Question

A straight steel wire of length $l$ has magnetic moment $m$. If the wire is bent in the form of a semicircle, the new value of the magnetic dipole moment is _______

• A. $m$
• B. $\frac{m}{2}$
• C. $\frac{m}{\pi }$
• D. $\frac{2m}{\pi }$

Answer 1

The correct option is D $\frac{2m}{\pi }$

Length $l$, long wire is twisted in the state of half circle of span $r$, at that point, length of the wire is calculated as follows:

Circuit of the half circle (i.e., $l=\pi r$)

$⟹r=\frac{l}{\pi }$

Now, attractive dipole=attractive dipole minute$\times$least detachment between two dipoles

Least division between the dipoles is=$2r=\frac{2l}{\pi }$

$\therefore$Attractive dipole=$M\times \frac{2l}{\pi }=\frac{2m}{\pi }$