Question

Two identical circular loops are arranged such that their centres are at the origin and the loops are in $xy$ and $yz$ planes. The loops carry currents in directions shown in the diagram. The net magnetic dipole moment of this combination is

• A. $I\pi R^2$
• B. $2I\pi R^2$
• C. $\sqrt{2}I\pi R^2$
• D. $\sqrt{3}I\pi R^2$

Answer 1

The correct option is C $\sqrt{2}I\pi R^2$

We know that, magnetic dipole moment is defined as,

$\overrightarrow{M}=I\overrightarrow{A}$

And from the diagram, we can see that loop $\left(1\right)$ is lying in $xy-$plane, so the direction of area vector will be in negative $z-$ direction (from right-hand curl rule).

$\therefore {\overrightarrow{A}}_1=\pi R^2(-\hat{k})$

Similarly, loop $\left(2\right)$ is lying in $yz-$ plane, so

${\overrightarrow{A}}_2=\pi R^2(-\hat{i})$

Therefore, magnetic dipole moment due to both loops will be

${\overrightarrow{M}}_1=I{\overrightarrow{A}}_1=I\pi R^2(-\hat{k})$

${\overrightarrow{M}}_2=I{\overrightarrow{A}}_2=I\pi R^2(-\hat{i})$

So, the net magnetic dipole moment will be

$\overrightarrow{M}={\overrightarrow{M}}_1+{\overrightarrow{M}}_2$

$\overrightarrow{M}=I\pi R^2(-\hat{i})+I\pi R^2(-\hat{k})=I\pi R^2(-\hat{i}-\hat{k})$

$|\overrightarrow{M}|=\sqrt{|{\overrightarrow{M}}_1{|}^2+|{\overrightarrow{M}}_2{|}^2}(\because {\overrightarrow{M}}_1\perp {\overrightarrow{M}}_2)$

$|\overrightarrow{M}|=\sqrt{\left(I\pi R^2{)}^2\right[(-1{)}^2+(-1{)}^2]}$

$\therefore |\overrightarrow{M}|=\sqrt{2}I\pi R^2$

Therefore, option $\left(C\right)$ is the right answer.