magnetic dipole moment

thenewbosco

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

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

thanks for the help on this one

Psi-String

Maybe you can divide the sphere in to many rings. Each ring is a magnetic dipole moment, which magnitude is

[tex] \mu = iA [/tex]

Then integrate them.

By the way, what's the given answer?
Is it

[tex] \frac{4}{15} \rho \omega \pi R^5 [/tex] ?

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 [tex]dV \rho = dq[/tex] and [tex]dt=\frac{2\pi}{\omega}[/tex] so that [tex]I=\frac{dq}{dt}=\frac{dV\omega\rho}{2\pi}[/tex]

now since magnetic moment is [tex]\mu=IA[/tex] i wrote [tex] d\mu = \frac{dV\omega\rho}{2\pi}dA[/tex] then for a sphere the differential volume element i used was [tex]dV= r^2 sin\theta dr d\theta d\phi[/tex]

putting this all together i have
[tex]\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[/tex]

where [tex]\int dA = volume of sphere = \frac{4}{3}\pi r^3[/tex]

Psi-String

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

[tex] r^2 sin^2 \phi \pi [/tex]

A bigger ring will involve a small one, so

[tex] \int dA \neq \frac{4}{3} \pi R^3 [/tex]

My solution is similar as yours:

A ring has charge

[tex] q=2 \pi r sin \phi r d \phi dr \rho [/tex]

so each ring has

[tex] i=2 \pi r sin \phi r d \phi dr \rho \frac{\omega}{2 \pi} [/tex]

magnetic moment [tex] \mu = iA [/tex] [tex]A =r^2 sin^2 \phi [/tex]

put this all together

[tex] \mu = \rho \omega \pi \int_{0}^{R}\int_{0}^{\pi} r^4 sin^3 \phi dr d \phi [/tex]

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

[tex] \theta_{yours} = \phi_{mine} [/tex]

[tex] \int_{0}^{2 \pi} d \phi_{yours} = 2 \pi_{mine} [/tex]

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

thenewbosco

just wondering how you have for area that

[tex]A=r^2 sin^2\phi[/tex]

otherwise your solution looks right to me

Psi-String

Oh I'm sorry! It should be

[tex] A= r^2 sin^2 \phi \pi [/tex]

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 [tex]\sigma[/tex]. I then obtain the final answer

[tex]4/3[/tex][tex]\pi[/tex]R[tex]\omega\sigma[/tex]

Which is equivalent to

Volume of Sphere x [tex]\omega\sigma[/tex] (+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 ....