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,

\[\text{ For  a  regure  polygon  of  n - sides,   the  angle  subtended  at  the  centre  is } \frac{2\pi}{n}\] 
\[\tan\theta = \frac{l}{2x}\] 

\[ \Rightarrow x = \frac{l}{2\tan\theta}\] 

\[\text{ considering  angle  to  be  small}  \] 

\[\frac{l}{2} = \frac{\pi r}{n}\] 

\[\text{ Using  Biot - Savart  law  for  one  side }\] 

\[B = \frac{\mu_0}{4\pi}\frac{\text{idl }\sin\theta}{x^2}\] 

\[ \Rightarrow B = \frac{\mu_0}{4\pi}\frac{i(\sin\theta + \sin\theta)}{x^2} = \frac{\mu_0 i2\left( \tan\theta \right)\left( 2\sin\theta \right)}{4\pi l}      \left( \text{ Putting  value  of  r }\right)\] 

\[ \Rightarrow B = \frac{\mu_0 i2n\left( \tan\left( \frac{\pi}{n} \right) \right)\left( 2\sin\left( \frac{\pi}{n} \right) \right)}{4\pi\left( 2\pi r \right)}              \left( \text{ Putting  value  of  l } \right)\] 

\[\text{ For  n - sided  polygon} \] 

\[B' = nB\]  

\[ \Rightarrow B' = \frac{\mu_0 i n^2 \tan\left( \frac{\pi}{n} \right)\sin\left( \frac{\pi}{n} \right)}{2 \pi^2 r}\] 

(b)
when n→ ∞, polygon becomes a circle with radius r
and magnetic field will become  \[B = \frac{\mu_0 i}{2r}\]