# What is the magnetic dipole moment of the sphere?

## Homework Statement

A uniformly charged solid sphere of radius R carries a total charge Q, and is set spinning with angular velocity $\omega$ about the z axis. (a) What is the magnetic dipole moment of the sphere?

## Homework Equations

$$\vec{m} = I \int d\vec{a}$$

## The Attempt at a Solution

having a lot of difficulty with this stuff

since we are talking about a solid sphere ... first find volume current density $J = \rho v$
$$\rho = \frac{Q}{\frac{4}{3} \pi R^3}$$
$$v = \omega \times r = \omega r sin\theta$$

so $$J = \frac{Q}{\frac{4}{3} \pi R^3} \omega r sin\theta$$
is this ok so far??

alright now to find the total current $I = \int J \cdot da$

$$I = \int \frac{Q}{\frac{4}{3} \pi R^3} \omega R^3 \sin^2\theta d\theta d\phi$$

this doesnt seem dimensionally correct since the radians do not cancel out ...

where have i gone wrong??? Is it the part of the angular momentum??

Hello,

There are different current for different point on the spining solid sphere.
The magnitude of current density $$J$$ you have is correct.
But the total magnetic moment should be
$$\vec{m}=\int d\vec{m}$$
,where $$d\vec{m}$$ is a thin ring with radius $$r\sin\theta$$ rotating along z-axis.
The cross section area of the thin ring is $$dr\cdot rd\theta$$.
The effective current due to the thin ring is $$J(dr\cdot rd\theta)$$.
Therefore, $$d\vec{m}=\hat{z}J(dr\cdot rd\theta)\pi(r\sin\theta)^2$$.
You can substitute the magnetic moment element into the first integral above and find the aim.

Good luck

Meir Achuz
Homework Helper
Gold Member
Use (in Gaussian units)
$${\vec m}=\frac{1}{2c}\int{\vec r}\times{\vec j}d^3r$$

ok so since $$\vec{m} = \frac{1}{2} \int \vec{r} \times \vec{J} d\tau$$
in SI units
$$J = \rho v = \rho (\omega \times r)$$

$$m = \frac{1}{2} \int r \times \rho (\omega \times r) d\tau$$
using the appropriate product rule

$$m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau$$

the second term in the integrand is zero since $$\int_{0}^{\pi} \cos\theta\sin\theta d\theta =0$$

so we hav
$$m =\frac{\rho}{2} \int \omega r^2 d\tau = \frac{\rho\omega}{2} \int_{0}^{R} r^4 dr (4\pi) = \frac{4\pi\rho\omega}{2} \frac{R^5}{5}$$
but $$\rho = \frac{Q}{\frac{4}{3} \pi R^3}$$

so $$m = \frac{2\pi\omega R^5}{5} \frac{Q}{\frac{4}{3} \pi R^3} = \frac{3}{10} Q\omega R^2$$

the answer is supposed to be $$\frac{1}{5} Q\omega R^2$$

maybe im missing some factors??

Meir Achuz
$$\oint d\Omega{\vec r}({\vec r}\cdot\omega) =4\pi r^2\omega/3.$$
$$m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau$$
$$m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau [tex] the correct answer is = 1/3*Q*R^2*w .... please correct your solution as soon as possible . the correct answer is = 1/3*Q*R^2*w .... please correct your solution as soon as possible . I disagree. I get [tex]\vec m = \frac 15 Q \omega R^2 \hat k$$. I see the mistake as having been made in the second integration above; it's NOT identically zero. The $$x$$ and $$y$$ components are because of the integration in $$\phi$$, but the $$z$$ component is nonzero. Don't forget that the first "r" that shows up in the second integration is $$\vec r$$, NOT $$|\vec r|$$. Since we're working in spherical coordinates, this has three components with complicated expressions in terms of $$\theta$$ and $$\phi$$.