Potential vector (A) of a disk with a surface current

[Cloruro de potasio]

Homework Statement

Consider a sheet of current K = K0ˆx flowing on the surface of a finite circular disk of radius R lying in the xy plane. (1) Calculate the vector potential A(r) along the zˆ-axis above and below the sheet. Plot the z dependence of the non-zero component of the vector potential. (2) Calculate the derivatives of the vector potential just above and just below the sheet. Verify that it follows equation 5.78 from the book, namely the discontinuity in the derivative of the vector potential normal to the current sheet is ∂Aabove/∂n − ∂Abelow/∂n = −µ0K. (3) Using the above results, calculate the magnetic field B(r) at an arbitrary point r = (0, 0, z) along the zˆ-axis above and below the sheet. (4) We now investigate what happens in the limit that the radius of the current-carrying circular disk R → ∞. Calculate the vector potential A(r) and the magnetic field B(r) in that limit. Comment on your result.

Relevant Equations

$B=\nabla x A$

Hi,

I've been stuck for a long time with this exercise. I am not able to calculate the potential vector, since I do not know very well how to pose the itegral, or how to decompose the disk to facilitate the resolution of the problem. I know that because the potential vector must be parallel to the current, it must have only an x component, but I don't know how to continue ...

Thank you very much in advance

 

[PhDeezNutz]

I would think for the vector potential along the z-axis

$\vec{A} \left( \vec{r} \right) = \frac{\mu_0}{4 \pi} \int \frac{\vec{K}}{\left|\vec{r} - \vec{r}' \right|}\,da'$

$= \frac{\mu_0}{4 \pi}\int \frac{\left(K_0,0,0 \right)}{z} \, dx' dy'$

That $\frac{1}{z}$ could be brought out of the integral and really the only thing you have to do is evaluate the $x$- component of the integral above. You can pretty much bring everything out of the integral besides $dx'dy'$ and evaluating that integral is trivial.

 

[TSny]

I would think for the vector potential along the z-axis

$\vec{A} \left( \vec{r} \right) = \frac{\mu_0}{4 \pi} \int \frac{\vec{K}}{\left|\vec{r} - \vec{r}' \right|}\,da'$

Ok.

$= \frac{\mu_0}{4 \pi}\int \frac{\left(K_0,0,0 \right)}{z} \, dx' dy'$

For a general element of area $da'$,
$|\vec{r} - \vec r'| \neq z$

 

[PhDeezNutz]

Ok.

For a general element of area $da'$,
$|\vec{r} - \vec r'| \neq z$

I was going to argue with you. Realized I was wrong.

$\vec{A} \left( z \right) = \frac{\mu_0}{4 \pi} \int \frac{(K_0,0,0)}{\sqrt{s'^2 + Z^2}}\, 2 \pi s' ds'$

I think this is right OP. My apologies.

 

[TSny]

I was going to argue with you. Realized I was wrong.

$\vec{A} \left( z \right) = \frac{\mu_0}{4 \pi} \int \frac{(K_0,0,0)}{\sqrt{s'^2 + Z^2}}\, 2 \pi s' ds'$

OK. But we don't want to give too much of the solution away. The OP needs to show some work.

 

[PhDeezNutz]

OK. But we don't want to give too much of the solution away. The OP needs to show some work.

Understood.