A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring carries a current of i = 4 A. A horizontal magnetic field B = 10 T is switched on at time t = 0 as shown. The initial angular acceleration of ring will be

Magnetic moment of loop  (Assuming direction of current fflowing in the loop is anticlockwise)

$\begin{array}{l}\overrightarrow{\mathrm{M}}=\mathrm{i}\overrightarrow{\mathrm{A}}=4\times \mathrm{\pi }{\left(0.5\right)}^2\left(-\widehat{\mathrm{k}}\right)\\ =-4\mathrm{\pi }{\left(0.5\right)}^2\widehat{\mathrm{k}}\\ \mathrm{\tau }=\overrightarrow{\mathrm{M}}\times \overrightarrow{\mathrm{B}}\\ \mathrm{\tau }=\left(-\mathrm{\pi }\widehat{\mathrm{k}}\right)\times \left(10\mathrm{i}\right)=-10\mathrm{\pi }\widehat{\mathrm{j}}\end{array}$

So, axis of rotation is along $\overrightarrow{\tau }$, i.e., along negative y-direction.

Moment of inertia of ring about y-axis

$\begin{array}{l}\mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2=\frac{1}{2}\times 2\times {\left(0.5\right)}^2=\frac{1}{4}\mathrm{kg}{\mathrm{m}}^2\\ \mathrm{\tau }=\mathrm{I\alpha }\mathrm{so}\mathrm{\alpha }=\frac{\mathrm{\tau }}{\mathrm{I}}=\frac{10\mathrm{\pi }}{1/4}=40\mathrm{\pi }\mathrm{rad}/{\mathrm{s}}^2\\ \mathrm{\alpha }=40\mathrm{\pi }\mathrm{rad}/{\mathrm{s}}^2\end{array}$