How To Write The Prolem Re Extr Reit Homework

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 1 This print-out shoul he 34 questions. Multiple-hoie questions my ontinue on the next olumn or pge fin ll hoies efore mking your seletion. The ue time is Centrl time. these prolem re extr reit homework, ut the mteril will e in the finl 001 (prt 1 of 2) 10 points A retngulr loop lote istne from long wire rrying urrent is shown in the figure. The wire is prllel to the longest sie of the loop. From Ampère s lw, we know tht the strength of the mgneti fiel rete y the urrent-rrying wire t istne r from the wire is (see figure.) = µ 0 I 2 π r, so the fiel ries oer the loop n is irete perpeniulr to the pge. Sine is prllel to A, we n express the mgneti flux through n re element A s Φ A = µ0 I 2 π r A. 0.0915 A 6.45 m 4.2 m 18 m Fin the totl mgneti flux through the loop. Corret nswer: 1.65187 10 9 W. Let : = 6.45 m, = 4.2 m, = 18 m, n I = 0.0915 A. r Note: is not uniform ut rther epens on r, so it nnot e remoe from the integrl. In orer to integrte, we express the re element she in the figure s A = r. Sine r is the only rile tht now ppers in the integrl, we otin for the mgneti flux Φ = µ 0 I + 2 π r r = µ 0 I 2 π ln r + = µ ( ) 0 I + 2 π ln = µ ( ) 0 (0.0915 A)(0.18 m) + ln 2 π = µ 0 (0.0915 A)(0.18 m) (0.50148) 2 π = 1.65187 10 9 W. I r 002 (prt 2 of 2) 10 points Wht is the iretion of the mgneti fiel through the retngulr loop? 1. out of the plne of the pper 2. into the plne of the pper 3. nnot e etermine with informtion gien orret

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 2 The iretion of the urrent is not gien, hene the solute iretion of the mgneti fiel nnot e etermine, lthough the mgneti fiel is perpeniulr to the plne of the pper. For exmple if the urrent flows upwr (ownwr) the mgneti fiel woul e into (out of) the plne of the pper. 003 (prt 1 of 1) 10 points A oil is wrppe with 198 turns of wire on the perimeter of squre frme of sies 34.4 m. Eh turn hs the sme re, equl to tht of the frme, n the totl resistne of the oil is 1.74 Ω. A uniform mgneti fiel is turne on perpeniulr to the plne of the oil. If the fiel hnges linerly from 0 to 0.908 W/m 2 in time of 1.13 s, fin the mgnitue of the inue emf in the oil while the fiel is hnging. Corret nswer: 18.8274 V. si Conept: Fry s Lw is E = Φ Solution: The mgneti flux through the loop t t = 0 is zero sine = 0. At t = 1.13 s, the mgneti flux through the loop is Φ = A = 0.107449 W. Therefore the mgnitue of the inue emf is E = N Φ t = (198 turns) [(0.107449 W) 0] (1.13 s) = 18.8274 V E = 18.8274 V. 004 (prt 1 of 1) 10 points A mgneti fiel of 0.297 T exists in the region enlose y solenoi tht hs 420 turns n imeter of 5.42 m. Within wht perio of time must the fiel e reue to zero if the erge mgnitue of the inue emf within the oil uring this time interl is to e 12.3 kv? Corret nswer: 2.33986 10 5 s.. si Conept: Fry s Lw: E = N Φ From Fry s Lw, we get E = N Φ = N A = N A t = N A t So, the time neee equls t = N A E = (420 turns) (0.00230722 m2 ) (0.297 T) (12.3 kv) = 2.33986 10 5 s. 005 (prt 1 of 1) 10 points A irulr onuting loop is hel fixe in uniform mgneti fiel tht ries in time oring to (t) = 0 exp( t) where t is in s, is in s 1 n is the fiel strength in T t t = 0. At t = 0, the emf inue in the loop is 0.0659 V. At t = 3.5 s, the emf is 0.0221 V,. Fin. Corret nswer: 0.31216 s 1. si onept Fry s Lw E = A Solution: Sine the emf is E = A, sine only the mgneti fiel is hnging n = exp ( t),

