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Magnetic Moment

  1. Feb 25, 2009 #1
    A small sphere is uniformly charged throughout its' volume and rotating with constant angular velocity [itex]\omega[/itex]. Show it's magnetic moment is given by
    [itex]m=\frac{1}{5}Q \omega a^2[/itex]. the question doesn't say but i'm assuming a is the radius.

    Anyway, so far I have:

    [itex]m=\frac{1}{2} \int_V dV (\mathbf{r \wedge J})[/itex]

    but [itex]\mathbf{J}= \rho \mathbf{v} = \rho \mathbf{\omega \wedge r}[/itex]

    and so [itex]m=\frac{\rho}{2} \int_V r^2 \mathbf{\omega} - (\mathbf{r \cdot \omega})\mathbf{r}[/itex]

    but i don't really know where to go from here?
     
  2. jcsd
  3. Feb 25, 2009 #2

    gabbagabbahey

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    Choose a coordinate system and carry out the integration. The easiest choice is probably spherical coordinates, with omega pointing in the z-direction.
     
  4. Feb 26, 2009 #3
    ok so [itex]\int (\mathbf{r \cdot \omega})\mathbf{r} dV=\int r \omega \cos{\theta} r^2 \sin{\theta} dr d \theta d \phi \frac{\mathbf{\hat{r}}}{r}=\omega \int_0^a r^2 dr \int_0^{\pi} \cos{\theta} \sin{\theta} d \theta \int_0^{2 \pi} d \phi \mathbf{\hat{r}}[/itex]
    giving
    [itex]\omega \frac{a^3}{3} 2 \pi \int_0^{\pi} \frac{1}{2} \sin{2 \theta} d \theta \mathbf{\hat{r}}= \frac{2}{3} \pi a^3 \left[ -\frac{1}{4} \cos{\theta} \right]_0^{\pi} \mathbf{\hat{r}}[/itex]

    which is

    [itex]\omega \frac{a^3}{12} \pi (-[-1-1])=\frac{\omega \pi a^3}{6} \mathbf{\hat{r}}[/itex]
     
  5. Feb 26, 2009 #4
    and the first integral is [itex]\omega \int_0^a r^4 dr \int_0^{\pi} \sin{\theta} d \theta \int_0^{2 \pi} d \phi \mathfb{\hat{z}}=\omega \frac{2a^5}{5} 2 \pi \mathbf{\hat{z}}=\frac{4 \omega \pi a^5}{5} \mathbf{\hat{z}} [/itex]
     
  6. Feb 26, 2009 #5
    reckon i've messed up somewhere though because they shouldn't have directions should they?
     
  7. Feb 26, 2009 #6

    gabbagabbahey

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    Magnetic moment is a vector, so it most definitely should have a direction.

    But you have messed up somewhere; is [itex]\hat{r}[/itex] really position independent? Because that's what you are assuming when you pull it out of the integral.
     
  8. Feb 26, 2009 #7
    ok. not sure what to do with it then?
     
  9. Feb 26, 2009 #8

    gabbagabbahey

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    Cartesian unit vectors are position independent, so rewrite [itex]\hat{r}[/itex] in terms of them.

    [tex]\hat{r}=\sin\theta\cos\phi\hat{x}+\ldots[/tex]
     
  10. Feb 26, 2009 #9
    ok. did that. both the [itex]\hat{x},\hat{y}[/itex] bits dropped out.

    and i got

    [itex]\omega \int_0^a r^2 dr \int_0^{\pi} \sin{\theta} \cos^2{\theta} d \theta \int_0^{2 \pi} d \phi \mathbf{\hat{z}}=-\frac{\omega a^3}{3} \int_{1}^{-1} (-u^2) du 2 \pi \mathbf{\hat{z}}[/itex] where [itex]u=\cos{\theta}[/itex]
    which simplifies to [itex]\frac{2 \omega \rho \pi a^3}{3} \mathbf{\hat{z}}[/itex] when you multiply in the factor of [itex]\frac{\rho}{2}[/itex] and so overall we get

    [itex]m=\left( \frac{2 \omega \rho \pi a^5}{5} -\frac{2 \omega \rho \pi a^3}{3} \right) \mathbf{\hat{z}}[/itex] which doesn't quite give me what i want. there must still be a mistake somewhere i guess.

    and i assume [itex]Q=\fac{4}{3} \pi \rho a^3[/itex]
     
  11. Feb 26, 2009 #10

    gabbagabbahey

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    What is 2/5-2/3?:wink:

    Edit why does your second terms have a^3 instead of a^5?
     
    Last edited: Feb 26, 2009
  12. Feb 26, 2009 #11
    kool. i got it.

    the next bit is to find the angular momentum L of the sphere of mass M and verify that [itex]\mathbf{m}=\frac{Q}{2M}\mathbf{L}[/itex]

    i was wanting to use the formula i'm meant to be verifying for finding L so since that's out the window, I was guessing [itex]\mathbf{L}=M(r^2\mathbf{\omega}-(\mathbf{r \cdot \omega})\mathbf{r}[/itex] but that's not getting me anywhere!
     
  13. Feb 26, 2009 #12

    gabbagabbahey

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    Use the definition of angular momentum: [tex]\vec{L}=M\vec{v}\times\vec{r}[/tex]

    P.S. \mathbf doesn't always show up too well here, so you might want to switch to \vec

    Edit: [tex]\vec{dL}=\rho_M\vec{v}\times\vec{r}dV[/tex] where [tex]\rho_M=\frac{M}{\frac{4}{3}\pi a^3}[/tex]:wink:
     
    Last edited: Feb 26, 2009
  14. Feb 26, 2009 #13
    do i sub in for v before i integrate?
     
  15. Feb 26, 2009 #14

    gabbagabbahey

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    did you sub in v before integrating when inding m?:wink:
     
  16. Feb 28, 2009 #15
    yes. so is it basically going to be the same integral again?
     
  17. Feb 28, 2009 #16

    gabbagabbahey

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    You tell me...
     
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