# CHAPTER 21: Electromagnetic Induction and Faraday s Law

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1 HAT : lecromagneic nducion and Faraday s aw Answers o Quesions. The advanage of using many urns (N = large number) in Faraday s experimens is ha he emf and induced curren are proporional o N, which makes i easier o experimenally measure hose quaniies.. Magneic flux is proporional o he oal number of magneic field lines passing hrough an enclosed loop area: A cos, and so he flux is proporional o he magniude of he magneic field. Thus, hey also have differen unis (magneic field = Tesla = T; magneic flux = Tm = Wb). Anoher difference is ha magneic field is a vecor (size and direcion), while magneic flux is a scalar (size).. Yes, a curren is induced in he ring when you bring he souh pole oward he ring. An emf and curren are induced in he ring due o he changing magneic flux (as he magne ges closer o he ring, more magneic field lines are going hrough he ring). No, a curren is no induced in he ring when he magne is held seady wihin he ring. An emf and curren are no induced in he ring since he magneic flux hrough he ring is no changing while he magne is held seady. Yes, a curren is induced in he ring when you wihdraw he magne. An emf and curren are induced in he ring due o he changing magneic flux (as you pull he magne ou of he ring oward you, fewer magneic field lines are going hrough he ring). Using enz s law and he igh Hand ule, he direcion of he induced curren when you bring he souh pole oward he ring is clockwise. n his case, he number of magneic field lines coming hrough he loop and poining oward you is increasing (remember, magneic field lines poin oward he souh pole of he magne). The induced curren in he loop is going o ry o oppose his change in flux and will aemp o creae magneic field lines hrough he loop ha poin away from you. A clockwise induced curren will provide his opposing magneic field. Using enz s law and he igh Hand ule again, he direcion of he induced curren when you wihdraw he souh pole from he ring is counerclockwise. n his case, he number of magneic field lines coming hrough he loop and poining oward you is decreasing. The induced curren in he loop is going o ry o oppose his change in flux and will aemp o creae more magneic field lines hrough he loop ha poin oward you. A counerclockwise induced curren will provide his opposing magneic field. 4. For he ring on he lef side of he curren-carrying wire here is no induced curren. As he ring moves along parallel o he wire, he magneic flux hrough he ring does no change, which means here is no induced emf and no induced curren. For he ring on he righ side of he curren-carrying wire, he induced curren is clockwise. As he ring moves away from he wire, he magneic flux hrough he ring is decreasing (here are fewer magneic field lines poining ino he loop). n an aemp o oppose his decrease (due o enz s law), an emf and curren will be induced in he ring in a clockwise direcion (using he igh Hand ule). 5. (a) Yes. As he baery is conneced o he fron loop and curren sars o flow, i will creae an increasing magneic field ha poins away from you and down hrough he wo loops. The second loop will ry o oppose his increase in magneic flux hrough i and an emf and curren will be induced. (b) The induced curren in he second loop sars o flow as soon as he curren in he fron loop sars o increase and creae a magneic field (basically, immediaely upon he connecion of he baery o he fron loop). 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 4

2 Giancoli hysics: rinciples wih Applicaions, 6 h diion (c) The curren in he second loop sops flowing as soon as he curren in he fron loop becomes seady. Once he baery has increased he curren in he fron loop from zero o is seady-sae value, hen he magneic field i creaes is also seady. Since he magneic flux hrough he second loop is no longer changing, he induced curren goes o zero. (d) The induced curren in he second loop is counerclockwise. Since he increasing clockwise curren in he fron loop is causing an increase in he number of magneic field lines down hrough he second loop, enz s law saes ha he second loop will aemp o oppose his change in flux. To oppose his change, he igh Hand ule says ha a counerclockwise curren will be induced in he second loop. (e) Yes. Since boh loops carry currens and creae magneic fields while he curren in he fron loop is increasing from he baery, hese wo magneic fields will inerac and pu a force on each of he loops. (f) The force beween he wo loops will repel each oher. The fron loop is creaing a magneic field poined oward he second loop. This changing magneic field induces a curren in he second loop o oppose he increasing magneic field and his induced curren creaes a magneic field poining oward he fron loop. These wo magneic fields will ac like wo norh poles poining a each oher and repel. 6. (a) The induced curren in A is o he righ as coil is moved oward coil A. As approaches A, he magneic flux hrough coil A increases (here are now more magneic field lines in coil A poining o he lef). oil A aemps o oppose his increase in flux, and he induced emf creaes a curren o produce a magneic field poining o he righ hrough he cener of he coil. A curren hrough A o he righ will produce his opposing field. (b) The induced curren in A is o he lef as coil is moved away from coil A. As recedes from A, he magneic flux hrough coil A decreases (here are now fewer magneic field lines in coil A poining o he lef). oil A aemps o oppose his decrease in flux, and he induced emf creaes a curren o produce a magneic field poining o he lef hrough he cener of he coil. A curren hrough A o he lef will produce his opposing field. (c) The induced curren in A is o he lef as in coil is increased. As increases, he curren in coil decreases, which also decreases he magneic field coil produces. As he magneic field from coil decreases, he magneic flux hrough coil A decreases (here are now fewer magneic field lines in coil A poining o he lef). oil A aemps o oppose his decrease in flux, and he induced emf creaes a curren o produce a magneic field poining o he lef hrough he cener of he coil. A curren hrough A o he lef will produce his opposing field. 7. As he signal in he wire varies in ime, i creaes changing magneic fields ha emanae from he wire. f here were exernal magneic fields nearby, hese could inerac wih he signal wire s magneic fields and cause inerference or noise in he signal. Wih he shield in place around he signal wire carrying he reurn curren, he ne curren in he wire, as seen by he ouside world, would be zero and, hus, he wire would no emanae a magneic field. n urn, he exernal magneic fields would no have any signal field o inerac wih and he inerference and noise in he signal would be reduced. 8. One advanage of placing he wo insulaed wires carrying ac close ogeher is ha whaever magneic fields are creaed by he changing curren moving one way in one wire is approximaely cancelled ou by he magneic field creaed by he curren moving in he opposie direcion in he second wire. Also, since large loops of wire in a circui can generae a large self-induced back emf, by placing he wo wires close o each oher, or even wising hem abou each oher, he effecive area of he curren loop is decreased and he induced curren is minimized. 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 5

