Question

A uniform ring of mass 2kg carrying a current 4A is placed on a smooth horizontal surface as shown in the figure. Now a uniform magnetic field of 10T is switched on in the plane of the ring horizontally. The initial angular acceleration of the ring will be (in $rad/s^2)$ :

• A. $40\pi$
• B. $20\pi$
• C. $80\pi$
• D. None of these

Answer 1

The correct option is A $40\pi$

Torque on the loop $|\overrightarrow{\mu }\times \overrightarrow{B}|=I\alpha$
where $\overrightarrow{\mu }$ is the magnetic moment of the loop, $\overrightarrow{B}$ is the magnetic field , $I$ is the moment of inertia of the loop about an axis passing through the center of the loop, $\alpha$ is the angular acceleration of the loop.
$\therefore i\left(\pi r^2\right)B=\frac{m{r}^2}{2}\alpha$
$\therefore 4\times 10\times \pi r^2=\frac{m{r}^2}{2}\alpha$
$\therefore \alpha =\frac{80\pi }{2}=40\pi rad/s^2$ since $m=2kg$