amperes_law - 1/6 Ampere’s Law Applications Consider the...

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Unformatted text preview: 1/6 Ampere’s Law - Applications Consider the situation shown schematically in the diagram. Currents [1,12,13,14,I5 are flowing out of (-)or (+) into the paper The corresponding B —f1elds swirl around their sources as shown. The main point is that the B field lines circulate around the currents Choose a closed loop (L) Start atA, measure B, choose small stop Al along loop Calculate Dot product (component of Balong Al multiplied by AZ) Kama 151m 0105 (E, M) .ILE-Uéf, Pagan; 0 p Rpeat this calculation at everystep as shown. BoAl —> —> Write out the sum ‘23c BaoAl; c: closed loop. This sum is called circulation of B around closed loop and Ampere says that it is determined . solely by currents threading through the surface on which the loop' 15 drawn & only currents within the loop contribute, i e. exclude 15. The mathematical Equation 13: T—m EcBwAl=yoEInuo= 475x 10'7—— A 2J6 Qu- In words, circulation of B aroundA closed 100p is proportional to the algebraic sum of the currents _) V threading the-m S werfw 69W w«V\ac/‘"<:.£~ def/L «(£32352 Li's dreamy. N016 lbs" t'm Cause ‘7,” éaus’f Lam, flmpeygég law grueg thcfu/d/{me/bui 0906 I?) {0 366’? /?> you raged Ltgp/Ewsgwmi/éfipl Applications 1.) swam; Cu £12m Single Wire with current I, there IS cylindrical symmetry so B can be a function of r only & must encircle]. EA, ’3 anyf (LE, m "Para/M “ID 63% Gaga—mad] Appropriate loop IS circle of radius r centered on the wire 3-2271? ,uOI J44!“ 30,13—2” ‘ .. ,, as claimed previously 2.) Next, we'begin by Showing that if current is outside the loop it contributes nothing to the ' circulation. Choose L MN 0 P Q with I at the center of the circles of radii r1 , and r2 rm in: 2/5 B( if” — r1 2727”: ELM 103‘? ”‘3' _#o . 2 #01 ’ ' #01 [The first” (firm as #:ch ggélzfizm+0+§g2flé+0 l3; awdflfl (32442, :L—)M—+N+N—)0+ OefaQ '+Q_,L $3,073,)??Y‘: fa m =0 munch/b 3.) y x Z J At y=0 there is a Current sheet of thickness 2‘ carrying current density J = — J2. Looking? it end— on we SEC sources as B _) m and we see that y—components of g: cancel out. gun? survives. Let us take loop of width 1 and height h. = 231 ’ T = ,quz‘l ' so, Jt Jt B = 502—353 = L)? y > o «57.4 _ 4.) Hollow Cylindrical Conductor— Radius R, carries uniform current. We want B at a distance r a a from its axis. Since there is a cylindrical symmetry we should use circles centered on the axis for our closed loop. For r<R. use loop 1. B . 2727" : 0 -, No Current threads through loop 1. fat «'2 >12 so, B. 27zr = ,u01,the entire current threadsloop 2. #01 A B:— SO’. a, 2”. . ZfiRf 9 Note: if cylinder has wall thickness f: I = J . Wand field at surface would be ,qut. . > i Again field would jump by yth on crossing a current sheet. 5.) SOLID CYLINDRICAL CONDUCTOR — with uniform current I Define (I: ”R2 i‘IlI-IFV R Now for r<R I: Jm‘2 - ' J A 1;: egg R 6/5 _ . . ' [“01 A For >R,'ent1re I contributes B = — —> 2717” R x i r 6.) Solenoid: Tightly wound, smallradius, length much larger than radius: N 1 Look at two neighboring turns , ' N turns, L longjn = =# of turns per meter. r—component cancels By inside survives. Long—narrow solenoid B m 0 just outside. ._) [:13 $bflll1’2éf “MM/St” Cam-e. out qui'DPleDl’ cfiflouwd 61"”58 Bl = ,uonIl 42mm cuff Take loop as shown B : #011 I boa NW ”NE V For ease shown g = you]? be. Vic/lo ] ...
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