A loop of wire enclosing n re A is plced in region where the mgnetic field is perpendiculr to the plne of the loop. The mgnitude of B ries in time ccording to the expression B 5 B mx e 2t, where is some constnt. Tht is, t t 5 0, the field is B mx, nd for t. 0, the field decreses exponentilly (Fig. 31.6). Find the induced emf in the loop s function of time. OLUTON Figure 31.6 (Exmple 31.2) Exponentil decrese in the mgnitude of the mgnetic field with time. The induced emf nd induced current ry with time in the sme wy. Conceptulize The physicl sitution is similr to tht in Exmple 31.1 except for two things: there is only one loop, nd the field ries exponentilly with time rther thn linerly. Ctegorize We will elute the emf using Frdy s lw from this section, so we ctegorize this exmple s substitution problem. Elute Eqution 31.1 for the sitution described here: e 52 df B B mx B 52 d 1AB mx e 2t 2 52AB mx d e 2t 5 AB mx e 2t This expression indictes tht the induced emf decys exponentilly in time. The mximum emf occurs t t 5 0, where e mx 5 AB mx. The plot of e ersus t is similr to the B-ersus-t cure shown in Figure t

6 898 CHAPTE 31 Frdy s Lw n stedy stte, the electric nd mgnetic forces on n electron in the wire re blnced. F e F B E Due to the mgnetic force on electrons, the ends of the conductor become oppositely chrged, which estblishes n electric field in the conductor. Figure 31.7 A stright electricl conductor of length, moing with elocity through uniform mgnetic field B directed perpendiculr to Motionl emf n Exmples 31.1 nd 31.2, we considered cses in which n emf is induced in sttionry circuit plced in mgnetic field when the field chnges with time. n this section, we describe motionl emf, the emf induced in conductor moing through constnt mgnetic field. The stright conductor of length, shown in Figure 31.7 is moing through uniform mgnetic field directed into the pge. For simplicity, let s ssume the conductor is moing in direction perpendiculr to the field with constnt elocity under the influence of some externl gent. The electrons in the conductor experience force FB 5 q 3 B tht is directed long the length,, perpendiculr to both nd B (Eq. 29.1). Under the influence of this force, the electrons moe to the lower end of the conductor nd ccumulte there, leing net positie chrge t the upper end. As result of this chrge seprtion, n electric field E is produced inside the conductor. The chrges ccumulte t both ends until the downwrd mgnetic force qb on chrges remining in the conductor is blnced by the upwrd electric force qe. The condition for equilibrium requires tht the forces on the electrons blnce: qe 5 qb or E 5 B The electric field produced in the conductor is relted to the potentil difference cross the ends of the conductor ccording to the reltionship DV 5 E, (Eq. 25.6). Therefore, for the equilibrium condition, DV 5 E, 5 B, (31.4) where the upper end of the conductor in Figure 31.7 is t higher electric potentil thn the lower end. Therefore, potentil difference is mintined between the ends of the conductor s long s the conductor continues to moe through the uniform mgnetic field. f the direction of the motion is reersed, the polrity of the potentil difference is lso reersed. A more interesting sitution occurs when the moing conductor is prt of closed conducting pth. This sitution is prticulrly useful for illustrting how chnging mgnetic flux cuses n induced current in closed circuit. Consider circuit consisting of conducting br of length, sliding long two fixed, prllel conducting rils s shown in Actie Figure For simplicity, let s ssume the br hs A counterclockwise current is induced in the loop. The mgnetic force F B on the br crrying this current opposes the motion. F B F pp e B ACTVE FGUE 31.8 () A conducting br sliding with elocity long two conducting rils under the ction of n pplied force Fpp. (b) The equilent circuit digrm for the setup shown in (). x b

7 31.2 Motionl emf 899 zero resistnce nd the sttionry prt of the circuit hs resistnce. A uniform nd constnt mgnetic field B is pplied perpendiculr to the plne of the circuit. As the br is pulled to the right with elocity under the influence of n pplied force F pp, free chrges in the br experience mgnetic force directed long the length of the br. This force sets up n induced current becuse the chrges re free to moe in the closed conducting pth. n this cse, the rte of chnge of mgnetic flux through the circuit nd the corresponding induced motionl emf cross the moing br re proportionl to the chnge in re of the circuit. Becuse the re enclosed by the circuit t ny instnt is,x, where x is the position of the br, the mgnetic flux through tht re is F B 5 B,x Using Frdy s lw nd noting tht x chnges with time t rte dx/ 5, we find tht the induced motionl emf is e 52 df B 52 d dx 1B,x2 52B, e 5 2B, (31.5) Motionl emf Becuse the resistnce of the circuit is, the mgnitude of the induced current is 5 0e0 5 B, (31.6) The equilent circuit digrm for this exmple is shown in Actie Figure 31.8b. Let s exmine the system using energy considertions. Becuse no bttery is in the circuit, you might wonder bout the origin of the induced current nd the energy deliered to the resistor. We cn understnd the source of this current nd energy by noting tht the pplied force does work on the conducting br. Therefore, we model the circuit s nonisolted system. The moement of the br through the field cuses chrges to moe long the br with some erge drift elocity; hence, current is estblished. The chnge in energy in the system during some time interl must be equl to the trnsfer of energy into the system by work, consistent with the generl principle of consertion of energy described by Eqution 8.2. Let s erify this equlity mthemticlly. As the br moes through the uniform mgnetic field B, it experiences mgnetic force F B of mgnitude,b (see ection 29.4). Becuse the br moes with constnt elocity, it is modeled s prticle in equilibrium nd the mgnetic force must be equl in mgnitude nd opposite in direction to the pplied force, or to the left in Actie Figure (f F B cted in the direction of motion, it would cuse the br to ccelerte, iolting the principle of consertion of energy.) Using Eqution 31.6 nd F pp 5 F B 5,B, the power deliered by the pplied force is P 5 F pp 5 1,B2 5 B 2, e2 (31.7) From Eqution 27.21, we see tht this power input is equl to the rte t which energy is deliered to the resistor. Quick Quiz 31.2 n Actie Figure 31.8, gien pplied force of mgnitude F pp results in constnt speed nd power input P. mgine tht the force is incresed so tht the constnt speed of the br is doubled to 2. Under these conditions, wht re the new force nd the new power input? () 2F nd 2P (b) 4F nd 2P (c) 2F nd 4P (d) 4F nd 4P

