A wire of total length 2π r carrying a current i is bent to form a...

Total length of the wire is $2\pi r$.

The polygon has $n$ sides, so the length of each side is $a = \dfrac{2\pi r}{n}$.

The perpendicular distance from the centre to any side is $d = \dfrac{a}{2} \cot \left(\dfrac{\pi}{n}\right) = \dfrac{\pi r}{n} \cot\left(\dfrac{\pi}{n}\right).<br><br>The magnetic field at the centre due to one side of the polygon is given by the formula for a finite straight wire:<br><br>$B_1 = $\dfrac{\mu_0 i}{4\pi d}\left[ \sin \left(\dfrac{\pi}{n}\right) + \sin\left(\dfrac{\pi}{n}\right) \right]<br><br>Substituting the value of$d$:<br><br>$B_1 = \dfrac{\mu_0 i}{4\pi \left( \dfrac{\pi r}{n} \cot \left(\dfrac{\pi}{n}\right) \right)} \left[2\sin\left(\dfrac{\pi}{n}\right) \right]

$B_1 =$\dfrac{\mu_0 i n \sin\left(\dfrac{\pi}{n}\right) \tan\left(\dfrac{\pi}{n}\right)}{2\pi^2 r}

Since the polygon has $n$ identical sides and the magnetic field due to each side is in the same direction, the total magnetic field at the centre is:

$B = n B_1 =$\dfrac{\mu_0 i n^2 \sin\left(\dfrac{\pi}{n}\right) \tan\left(\dfrac{\pi}{n}\right)}{2\pi^2 r}