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 magnetic moment.

• A. $(-0.24\hat{i}+0.32\hat{j}){\text{Am}}^2$
• B. $(0.24\hat{i}+0.32\hat{j}){\text{Am}}^2$
• C. $(-0.24\hat{i}-0.32\hat{j}){\text{Am}}^2$
• D. $(0.24\hat{i}-0.32\hat{j}){\text{Am}}^2$

Answer 1

The correct option is A $(-0.24\hat{i}+0.32\hat{j}){\text{Am}}^2$

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,

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

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

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