8 900 CHAPTE 31 Frdy s Lw Exmple 31.3 Mgnetic Force Acting on liding Br The conducting br illustrted in Figure 31.9 moes on two frictionless, prllel rils in the presence of uniform mgnetic field directed into the pge. The br hs mss m, nd its length is,. The br is gien n initil elocity i to the right nd is relesed t t 5 0. (A) Using Newton s lws, find the elocity of the br s function of time. OLUTON Conceptulize As the br slides to the right in Figure 31.9, counterclockwise current is estblished in the circuit consisting of the br, the rils, nd the resistor. The upwrd current in the br results in mgnetic force to the left on the br s shown in the figure. Therefore, the br must slow down, so our mthemticl solution should demonstrte tht. Ctegorize The text lredy ctegorizes this problem s one tht uses Newton s lws. We model the br s prticle under net force. x F B i Figure 31.9 (Exmple 31.3) A conducting br of length, on two fixed conducting rils is gien n initil elocity i to the right. Anlyze From Eqution 29.10, the mgnetic force is F B 5 2,B, where the negtie sign indictes tht the force is to the left. The mgnetic force is the only horizontl force cting on the br. Apply Newton s second lw to the br in the horizontl direction: ubstitute 5 B,/ from Eqution 31.6: errnge the eqution so tht ll occurrences of the rible re on the left nd those of t re on the right: ntegrte this eqution using the initil condition tht 5 i t t 5 0 nd noting tht (B 2, 2 /m) is constnt: Define the constnt t 5 m/b 2, 2 nd sole for the elocity: F x 5 m 5 m d 52,B m d 2 52B, 2 d 52B 2, 2 m b 3 i d 2 52B, 2 m 3 (1) 5 i e 2t/t t 0 ln b 52 B 2, 2 i m bt Finlize This expression for indictes tht the elocity of the br decreses with time under the ction of the mgnetic force s expected from our conceptuliztion of the problem. (B) how tht the sme result is found by using n energy pproch. OLUTON Ctegorize The text of this prt of the problem tells us to use n energy pproch for the sme sitution. We model the entire circuit in Figure 31.9 s n isolted system. Anlyze Consider the sliding br s one system component possessing kinetic energy, which decreses becuse energy is trnsferring out of the br by electricl trnsmission through the rils. The resistor is nother system component possessing internl energy, which rises becuse energy is trnsferring into the resistor. Becuse energy is not leing the system, the rte of energy trnsfer out of the br equls the rte of energy trnsfer into the resistor. Equte the power entering the resistor to tht leing the br: P resistor 5 2P br

9 31.2 Motionl emf cont. ubstitute for the electricl power deliered to the resistor nd the time rte of chnge of kinetic energy for the br: Use Eqution 31.6 for the current nd crry out the deritie: errnge terms: 2 52 d 1 1 2m 2 2 B 2, m d d 52B 2, 2 m b Finlize This result is the sme expression to be integrted tht we found in prt (A). WHAT F? uppose you wished to increse the distnce through which the br moes between the time it is initilly projected nd the time it essentilly comes to rest. You cn do so by chnging one of three ribles i,, or B by fctor of 2 or 1 2. Which rible should you chnge to mximize the distnce, nd would you double it or hle it? Answer ncresing i would mke the br moe frther. ncresing would decrese the current nd therefore the mgnetic force, mking the br moe frther. Decresing B would decrese the mgnetic force nd mke the br moe frther. Which method is most effectie, though? Use Eqution (1) to find the distnce the br moes by integrtion: 5 dx 5 i e 2t/t x 5 3 ` 0 i e 2t/t 52 i te 2t/t` ` 0 52 i t i t5 i m B 2, 2b This expression shows tht doubling i or will double the distnce. Chnging B by fctor of 1 2, howeer, cuses the distnce to be four times s gret! Exmple 31.4 Motionl emf nduced in otting Br A conducting br of length, rottes with constnt ngulr speed bout piot t one end. A uniform mgnetic field B is directed perpendiculr to the plne of rottion s shown in Figure Find the motionl emf induced between the ends of the br. Figure (Exmple 31.4) A OLUTON conducting br rotting round piot t one end in uniform Conceptulize The rotting br is different in nture from the sliding br in Actie Figure Consider smll segment of the br, howeer. t is short length of conductor moing in mgnetic field nd hs n emf generted in it like the sliding mgnetic field tht is perpendiculr to the plne of rottion. A motionl emf is induced between the ends of the br. Piot br. By thinking of ech smll segment s source of emf, we see tht ll segments re in series nd the emfs dd. Ctegorize Bsed on the conceptuliztion of the problem, we pproch this exmple s we did Exmple 31.3, with the dded feture tht the short segments of the br re treling in circulr pths. r dr Anlyze Elute the mgnitude of the emf induced in segment of the br of length dr hing elocity from Eqution 31.5: de 5 B dr continued

