Question

A uniform, constant magnetic field $\overrightarrow{B}$ directed at an angle of ${45}^o$ to the x-axis in the x-y plane. PQRS is a rigid, square wire frame carrying a steady current $I_0$, with its center at the origin O. At time $t=0$, the frame is at rest in the position shown in fig, with its sides parallel to the x- and y-axes. Each side of the frame is of mass M and length L.

If the angle by which the frame rotates under the action of this torque in a short interval of time $\Delta t$, and the axis about which this rotation occurs is $\theta =\frac{3}{X}\frac{I_{0}B}{M}\Delta t^2$. Find X?

Given, moment of inertia of the frame about an axis through its center perpendicular to its plane is $\left(4/3\right)M{L}^2$

Answer 1

By the theorem of perpendicular axis, moment of inertia of the frame about QS,
$I_{QS}=\frac{1}{2}{I}_z=\frac{1}{2}\left(\frac{4}{3}M{L}^2\right)=\frac{2}{3}M{L}^2$
As $\tau =I\alpha \Rightarrow =\frac{\tau }{I}=\frac{I_0{L}^{2}B\times 3}{2L^{2}M}=\frac{3}{2}\frac{I_{0}B}{M}$
As here $\alpha$ is constant, equations of circular motion are valid. Hence, from $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$, with ${\omega }_0=0$, we have
$\theta =\frac{1}{2}\alpha t^2=\frac{1}{2}\left(\frac{3}{2}\frac{I_{0}B}{M}\right)(\Delta t{)}^2=\frac{3}{4}\frac{I_{0}B}{M}\Delta t^2$