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Current of Spinning sphere of constant voltage

  1. Dec 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the magnetic dipole moment of a spinning sphere of voltage ##V## and radius ##R## with angular frequency ##\omega##.

    2. Relevant equations


    3. The attempt at a solution
    To find the dipole moment, we need to do ##I \int d \vec{a}##, which would be ##I 4 \pi R^2 \hat{r}##, but I need to find the current knowing only the voltage and angular velocity. I'm not sure how to proceed.

    Edit: Thought about this again and noticed a mistake. What I need to do is integrate to add up all the current rings on the sphere. So I want to be able to do something like (the sphere is spinning along the z axis):

    $$\vec{m} = \sigma R^3 \omega \int ^{2 \pi} _{0} \int ^{\pi} _{0} \sin^3 \theta d \theta \phi \hat{z}$$ .

    Where ##I = \sigma R \omega \sin \theta## and ##\vec{A} = R^2 \sin^2 theta##
     
    Last edited: Dec 14, 2016
  2. jcsd
  3. Dec 14, 2016 #2

    Orodruin

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    Your expression for the dipole moment is the dipole moment of a conductor carrying a current I. You have a more general current density. It is also unclear what you mean by ##\hat r##.

    Hint: You know the potential so you should be able to find the charge density on the sphere. From the charge density and rotation of the sphere, you should be able to deduce the current density.
     
  4. Dec 14, 2016 #3
    I corrected my mistake I think, but I still am not sure how to find the surface charge density from only the voltage. I think my brain stopped working.
     
  5. Dec 14, 2016 #4

    Orodruin

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    Did you perhaps do electrostatics before magnetostatics?
     
  6. Dec 14, 2016 #5
    Yeah, I think I was just having a brain malfunction... it's pretty late here.

    I think this is the right approach: For a surface charged sphere: ##\vec{E} = \frac{R^2 \sigma}{\epsilon _{0}r^2}## then we have ##V = - \int ^{R} _{\infty} \frac{R^2 \sigma}{\epsilon _{0}r^2} dr =\frac{R^2 \sigma}{\epsilon _{0}R}## which gives ##\sigma = \frac{V \epsilon _{0}}{R}##.

    Edit: Just heard from a friend this is the correct solution, although my answer is missing a factor of R somehow. I have ##I = \sigma \omega \sin \theta R## and ## da = \pi R^2 \sin ^2 \theta ##, putting them together I get ##\vec{m} = \sigma \omega R^3 \int ^{\pi} _{0} \sin ^3 \theta d\theta##
     
    Last edited: Dec 14, 2016
  7. Dec 14, 2016 #6

    Orodruin

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    What is the width of the circular loop between ##\theta## and ##\theta + d\theta##?

    Edit: Dimensional analysis is also a powerful tool to figure out where a factor of ##R## is missing ...
     
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