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Griffiths 5.58

  • Thread starter ehrenfest
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  • #1
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[SOLVED] Griffiths 5.58

Homework Statement


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.


Homework Equations





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?
 

Answers and Replies

  • #2
980
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I'd imagine that the sigma in the dipole moment of a disk is a charge density per area.
 
  • #3
2,012
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I'd imagine that the sigma in the dipole moment of a disk is a charge density per area.
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:
  • #4
980
2
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.
 
  • #5
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isn't dz = Rd\theta ?
 
  • #6
980
2
No. Try [tex]r \sin \theta\,d\theta[/tex]
 
  • #7
2,012
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[tex]z = R\cos \theta [/tex] so [tex] dz = - R \sin \theta d\theta[/tex] so the correct integral is

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

right?
 
Last edited:

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