10 902 CHAPTE 31 Frdy s Lw 31.4 cont. Find the totl emf between the ends of the br by dding the emfs induced cross ll segments: The tngentil speed of n element is relted to the ngulr speed through the reltionship 5 r (Eq ); use tht fct nd integrte: e 5 3 B dr, e 5 B 3 dr 5 B 3 r dr B, 2 0 Finlize n Eqution 31.5 for sliding br, we cn increse e by incresing B,,, or. ncresing ny one of these ribles by gien fctor increses e by the sme fctor. Therefore, you would choose whicheer of these three ribles is most conenient to increse. For the rotting rod, howeer, there is n dntge to incresing the length of the rod to rise the emf becuse, is squred. Doubling the length gies four times the emf, wheres doubling the ngulr speed only doubles the emf. WHAT F? uppose, fter reding through this exmple, you come up with brillint ide. A Ferris wheel hs rdil metllic spokes between the hub nd the circulr rim. These spokes moe in the mgnetic field of the Erth, so ech spoke cts like the br in Figure You pln to use the emf generted by the rottion of the Ferris wheel to power the lightbulbs on the wheel. Will this ide work? Answer Let s estimte the emf tht is generted in this sitution. We know the mgnitude of the mgnetic field of the Erth from Tble 29.1: B T. A typicl spoke on Ferris wheel might he length on the order of 10 m. uppose the period of rottion is on the order of 10 s. Determine the ngulr speed of the spoke: 5 2p T 5 2p 10 s s21, 1 s 21 Assume the mgnetic field lines of the Erth re horizontl t the loction of the Ferris wheel nd perpendiculr to the spokes. Find the emf generted: e 5 1 2B, T211 s m V, 1 mv This lue is tiny emf, fr smller thn tht required to operte lightbulbs. An dditionl difficulty is relted to energy. Een ssuming you could find lightbulbs tht operte using potentil difference on the order of milliolts, spoke must be prt of circuit to proide oltge to the lightbulbs. Consequently, the spoke must crry current. Becuse this current-crrying spoke is in mgnetic field, mgnetic force is exerted on the spoke in the direction opposite its direction of motion. As result, the motor of the Ferris wheel must supply more energy to perform work ginst this mgnetic drg force. The motor must ultimtely proide the energy tht is operting the lightbulbs, nd you he not gined nything for free! 31.3 Lenz s Lw Frdy s lw (Eq. 31.1) indictes tht the induced emf nd the chnge in flux he opposite lgebric signs. This feture hs ery rel physicl interprettion tht hs come to be known s Lenz s lw: 1 The induced current in loop is in the direction tht cretes mgnetic field tht opposes the chnge in mgnetic flux through the re enclosed by the loop. Tht is, the induced current tends to keep the originl mgnetic flux through the loop from chnging. We shll show tht this lw is consequence of the lw of consertion of energy. To understnd Lenz s lw, let s return to the exmple of br moing to the right on two prllel rils in the presence of uniform mgnetic field (the externl mgnetic field; Fig ) As the br moes to the right, the mgnetic flux through 1 Deeloped by Germn physicist Heinrich Lenz ( ).

11 31.3 Lenz s Lw 903 As the conducting br slides to the right, the mgnetic flux due to the externl mgnetic field into the pge through the re enclosed by the loop increses in time. Figure () Lenz s lw cn be used to determine the direction of the induced current. (b) When the br moes to the left, the induced current must be clockwise. Why? By Lenz s lw, the induced current must be counterclockwise to produce countercting mgnetic field directed out of the pge. b the re enclosed by the circuit increses with time becuse the re increses. Lenz s lw sttes tht the induced current must be directed so tht the mgnetic field it produces opposes the chnge in the externl mgnetic flux. Becuse the mgnetic flux due to n externl field directed into the pge is incresing, the induced current if it is to oppose this chnge must produce field directed out of the pge. Hence, the induced current must be directed counterclockwise when the br moes to the right. (Use the right-hnd rule to erify this direction.) f the br is moing to the left s in Figure 31.11b, the externl mgnetic flux through the re enclosed by the loop decreses with time. Becuse the field is directed into the pge, the direction of the induced current must be clockwise if it is to produce field tht lso is directed into the pge. n either cse, the induced current ttempts to mintin the originl flux through the re enclosed by the current loop. Let s exmine this sitution using energy considertions. uppose the br is gien slight push to the right. n the preceding nlysis, we found tht this motion sets up counterclockwise current in the loop. Wht hppens if we ssume the current is clockwise such tht the direction of the mgnetic force exerted on the br is to the right? This force would ccelerte the rod nd increse its elocity, which in turn would cuse the re enclosed by the loop to increse more rpidly. The result would be n increse in the induced current, which would cuse n increse in the force, which would produce n increse in the current, nd so on. n effect, the system would cquire energy with no input of energy. This behior is clerly inconsistent with ll experience nd ioltes the lw of consertion of energy. Therefore, the current must be counterclockwise. Quick Quiz 31.3 Figure shows circulr loop of wire flling towrd wire crrying current to the left. Wht is the direction of the induced current in the loop of wire? () clockwise (b) counterclockwise (c) zero (d) impossible to determine Figure (Quick Quiz 31.3) Conceptul Exmple 31.5 Appliction of Lenz s Lw A mgnet is plced ner metl loop s shown in Figure (pge 904). (A) Find the direction of the induced current in the loop when the mgnet is pushed towrd the loop. continued

12 904 CHAPTE 31 Frdy s Lw 31.5cont. When the mgnet is moed towrd the sttionry conducting loop, current is induced in the direction shown. The mgnetic field lines re due to the br mgnet. This induced current produces its own mgnetic field directed to the left tht countercts the incresing externl flux. When the mgnet is moed wy from the sttionry conducting loop, current is induced in the direction shown. This induced current produces mgnetic field directed to the right nd so countercts the decresing externl flux. N N N N b c d Figure (Conceptul Exmple 31.5) A moing br mgnet induces current in conducting loop. OLUTON As the mgnet moes to the right towrd the loop, the externl mgnetic flux through the loop increses with time. To counterct this increse in flux due to field towrd the right, the induced current produces its own mgnetic field to the left s illustrted in Figure 31.13b; hence, the induced current is in the direction shown. Knowing tht like mgnetic poles repel ech other, we conclude tht the left fce of the current loop cts like north pole nd the right fce cts like south pole. (B) Find the direction of the induced current in the loop when the mgnet is pulled wy from the loop. OLUTON f the mgnet moes to the left s in Figure 31.13c, its flux through the re enclosed by the loop decreses in time. Now the induced current in the loop is in the direction shown in Figure 31.13d becuse this current direction produces mgnetic field in the sme direction s the externl field. n this cse, the left fce of the loop is south pole nd the right fce is north pole. Conceptul Exmple 31.6 A Loop Moing Through Mgnetic Field A rectngulr metllic loop of dimensions, nd w nd resistnce moes with constnt speed to the right s in Figure The loop psses through uniform mgnetic field B directed into the pge nd extending distnce 3w long the x xis. Define x s the position of the right side of the loop long the x xis. (A) Plot the mgnetic flux through the re enclosed by the loop s function of x. OLUTON Figure 31.14b shows the flux through the re enclosed by the loop s function of x. Before the loop enters the field, the flux through the loop is zero. As the loop enters the field, the flux increses linerly with position until the left edge of the loop is just inside the field. Finlly, the flux through the loop decreses linerly to zero s the loop lees the field. (B) Plot the induced motionl emf in the loop s function of x. OLUTON Before the loop enters the field, no motionl emf is induced in it becuse no field is present (Fig c). As the right side of the loop enters the field, the mgnetic flux directed into the pge increses. Hence, ccording to Lenz s lw, the induced current is counterclockwise becuse it must produce its own mgnetic field directed out of the pge. The motionl emf 2B, (from Eq. 31.5) rises from the mgnetic force experienced by chrges in the right side of the loop.

