JEE Main 2026 Apr 8 Shift 2, Physics Q14: Torque and Work Done in a Magnetic Field

Magnetic moment: $\vec{M}=IA\hat{n}=I(\pi r^2)\hat{n}$ with $r=0.02$ m.

$\vec{M}=(100\sqrt{2})(3.14)(4\times 10^{-4})\cdot \frac{\hat{i}+\hat{k}}{\sqrt{2}}=100\cdot 3.14\cdot 4\times 10^{-4}(\hat{i}+\hat{k})=0.1256(\hat{i}+\hat{k})$.

Torque: $\vec{\tau }=\vec{M}\times \vec{B}$ with $\vec{B}=(4\times 10^{-3})(3\hat{i}+2\hat{k})$.

Pulling out the scalar factors: $\vec{\tau }=(100\sqrt{2})(3.14)(4\times 10^{-4})(4\times 10^{-3})\cdot \frac{(\hat{i}+\hat{k})}{\sqrt{2}}\times (3\hat{i}+2\hat{k})$

The cross product: $(\hat{i}+\hat{k})\times (3\hat{i}+2\hat{k})=2(\hat{i}\times \hat{k})+3(\hat{k}\times \hat{i})=-2\hat{j}+3\hat{j}=\hat{j}$.

Scalar factor: $\frac{100\sqrt{2}}{\sqrt{2}}\cdot 3.14\cdot 4\times 10^{-4}\cdot 4\times 10^{-3}=100\cdot 3.14\cdot 16\times 10^{-7}=5024\times 10^{-7}$.

$\therefore$ $\vec{\tau }=5024\times 10^{-7}\hat{j}$.