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106-19992-FIN - PROBLEM{I(25 points 09 Two infinitelyr...

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Unformatted text preview: PROBLEM {I} (25 points) 09 - Two infinitelyr large sheets of charge are placed parallel toeach other. as ahowninthefigure. The surfncccharge densities or; = 4 x liloE Cimz and oz: '4 x 10-3 Cintl'er'e' uniforme distributed overthc shccts {Take £n= —9>< 10'” CZ!” 4mg}. ' (it) Find the direction and magnitude of the electric field E in regions I H end H! State the direction according to the axes given. (a) (JHE‘idl-rt' t1. cerimdrtlfli Gflufllmfi Wfiq. w'fl‘h End Cars, office: A- M GMSL' inn-u fiw Gout-5.1m wfinqfi) 23%»: can .3. E5- Ear L1 Ettgt — r141 "Ha: I ‘1': E—Lu—ln- :- ' Ite. "I ' 22.. . -'-i .. H EI=_ Klti' inti- =_2___ mu‘tfié 2:15:19 ) "l . (13} Find the potential difference between points A and B E r a t. q \i' 5-"le '15 -——> v —v =~f€ ds=(~J‘€z-ds)+{-IE3-ds) 1" c r II’L“ U“ t'EZ £15 +£15.25 =—-—-J:-I..I'tfl'!i'{Fifi}+v1 THU qrfifi) Sirocc— EE 7-” 314L- Pltjlr 1'95 FinniEm Page” June LEW 1-0 - ' PROBLEM {2] {15 points) ' Two erieal. thin conducting shells. with retlii e and b are placed fer . but ennnected to each other 1 thin. long ennduettng 1wire 15 shown in the figure. The III are uncharged. A point charge+qietben planed heme eenternfthe largerephere. Find the chargedien'ibtttinne, intlueednn the inner and outer etlri'mee of bath spheres. in terms of we parameters given. M3 cling: . 5L; 1:! Phylflbfinflfim Pup-#9 M5,!“ 1 1 PROBLEM (3} (15 points} '- t“ The circuit shows an ideal battery, two cspuci- - totsC.andCz,sresistorRsndaswit¢hS.With _ ' R capacitors completely uncharged, switch 5 is closed on point a. When C1 is completely I | I charged,3isthrownfromatobsttintct=fl. V Cl ('4 (a) When equilibrium is reached (i.c.. 1—) on) - what is the potential difference across each 5 capacitor? a b '- "' ' - :- (‘Ll WW C1 v}. MPHM chawfitdj Moi-«Mac MI'E is ch = (LIV If 5 s Wm M w I» swam-I firm/5‘s m . Mi‘r‘anlcd Lew-em ....C', Ml. a3 sum W M-f-iutal LWWQQ but: .n ,. . - awed-”It =c,v I can" +c',_UJ = C, V 'I' .r I' Up =Ut. =i/ Ill V; - .—c—._'u- V ¢I+¢L {b} How much energy is dissipated as heat in resistor R from r = D to r: on? q——l—I-C| WI .> t a |I :FLC I. L .. MWfi domains z. “cht V new we: amt can mass rm 5. arm PROBLEM {4) (15 points} A battery of V vults is connected across the ends of a cylindrical cantiuctcr of length L and resistivity p. as shown in the figure. The number of free electrons per unit volume of this conductor is n. In terms of these parame- ters, find the magnitude and direction {accord- ing to the given axes} of: [a] the current density J in the conductor: (h) the drift velocity up of the free electrons in the conductor; --a [c] the magnetic field B at C. inside the conductor. at radial distance r from the axis. Usnhj Mpm's (m, we. Md 39E d; = 33 [Sec-adds ya the Phys ME Fina-t Emu: JHMIJDOO 13-. . PROBLEM (5) (15 points) Apointlchsrgcq .widtmassmispro‘jected' intoare» on cfnniforntq magnetic field B,w1th an initial Irc- *3 f - ocity 1'. making an angle of s 1with the magnetic _3... field, asshowninthefigure. Sincetheveloci ofthe (\“ chargedpartlclehas'acomoonen tparallelto Ina-g “‘4‘" s I l netic field. we particle 1null more in a helical path about the direction of the field vector. (a) Does the horizontal velocity component affect the radius of the helical path? Explain. No. TLu, mediwiar @94ng dfl‘ibrwtmce ‘i‘m. mm o{‘ M LIA-“Ml W. L5: "fee:- 9 “-5 saucy am end—«m Drum”. an, —. are; e __3 we... await” mm {vi-M) Helimlmmlr = ULM + 54.24641 W ' ~ ' acMucM dhmb‘l Thar/“aw OFM Miimlwhmdiafi ”Tmfil-L (h) Express R, the radius of the helical path, in terms of the parameters given. (qu _MUPMr:-4§ R:i‘{T§— flig“ £115 E (c) Find the distance x, where the point charge crosses the horizontal line that passes through the point of projection. PhysIDGFtnaiEnm Mei-’9 JaneJthiti PROBLEM {6) (15 points) '1 4 A thin wire of resistance R is bent to fon'n a square loop with sides of length L. This loop travels at constant velocity 1' through a' re ion of width It in which a uniform magnetic fie d B is resent, as shown in the figure. Drew the gm 1: of t induced current in the loop as s function e x. which denotes the position of the right hand side of the leap. Lshei the clockwise direction of the current as positive, and the counterclockwise direction as negative. in your graph. Give the magnitudes of the current on the graph as well. :xxxxxx{ XXXXXXI )4 I X ‘JIE 5t: 1‘ I X X K K X X: r- :x x x x x x: Btinlothepsaei TL“- iCiLU‘ it“? zero whim Wimp r}. not vii "I‘m field; H: is BL; tutu“ W— lonw is» assets is w ecu) it c sL-x. wm'mi'oefw PM'H muj Behring Mf—iefldi Md W H: [5 BLEL— (x—zcjj when M19011 'gaxi-iqfiflo lamina M5113 W; 435 V5 xgwit” 4%. i-=—d¢‘=- 3::— Phys ms Fm: 5mm Fuse $9 die. : 1%0‘ = — (slmof 4’5‘5 law-i") U" Jm5,2000 PROBLEM (7) {15 points) 1 5 f The circuit shcws three ideal batteries. three resistm-s, an ideal inductor. Initially switch 3 is open. but closed at time t: D. (It) Find the potential difference across the inductor, just after the switch 5 is closed. [mil-3&3! aw. iclul thunder sari-r? W QWVh 4w; WMWH a rt} itadsmmcpucfirmdtct 13:5 Ma WM dawn/Mi fem): Mli'l MM Ill-1.40M». FW- I'wrl bffi-b : +3+tht=°)—?=o _—:. Mitwfieéfil Malice.“ 4344. and a? Wham “nadir-la S'L‘Ai dot W Wfiafi WM ‘l‘vu, End “gawk M 2—4 MW {b} Find the current through the inductor. a "long' time after the switch 3 is classed (ism. t —> on}. A L91»: M [diam J an I'M Wdutfior ads llltfl. asW‘l’ film. Tia—whys. VLitsNJ=OIILGI15W hm. cur-N41 v34 loo-p baf-b $stme WW9 Imim ...
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