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

Let the regular polygon have $n$ sides. The total length of the wire is $2\pi r$, so the length of each side is $a = \dfrac{2\pi r}{n}$.

The perpendicular distance from the center to any side is $d = \dfrac{a}{2} \cot$\left(\dfrac{\pi}{n}\right) = \dfrac{\pi r}{n} \cot\left(\dfrac{\pi}{n}\right).

The magnetic field at the center due to one side of the polygon is given by:

$B_1 =$\dfrac{\mu_0 i}{4\pi d} \left[ \sin\left(\dfrac{\pi}{n}\right) + \sin \left(\dfrac{\pi}{n}\right) \right] = \dfrac{\mu_0 i}{2\pi d} \sin\left(\dfrac{\pi}{n}\right)$<br><br>The total magnetic field due to all$n$sides is:<br><br>$B = n B_1 = \dfrac{n \mu_0 i}{2\pi d} \sin $\left(\dfrac{\pi}{n}\right)<br><br>Substituting the value of$d$:<br><br>$B = $\dfrac{n \mu_0 i}{2\pi \left[ \dfrac{\pi r}{n} \cot\left(\dfrac{\pi}{n}\right) \right]} \sin\left(\dfrac{\pi}{n}\right) = \dfrac{\mu_0 i}{2r} \left( \dfrac{n}{\pi} \right)^2 \sin \left(\dfrac{\pi}{n}\right) \tan\left(\dfrac{\pi}{n}\right)$

Taking the limit as $n \to \infty$, we can substitute $\theta = \dfrac{\pi}{n}$, where $\theta \to 0$:

B = \lim_{\theta \to 0} \dfrac{\mu_0 i}{2r} \left(\dfrac{\sin \theta}{\theta} \right) \left( \dfrac{\tan \theta}{\theta} \right)

Using the standard limits $\lim_{\theta \to 0}$\dfrac{\sin \theta}{\theta} =$1$ and $\lim_{\theta \to 0}$\dfrac{\tan \theta}{\theta} =$1$, we get:

$B =$\dfrac{\mu_0 i}{2r}

Alternatively, as $n \to \infty$, the polygon approaches a circle of circumference $2\pi r$ and radius $r$. The magnetic field at the center of a circular loop is \dfrac{\mu_0 i}{2r}.