Question

A square loop of mass M side 'a', and carrying current $i_0$ lies in the x-y plane as shown in the figure. It is hinged at y-axis so that it can freely route about it. Moment of inertia of the loop about an axis through its c.m. and normal to its plane is equal to $\gamma M{a}^2$. At t=0, an external magnetic field of induction B is applied along +ve x axis. The initial angular acceleration of the loop is equal to

• A. $\frac{4i_{0}B}{M(2\gamma +1)}$
• B. $\frac{i_{0}B}{M\gamma }$
• C. $\frac{4i_0\pi B}{M(2\gamma +1)}$
• D. $\frac{4i_{0}B}{(2\gamma +1)}$

Answer 1

The correct option is A $\frac{4i_{0}B}{M(2\gamma +1)}$

$\begin{array}{c}\tau =B{i}_0{a}^2\\ \tau =I\alpha \\ \Rightarrow B{i}_0{a}^2=\left(\frac{\gamma M{a}^2}{2}+\frac{M{a}^2}{4}\right)\alpha \\ \Rightarrow B{i}_0{a}^2=M{a}^2\left[\frac{\gamma }{2}+\frac{1}{4}\right]\alpha \\ \Rightarrow B{i}_0=\frac{M}{4}\left[2\gamma +1\right]\alpha \\ \Rightarrow \alpha =\frac{4B{i}_0}{M\left(2\gamma +1\right)}\end{array}$

Hence,

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