# The Unapologetic Mathematician

## The Biot-Savart Law

What I’m going to present is slightly different from what usually gets called the Biot-Savart law, but I think it’s the most natural parallel to the Coulomb law. As far as I can tell, it doesn’t get stressed all that much in modern coverage; in the first course I took on electromagnetism way back in the summer of 1994 I didn’t even see the name written down and parsed what I heard as “bee-ohs of r”. So at least you know how it’s supposed to be pronounced.

If we have two charged particles that are both moving, then they also feel a different force than the electric one. We call the excess the magnetic force. In magnitude it’s proportional to both the magnitudes of the charges and their speeds, and inversely proportional to the square of the distance between them, and then it gets complicated. It’s probably going to be easier to write this down as a formula first. $\displaystyle F=\frac{\mu_0}{4\pi}\frac{qv\times(q'v'\times\hat{r})}{\lvert r^2\rvert}=\frac{\mu_0}{4\pi}\frac{qv\times(q'v'\times r)}{\lvert r^3\rvert}$

So many cross products! Like last time, this is the force exerted on the first particle by the second; $r$ is the vector pointing from the second particle to the first, and $\hat{r}=\frac{r}{\lvert r\rvert}$ is the unit vector pointing in the same direction. From this formula, we see that the force is perpendicular to the direction the first particle is moving — the magnetic force can only turn a particle, not speed it up or slow it down — and in the plane spanned by the direction the second particle is moving and the displacement vector between them.

Again, we’ve written the constant of proportionality in a weird way. The “magnetic constant” $\mu_0$ now appears in the numerator, so its units are almost like the inverse of those on the electric constant. But we’ve also got two velocities to contend with; these lead to a factor of time squared over area, resulting in mass times distance over charge squared. In the SI system we have another convenience unit called the “henry”, with symbol $H$, defined by $\displaystyle\mathrm{H}=\frac{\mathrm{m}^2\cdot\mathrm{kg}}{\mathrm{C}^2}$

which lets us write $\mu_0$ in units of henries per meter. Specifically, the SI units give it a value of $\mu_0=4\pi\times10^{-7}\frac{\mathrm{H}}{\mathrm{m}}\approx1.2566370614\times10^{-6}\frac{\mathrm{H}}{\mathrm{m}}$. Yes, I know that this makes the proportionality constant look that much weirder, since it just works out to $10^{-7}$, but still.

January 4, 2012