ele401TermTest_2

ele401TermTest_2 - Vania" “B” “A ELE401:...

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Unformatted text preview: Vania" “B” “A ELE401: Field Theory Term Test #2 Examiner: flirt Silmberg Sections: 01 - [13 Paul KanltrrEk. Timu '- LTE hour Instructions: D Cleaned heel: test. Aide swell as VEETDI' upemtere and integral tables will be provided ifneedetl. Ne queetiens are he he asked during the teeL Ifseme aspect ufa quesljen i3 endear make euitnhie nssmnptions — mite down the flS'IJIIlPIiDUS - and continue seiv'teg the prehlm Calculators wiil not be require-'1 and are not permitted. e Answer all three {3} queatiens. Weight of each qua-slim: is indicated. Part marks will be given. Note that there are 6|} marks available in total, but only it] marks are needed for a cemplele paper: Le. 20 bonus marks are pumible. e IIEIeatI}r Show all work in the space previded. If: "Rigel!" answer appears without any htdieelieu. on how it Will derived that answer will be considered [a be "Wt-neg". Use the back sides efpagee ifmcrre weridng. space is needed. Underline or your simplified inmate and shew proper units. Cirele peerseelien number. [11. 02, 113. D4, [15. DE, 0?. +33 Page 1 mm: mm: In m :2 mm “B” Mill. 2N3 Miami-'Mmlflml [flMngnEflchctaniddHDueTuAFhfleUncflf Current: [Emilia] Use the exprusiun Earths]! -fic1d due to 3511911 lineul'mln'eut “I " located on the z— axis, i.e. H = J—{majrcmul} a‘ [Aim] 47m to win: for the E - Enid caused by a short line ufcun'entl [A] flaming fi-om point M 5, D. {I} m p013 E{- 5 43, o, u} as seam a: pain! rm. 5, a). " I‘le‘ll- H}- —-"‘-" “‘l'_- —--:"—1 {‘5' (I mail: ‘5"; 2 4E "~ ELM)???“ l r.“ J,” 2 I. a '- . K EchJC-‘fl': “"31 'E > J . " I Pap: fl mm.- mnmy: nm rm to rm 3” my, ma We Pro-Elm: I: {h} Magnetization mi Bound Volume and Sol-fie: Current Definition: [6 merits] Ifthe nmgoetie inteufity vector field H ineide n hfllin‘w infinitely long conducting oylinqtller, neutered nil- on the maximisg'nren by: 3E :ejfiflh- H = 1 a. we] A2" write?) 9 lee—i oflefifi The inner radius of the cylinder is “3” [m1 and the outer is "1:" [In] and p ,= 34E] for the mil-[filial ,, inside the eyiinder. 'x/ m N A ["1] Develop the expreeaions for the magnetization l.revetor field M. [Aha]. all-’1' xm-vur—f = 3 _' i : 5:"? ____fi__ fiimm for the volume Current density J b. and the magnefinlion eurfaee content density K I, on both minute of the cylinder. _u_ __ a d at? “a F“ = J I” to 3;: \j? x m {a a 3 L Hut-$31: f;— Qfl“ (“EL {iiiJDeteuuine the expression for the auto of Ii... + I... where “13., " is the honed volume current and "h. " is the bound snrl'nee current. a”. jb: [:3 .clna t C} .L. o :55 = §ZK§L (eajegaaji eggxecejrmfle {ii-1‘6. j c4433 0 "Mb—":1 H; '51? __n_ -_*-- h "A I g; "" _'_" Ci 'Qg’ Tijifiidéx .liacfpéfi‘g‘ +§rfiwfifij ¢=e at! time o sue a -6 (L?) + (g) 21(25)::3/1: ref} CD Mfll:flflimw: mm '3 MIMI: [c]: Magnum-Romanflfinuflfimuhj Thehaundaryb-amBn two difiemntmaguefinmamialsisgiwnby thncquafiuny=fl maimhasmlnfiwpeuneahflityp” = 1.93m H; llax+2Iy+3u[Mn]. mfibflfluug=1flanflasmfimmrtmldmfilynnthebuunflaryfl=+4afiNmL Solve fDrB|,B:.H:.andKainh-Dth regions. W- Pfifif W ELEEIGMGD: 1;? a; :gufzj, ____,_._-—--"—" m a: #1 Q = "I 313:2; jian KLEIN: MM: Tm Tut I: Han-ck”, 2M5 anti-under: Milan 1: [d] Induced Current: [2 math] Fur Ihe diagram show below drain game correct mars which would cause flu: nun-en! to be flowing in the mndufling ring in the dimefiern as shown. (i) The switch has been denied for a lung lime. Elfin?! new: r [fiThe mtch has been upen for I. lung urne. T: .' 