Vector potential with current density

[Faust90]

Homework Statement

Hey,
I got the current density \vec{j}=\frac{Q}{4\pi R^2}\delta(r-R)\vec{\omega}\times\vec{r} 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
I'm really not sure how to go on now. Is this right what I wrote there?

 

[TSny]

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

[EDIT: Also, do \theta and \phi refer to the unprimed position or the primed position?]

 

[Faust90]

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

[EDIT: Also, do \theta and \phi 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 don't know how this can be helpful.

 

[TSny]

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!

 

[Faust90]

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?

 

[TSny]

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.