13 31.4 nduced emf nd Electric Fields cont. When the loop is entirely in the field, the chnge in mgnetic flux through the loop is zero; hence, the motionl emf nishes. Tht hppens becuse once the left side of the loop enters the field, the motionl emf induced in it cncels the motionl emf present in the right side of the loop. As the right side of the loop lees the field, the flux through the loop begins to decrese, clockwise current is induced, nd the induced emf is B,. As soon s the left side lees the field, the emf decreses to zero. (C) Plot the externl pplied force necessry to counter the mgnetic force nd keep constnt s function of x. OLUTON The externl force tht must be pplied to the loop to mintin this motion is plotted in Figure 31.14d. Before the loop enters the field, no mgnetic force cts on it; hence, the pplied force must be zero if is constnt. When the right side of the loop enters Bw b B 0 w 3w 4w the field, the pplied force necessry to mintin constnt speed must be equl in mgnitude nd opposite in direction to the mgnetic force exerted on tht side. When the loop is entirely in the field, the flux through the loop is not chnging with time. Hence, the net emf induced in the loop is zero nd the current lso is zero. Therefore, no externl force is needed to mintin the motion. Finlly, s the right side lees the field, the pplied force must be equl in mgnitude nd opposite in direction to the mgnetic force cting on the left side of the loop. From this nlysis, we conclude tht power is supplied only when the loop is either entering or leing the field. Furthermore, this exmple shows tht the motionl emf induced in the loop cn be zero een when there is motion through the field! A motionl emf is induced only when the mgnetic flux through the loop chnges in time. w 0 3w x x B B c B 2 2 d e F x 0 w 3w 4w Figure (Conceptul Exmple 31.6) () A conducting rectngulr loop of wih w nd length, moing with elocity through uniform mgnetic field extending distnce 3w. (b) Mgnetic flux through the re enclosed by the loop s function of loop position. (c) nduced emf s function of loop position. (d) Applied force required for constnt elocity s function of loop position. x x 31.4 nduced emf nd Electric Fields We he seen tht chnging mgnetic flux induces n emf nd current in conducting loop. n our study of electricity, we relted current to n electric field tht pplies electric forces on chrged prticles. n the sme wy, we cn relte n induced current in conducting loop to n electric field by climing tht n electric field is creted in the conductor s result of the chnging mgnetic flux. We lso noted in our study of electricity tht the existence of n electric field is independent of the presence of ny test chrges. This independence suggests tht een in the bsence of conducting loop, chnging mgnetic field genertes n electric field in empty spce. This induced electric field is nonconsertie, unlike the electrosttic field produced by sttionry chrges. To illustrte this point, consider conducting loop of rdius r situted in uniform mgnetic field tht is perpendiculr to the plne of the loop s in Figure f the mgnetic field chnges with time, n emf e 5 2dF B / is, ccording to Frdy s lw (Eq. 31.1), induced in the loop. The induction of current in the loop implies the presence of n induced electric field E, which must be tngent to the loop becuse tht is the direction in which the chrges in the wire moe in response to the electric force. The work done by the electric field in moing test chrge q once round the loop is equl to qe. Becuse f B chnges in time, n electric field is induced in direction tngent to the circumference of the loop. E E r E E Figure A conducting loop of rdius r in uniform mgnetic field perpendiculr to the plne of the loop.

14 906 CHAPTE 31 Frdy s Lw Pitfll Preention 31.2 nduced Electric Fields The chnging mgnetic field does not need to exist t the loction of the induced electric field. n Figure 31.15, een loop outside the region of mgnetic field experiences n induced electric field. Frdy s lw in generl form the electric force cting on the chrge is qe, the work done by the electric field in moing the chrge once round the loop is qe(2pr), where 2pr is the circumference of the loop. These two expressions for the work done must be equl; therefore, qe 5 qe(2pr) e E 5 2pr Using this result long with Eqution 31.1 nd tht F B 5 BA 5 Bpr 2 for circulr loop, the induced electric field cn be expressed s E pr df B 52 r 2 db (31.8) f the time rition of the mgnetic field is specified, the induced electric field cn be clculted from Eqution The emf for ny closed pth cn be expressed s the line integrl of E? d s oer tht pth: e 5 r E? d s. n more generl cses, E my not be constnt nd the pth my not be circle. Hence, Frdy s lw of induction, e 52dF B /, cn be written in the generl form C E? d s 52 df B (31.9) The induced electric field E in Eqution 31.9 is nonconsertie field tht is generted by chnging mgnetic field. The field E tht stisfies Eqution 31.9 cnnot possibly be n electrosttic field becuse were the field electrosttic nd hence consertie, the line integrl of E? d s oer closed loop would be zero (ection 25.1), which would be in contrdiction to Eqution Exmple 31.7 Electric Field nduced by Chnging Mgnetic Field in olenoid A long solenoid of rdius hs n turns of wire per unit length nd crries timerying current tht ries sinusoidlly s 5 mx cos t, where mx is the mximum current nd is the ngulr frequency of the lternting current source (Fig ). (A) Determine the mgnitude of the induced electric field outside the solenoid t distnce r. from its long centrl xis. Pth of integrtion r OLUTON Conceptulize Figure shows the physicl sitution. As the current in the coil chnges, imgine chnging mgnetic field t ll points in spce s well s n induced electric field. Ctegorize Becuse the current ries in time, the mgnetic field is chnging, leding to n induced electric field s opposed to the electrosttic electric fields due to sttionry electric chrges. Anlyze First consider n externl point nd tke the pth for the line integrl to be circle of rdius r centered on the solenoid s illustrted in Figure Figure (Exmple 31.7) A long solenoid crrying timerying current gien by 5 mx cos t. An electric field is induced both inside nd outside the solenoid. Elute the right side of Eqution 31.9, noting tht B is perpendiculr to the circle bounded by the pth of integrtion nd tht this mgnetic field exists only inside the solenoid: (1) 2 df B 52 d 1Bp p 2 db

