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A magnet of length L and moment M is cut into two halves (A and B) perpendicular to tis 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

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$\frac{M}{π}$

$\frac{M (2 + π)}{2 π}$

$\frac{M}{2 π}$

$\frac{M π}{2 + π}$

answer is C.

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Detailed Solution

$M^{1} = \frac{2 (\frac{M}{2}) \sin 90^{0}}{π} + \frac{M}{2} = \frac{M}{π} + \frac{M}{2}$

$M^{1} = M [\frac{2 + π}{2 π}]$

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