hw6 - 8 III" u...—-———-._._.——--—-- -r-...

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Unformatted text preview: 8 III" u...—-———-._._.——--—-- -r- ' C' 3 X 10'“ x 5 9 Force per Cm: IB/c = 6x lo'°x 0.3/3 x lo'°= L6 dyne/cm 8000 amp = 2.1+ x I0I2 esu/sec curren+ insl'de r=l is 6xlo‘zesulsec “a; _ exexrolz _ Bl " e " 3x lo'°xl ‘ “go gauss __ 2 K Zl-l x lo'2 _ Ba _ -——.—_._3MO‘HZ _ 800 gauss _ 2 x 24 x «3'2 _ 53 ._ __._____3 x [0li 3 .. 533 gauss z: Zwb‘I [E’de zznb‘I °° dz 3 a 9 30324-22)" -o:: C _w(bz+z)/‘ _ anle: Z M_ 2114321 47rI ‘ c b‘(b1+1‘)"2 " c b2 = c. _m — To see why The return pa'rh ,"' “\K can be ignored in The “mil 1/ \ z -* W , consider +he Flnll’e [I b < /+\\ . pa-lh ou+ fa z =1", refurning_L_fi_9_L_-‘J:-(_)___,_ by way 0F +he large semicircle. T - Z I On The axis 32 ~ 33 , For Z >> b and we may aner +ha1', going Ou‘l' in any direcfion From The Flag, [Bl H "1:13 as r -+ 0°. Since ‘l‘he leng‘l‘h 0? H16: semicircle f5 proporfional +0 r‘, The infegral IQ -dg over +he semicircle mus+ vanish a+ least as Fas’r as I/I‘1 as r-) m. [In Fad, i+ vanishes Jus+ Thal- Fas’r] Each Sfruighf secfion confribuhes HaIF fhe FIE-id 01cc: half: We ‘Fl'eld O'F an fnfl'n'rfe. wire Complefe ring B=(2+1T)'r;% = 5.!4l6F-2'E5 - Af F; The. Field 01" wires A and C 21 I cancel. Field 01“ wire 3 0+ 3914‘?) QC . 2x21 _ WEI z I R l$ d -— d \x\ :d C(T‘f) C- t<< : metfi'5a+g==2§§l Fieidam‘ ‘ Curran? ring — I x 2n(R/2) _ (R12) B ~— ._.._____..._ / com): (xi-53R) :1 I z 0.5 gauss R I = 9.2m; = 5?? 3x Io'°>< 6 xso'esu/sec = 9X I0I8 esu/Sec = 3x 109 amp - Eviden'l'l): +he magnel'lc Field OF The curren‘l‘ in +he wire was .2 gauss al' a disfance 0F roughly 2 cm. The curren‘l musf have been aboa+ Z amperes. - For 107 wa‘l‘l's al 5 x lo“ walls’ I = 200 amp. . . Ik— Field In ‘l’eslas al I mefer = %—erfl'2§er§ -1 _ 4Trxio 200 _ -s - 2n X—r-—’+xIOT=.‘lgauss O+her wire cauSes equal Field 1 B = 8 x l0" T = 0.8 gauss m Average diamefer 0F +um = 8+2 x 0.l63 = 8.3 cm Tofal lengl'h 0F wire = 1r x 8.53 x8 x 32 = 67 me'l‘er's Resislance = 0.6! ohm I: 50/ .67 =75 amps Power = 50 x 15 = 3750 wa‘l'l's W 8 K '75 amp’l’ums/Cm .._ 45-21791 Field I'n inFini‘l'ely long Coll would be afi‘m—lml'e') .Lhr x 600 or 75‘:t gauss. B = 75"L x C05 flan" 4;) = 731 gauss - B a. 7% d: ’ Y 7 coils For" apar+ coils close +u9e+her K—-b-——!i biz—x biz 4" . . B a: + y Differenl-im‘e Y [01+(blz_y)z]3/1 [G2 + (b): +Y)2]3/z 2. 'l‘wice and set :3: = 0 of y=0. This gives: b=a Y daB Nola +ha+ dyay = 0 of y=o Jusf From Symmetry. Wl'l'h '3 = 0 We have 3, = Sim) 4' consTan‘l' x y" + r- Two Coils +hus arranged are called Helmhon Coils. m IF ‘l'he hole were. Filled Will! a copper rod Carrying a curren+ OF 300 amperes, compie+e symme‘rry would be resi'ored and The Field 0+ P would Surely be zero. 50 +he ac+ual Field 01' P mus’r be fine negafive 0F +he Field OF The rod Jus’r described. I+s magnii'ude in gauss rs (z/io)I/r~ wi+h I = 300 amperea and r = 2 cm , or 30 gauss, and il’ poin'ls +0 The leF+. A more remarkable Fact noi' 1'00 hard +0 prove : The. Field is 30 gauss poin‘l'ing *l'o‘rhe lei-‘1' nof only 01“ P bu+ everywhere w'd'hin file; cylindrical hole [ ...
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This note was uploaded on 04/25/2010 for the course PHYSICS 200 taught by Professor Davidbrown during the Spring '10 term at UCSD.

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hw6 - 8 III" u...—-———-._._.——--—-- -r-...

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