Question

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

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

Answer 1

Answer: (B)

Let $\overrightarrow{B}=B\hat{i}$
Magnetic moment $\mu =iA(-\hat{k})=i\pi r^2(-\hat{k})$

$\therefore \overrightarrow{\tau }=\mu \times \overrightarrow{B}=i\pi r^2(-\hat{k})\times B\hat{i}=i\pi r^{2}B(-\hat{j}).......................................\left(1\right)$

But, $\tau =I\alpha$ where $I=m{r}^2$ is the moment of inertia and $\alpha$ is the angular acceleration

$\therefore \tau =m{r}^2\alpha ..............................\left(2\right)$

Equating eqns $\left(1\right)$ and $\left(2\right),i\pi r^{2}B=m{r}^2\alpha$

$\Rightarrow \alpha =\frac{i\pi B}{m}=\frac{4\times \pi \times 10}{2}=20\pi rad{s}^{-2}$