3 haper lecromagneic nducion and Faraday s aw 9. As he refrigeraor moor firs sars up, here is a small back emf in he circui (back emf is proporional o he roaion speed of he moor), which allows for a large amoun of curren o flow o he refrigeraor. This large curren overaxes he source and diminishes he curren o all of he devices on he circui, including he lighs. As he refrigeraor moor speeds up o is normal operaional speed, he back emf increases o is normal level and he curren delivered o he refrigeraor is now limied o is usual amoun. This curren is no longer enough o overburden he source and all of he devices in he room go back o normal. Thus, i appears ha he lighs flicker jus when he refrigeraor moor firs sars up. A heaer, on he oher hand, draws a large amoun of curren (i is a very low-resisance device) a all imes. The source is hen coninually overaxed and he oher devices on he same circui remain dimmed all he while ha he heaer is operaing. n an ideal siuaion, he source could provide any amoun of curren o he whole circui in eiher siuaion. n realiy, hough, he higher currens in he wires causes bigger losses of energy along he way o he devices and he lighs dim.. Figure -7 shows ha he induced curren in he upper armaure segmen poins ino he page. This can be shown using a igh Hand ule: The charges in he op meal armaure segmen are moving in he direcion of he velociy shown wih he green arrow (up and o he righ) and hese moving charges are in a magneic field shown wih he blue arrows (o he righ), and hen he igh Hand ule says ha he charges experience a force ino he page producing he induced curren. This induced curren is also in he magneic field. Using anoher igh Hand ule, a currencarrying wire, wih he curren going ino he page (as in he upper armaure segmen), in a magneic field poined o he righ, will experience a force in a downward direcion. Thus, here is a counerclockwise orque on he armaure while i is roaion in a clockwise direcion. The back emf is opposing he moion of he armaure during is operaion.. Yes, eddy curren brakes will work on meallic wheels, such as copper and aluminum. ddy curren brakes do no need o ac on ferromagneic wheels. The exernal magneic field of he eddy brake jus needs o inerac wih he free conducion elecrons in he meal wheels o make i work. Firs, he magneic field creaes eddy currens in he moving meal wheel using he free conducion elecrons (he igh Hand ule says moving charges in a magneic field will experience a magneic force, making hem move, and creaing an eddy curren). This eddy curren is also in he braking magneic field. The igh Hand ule says hese currens will experience a force opposing he original moion of he piece of meal and he eddy curren brake will begin o slow he wheel. Good conducors, such as copper and aluminum, have many free conducion elecrons and will allow large eddy currens o be creaed, which in urn will provide good braking resuls.. As he pieces of conducing maerials slide down he incline pas permanen magnes, he free conducion elecrons in he pieces are forced o creae eddy currens (he igh Hand ule says moving charges in a magneic field will experience a magneic force, making hem move, and creaing an eddy curren). These eddy currens are no creaed in he non-meallic pieces. These eddy currens are also moving hrough he magneic fields of he permanen magnes and he igh Hand ule says hese currens will experience a magneic force opposing he original moion of he pieces of meal. Thus, he meallic pieces will be slowed as hey slide down he incline, while he non-meallic pieces will coninue o accelerae he enire ime. Since he meallic pieces have been slowed, hey will basically jus drop off of he end of he incline and you would pu a meal bin direcly below he end o cach he meal pieces. The non-meallic pieces, on he oher hand, will come off he end of he incline wih a high speed and overshoo he meal bin and be caugh in a non-meal bin siing farher away from he end. 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 6

