HW9_sol

# 4 nowpoints griffiths problem 532 source element on 10

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Unformatted text preview: n()(iJ,so the right .side of Eq. 5.76 is K 9-.Problemsin5.25 and the equation is satisfied. 4 (Nowpoints) Griffiths Problem 5.32 source element on 10 consider /l0(J1J.) ()(iJ, the symmetrically placed R z . #2, 5.32 Problem points y' - z'). Since z' changes sign, function only loop(a) A at (x', in ,the same direction as I; and is awhile every- of s (the distance from the wire). In cylindrical ( )( )( ) I I I = -poln2 B ~ I I I I I ~ I l I I l I l I I a I I - - I ( I - ~ - -~V _ a ( I ) I 1 ~ a a thing else is the same, = A(s)and so components A from- dB1 ~ = fL o1 (the field of an infinite wire). Therefore A and 27rS coordinates,then, A the x Z, y B = V x = s Because Aabove only a z component. on dB2 cancel, leaving = Abelow at every pointqed the surface, it follows that ~~ and ~~ are the same above and With this, Ampere's law is confined to the normal derivative. Abelow;oI, and A(r) = _fL o1 In(s/a) z (the constant a is arbitrary; you could use 1, but then the units = _fL any discontinuity yields immediately: a B s - 2~ B = - 8AYabove27r+ 8AYbelow + 8Axabove...
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## This note was uploaded on 04/23/2013 for the course PHSX 531 taught by Professor Chiu during the Fall '12 term at Kansas.

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