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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 103 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‘idlrt' t1. cerimdrtlﬂi Gﬂuﬂlmﬁ Wﬁq. w'ﬂ‘h End Cars, ofﬁce: A M GMSL' innu
ﬁw Gout5.1m 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‘€zds)+{IE3ds)
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 10  '
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» swamI ﬁrm/5‘s m .
Mi‘r‘anlcd Lewem ....C', Ml. a3 sum W Mfiutal 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—IC
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 [Secadds ya the Phys ME Finat 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 Inag “‘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 {viM) Helimlmmlr = ULM + 54.24641 W ' ~ ' acMucM dhmb‘l
Thar/“aw OFM Miimlwhmdiaﬁ ”TmﬁlL (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 sLx. wm'mi'oefw PM'H muj Behring Mf—ieﬂdi Md W H: [5 BLEL— (x—zcjj when
M19011 'gaxiiqﬁﬂo lamina M5113 W; 435 V5 xgwit” 4%. i=—d¢‘= 3::— Phys ms Fm: 5mm Fuse $9 die. : 1%0‘ = — (slmof 4’5‘5 lawi") U" Jm5,2000 PROBLEM (7) {15 points) 1 5 f
The circuit shcws three ideal batteries. three resistms,
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. [mil3&3! aw. iclul thunder sarir?
W QWVh 4w; WMWH a
rt} itadsmmcpucﬁrmdtct
13:5 Ma WM dawn/Mi fem): Mli'l MM Ill1.40M». FW I'wrl bfﬁb : +3+tht=°)—?=o _—:. Mitwﬁeéﬁl Malice.“ 4344. and a? Wham “nadirla 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. curN41 v34 loop bafb $stme WW9 Imim ...
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 Spring '12
 OnurFen
 Tn, 4w, Winona, Mc prmaimm mmﬁm, euj Behring

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