the circuit changes. This change triggers a signal in the earphones with the magnetic field.

Transcription

1 chpter 32 nductnce 32.1 elf-nduction nd nductnce 32.2 Circuits 32.3 Energy in Mgnetic Field 32.4 Mutul nductnce 32.5 Oscilltions in n C Circuit 32.6 The C Circuit n Chpter 31, we sw tht n emf nd current re induced in loop of wire when the mgnetic flux through the re enclosed y the loop chnges with time. This phenomenon of electromgnetic induction hs some prcticl consequences. n this chpter, we first descrie n effect known s self-induction, in which time-vrying current in circuit produces n induced emf opposing the emf tht initilly set up the time-vrying current. elfinduction is the sis of the inductor, n electricl circuit A tresure hunter uses metl detector to serch for uried ojects t ech. At the end of the metl detector is coil of element. We discuss the energy stored in the mgnetic wire tht is prt of circuit. When the coil comes ner metl oject, the inductnce of the coil is ffected nd the current in field of n inductor nd the energy density ssocited the circuit chnges. This chnge triggers signl in the erphones with the mgnetic field. worn y the tresure hunter. We investigte inductnce in this chpter. (Andy yn/tone/getty mges) Next, we study how n emf is induced in coil s result of chnging mgnetic flux produced y second coil, which is the sic principle of mutul induction. Finlly, we exmine the chrcteristics of circuits tht contin inductors, resistors, nd cpcitors in vrious comintions elf-nduction nd nductnce n this chpter, we need to distinguish crefully etween emfs nd currents tht re cused y physicl sources such s tteries nd those tht re induced y chnging mgnetic fields. When we use term (such s emf or current) without n djective, we re descriing the prmeters ssocited with physicl source. We use 927

2 928 CHAPTE 32 nductnce North Wind/North Wind Picture Archives Joseph Henry Americn Physicist ( ) Henry ecme the first director of the mithsonin nstitution nd first president of the Acdemy of Nturl cience. He improved the design of the electromgnet nd constructed one of the first motors. He lso discovered the phenomenon of self-induction, ut he filed to pulish his findings. The unit of inductnce, the henry, is nmed in his honor. the djective induced to descrie those emfs nd currents cused y chnging mgnetic field. Consider circuit consisting of switch, resistor, nd source of emf s shown in Figure The circuit digrm is represented in perspective to show the orienttions of some of the mgnetic field lines due to the current in the circuit. When the switch is thrown to its closed position, the current does not immeditely jump from zero to its mximum vlue e/. Frdy s lw of electromgnetic induction (Eq. 31.1) cn e used to descrie this effect s follows. As the current increses with time, the mgnetic flux through the circuit loop due to this current lso increses with time. This incresing flux cretes n induced emf in the circuit. The direction of the induced emf is such tht it would cuse n induced current in the loop (if the loop did not lredy crry current), which would estlish mgnetic field opposing the chnge in the originl mgnetic field. Therefore, the direction of the induced emf is opposite the direction of the emf of the ttery, which results in grdul rther thn instntneous increse in the current to its finl equilirium vlue. Becuse of the direction of the induced emf, it is lso clled ck emf, similr to tht in motor s discussed in Chpter 31. This effect is clled self-induction ecuse the chnging flux through the circuit nd the resultnt induced emf rise from the circuit itself. The emf e set up in this cse is clled self-induced emf. To otin quntittive description of self-induction, recll from Frdy s lw tht the induced emf is equl to the negtive of the time rte of chnge of the mgnetic flux. The mgnetic flux is proportionl to the mgnetic field, which in turn is proportionl to the current in the circuit. Therefore, self-induced emf is lwys proportionl to the time rte of chnge of the current. For ny loop of wire, we cn write this proportionlity s e 52 d (32.1) where is proportionlity constnt clled the inductnce of the loop tht depends on the geometry of the loop nd other physicl chrcteristics. f we consider closely spced coil of N turns ( toroid or n idel solenoid) crrying current nd contining N turns, Frdy s lw tells us tht e 5 2N df B /. Comining this expression with Eqution 32.1 gives nductnce of n N-turn coil After the switch is closed, the current produces mgnetic flux through the re enclosed y the loop. As the current increses towrd its equilirium vlue, this mgnetic flux chnges in time nd induces n emf in the loop. B Figure 32.1 elf-induction in simple circuit. e 5 NF B (32.2) where it is ssumed the sme mgnetic flux psses through ech turn nd is the inductnce of the entire coil. From Eqution 32.1, we cn lso write the inductnce s the rtio e 52 (32.3) d/ ecll tht resistnce is mesure of the opposition to current ( 5 DV/); in comprison, Eqution 32.3 shows us tht inductnce is mesure of the opposition to chnge in current. The unit of inductnce is the henry (H), which s we cn see from Eqution 32.3 is 1 volt-second per mpere: 1 H 5 1 V? s/a. As shown in Exmple 32.1, the inductnce of coil depends on its geometry. This dependence is nlogous to the cpcitnce of cpcitor depending on the geometry of its pltes s we found in Chpter 26. nductnce clcultions cn e quite difficult to perform for complicted geometries, ut the exmples elow involve simple situtions for which inductnces re esily evluted. Quick Quiz 32.1 A coil with zero resistnce hs its ends leled nd. The potentil t is higher thn t. Which of the following could e consistent with this sitution? () The current is constnt nd is directed from to.

3 32.2 Circuits 929 () The current is constnt nd is directed from to. (c) The current is incresing nd is directed from to. (d) The current is decresing nd is directed from to. (e) The current is incresing nd is directed from to. (f) The current is decresing nd is directed from to. Exmple 32.1 nductnce of olenoid Consider uniformly wound solenoid hving N turns nd length,. Assume, is much longer thn the rdius of the windings nd the core of the solenoid is ir. (A) Find the inductnce of the solenoid. OUTON Conceptulize The mgnetic field lines from ech turn of the solenoid pss through ll the turns, so n induced emf in ech coil opposes chnges in the current. Ctegorize Becuse the solenoid is long, we cn use the results for n idel solenoid otined in Chpter 3. Anlyze Find the mgnetic flux through ech turn of re A in the solenoid, using the expression for the mgnetic field from Eqution 3.17: ustitute this expression into Eqution 32.2: F B 5 BA 5m na 5m N, A 5 NF B 5 m N 2, A (32.4) (B) Clculte the inductnce of the solenoid if it contins 3 turns, its length is 25. cm, nd its cross-sectionl re is 4. cm 2. OUTON ustitute numericl vlues into Eqution 32.4: p T? m/a m m T? m 2 /A mh (C) Clculte the self-induced emf in the solenoid if the current it crries decreses t the rte of 5. A/s. OUTON ustitute d/ A/s nd the nswer to prt (B) into Eqution 32.1: e 52 d H2125. A/s mv Finlize The result for prt (A) shows tht depends on geometry nd is proportionl to the squre of the numer of turns. Becuse N 5 n,, we cn lso express the result in the form 5m 1n,22 A 5m n 2 A, 5m n 2 V (32.5), where V 5 A, is the interior volume of the solenoid Circuits f circuit contins coil such s solenoid, the inductnce of the coil prevents the current in the circuit from incresing or decresing instntneously. A circuit element tht hs lrge inductnce is clled n inductor nd hs the circuit symol. We lwys ssume the inductnce of the reminder of circuit is