4 Giancoli hysics: rinciples wih Applicaions, 6 h diion. The slos cu ino he piece of pivoed meal bar confine he induced eddy currens o many very small loops, insead of jus one large loop. These smaller curren loops have less magneic flux due o heir smaller areas, which creaes less emf and smaller eddy currens. The smaller eddy currens hen experience a smaller opposing force o he moion of he meal bar. Thus, he sloed bar falls more quickly hrough he magneic field. 4. As you ry o move he aluminum shee ou of he magneic field, areas of he shee ha used o be in = regions sar o gain magneic flux. This changing flux will induce currens in he free conducion elecrons of he aluminum o oppose he change. These eddy currens are hen aced on by he magneic field and he resuling force opposes he moion of he aluminum shee. Thus, i requires some amoun of force o remove he shee from beween he poles. 5. As a magne falls hrough a meal ube, an increase in he magneic flux is creaed in he areas ahead of i in he ube. This flux change induces a curren o flow around he ube walls o creae an opposing magneic field in he ube (enz s law). This induced magneic field pushes agains he falling magne and causes i o slow down. The opposing magneic field canno cause he magne o acually come o a sop, since hen he flux would become a consan and he induced curren would disappear and so would he opposing magneic field. Thus, he magne reaches a sae of equilibrium and falls a a consan erminal velociy. 6. As he meal bar eners (or leaves) he magneic field during he swinging moion, areas of he meal bar experience a change in magneic flux. This changing flux induces eddy currens wih he free conducion elecrons in he meal bar. These eddy currens are hen aced on by he magneic field and he resuling force opposes he moion of he swinging meal bar. This opposing force acs on he bar no maer which direcion i is swinging hrough he magneic field, hus damping he moion during boh he o and fro porions of he swing. 7. To deermine he raio of urns on he wo coils of a ransformer wihou aking i apar, apply a known ac inpu source volage o one pair of leads and carefully measure he oupu volage across he oher wo leads. Then, source / oupu = N s /N o, which provides us wih he raio of urns on he wo coils. To deermine which leads are paired wih which, you could use an ohmmeer, since he wo source wires are in no way elecrically conneced o he wo oupu wires. f he resisance beween wo wires is very small, hen hose wo wires are a pair. f he resisance beween wo wires is infinie, hen hose wo wires are no a pair. 8. Higher volages, such as 6 or, would be dangerous if hey were used in household wires! Such a large poenial difference beween household wires and anyhing ha is grounded (oher wires, people, ec.) would more easily cause elecrical breakdown of he air and hen much more sparking would occur. asically, his would supply each of he charges in he household wires wih much more energy han he lower volages, which would allow hem o arc o oher conducors. This would increase he possibiliy of more shor circuis and accidenal elecrocuions. 9. When dc is applied o he ransformer, here is no induced back emf ha would usually occur wih ac. This means ha he dc encouners much less resisance han he ac, which would cause oo much curren o flow hrough he ransformer. This large amoun of curren would overhea he coils, which are usually wound wih many loops of very fine wire, and could mel he insulaion and burn ou or shor ou he ransformer.. (a) To creae he larges amoun of muual inducance wih wo fla circular coils of wire, you would place hem face-o-face and very close o each oher. This way, almos all of he magneic flux from one coil also goes hrough he oher coil. 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 7

5 haper lecromagneic nducion and Faraday s aw (b) To creae he leas amoun of muual inducance wih wo fla circular coils, you would place hem wih heir faces a righ angles. This way, almos none of he magneic flux from one coil goes hrough he oher coil.. (a) No. Alhough he curren hrough an circui is described by subsiue. Thus, a given fracion of a imum possible curren, e and is independen of he baery emf. (b) Yes. Since value of he baery emf. e e, we can, is equal o, if a given value of he curren is desired, i is dependen on he. (a) Yes. (b) Yes. The volages across eiher an inducor or a capacior of an circui can be greaer han he source volage because he differen volages are ou of phase wih each oher. A any given insan, he volage across eiher he inducor or he capacior could be negaive, for example, hus allowing for a very large posiive volage on he oher device. (The volages, however, are always posiive by definiion.). (a) The frequency of he source emf does no affec he impedance of a pure resisance. The impedance of a pure resisance is independen of he source emf frequency. (b) The impedance of a pure capaciance varies inversely wih he frequency of he source emf according o X f. As he source frequency ges very small, he impedance of he capacior ges very large, and as he source frequency ges very large, he impedance of he capacior ges very small. (c) The impedance of a pure inducance varies direcly wih he frequency of he source emf according o X f. Thus, as he source frequency ges very small, he impedance of he inducor ges very small and as he source frequency ges very large, he impedance of he inducor ges very large. (d) The impedance of an circui, wih a small, near resonance is very sensiive o he frequency of he source emf. f he frequency is se a resonance exacly, where X X, hen he circui s impedance is very small and equal o. The impedance increases rapidly as he source frequency is eiher increased or decreased a small amoun from resonance. (e) The impedance of an circui, wih a small, very far from resonance depends on wheher he source frequency is much higher or much lower han he resonance frequency. f he source frequency is much higher han resonance, hen he impedance is direcly proporional o he frequency of he source emf. asically, a exremely high frequencies, he circui impedance is equal o f. f he source frequency is much lower han resonance (nearly zero), hen he impedance is inversely proporional o he frequency of he source emf. asically, a exremely low frequencies, he circui impedance is equal o f. 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 8