15 31.5 Genertors nd Motors cont. Elute the mgnetic field in the solenoid from Eqution 30.17: ubstitute Eqution (2) into Eqution (1): Elute the left side of Eqution 31.9, noting tht the mgnitude of E is constnt on the pth of integrtion nd E is tngent to it: ubstitute Equtions (3) nd (4) into Eqution 31.9: (2) B 5 m 0 n 5 m 0 n mx cos t (3) 2 df B (4) C E? d s 5 E 12pr2 5 2p 2 m 0 n mx d 1cos t2 5p 2 m 0 n mx sin t E(2pr) 5 p 2 m 0 n mx sin t ole for the mgnitude of the electric field: E 5 m 0n mx 2 2r sin t (for r. ) Finlize This result shows tht the mplitude of the electric field outside the solenoid flls off s 1/r nd ries sinusoidlly with time. As we will lern in Chpter 34, the time-rying electric field cretes n dditionl contribution to the mgnetic field. The mgnetic field cn be somewht stronger thn we first stted, both inside nd outside the solenoid. The correction to the mgnetic field is smll if the ngulr frequency is smll. At high frequencies, howeer, new phenomenon cn dominte: The electric nd mgnetic fields, ech re-creting the other, constitute n electromgnetic we rdited by the solenoid s we will study in Chpter 34. (B) Wht is the mgnitude of the induced electric field inside the solenoid, distnce r from its xis? OLUTON Anlyze For n interior point (r, ), the mgnetic flux through n integrtion loop is gien by F B 5 Bpr 2. Elute the right side of Eqution 31.9: ubstitute Eqution (2) into Eqution (5): ubstitute Equtions (4) nd (6) into Eqution 31.9: ole for the mgnitude of the electric field: (5) 2 df B (6) 2 df B 52 d 1Bpr pr 2 db 5 2pr 2 m 0 n mx d 1cos t2 5pr 2 m 0 n mx sin t E(2pr) 5 pr 2 m 0 n mx sin t E 5 m 0n mx 2 r sin t (for r, ) Finlize This result shows tht the mplitude of the electric field induced inside the solenoid by the chnging mgnetic flux through the solenoid increses linerly with r nd ries sinusoidlly with time Genertors nd Motors Electric genertors tke in energy by work nd trnsfer it out by electricl trnsmission. To understnd how they operte, let us consider the lternting-current (AC) genertor. n its simplest form, it consists of loop of wire rotted by some externl mens in mgnetic field (Actie Fig , pge 908). n commercil power plnts, the energy required to rotte the loop cn be deried from riety of sources. For exmple, in hydroelectric plnt, flling wter directed ginst the bldes of turbine produces the rotry motion; in colfired plnt, the energy relesed by burning col is used to conert wter to stem, nd this stem is directed ginst the turbine bldes.

16 908 CHAPTE 31 Frdy s Lw ACTVE FGUE () chemtic digrm of n AC genertor. (b) The lternting emf induced in the loop plotted s function of time. An emf is induced in loop tht rottes in mgnetic field. lip rings N e e mx Externl circuit t Brushes b As loop rottes in mgnetic field, the mgnetic flux through the re enclosed by the loop chnges with time, nd this chnge induces n emf nd current in the loop ccording to Frdy s lw. The ends of the loop re connected to slip rings tht rotte with the loop. Connections from these slip rings, which ct s output terminls of the genertor, to the externl circuit re mde by sttionry metllic brushes in contct with the slip rings. nsted of single turn, suppose coil with N turns ( more prcticl sitution), with the sme re A, rottes in mgnetic field with constnt ngulr speed. f u is the ngle between the mgnetic field nd the norml to the plne of the coil s in Figure 31.18, the mgnetic flux through the coil t ny time t is F B 5 BA cos u 5 BA cos t where we he used the reltionship u 5 t between ngulr position nd ngulr speed (see Eq. 10.3). (We he set the clock so tht t 5 0 when u 5 0.) Hence, the induced emf in the coil is e 52N df B 52NAB d 1cos t2 5 NAB sin t (31.10) This result shows tht the emf ries sinusoidlly with time s plotted in Actie Figure 31.17b. Eqution shows tht the mximum emf hs the lue B e mx 5 NAB (31.11) Norml u which occurs when t or n other words, e 5 e mx when the mgnetic field is in the plne of the coil nd the time rte of chnge of flux is mximum. Furthermore, the emf is zero when t 5 0 or 1808, tht is, when B is perpendiculr to the plne of the coil nd the time rte of chnge of flux is zero. The frequency for commercil genertors in the United ttes nd Cnd is 60 Hz, wheres in some Europen countries it is 50 Hz. (ecll tht 5 2pf, where f is the frequency in hertz.) Figure A cutwy iew of loop enclosing n re A nd contining N turns, rotting with constnt ngulr speed in mgnetic field. The emf induced in the loop ries sinusoidlly in time. Quick Quiz 31.4 n n AC genertor, coil with N turns of wire spins in mgnetic field. Of the following choices, which does not cuse n increse in the emf generted in the coil? () replcing the coil wire with one of lower resistnce (b) spinning the coil fster (c) incresing the mgnetic field (d) incresing the number of turns of wire on the coil

