Q 12-04-165JEE MainJEE Main 2021 (31 Aug, Shift 2)Medium
A current of $1.5$ A is flowing through a triangle, of side $9$ cm each. The magnetic field at the centroid of the triangle is: (Assume that the current is flowing in the clockwise direction.)
Answer: (D) $3\times10^{-5}$ T, inside the plane of triangle
Distance of the centroid from each side: $d = \dfrac{a}{2\sqrt3} = \dfrac{9}{2\sqrt3} = \dfrac{3\sqrt3}{2}$ cm.
Each side subtends $60^\circ$ on either side of the perpendicular:
$$B_1 = \frac{\mu_0I}{4\pi d}(\sin60^\circ + \sin60^\circ) = \frac{10^{-7}\times1.5\times\sqrt3}{1.5\sqrt3\times10^{-2}} = 10^{-5}\ \text{T}$$
Total $B = 3\times10^{-5}$ T. A clockwise current gives a field into the plane (inside).
Solution by Sreeraj P, M.Sc Physics