Question

A uniform constant magnetic field $B$ is directed at an angle ${45}^o$ to the $x$-axis in the $xy$ plane. $\text{PQRS}$ is a rigid square frame carrying a constant current $I_o$, with its centre at origin $\text{O}$. At time $t=0$, the frame is at rest in the position shown in figure with its sides parallel to $X$ & $Y$ axes. The sides of the square has length $L$. The torque $\overrightarrow{\tau }$ about $\text{O}$ acting on the frame will be:
(Given, the coil has single turn)

• A. $\overrightarrow{\tau }=\frac{B{I}_o{L}^2}{\sqrt{2}}(-\hat{i}+\hat{j})$
• B. $\overrightarrow{\tau }=\frac{B{I}_o{L}^2}{\sqrt{2}}(\hat{i}-\hat{j})$
• C. $\overrightarrow{\tau }=\frac{B{I}_o{L}^2}{\sqrt{2}}(\hat{i}+\hat{j})$
• D. $\overrightarrow{\tau }=\frac{B{I}_o{L}^2}{\sqrt{2}}(-\hat{i}-\hat{j})$

Answer 1

The correct option is A $\overrightarrow{\tau }=\frac{B{I}_o{L}^2}{\sqrt{2}}(-\hat{i}+\hat{j})$

The area vector is perpendicular to plane of coil and is pointing outwards.

$\overrightarrow{A}=L^2\hat{k}$

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

$\overrightarrow{B}=B\cos {45}^o\hat{i}+B\sin {45}^o\hat{j}$

or $\overrightarrow{B}=\frac{B}{\sqrt{2}}\hat{i}+\frac{B}{\sqrt{2}}\hat{j}$

Thus, torque on frame is given by,

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

or, $\overrightarrow{\tau }=I_o{L}^2\left(\hat{k}\right)\times \frac{B}{\sqrt{2}}(\hat{i}+\hat{j})$

$\overrightarrow{\tau }=\frac{I_{0}B{L}^2}{\sqrt{2}}\left[\right(\hat{k}\times \hat{i})+(\hat{k}\times \hat{j}\left)\right]$

$\overrightarrow{\tau }=\frac{I_{0}B{L}^2}{\sqrt{2}}(\hat{j}-\hat{i})$

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