This preview shows page 1. Sign up to view the full content.
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...
View
Full
Document
This note was uploaded on 04/23/2013 for the course PHSX 531 taught by Professor Chiu during the Fall '12 term at Kansas.
 Fall '12
 chiu

Click to edit the document details