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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

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$\frac{M}{2 π} (2 + π)$

$\frac{M}{π}$

$\frac{M}{2 π}$

$\frac{Mπ}{2 + π}$

answer is A.

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Detailed Solution

$\begin{aligned} M^{1} = M_{1} + M_{2} \\ = \frac{M}{π} + \frac{M}{2} \end{aligned}$
$= \frac{M}{2 π} [2 + π]$

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