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Vector potential with current density

  1. Dec 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey,
    I got the current density [itex]\vec{j}=\frac{Q}{4\pi R^2}\delta(r-R)\vec{\omega}\times\vec{r}[/itex] and now I should calculate the vector potential:
    \vec{A}(\vec{r})=\frac{1}{4\pi}\int\frac{j(\vec{r})}{|r-r'|}.


    2. Relevant equations
    3. The attempt at a solution
    here my attempt till now:
    http://phymat.de/physics.png [Broken]
    I'm really not sure how to go on now. Is this right what I wrote there?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 14, 2012 #2

    TSny

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    In the second equation of the notes, should the vector r in the numerator have a prime?

    [EDIT: Also, do [itex]\theta[/itex] and [itex]\phi[/itex] refer to the unprimed position or the primed position?]
     
    Last edited: Dec 14, 2012
  4. Dec 14, 2012 #3
    Hi,

    thanks for your answer. I think they should all have primes.

    Edit: I should only calculate the potential A(r) on the z-axis (for r=z e_z). But I dont know how this can be helpful.
     
    Last edited: Dec 14, 2012
  5. Dec 14, 2012 #4

    TSny

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    Ahh. If ##\vec{\omega}## is also along the z-axis, then life is looking Wunderbar!
     
  6. Dec 15, 2012 #5
    Hi, thanks,
    Yes Omega is also along the z-Axis, but I don't know how this could help me. Didn't I have to solve the integral first?
     
  7. Dec 15, 2012 #6

    TSny

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    I don't think so. You can get the answer from symmetry considerations. Note that the current ##\vec{j}(r') d^3r'## in a small volume element makes an infinitesitmal contribution to the vector potential at ##r## of amount ##d\vec{A}(r)##, and ##d\vec{A}(r)## has the same direction as ##\vec{j}(r')##.

    Do you see that each nonzero value of ##\vec{j}(r') d^3r'## is parallel to the xy plane? So, the total value of ##\vec{A}(r)## must be parallel to the xy plane at any observation point ##r##.

    For this problem the observation point is at some z on the z-axis. Note that the current distribution ##\vec{j}(r')## is axially symmetric about the z axis. So ##\vec{A}(z)## would have to point perpendicular to the z-axis and yet be rotationally invariant about the z-axis. There is only one possibility for the value of ##\vec{A}## at z.
     
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