17 31.5 Genertors nd Motors 909 Exmple 31.8 emf nduced in Genertor The coil in n AC genertor consists of 8 turns of wire, ech of re A m 2, nd the totl resistnce of the wire is 12.0 V. The coil rottes in T mgnetic field t constnt frequency of 60.0 Hz. (A) Find the mximum induced emf in the coil. OLUTON Conceptulize tudy Actie Figure to mke sure you understnd the opertion of n AC genertor. Ctegorize We elute prmeters using equtions deeloped in this section, so we ctegorize this exmple s substitution problem. Use Eqution to find the mximum induced emf: e mx 5 NAB 5 NAB(2pf ) ubstitute numericl lues: e mx 5 8( m 2 )(0.500 T)(2p)(60.0 Hz) V (B) Wht is the mximum induced current in the coil when the output terminls re connected to low-resistnce conductor? OLUTON Use Eqution 27.7 nd the result to prt (A): mx 5 e mx V 12.0 V A The direct-current (DC) genertor is illustrted in Actie Figure uch genertors re used, for instnce, in older crs to chrge the storge btteries. The components re essentilly the sme s those of the AC genertor except tht the contcts to the rotting coil re mde using split ring clled commuttor. n this configurtion, the output oltge lwys hs the sme polrity nd pulstes with time s shown in Actie Figure 31.19b. We cn understnd why by noting tht the contcts to the split ring reerse their roles eery hlf cycle. At the sme time, the polrity of the induced emf reerses; hence, the polrity of the split ring (which is the sme s the polrity of the output oltge) remins the sme. A pulsting DC current is not suitble for most pplictions. To obtin stedier DC current, commercil DC genertors use mny coils nd commuttors distributed so tht the sinusoidl pulses from the rious coils re out of phse. When these pulses re superimposed, the DC output is lmost free of fluctutions. A motor is deice into which energy is trnsferred by electricl trnsmission while energy is trnsferred out by work. A motor is essentilly genertor operting in reerse. nsted of generting current by rotting coil, current is supplied to the coil by bttery, nd the torque cting on the current-crrying coil (ection 29.5) cuses it to rotte. Commuttor N e t Brush b ACTVE FGUE () chemtic digrm of DC genertor. (b) The mgnitude of the emf ries in time, but the polrity neer chnges.

18 910 CHAPTE 31 Frdy s Lw John W. Jewett, Jr. Figure The engine comprtment of Toyot Prius, hybrid ehicle. Useful mechnicl work cn be done by ttching the rotting coil to some externl deice. As the coil rottes in mgnetic field, howeer, the chnging mgnetic flux induces n emf in the coil; this induced emf lwys cts to reduce the current in the coil. f tht were not the cse, Lenz s lw would be iolted. The bck emf increses in mgnitude s the rottionl speed of the coil increses. (The phrse bck emf is used to indicte n emf tht tends to reduce the supplied current.) Becuse the oltge ilble to supply current equls the difference between the supply oltge nd the bck emf, the current in the rotting coil is limited by the bck emf. When motor is turned on, there is initilly no bck emf, nd the current is ery lrge becuse it is limited only by the resistnce of the coil. As the coil begins to rotte, the induced bck emf opposes the pplied oltge nd the current in the coil decreses. f the mechnicl lod increses, the motor slows down, which cuses the bck emf to decrese. This reduction in the bck emf increses the current in the coil nd therefore lso increses the power needed from the externl oltge source. For this reson, the power requirements for running motor re greter for hey lods thn for light ones. f the motor is llowed to run under no mechnicl lod, the bck emf reduces the current to lue just lrge enough to oercome energy losses due to internl energy nd friction. f ery hey lod jms the motor so tht it cnnot rotte, the lck of bck emf cn led to dngerously high current in the motor s wire. This dngerous sitution is explored in the Wht f? section of Exmple A modern ppliction of motors in utomobiles is seen in the deelopment of hybrid drie systems. n these utomobiles, gsoline engine nd n electric motor re combined to increse the fuel economy of the ehicle nd reduce its emissions. Figure shows the engine comprtment of Toyot Prius, one of the hybrids ilble in the United ttes. n this utomobile, power to the wheels cn come from either the gsoline engine or the electric motor. n norml driing, the electric motor ccelertes the ehicle from rest until it is moing t speed of bout 15 mi/h (24 km/h). During this ccelertion period, the engine is not running, so gsoline is not used nd there is no emission. At higher speeds, the motor nd engine work together so tht the engine lwys opertes t or ner its most efficient speed. The result is significntly higher gsoline milege thn tht obtined by trditionl gsoline-powered utomobile. When hybrid ehicle brkes, the motor cts s genertor nd returns some of the ehicle s kinetic energy bck to the bttery s stored energy. n norml ehicle, this kinetic energy is not recoered becuse it is trnsformed to internl energy in the brkes nd rodwy. Exmple 31.9 The nduced Current in Motor A motor contins coil with totl resistnce of 10 V nd is supplied by oltge of 120 V. When the motor is running t its mximum speed, the bck emf is 70 V. (A) Find the current in the coil t the instnt the motor is turned on. OLUTON Conceptulize Think bout the motor just fter it is turned on. t hs not yet moed, so there is no bck emf generted. As result, the current in the motor is high. After the motor begins to turn, bck emf is generted nd the current decreses. Ctegorize We need to combine our new understnding bout motors with the reltionship between current, oltge, nd resistnce. Anlyze Elute the current in the coil from Eqution 27.7 with no bck emf generted: 5 e V 10 V 5 12 A

