An iron wire of length l is magnetized have a magnetic moment m . If it is bent into a semicircle of radius r , its magnetic moment is.

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$m \frac{1}{r}$

$\frac{m}{π}$

$\frac{2 m}{π}$

answer is B.

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Detailed Solution

The magnetic moment of straight wire, $m = (q_{m}) (l)$
The magnetic moment of the semicircular wire, $m^{'} = (q_{m}) (2 r)$
$∴ \frac{m^{'}}{m} = \frac{2 r}{l} = \frac{2 r}{πr} = \frac{2}{π} ∴ m^{'} = \frac{2 m}{π}$

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