Question

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

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

Answer 1

The correct option is A

$40\pi rad{s}^{-2}$

$\overrightarrow{M}=IA(-\hat{k})=-\left(4\right)\left(\pi \right)(0.5{)}^2\hat{k}$

$\Rightarrow \overrightarrow{M}=-\pi \hat{k}\left(A{m}^2\right)$

Since, $\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}=(-\pi \hat{k})\times \left(10\hat{i}\right)=-\left(10\pi \right)\hat{j}$

The axis of rotation is along $\overrightarrow{\tau }$ i.e., axis of rotation is the y – axis, moment of inertia about which is
$I=\frac{1}{2}m{R}^2=\frac{1}{2}\left(2\right)(0.5{)}^2=\frac{1}{4}kg{m}^2$

$\Rightarrow \alpha =\frac{|\overrightarrow{\tau }|}{I}=\frac{10\pi }{\left(\frac{1}{4}\right)}=40\pi rad{s}^{-2}$