HW+7+solutions

HW+7+solutions - - _ 2]: '0 m B“ ‘— 5 «Lame = 0.8...

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Unformatted text preview: - _ 2]: '0 m B“ ‘— 5 «Lame = 0.8 gauss r 3 x {0‘0 x 5 Force per cm: IB/c = 6xlo'0x 0.8/3 x 10'”: L6 dyne/cm - 8000 amp = 24 X lo'2 esfl/Sfic currenl’ inside r=| is 6:4- 10‘2 esulsec _ 21 _ z'xex'ro'2 81-??? _m : 400 gauss _ 2 K 24 x 10*2 Ba _ Max loiOxZ : 800 gquss z x 2'4 ><lCl12 : _____.._.___ : 5 53 3 X mm x 3 33 gouss I 6.5! ' A’r P, The Field 0F wires A andC ®2I IQPC Cancel. Fs'eld 0F wire 8 a+- B d \ngi) | P, is “:1 = 1%}. C E \\ :d P [\l 5' Field OF B of Pa = chm I FieldolBX 21 \ . _ A©_‘——d—.X}<_L_F|8!d O‘F A-h—d— I . PERFE'CId OFC The veclor sum 0? +he 3 Fields 0+ P2 is zero. Curren‘l rl n9 5 _ I x 2n(R/2), (R/zl / ‘ cN-ERDZ HER) : HI : 0.5 gauss c R I : QgcR : 9|? 3x IO'ox 6,>< logesu/sec R: 6 X loscm :; qx 1018 esu/sec : 3 X lo? amp For Io’wafis m 5 x10” volTS, I = 200 amp- fiolkflamp Field in Teslas al I meTer : mire/mews —1 _ Ll'rrx l0 200 : ~5 _ , 2fi__.x I 7L+><l0 T — .‘lgauss O‘rher wire causes equal Field 1 B : 8 x IO'ST = 0.8 gauss Average diamefer 0F +urn = 8+2 x 0.163 : 8.3 cm ToTal lengfh 0? wire in x 8.53 x8 X 32 = 6'! meTers Resismnce = 0.67 ohm I: 50/.67 :75 amps Power = 50 x '15 : 3750 wa’r‘rs {V 8 >_< ‘15 amp’rums/Cm B /e:__ Field in inFin'tfely Iong Coii Would be I“ 4'“ _. . 93m” (4H6) .LHT x 600 or 75% gauss. B = “154 X Cos (Tarr' ~15) = 7'31 QGUSS Capoc’d’ance 0F coax'lal cylindersJ hang-H“: L: L . Charge per unH Eengfh on C : 21mg. inner Cylinder (assumed posifive) = CPC/l— : 43/2 2W1;— : So/zbuM/a) : ‘37 esu/cm IF inner Cyiinder ro‘rdfes’l 0+ 30 rev/sec olenoidal surFace curren‘r 0F dens‘lfy } = 30x 87 Inside #1ch cylinder B = Ling/c B = O, r>a . i+ £5 0 s : 2610 esu cm’l Sec". = 1.09 x to"" gduss (Em‘o paper), N G: I? boTh cylinders rofa’re 11 0+ 30 rev/ sec; 5 : LOCJ x K)" gauss (001%)? paper) a 4 r < b and B20} ['(O, r>b. The analysis in Sec. 59 showed Thaw“ cl TeST charge maving parallel To a Wire carrying currenT experiences a Force. which, as observed in The res+ Frame oF The TesT Charge , is due To an elecTric Field. To undersTand why The inTroducTion 0F 0 conducTor, such as a m‘eTai plaTe, beTween The wire and The TesT charge has no eFFeeF, leT us view The siTaaTion From The res+ Frame OF The TesT charge. In ThaT Frame, The Conduchng ploTe, which is STaTionary in The lab Frame, is moving. IT is moving Through a magneTic Field $131 an elecTri‘c Field and These are relaTed precisely So as To make The Ton' Force on any charge in The plaTe zero. Hence There is no redisTri- buTion whaTever OF The eleCTrons In The plaTe. On The oTher hand, iF we caused The piaTe To move along wiTh The same veloCiTy as The TesT Charge, 1T M make a diFFerence. An obServer in The TesT-eharge Frame would say ThaT we have inTroduced a sTaTionary plaTe 'mTo on elecTrosTaTic Field, wiTh a consecLuenT redisTrlbuTion oF charge on The plaTe and a resulh'ng al+eraTion OF The ToTal elecTric Field. An observer in The lab Frame, where There was no elecTric Field beFore, would say ThaT The eieeTrons in The pIaTe have redisTrib— uTed Themselves under The inFlaence OFThe CL%X§ Force, and The new charge disTribuTion iTselF produces an . elecTric Field. ...
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This note was uploaded on 03/01/2012 for the course PHY 9HD taught by Professor Curron during the Fall '11 term at UC Davis.

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HW+7+solutions - - _ 2]: '0 m B“ ‘— 5 «Lame = 0.8...

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