Question

The figure shows a current-carrying square loop $ABCD$ of side $10cm$ and current $i=10A$ . The magnetic moment $\overrightarrow{M}$ of the loop is

• A. $(0.05)(\hat{i}-\sqrt{3}\hat{k}){\mathrm{Am}}^2$
• B. $(0.05)(\hat{j}+\hat{k})A{m}^2$
• C. $(0.05)(\sqrt{3}\hat{i}+\hat{k})\mathrm{A}{\mathrm{m}}^2$
• D. $(\hat{i}+\hat{k})A^2{m}^2$

Answer 1

Answer: (A)

Areal vectors will be ⊥ to the plane.
Area vector is at $60^{\text{o}}$ angle with x-axis and $30^{\text{o}}$ angle with -z-axis.
$\overrightarrow{M}=\text{NI}\overrightarrow{A}$
$\overrightarrow{M}=10\times 10^{-2}\text{A }{\text{m}}^2$
$\text{M}=0\cdot 1\text{A }{\text{m}}^2$
$\overrightarrow{M}={\text{M}}_x\hat{\text{i}}-{\text{M}}_z\hat{\text{k}}=\text{Mcos}60\hat{\text{i}}-\text{Mcos}30^{\text{o}}\hat{\text{k}}$
$\overrightarrow{M}=\text{0.05}\left(\hat{\text{i}}-\sqrt{3}\hat{\text{k}}\right)$