HW9_sol

# We may as 515 pick c2 0 clyz lyl fxoyzdy and

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Unformatted text preview: , 1m z > -a. -1 z{~ampedan loop 98~1(158~2 = Fx(O,y,z). We may as 5.15 pick C2 = 0, Cl(y,z) = lYl Fx(O,y',z)dy', and we're done, with .1 - points) Griffiths Problem well ~ l { 0 so IB 8 Y = I I ~y ~y l Problem 5.15 Problem 5.15 =0; a solenoid is ponI, and outside it is zero. The outer solenoid's field points to the left (-2), The fieldWx inside Wy= lx Fz(x',y,z)dx'; Wz = lY Fx(O,y',z)dy' -lx Fy(x',y,z)dx'. The the inner one solenoid is ponI, and outside (i) isBzero. The outer solenoid'sB field points z, the left (-2), field inside a points to the right (+z). So: it whereas = poI(n1 - n2) z, (ii) = -poln2 to (iii) B = 0.1 I I I Problem oints) one points to the + .16 9whereas ptheW = Griffiths Problem right8Wx - So: (i) y + poI(n1 - - 8Wx z, Z(ii) .2 (b) V x 5.16 (15 inner 8Wz - 8Wy x 5 (+z). 8Wz B = 8Wy n2) I I z, (iii) B = 0.1 8y 8z 8x 8x 8y Problem Ex. 5.8, the top plate produces 8z field poK/2 (aiming out of the page, for points above it, and into From 5.16 a . r8Fy(x',y,z) ' the page, for )points the t...
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