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Magnetic Moment of a Charged, Rotating Sphere

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Show for a solid spherical ball of mass m rotating about an axis through its center with a charge q uniformly distributed on the surface of the ball that the magnetic moment [itex]\mu[/itex] is related to the angular momentum

    [itex] \vec{\mu}={\frac{5q}{6mc}}\vec{L} [/itex]

    2. Relevant equations

    [itex]\mu = \frac{IA}{c}[/itex]

    [itex]I_{sphere} = \frac{2mr^2}{5}
    [/itex]

    3. The attempt at a solution
    My initial thought was to think of the sphere as rings of charge, each of which is comprised of point particles of charge [itex] dq=\frac{q}{4{\pi}r^2} [/itex] which trace out current rings of area [itex] A={\pi}r^2 [/itex]. Then I could integrate dI w.r.t dr, but I got a factor of 1/4 instead of 5/6, making me think that I'm going about the integration the wrong way.

    Any ideas?
     
  2. jcsd
  3. Sep 6, 2011 #2

    diazona

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    Not all the rings have [itex]A = \pi r^2[/itex] - if they did, it would be a cylinder, not a sphere. See if that helps.
     
  4. Sep 6, 2011 #3
    Okay, so for the area, I could integrate from 0->R twice to sum all the cross-sectional areas of the sphere:

    [itex] A = \pi r^2 [/itex] therefore [itex] dA = 2{\pi}rdr [/itex]

    and
    [tex] A_{total}=2\int_0^R \! 2{\pi}rdr [/tex]

    and [itex]A= 2{\pi}R^2 [/itex]

    Now we have: [tex] \vec{\mu} = \frac{q}{T} \frac{2{\pi}R^2}{c} = \frac{q{\omega}}{2{\pi}} \frac{2{\pi}R^2}{c} = \frac{q{\omega}R^2}{c} \frac{m}{m}
    [/tex]

    Since the moment of inertia of a solid sphere is [itex]I_{sphere} = \frac{2mr^2}{5}
    [/itex], we have [itex] \frac{5}{2} I_{sphere} [/itex] in the equation. It follows that:
    [tex] \vec{\mu} = \frac{5}{2}I \frac{q{\omega}}{c} = \frac{5q}{2mc} {\vec{L}}[/tex]
    But now I'm off by a factor of 1/3, so I still think I'm missing something in the integration...
     
  5. Sep 6, 2011 #4
    I'm a little confused. I realize integration has to be done at some point, but isn't the surface area of the sphere 4 \pi r^{2} regardless of the charge?
     
  6. Sep 6, 2011 #5
    Originally I thought that surface area would come into play, but I think that since we're worried about Current Areas maybe not? I feel like there should be a dq in there somewhere. So maybe two integrals?
     
  7. Sep 7, 2011 #6

    diazona

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    Ooh... I have to apologize because what I posted earlier doesn't actually make any sense. Not sure what I was thinking there :blushing:
    I'm not really sure what you're doing here. You were right to think about splitting the sphere up into "slices" (individual current loops), but beyond that I think you might be confusing several different areas.

    Let's think about the computation of the magnetic moment of a single current loop, which corresponds to a single slice of the sphere as shown in this picture:
    sectioning.png
    (the label got a little messed up, but [itex]\mathrm{d}a[/itex] is the blue area). There are two areas involved that you need to deal with:
    • Each current loop consists of a certain amount of charge [itex]\mathrm{d}q[/itex]. Since the charge is spread out over a surface, in order to find [itex]\mathrm{d}q[/itex], you need to multiply the surface charge density by the surface area over which the charge is spread. For one "slice"/current loop, the area to use for this is [itex]\mathrm{d}a[/itex], the blue area, as shown in the figure. (I'm using lowercase a to avoid confusion)
    • In order to find the magnetic moment of the current loop, you will need to multiply the current by the area enclosed by the current. Since the current runs around the edge of the "slice" of the sphere, you need the area of the slice, designated [itex]A[/itex] in the picture. Note that the radius of this circle is not [itex]r[/itex], the radius of the sphere.
    See if that enables you to find the current of a single "slice" [itex]\mathrm{d}I[/itex], the enclosed area [itex]A[/itex], and from that the magnetic moment of the slice [itex]\mathrm{d}\mu[/itex]. Once you do that, you can integrate it to get the total magnetic moment of the sphere.
     
  8. Sep 7, 2011 #7
    Bah. I don't know why this is eluding me.
    Here's where I am now:

    [tex]dq= \frac{q}{4{\pi}r^2}da , da=2{\pi}rdr [/tex]

    therefore:

    [tex] dq = \frac{q}{2r}dr [/tex]

    (Here r is the radius of the slice, I'll use R for radius of the sphere).
    However I don't think that I have my da right. What's the "thickness" of the slice that gives me my "d" term? Do I need to do this in terms of [itex]\theta[/itex] and hope that it cancels out?

    Then if I try to find A (area enclosed by the loop) I have:

    [tex]A={\pi}r^2 , dA = 2{\pi}rdr[/tex]

    However this is the same as my da used to find q, and that can't be right...

    Sorry for being so slow on this one! It's always the math that gets me, not the physics.
     
  9. Sep 7, 2011 #8

    diazona

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    Think carefully about which radius you need to use in each case. One of them is wrong.
    Yes, you need to express it in terms of [itex]\mathrm{d}\theta[/itex]. Don't worry about stuff canceling out just yet.
     
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