JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Magnetic Effects of Current

The centroid of an equilateral triangle of side $a=4\sqrt{3}$ cm lies a perpendicular distance $d=\frac{a}{2\sqrt{3}}=2$ cm $=0.02$ m from each side, and each side subtends $60^{\circ }$ on either side of the foot of that perpendicular. A finite straight wire gives $B=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$, so one side contributes $\frac{{\mu }_{0}I}{4\pi d}(2\sin 60^{\circ })=\frac{10^{-7}\times 2}{0.02}\times \sqrt{3}=\sqrt{3}\times 10^{-5}$ T. The three sides produce fields in the same sense at $O$, so the net field at the centroid is $3\sqrt{3}\times 10^{-5}$ T.