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.
Solution
(a)
Using figure,
\[\text{ For a regure polygon of n - sides, the angle subtended at the centre is } \frac{2\pi}{n}\]
\[\tan\theta = \frac{l}{2x}\]
\[ \Rightarrow x = \frac{l}{2\tan\theta}\]
\[\text{ considering angle to be small} \]
\[\frac{l}{2} = \frac{\pi r}{n}\]
\[\text{ Using Biot - Savart law for one side }\]
\[B = \frac{\mu_0}{4\pi}\frac{\text{idl }\sin\theta}{x^2}\]
\[ \Rightarrow B = \frac{\mu_0}{4\pi}\frac{i(\sin\theta + \sin\theta)}{x^2} = \frac{\mu_0 i2\left( \tan\theta \right)\left( 2\sin\theta \right)}{4\pi l} \left( \text{ Putting value of r }\right)\]
\[ \Rightarrow B = \frac{\mu_0 i2n\left( \tan\left( \frac{\pi}{n} \right) \right)\left( 2\sin\left( \frac{\pi}{n} \right) \right)}{4\pi\left( 2\pi r \right)} \left( \text{ Putting value of l } \right)\]
\[\text{ For n - sided polygon} \]
\[B' = nB\]
\[ \Rightarrow B' = \frac{\mu_0 i n^2 \tan\left( \frac{\pi}{n} \right)\sin\left( \frac{\pi}{n} \right)}{2 \pi^2 r}\]
(b)
when n→ ∞, polygon becomes a circle with radius r
and magnetic field will become \[B = \frac{\mu_0 i}{2r}\]
