Irodov 3.220: A Polygon Closing In on a Circle – Amit & Deepali Sinha Foundation

Problem Irodov 3.220

A thin wire bent into a regular polygon with $n$ sides, whose circumscribed circle has radius $R$ , carries a current $I$ . Find the magnetic induction at the center of the polygon, and examine the result as $n \to \infty$ .

The regular polygon (drawn for n = 6) inscribed in a circle of radius R, with the current I (sense chosen: counter-clockwise) and the center O. — The regular polygon (drawn for $n = 6$ ) inscribed in a circle of radius $R$ , with the current $I$ (sense chosen: counter-clockwise) and the center $O$ .

Given

$I$
$n$
$R$ (circumscribed radius)

Find

B

at the center

O

; its limit as

n \to \infty

Solution

Step 1The field of one straight piece

One side seen from O: the perpendicular from O meets it at its middle and makes the half angle π/n with the radius to either end. For a counter-clockwise current B at O is out of the page. — One side seen from $O$ : the perpendicular from $O$ meets it at its middle and makes the half angle $\pi/n$ with the radius to either end. For a counter-clockwise current $\vect B$ at $O$ is out of the page.

Take one side, at perpendicular distance $b$ from $O$ , and measure $x$ along it from the foot of that perpendicular. An element $dx$ sits at distance $r = (x^2 + b^2)^{1/2}$ , and the sine of the angle between $d\vect l$ and $\vect r$ is $b/r$ , so the Biot-Savart law gives

dB = \frac{\mu_0 I}{4\pi}\,\frac{b\,dx}{(x^2 + b^2)^{3/2}}

If each end of the side is seen from $O$ at the angle $\alpha$ from the perpendicular, the side runs from $x = -b\tan\alpha$ to $x = b\tan\alpha$ :

\begin{aligned} B_1 &= \frac{\mu_0 I}{4\pi}\int_{-b\tan\alpha}^{b\tan\alpha}\frac{b\,dx}{(x^2 + b^2)^{3/2}} = \frac{\mu_0 I}{4\pi}\left[\frac{x}{b\,(x^2 + b^2)^{1/2}}\right]_{-b\tan\alpha}^{b\tan\alpha}\\ &= \frac{\mu_0 I}{4\pi}\cdot\frac{2}{b}\cdot\frac{\tan\alpha}{(1 + \tan^2\alpha)^{1/2}} \end{aligned}

B_1 = \frac{\mu_0 I}{4\pi}\,\frac{2\sin\alpha}{b}

(1)

Step 2The polygon’s numbers

Each side subtends $2\pi/n$ at the center, so its ends are seen at $\alpha = \pi/n$ on either side of the perpendicular, which is the apothem:

\begin{aligned} \alpha &= \frac{\pi}{n}\\ b &= R\cos\frac{\pi}{n} \end{aligned}

Every side gives $\vect B$ the same sense at $O$ (the current runs the same way round $O$ along every side), so the $n$ equal values (1) add:

B = n\cdot\frac{\mu_0 I}{4\pi}\,\frac{2\sin(\pi/n)}{R\cos(\pi/n)}

B = \frac{n\mu_0 I}{2\pi R}\,\tan\frac{\pi}{n}

(2)

$\vect B$ is normal to the plane of the polygon, out of the page for the counter-clockwise current drawn.

Step 3Many sides: the circle

As $n \to \infty$ the angle $\pi/n$ becomes small and $\tan(\pi/n) \to \pi/n$ , so

B \to \frac{n\mu_0 I}{2\pi R}\cdot\frac{\pi}{n}

B_{\infty} = \frac{\mu_0 I}{2R}

(3)

which is the field at the center of a circular loop of radius $R$ : the polygon has become its circumscribed circle.

Check

The other end of the range: at $n = 2$ the polygon collapses into a diameter traversed twice, a wire passing through $O$ itself, and (2) correctly diverges there, since $\tan(\pi/2)$ is infinite. Between the two ends the field decreases steadily with $n$ toward the circle’s value. The units are those of $\mu_0 I/R$ , tesla.

Remark

Since $\tan x > x$ for $0 < x < \pi/2$ , the polygon’s field always exceeds the circle’s, $B/B_\infty = \tan(\pi/n)/(\pi/n)$ : 1.65 for a triangle, 1.27 for a square, 1.10 for a hexagon. The polygon’s sides pass closer to the center than the circle does, and the near parts of a wire dominate its field.