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?


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.

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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*}

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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*}

In the limit of large N, the angle \pi/N becomes very small and we can approximate \tan (\pi/N) \approx \pi/N (see figure). Then,

(9)   \begin{equation*}   \lim_{N \to \infty} \vec{B}_{\rm polygon} = \frac{\mu_0 I}{2 \pi R} (-\hat{z}) , \end{equation*}

which is the field generated by a ring of current of radius R, at its center.

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As N is made larger the polygon looks more and more like a ring of current.

Note: We could also have related the differential length {\rm d} l to {\rm d} \theta with geometry (see figure),

(10)   \begin{equation*}   {\rm d} l \sin \theta = r {\rm d} \theta . \end{equation*}

Substituting r \sin \theta = R \cos(\pi/N) leads us to (5).

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Bonus problem

Find the magnetic field at point O in the figure below.

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