# Calculating the magnetic flux density of a thin, flat conductor

I am currently preparing for a physics exam and I have troubles understanding how to solve the following exercise:

A current that has the constant current density $\vec{j}$ flows through a thin, flat conducter of width $w$, height $h$ ($h \ll w$) and very large length. Calculate the magnetic flux density $\vec{B}$ in a point $P$ which has distance $d$ to the midline of the conductor (see this figure).

The standard solution suggests to divide the conductor into infinitesimal current filaments and then to integrate. They find that a infinitesimal current filament with current $\mathrm dI$ has the magnitude:

$$\mathrm dB = \frac{\mu_0 \mathrm dI}{2 \pi r} \cos(\alpha),$$

where $r = \sqrt{x^2 + d^2}$ is the distance between current filament and $P$ and $\alpha$ is the angle between $r$ and the $y$-axis. I actually thought that starting with Biot-Savart's law would be a good idea, but this doesn't look like Biot-Savart's law. How do I arrive at this formula? If it is Biot-Savart's law, where should I get the factor $2$ and the cosine from?

Thanks for any help in advance.

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If I'm not mistaken, it is $B = \frac{\mu_0 I}{2 \pi r}$. This looks pretty much like the expression in the standard solution, but why the cosine? –  Huy Aug 5 '12 at 18:33