Vector potential with current density

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Homework Statement


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'|}.


Homework Equations


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?
 
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Answers and Replies

  • #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?]
 
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  • #3
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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?]

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.
 
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  • #4
TSny
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I should only calculate the potential A(r) on the z-axis (for r=z e_z).

Ahh. If ##\vec{\omega}## is also along the z-axis, then life is looking Wunderbar!
 
  • #5
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Ahh. If ##\vec{\omega}## is also along the z-axis, then life is looking Wunderbar!

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?
 
  • #6
TSny
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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?

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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