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

Explanation

As `/_ AOB = (2pi)/(n), OC` or perpendicular distance of any segment from centre equls `R" cos"(pi)/(n)`. Now magnectic induction at `O` due to the right current carrying element `AB`
`= (mu_(0))/(4pi) (i)/(R" cos"(pi)/(n)) 2" sin"(pi)/(n)`
(From Biot- Savart's the magentic field at `O` due to any section such as `AB` perpendicular to the plane of the figure and has the magnitude).
`B int (mu_(0))/(4pi)i (dx)/(r^(2))cos theta = int_(- (pi)/(n))^((pi)/(n)) (mu_(0) i)/(4pi) (R"cos"(pi)/(n) sec^(2) theta d theta)/(R^(2)"cos"(2pi)/(n) sec^(2) theta) = (mu_(0) i)/(4pi) (1)/(R"cos"(pi)/(n)) 2"sin"(pi)/(n)`
As there are `n` number of sides and magnitude induction vectors, due to each side at `O` are equal is magnitude and direction. So,
`B_(0) = (mu_(0))/(4pi) (ni)/(R"cos"(pi)/(n)) 2"sin"(pi)/(n). n`
`= (mu_(0))/(2pi) (ni)/(R) "tan" (pi)/(n)` and fro `n rarr oo`
`B_(0) = (mu_(0))/(2) (i)/(R) underset(nrarroo)Lt(("tan" (x)/(n))/(pi//n)) = (mu_(0))/(2) (i)/(R)`