Magnetic dipole moment homework

[thenewbosco]

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

 

[Psi-String]

Maybe you can divide the sphere into 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 ?

 

[thenewbosco]

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

 

[Psi-String]

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

 

[thenewbosco]

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

 

[Psi-String]

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

 

[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

 

[Psi-String]

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}

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, please tell me.
Thanks a lot

 

[thenewbosco]

just wondering how you have for area that

A=r^2 sin^2\phi

otherwise your solution looks right to me

 

[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.

 

[eyenkay]

(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\piR\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.

 

[klp_l123]

the correct answer is 1/3*q*(R^2)*w ... now calculate it correctly ...