Question

The wire loop $\text{PQRSP}$ formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current $I$, as shown in the figure. The magnetic moment of the loop is

• A. $\pi I(R_2^2-R_1^2)$
• B. $\frac{\pi I}{2}(R_2^2-R_1^2)$
• C. $\pi I(R_2^2+R_1^2)$
• D. $\frac{\pi I}{2}(R_2^2+R_1^2)$

Answer 1

The correct option is B $\frac{\pi I}{2}(R_2^2-R_1^2)$

We know that,
$\overrightarrow{M}=I\overrightarrow{A}$

For the loop having radius $R_1$, the magnetic moment is given by,

$M_1=I\times \frac{\pi R_1^2}{2}$

The direction of $M_1$ will be normally outwards.

For the loop having radius $R_2$

$M_2=I\times \frac{\pi R_2^2}{2}$

The direction of $M_2$ will be normally inwards.

So, the net magnetic dipole moment will be

$M_{net}=M_2-M_1=\frac{I\pi R_2^2}{2}-\frac{I\pi R_1^2}{2}$

$M_{net}=\frac{I\pi }{2}(R_2^2-R_1^2)$

Hence, option $\left(B\right)$ is the correct answer.