Biot-Savart law on a polygon

A current $I$ flows along a thin wire, shaped as a regular polygon with $N$ sides, which can be inscribed intro a circle of radius $R$ . Find the magnetic field at the center of the polygon. What happens if $N$ is made very large?

Solution:

The polygon described in the problem is illustrated in the figure below, for $N=8$ . The magnetic field created by this structure is the superposition of fields from $N$ straight current carrying wires of length $2 R \sin (\pi/N)$ . Therefore, we will need to find the magnetic field due to a wire of finite length first.

The Biot-Savart law gives the magnetic field contribution from a small circuit element ${\rm d} \vec{l}$ carrying current $I$ , at a point $\vec{r}$ away from it:

(1) $\begin{equation*} {\rm d} \vec{B} = \frac{\mu_0 I}{4 \pi} \frac{{\rm d} \vec{l} \times \hat{r}}{r^2} . \end{equation*}$

To find the field generated by a straight wire, we break it up into small segments ${\rm d} \vec{l} \equiv \hat{y} {\rm d} l$ as shown in the figure. We are interested in the field at a point $R \cos (\pi/N)$ away from the middle of the wire. Using the right hand rule,

(2) $\begin{equation*} {\rm d} \vec{l} \times \hat{r} = (-\hat{z}) \, {\rm d} l \sin \theta = (-\hat{z}) \, \frac{R \cos (\pi/N) {\rm d} l}{r} , \end{equation*}$

which points into the plane of the paper. Therefore, the field due to the differential element ${\rm d} \vec{l}$ is

(3) $\begin{equation*} {\rm d} \vec{B} = \frac{\mu_0 I}{4 \pi} \frac{R \cos (\pi/N) {\rm d} l}{r^3} (-\hat{z}) . \end{equation*}$

We can also find the following relationship between different lengths from the above figure,

(4) $\begin{equation*} R^2 \cos^2 \left( \frac{\pi}{N} \right) + \left( R \sin \left( \frac{\pi}{N} \right) - l \right)^2 = r^2 = \frac{R^2 \cos^2 (\pi/N)}{\sin^2 \theta} \end{equation*}$

Differentiating this leads us to a relationship between ${\rm d} l$ and ${\rm d} \theta$ ,

(5) $\begin{equation*} \left( R \sin \left( \frac{\pi}{N} \right) - l \right) {\rm d} l = R^2 \cos^2 \left( \frac{\pi}{N} \right) \frac{\cos \theta {\rm d} \theta}{\sin^3 \theta} \implies {\rm d} l = R \cos \left( \frac{\pi}{N} \right) \frac{{\rm d} \theta}{\sin^2 \theta} , \end{equation*}$

where we have used $\tan \theta = R \cos(\pi/N)/(R \sin(\pi/N) - l)$ in the last step (another way of deriving the above relation is discussed at the end). Plugging this into (3),

(6) $\begin{equation*} {\rm d} \vec{B} = \frac{\mu_0 I (-\hat{z})}{4 \pi R \cos (\pi/N)} \sin \theta {\rm d} \theta . \end{equation*}$

Integrating this expression gives the magnetic field due to a single side of the polygon at its center,

(7) $\begin{equation*} \vec{B}_{\rm side} = \frac{\mu_0 I (-\hat{z})}{4 \pi R \cos (\pi/N)} \int_{\pi/2-\pi/N}^{\pi/2+\pi/N} \sin \theta {\rm d} \theta = \frac{\mu_0 I (-\hat{z})}{2 \pi R} \tan \left( \frac{\pi}{N} \right) . \end{equation*}$

The total field at the center is

(8) $\begin{equation*} \boxed{ \vec{B}_{\rm polygon} = N \vec{B}_{\rm side} = \frac{\mu_0 I N (-\hat{z})}{2 \pi R} \tan \left( \frac{\pi}{N} \right) . } \end{equation*}$