## magnetic dipole moment

Hello, i have absolutely no clue on how to start this one:

a sphere of radius R has a uniform volume charge density $$\rho$$.
Determine the magnetic dipole moment of the sphere when it rotates as a rigid body with angular velocity $$\omega$$ about an axis through its center.

thanks for the help on this one
 Maybe you can divide the sphere in to many rings. Each ring is a magnetic dipole moment, which magnitude is $$\mu = iA$$ Then integrate them. By the way, what's the given answer? Is it $$\frac{4}{15} \rho \omega \pi R^5$$ ?
 hello, i do not have the answer at this time, but i tried your method you describe and got the same as you without the 15 in the denominator. can you explain how you ended up with 4/15? thanks

## magnetic dipole moment

Hi. There was R^5 and sin^3 in my integration so it ended up with 4/15. It seems that we have some difference from the start. If we use the same idea, it should come out the same result. Can you post you equation?

Thanks
 heres what i have done: since $$dV \rho = dq$$ and $$dt=\frac{2\pi}{\omega}$$ so that $$I=\frac{dq}{dt}=\frac{dV\omega\rho}{2\pi}$$ now since magnetic moment is $$\mu=IA$$ i wrote $$d\mu = \frac{dV\omega\rho}{2\pi}dA$$ then for a sphere the differential volume element i used was $$dV= r^2 sin\theta dr d\theta d\phi$$ putting this all together i have $$\mu = \frac{\omega\rho}{2\pi}\int_{0}^{R}r^2 dr\int_{0}^{\pi}sin\theta d\theta\int_{0}^{2\pi}d\phi\int dA$$ where $$\int dA = volume of sphere = \frac{4}{3}\pi r^3$$

 Quote by thenewbosco where $$\int dA = volume of sphere = \frac{4}{3}\pi r^3$$
This is the difference.

I think that we can obtain a circle by intersecting a plan and the sphere. There are many many diffrential rings on the circle, each has area

$$r^2 sin^2 \phi \pi$$

A bigger ring will involve a small one, so

$$\int dA \neq \frac{4}{3} \pi R^3$$

My solution is similar as yours:

A ring has charge

$$q=2 \pi r sin \phi r d \phi dr \rho$$

so each ring has

$$i=2 \pi r sin \phi r d \phi dr \rho \frac{\omega}{2 \pi}$$

magnetic moment $$\mu = iA$$ $$A =r^2 sin^2 \phi$$

put this all together

$$\mu = \rho \omega \pi \int_{0}^{R}\int_{0}^{\pi} r^4 sin^3 \phi dr d \phi$$
 i am wondering what angle phi is on yours, i have used phi and theta as in spherical polar coordinates, while you have only theta. can you describe what this is. thanks

 Quote by thenewbosco i am wondering what angle phi is on yours, i have used phi and theta as in spherical polar coordinates, while you have only theta. can you describe what this is. thanks $$\mu = \frac{\omega\rho}{2\pi}\int_{0}^{R}r^2 dr\int_{0}^{\pi}sin\theta d\theta\int_{0}^{2\pi}d\phi\int dA$$
$$\theta_{yours} = \phi_{mine}$$

 Quote by Psi-String $$q=2 \pi r sin \phi r d \phi dr \rho$$
$$\int_{0}^{2 \pi} d \phi_{yours} = 2 \pi_{mine}$$

I'm not 100% sure whether my solution and answer are right or not.
If you have other ideas or know the correct answer, plz tell me.
Thanks a lot
 just wondering how you have for area that $$A=r^2 sin^2\phi$$ otherwise your solution looks right to me

 Quote by Psi-String magnetic moment $$\mu = iA$$ $$A=r^2 sin^2 \phi$$
Oh I'm sorry! It should be

$$A= r^2 sin^2 \phi \pi$$

Sorry for mistake.
 (sorry for all the greek letters in superscript, I don't know why it's doing that...) I am doing this same question except the sphere only carries a uniform surface charge $$\sigma$$. I then obtain the final answer $$4/3$$$$\pi$$R$$\omega\sigma$$ Which is equivalent to Volume of Sphere x $$\omega\sigma$$ (+z direction) Would this be the correct answer? The only real difference when doing the question is that I only needed to integrate wrt theta instead of over the whole volume since all the charge lies on the surface.
 the correct answer is 1/3*q*(R^2)*w ... now calculate it correctly ....