Question

The square loop in the figure has sides of length $20\text{cm}$. It has $5$ turns and carries a current of $2\text{A}$. The normal to the loop is at ${37}^{\circ }$ to a uniform field $\overrightarrow{B}=0.5\hat{j}\text{T}$. Find the torque on the loop.

• A. $0.12\hat{k}\text{Nm}$
• B. $-0.12\hat{k}\text{Nm}$
• C. $0.12\hat{j}\text{Nm}$
• D. $-0.12\hat{j}\text{Nm}$

Answer 1

The correct option is B $-0.12\hat{k}\text{Nm}$

We redraw the diagram from a different perspective, the top view shown in figure.


From the figure, we see that the unit vector normal to loop,

$\hat{n}=-\sin {37}^{\circ }\hat{i}+\cos {37}^{\circ }\hat{j}=-0.6\hat{i}+0.8\hat{j}$

The magnetic moment is,

$\overrightarrow{\mu }=NIA\hat{n}=\left(5\right)\times \left(2\right)\times \left(0.2{)}^2\right(-0.6\hat{i}+0.8\hat{j})$

$\therefore \overrightarrow{\mu }=(-0.24\hat{i}+0.32\hat{j}){\text{Am}}^2$

The torque,

$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}=(-0.24\hat{i}+0.32\hat{j})\times \left(0.5\hat{j}\right)=-0.12\hat{k}\text{Nm}$

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