Q 12-04-193JEE MainJEE Main 2020 (3 Sep, Shift 1)Medium
Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side $10\ \text{cm}$, $50$ turns and carrying current $I$ (Ampere) in units of $\dfrac{\mu_0 I}{\pi}$ is:
Answer: (C) $500\sqrt3$
Each side subtends $60^\circ$ at the centre and is at distance $d = \dfrac{\sqrt3}{2}a = 0.05\sqrt3$ m from it. Field due to one side:
$$B_1 = \frac{\mu_0I}{4\pi d}(\sin30^\circ + \sin30^\circ) = \frac{\mu_0I}{4\pi d}$$
For $6$ sides and $50$ turns:
$$B = 300\cdot\frac{\mu_0I}{4\pi\times0.05\sqrt3} = \frac{\mu_0I}{\pi}\cdot\frac{1500}{\sqrt3} = 500\sqrt3\,\frac{\mu_0I}{\pi}$$
Solution by Sreeraj P, M.Sc Physics