1 smears: a 90 I £5: l f: ‘U A ._ . {EiiJThe Mich is just being timed. f,”— E: [iflThJe switch isjus’: [min—g opening: E} THE wMLCEB FIELD LE. Saflflffflfi THE Arvin-5?: Pan-5553 *' 7?”: fiPflJE-fll «WE-iii F: L mug}- 3:3 EécfigfixS/AJG -._.-..._.- If fir—(H {5 J6! ST BL?HJC'§ f .o—— '--.__ _.—-'--._ _.--'—-..._ __\_H_ (:LLLJ EFJE fife-Life?“ x; H .1} .1; {3 {11:22:55 _._|___.—..._ _ ,2. THE 5M {a} A Dam: of Surface Current Dtnlil'jl': [11} mark] Admnejefinedbywa; flsflsfifi; D£¢slmearriesamfaeecmentdemity K = Kn shfl a‘ [Aim] [iJSelvefurthemulem-rent'T [A3flowiuguulhedem1e.{flfigme I= IKxnlrd'I } 5’32 " I= jzgefificflxflfi aficzféczg 9:” 2% Ha fir; z: “I I P4315 ELENJ: mm mm.- M 'Ihi' #2 um “B” “with 29,. ENE Continuation my 2: (fi}Der1ve flu: magnetic utensil}! mm: field expression H at the origin due m the surface cumnt dnuslw K. Use the inmnntal version Dfihe Biol-flay” Lam and Clth indicate what are the canyonan r, I", [r - 1"], Ir v r‘1: and what '55 ".113" in dis case. tint-z Isaa’aae=—jm“adma=j{m19—1}dcoss3 _W . Ax}; fJSfiri'é ; Itfl; Tara-fl; «cf-FF £3513 - I: ’: JUL“; 1:? . (grmmmmyzfi = “diam e95!" 6519¢ ' I HI. u TE._-'LL, Egg-“f: awake fix; L. _ '_;_'_;'_ _-- J“:— flr I? J?” a; 352’ #1 ° 535 H = : rm flmrgficgg 3 52%; I f} ‘5" m fir': r @LU 93-0 ,- Jae E; 45 2‘— 1 "F ' dry: t'H I'I—r; wfi‘fig F flflcufr-EEIVL‘FCL-flfl-‘f—r’ 41”” I. n: 1 _ A __:-_ _ I _ {a LIV, _ _ 3/3 ‘ £164; LL11 44m 53?. - find a? - (m5! at; 4» £15; _';. _ " '_ :_ fly“ (I; : {Rig Cfi-fiqfl gag. 1n. a; F}; FAQ-“3 1:151:5- 7%.! am am?! a? 9*???“ Fl *4“ "5 I __I f‘ JEMJMmE Elf—TELL. 4:3; %; fcflwu h fl; mint; {MAL {'3 35—; A If (II-J lg,- Egg (21310 Jpn-Sf? - :1 53$. :- — Q-r-‘V‘LE g d; a: “A .nfl. Hr: +__% g {4m fligdqfi Caz) .- 6 .~ ~ (= 15:: r F, 1:51 flit] F" Eu- 2 59:0 Wazw : (a; -mé] 6%; :1 L1” 5’ ,4 =52- [éfia - M H 3% if: JEE-flrifllflg If "fil%4§mh 3" {EL Widen “3” m1.- Field W: Elm Tflf H3 film-FIJI, 2W3 Centaur-fin: Problem 2‘.- (IJ} flu: Infinitely bung Currant Carrying Cylinder centred at: the I'. — uh]: [ 10 math] {,4 544:. An infinitely Ionng cylinder nfradiue "h " [m1 centered on the z—nxis supports a volume current Intensity.r given by: F5- J = Japan; [Aim] If. 4:15-73»? | [i] Solve for the total current "T" [A] flowing in the cylinder. \ :l _l I; an .e -— _.—I "_1. _,_. '1 __.-' J II \ I: {3,593 = “he? (6369904?! ‘ a 5- . n- .I _ g? r'l—rrl- Ufa 1|— K _t [TW— {iflDeme' the expressinnn for the B field inside amt outside of the conductor. F '—” ‘: J I T- ' 12-: ._': f 1/} J at": “3:111: H “5*” = 3*“?st I? l m «11 - 3 {_JE; {M 3 HF 239g re RA _ _ _ I. _ «Lt/gt: all; gig—£3525 [#1ng A161 filtrate; r 3.4357; Jet-Me. (flee) H? fiffl : @Qea #9: : gm _i/fl _ __ ___ f4 = {*“J: 5; We] 4: f3 mnmmw: mmu mm “B” M 19‘, M3 Cam: Haiku 2.- (‘ftilLSolvn fur the vector magnetic pnteutial A Enid: affine conductor if}; = C! a]. p = b [m] using the vector Poisson Equafiun. 