Question

In the shown figure a conducting ring of mass $m=2\text{kg}$ and radius $R=0.5\text{m}$ lies on a smooth horizontal plane with its plane vertical. The ring carries a current of $I=\frac{1}{\pi }A$. A horizontal uniform magnetic field of $B=12\text{T}$ is switched on at $t=0$. The initial angular acceleration $\alpha$ in ${\text{rad/sec}}^2$ of the ring will be $4x$,then $x$ is

Answer 1

Torque due to magnetic field is $\tau =\overrightarrow{M}\times \overrightarrow{B}$
$\tau =IA\hat{k}\times B\hat{i}$
$\tau =IAB\hat{j}$
As net torque is about y axis,
Applying rotatory version of newtons second law,
$\tau =I_{y-axis}\alpha$
$(I.\pi r^2)B=\frac{1}{2}m{r}^2\alpha$
$\alpha =12{\text{rad/sec}}^2$
$x=3$