6 Giancoli hysics: rinciples wih Applicaions, 6 h diion 4. The way o make he impedance of an circui a minimum is o make he resisance very small and make he reacance of he capacior equal o he reacance of he inducor: X X, or f f f. Soluions o roblems. The average induced emf is given by q. -b. 8 Wb 5 Wb N 4.4s. As he coil is pushed ino he field, he magneic flux hrough he coil increases ino he page. To oppose his increase, he flux produced by he induced curren mus be ou of he page, so he induced curren is counerclockwise.. As he magne is pushed ino he coil, he magneic flux increases o he righ. To oppose his increase, flux produced by he induced curren mus be o he lef, so he induced curren in he resisor will be from righ o lef. 4. We assume he plane of he coil is perpendicular o he magneic field. The magniude of he average induced emf is given by q. -a. A.48 m. T.5s The flux changes because he loop roaes. The angle changes from o o 9 o. The magniude of he average induced emf is given by q. -a. A cos.6 m.5t.s 8.5 o o.6 m.5t cos 9 cos.s 6. We choose up as he posiive direcion. The average induced emf is given by q. -a. A.5m.5 T.6T.5s (a) When he plane of he loop is perpendicular o he field lines, he flux is given by he imum of q. -. (b) The angle is (c) Use q. -. A r.5 T.75 m 8.8 Wb o 55 o cos.5 T.75 m cos Wb A r 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 9

7 haper lecromagneic nducion and Faraday s aw 8. (a) As he resisance is increased, he curren in he ouer loop will decrease. Thus he flux hrough he inner loop, which is ou of he page, will decrease. To oppose his decrease, he induced curren in he inner loop will produce a flux ou of he page, so he direcion of he induced curren will be counerclockwise. (b) f he small loop is placed o he lef, he flux hrough he small loop will be ino he page and will decrease. To oppose his decrease, he induced curren in he inner loop will produce a flux ino he page, so he direcion of he induced curren will be clockwise. 9. (a) The increasing curren in he wire will cause an increasing field ou of he page hrough he loop. To oppose his increase, he induced curren in he loop will produce a flux ino he page, so he direcion of he induced curren will be clockwise. (b) The decreasing curren in he wire will cause a decreasing field ou of he page hrough he loop. To oppose his decrease, he induced curren in he loop will produce a flux ou of he page, so he direcion of he induced curren will be counerclockwise. (c) The decreasing curren in he wire will cause a decreasing field ino he page hrough he loop. To oppose his decrease, he induced curren in he loop will produce a flux ino he page, so he direcion of he induced curren will be clockwise. (d) ecause he curren is consan, here will be no change in flux, so he induced curren will be zero.. As he solenoid is pulled away from he loop, he magneic flux o he righ hrough he loop decreases. To oppose his decrease, he flux produced by he induced curren mus be o he righ, so he induced curren is counerclockwise as viewed from he righ end of he solenoid.. (a) The average induced emf is given by q. -a. A 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he.6 m.45 T.5 T.8s 6. (b) The posiive resul for he induced emf means he induced field is away from he observer, so he induced curren is clockwise.. (a) ecause he velociy is perpendicular o he magneic field and he rod, we find he induced emf from q. -. v.8 T. m.5 m s.44 (b) ecause he upward flux is increasing, he induced flux will be ino he page, so he induced curren is clockwise. Thus he induced emf in he rod is down, which means ha he elecric field will be down. The elecric field is he induced volage per uni lengh. l.44. m, down. m. (a) The magneic flux hrough he loop is ino he paper and decreasing, because he area is decreasing. To oppose his decrease, he induced curren in he loop will produce a flux ino he paper, so he direcion of he induced curren will be clockwise. (b) The average induced emf is given by q. -a. A T. m. m.5s

8 Giancoli hysics: rinciples wih Applicaions, 6 h diion (c) We find he average induced curren from Ohm s law A.5 4. (a) The velociy is found from q. -.. lv v l.9 T. m. m s (b) ecause he ouward flux is increasing, he induced flux will be ino he page, so he induced curren is clockwise. Thus he induced emf in he rod is down, which means ha he elecric field will be down. The magniude of he elecric field is he induced volage per uni lengh...9 m, down l. m 5. As he loop is pulled from he field, he flux hrough he loop decreases, causing an induced MF whose magniude is given by q. -, lv. ecause he inward flux is decreasing, he induced flux will be ino he page, so he induced curren is clockwise, given by. ecause his curren in he lef-hand side of he loop is in a downward magneic field, here will be a magneic force o he lef. To keep he rod moving, here mus be an equal exernal force o he righ, given by F l. lv F l l l l v.55 T.5 m.4 m s..548 N 6. The emf induced in he shor coil is given by q. -b, where N is he number of loops in he shor coil, and he flux change is measured over he area of he shor coil. The magneic flux comes from he -field creaed by he solenoid. The field in a solenoid is given by q. -8, N l, and he changing curren in he solenoid causes he field o change. solenoid solenoid N solenoid N A shor shor N A l N N A shor shor solenoid shor solenoid shor l 7 4 T m A 5.5 m 5. A.5 m.6s 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he solenoid (a) ecause he velociy is perpendicular o he magneic field and he rod, we find he induced emf from q. -. lv.5t. m.6 m s.68.7 (b) Find he induced curren from Ohm s law A 6. A 7.5 (c) The induced curren in he rod will be down. ecause his curren is in an upward magneic field, here will be a magneic force o he lef. To keep he rod moving, here mus be an equal exernal force o he righ, given by q. -. F l 4 6. A. m.5 T 6.4 N

