Calculating Magnetic Field of Long Sheet with Uniformly Distributed Current

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Homework Statement



A long metallic sheet of width 'a' and negligible thickness has a current uniformly distributed along its length. Find the magnetic field on the plane of the sheet a distance 'b' from its axis.

The current flows in the x direction; the z direction is perpendicular to the surface.

Homework Equations



The Biot-Savart law:
[itex] \vec{B}=\frac{\mu_0}{4\pi}\int_S \frac{J_s \times \hat{r-r'}}{|r-r'|^2} dS'[/itex]

The Attempt at a Solution


Since J_s is constant, it can be put outside the integral.[itex] \vec{B}=\frac{\mu_0 J_s}{4\pi}\int_S \frac{\hat{x} \times \hat{r-r'}}{|r-r'|^2} dS'.[/itex]

Considering a finite region of length L we have:

[itex] \vec{B}=\frac{\mu_0 J_s}{4\pi}\int_{-a/2}^{a/2} \int_{-L/2}^{L/2} \frac{\hat{x} \times \hat{r-r'}}{|r-r'|^2} dx dy[/itex]

or[itex] B=\frac{\mu_0 J_s}{4\pi}\int_{-a/2}^{a/2} \int_{-L/2}^{L/2} \frac{1}{|r-r'|^2} dx dy[/itex]

in the z direction. But [itex]|r-r'|=\sqrt(x^2+(b-y)^2)[/itex]Substituting the denominator, we have[itex] B=\frac{\mu_0 J_s}{4\pi}\int_{-a/2}^{a/2} \int_{-L/2}^{L/2} \frac{1}{x^2+(b-y)^2} dx dy[/itex]

Is it correct?
How resolve this integral?
 
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I take it that the sheet isn't long enough to use ampere's law? Umm, I am having trouble looking at your work as you've typed it, and I'm not smart enough to do this in my head. I would make the curly'r unit vector into curly'r_vector/curly'r from the start. That's always the best way to do these types of problems, in my opinion. If you need some more help and can't get the LaTeX right, then I'll work it out myself at some point and relate mine to yours.