Question

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

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

Answer 1

The correct option is D $8A{m}^2$

Given:

The dipole moment of the wire is $4\pi A{m}^2$.

The Dipole moment of the wire is represented as:

$M_i=mL$

$m=\frac{4\pi }{L}$

Now, when the wire becomes a semicircle, the new magnetic moment will be:

$M_f=m{L}^{\prime }$

The new distance between two poles is:

$L^{\prime }=2r\Rightarrow 2\frac{L}{\pi }$

So, new magnetic moment will be:

$M_f=m\times \frac{2L}{\pi }$

$M_f=\frac{4\pi }{L}\times \frac{2L}{\pi }$

$=8A{m}^2$