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Magnetic dipole moment 
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#1
Jan2906, 06:30 PM

P: 187

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 


#2
Jan3006, 05:16 AM

P: 79

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] ? 


#3
Jan3006, 01:16 PM

P: 187

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 


#4
Jan3006, 09:45 PM

P: 79

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 


#5
Jan3006, 11:05 PM

P: 187

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] 


#6
Jan3106, 12:00 AM

P: 79

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] 


#7
Jan3106, 12:11 AM

P: 187

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



#8
Jan3106, 01:22 AM

P: 79

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 


#9
Jan3106, 08:56 PM

P: 187

just wondering how you have for area that
[tex]A=r^2 sin^2\phi[/tex] otherwise your solution looks right to me 


#10
Jan3106, 10:54 PM

P: 79

[tex] A= r^2 sin^2 \phi \pi [/tex] Sorry for mistake. 


#11
Mar309, 01:15 PM

P: 7

(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. 


#12
May209, 09:18 PM

P: 9

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



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