An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is :

The correct answer is $\frac{M}{2}$.

Let the original straight iron bar have length $L$ and magnetic pole strength $p$. Its magnetic moment is defined as the product of the pole strength and the separation of the poles:

$M=pL$

When the bar is bent at its midpoint, each arm has length $\frac{L}{2}$. The pole strength of each arm remains $p$, so the magnetic moment associated with each arm is

$M$_arm=p⁢L2=M2

The two arms make an angle of 60° with each other. Their magnetic‑moment vectors point along the directions of the arms. The components perpendicular to the line joining the two outer poles are equal in magnitude and opposite in direction, so they cancel each other out. Only the components along the line joining the outer poles add together.

Since the perpendicular components cancel, the net magnetic moment of the bent bar is simply the magnitude of one arm’s moment:

$M$_net=M2

Thus the magnetic moment of the new V‑shaped magnet is reduced to one‑half of the original moment, i.e., $\frac{M}{2}$.