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 3 we he 2 equtions from the 2 ifferent times. They re t t = 0, A = 0.0659 V, t t = 1.5 s, A exp[ (3.5 s)] = 0.0221 V. n Diiing the seon eqution y the first n then tking the nturl logrithm, we he [ ] (0.0221 V) ln (0.0659 V) = = 0.31216 s 1. (3.5 s) 006 (prt 1 of 1) 10 points A(n) 43.2 m length of wire when use s resistor hs resistne of 0.00365 Ω. The ens of the wire re onnete to form irulr loop, n the plne of the loop is positione t right ngles to uniform mgneti fiel tht is inresing t the rte of 0.0726 T/s. At wht rte is therml energy generte in the wire? Corret nswer: 318.49 µw. The hnging mgneti fiel genertes urrent in the wire. The inue potentil is V = A. The rius is foun from the irumferene, (C = l), to e: r = C 2π Hene the re is gien y 43.2 m =. 2π ( ) 2 C A = πr 2 = π = 1 2π π Then the inue potentil is V = 1 ( ) 2 C π 2 = 0.00107819 V ( ) 2 C 2 The power issipte is then P = V 2 = (0.00107819 V)2 (10 6 ) 0.00365 Ω = 318.49 µw 007 (prt 1 of 3) 10 points Assume: The inue emf for the lose loop otgonl CXDY C is E. A solenoi (with mgneti fiel ) proues steily inresing uniform mgneti flux through its irulr ross setion. A otgonl iruit surrouns the solenoi s shown in the figure. The wires onneting in the iruit re iel, hing no resistne. The iruit onsists of two ientil light uls (lele X n Y ) in series. A wire onnets points C n D. The rtio of the solenoi s re A L left of the wire CD n the solenoi s re A right of the wire CD is A L = 4. A Y A L i 2 i 3 A The equtions for the (right) loop CXDC n the (left) loop CDY C re respetiely gien y 1. E 5 + i 1 = 0 n 4 E 5 i 2 = 0. 2. 4 E 5 i 1 = 0 n E 5 + i 2 = 0. 3. 4 E 5 i 1 = 0 n E 5 i 2 = 0. 4. E 5 i 1 = 0 n 4 E 5 + i 2 = 0. 5. 4 E 5 + i 1 = 0 n E 5 + i 2 = 0. 6. E 5 i 1 = 0 n 4 E 5 i 2 = 0. D C i 1 X

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 4 7. E 5 + i 1 = 0 n 4 E 5 + i 2 = 0. orret 8. 4 E 5 + i 1 = 0 n E 5 i 2 = 0. y efinition, the res of the left n right loops re relte y A = A L + A. Sine A L A = 4, we n sole for A L n A in terms of A. A L = 4 A 5 A = A 5. Then we n ompute the mgnitue of the inue emf roun the right n left loops. E = A = A 5 = 1 5 E E L = A L = 4 A 5 = 4 5 E. The inue emf n the hnging mgneti flux re relte y E = Φ = A. Sine the mgneti flux is inresing, the inue emf is in the lokwise iretion n the iretion of the urrent is ounter-lokwise, s shown in the figure. From Kirhoff s lws, the loop equtions for the right n left loops respetiely re right loop : left loop : 1 5 E + i 1 = 0 (1) 4 5 E + i 2 = 0 (2) 008 (prt 2 of 3) 10 points Note: i 3 is efine s positie if it flows in the sme iretion s shown in the figure. Wht is the urrent i 3? 1. i 3 = 3 E 4 2. i 3 = 3 E 5 orret 3. i 3 = 4 E 5 4. i 3 = + 3 E 4 5. i 3 = + E 4 6. i 3 = E 4 7. i 3 = + 3 E 5 8. i 3 = + 4 E 5 9. i 3 = 0 From the loop Eqs. 1 n 2 in Prt 1, we n sole for the urrents i 1 n i 2, i 1 = 1 E 5 i 2 = 4 E 5. Sine hrge is onsere t juntion we he i 2 = i 1 + i 3 i 3 = i 2 i 1, i 3 = 4 E 5 + 1 5 = 3 E 5. E 009 (prt 3 of 3) 10 points The rtio of the rightness of ul Y to tht of ul X, rightness Y rightness X, is 1. rightness Y rightness X = 3 2. rightness Y rightness X = 9 3. rightness Y rightness X = 4 4. rightness Y rightness X = 16 orret