19 31.6 Eddy Currents cont. (B) Find the current in the coil when the motor hs reched mximum speed. OLUTON Elute the current in the coil with the mximum bck emf generted: 5 e 2 e bck V 2 70 V 10 V 5 50 V 10 V A Finlize The current drwn by the motor when operting t its mximum speed is significntly less thn tht drwn before it begins to turn. WHAT F? uppose this motor is in circulr sw. When you re operting the sw, the blde becomes jmmed in piece of wood nd the motor cnnot turn. By wht percentge does the power input to the motor increse when it is jmmed? Answer You my he eerydy experiences with motors becoming wrm when they re preented from turning. Tht is due to the incresed power input to the motor. The higher rte of energy trnsfer results in n increse in the internl energy of the coil, n undesirble effect. et up the rtio of power input to the motor when jmmed, using the current clculted in prt (A), to tht when it is not jmmed, prt (B): ubstitute numericl lues: P jmmed P jmmed P not jmmed A P not jmmed 2 B 5 A 112 A A B 2 Tht represents 476% increse in the input power! uch high power input cn cuse the coil to become so hot tht it is dmged Eddy Currents As we he seen, n emf nd current re induced in circuit by chnging mgnetic flux. n the sme mnner, circulting currents clled eddy currents re induced in bulk pieces of metl moing through mgnetic field. This phenomenon cn be demonstrted by llowing flt copper or luminum plte ttched t the end of rigid br to swing bck nd forth through mgnetic field (Fig ). As the plte enters the field, the chnging mgnetic flux induces n emf in the plte, which in turn cuses the free electrons in the plte to moe, producing the swirling eddy currents. According to Lenz s lw, the direction of the eddy currents is such tht they crete mgnetic fields tht oppose the chnge tht cuses the currents. For this reson, the eddy currents must produce effectie mgnetic poles on the plte, which re repelled by the poles of the mgnet; this sitution gies rise to repulsie force tht opposes the motion of the plte. (f the opposite were true, the plte would ccelerte nd its energy would increse fter ech swing, in ioltion of the lw of consertion of energy.) As indicted in Actie Figure (pge 912), with B directed into the pge, the induced eddy current is counterclockwise s the swinging plte enters the field t position 1 becuse the flux due to the externl mgnetic field into the pge through the plte is incresing. Hence, by Lenz s lw, the induced current must proide its own mgnetic field out of the pge. The opposite is true s the plte lees the field t position 2, where the current is clockwise. Becuse the induced Piot N As the plte enters or lees the field, the chnging mgnetic flux induces n emf, which cuses eddy currents in the plte. Figure Formtion of eddy currents in conducting plte moing through mgnetic field.

20 912 CHAPTE 31 Frdy s Lw ACTVE FGUE When conducting plte swings through mgnetic field, the mgnetic force F B opposes its elocity, nd it eentully comes to rest. As the conducting plte enters the field, the eddy currents re counterclockwise. As the plte lees the field, the currents re clockwise. When slots re cut in the conducting plte, the eddy currents re reduced nd the plte swings more freely through the mgnetic field. Piot 1 2 Piot 1 2 F B F B F B F B b eddy current lwys produces mgnetic retrding force F B when the plte enters or lees the field, the swinging plte eentully comes to rest. f slots re cut in the plte s shown in Actie Figure 31.22b, the eddy currents nd the corresponding retrding force re gretly reduced. We cn understnd this reduction in force by relizing tht the cuts in the plte preent the formtion of ny lrge current loops. The brking systems on mny subwy nd rpid-trnsit crs mke use of electromgnetic induction nd eddy currents. An electromgnet ttched to the trin is positioned ner the steel rils. (An electromgnet is essentilly solenoid with n iron core.) The brking ction occurs when lrge current is pssed through the electromgnet. The reltie motion of the mgnet nd rils induces eddy currents in the rils, nd the direction of these currents produces drg force on the moing trin. Becuse the eddy currents decrese stedily in mgnitude s the trin slows down, the brking effect is quite smooth. As sfety mesure, some power tools use eddy currents to stop rpidly spinning bldes once the deice is turned off. Eddy currents re often undesirble becuse they represent trnsformtion of mechnicl energy to internl energy. To reduce this energy loss, conducting prts re often lminted; tht is, they re built up in thin lyers seprted by nonconducting mteril such s lcquer or metl oxide. This lyered structure preents lrge current loops nd effectiely confines the currents to smll loops in indiidul lyers. uch lminted structure is used in trnsformer cores (see ection 33.8) nd motors to minimize eddy currents nd thereby increse the efficiency of these deices. Quick Quiz 31.5 n n equl-rm blnce from the erly 20th century (Fig ), n luminum sheet hngs from one of the rms nd psses between the poles of mgnet, cusing the oscilltions of the blnce to decy rpidly. n the bsence of such mgnetic brking, the oscilltion might continue for long time, nd the experimenter would he to wit to tke reding. Why do the oscilltions decy? () becuse the luminum sheet is ttrcted

21 Objectie Questions 913 to the mgnet (b) becuse currents in the luminum sheet set up mgnetic field tht opposes the oscilltions (c) becuse luminum is prmgnetic John W. Jewett, Jr. Figure (Quick Quiz 31.5) n n old-fshioned equl-rm blnce, n luminum sheet hngs between the poles of mgnet. Concepts nd Principles ummry Frdy s lw of induction sttes tht the emf induced in loop is directly proportionl to the time rte of chnge of mgnetic flux through the loop, or e 52 df B (31.1) where F B 5 r B? da is the mgnetic flux through the loop. When conducting br of length, moes t elocity through mgnetic field B, where B is perpendiculr to the br nd to, the motionl emf induced in the br is e 5 2B, (31.5) Lenz s lw sttes tht the induced current nd induced emf in conductor re in such direction s to set up mgnetic field tht opposes the chnge tht produced them. A generl form of Frdy s lw of induction is C E? d s 52 df B (31.9) where E is the nonconsertie electric field tht is produced by the chnging mgnetic flux. Objectie Questions denotes nswer ilble in tudent olutions Mnul/tudy Guide 1. Figure OQ31.1 is grph of the mgnetic flux through certin coil of wire s function of time during n interl while the rdius of the coil is incresed, the coil is rotted through 1.5 reolutions, nd the externl source of the mgnetic field is turned off, in tht order. nk the emf induced in the coil t the instnts mrked A through E from the lrgest positie lue to the lrgest-mgnitude negtie lue. n your rnking, note ny cses of equlity nd lso ny instnts when the emf is zero. B A B C D E Figure OQ31.1 t