4 Jam EH45 {g qfljaéf' 4 " j {'34: Eigg= yaw I II' I ll _ H _ _ .2. " =- - "' Fir—fi'flrfifléah #/{"‘Z/5 1‘3 flair“) - ryj1mlfnfg..mm€ fawbtlaflq. fr: : _ _ 3 ’g—fi(r”%}JflJJ‘/fiZ/fl «ii-cw}: wane-EL? an». {3? f“: 1- L-‘g‘n-Ld: ‘/ .a’ . «- tib’filj'dfbffi'pfi fmxigfi w(.2’./fl flair: c1 «9C".Crzr:=f I' I“; = ymflcfifi + ‘3 I KL "4‘ ‘1 ?}{LA—L/-:EmLI f2 (Mg-f hi. 3” - gfzfiv-{y‘gfiéfi +3! _ J _ _. _‘__ _ _n '3 —._.. ' w A" : “J: EME— E FI/lflgf £25“ 1?: _. mm Wigwam/Ag E}: =f1’p 32:33:33.; AM Zia/mi Mm“ CE = r3. - 745':r_ Cm {mefrf {3'2 5% 4-5-3. £flw.{ufl_.n{:£/g4fi1_~ _-5a- Wag-£¥+ CZ”; fl-L Cl: 4/00 zé—_ "x Hi? “ flmlfpm r: ydafifii‘?’ .4 £131: v KG / KG KLEIN.- Fia‘ld' We Ti-ml Tun! H may, .3903 Communion mm 2: {ME the nuguetic energyr density inside the conductor is gitreu by “in __- calculate Lhe interns] self-inductance. Lu: fur turf“! ",[Lm]iength cf the conductor. . Wm = \Mm diet)" : BS” degfgijdé, ale- 1. 4,; 5‘°d-=or“=“ F t 1 A5 2, HE _,_. .59. —- _ =3G¥1 p3 ¢‘#gl ;a;aJafb§ff ' 3.1 ‘57 a ,.. “3:; m) _ ._,1 EJ f MM 1 ’f‘afl Jo Eff}:qu : r” 5’. S” 2 i _r *- .: ,1; F get/it, fig I= J6 TM M nu? _____ _ I 5.1 ’i€_fl_- =‘R fig a 6%? 7 - _ f? .I' U1) 1'1 if“ I Pm“!- 3: MM: [10 math} A triangular conducting Imp ED‘MECIS the points M4. 0. D}. EH), 3. III}!1 and CH. 3, D). [e] A currentlc [A] is flaw-Eng in IheABC directicu in the eituiit in the presence ufnn men-cal magnetic fieldB=Eue*Ix [T]. [i] Detenttine the force Fm [N] on the current flowing in the flame cfthe triangulir loop. [lime-Ital 4g"? 655:4 - In $3133 XE: __ _3.—u = E“. {bar (/3 5., I 'I I l _t_ *-_._c .1" ..-. E )3. _.—o-"""F 1. :1}. Eu— _Ifl .— F," ,C Fm: = if: 2’1 3' {iiJCaletflnte the value of line tubal fume F:- = Fm + F3; + Ft... on the circuit. [2 math] ' "A " - Jfiieeg 51AM E: M 41W {J .I Wu city-rle 31:13. xii-31139:} __ r" A; r .—1_ fl 31F-— C} a \ij— Q E 4521123222 A ___ fl 1—— C/ UEEfltrJAI Xi _ mg :fTefi'i‘E M Pup? H Vain“ “B” m1.- Il._____.___________.T WIMMJ: [iii]EeI-I.re for IJte torque Tu [Mn] enthe side CAmken about the edginfi math] drfijx‘g’f r—Hx. «it: MLMJXfi-fifimtm‘} W . __ A _.. _.. _. {Y1 =—Jflat(é’erxcza+3mdxg)fif 1 w: _\/ h _._ e'tg] fig-i; 1%; e —_'E” ‘ — _' 6 "*3 : a E“ g 6‘ (if €633) 0V Eh; ‘3“ 5 — fifiajfié‘fi" ‘mejcié" 3 W3 3:3 v: “ - e If“? ‘ __ [iv}Detennine the total torque T: = T“ + T”; + Tu telnet: about the 130' E. [1 math] _._'-I-. _-—h 2 J -._h I H _ S _. - : _.. a : r I.th- tl-c—IE: T1— : I” S) Kb “rap-NIL E "R g) Elma. I '—' . a a re 35¢ CL!- <33 32:1; eaia‘x fleece; _ e —t —--—-—-——':..‘ ‘ . “a \?‘ = eLEofia 1M1 ? L.- —'- —---'J'-’—£-|'—h- [It] The himgule: 10011 is in the presence of a lime-WE magnetic field B = En t I: [T]. [i] Calculate the voltage. V“. induced in the leap.“ math] of ( a . e 1U : __'—." I “:3: cfiE— _ __ q —_— 3g .91 i r E- mretflefimfi a; FL d: , Arr- aw [LE ___, ‘1‘ 4. 5J— . _ 41.? j} ' gar/ad: lent-a HI. Hal-A13ch E r): :2: {fiHndi-ente the direction that an induced current would he fluwiug.[2 math] r31; Eli—tux} :3} THE W“ fig A r; 3: r2 e—Lffibu _.—L_..-\..__- —_ — — P113310 ‘ ...
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ele401TermTest_2 - Vania&amp;quot; “B” “A ELE401:...

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