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Magnetic moment of a solid, uniformly charged ball

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Show for a solid spherical ball of mass m rotating through its centre with a charge q uniformly distributed on the surface of the ball that the magnetic moment μ is related to the angular momentum L by the relation:

    [itex]\textbf{μ} = \frac{5q}{6mc}\textbf{L}[/itex]


    2. Relevant equations

    [itex]μ = IA [/itex]
    [itex]I = \frac{2}{5}MR^2[/itex]
    [itex]σ = \frac{q}{4πR^2}[/itex]
    [itex]\frac{1}{T} = \frac{ω}{2π}[/itex]


    3. The attempt at a solution

    Split sphere into thin segments, each one has magnetic moment:

    [itex]dμ = dI \cdot dA [/itex]

    where dA is the area enclosed by the segment:

    [itex]dA = 2π r dr[/itex]

    The current dI on each segment is:

    [itex]dI = \frac{dq}{T} = \frac{ω}{2π} dq[/itex]

    The charge on each segment is:

    [itex]dq = σ da[/itex]

    Where da is the surface area of the segment:

    [itex]da = r^2\sinθ dθ d\phi[/itex]

    This results in
    [itex]dq = σ \cdot r^2\sinθ dθ d\phi = \frac{q}{4πR^2} r^2\sinθ dθ d\phi[/itex]

    We now have an expression for dI:

    [itex]dI = \frac{ωqr^2\sinθ}{8π^2R^2} dθ d\phi[/itex]

    Now we can calculate the moment of each segment:

    [itex]dμ = \frac{ωqr^2\sinθ}{8π^2R^2} 2πr dr d\phi dθ[/itex]

    Integrating this over the whole sphere gives

    [itex]μ = \frac{ωq}{4πR^2} \int^{R}_{0} r^3 dr \int^{π}_{0} \sinθ dθ \int^{2π}_{0} d\phi [/itex]

    [itex]μ = \frac{wqR^2}{8}[/itex]

    Obviously this is not correct, the answer I should be getting is

    [itex]μ = \frac{wqR^2}{3}[/itex]

    Where am I going wrong?
     
    Last edited: Sep 10, 2013
  2. jcsd
  3. Sep 10, 2013 #2

    TSny

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    Is that the correct expression for the area enclosed? Can you describe the area that you are talking about? Did you make a sketch?
     
  4. Sep 10, 2013 #3
    For a normal loop of wire, A would be the area inside the loop so in this case dA is the area bounded by the surface, da, of the thin segment. I have tried using the volume of the segments but the equation for magnetic moments requires an area, so I couldn't see how this would work.

    Edit: I have seen on similar problems that the radius of each segment is rsinθ making the area enclosed:

    [itex]A = πr^2\sin^2θ[/itex]

    This may be where I am going wrong but I am struggling to see why the radius is rsinθ
     
    Last edited: Sep 10, 2013
  5. Sep 10, 2013 #4

    TSny

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    The area enclosed by a circular strip on the sphere is shown in yellow in the figure. Can you express that area in terms of R and θ?
     

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  6. Sep 10, 2013 #5
    I see it now, thank you!

    The radius of the segment is just simple trigonometry,

    [itex]r = R\sinθ [/itex]

    So the area now becomes:

    [itex]A = πR^2 \sin^2θ [/itex]
     
  7. Sep 10, 2013 #6

    TSny

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    Good.
     
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