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A magnet of length L and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radius R and is joined to the other piece at the poles as shown in the figure below.
Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting magnet is given by $\frac{M (x + π)}{x π}$ then ‘x’ is

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answer is 2.

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Detailed Solution

Magnetic moment of each part is $\frac{M}{2}$ for semi circular piece A

$M_{A} = \frac{2 (M / 2) \sin \frac{180^{\circ}}{2}}{π} = \frac{M \sin 90^{\circ}}{π} = \frac{M}{π}$

$M_{C} = M_{A} + M_{B} = \frac{M}{π} + \frac{M}{2}$

$= M [\frac{2 + π}{2 π}]$ $∴ x = 2$

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