4 93 CHAPTE 32 nductnce e 1 When switch 1 is thrown closed, the current increses nd n emf tht opposes the incresing current is induced in the inductor. 2 When the switch 2 is thrown to position, the ttery is no longer prt of the circuit nd the current decreses. ACTVE FGUE 32.2 An circuit. When switch 2 is in position, the ttery is in the circuit. negligile compred with tht of the inductor. Keep in mind, however, tht even circuit without coil hs some inductnce tht cn ffect the circuit s ehvior. Becuse the inductnce of n inductor results in ck emf, n inductor in circuit opposes chnges in the current in tht circuit. The inductor ttempts to keep the current the sme s it ws efore the chnge occurred. f the ttery voltge in the circuit is incresed so tht the current rises, the inductor opposes this chnge nd the rise is not instntneous. f the ttery voltge is decresed, the inductor cuses slow drop in the current rther thn n immedite drop. Therefore, the inductor cuses the circuit to e sluggish s it rects to chnges in the voltge. Consider the circuit shown in Active Figure 32.2, which contins ttery of negligile internl resistnce. This circuit is n circuit ecuse the elements connected to the ttery re resistor nd n inductor. The curved lines on switch 2 suggest this switch cn never e open; it is lwys set to either or. (f the switch is connected to neither nor, ny current in the circuit suddenly stops.) uppose 2 is set to nd switch 1 is open for t, nd then thrown closed t t 5. The current in the circuit egins to increse, nd ck emf (Eq. 32.1) tht opposes the incresing current is induced in the inductor. With this point in mind, let s pply Kirchhoff s loop rule to this circuit, trversing the circuit in the clockwise direction: e 2 2 d 5 (32.6) where is the voltge drop cross the resistor. (Kirchhoff s rules were developed for circuits with stedy currents, ut they cn lso e pplied to circuit in which the current is chnging if we imgine them to represent the circuit t one instnt of time.) Now let s find solution to this differentil eqution, which is similr to tht for the C circuit (see ection 28.4). A mthemticl solution of Eqution 32.6 represents the current in the circuit s function of time. To find this solution, we chnge vriles for convenience, letting x 5 (e/) 2, so dx 5 2d. With these sustitutions, Eqution 32.6 ecomes x 1 dx 5 errnging nd integrting this lst expression gives x dx 3 x x 52 3 ln x x 52 t where x is the vlue of x t time t 5. Tking the ntilogrithm of this result gives x 5 x e 2t/ Becuse 5 t t 5, note from the definition of x tht x 5 e/. Hence, this lst expression is equivlent to e 2 5 e e 2t/ 5 e 11 2 e 2t/ 2 This expression shows how the inductor ffects the current. The current does not increse instntly to its finl equilirium vlue when the switch is closed, ut insted increses ccording to n exponentil function. f the inductnce is removed from the circuit, which corresponds to letting pproch zero, the exponentil term t

5 32.2 Circuits 931 ecomes zero nd there is no time dependence of the current in this cse; the current increses instntneously to its finl equilirium vlue in the sence of the inductnce. We cn lso write this expression s After switch 1 is thrown closed t t, the current increses towrd its mximum vlue e/. e e 2t/t 2 (32.7) where the constnt t is the time constnt of the circuit: t5 (32.8) e e.632 t Physiclly, t is the time intervl required for the current in the circuit to rech (1 2 e 21 ) % of its finl vlue e/. The time constnt is useful prmeter for compring the time responses of vrious circuits. Active Figure 32.3 shows grph of the current versus time in the circuit. Notice tht the equilirium vlue of the current, which occurs s t pproches infinity, is e/. Tht cn e seen y setting d/ equl to zero in Eqution 32.6 nd solving for the current. (At equilirium, the chnge in the current is zero.) Therefore, the current initilly increses very rpidly nd then grdully pproches the equilirium vlue e/ s t pproches infinity. et s lso investigte the time rte of chnge of the current. Tking the first time derivtive of Eqution 32.7 gives d e 5 e 2t/t (32.9) This result shows tht the time rte of chnge of the current is mximum (equl to e/) t t 5 nd flls off exponentilly to zero s t pproches infinity (Fig. 32.4). Now consider the circuit in Active Figure 32.2 gin. uppose switch 2 hs een set t position long enough (nd switch 1 remins closed) to llow the current to rech its equilirium vlue e/. n this sitution, the circuit is descried y the outer loop in Active Figure f 2 is thrown from to, the circuit is now descried y only the right-hnd loop in Active Figure Therefore, the ttery hs een eliminted from the circuit. etting e 5 in Eqution 32.6 gives 1 d 5 t is left s prolem (Prolem 18) to show tht the solution of this differentil eqution is 5 e e 2t/t 5 i e 2t/t (32.1) where e is the emf of the ttery nd i 5 e/ is the initil current t the instnt the switch is thrown to. f the circuit did not contin n inductor, the current would immeditely decrese to zero when the ttery is removed. When the inductor is present, it opposes the decrese in the current nd cuses the current to decrese exponentilly. A grph of the current in the circuit versus time (Active Fig. 32.5) shows tht the current is continuously decresing with time. Quick Quiz 32.2 Consider the circuit in Active Figure 32.2 with 1 open nd 2 t position. witch 1 is now thrown closed. (i) At the instnt it is closed, cross which circuit element is the voltge equl to the emf of the ttery? () the resistor () the inductor (c) oth the inductor nd resistor (ii) After very long time, cross which circuit element is the voltge equl to the emf of the ttery? Choose from mong the sme nswers. ACTVE FGUE 32.3 t Plot of the current versus time for the circuit shown in Active Figure The time constnt t is the time intervl required for to rech 63.2% of its mximum vlue. The time rte of chnge of current is mximum t t, which is the instnt t which switch 1 is thrown closed. e d Figure 32.4 Plot of d/ versus time for the circuit shown in Active Figure The rte decreses exponentilly with time s increses towrd its mximum vlue. At t, the switch is thrown to position nd the current hs its mximum vlue e/. e ACTVE FGUE 32.5 Current versus time for the righthnd loop of the circuit shown in Active Figure For t,, switch 2 is t position. t t t