9 haper lecromagneic nducion and Faraday s aw 8. (a) There is an emf induced in he coil since he flux hrough he coil changes. The curren in he coil is he induced emf divided by he resisance of he coil. The resisance of he coil is found from q. 8-. NA coil A NA A coil wire wire NA A coil wire.m. m 8.65 T s 8.68 m..54 A.5 A (b) The rae a which hermal energy is produced in he wire is he power dissipaed in he wire m A 9.9 W Awire. m 9. The charge ha passes a given poin is he curren imes he elapsed ime, Q. The curren will be he emf divided by he resisance,. The resisance is given by. 8-, Awire, and he emf is given by q. -b. ombine hese equaions o find he charge during he operaion. A loop A loop A A loop wire ; ; A wire A Q loop wire loop wire loop wire r loop wire A A r r r r.66 m.5 m.75 T 8.68 m We find he number of urns from q. -5. The facor muliplying he sine erm is he peak oupu volage. peak 4. N A N 4. 4 loops peak A.4T rad rev 6 rev s.6 m. From q. -5, he induced volage is proporional o he angular speed. Thus heir quoien is a consan. 5 rpm.4 8. rpm 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he

10 Giancoli hysics: rinciples wih Applicaions, 6 h diion. From q. -5, he peak volage is N A. The volage is he peak volage divided by peak, and so = N A. peak. From q. -5, he peak volage is N A. Solve his for he roaion speed. peak f N A peak peak.8 rad s rad rev NA.8 rev s.65 T. m.8 rad s 4. (a) The volage is found from he peak induced emf. The peak induced emf is calculaed from q. -5. N A peak peak N A T rad rev rev s.4m (b) To double he oupu volage, you mus double he roaion frequency o 4 rev/s. 5. (a) The peak curren is found from he curren. peak 7. A 99. (b) The area can be found from q. -5. A peak N A N 5. T 85 rev s rad rev. m 6. When he moor is running a full speed, he back emf opposes he applied emf, o give he ne across he moor. 8. A applied back back applied 7. From q. -5, he induced volage (back emf) is proporional o he angular speed. Thus heir quoien is a consan. 5 rpm 7 8 rpm 8. The magniude of he back emf is proporional o boh he roaion speed and he magneic field, from q. -5. Thus is consan. 65 rpm 5 rpm 95 So reduce he magneic field o 7% of is original value..7 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he

11 haper lecromagneic nducion and Faraday s aw 9. The back emf is proporional o he roaion speed (q. -5). Thus if he moor is running a half speed, he back emf is half he original value, or 54. Find he new curren from wriing a loop equaion for he moor circui, from Figure -9. back 54 A back 5.. We find he number of urns in he secondary from q. -6. N S S S, rpm N N 64, 7 urns S N. ecause N N, his is a sep-down ransformer. Use q. -6 o find he volage raio, and q. S -7 o find he curren raio. N S S urns S N urns N urns N urns S. Use qs. -6 and -7 o relae he volage and curren raios. N S S S N S S 5 ;. N N S S S. We find he raio of he number of urns from q. -6. N S S 5 N 4 f he ransformer is conneced backward, he role of he urns will be reversed: N S S S S N (a) Use qs. -6 and -7 o relae he volage and curren raios. N S S S N S.5 A ; 5.6 S N N 7.5 A (b) ecause S S S S, his is a sep-down ransformer. 5. (a) We assume % efficiency, and find he inpu volage from. 95 W 4.8 A (b) Since, his is a sep-up ransformer. S S Use qs. -6 and -7 o relae he volage and curren raios. N N S S S 4 urns 487 S N N urns 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 4

12 Giancoli hysics: rinciples wih Applicaions, 6 h diion S N N S N N S S 4 urns 5. A 6.9 A urns 7. (a) The curren in he ransmission lines can be found from q. 8-9a, and hen he emf a he end of he lines can be calculaed from Kirchhoff s loop rule. 6 W own own 45 oupu 6 own W oupu k (b) The power loss in he lines is given by Fracion wased 45. loss 667 A loss loss 6 oal own loss own A 4. W 667 A Wihou he ransformers, we find he delivered curren, which is he curren in he ransmission lines, from he delivered power, and he power los in he ransmission lines. ou 65 W ou ou line line 54.7 A los line line ou 54.7 A W Thus here mus be 65 W 5869 W 69W 4 kw of power generaed a he sar of he process. Wih he ransformers, o deliver he same power a, he delivered curren from he sep-down ransformer mus sill be 54.7 A. Using he sep-down ransformer efficiency, we calculae he curren in he ransmission lines, and he loss in he ransmission lines A ou ou A ou line ou ou line line line end los line line 54.7 A. 599 W The power o be delivered is 65 W. The power ha mus be delivered o he sep-down ransformer is 65 W W. The power ha mus be presen a he sar of he ransmission.99 mus be W 599 W 6656 W o compensae for he ransmission line loss. The power ha mus ener he ransmission lines from he 99% efficien sep-up ransformer is 6656 W kw. So he power saved is 4 kw 67 kw 54 kw Find he induced emf from q A 5. A.8 H s line 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 5

