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Magnetism

Name: Anonymous 2009-06-17 22:32

I'm having some problems finding an expression to the magnetic moment of a rotating sphere of m mass with a uniformly charged surface and an angular speed w, the rotation axis is z through the center. It's supposed to be (5Q/6m)*L, where L is the angular momentum (mwr^2).
The integrals just won't lead me to it, I must be missing something crucial. So I would like some guidance on how to solve it, since this is pretty much freaking me out.

Name: Anonymous 2009-06-18 9:00

Anyone?

Name: Anonymous 2009-06-18 10:45

i doubt i'll be able to help but post your integrals maybe?

Name: Anonymous 2009-06-18 13:32

Protip:  Don't look for an answer to a sophomore or junior level question on 4chan.  It doesn't work.

Name: Anonymous 2009-06-18 13:58

>>3
Actually, I just realized they make no sense. I began finding an expression for a differential amount of current I in the sphere's surface. Where dI=(dQ*v)/(2pi*R*Sin[@]), where @ is the angle between the z axis and the radius R. Now v/R*Sin[@]=w since it's rotating in the z axis. so dI=dQ*w/2pi, Now dQ=p*dA, where p is the charge density, it's constant. But now dA=2pi*R^2*Sin[@]*d@ (?), and p=Q/(4pi*R^2), so now dI=(Q*w*Sin[@]*d@)/4pi, Now, an infinitesimal amount of magnetic moment should be the current times an infinitesimal amount of vector area da, which in this case would be all a areas of radius R*Sin[@]. So, du=Ida. And well, first I integrated dI, then found da, and integrated the product. I gives a similar result, but it's off by the scalar accompanying (Q/m)*L, which should be 5/6.

Name: Anonymous 2009-06-18 13:59

>>4
Then what does work?

Name: Anonymous 2009-06-18 15:55

>>5
Just realized dA is wrong since the Jacobian determinant is just R*d@*d@', where d@'=2Pi, so dA=2Pi*R*d@, but this renders an even weirder result.

Name: Anonymous 2009-06-18 16:14

No help? 4tran?

Name: Anonymous 2009-06-18 22:58

Well, made the same problem but in the case of sphere fully charged (Q=p*(4/3)pi*(R^3)), the Jacobian being (R^2)*Sin[@] and not integrating dI but du=dI*a renders the correct result for this case, which is (Q/5m)L, But changing the charge to a surface one won't, don't know what's going on.

Name: Anonymous 2009-06-19 7:33

>>9
Fuck no, the Jacobian is (R^2)*Sin[@], so dA really is 2pi*R^2*Sin[@]*d@, Integrating dI*a gives (Q/3m)*L.

Name: Anonymous 2009-06-19 12:29

>>10
Alright, so this was correct, I ended up proving the angular momentum of a solid sphere is (2/5)mw(R^2)=L, so mw(R^2)=(5/2)*L, so u=(Q/3m)*(5/2)*L=(5Q/6m)*L.
It was correct all the time, just some stupid thing at the end, so it was actually rather easy.
Thanks for the "help".

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