6 932 CHAPTE 32 nductnce Exmple 32.2 Time Constnt of n Circuit Consider the circuit in Active Figure 32.2 gin. uppose the circuit elements hve the following vlues: e V, 5 6. V, nd 5 3. mh. (A) Find the time constnt of the circuit. OUTON Conceptulize You should understnd the ehvior of this circuit from the discussion in this section. Ctegorize We evlute the results using equtions developed in this section, so this exmple is sustitution prolem. Evlute the time constnt from Eqution 32.8: t H 6. V 5 5. ms (B) witch 2 is t position, nd switch 1 is thrown closed t t 5. Clculte the current in the circuit t t 5 2. ms. OUTON Evlute the current t t 5 2. ms from Eqution 32.7: 5 e 11 2 e 2t/t V 6. V 11 2 e 22. ms/5. ms A 11 2 e A (C) Compre the potentil difference cross the resistor with tht cross the inductor. V (V) 12 OUTON At the instnt the switch is closed, there is no current nd therefore no potentil difference cross the resistor. At this instnt, the ttery voltge ppers entirely cross the inductor in the form of ck emf of 12. V s the inductor tries to mintin the zero-current condition. (The top end of the inductor in Active Fig is t higher electric potentil thn the ottom end.) As time psses, the emf cross the inductor decreses nd the current in the resistor (nd hence the voltge cross it) increses s shown in Figure The sum of the two voltges t ll times is 12. V WHAT F? n Figure 32.6, the voltges cross the resistor nd inductor re equl t 3.4 ms. Wht if you wnted to dely the condition in which the voltges re equl to some lter instnt, such s t 5 1. ms? Which prmeter, or, would require the lest djustment, in terms of percentge chnge, to chieve tht? Answer Figure 32.6 shows tht the voltges re equl when the voltge cross the inductor hs fllen to hlf its originl vlue. Therefore, the time intervl required for the voltges to ecome equl is the hlf-life t 1/2 of the decy. We introduced the hlf-life in the Wht f? section of Exmple 28.1 to descrie the exponentil decy in C circuits, where t 1/ t. 8 4 V V t (ms) 1 Figure 32.6 (Exmple 32.2) The time ehvior of the voltges cross the resistor nd inductor in Active Figure 32.2 given the vlues provided in this exmple. From the desired hlf-life of 1. ms, use the result from Exmple 28.1 to find the time constnt of the circuit: t5 t 1/2 1. ms ms Hold fixed nd find the vlue of tht gives this time constnt: t5 5 t H 14.4 ms V Now hold fixed nd find the pproprite vlue of : t5 5t ms216. V H The chnge in corresponds to 65% decrese compred with the initil resistnce. The chnge in represents 188% increse in inductnce! Therefore, much smller percentge djustment in cn chieve the desired effect thn would n djustment in.

7 32.3 Energy in Mgnetic Field Energy in Mgnetic Field A ttery in circuit contining n inductor must provide more energy thn in circuit without the inductor. Prt of the energy supplied y the ttery ppers s internl energy in the resistnce in the circuit, nd the remining energy is stored in the mgnetic field of the inductor. Multiplying ech term in Eqution 32.6 y nd rerrnging the expression gives e d (32.11) ecognizing e s the rte t which energy is supplied y the ttery nd 2 s the rte t which energy is delivered to the resistor, we see tht (d/) must represent the rte t which energy is eing stored in the inductor. f U is the energy stored in the inductor t ny time, we cn write the rte du/ t which energy is stored s du d 5 To find the totl energy stored in the inductor t ny instnt, let s rewrite this expression s du 5 d nd integrte: U 5 3 du 5 3 d 5 3 d U (32.12) where is constnt nd hs een removed from the integrl. Eqution represents the energy stored in the mgnetic field of the inductor when the current is. t is similr in form to Eqution for the energy stored in the electric field of cpcitor, U 5 1 2C(DV) 2. n either cse, energy is required to estlish field. We cn lso determine the energy density of mgnetic field. For simplicity, consider solenoid whose inductnce is given y Eqution 32.5: 5 m n 2 V The mgnetic field of solenoid is given y Eqution 3.17: B 5 m n ustituting the expression for nd 5 B/m n into Eqution gives U m n 2 V B 2 m n 5 B 2 V (32.13) 2m The mgnetic energy density, or the energy stored per unit volume in the mgnetic field of the inductor, is u B 5 U V 5 B 2 (32.14) 2m Although this expression ws derived for the specil cse of solenoid, it is vlid for ny region of spce in which mgnetic field exists. Eqution is similr in form to Eqution for the energy per unit volume stored in n electric field, u E 5 1 2P E 2. n oth cses, the energy density is proportionl to the squre of the field mgnitude. Pitfll Prevention 32.1 Cpcitors, esistors, nd nductors tore Energy Differently Different energy-storge mechnisms re t work in cpcitors, inductors, nd resistors. A chrged cpcitor stores energy s electricl potentil energy. An inductor stores energy s wht we could cll mgnetic potentil energy when it crries current. Energy delivered to resistor is trnsformed to internl energy. Energy stored in n inductor Mgnetic energy density Quick Quiz 32.3 You re performing n experiment tht requires the highest-possile mgnetic energy density in the interior of very long current-crrying solenoid. Which of the following djustments increses the energy density? (More thn one choice my e correct.) () incresing the numer of turns per unit length on the solenoid () incresing the crosssectionl re of the solenoid (c) incresing only the length of the solenoid while keeping the numer of turns per unit length fixed (d) incresing the current in the solenoid

