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Unformatted text preview: a given surface current K , the magnetic moment m is given by (in gaussian units) m = 1 2 c i r × K d 2 r, (4) where the integral is over the surface of the sphere. Now r × K = rK ˆ θ , where ˆ θ is a unit vector in the θ direction. However, only the z component of m is non-zero; the transverse parts cancel out. So we can replace ˆ θ by ( ˆ θ · ˆ z )ˆ z = − cos θ ˆ z , and m becomes m = − ˆ z B 1 2 c c 4 π i r cos θd 2 r. (5) Writing d 2 r = r 2 sin θdθdφ , and setting r = a, we get m = − ˆ z B a 3 8 π i π cos θ sin θdθ i π dφ. (6) 1 Carrying out the two integrals, we get m = − B a 3 2 ˆ z (7) as obtained by the other two methods. (Note that B = H .) 2...
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This note was uploaded on 07/28/2011 for the course PHYSICS 880.06 taught by Professor Stroud during the Fall '10 term at Ohio State.
- Fall '10