Question: 40. 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: { }^{\circ}
(1) M
(2) \frac{\mathrm{M}}{2}
(3) 2 M
(4) \frac{\mathrm{M}}{\sqrt{3}}
Answer: Option (2)
Explanation:
The magnetic moment of a bar magnet is given by M = mL,
where m is the pole strength and L is the length of the magnet.
When the bar is bent at the middle, it is divided into two equal parts,
each of length \frac{L}{2}.
The magnetic moment of each half is therefore \frac{M}{2}.
The two magnetic moment vectors of magnitude \frac{M}{2} make an angle of 60^{\circ} with each other.
The resultant magnetic moment is given by vector addition:
M' = \sqrt{\left(\frac{M}{2}\right)^2 + \left(\frac{M}{2}\right)^2 + 2\left(\frac{M}{2}\right)\left(\frac{M}{2}\right)\cos 60^{\circ}}.
Since \cos 60^{\circ} = \frac{1}{2},
M' = \sqrt{\frac{M^2}{4} + \frac{M^2}{4} + \frac{M^2}{4}} = \sqrt{\frac{3M^2}{4}}.
This simplifies to
M' = \frac{M\sqrt{3}}{2}.
The magnetic moment of the bent magnet is equal to the component of this resultant along the original direction, which is
M_{\text{effective}} = M' \cos 30^{\circ} = \frac{M\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{M}{2}.
Hence, the magnetic moment of the new magnet is \frac{M}{2}.