# Griffiths 5.58

[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.

## The Attempt at a Solution

The magnetic dipole moment of a disk is

$$(1/4) r^4 \sigma \pi \omega$$

Why does

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

with $\sigma$ replaced by $3Q/(4 \pi R^3)$ not give the correct answer?

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I'd imagine that the sigma in the dipole moment of a disk is a charge density per area.

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!

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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.

isn't dz = Rd\theta ?

No. Try $$r \sin \theta\,d\theta$$

$$z = R\cos \theta$$ so $$dz = - R \sin \theta d\theta$$ so the correct integral is

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

right?

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