8 934 CHAPTE 32 nductnce Exmple 32.3 Wht Hppens to the Energy in the nductor? Consider once gin the circuit shown in Active Figure 32.2, with switch 2 t position nd the current hving reched its stedy-stte vlue. When 2 is thrown to position, the current in the right-hnd loop decys exponentilly with time ccording to the expression 5 i e 2t/t, where i 5 e/ is the initil current in the circuit nd t 5 / is the time constnt. how tht ll the energy initilly stored in the mgnetic field of the inductor ppers s internl energy in the resistor s the current decys to zero. OUTON Conceptulize Before 2 is thrown to, energy is eing delivered t constnt rte to the resistor from the ttery nd energy is stored in the mgnetic field of the inductor. After t 5, when 2 is thrown to, the ttery cn no longer provide energy nd energy is delivered to the resistor only from the inductor. Ctegorize We model the right-hnd loop of the circuit s n isolted system so tht energy is trnsferred etween components of the system ut does not leve the system. Anlyze The energy in the mgnetic field of the inductor t ny time is U. The rte du/ t which energy leves the inductor nd is delivered to the resistor is equl to 2, where is the instntneous current. ustitute the current given y Eqution 32.1 into du/ 5 2 : olve for du nd integrte this expression over the limits t 5 to t `: The vlue of the definite integrl cn e shown to e /2 (see Prolem 32). Use this result to evlute U : du i e 2t/ i 2 e 22t/ ` ` U i e 22t/ 5 2 i 3 e 22t/ U 5 i i 2 Finlize This result is equl to the initil energy stored in the mgnetic field of the inductor, given y Eqution 32.12, s we set out to prove. Exmple 32.4 The Coxil Cle Coxil cles re often used to connect electricl devices, such s your video system, nd in receiving signls in television cle systems. Model long coxil cle s thin, cylindricl conducting shell of rdius concentric with solid cylinder of rdius s in Figure The conductors crry the sme current in opposite directions. Clculte the inductnce of length, of this cle. dr OUTON Conceptulize Consider Figure Although we do not hve visile coil in this geometry, imgine thin, rdil slice of the coxil cle such s the light gold rectngle in Figure f the inner nd outer conductors re connected t the ends of the cle (ove nd elow the figure), this slice represents one lrge conducting loop. The current in the loop sets up mgnetic field etween the inner nd outer conductors tht psses through this loop. f the current chnges, the mgnetic field chnges nd the induced emf opposes the originl chnge in the current in the conductors. Ctegorize We ctegorize this sitution s one in which we must return to the fundmentl definition of inductnce, Eqution B Figure 32.7 (Exmple 32.4) ection of long coxil cle. The inner nd outer conductors crry equl currents in opposite directions. r Anlyze We must find the mgnetic flux through the light gold rectngle in Figure Ampère s lw (see ection 3.3) tells us tht the mgnetic field in the region etween the conductors is continued

9 32.4 Mutul nductnce cont. due to the inner conductor lone nd tht its mgnitude is B 5 m /2pr, where r is mesured from the common center of the cylinders. A smple circulr field line is shown in Figure The mgnetic field is zero outside the outer shell ecuse the net current pssing through the re enclosed y circulr pth surrounding the cle is zero; hence, from Ampère s lw, r B? d s 5. The mgnetic field is perpendiculr to the light gold rectngle of length, nd wih 2, the cross section of interest. Becuse the mgnetic field vries with rdil position cross this rectngle, we must use clculus to find the totl mgnetic flux. Divide the light gold rectngle into strips of wih dr such s the drker strip in Figure Evlute the mgnetic flux through such strip: ustitute for the mgnetic field nd integrte over the entire light gold rectngle: Use Eqution 32.2 to find the inductnce of the cle: F B 5 3 B da 5 3 B, dr F B F B m 2pr, dr 5 m, 2p 3 5 m, 2p ln dr r 5 m, 2p ln Finlize The inductnce increses if, increses, if increses, or if decreses. This result is consistent with our conceptuliztion: ny of these chnges increses the size of the loop represented y our rdil slice nd through which the mgnetic field psses, incresing the inductnce Mutul nductnce Very often, the mgnetic flux through the re enclosed y circuit vries with time ecuse of time-vrying currents in nery circuits. This condition induces n emf through process known s mutul induction, so nmed ecuse it depends on the interction of two circuits. Consider the two closely wound coils of wire shown in cross-sectionl view in Figure The current 1 in coil 1, which hs N 1 turns, cretes mgnetic field. ome of the mgnetic field lines pss through coil 2, which hs N 2 turns. The mgnetic flux cused y the current in coil 1 nd pssing through coil 2 is represented y F 12. n nlogy to Eqution 32.2, we cn identify the mutul inductnce M 12 of coil 2 with respect to coil 1: M 12 5 N 2F 12 (32.15) 1 Mutul inductnce depends on the geometry of oth circuits nd on their orienttion with respect to ech other. As the circuit seprtion distnce increses, the mutul inductnce decreses ecuse the flux linking the circuits decreses. f the current 1 vries with time, we see from Frdy s lw nd Eqution tht the emf induced y coil 1 in coil 2 is A current in coil 1 sets up mgnetic field, nd some of the mgnetic field lines pss through coil 2. Coil 1 Coil 2 e 2 52N 2 df 12 52N d 2 M M d 1 N 12 2 (32.16) n the preceding discussion, it ws ssumed the current is in coil 1. et s lso imgine current 2 in coil 2. The preceding discussion cn e repeted to show tht there is mutul inductnce M 21. f the current 2 vries with time, the emf induced y coil 2 in coil 1 is N 1 1 N 2 2 e 1 52M 21 d 2 (32.17) Figure 32.8 A cross-sectionl view of two djcent coils.

10 936 CHAPTE 32 nductnce n mutul induction, the emf induced in one coil is lwys proportionl to the rte t which the current in the other coil is chnging. Although the proportionlity constnts M 12 nd M 21 hve een treted seprtely, it cn e shown tht they re equl. Therefore, with M 12 5 M 21 5 M, Equtions nd ecome e 2 52M d 1 nd e 1 52M d 2 These two equtions re similr in form to Eqution 32.1 for the self-induced emf e 5 2 (d/). The unit of mutul inductnce is the henry. Quick Quiz 32.4 n Figure 32.8, coil 1 is moved closer to coil 2, with the orienttion of oth coils remining fixed. Becuse of this movement, the mutul induction of the two coils () increses, () decreses, or (c) is unffected. Exmple 32.5 Wireless Bttery Chrger An electric toothrush hs se designed to hold the toothrush hndle when not in use. As shown in Figure 32.9, the hndle hs cylindricl hole tht fits loosely over mtching cylinder on the se. When the hndle is plced on the se, chnging current in solenoid inside the se cylinder induces current in coil inside the hndle. This induced current chrges the ttery in the hndle. We cn model the se s solenoid of length, with N B turns (Fig. 32.9), crrying current, nd hving cross-sectionl re A. The hndle coil contins N H turns nd completely surrounds the se coil. Find the mutul inductnce of the system. OUTON Conceptulize Be sure you cn identify the two coils in the sitution nd understnd tht chnging current in one coil induces current in the second coil. Ctegorize We will determine the result using concepts discussed in this section, so we ctegorize this exmple s sustitution prolem. Use Eqution 3.17 to express the mgnetic field in the interior of the se solenoid: Find the mutul inductnce, noting tht the mgnetic flux F BH through the hndle s coil cused y the mgnetic field of the se coil is BA:. y Brun GmH, Kronerg Coil 1 (se) Coil 2 (hndle) Figure 32.9 (Exmple 32.5) () This electric toothrush uses the mutul induction of solenoids s prt of its tterychrging system. () A coil of N H turns wrpped round the center of solenoid of N B turns. B 5m N B, M 5 N HF BH 5 N H BA 5 m N BN H, Wireless chrging is used in numer of other cordless devices. One significnt exmple is the inductive chrging used y some mnufcturers of electric crs tht voids direct metl-to-metl contct etween the cr nd the chrging pprtus. A N B N H 32.5 Oscilltions in n C Circuit When cpcitor is connected to n inductor s illustrted in Figure 32.1, the comintion is n C circuit. f the cpcitor is initilly chrged nd the switch is

