Magnetic Moment of Uniformly Charged Rotating Sphere

[latentcorpse]

A small sphere is uniformly charged throughout its' volume and rotating with constant angular velocity $\omega$. Show it's magnetic moment is given by
$m=\frac{1}{5}Q \omega a^2$. the question doesn't say but I'm assuming a is the radius.

Anyway, so far I have:

$m=\frac{1}{2} \int_V dV (\mathbf{r \wedge J})$

but $\mathbf{J}= \rho \mathbf{v} = \rho \mathbf{\omega \wedge r}$

and so $m=\frac{\rho}{2} \int_V r^2 \mathbf{\omega} - (\mathbf{r \cdot \omega})\mathbf{r}$

but i don't really know where to go from here?

 

[gabbagabbahey]

Choose a coordinate system and carry out the integration. The easiest choice is probably spherical coordinates, with omega pointing in the z-direction.

 

[latentcorpse]

ok so $\int (\mathbf{r \cdot \omega})\mathbf{r} dV=\int r \omega \cos{\theta} r^2 \sin{\theta} dr d \theta d \phi \frac{\mathbf{\hat{r}}}{r}=\omega \int_0^a r^2 dr \int_0^{\pi} \cos{\theta} \sin{\theta} d \theta \int_0^{2 \pi} d \phi \mathbf{\hat{r}}$
giving
$\omega \frac{a^3}{3} 2 \pi \int_0^{\pi} \frac{1}{2} \sin{2 \theta} d \theta \mathbf{\hat{r}}= \frac{2}{3} \pi a^3 \left[ -\frac{1}{4} \cos{\theta} \right]_0^{\pi} \mathbf{\hat{r}}$

which is

$\omega \frac{a^3}{12} \pi (-[-1-1])=\frac{\omega \pi a^3}{6} \mathbf{\hat{r}}$

 

[latentcorpse]

and the first integral is $\omega \int_0^a r^4 dr \int_0^{\pi} \sin{\theta} d \theta \int_0^{2 \pi} d \phi \mathfb{\hat{z}}=\omega \frac{2a^5}{5} 2 \pi \mathbf{\hat{z}}=\frac{4 \omega \pi a^5}{5} \mathbf{\hat{z}}$

 

[latentcorpse]

reckon I've messed up somewhere though because they shouldn't have directions should they?

 

[gabbagabbahey]

reckon I've messed up somewhere though because they shouldn't have directions should they?

Magnetic moment is a vector, so it most definitely should have a direction.

But you have messed up somewhere; is $\hat{r}$ really position independent? Because that's what you are assuming when you pull it out of the integral.

 

[latentcorpse]

ok. not sure what to do with it then?

 

[gabbagabbahey]

Cartesian unit vectors are position independent, so rewrite $\hat{r}$ in terms of them.

$$\hat{r}=\sin\theta\cos\phi\hat{x}+\ldots$$

 

[latentcorpse]

ok. did that. both the $\hat{x},\hat{y}$ bits dropped out.

and i got

$\omega \int_0^a r^2 dr \int_0^{\pi} \sin{\theta} \cos^2{\theta} d \theta \int_0^{2 \pi} d \phi \mathbf{\hat{z}}=-\frac{\omega a^3}{3} \int_{1}^{-1} (-u^2) du 2 \pi \mathbf{\hat{z}}$ where $u=\cos{\theta}$
which simplifies to $\frac{2 \omega \rho \pi a^3}{3} \mathbf{\hat{z}}$ when you multiply in the factor of $\frac{\rho}{2}$ and so overall we get

$m=\left( \frac{2 \omega \rho \pi a^5}{5} -\frac{2 \omega \rho \pi a^3}{3} \right) \mathbf{\hat{z}}$ which doesn't quite give me what i want. there must still be a mistake somewhere i guess.

and i assume $Q=\fac{4}{3} \pi \rho a^3$

 

[gabbagabbahey]

What is 2/5-2/3?

Edit why does your second terms have a^3 instead of a^5?

 

[latentcorpse]

kool. i got it.

the next bit is to find the angular momentum L of the sphere of mass M and verify that $\mathbf{m}=\frac{Q}{2M}\mathbf{L}$

i was wanting to use the formula I'm meant to be verifying for finding L so since that's out the window, I was guessing $\mathbf{L}=M(r^2\mathbf{\omega}-(\mathbf{r \cdot \omega})\mathbf{r}$ but that's not getting me anywhere!

 

[gabbagabbahey]

Use the definition of angular momentum: $$\vec{L}=M\vec{v}\times\vec{r}$$

P.S. \mathbf doesn't always show up too well here, so you might want to switch to \vec

Edit: $$\vec{dL}=\rho_M\vec{v}\times\vec{r}dV$$ where $$\rho_M=\frac{M}{\frac{4}{3}\pi a^3}$$

 

[latentcorpse]

do i sub in for v before i integrate?

 

[gabbagabbahey]

did you sub in v before integrating when inding m?

 

[latentcorpse]

yes. so is it basically going to be the same integral again?

 

[gabbagabbahey]

You tell me...