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 $L$. What is the magnitude of torque $\tau$ about $O$ acting on the frame due to magnetic field?

• A. $I_o{L}^{2}B$
• B. $2I_o{L}^{2}B$
• C. $\frac{I_o{L}^{2}B}{2}$
• D. $\frac{I_o{L}^{2}B}{\sqrt{2}}$

Answer 1

The correct option is A $I_o{L}^{2}B$


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,

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

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