Question

Magnitude of magnetic field (in S.I Unit) at the centre of a hexagonal shape coil of side $10\text{cm}$, $50$ turns and carrying a current $I$ (Ampere ) in units of $\frac{{\mu }_{0}I}{\pi }$ is

• A. $50\sqrt{3}$
• B. $5\sqrt{3}$
• C. $250\sqrt{3}$
• D. $500\sqrt{3}$

Answer 1

The correct option is D $500\sqrt{3}$

Side $=10\text{cm}$

From figure,

$\tan {30}^{\circ }=\frac{x}{d}\Rightarrow d=\frac{x}{\tan 30}=\frac{5\times {10}^{-2}}{\left(\frac{1}{\sqrt{3}}\right)}$

$\therefore d=5\sqrt{3}\times {10}^{-2}\text{m}$

Magnetic field at the centre is , $B=\frac{6{\mu }_{0}IN}{4\pi d}[\sin {\theta }_1+\sin {\theta }_2]$

$B=\frac{6\times 50}{4\times 5\sqrt{3}\times {10}^{-2}}\left(\frac{{\mu }_{0}I}{\pi }\right)({\theta }_1={30}^{\circ };{\theta }_2={30}^{\circ })$

$B=500\sqrt{3}\left(\frac{{\mu }_{0}I}{\pi }\right)$

Hence, option (d) is the correct answer.