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 5 5. rightness Y rightness X = 2 The rightness of ul is proportionl to the power issipte y it. If the resistne of the ul is, then P Y P X = = = ECDY 2 C ECXDC 2 A L A ( ) 2 AL A = (4) 2 = 16. 010 (prt 1 of 2) 10 points Gien: Assume the r n rils he negligile resistne n frition. In the rrngement shown in the figure, the resistor is 9 Ω n 2 T mgneti fiel is irete into the pper. The seprtion etween the rils is 7 m. Neglet the mss of the r. An pplie fore moes the r to the right t onstnt spee of 9 m/s. 2 T 2 T 7 m I 9 Ω m 1 g 2 9 m/s Clulte the pplie fore require to moe the r to the right t onstnt spee of 9 m/s. Corret nswer: 196 N. si Conept: Motionl emf E = l. Mgneti fore on urrent Ohm s Lw F = I l. I = V. Solution: The motionl emf inue in the iruit is E = l = (2 T) (7 m) (9 m/s) = 126 V. From Ohm s lw, the urrent flowing through the resistor is I = E = 126 V 9 Ω = 14 A. Thus, the mgnitue of the fore exerte on the r ue to the mgneti fiel is F = I l = (14 A)(7 m)(2 T) = 196 N. To mintin the motion of the r, fore must e pplie on the r to lne the mgneti fore F = F = 196 N 011 (prt 2 of 2) 10 points At wht rte is energy issipte in the resistor? Corret nswer: 1764 W. The power issipte in the resistor is P = I 2 = (14 A) 2 (9 Ω) = 1764 W.

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 6 Note: Seon of four ersions. 012 (prt 1 of 4) 10 points A r of negligile resistne n mss m = 38 kg in the figure elow is pulle horizontlly ross fritionless prllel rils, lso of negligile resistne, y mssless string tht psses oer n iel pulley n is tthe to suspene mss M = 210 g. The uniform mgneti fiel hs mgnitue = 640 mt, n the istne etween the rils is l = 91 m. The rils re onnete t one en y lo resistor = 71 mω. Use g = 9.8 m/s 2. M Wht is the mgnitue of the terminl eloity (i.e., the eentul stey-stte spee ) rehe y the r? Corret nswer: 0.430785 m/s. si Conepts: F g = M g m F m = I l F net = (M + m) = F g F m E = I = Φ Φ = A E = l Solution: It follows from Lenz s lw tht the mgneti fore opposes the motion of the r. When the wire quires stey-stte spee, the grittionl fore F g is ounter-lne y the mgneti fore F m (see figure elow) M T F g T m F m I = M g (2) l To fin the inue urrent, we use Ohm s lw n sustitute in the inue emf, E = Φ I = E = 1 Φ (3) Note: We he ignore the minus sign from the inue emf E euse we will eentully elute the mgnitue of the terminl eloity. The flux is Φ = A. So Φ = A = l (4) I = l. (5) Using (2) n (5) n noting tht is the terminl eloity M g l = l. (6) Soling for the mgnitue of the terminl eloity = M g l 2 2 (7) = (0.21 kg)(9.8 m/s2 )(0.071 Ω) (0.91 m) 2 (640 mt) 2 = 0.430785 m/s. 013 (prt 2 of 4) 10 points Wht is the elertion when the eloity = 0.75 s? Corret nswer: 0.0399108 m/s 2. To get the eloity s funtion of time we nee the elertion (). (8) Apply Newton s seon lw to the r n the suspene mss seprtely F g = M g = F m = l I (1) m = T F m M = F g T, n

