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Question

(A) The magnetic moment $\vec{M}$ of a current loop is given by $\vec{M} = I\vec{A}$,where $\vec{A}$ is the area vector.For the loop $A, B, C, D, E, F

Vedclass · Accepted Answer

$\sqrt{2} abI$,along $\left(\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}\right)$

Vedclass · Answer

(A) The magnetic moment $\vec{M}$ of a current loop is given by $\vec{M} = I\vec{A}$,where $\vec{A}$ is the area vector.For the loop $A, B, C, D, E, F, A$,we can consider it as two loops $ABCD$ and $DEFA$ joined together.The magnetic moment of loop $ABCD$ (in the $XY$-plane) is $\vec{M}_1 = I(ab)\hat{k} = abI\hat{k}$.The magnetic moment of loop $DEFA$ (in the $XZ$-plane) is $\vec{M}_2 = I(ab)\hat{j} = abI\hat{j}$.The total magnetic moment is $\vec{M} = \vec{M}_1 + \vec{M}_2 = abI(\hat{j} + \hat{k})$.The magnitude is $|\vec{M}| = abI\sqrt{1^2 + 1^2} = \sqrt{2}abI$.The direction is given by the unit vector $\frac{\vec{M}}{|\vec{M}|} = \frac{abI(\hat{j} + \hat{k})}{\sqrt{2}abI} = \left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$.Thus,the magnetic moment is $\sqrt{2}abI$ along $\left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$.

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