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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 $60^\circ$ with each other. The magnetic moment of this new magnet is:
A
$2M$
B
$\frac{M}{\sqrt{3}}$
C
$M$
D
$\frac{M}{2}$
Detailed Solution
Initially, M = ml (m = pole strength, l = length).
When the rod is bent at the middle, each arm is $\frac{l}{2}$ and the arms make $60^\circ$ with each other.
[add image: bar bent at the middle into two arms of length l/2 at $60^\circ$]
Effective length $l_{eff} = 2 \times \frac{l}{2}\sin 30^\circ = \frac{l}{2}$
$M' = m \times \frac{l}{2} = \frac{M}{2}$