Question

A conducting ring of mass $2\text{kg}$ and radius $0.5\text{m}$ is placed on a smooth horizontal plane. The ring carries a current $i=4\text{A}$.
A horizontal magnetic field $B=10\text{T}$ is switched on at time $t=0$ as shown in figure.

The initial angular accelaration of the ring will be:

• A. $40\pi {\text{rad/s}}^2$
• B. $20\pi {\text{rad/s}}^2$
• C. $5\pi {\text{rad/s}}^2$
• D. $15\pi {\text{rad/s}}^2$

Answer 1

The correct option is A $40\pi {\text{rad/s}}^2$

The area vector of the ring will be perpendicular to the direction of $\overrightarrow{B}$, because ring is placed in such a way that plane of ring contains $\overrightarrow{B}$.

The angle between the magnetic moment $\overrightarrow{M}$ & magnetic field $\overrightarrow{B}$ is,
$\theta ={90}^{\circ }$

Magnitude of torque is given by,

$\tau =MBsin{90}^{\circ }$

$\tau =i(\pi R{)}^2\times B$

The direction of torque is shown below;

Thus, moment of inertia of ring about given axis by applying perpendicular axis theorem,

$I_x+I_y=I_z$

$2I_y=m{R}^2(\because I_x=I_y)$

$I_y=\frac{m{R}^2}{2}$

Using,

$\tau =I\alpha$

$\Rightarrow i\left(\pi R^2\right)B=\frac{m{R}^2}{2}\times \alpha$

or, $\alpha =\frac{2i\pi R^{2}B}{m{R}^2}$

$\alpha =\frac{2iB\pi }{m}$

$\alpha =\frac{2\times 4\times 10\times \pi }{2}$

$\therefore \alpha =40\pi {\text{rad/s}}^2$

Hence, option $\left(A\right)$ is the correct answer.