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 7 where T is the tension in the string. Comine n sole for = F g F m m + M, (9) where F g = M g n F m = I l. the inue urrent I = E n Sine, from (1) I = l Further, n E = l, so F m = 2 l 2. (10) F g = M g. (11) Thus, using Eqs. (9), (10), n (11), Eq. (8) reues to the ifferentil eqution or = M g (m + M) 2 l 2 (m + M), (12) = = M g 2 l 2. (13) (m + M) 2 l 2 014 (prt 3 of 4) 10 points Wht is the time onstnt τ? Corret nswer: 7.99821 s. ewriting the ifferentil eqution, Eq. (12), in imensionless form n isolting t on one sie n on the other, we get or M g 2 l 2 = = τ (m + M) 2 l 2 (14) (15) where is efine in Prt 1, (7), n the time onstnt τ is (m + M) τ = 2 l 2 (16) (0.071 Ω)(38 kg + 0.21 kg) = (640 mt) 2 (0.91 m) 2 = 7.99821 s. 015 (prt 4 of 4) 10 points Wht is the horizontl spee of the r t time t = 4.15907 s, ssuming tht the r ws t rest t t = 0 s? Corret nswer: 0.174675 m/s. Integrting (14) we he = 0 t = 0 τ ( ) ( ) 1 1 ln ln = t τ 0 ( ) ln = t τ ( t 1 e τ ( ) t = e τ ) (17) (18) ( ) 4.15907 s = (0.430785 m/s) 1 e 7.99821 s = 0.174675 m/s. 016 (prt 1 of 1) 10 points A light ul is onnete to ttery, turne on, n is isily lit. An iron ore is first rpily thrust into the oil, then rpily withrwn. S iron ore light ul These two tions will temporrily

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 8 1. righten the ul oth wys. 2. he no effet on the ul s rightness. 3. im the ul oth wys. 4. righten one wy, im the other. orret While the ore moes, urrent will e inue. Moing in one iretion will inrese the DC urrent, while moing in the other iretion will erese the DC urrent. 017 (prt 1 of 8) 10 points A retngulr loop with resistne hs N turns, eh of length L n wih W s shown in the figure. The loop moes into uniform mgneti fiel (into the pge) with spee. 0 x 0 0 x 0 L W Wht is Φtotl (the time eritie of the flux for ll turns of the loop) just fter the front ege (sie ) of the loop enters the fiel? 1. Φtotl 2. Φtotl = zero = W 3. Φtotl 4. Φtotl 5. Φtotl 6. Φtotl 7. Φtotl 8. Φtotl 9. Φtotl = N W = N L = L = N W L = W L = N L = N W orret 10. Φtotl = N W L si Conepts: Mgneti flux is efine s: Φ = A Φ totl = N Φ = N [ A] Gien: =the resistne of the loop, N=the numer of turns, L=the length of eh loop, W =the wih of the loop. Fin: (1) Φtotl, (2) The urrent in the loop, (3) The fore on the loop s it enters the fiel, (4) The fore on the loop s it moes within the fiel. Solution: The mgneti flux is gien y Φ totl = N ( A), where N n re onstnt, ut the re is hnging. A = W ( t) initilly, so Φ totl = N W t Φ totl = N W.

