advertisement

February 28, 2000 Notes for Lecture #19 Magnetic dipole eld These notes are very similar to the notes for Lecture #13 on the electric dipole eld. The magnetic dipole moment is dened by Z 1 m= 2 with the corresponding potential d3 r 0 r0 A(r) = and magnetostatic eld B(r) = ( 0 4 J(r0); m ^r 0 4 r2 (1) (2) ; ) 3^r(m ^r) m + 8 mÆ3(r) r3 3 (3) : The last term of the eld expression follows from the following derivation. We note that Eq. (3) is poorly dened as r ! 0, and consider the value of a small integral of B(r) about zero. (For this purpose, we are supposing that the dipole m is located at r = 0.) In this case we will approximate Z 3 B(r 0) B(r)d r Æ 3 (r): (4) sphere First we note that Z rR B(r)d3 r = R2 Z r=R ^r A(r) d : (5) VdA: (6) This result follows from the divergence theorm: Z r Vd r = 3 vol Z surface In our case, this theorem can be used to prove Eq. (5) for each cartesian coordinate of r A since r A = x^ (x^ (r A)) + y^ (y^ (r A)) + ^z (^z (r A)): Note that x^ (r A) = r (x^ A) and that we can use the Divergence theorem with V x^ A(r) for the x component for example: Z Therefore, Z rR r (x^ A)d3r = vol (r A)d3r = Z r=R Z surface (x^ A) ^rdA = Z surface (A ^r) x^dA: (A ^r) (x^x^ + y^y^ + ^z^z)dA = R2 Z r=R (^r A)d (7) (8) which is identical to Eq. (5). Now, expressing the vector potential in terms of the current density: Z J(r0 ) A(r) = 0 d3 r (9) 4 jr r0j ; we can use the identity, Therefore, R2 Z Z r =R jr ^r r0j = 43 rr<2 r^0: > (10) d (^r A)d = 4R 3 2 Z d3 r 0 r< r2 > r^0 J(r0 ): (11) If the sphere R contains the entire current distribution, then r> = R and r< = r0 so that (11) becomes Z Z 4 2 (12) R (^r A)d = 3 d3r0 r0 J(r0) 83 m: r=R