Question

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

Find the angle by which the frame rotates under the action of this torque in a short interval of time $\Delta t$. $(\Delta t$ is short, so that any variation in the torque during this interval may be neglected$)$.

Given: The moment of inertia of the frame about an axis through its centre perpendicular to its plane is $\frac{4}{3}M{L}^2$.

• A. $\frac{3}{2}\frac{I_{0}B}{M}(\Delta t{)}^2$
• B. $\frac{3}{2}\frac{I_{0}B}{M}\left(\Delta t\right)$
• C. $\frac{3}{4}\frac{I_{0}B}{M}\left(\Delta t\right)$
• D. $\frac{3}{4}\frac{I_{0}B}{M}(\Delta t{)}^2$

Answer 1

The correct option is D $\frac{3}{4}\frac{I_{0}B}{M}(\Delta t{)}^2$


The magnetic field is directed at an angle ${45}^{\circ }$ with $\text{x-axis}$.

So, $\overrightarrow{B}=B\cos {45}^{\circ }\hat{i}+B\sin {45}^{\circ }\hat{j}$

The magnetic moment of the loop is,

$\overrightarrow{\mu }=I_0{L}^2\hat{k}$

The torque on the loop is,

$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$

$=\left(I_0{L}^2\hat{k}\right)\times (B\cos {45}^{\circ }\hat{i}+B\sin {45}^{\circ }\hat{j})$

$=\frac{I_0{L}^{2}B}{\sqrt{2}}(\hat{j}-\hat{i})$

Magnitude of torque is,

$|\overrightarrow{\tau }|=\frac{I_0{L}^{2}B}{\sqrt{2}}\times \sqrt{1^2+(-1{)}^2}=I_0{L}^{2}B$

The direction of vector $(\hat{j}-\hat{i})$ is diagonal $QS$ of the loop, and $QS$ is axis of rotation.

From Newton's second law for rotation,

$\tau =I\alpha ...\left(1\right)$

From perpendicular axis theorem,

$I_1+I_2=2I=\frac{4}{3}M{L}^2$

$\Rightarrow I=\frac{4}{6}M{L}^2$


So, $\alpha =\frac{\tau }{I}=\frac{3}{2}\frac{I_{0}B}{M}$

The angle turned by loop,

$\Delta \theta =\frac{1}{2}\alpha (\Delta t{)}^2=\frac{3}{4}\frac{I_{0}B}{M}(\Delta t{)}^2$

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