11 32.5 Oscilltions in n C Circuit 937 then closed, oth the current in the circuit nd the chrge on the cpcitor oscillte etween mximum positive nd negtive vlues. f the resistnce of the circuit is zero, no energy is trnsformed to internl energy. n the following nlysis, the resistnce in the circuit is neglected. We lso ssume n idelized sitution in which energy is not rdited wy from the circuit. This rdition mechnism is discussed in Chpter 34. Assume the cpcitor hs n initil chrge Q mx (the mximum chrge) nd the switch is open for t, nd then closed t t 5. et s investigte wht hppens from n energy viewpoint. When the cpcitor is fully chrged, the energy U in the circuit is stored in the cpcitor s electric field nd is equl to Q 2 mx /2C (Eq ). At this time, the current in the circuit is zero; therefore, no energy is stored in the inductor. After the switch is closed, the rte t which chrges leve or enter the cpcitor pltes (which is lso the rte t which the chrge on the cpcitor chnges) is equl to the current in the circuit. After the switch is closed nd the cpcitor egins to dischrge, the energy stored in its electric field decreses. The cpcitor s dischrge represents current in the circuit, nd some energy is now stored in the mgnetic field of the inductor. Therefore, energy is trnsferred from the electric field of the cpcitor to the mgnetic field of the inductor. When the cpcitor is fully dischrged, it stores no energy. At this time, the current reches its mximum vlue nd ll the energy in the circuit is stored in the inductor. The current continues in the sme direction, decresing in mgnitude, with the cpcitor eventully ecoming fully chrged gin ut with the polrity of its pltes now opposite the initil polrity. This process is followed y nother dischrge until the circuit returns to its originl stte of mximum chrge Q mx nd the plte polrity shown in Figure The energy continues to oscillte etween inductor nd cpcitor. The oscilltions of the C circuit re n electromgnetic nlog to the mechnicl oscilltions of the lock spring system studied in Chpter 15. Much of wht ws discussed there is pplicle to C oscilltions. For exmple, we investigted the effect of driving mechnicl oscilltor with n externl force, which leds to the phenomenon of resonnce. The sme phenomenon is oserved in the C circuit. (ee ection 33.7.) A representtion of the energy trnsfer in n C circuit is shown in Active Figure on pge 938. As mentioned, the ehvior of the circuit is nlogous to tht of the oscillting lock spring system studied in Chpter 15. The potentil energy 1 2kx 2 stored in stretched spring is nlogous to the potentil energy Q 2 mx /2C stored in the cpcitor. The kinetic energy 1 2mv 2 of the moving lock is nlogous to the mgnetic energy stored in the inductor, which requires the presence of moving chrges. n Active Figure 32.11, ll the energy is stored s electric potentil energy in the cpcitor t t 5 (ecuse 5 ), just s ll the energy in lock spring system is initilly stored s potentil energy in the spring if it is stretched nd relesed t t 5. n Active Figure 32.11, ll the energy is stored s mgnetic energy mx in the inductor, where mx is the mximum current. Active Figures 32.11c nd 32.11d show susequent qurter-cycle situtions in which the energy is ll electric or ll mgnetic. At intermedite points, prt of the energy is electric nd prt is mgnetic. Consider some ritrry time t fter the switch is closed so tht the cpcitor hs chrge Q, Q mx nd the current is, mx. At this time, oth circuit elements store energy, s shown in Figure 32.11e, ut the sum of the two energies must equl the totl initil energy U stored in the fully chrged cpcitor t t 5 : U 5 U C 1 U 5 Q 2 2C (32.18) Becuse we hve ssumed the circuit resistnce to e zero nd we ignore electromgnetic rdition, no energy is trnsformed to internl energy nd none is trnsferred out of the system of the circuit. Therefore, the totl energy of the system must remin constnt in time. We descrie the constnt energy of the system mthemticlly C Q mx Figure 32.1 A simple C circuit. The cpcitor hs n initil chrge Q mx, nd the switch is open for t, nd then closed t t 5. Totl energy stored in n C circuit

12 938 CHAPTE 32 nductnce +Q = mx E Q mx C % 1 5 Energy in inductor Energy in cpcitor Totl energy k v m mx C Q B % 1 5 Energy in inductor Energy in cpcitor Totl energy k m v mx c Q mx E Q mx C % 1 5 Energy in inductor Energy in cpcitor Totl energy k v m d mx Q C B % 1 5 Energy in inductor Energy in cpcitor Totl energy k v mx m e Q Q C E B % 1 5 Energy in inductor Energy in cpcitor Totl energy k A A x m v x ACTVE FGUE Energy trnsfer in resistnceless, nonrditing C circuit. The cpcitor hs chrge Q mx t t 5, the instnt t which the switch in Figure 32.1 is closed. The mechnicl nlog of this circuit is lock spring system. () through (d) At these specil instnts, ll of the energy in the circuit resides in one of the circuit elements. (e) At n ritrry instnt, the energy is split etween the cpcitor nd the inductor. y setting du/ 5. Therefore, y differentiting Eqution with respect to time while noting tht Q nd vry with time gives du 5 d 2 Q 2C Q C dq 1 d 5 (32.19) We cn reduce this result to differentil eqution in one vrile y rememering tht the current in the circuit is equl to the rte t which the chrge on the cpcitor chnges: 5 dq/. t then follows tht d/ 5 d 2 Q/ 2. ustitution of these reltionships into Eqution gives Q C 1 d 2 Q 5 2 d 2 Q C Q (32.2) et s solve for Q y noting tht this expression is of the sme form s the nlogous Equtions 15.3 nd 15.5 for lock spring system: d 2 x 52k 2 m x 5 2v2 x

