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Magnetic Dipole moment

by siddharth
Tags: dipole, magnetic, moment
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siddharth
#1
Mar23-06, 12:49 PM
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For a current loop, the magnetic dipole moment is defined as [tex]i \int d\vec{a}[/tex]

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?
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lightgrav
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Mar23-06, 10:32 PM
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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.
siddharth
#3
Mar24-06, 03:17 AM
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Quote 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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