sample_problems7_a

sample_problems7_a - /, A Cop/957’” coma/Md?” 00/51...

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Unformatted text preview: /, A Cop/957’” coma/Md?” 00/51 condmchw/y 0”; 5-7 X/01 S/m carries a. Cum/cal; Op K74. 7;“: 00/356240,vK Aas 0L Man/76:er of 2mm. Ca/ca/al‘e (a) E/ch’ic Kr‘a/o/ mfg/>97 )ns/a/e €h¢ c'onQ’MCr‘or (6) W16, achaL?e dr/‘Fé Ve/oaéy ofi C/cchans m (426 Conoéucwr assuming a. Free c/eu’von dens/‘6? 013 8X mu elecHons/MS’ So/uh‘on: (a) ()5 Wu? fiawczmm 7‘0"; Yaobiuo=IMM Also f3; :IXIO'g A$ar¢0 1 Ana A : IrZ : 3/4 XQXIO'a 23/4 x (0'6 m2 : IO , E flM—Q : 5~554Xl02 w), 5-7X/o1x gum/0' (b). Bur/H vc/ocxfiy 4:5 flue Vavg ofl fhg glee/Won I- HAVavgxc’c’), Van/g ‘1 — q 890/191’36 pd dad/Hon : lvéXIOI C 6 n-a nmbefi Dfi‘: Afi amok = 2/4 Xio‘ ml Von/3 ; IO. v 40: v6 gx/015X3KI#><’O x 3 2.487 x10” m/s. i. The plane x = 0 (in Cartesian co-ordinates) is the interface between 2. air and a ferromagnetic material of relative permeability, u, = 100. The magnetic field on the air side of the interface is H = 1000)? + 200§ + 100? (A/m), where 9;, § and ’z‘ are unit vectors in the direction of the X, Y and Z axes respectively. Find the magnetic field in the ferromagnetic material. Assume there is no current in the interface. ;/ On the Plane, 1:0, éqp'agg Y L Z components y are (ngenfiafl & x X Componenf' is normed. 13mm“sz H2® HI; {0001+ xgog 77:23) Puma. +lOOZ (A/m) . l 4 " Gal m onenrsz>ZOO 8" (He!) Imaafl Co P 4‘00 Z A (Hm) NOWY'WQ meonaM'Q. 10007, New using our boundlargflkciondxfiong (we Lwrfe ’ _- HE ’Ht "2 K550 I, Hex: Him \302' 3m 5 O i) la; HnL5f4an| —- (at *3 enlfioJ COMPODCAES, lav/1% A A He: Ht} .200 5mm); A. Hnl: Hn' ’ Hr loo Final? Soiwiion ‘0‘) TokoJ R'ch ‘6 sum 09 tangent-L Z wormed mmPonenrs. H : H,, + Ht; 9_I07‘c+2009+t002 a (A/m) xix. Magnetic, field if) Hne ~ te ‘ \ gevmwwsaflahc mot Y‘OL (a) K5: SUTPQQ€\ J> gt\' (0 a? 33,— 20 2 TV”: boundargj condkhons 00rd, ——> ~—> -—> E gm 1:) BnL' h, 7- y g A : QOZ «. Curranf . I’vc a$5mmad {4001' OKAY (g log/bow 2:0 2 m makehafl Rs above 1:0} 5% 50% com a33ume ’H/xd OHheY (200% [’00 O& mm?one,nk M6304, The worm Mr (AM vale «Cor Hm. AP ‘ We ~fl Cmfrérxll— and3%0\1\ uo‘x“ ¢§nd ’(hok’r Hui 96d due h: currenk \s ‘mvkhe. E1. :0 M l> “Ht, 6—»; fir? 00o H4917“ Ht‘: KS 6—? .17 \K/ L‘ g5 :QOO + ZXKO \AD 2. A o + LMOL‘t—Q 7(— O o n g2: 3 A LC) {45:2xxo 9c 1 Z Mal’cwo} My 7/ V y mad/MM} XL PK“ {(05%) Zwren/F ‘ all A “V a wmponenk a? H\ aka/13 \/ oLwe o I n \\ H II n H H’z i) 7, w Mow Hy =0 (4 wmponenk —o) a l he rule par H’AQ— E‘Nfl WW} ’F‘Ni vaverxf WW} OPFOses Hy; flPPldma Wu: via awrrcnt- ) we can PVCdu/tces 6L 9' ad H \' OVQJO H7 . “’25 X 5 A3 ‘ elf‘ A \/ M HVI> Hyl' Way,” H — — 2 K v, Hy‘ My2 s \O ’ Ev; : 2><\o3 ‘2 % Eyz — JZX \D h Byl’”:27k \CDLf (Rom T:ka C00 00¢ (Aowe. #7 A Li: 0&3 En : 20 Z A onenl’ m ‘ HM; 2t wmp 810cc K9 :0 “a a , ’ Bx}: Bx‘ f: HXZ’ \4X‘ " Pf ‘ I:4OOA a a; ‘0 mm The two vaa'wng “ad “ad loo onV& about avg g 3 :enc @ HM = H 1‘1“ z _ ; TA (WU: T ‘S Alana ' 3’ ok$ MMQoY‘M) T: (/wrro/‘k depngg : ’1’” TVoC” o ' 0 Kane :‘C: "IT—a” >CT r1) 9 AYCO‘ eflcbosecq I ‘00? :. I TL 0."; y‘4 H T: a7— (oLréa) H : 1+ :EPL/ 1 7, 2W0“ ZXTYX(0.0D = 6'37XIDE’ v LA/ND Th5 ouw‘id'ion 09 Qx‘do‘ ‘9 We; sama as Park (on £33 H‘w : Iemc LHJJWY) ,- SMce We 6J1ch W‘H‘C t; Ian. 3 .L C U9in H)”; we? 3 Wre \s m: Curved!- bemw agr) HQfiY : 1 H; I - DO _ ...
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