Question

Each side of the cube is $L$ then the magnitude of magnetic dipole moment of the loop $oabcdeo$ carrying current $i$ as shown in figure is

• A. $i{L}^2$
• B. $\sqrt{2}i{L}^2$
• C. $\sqrt{3}i{L}^2$
• D. $0$

Answer 1

The correct option is B $\sqrt{2}i{L}^2$


For loop $oabeo$, magnetic dipole moment,
${\overrightarrow{\mu }}_1=i{L}^2(-\hat{j})$
For loop $cdebc$, magnetic dipole moment,
${\overrightarrow{\mu }}_2=i{L}^2\left(\hat{k}\right)$

So, net magnetic dipole moment, $\overrightarrow{\mu }={\overrightarrow{\mu }}_1+{\overrightarrow{\mu }}_2=-i{L}^2\hat{j}+i{L}^2\hat{k}$

Further, its magnitude, $\mu =\sqrt{2}i{L}^2$