Question

A square loop $\text{OABCO}$ of side length $l$ carries a current $i$. It is placed as shown in the figure. The magnetic moment of the loop will be :

• A. $\frac{i{l}^2}{2}(\hat{i}-\sqrt{3}\hat{k})$
• B. $i{l}^2(\hat{j}-\sqrt{2}\hat{i})$
• C. $\frac{i{l}^2}{5}(\hat{j}-\sqrt{5}\hat{i})$
• D. $\frac{i{l}^2}{2}(\hat{j}+\sqrt{3}\hat{i})$

Answer 1

The correct option is A $\frac{i{l}^2}{2}(\hat{i}-\sqrt{3}\hat{k})$

The magnetic moment of the loop can be written as,

$\overrightarrow{M}=i\overrightarrow{A}$

$\overrightarrow{M}=i(\overrightarrow{OA}\times \overrightarrow{OC})$

(using the concept of cross product for area of parallelogram)

Here, $\overrightarrow{OA}=l\hat{j}$

$\overrightarrow{OC}=l\sin {60}^{\circ }\hat{i}+l\cos {60}^{\circ }\hat{k}$

$\overrightarrow{OC}=\frac{\sqrt{3}l}{2}\hat{i}+\frac{l}{2}\hat{k}$

Therefore,

$\overrightarrow{M}=i\left[\right(l\hat{j})\times (\frac{\sqrt{3}l}{2}\hat{i}+\frac{l}{2}\hat{k})]$

$\therefore \overrightarrow{M}=\frac{i{l}^2}{2}(\hat{i}-\sqrt{3}\hat{k})$

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