Q 12-04-035JEE MainJEE Main 2026 (8 Apr, Shift 2)Medium
A current carrying circular loop of radius $2$ cm with unit normal $\hat{n} = \dfrac{\hat{k} + \hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\vec{B} = B_0(3\hat{i} + 2\hat{k})$. If $B_0 = 4 \times 10^{-3}$ T and current $I = 100\sqrt{2}$ A, the torque experienced by the loop is ______ Wb.A. ($\pi = 3.14$)
Answer: (D) $5024 \times 10^{-7}\hat{j}$
Concept. $\vec\tau=\vec m\times\vec B$ with $\vec m=IA\hat n$.
Step 1. Moment. $\vec m=100\sqrt2\times\pi\times4\times10^{-4}\times\dfrac{\hat i+\hat k}{\sqrt2}=0.1256(\hat i+\hat k)$.
Step 2. Torque. $0.1256\times4\times10^{-3}\,[(\hat i+\hat k)\times(3\hat i+2\hat k)]=5.024\times10^{-4}\,(3-2)\hat j$.
Answer. $5024\times10^{-7}\hat j$ (Option D).
Solution by Sreeraj P, M.Sc Physics