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 9 018 (prt 2 of 8) 10 points ememer: E = Φtotl = N Φ. Wht is the mgnitue of the urrent I in the loop just fter the front ege (sie ) of the loop enters the mgneti fiel? 1. I = E 2. I = E 3. I = E2 4. I = E2 3 5. I = E 2 6. I = zero 7. I = E 2 8. I = E 2 9. I = E2 2 10. I = E orret E = Φtotl = N W I = E = N W so the urrent I = N w is in the ounterlokwise iretion (from up to ) in orer tht flux is rete in the loop whih opposes the inrese of flux in the loop of the uniform mgneti fiel (into the pge) s the loop moes into this mgneti fiel. Note: The minus sign merely inites tht the iretion of the urrent will e set up in suh wy so s to resist the inresing mgneti flux. 019 (prt 3 of 8) 10 points Wht is the iretion of the urrent I in the loop just fter it enters the mgneti fiel? 1. ounter-lokwise orret 2. lokwise 3. No urrent The iretion is ounter-lokwise. 020 (prt 4 of 8) 10 points Gien: I is the urrent s foun in Prt 2. Wht is the mgnitue of the fore F on the loop just fter the front ege (sie ) of the loop enters the fiel? 1. F = N 2 2 W 2 L 2 2. F = N 2 2 W 2 3. F = N 2 2 W 4. F = zero 5. F = N 2 2 W 2 orret 6. F = N 2 2 W 2 7. F = N 2 2 W L 8. F = N 2 W 2 2 9. F = N 2 2 W 10. F = N 2 2 W 2 L The fore on the loop is gien y F = N I W = N 2 2 W 2 s the fores t on the urrent within the fiel, n the horizontl urrents he equl n opposite fores. Thus only the right hn ertil loops he fore ting on them. This fore ts to oppose the moement of the oils, n must point right.

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 10 021 (prt 5 of 8) 10 points Wht is the iretion of the fore F on the loop just fter the front ege (sie ) of the loop enters the fiel? 1. ineterminte, sine the fore is zero 2. towrs the ottom of the pge 3. left 4. right orret 5. towrs the top of the pge Sine F = W [ I ], ˆF = Î ˆ. Use right hn rule of ross prout, we see tht the iretion of the fore is pointing towrs right. 022 (prt 6 of 8) 10 points 0 x 0 Wht is the mgnitue of the fore F on the loop s it moes within the fiel? 1. F = N 2 2 W 2 2. F = N 2 2 W 3. F = N 2 2 W 4. F = N 2 2 W L 5. F = N 2 2 W 2 L 6. F = 0 orret 7. F = N 2 2 W 2 L 2 8. F = N 2 W 2 2 9. F = N 2 2 W 2 10. F = N 2 2 W 2 Within the fiel, the mgneti flux is onstnt, so Φtotl = 0. Thus, E = 0, I = 0, n no fore opposes the motion. 023 (prt 7 of 8) 10 points Wht is the iretion of the fore on the loop s it moes within the fiel? 1. towrs the top of the pge 2. right 3. towrs the ottom of the pge 4. left 5. ineterminte, sine the fore is zero orret 024 (prt 8 of 8) 10 points 0 x 0 Wht is the iretion of the fore on the loop s it moes out of the fiel (ege is in the fiel while ege is out of the fiel)? 1. left

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 11 2. right orret 3. ineterminte, sine the fore is zero 4. towrs the top of the pge 5. towrs the ottom of the pge The fore on the loop is gien y F = N I W = N 2 2 W 2 s the fores t on the urrent within the fiel, n the horizontl urrents he equl n opposite fores. Thus only the right hn ertil loops he fore ting on them. This fore ts to oppose the moement of the oils, n must point right. 025 (prt 1 of 5) 10 points A retngulr loop with resistne 13.1 Ω hs 62 turns. The loop s length is 9.3 m n wih is 2 m (s shown in the figure). The loop moes with spee of 4.1 m/s into region with uniform mgneti fiel of 5.2 T (into the pge). The fiel exists in the region 0 < x < 10.6 m. 5.2 T 0 10.6 m 9.3 m 4.1 m/s 13.1 Ω 2 m 0 10.6 m When the ege of the loop just enters the fiel n is etween 0 n 10.6 m, wht is the mgnitue of the inue urrent in the loop? Corret nswer: 201.808 A. Gien: = 13.1 Ω, the resistne of the loop, N = 62, the numer of turns, l = 9.3 m, the length of the loop, n w = 2 m, the wih of the loop. 0 10.6 m si Conepts: Mgneti flux is efine s: Φ n = N ( A) l = 13.1 Ω w Solution: We will fin: (1) Φ N, (2) The urrent in the loop, (3) The fore on the loop s it enters the fiel, (4) The fore on the loop s it moes into the fiel. The mgneti flux is gien y Φ N = N ( A) where N n re onstnt, ut the re is hnging. A = w ( t) initilly. Φ N = N w t Φ N = N w E = Φ N = N w I = E = N w (62) (5.2 T) (2 m) (4.1 m/s) = (13.1 Ω) = 201.808 A, so the urrent I = N w is in the ounterlokwise iretion (from up to ) in orer tht flux is rete in the loop whih opposes the inrese of flux in the loop of the uniform mgneti fiel of 5.2 T (into the pge) s the loop moes into this mgneti fiel.

