Question

Figure shows a square current-carrying loop $ABCD$ of side $2m$ and current $I=\frac{1}{2}A$. The magnetic moment $\overrightarrow{M}$ of the loop is:

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

Answer 1

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

$\overrightarrow{DA}=-2\cos 30\hat{i}-2\sin 30\hat{k}=(-\sqrt{3}\hat{i}-\hat{k})$

$\overrightarrow{AB}=2\hat{j}$

$\therefore \overrightarrow{M}=i(\overrightarrow{DA}\times \overrightarrow{AB})$

$=\frac{1}{2}\left]\right(-\sqrt{3}\hat{i}-\hat{k})\times (2\hat{j}\left)\right]$

$=-\sqrt{3}\hat{k}+i$

$=(i-\sqrt{3}\hat{k})A-m^2$