# Magnetic Dipole moment

by siddharth
Tags: dipole, magnetic, moment
 HW Helper PF Gold P: 1,198 For a current loop, the magnetic dipole moment is defined as $$i \int d\vec{a}$$ Now, if I'm given a surface current 'K', how can I find the magnetic dipole moment? For example, if a disc of surface charge [itex] \sigma [/tex] is rotating with angular velocity [itex] \omega [/tex], what is it's magnetic dipole moment? I thought about taking the dipole term from the multipole expansion of the magnetic vector potential, and then trying to separate it into something dependent on current distribution, and something dependent on the point where I'm trying to find the potential, could lead to an expression for m. But I'm getting nowhere that way. So, is it possible to get a term for m in terms of K (and maybe r' ?) ,analogous to a closed current loop?
 HW Helper PF Gold P: 1,123 Qv = iL = j HW L . For a disk that's rotating, each dQ moves around a ring with velocity v = omega x r. just add them up.
HW Helper
PF Gold
P: 1,198
 Quote by lightgrav Qv = iL = j HW L . For a disk that's rotating, each dQ moves around a ring with velocity v = omega x r. just add them up.
Yeah, that would work in this case. Thanks.

But for a general surface current distribution K(r') is there a way to find the magnetic dipole moment 'm'? For that matter, what if there is a volume current distribution J(r') ?

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