13 32.5 Oscilltions in n C Circuit 939 where k is the spring constnt, m is the mss of the lock, nd v 5!k/m. The solution of this mechnicl eqution hs the generl form (Eq. 15.6): x 5 A cos (vt 1 f) where A is the mplitude of the simple hrmonic motion (the mximum vlue of x), v is the ngulr frequency of this motion, nd f is the phse constnt; the vlues of A nd f depend on the initil conditions. Becuse Eqution 32.2 is of the sme mthemticl form s the differentil eqution of the simple hrmonic oscilltor, it hs the solution Q 5 Q mx cos (vt 1 f) (32.21) where Q mx is the mximum chrge of the cpcitor nd the ngulr frequency v is v5 1 "C (32.22) Note tht the ngulr frequency of the oscilltions depends solely on the inductnce nd cpcitnce of the circuit. Eqution gives the nturl frequency of oscilltion of the C circuit. Becuse Q vries sinusoidlly with time, the current in the circuit lso vries sinusoidlly. We cn show tht y differentiting Eqution with respect to time: 5 dq 5 2vQ mx sin 1vt 1f2 (32.23) To determine the vlue of the phse ngle f, let s exmine the initil conditions, which in our sitution require tht t t 5, 5, nd Q 5 Q mx. etting 5 t t 5 in Eqution gives 5 2vQ mx sin f which shows tht f 5. This vlue for f lso is consistent with Eqution nd the condition tht Q 5 Q mx t t 5. Therefore, in our cse, the expressions for Q nd re Q 5 Q mx cos vt (32.24) 5 2vQ mx sin vt 5 2 mx sin vt (32.25) Grphs of Q versus t nd versus t re shown in Active Figure The chrge on the cpcitor oscilltes etween the extreme vlues Q mx nd 2Q mx, nd the current oscilltes etween mx nd 2 mx. Furthermore, the current is 98 out of phse with the chrge. Tht is, when the chrge is mximum, the current is zero, nd when the chrge is zero, the current hs its mximum vlue. et s return to the energy discussion of the C circuit. ustituting Equtions nd in Eqution 32.18, we find tht the totl energy is U 5 U C 1 U 5 Q 2 mx 2C cos2 vt mx sin 2 vt (32.26) This expression contins ll the fetures descried qulittively t the eginning of this section. t shows tht the energy of the C circuit continuously oscilltes etween energy stored in the cpcitor s electric field nd energy stored in the inductor s mgnetic field. When the energy stored in the cpcitor hs its mximum vlue Q 2 mx /2C, the energy stored in the inductor is zero. When the energy stored in the inductor hs its mximum vlue mx, the energy stored in the cpcitor is zero. Plots of the time vritions of U C nd U re shown in Figure on pge 94. The sum U C 1 U is constnt nd is equl to the totl energy Q 2 mx /2C, or mx. Anlyticl verifiction is strightforwrd. The mplitudes of the two grphs in Figure must e equl ecuse the mximum energy stored in the cpcitor Chrge s function of time for n idel C circuit Angulr frequency of oscilltion in n C circuit Current s function of time for n idel C current The chrge Q nd the current re 9 out of phse with ech other. Q mx Q mx T T 3T 2T 2 2 ACTVE FGUE Grphs of chrge versus time nd current versus time for resistnceless, nonrditing C circuit. t t

14 94 CHAPTE 32 nductnce The sum of the two curves is constnt nd is equl to the totl energy stored in the circuit. U U C T 2 T 3T 2 Q 2 mx 2C t 2 mx 2 t 2T Figure Plots of U C versus t nd U versus t for resistnceless, nonrditing C circuit. Exmple 32.6 (when 5 ) must equl the mximum energy stored in the inductor (when Q 5 ). This equlity is expressed mthemticlly s Q 2 mx 2C 5 2 mx 2 Using this expression in Eqution for the totl energy gives U 5 Q 2 mx 2C 1cos2 vt 1 sin 2 vt2 5 Q 2 mx (32.27) 2C ecuse cos 2 vt 1 sin 2 vt 5 1. n our idelized sitution, the oscilltions in the circuit persist indefinitely; the totl energy U of the circuit, however, remins constnt only if energy trnsfers nd trnsformtions re neglected. n ctul circuits, there is lwys some resistnce nd some energy is therefore trnsformed to internl energy. We mentioned t the eginning of this section tht we re lso ignoring rdition from the circuit. n relity, rdition is inevitle in this type of circuit, nd the totl energy in the circuit continuously decreses s result of this process. Quick Quiz 32.5 (i) At n instnt of time during the oscilltions of n C circuit, the current is t its mximum vlue. At this instnt, wht hppens to the voltge cross the cpcitor? () t is different from tht cross the inductor. () t is zero. (c) t hs its mximum vlue. (d) t is impossile to determine. (ii) At n instnt of time during the oscilltions of n C circuit, the current is momentrily zero. From the sme choices, descrie the voltge cross the cpcitor t this instnt. Oscilltions in n C Circuit n Figure 32.14, the ttery hs n emf of 12. V, the inductnce is 2.81 mh, nd the cpcitnce is 9. pf. The switch hs een set to position for long time so tht the cpcitor is chrged. The switch is then thrown to position, removing the ttery from the circuit nd connecting the cpcitor directly cross the inductor. (A) Find the frequency of oscilltion of the circuit. e C OUTON Conceptulize When the switch is thrown to position, the ctive prt of the circuit is the right-hnd loop, which is n C circuit. Ctegorize We use equtions developed in this section, so we ctegorize this exmple s sustitution prolem. Figure (Exmple 32.6) First the cpcitor is fully chrged with the switch set to position. Then the switch is thrown to position, nd the ttery is no longer in the circuit. Use Eqution to find the frequency: f 5 v 2p 5 1 2p "C 1 ustitute numericl vlues: f 5 2p H F /2 16 Hz (B) Wht re the mximum vlues of chrge on the cpcitor nd current in the circuit? OUTON Find the initil chrge on the cpcitor, which equls the mximum chrge: Use Eqution to find the mximum current from the mximum chrge: Q mx 5 C DV 5 ( F)(12. V) C mx 5 vq mx 5 2pfQ mx 5 (2p s 21 )( C) A

