Q 12-04-206JEE MainJEE Main 2019 (10 Apr, Shift 2)Medium
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side $1\ \text{m}$ which is carrying a current of $10\ \text{A}$ is: [Take $\mu_0 = 4\pi\times10^{-7}\ \text{N A}^{-2}$]
Answer: (C) $18\ \mu\text{T}$
The centre is at distance $d = \dfrac{l}{2\sqrt3}$ from each side, and each side subtends $60^\circ$ on either side of the perpendicular:
$$B_1 = \frac{\mu_0I}{4\pi d}(\sin60^\circ + \sin60^\circ) = \frac{\mu_0I}{4\pi}\cdot\frac{2\sqrt3}{l}\cdot\sqrt3 = \frac{\mu_0I}{4\pi}\cdot\frac6l$$
All three sides add:
$$B = \frac{\mu_0I}{4\pi}\cdot\frac{18}{l} = 10^{-7}\times\frac{18\times10}{1} = 18\ \mu\text{T}$$
Solution by Sreeraj P, M.Sc Physics