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Magnetic Moment of a spinning shell

  1. Feb 25, 2008 #1
    The problem reads:

    Find the magnetic moment of a spinning shell of charge Q, radius R, and angular velocity [tex]\vec{\omega}[/tex].

    My solution:

    I split the sphere into infinitely small loops of current (from top to bottom) and add them up. I set my origin on the center of the sphere and integrate with respect to y from 0 to R times two to get the entire sphere.

    The start of the problem:

    [tex]\vec{\mu} \,= A\vec{I}[/tex]
    [tex]d\vec{\mu} \,= A(d\vec{I})[/tex]

    Solving for [tex]d\vec{I}[/tex]:

    [tex]d\vec{I} = \frac{dQ}{t}[/tex]
    [tex]d\vec{I} = \frac{\sigma dA}{t}[/tex]
    [tex]d\vec{I} = \frac{\sigma dA}{(\frac{2\pi}{\omega})}[/tex]
    [tex]d\vec{I} = \frac{\sigma \omega dA}{2\pi}[/tex]

    Plugging [tex]d\vec{I}[/tex] back:

    [tex]d\vec{\mu} \,= \frac{\sigma \omega A dA}{2\pi}[/tex]

    Changing A and dA, so I could integrate in terms of dy:

    [tex]A \,= \pi r^2[/tex]
    [tex]A \,= \pi (R^2-y^2)[/tex]
    [tex]dA \,= -2 \pi y dy[/tex]

    Then setting up the integral:
    [tex]d\vec{\mu} \,= - \frac{2 \sigma \omega \pi^2}{2\pi} y (R^2-y^2) dy[/tex]
    [tex]d\vec{\mu} \,= - \sigma \omega \pi y (R^2-y^2) dy[/tex]
    [tex]\vec{\mu} \,= - 2 \sigma \omega \pi \int_{0}^{R} y (R^2-y^2) dy[/tex]

    Without sigma:
    [tex]\vec{\mu} \,= - \frac {Q \omega}{2R^2} \int_{0}^{R} y (R^2-y^2) dy[/tex]

    This is incorrect, and I have a suspicion it has to do with the "dA" part.

    If you don't mind his handwriting, here is the correct way to do the problem. I am rather uncomfortable integrating over angles, so I opted for a different way.

    I would like to know why my way does not work.

    Last edited: Feb 25, 2008
  2. jcsd
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