15 32.6 The C Circuit The C Circuit et s now turn our ttention to more relistic circuit consisting of resistor, n inductor, nd cpcitor connected in series s shown in Active Figure We ssume the resistnce of the resistor represents ll the resistnce in the circuit. uppose the switch is t position so tht the cpcitor hs n initil chrge Q mx. The switch is now thrown to position. At this instnt, the totl energy stored in the cpcitor nd inductor t ny time is given y Eqution This totl energy, however, is no longer constnt s it ws in the C circuit ecuse the resistor cuses trnsformtion to internl energy. (We continue to ignore electromgnetic rdition from the circuit in this discussion.) Becuse the rte of energy trnsformtion to internl energy within resistor is 2, du 52 2 where the negtive sign signifies tht the energy U of the circuit is decresing in time. ustituting this result into Eqution gives d 1 Q dq C 522 (32.28) To convert this eqution into form tht llows us to compre the electricl oscilltions with their mechnicl nlog, we first use 5 dq/ nd move ll terms to the left-hnd side to otin Now divide through y : d 2 Q Q C 5 d 2 Q Q C 5 d 2 Q 2 1 dq 1 Q C 5 (32.29) The C circuit is nlogous to the dmped hrmonic oscilltor discussed in ection 15.6 nd illustrted in Figure The eqution of motion for dmped lock spring system is, from Eqution 15.31, m d 2 x 1 dx 1 kx 5 (32.3) 2 Compring Equtions nd 32.3, we see tht Q corresponds to the position x of the lock t ny instnt, to the mss m of the lock, to the dmping coefficient, nd C to 1/k, where k is the force constnt of the spring. These nd other reltionships re listed in Tle 32.1 on pge 942. Becuse the nlyticl solution of Eqution is cumersome, we give only qulittive description of the circuit ehvior. n the simplest cse, when 5, Eqution reduces to tht of simple C circuit s expected, nd the chrge nd the current oscillte sinusoidlly in time. This sitution is equivlent to removing ll dmping in the mechnicl oscilltor. When is smll, sitution tht is nlogous to light dmping in the mechnicl oscilltor, the solution of Eqution is Q 5 Q mx e 2t/2 cos v d t (32.31) where v d, the ngulr frequency t which the circuit oscilltes, is given y v d 5 c 1 C 2 2 1/2 2 d (32.32) e The switch is set first to position, nd the cpcitor is chrged. The switch is then thrown to position. ACTVE FGUE A series C circuit. C

16 942 CHAPTE 32 nductnce TABE 32.1 Anlogies Between Electricl nd Mechnicl ystems One-Dimensionl Electric Circuit Mechnicl ystem Chrge Q 4 x Position Current 4 v x Velocity Potentil difference DV 4 F x Force esistnce 4 Viscous dmping coefficient Cpcitnce C 4 1/k (k 5 spring constnt) nductnce 4 m Mss Current 5 time derivtive of chrge te of chnge of current 5 second time derivtive of chrge 5 dq 4 v x 5 dx d 5 d 2 Q 4 2 x 5 dv x 5 d 2 x 2 Velocity 5 time derivtive of position Accelertion 5 second time derivtive of position Energy in inductor U K 5 1 2mv 2 Kinetic energy of moving oject Energy in cpcitor U C 5 Q C 4 U kx 2 Potentil energy stored in spring te of energy loss due 2 4 v 2 te of energy loss due to resistnce to friction C circuit d 2 Q 2 1 dq 1 Q C 5 4 m d 2 x 2 1 dx 1 kx 5 Dmped oject on spring Tht is, the vlue of the chrge on the cpcitor undergoes dmped hrmonic oscilltion in nlogy with lock spring system moving in viscous medium. Eqution shows tht when,,!4/c (so tht the second term in the rckets is much smller thn the first), the frequency v d of the dmped oscilltor is close to tht of the undmped oscilltor, 1/!C. Becuse 5 dq/, it follows tht the current lso undergoes dmped hrmonic oscilltion. A plot of the chrge versus time for the dmped oscilltor is shown in Active Figure 32.16, nd n oscilloscope trce for rel C circuit is shown in Active Figure The mximum vlue of Q decreses fter ech oscilltion, just s the mplitude of dmped lock spring system decreses in time. For lrger vlues of, the oscilltions dmp out more rpidly; in fct, there exists criticl resistnce vlue c 5!4/C ove which no oscilltions occur. A system with 5 c is sid to e criticlly dmped. When exceeds c, the system is sid to e overdmped. The Q-versus-t curve represents plot of Eqution Q Q mx ACTVE FGUE () Chrge versus time for dmped C circuit. The chrge decys in this wy when,!4/c. () Oscilloscope pttern showing the decy in the oscilltions of n C circuit. t Courtesy of J. udmin

17 ummry 943 Concepts nd Principles ummry When the current in loop of wire chnges with time, n emf is induced in the loop ccording to Frdy s lw. The self-induced emf is e 52 d (32.1) where is the inductnce of the loop. nductnce is mesure of how much opposition loop offers to chnge in the current in the loop. nductnce hs the unit of henry (H), where 1 H 5 1 V? s/a. The inductnce of ny coil is 5 NF B (32.2) where N is the totl numer of turns nd F B is the mgnetic flux through the coil. The inductnce of device depends on its geometry. For exmple, the inductnce of n ir-core solenoid is 5m N 2, A (32.4) where, is the length of the solenoid nd A is the cross-sectionl re. f resistor nd inductor re connected in series to ttery of emf e t time t 5, the current in the circuit vries in time ccording to the expression 5 e 11 2 e 2t/t 2 (32.7) where t 5 / is the time constnt of the circuit. f we replce the ttery in the circuit y resistnceless wire, the current decys exponentilly with time ccording to the expression where e/ is the initil current in the circuit. 5 e e 2t/t (32.1) The energy stored in the mgnetic field of n inductor crrying current is U (32.12) This energy is the mgnetic counterprt to the energy stored in the electric field of chrged cpcitor. The energy density t point where the mgnetic field is B is u B 5 B 2 2m (32.14) The mutul inductnce of system of two coils is M 12 5 N 2F 12 5 M 21 5 N 1F 21 5 M (32.15) 1 2 This mutul inductnce llows us to relte the induced emf in coil to the chnging source current in nery coil using the reltionships e 2 52M 12 d 1 nd e 1 52M 21 d 2 (32.16, 32.17) n n C circuit tht hs zero resistnce nd does not rdite electromgneticlly (n ideliztion), the vlues of the chrge on the cpcitor nd the current in the circuit vry sinusoidlly in time t n ngulr frequency given y v5 1 (32.22) "C The energy in n C circuit continuously trnsfers etween energy stored in the cpcitor nd energy stored in the inductor. n n C circuit with smll resistnce, the chrge on the cpcitor vries with time ccording to where Q 5 Q mx e 2t/2 cos v d t (32.31) v d 5 c 1 C 2 2 1/2 2 d (32.32)

