Question

A regular polygon of n sides is formed by bending a wire of total length 2πr which carries a current i. (a) Find the magnetic filed B at the centre of the polygon. (b) By letting n → ∞, deduce the expression for the magnetic field at the centre of a circular current.

Answer 1

(a)
Using figure,
$$\begin{gathered} \mathrm{For}\mathrm{a}\mathrm{regure}\mathrm{polygon}\mathrm{of}\mathrm{n}-\mathrm{sides},\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is}\frac{2\mathrm{\pi }}{n} \\ \tan \theta =\frac{l}{2x} \\ \Rightarrow x=\frac{l}{2\tan \theta } \\ \mathrm{considering}\mathrm{angle}\mathrm{to}\mathrm{be}\mathrm{small} \\ \frac{l}{2}=\frac{\mathrm{\pi r}}{n} \\ \mathrm{Using}\mathrm{Biot}-\mathrm{Savart}\mathrm{law}\mathrm{for}\mathrm{one}\mathrm{side} \\ B=\frac{{\mathrm{\mu }}_0}{4\pi }\frac{idl\sin \theta }{x^2} \\ \Rightarrow B=\frac{{\mathrm{\mu }}_0}{4\pi }\frac{i(\sin \theta +\sin \theta )}{x^2}=\frac{{\mathrm{\mu }}_{0}i2\left(\tan \theta \right)\left(2\sin \theta \right)}{4\pi l}\left(\mathrm{Putting}\mathrm{value}\mathrm{of}r\right) \\ \Rightarrow B=\frac{{\mathrm{\mu }}_{0}i2n\left(\tan \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)\right)\left(2\sin \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)\right)}{4\mathrm{\pi }\left(2\mathrm{\pi r}\right)}\left(\mathrm{Putting}\mathrm{value}\mathrm{of}l\right) \\ \mathrm{For}\mathrm{n}-\mathrm{sided}\mathrm{polygon} \\ B\text{'}=nB \\ \Rightarrow B\text{'}=\frac{{\mathrm{\mu }}_{0}i{n}^2\tan \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{n}}\right)}{2{\pi }^{2}r} \end{gathered}$$
(b)
when n→ ∞, polygon becomes a circle with radius r
and magnetic field will become
$B=\frac{{\mathrm{\mu }}_{0}i}{2r}$