13 haper lecromagneic nducion and Faraday s aw 4. ecause he curren in increasing, he emf is negaive. We find he self-inducance from q s.5.58 H. A.8 A 4. Use he relaionship for he inducance of a solenoid, as given in xample -4. N A l 7 4 T m A.45 m.6 m.4 H 4. Use he relaionship for he inducance of a solenoid, as given in xample -4. N A l N l A. H. m 7 4 T m A.6 m 8 urns 4. (a) The number of urns can be found from he inducance of a solenoid, which is derived in xample -4. N A l 7 4 T m A 8.5 m.8 m (b) Apply he same equaion again, solving for he number of urns. N A l N l A.7 H.7 H.8 m 7 4 T m A.5 m : 8urns 44. We draw he coil as wo elemens in series, and pure resisance and a pure inducance. There is a volage drop due o he resisance of he coil, given by Ohm s law, and an induced emf due o he inducance of he coil, given by q. -9. Since he curren is increasing, he inducance will creae a poenial difference o oppose he increasing b curren, and so here is a drop in he poenial due o he inducance. The poenial difference across he coil is he sum of he wo poenial drops.. A.5.44 H.5 A s 8.9 ab increasing induced a 45. We assume ha boh he solenoid and he coil have he same cross-secional area. The magneic N field of he solenoid (which passes hrough he coil) is. When he curren in he l solenoid changes, he magneic field of he solenoid changes, and hus he flux hrough he coil changes, inducing an emf in he coil. N N A N A l N N A l As in q. -8a, he muual inducance is he proporinaliy consan in he above relaionship. N N A N N A M l l 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 6

14 Giancoli hysics: rinciples wih Applicaions, 6 h diion 46. The inducance of he solenoid is given by N l. The (consan) lengh of he wire is given by l N d, and so since d d, we also know ha N N. The fac ha he wire is wire sol sol sol ighly wound gives l sol Nd wire. Find he raio of he wo inducances. wire d d sol sol 4 sol sol sol sol wire N N lwire sol wire d d sol sol 4 N N l l l l N d l N d N l l l sol sol sol 47. The magneic energy in he field is derived from q. -. nergy sored u olume r 7 l.8t nergy olume. m.6 m 9 J 4 T m A 48. The iniial energy sored in he inducor is found from an equaion in Secion -. 5 U 6. H 5. A 7.5 J The final curren is found from he final energy, which is imes he iniial energy. U U f f f The curren increases a a consan rae. f.5 A s 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 7 N 5. A f.94 s.5 A s.5 A s.5 A s.5 A s 49. The magneic energy in he field is derived from q. -. The volume of a relaively hin spherical shell, like he firs km above he arh s surface, is he surface area of he sphere imes is hickness. u 4.5 T arh 7 u volume 4 h m m 5. J 4 T m A 5. When he swich is iniially closed, he inducor prevens curren from flowing, and so he iniial curren is, as shown in Figure -4. f he curren is, here is no volage drop across he resisor (since, and so he enire baery volage appears across he inducor. Apply q. -9 o find he iniial rae of change of he curren. The imum value of he curren is reached afer a long ime, when here is no volage across he inducor, and so he enire baery volage appears across he resisor. Apply Ohm s law.

15 haper lecromagneic nducion and Faraday s aw. Find he ime o reach he imum curren if he rae of curren change remained a. elapsed ime elapsed ime 5. For an circui, we have e. Solve for. (a) (b) (c) e e ln ln ln.9. ln ln ln ln N A 5. We use he inducance of a solenoid, as derived in xample -4:. sol l (a) oh solenoids have he same area and he same lengh. ecause he wire in solenoid is half as hick as he wire in solenoid, solenoid will have wice he number of urns as solenoid. N A N N 4 4 N N l N A l (b) To find he raio of he ime consans, boh he inducance and resisance raios need o be known. Since solenoid has wice he number of urns as solenoid, he lengh of wire used o make solenoid is wice ha used o make solenoid, or l wire l wire, and he diameer of he wire in solenoid is half ha in solenoid, or d d. Use his o find heir relaive resisances, and hen he raio of ime consans. l l l wire wire wire wire wire A d wire wire d l d wire wire wire 8 8 l l l l d wire wire wire wire wire Awire d d wire wire 4 = 8 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 8

16 Giancoli hysics: rinciples wih Applicaions, 6 h diion 5. The reacance of a capacior is given by q. -b, (a) X 68 6 f 6. Hz 7. F (b) X 6 6 f. Hz 7. F X f We find he frequency from q. -b for he reacance of an inducor. X 66 X f f 477 Hz. H 55. We find he frequency from q. -b for he reacance of a capacior. X f 6 f X F 9.9 Hz 56. The impedance is X f. mpedance (kilohms) apacior eacance Frequency (Hz) 57. The impedance is X f. 8 nducor eacance mpedance (ohms) Frequency (Hz) 58. We find he reacance from q. -b, and he curren from Ohm s law. X f. Hz.6 H 5. k X X A 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 9

