Question

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

• A. $\left(0 .05\right)\left(\hat{i} - \sqrt{3} \hat{k}\right)Am^{2}$
• B. $\left(0 .05\right)\left(\hat{j} + \hat{k}\right)Am^{2}$
• C. $\left(0 .05\right)\left(\sqrt{3} \hat{i} + \hat{k}\right)Am^{2}$
• D. $\left(\hat{i} + \hat{k}\right)Am^{2}$

Answer 1

Answer: (A)

Areal vectors will be ⊥ to the plane.
Area vector is at $6 0^{\text{o}}$ angle with x-axis and $3 0^{\text{o}}$ angle with -z-axis.
$\overset{ \rightarrow }{\text{M}} = \text{NI} \overset{ \rightarrow }{\text{A}}$
$\overset{ \rightarrow }{\text{M}} = 1 0 \times 1 0^{- 2} \text{A } \text{m}^{2}$
$\text{M} = 0 \cdot 1 \text{A } \text{m}^{2}$
$\overset{ \rightarrow }{\text{M}}=\text{M}_{x}\hat{\text{i}}-\text{M}_{z}\hat{\text{k}}=\text{Mcos}60\hat{\text{i}}-\text{Mcos}30^{\text{o}}\hat{\text{k}}$
$\overset{ \rightarrow }{\text{M}} = \text{0.05} \left(\hat{\text{i}} - \sqrt{3} \hat{\text{k}}\right)$