An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment will be

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\frac{2 M}{π}

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\frac{M}{π}

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M π

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Solution

The correct option is B $\frac{2 M}{π}$
On bending a rod it’s pole strength remains unchanged where as its magnetic moment changes.
New magnetic moment $M^{'} = m (2 R) = m (\frac{2 L}{π}) = \frac{2 M}{π}$

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