Magnetic Dipole moment

  1. siddharth

    siddharth 1,191
    Homework Helper
    Gold Member

    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?
    Last edited: Mar 23, 2006
  2. jcsd
  3. lightgrav

    lightgrav 1,247
    Homework Helper

    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.
  4. siddharth

    siddharth 1,191
    Homework Helper
    Gold Member

    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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