Question

A regular polygon of $n$ sides is formed by bending a wire of length ${}^{\prime }{L}^{\prime }$ which carries a current $i$, if the magnetic field $B$ at the centre is $\frac{{\mu }_{0}i{n}^2\sin \frac{\pi }{n}\tan \frac{\pi }{n}}{N\pi L}$, then $N$ will be______(Answer opto a maximum of two decimal points is allowed)

Answer 1

Let us find magnetic field at ${}^{\prime }{C}^{\prime }$ due to a wire ${}^{\prime }{\text{PQ}}^{\prime }$ (using the geometry)

$B_{PQ}=\frac{{\mu }_{0}i}{4\pi d}(\sin \frac{\pi }{n}+\sin \frac{\pi }{n})$ and $d=\frac{L}{2n}\cot \frac{\pi }{n}$
$\Rightarrow B_{PQ}=\frac{{\mu }_{0}i}{4\pi }\frac{(2\sin \frac{\pi }{n})}{\frac{L}{2n}\cot \frac{\pi }{n}}$

$=\frac{{\mu }_{0}in}{\pi L}\tan \frac{\pi }{n}\cdot \sin \frac{\pi }{n}$

$\Rightarrow B_{net}=n{B}_{pQ}=\frac{{\mu }_{0}i{n}^2}{\pi L}[\tan \frac{\pi }{n}\times \sin \frac{\pi }{n}]$
On comparing with given expression we get, $N=1$

Accepted answer : $1,1.0,1.00$