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 12 Note: The minus sign merely inites tht the iretion of the urrent will e set up in suh wy so s to resist the inresing mgneti flux. 026 (prt 2 of 5) 10 points When the ege of the loop just enters the fiel n is etween 0 n 10.6 m, wht fore is require to keep the loop moing with onstnt spee? Corret nswer: 130126 N. The fore on the loop is gien y F = N I w = N 2 2 w 2 s the fores t on the urrent within the fiel, n the horizontl urrents he equl n opposite fores. Thus only the right hn ertil loops he fore ting on them. This fore ts to oppose the moement of the oils, n must point left. 0 10.6 m Just fter psses 10.6 m, while the oil is within the region of the fiel, in wht iretion oes the urrent flow etween n? 1. There is no urrent orret 2. up to 3. own to There is no hnge in flux, onsequently the urrent is zero. F = N 2 2 w 2 = (62)2 (5.2 T) 2 (2 m) 2 (4.1 m/s) (13.1 Ω) = 130126 N. 029 (prt 5 of 5) 10 points 027 (prt 3 of 5) 10 points When the ege of the loop just enters the fiel n is etween 0 n 10.6 m, wht is the iretion of the inue urrent flow etween n? 1. up to 2. own to orret 3. There is no urrent See Prt 1, the urrent is ounter-lokwise. 028 (prt 4 of 5) 10 points 0 10.6 m Just fter ege exits the fiel, in wht iretion oes the urrent flow etween n? 1. own to 2. There is no urrent 3. up to orret See Prt 1, the urrent is now lokwise; i.e., the hnge in flux is opposite to tht in Prt 1. 030 (prt 1 of 2) 10 points Gien: g = 9.8 m/s 2.

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 13 Assume: The ro remins in ontt with the rils s it slies own the rils. The ro experienes no frition or ir rg. The rils t eh sie n on the ottom he negligile resistne. A stright, horizontl ro slies long prllel onuting rils t n ngle with the horizontl, s shown elow. The rils re onnete t the ottom y horizontl ril so tht the ro n rils forms lose retngulr loop. A uniform ertil fiel exists throughout the region. sliing ro 4.5 m/s 1 m 58 g 6.1 Ω sliing ro l m 0 Ω Viewe from oe 0.42 T 0 Ω Viewe from oe si Conepts: θ Viewe from the sie 0.42 T 4.5 m/s 20 Viewe from the sie If the eloity of the ro is 4.5 m/s, wht is the urrent through the resistor? Corret nswer: 291.151 ma. Let : l = 1 m, m = 58 g, = 6.1 Ω, = 4.5 m/s, = 0.42 T. n E = Φ The moement of the ro ereses the re of the loop, so the flux through the loop is hnging in time, n there is n inue emf E. If we enote the re y A, this inue emf is E = Φ = ( A os θ) = os θ A. sine the flux is A = A os θ, where θ is the ngle etween the mgneti fiel n the norml etor to the re. The mgneti fiel n the ngle re oth onstnt n were pulle out of the ifferentition. Now, if we ll the istne from the ro to the resistor x, the emf eomes (l x) E = os θ = l os θ x = l os θ.

