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Homework Help: Griffiths 5.58

  1. Apr 9, 2008 #1
    [SOLVED] Griffiths 5.58

    1. The problem statement, all variables and given/known data
    A uniformly charged solid sphere of radius R carries a total charge Q, and is set spinning with angular velocity \omega.

    What is the magnetic dipole moment of the sphere.

    2. Relevant equations

    3. The attempt at a solution
    The magnetic dipole moment of a disk is

    [tex](1/4) r^4 \sigma \pi \omega[/tex]

    Why does

    [tex]\int_0^{\pi} (Rd\theta) ((1/4) R^4 \sin^4 \theta \sigma \pi \omega)[/tex]

    with [itex]\sigma[/itex] replaced by [itex]3Q/(4 \pi R^3)[/itex] not give the correct answer?
  2. jcsd
  3. Apr 9, 2008 #2
    I'd imagine that the sigma in the dipole moment of a disk is a charge density per area.
  4. Apr 9, 2008 #3
    So what is \sigma replaced with here? How do I get the moment for a solid sphere from the moment for a disk?

    The analogous idea worked to get the moment of a disk from the moment of a ring which is kind of confusing!
    Last edited: Apr 9, 2008
  5. Apr 9, 2008 #4
    There's nothing wrong with the idea. Here's a hint: each disc contains dz * \rho charge, where \rho is the charge per unit volume.
  6. Apr 9, 2008 #5
    isn't dz = Rd\theta ?
  7. Apr 9, 2008 #6
    No. Try [tex]r \sin \theta\,d\theta[/tex]
  8. Apr 9, 2008 #7
    [tex]z = R\cos \theta [/tex] so [tex] dz = - R \sin \theta d\theta[/tex] so the correct integral is

    = - \int_0^{\pi} ( \rho R \sin \theta d\theta) ((1/4) R^4 \sin^4 \theta \pi \omega)

    Last edited: Apr 9, 2008
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