Question

A magnetic wire of dipole moment $4\pi {\mathrm{Am}}^2$ is bent in the form of a semi-circle. What is the new magnetic moment is?

• A. ${\mathrm{\pi Am}}^2$
• B. $2{\mathrm{\pi Am}}^2$
• C. $8A{m}^2$
• D. $4{\mathrm{\pi Am}}^2$

Answer 1

The correct option is C

$8A{m}^2$

Step 1: Given Data

The dipole moment of wire is $M=4{\mathrm{\pi Am}}^2$.

r is the radius of the semicircle.

Step 2: Formula Used

1. $M=mL$, where m is the magnitude of charge and L is the distance between positive and negative charges.
2. Distance between poles is $L=\mathrm{\pi r}\Rightarrow r=\frac{L}{\mathrm{\pi }}$.

Step 3: Solution

Now as the wire is in semicircle, the new distance between the poles is $L\text{'}=2r=2\frac{L}{\mathrm{\pi }}$

And, the new dipole moment is $M\text{'}=mL\text{'}$.

Substitute the values of M and L in the equation $M=mL$:

$$\begin{gathered} 4{\mathrm{\pi Am}}^2=\mathrm{mL} \\ \Rightarrow \mathrm{m}=\frac{4{\mathrm{\pi Am}}^2}{\mathrm{L}} \end{gathered}$$

Substitute the values of m and L' in M'.

$$\begin{gathered} M\text{'}=\frac{4{\mathrm{\pi Am}}^2}{\mathrm{L}}\times 2\frac{L}{\mathrm{\pi }} \\ =8{\mathrm{Am}}^2 \end{gathered}$$

Hence, the dipole moment is $8A{m}^2$.