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) $5024 \times 10^{-7}\hat{j}$
Magnetic moment: $\vec{m} = I\pi r^2\hat{n} = 100\sqrt{2} \times 3.14 \times 4 \times 10^{-4} \times \dfrac{\hat{i} + \hat{k}}{\sqrt{2}} = 0.1256(\hat{i} + \hat{k})$.
$$\vec{\tau} = \vec{m} \times \vec{B} = 0.1256 \times 4 \times 10^{-3}\,(\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k})$$
$(\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k}) = 2(\hat{i} \times \hat{k}) + 3(\hat{k} \times \hat{i}) = -2\hat{j} + 3\hat{j} = \hat{j}$.
$\vec{\tau} = 5.024 \times 10^{-4}\,\hat{j} = 5024 \times 10^{-7}\,\hat{j}$.
Solution by Sreeraj P, M.Sc Physics