17 haper lecromagneic nducion and Faraday s aw 59. We find he reacance from Ohm s law, and he inducance by q. -b. X X X 4 X f 4.97 H f f 6. Hz.8 A 6. (a) We find he reacance from q. -b. X 8 f 7 Hz. F (b) We find he peak value of he curren from Ohm s law...8 A peak X The impedance of he circui is given by q. -5 wihou a capaciive reacance. The reacance of he inducor is given by q. -b. (a) Hz 45 H. Z X f (b) Hz 45 H Z X f The impedance of he circui is given by q. -5 wihou an inducive reacance. The reacance of he capacior is given by q. -b. (a).5 6 Z X 4 f 4 6 Hz 4. F (b) Z X.5 4 f 4 6 Hz 4. F 6. We find he impedance from q. -4. Z.7 7 A 64. Use q. -5, wih no capaciive reacance. Z X Z X The oal impedance is given by q. -5. Z X X f f 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 4

18 Giancoli hysics: rinciples wih Applicaions, 6 h diion Hz. H 4 9. Hz 6.5 F k The phase angle is given by q. -6a. f X X f an an 4. Hz. H 4 9. Hz 6.5 F an 64 o an The volage is lagging he curren, or he curren is leading he volage. The curren is given by q A Z The impedance is given by q. -5 wih no capaciive reacance. Z X f. 6 Z Z 4 f 4 6 Hz f 4 f Hz Hz f 6 6 Hz Hz 4 6 Hz H 645 Hz.6 khz 67. (a) The curren is he rmv volage divided by he impedance. The impedance is given by q. -5 wih no capaciive reacance, Z X f. Z 4 f Hz.5 H 6.65 A 85 (b) The phase angle is given by q. -6a wih no capaciive reacance. X f 6. Hz.5 H an an an The curren is lagging he source volage. o 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 4

19 haper lecromagneic nducion and Faraday s aw (c) The volage reading is he curren imes he resisance or reacance of he elemen. X f 6.65 A A 6. Hz.5 H 8.77 Noe ha, because he imum volages occur a differen imes, he wo readings do no add o he applied volage of. 68. (a) The curren is he rmv volage divided by he impedance. The impedance is given by q. -5 wih no inducive reacance, Z Z X f f Hz.8 F.4 A 89 (b) The phase angle is given by q. -6a wih no inducive reacance. 6 X 6. Hz.8 F f o an an an The curren is leading he source volage. (c) The volage reading is he curren imes he resisance or reacance of he elemen. X.4 A A f 6. Hz.8 F Noe ha, because he imum volages occur a differen imes, he wo readings do no add o he applied volage of. 69. The resonan frequency is given by q. -9. f H 5 F 5.6 Hz 7. (a) The resonan frequency is given by q. -9. f f 58 58kHz f 6 6kHz 58 khz 6 khz 6 khz 58 khz pf 67.9 pf 7 pf 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 4

20 Giancoli hysics: rinciples wih Applicaions, 6 h diion (b) The inducance can be found from he resonan frequency and he capaciance. f 5 4 f Hz 8 F 5.7 H 7. (a) We find he capaciance from he resonan frequency, q. -9. f 4 f H 6 Hz 7. F (b) A resonance he impedance is he resisance, so he curren is given by Ohm s law. peak 5 4.A peak (a) The clockwise curren in he lef-hand loop produces a magneic field which is ino he page wihin he loop and ou of he page ouside he loop. Thus he righ-hand loop is in a magneic field ha is direced ou of he page. efore he curren in he lef-hand loop reaches is seady sae, here will be an induced curren in he righ-hand loop ha will produce a magneic field ino he page o oppose he increase of he field from he lef-hand loop. Thus he induced curren will be clockwise. (b) Afer a long ime, he curren in he lef-hand loop is consan, so here will be no induced curren in he righ-hand coil. (c) f he second loop is pulled o he righ, he magneic field ou of he page from he lef-hand loop hrough he second loop will decrease. During he moion, here will be an induced curren in he righ-hand loop ha will produce a magneic field ou of he page o oppose he decrease of he field from he lef-hand loop. Thus he induced curren will be counerclockwise. 7. The elecrical energy is dissipaed because here is curren flowing in a resisor. The power dissipaion by a resisor is given by, and so he energy dissipaed is. The curren is creaed by he induced emf caused by he changing -field. The emf is calculaed by q. -a. A A A A 7.5 J.4 m.665 T 5..4s 74. (a) ecause, his is a sep-down ransformer. S (b) Assuming % efficiency, he power in boh he primary and secondary is 45 W. Find he curren in he secondary from he relaionship. S 45 W S S S S.8 A (c) S 45 W.8 A 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 4

21 haper lecromagneic nducion and Faraday s aw (d) Find he resisance of he bulb from Ohm s law. The bulb is in he secondary circui. S. S S.75 A S 75. ecause here are perfec ransformers, he power loss is due o resisive heaing in he ransmission 5 MW lines. Since he own requires 5 MW, he power a he generaing plan mus be MW. Thus he power los in he ransmission is.76 MW. This can be used o deermine he curren in he ransmission lines W 8 km. km 79.6 A To produce 5.76 MW of power a 79.6 A requires he following volage W k 79.6 A 76. (a) From he efficiency of he ransformer, we have.8. Use his o calculae he curren S in he primary. S 75 W A.85 A S.8.8 (b) The volage in boh he primary and secondary is proporional o he number of urns in he respecive coil. The secondary volage is calculaed from he secondary power and resisance since. N N S S S S 75 W (a) The volage drop across he lines is due o he resisance A k ou in (b) The power inpu is given by in in A 4.8 W. W in in (c) The power loss in he lines is due o he curren in he resisive wires A W 8.8 W loss (d) The power oupu is given by ou ou A 486. W. W ou ou This could also be found by subracing he power los from he inpu power W 8.76 W. W. W ou in loss 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 44

