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Magnetic dipole moment homework

  1. Jan 29, 2006 #1
    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. jcsd
  3. Jan 30, 2006 #2
    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] ?
    Last edited: Jan 30, 2006
  4. Jan 30, 2006 #3
    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?

  5. Jan 30, 2006 #4
    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?

    Last edited: Jan 30, 2006
  6. Jan 30, 2006 #5
    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]
  7. Jan 31, 2006 #6
    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]
    Last edited: Jan 31, 2006
  8. Jan 31, 2006 #7
    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
  9. Jan 31, 2006 #8
    [tex] \theta_{yours} = \phi_{mine} [/tex]:smile:

    [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
    Last edited: Jan 31, 2006
  10. Jan 31, 2006 #9
    just wondering how you have for area that

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

    otherwise your solution looks right to me
  11. Jan 31, 2006 #10
    Oh I'm sorry! It should be

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

    Sorry for mistake.
  12. Mar 3, 2009 #11
    (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


    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.
  13. May 2, 2009 #12
    the correct answer is 1/3*q*(R^2)*w ... now calculate it correctly ....
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