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 14 Thus the urrent in the resistor is I = E = l os θ = = (0.42 T) (1 m) (4.5 m/s) os(20 ) (6.1 Ω) = 0.291151 A = 291.151 ma. 031 (prt 2 of 2) 10 points Wht is the terminl eloity of the ro? Corret nswer: 7.61317 m/s. The terminl eloity is rehe when the fores on the ro nel, so it feels no more elertion. The fore from the inue urrent is, sine the ro is perpeniulr to the mgneti fiel, F,totl = I l. Howeer, this fore is irete prllel to the groun. We nee the omponent of this fore prllel to the trks, whih is F = I l os θ. The omponent of the fore of grity prllel to the trks is F g = m g sin θ. At the terminl eloity, these fores re in equilirium, F = F g, whih yiels m g sin θ = l 0 os θ l os θ where the expression for I from prt 1 ws use. We proee to sole for 0 0 = m g sin θ [l os θ] 2 = (6.1 Ω) (0.058 kg) (9.8 m/s2 ) sin(20 ) [(1 m) (0.42 T) os(20 )] 2 = 7.61317 m/s. ther thn worrying out fore omponents, it might e esier to use slr quntity, suh s power. When the ro is t its terminl eloity, the power eing lost in the resistor must equl the power eing gine ue to grity. Thus P = U gr E 2 = m g z ( l 0 os θ) 2 = m g 0 sin θ. Soling this for 0 gies the sme result s oe. 032 (prt 1 of 1) 10 points In the figure shown, the north pole of the mgnet is first moe own towr the loop of wire, then is withrwn upwr. Clokwise N Counterlokwise As iewe from oe, the inue urrent in the loop is 1. for oth ses lokwise with inresing mgnitue. 2. for oth ses ounterlokwise with eresing mgnitue. 3. for oth ses ounterlokwise with inresing mgnitue. 4. for oth ses lokwise with eresing mgnitue. 5. first lokwise, then ounterlokwise. 6. first ounterlokwise, then lokwise. orret From Ohm s lw n Fry s lw, the urrent in mgnitue is I = V = 1 Φ, where Φ is the mgneti flux through the loop. We know the sign of the rte of hnge of the mgneti flux is hnge when the mgnet is

Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 15 withrwn upwr, whih, oring to the eqution the iretion of the urrent is lso hnge. From Lenz s lw, we know when the mgnet is moe own towr the loop, the urrent in the loop is ounterlokwise s iewe from oe. 033 (prt 1 of 2) 10 points The ounter-lokwise irulting urrent in solenoi is inresing t rte of 8.39 A/s. The ross-setionl re of the solenoi is 3.14159 m 2, n there re 163 turns on its 18.4 m length. Wht is the mgnitue of the inue E proue y the inresing urrent? Corret nswer: 0.478277 mv. si Conepts: Fry s Lw for solenoi E = N Φ = N A. Mgneti fiel proue y the hnging urrent is = µ 0 N I L t = µ 0 N I L t. Fry s Lw for solenoi E = N Φ = N (A) = N 2 A L µ 0 I t. Mgneti fiel inue y urrent = (1.25664 10 6 N/A 2 ) (163) 2 18.4 m (3.14159 m 2 )(8.39 A/s) ( 10 3 mv V = 0.478277 mv. ) ( 1 10 2 m m 034 (prt 2 of 2) 10 points Choose the orret sttement 1. The E tries to keep the urrent in the solenoi flowing in the ounter-lokwise iretion 2. The E oes not effet the urrent in the solenoi 3. Not enough informtion is gien to etermine the effet of the E 4. y the right hn rule, the E proues mgneti fiels in iretion perpeniulr to the preiling mgneti fiel 5. The E ttempts to moe the urrent in the solenoi in the lokwise iretion orret As the urrent is inresing in the ounterlokwise iretion, y Lenz s lw, the E will ttempt to retr the urrent, whih estlishes n E tht tries to ounter the flow of the urrent, whih in this se woul e in the lokwise iretion. ) = µ 0 N I L. Solution: Thus, the inue E is E = µ 0 N 2 L A I