Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

Do you know the formula for the magnetic field generate by a (infinitely thin) wire?

share|improve this answer
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
The cosine is due to the fact that you are interested in the y-component of the magnetic field. –  Fabian Aug 5 '12 at 19:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.