Magnitude of magnetic field at the center of a hexagonal shape coil of side 5 cm,100 turns and carrying current I(ampere) in units of $2000\sqrt{x}\frac{{\mu }_{0}I}{\pi }$. Find ‘$x$’

$B_{each}=\frac{{\mu }_{0}In}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

$B=\frac{{\mu }_{0}In}{4\pi \frac{\sqrt{3}a}{2}}(Sin{30}^0+Sin{30}^0) [\because d=\frac{\sqrt{3}a}{2}]=\frac{{\mu }_0\mathrm{I}n}{4\sqrt{3}\pi a}2\times 2\times \frac{1}{2}$

$B_{each}=\frac{{\mu }_{0}In}{2\sqrt{3}\pi a}$

Now, magnetic field due to one hexagon coil $B_{net}=6B_{each}$

$B=6\times \frac{{\mu }_{0}In}{2\sqrt{3}\pi a}=\frac{{\mu }_{0}I}{\pi a}.\frac{\sqrt{3}\times 100\times 100}{5}=2000\sqrt{3}\frac{{\mu }_{0}I}{\pi }.$