HW#8_Solutions

# HW#8_Solutions - Problem 6 12(a M ksi Jb = VXM =...

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Unformatted text preview: Problem 6. 12 (a) M: ksi; Jb = VXM = —k<z>; K, = M x ﬁszé. B is in the z direction (this is essentially a superposition of solenoids). So Use the amperian loop shown (shaded)——inner side at radius .9: fB - d1 = Bl: p018nc = p0 [fdea + K51] = no [—kzuz — s) + 1:121] = pokls. B = poksi inside. (b) By symmetry. H points in the z direction. That same amperian loop gives fH - d1 = H l = “01"”: = 0, since there is no free current here. So , and hence Outside M = 0, so B = 0; inside M = 1:32, so B = poksi. Problem 6.17 From Eq. 6.20: fI-I . d1 = Hm”) = II”: = {1(39/02), (s < a); VI (3 > a). I: no 1+xm 1! (3 < a), l H: W: 09(0)}, SOB: H: 21m ’ ’ 1 l (s>a> " (s>a>. 5 I . . Jb = Xme (Eq. 6.33), and J, = #5, so J5 = 29"— (same direction as I). 1ra2 11, = Jb(1'ra2) + Kb(21m) = me — me = m (as it should be, of course). Problem 6.18 By the method of Prob. 6.15: For large 1', we want B(r,9) —> B0 = Boi, so H = LIB-B —> 313305, and hence W —> —%Boz = — ElaBor cos 0. “Potentials”: mums) = zAir'mcoso), (-r < R); { Wou,(r, 0) = —‘+°Borc050 +2 ;%1-Px(cos0), (r > R). Boundary Conditions: { vvin (R: = Wout(R90)s (ii) *ﬂowginln 'H‘gg-hln = 0- (The latter follows from Eq. 6.26.) (ii) => no [LIB-Bo c090 + 20 + 1) RﬁizPAcos 9)] + leAlR"IH(cos6) = 0. For I 75 1, (i) => B; = R2I’HA1, so [,uo(l + 1) + pl]A¢IZ“1 = 0, and hence A; = 0. For 1 = 1, => AlR = —ul—oBoR+B1/R2, and (ii) => Bo+2ﬂoBl/R3+pA1 = 0, SO A1 = -3Bo/(2[lo+ﬂ). 3Bo rcoso _ - 3302 330 i = 330 (2% + u) — (2110 + u)' (2110 + [1) (2m + p)' B: H=——= —— B. “ (2mm 1+xm/3 ° “linen = Hin = "Vvvin = By the method of Prob. 4.23: Step 1: Bo magnetizes the sphere: Mo = meo = WBO. This magnetization sets up a ﬁeld within the sphere given by Eq. 6.16: EicBo (where n = A). B1 = ~quo = " Bo - 1+x... 31+Xm 3 Step 2: Bl magnetizes the sphere an additional amount M1 = ﬁBl. This sets up an additional ﬁeld in the sphere: 2 2 2n 2 32 — 5,101“; — 35B} — Bo, etc. The total field is: B B=Bo+Bi+Bg+---=Bo+(2n/3)Bo+(2n/3)2Bo+---=[1+(2n/3)+(2n/3)2+~-]Bo=m3? 1 _ 3 _ 3+3Xm _3(1+xm) so B_ 1+xm B 1—2n/3‘3—2Xm/(HXm)‘3+3Xm—2x,n— 3+Xm’ ‘ 1+xm/3 0' Prublcru 0.20 At the interface, the perpendicular component of B is continuous (Eq. 6.26), and the parallel component of H is continuous (Eq. 6.25 with K , = 0). So 13l = 132*, H'll = Hg. But B = pH (Eq. 6.31), so Ell-B'I' = aka-Bl. Now tanoi = Bil/BIL, and tan02 = Blal/Bgi, so tan02 _ B! 3,1 _ B'zl _p.2 tanei-B%'1_3—llr—El—l_p1 ...
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## This note was uploaded on 02/02/2012 for the course ELECTRICAL ENEE380 taught by Professor D.romeo during the Fall '11 term at Maryland.

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HW#8_Solutions - Problem 6 12(a M ksi Jb = VXM =...

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