A wire shaped to a regular hexagon of side $2 cm$ carries a current of $2 A$. Find the magnetic field at the center of the hexagon.

$$\begin{gathered} \because \theta =30° \\ \frac{BC}{OC}=\tan \theta \\ \therefore \frac{1}{r}=\tan 30°=\frac{1}{\sqrt{3}} \\ \therefore r=\sqrt{3}cm \end{gathered}$$

Net magnetic field at $O$ is $6$ times the magnetic field due to one side.

$$\begin{gathered} \therefore B=6\left[\frac{{\mu }_0}{4\mathrm{\pi }}\frac{i}{r}\left(\sin \theta +\cos \theta \right)\right] \\ =\frac{6\left({10}^{-7}\right)\left(2\right)}{\sqrt{3}\times {10}^{-2}}\left(\frac{1}{2}+\frac{1}{2}\right) \\ =6.9\times {10}^{-5 }T \end{gathered}$$