22 914 CHAPTE 31 Frdy s Lw 2. A circulr loop of wire with rdius of 4.0 cm is in uniform mgnetic field of mgnitude T. The plne of the loop is perpendiculr to the direction of the mgnetic field. n time interl of 0.50 s, the mgnetic field chnges to the opposite direction with mgnitude of T. Wht is the mgnitude of the erge emf induced in the loop? () 0.20 V (b) V (c) 5.0 mv (d) 1.0 mv (e) 0.20 mv 3. A rectngulr conducting loop is plced ner long wire crrying current s shown in Figure OQ31.3. f decreses in time, wht cn be sid of the current induced in the loop? () The direction of the current depends on the size of the loop. (b) The current is clockwise. (c) The current is counterclockwise. (d) The current is zero. (e) Nothing cn be sid bout the current in the loop without more informtion. Figure OQ A flt coil of wire is plced in uniform mgnetic field tht is in the y direction. (i) The mgnetic flux through the coil is mximum if the plne of the coil is where? More thn one nswer my be correct. () in the xy plne (b) in the yz plne (c) in the xz plne (d) in ny orienttion, becuse it is constnt (ii) For wht orienttion is the flux zero? Choose from the sme possibilities s in prt (i). 5. A squre, flt loop of wire is pulled t constnt elocity through region of uniform mgnetic field directed perpendiculr to the plne of the loop s shown in Figure OQ31.5. Which of the following sttements re correct? More thn one sttement my be correct. () Current is induced in the loop in the clockwise direction. (b) Current is induced in the loop in the counterclockwise direction. (c) No current is induced in the loop. (d) Chrge seprtion occurs in the loop, with the top edge positie. (e) Chrge seprtion occurs in the loop, with the top edge negtie. current in the loop is counterclockwise. (d) An externl force is required to keep the br moing t constnt speed. (e) No force is required to keep the br moing t constnt speed. B out Figure OQ A br mgnet is held in erticl orienttion boe loop of wire tht lies in the horizontl plne s shown in Figure OQ31.7. The south end of the mgnet is towrd the loop. After the mgnet is dropped, wht is true of the induced current in the loop s iewed from boe? () t is clockwise s the mgnet flls towrd the loop. (b) t is counterclockwise s the mgnet flls towrd the loop. (c) t is clockwise fter the mgnet hs moed through the loop nd moes wy from it. (d) t is lwys clockwise. (e) t is first counterclockwise s the mgnet pproches the loop nd then clockwise fter it hs pssed through the loop. N Figure OQ Wht hppens to the mplitude of the induced emf when the rte of rottion of genertor coil is doubled? () t becomes four times lrger. (b) t becomes two times lrger. (c) t is unchnged. (d) t becomes one-hlf s lrge. (e) t becomes one-fourth s lrge. Figure OQ The br in Figure OQ31.6 moes on rils to the right with elocity, nd uniform, constnt mgnetic field is directed out of the pge. Which of the following sttements re correct? More thn one sttement my be correct. () The induced current in the loop is zero. (b) The induced current in the loop is clockwise. (c) The induced 9. Two coils re plced ner ech other s shown in Figure OQ31.9. The coil on the left is connected to bttery nd switch, nd the coil on the right is connected to resistor. Wht is the direction of the current in the resistor (i) t n instnt immeditely fter the switch is thrown closed, (ii) fter the switch hs been closed for seerl sec- e Figure OQ31.9

23 Conceptul Questions 915 onds, nd (iii) t n instnt fter the switch hs then been thrown open? Choose ech nswer from the possibilities () left, (b) right, or (c) the current is zero. 10. A circuit consists of conducting moble br nd lightbulb connected to two conducting rils s shown in Figure OQ An externl mgnetic field is directed perpendiculr to the plne of the circuit. Which of the following ctions will mke the bulb light up? More thn one sttement my be correct. () The br is moed to the left. (b) The br is moed to the right. (c) The mgnitude of the mgnetic field is incresed. (d) The mgnitude of the mgnetic field is decresed. (e) The br is lifted off the rils. 11. Two rectngulr loops of wire lie in the sme plne s shown in Figure OQ f the current in the outer loop is counterclockwise nd increses with time, wht is true of the current induced in the inner loop? More thn one sttement my be correct. () t is zero. (b) t is clockwise. (c) t is counterclockwise. (d) ts mgnitude depends on the dimensions of the loops. (e) ts direction depends on the dimensions of the loops. Figure OQ31.10 Figure OQ31.11 Conceptul Questions 1. Wht is the difference between mgnetic flux nd mgnetic field? 2. A spcecrft orbiting the Erth hs coil of wire in it. An stronut mesures smll current in the coil, lthough there is no bttery connected to it nd there re no mgnets in the spcecrft. Wht is cusing the current? 3. A circulr loop of wire is locted in uniform nd constnt mgnetic field. Describe how n emf cn be induced in the loop in this sitution. 4. A br mgnet is dropped towrd conducting ring lying on the floor. As the mgnet flls towrd the ring, does it moe s freely flling object? Explin. 5. n hydroelectric dm, how is energy produced tht is then trnsferred out by electricl trnsmission? Tht is, how is the energy of motion of the wter conerted to energy tht is trnsmitted by AC electricity? 6. A piece of luminum is dropped erticlly downwrd between the poles of n electromgnet. Does the mgnetic field ffect the elocity of the luminum? 7. n ection 7.7, we defined consertie nd nonconsertie forces. n Chpter 23, we stted tht n electric chrge cretes n electric field tht produces consertie force. Argue now tht induction cretes n electric field tht produces nonconsertie force. 8. When the switch in Figure CQ31.8 is closed, current is set up in the coil nd the metl ring springs upwrd (Fig. CQ31.8b). Explin this behior. ron core Metl ring denotes nswer ilble in tudent olutions Mnul/tudy Guide Figure CQ31.8 Conceptul Questions 8 nd Assume the bttery in Figure CQ31.8 is replced by n AC source nd the switch is held closed. f held down, the metl ring on top of the solenoid becomes hot. Why? 10. A loop of wire is moing ner long, stright wire crrying constnt current s shown in Figure CQ () Determine the direction of the induced current in the loop s it moes wy from the wire. (b) Wht would be the direction of the induced current in the loop if it were moing towrd the wire? Figure CQ31.10 b Cengge Lerning/Chrles D. Winters