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What is the magnetic dipole moment of the sphere?

  1. May 29, 2007 #1
    1. The problem statement, all variables and given/known data
    A uniformly charged solid sphere of radius R carries a total charge Q, and is set spinning with angular velocity [itex] \omega [/itex] about the z axis. (a) What is the magnetic dipole moment of the sphere?


    2. Relevant equations
    [tex] \vec{m} = I \int d\vec{a} [/tex]

    3. The attempt at a solution
    having a lot of difficulty with this stuff

    since we are talking about a solid sphere ... first find volume current density [itex] J = \rho v[/itex]
    [tex] \rho = \frac{Q}{\frac{4}{3} \pi R^3} [/tex]
    [tex] v = \omega \times r = \omega r sin\theta [/tex]

    so [tex] J = \frac{Q}{\frac{4}{3} \pi R^3} \omega r sin\theta [/tex]
    is this ok so far??

    alright now to find the total current [itex] I = \int J \cdot da [/itex]

    [tex] I = \int \frac{Q}{\frac{4}{3} \pi R^3} \omega R^3 \sin^2\theta d\theta d\phi [/tex]

    this doesnt seem dimensionally correct since the radians do not cancel out ...

    where have i gone wrong??? Is it the part of the angular momentum??

    thansk for your help
     
  2. jcsd
  3. May 29, 2007 #2
    Hello,

    There are different current for different point on the spining solid sphere.
    The magnitude of current density [tex]J[/tex] you have is correct.
    But the total magnetic moment should be
    [tex]\vec{m}=\int d\vec{m}[/tex]
    ,where [tex]d\vec{m}[/tex] is a thin ring with radius [tex]r\sin\theta[/tex] rotating along z-axis.
    The cross section area of the thin ring is [tex]dr\cdot rd\theta[/tex].
    The effective current due to the thin ring is [tex]J(dr\cdot rd\theta)[/tex].
    Therefore, [tex]d\vec{m}=\hat{z}J(dr\cdot rd\theta)\pi(r\sin\theta)^2[/tex].
    You can substitute the magnetic moment element into the first integral above and find the aim.


    Good luck
     
  4. May 29, 2007 #3

    Meir Achuz

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    Use (in Gaussian units)
    [tex]{\vec m}=\frac{1}{2c}\int{\vec r}\times{\vec j}d^3r[/tex]
     
  5. May 29, 2007 #4
    ok so since [tex] \vec{m} = \frac{1}{2} \int \vec{r} \times \vec{J} d\tau [/tex]
    in SI units
    [tex] J = \rho v = \rho (\omega \times r) [/tex]

    [tex] m = \frac{1}{2} \int r \times \rho (\omega \times r) d\tau [/tex]
    using the appropriate product rule

    [tex] m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau [/tex]

    the second term in the integrand is zero since [tex] \int_{0}^{\pi} \cos\theta\sin\theta d\theta =0 [/tex]

    so we hav
    [tex] m =\frac{\rho}{2} \int \omega r^2 d\tau = \frac{\rho\omega}{2} \int_{0}^{R} r^4 dr (4\pi) = \frac{4\pi\rho\omega}{2} \frac{R^5}{5} [/tex]
    but [tex] \rho = \frac{Q}{\frac{4}{3} \pi R^3} [/tex]

    so [tex] m = \frac{2\pi\omega R^5}{5} \frac{Q}{\frac{4}{3} \pi R^3} = \frac{3}{10} Q\omega R^2 [/tex]

    the answer is supposed to be [tex] \frac{1}{5} Q\omega R^2[/tex]

    maybe im missing some factors??
     
  6. May 29, 2007 #5

    Meir Achuz

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    [tex]\oint d\Omega{\vec r}({\vec r}\cdot\omega)
    =4\pi r^2\omega/3.[/tex]
     
    Last edited: May 29, 2007
  7. Apr 19, 2009 #6
    i think there is a mistake in tis integral. but i dont know where. XD
    [tex]
    m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau
    [/tex]
     
  8. Apr 19, 2009 #7
    [tex]
    m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau [tex]
     
  9. May 2, 2009 #8
    the correct answer is = 1/3*Q*R^2*w .... please correct your solution as soon as possible .
     
  10. Jan 27, 2010 #9
    I disagree. I get [tex]\vec m = \frac 15 Q \omega R^2 \hat k[/tex]. I see the mistake as having been made in the second integration above; it's NOT identically zero. The [tex]x[/tex] and [tex]y[/tex] components are because of the integration in [tex]\phi[/tex], but the [tex]z[/tex] component is nonzero. Don't forget that the first "r" that shows up in the second integration is [tex]\vec r[/tex], NOT [tex]|\vec r|[/tex]. Since we're working in spherical coordinates, this has three components with complicated expressions in terms of [tex]\theta[/tex] and [tex]\phi[/tex].
     
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