18 944 CHAPTE 32 nductnce Ojective Questions 1. nitilly, n inductor with no resistnce crries constnt current. Then the current is rought to new constnt vlue twice s lrge. After this chnge, when the current is constnt t its higher vlue, wht hs hppened to the emf in the inductor? () t is lrger thn efore the chnge y fctor of 4. () t is lrger y fctor of 2. (c) t hs the sme nonzero vlue. (d) t continues to e zero. (e) t hs decresed. 2. A long, fine wire is wound into coil with inductnce 5 mh. The coil is connected cross the terminls of ttery, nd the current is mesured few seconds fter the connection is mde. The wire is unwound nd wound gin into different coil with 5 1 mh. This second coil is connected cross the sme ttery, nd the current is mesured in the sme wy. Compred with the current in the first coil, is the current in the second coil () four times s lrge, () twice s lrge, (c) unchnged, (d) hlf s lrge, or (e) one-fourth s lrge? 3. Two solenoids, A nd B, re wound using equl lengths of the sme kind of wire. The length of the xis of ech solenoid is lrge compred with its dimeter. The xil length of A is twice s lrge s tht of B, nd A hs twice s mny turns s B. Wht is the rtio of the inductnce of solenoid A to tht of solenoid B? () 4 () 2 (c) 1 (d) 1 2 (e) n Figure OQ32.4, the switch is left in position for long time intervl nd is then quickly thrown to position. nk the mgnitudes of the voltges cross the four circuit elements short time therefter from the lrgest to the smllest. 12. V denotes nswer ville in tudent olutions Mnul/tudy Guide H 5. A solenoidl inductor for printed circuit ord is eing redesigned. To sve weight, the numer of turns is reduced y one-hlf, with the geometric dimensions kept the sme. By how much must the current chnge if the energy stored in the inductor is to remin the sme? () t must e four times lrger. () t must e two times lrger. (c) t should e left the sme. (d) t should e one-hlf s lrge. (e) No chnge in the current cn compenste for the reduction in the numer of turns. 6. f the current in n inductor is douled, y wht fctor is the stored energy multiplied? () 4 () 2 (c) 1 (d) 1 2 (e) The centers of two circulr loops re seprted y fixed distnce. (i) For wht reltive orienttion of the loops is their mutul inductnce mximum? () coxil nd lying in prllel plnes () lying in the sme plne (c) lying in perpendiculr plnes, with the center of one on the xis of the other (d) The orienttion mkes no difference. (ii) For wht reltive orienttion is their mutul inductnce minimum? Choose from the sme possiilities s in prt (i). Figure OQ32.4 Conceptul Questions 1. The current in circuit contining coil, resistor, nd ttery hs reched constnt vlue. () Does the coil hve n inductnce? () Does the coil ffect the vlue of the current? 2. () Wht prmeters ffect the inductnce of coil? () Does the inductnce of coil depend on the current in the coil? 3. A switch controls the current in circuit tht hs lrge inductnce. The electric rc t the switch (Fig. CQ32.3) denotes nswer ville in tudent olutions Mnul/tudy Guide cn melt nd oxidize the contct surfces, resulting in high resistivity of the contcts nd eventul destruction of the switch. s sprk more likely to e produced t the switch when the switch is eing closed, when it is eing opened, or does it not mtter? 4. Consider the four circuits shown in Figure CQ32.4, ech consisting of ttery, switch, lightul, resistor, nd Alexndr Héder c d Figure CQ32.3 Figure CQ32.4

19 Prolems 945 either cpcitor or n inductor. Assume the cpcitor hs lrge cpcitnce nd the inductor hs lrge inductnce ut no resistnce. The lightul hs high efficiency, glowing whenever it crries electric current. (i) Descrie wht the lightul does in ech of circuits () through (d) fter the switch is thrown closed. (ii) Descrie wht the lightul does in ech of circuits () through (d) when, hving een closed for long time intervl, the switch is opened. 5. Consider this thesis: Joseph Henry, Americ s first professionl physicist, cused sic chnge in the humn view of the Universe when he discovered self-induction during school vction t the Alny Acdemy out 183. Before tht time, one could think of the Universe s composed of only one thing: mtter. The energy tht temporrily mintins the current fter ttery is removed from coil, on the other hnd, is not energy tht elongs to ny chunk of mtter. t is energy in the mssless mgnetic field surrounding the coil. With Henry s discovery, Nture forced us to dmit tht the Universe consists of fields s well s mtter. () Argue for or ginst the sttement. () n your view, wht mkes up the Universe? 6. Discuss the similrities etween the energy stored in the electric field of chrged cpcitor nd the energy stored in the mgnetic field of C e current-crrying coil. 7. The open switch in Figure CQ32.7 is thrown closed t t 5. Before the Figure CQ32.7 switch is closed, the cpcitor is unchrged nd ll currents re zero. Determine the currents in, C, nd, the emf cross, nd the potentil differences cross C nd () t the instnt fter the switch is closed nd () long fter it is closed. 8. After the switch is closed in the C circuit shown in Figure CQ32.8, the chrge on the cpcitor is sometimes zero, ut t such instnts the current in the circuit is not zero. How is this ehvior possile? C Q mx Figure CQ32.8 Conceptul Question 8 nd Prolems 46, 48, nd How cn you tell whether n C circuit is overdmped or underdmped? 1. () Cn n oject exert force on itself? () When coil induces n emf in itself, does it exert force on itself? Prolems The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. denotes strightforwrd prolem; 2. denotes intermedite prolem; 3. denotes chllenging prolem 1. full solution ville in the tudent olutions Mnul/tudy Guide 1. denotes prolems most often ssigned in Enhnced WeAssign; these provide students with trgeted feedck nd either Mster t tutoril or Wtch t solution video. denotes sking for quntittive nd conceptul resoning denotes symolic resoning prolem denotes Mster t tutoril ville in Enhnced WeAssign denotes guided prolem shded denotes pired prolems tht develop resoning with symols nd numericl vlues ection 32.1 elf-nduction nd nductnce 1. The current in coil chnges from 3.5 A to 2. A in the sme direction in.5 s. f the verge emf induced in the coil is 12. mv, wht is the inductnce of the coil? 2. A technicin wrps wire round tue of length 36. cm hving dimeter of 8. cm. When the windings re evenly spred over the full length of the tue, the result is solenoid contining 58 turns of wire. () Find the inductnce of this solenoid. () f the current in this solenoid increses t the rte of 4. A/s, find the self-induced emf in the solenoid. 3. A 2.-H inductor crries stedy current of.5 A. When the switch in the circuit is opened, the current is effectively zero fter 1. ms. Wht is the verge induced emf in the inductor during this time intervl? 4. A solenoid of rdius 2.5 cm hs 4 turns nd length of 2. cm. Find () its inductnce nd () the rte t which current must chnge through it to produce n emf of 75. mv. 5. An emf of 24. mv is induced in 5-turn coil when the current is chnging t the rte of 1. A/s. Wht is the mgnetic flux through ech turn of the coil t n instnt when the current is 4. A? 6. A 4.-mA current is crried y uniformly wound ir-core solenoid with 45 turns, 15.-mm dimeter, nd 12.-cm length. Compute () the mgnetic field inside