22 Giancoli hysics: rinciples wih Applicaions, 6 h diion 78. A side view of he rail and bar is shown in he figure. From he F discussion in Secion -, he emf in he bar is produced by he N componens of he magneic field, he lengh of he bar, and he velociy of he bar, which are all muually perpendicular. The v F magneic field and he lengh of he bar are already perpendicular. The s componen of he velociy of he bar ha is perpendicular o he mg magneic field is v cos, and so he induced emf is given by he following. lv cos This produces a curren in he wire, which can be found by Ohm s law. Tha curren is poining ino he page on he diagram. lv cos ecause he curren is perpendicular o he magneic field, he force on he wire from he magneic field can be calculaed from q. -, and will be horizonal, as shown in he diagram. lv cos l v cos F l l For he wire o slide down a a seady speed, he ne force along he rail mus be zero. Wrie Newon s second law for forces along he rail, wih up he rail being posiive. l v cos F F cos mg sin mg sin ne v mg sin l.6.4 kg 9.8 m s sin 6. cos.55 T. m cos 6. o o.8 m s 79. We find he curren in he ransmission lines from he power ransmied o he user, and hen find he power loss in he lines. T T T T 8. The induced curren in he coil is he induced emf divided by he resisance. The induced emf is found from he changing flux by q. -b. The magneic field of he solenoid, which causes he flux, is given by q. -8. For he area used in q. -b, he cross-secional area of he solenoid (no he coil) mus be used, because all of he magneic flux is inside he solenoid. N ind sol sol sol N N A = ind coil coil sol sol l N A coil sol N sol sol l N A N sol coil sol sol sol l 7 5 urns.45 m 4 T m A urns. A sol.m. s sol 4.6 A As he curren in he solenoid increases, a magneic field from righ o lef is creaed in he solenoid and he loop. The induced curren will flow in such a direcion as o oppose ha field, and so mus flow from lef o righ hrough he resisor. 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 45

23 haper lecromagneic nducion and Faraday s aw 8. uing an inducor in series wih he device will proec i from sudden surges in curren. The growh of curren in an circui is given is secion -. e e The imum curren is 55 ma, and he curren is o have a value of 7.5 ma afer a ime of microseconds. Use his daa o solve for he inducance. e e ln 4. sec 5 ln 7.5 ma 55 ma.8 H. H in series 8. The emf is relaed o he flux change by q. -b. The flux change is caused by he changing magneic field. A N N 9T s NA m 8. We find he peak emf from q. -5. N A 55. T rad rev rev s 6.6 m peak 84. (a) The elecric field energy densiy is given by q. 7-, and he magneic field energy densiy is given by q. -. u u N m. m 4.45 J m 4.4 J m 4 4.T 4 T m A.59 J m.6 J m u.59 J m u 4.45 J m The energy densiy in he magneic field is.6 billion imes greaer han he energy densiy in he elecric field. (b) Se he wo densiies equal and solve for he magniude of he elecric field. u u.t N m 4 T m A 8 6. m 85. The impedance is found from q. -4, and hen he inducance is found from q. -b. X Z X X f. H f f 6. Hz 5.8 A 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 46

24 Giancoli hysics: rinciples wih Applicaions, 6 h diion 86. For he curren and volage o be in phase, he ne reacance of he capacior and inducor mus be zero, which means ha he circui is a resonance. Thus q. -9 applies. f 4 f 4. H 6 Hz 7.5 F 87. The resisance of he coil is found from he dc curren via Ohm s law. dc 6 dc dc A dc The impedance of he coil is found from he ac curren by q. -4, and hen he inducance can be found from he impedance by q. -5a. ac Z Z X 4 f ac ac ac ac.8 A ac 4 f 4 6 Hz H 7.5 H 88. (a) The curren is found from he volage and impedance. The impedance is given by q. -5. Z X X f. 6 Hz.5 H A Z 8.76 (b) Use q. -6a o find he phase angle. f X X f an an 6 Hz.5 H f 6 6 Hz F 6 6 Hz F 8.7 o an an (a) From he ex of he problem, he Q facor is he raio of he volage across he capacior or inducor o he volage across he resisor, a resonance. The resonan frequency is given by q. -9. X f res res Q res 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 47

25 haper lecromagneic nducion and Faraday s aw (b) Find he inducance from he resonan frequency, and he resisance from he Q facor. f res 4 f 8 6 res 4. F. Hz H.5 H Q 6.5 H Q F.9 5 earson ducaion, nc., Upper Saddle iver, NJ. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion of his maerial may be reproduced, in any form or by any means, wihou permission in wriing from he 48

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