Current ${}^{\text{'}}i^{\text{'}}$ is flowing in hexagonal coil of side $\mathrm{a}$. The magnetic induction at the centre of the coil will be

Given data: Current i is flowing in a hexagonal coil

Formula used: $B=\frac{{\mu }_{0}i}{2\pi r}$

Detailed Solution:

From the figure, we get 

$r=a\frac{\sqrt{3}}{2}$

As we know that the magnetic field due to straight line at a distance (x) is $\frac{{\mu }_{0}I}{4\pi x}\left(\sin {\theta }_2-\sin {\theta }_1\right)$

$=\frac{2{\mu }_{0}I}{4\pi \left(a\sqrt{3}\right)}\left(\sin \left({30}^0\right)+\sin \left({30}^0\right)\right)$

So, due to the all wires of Hexagon will be 

$$\begin{gathered} =\frac{12{\mu }_{0}I}{4\pi \left(a\sqrt{3}\right)}\left(\frac{1}{2}+\frac{1}{2}\right) \\ =\frac{\sqrt{3}{\mu }_{0}I}{\pi a} \end{gathered}$$

Therefore, $\frac{\sqrt{3}{\mu }_{0}I}{\pi a}$ is the magnetic induction at the center of the coil.