An iron rod of length L and magnetic moment M | vedclass

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(B) The magnetic moment of a rod is given by $M = m 	imes L$,where $m$ is the pole strength and $L$ is the length of the rod.When the rod is bent int

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$\frac{2M}{\pi}$

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(B) The magnetic moment of a rod is given by $M = m 	imes L$,where $m$ is the pole strength and $L$ is the length of the rod.When the rod is bent into a semicircle,the length $L$ becomes the arc length of the semicircle,so $L = \pi R$,where $R$ is the radius of the semicircle.Thus,the radius $R = \frac{L}{\pi}$.The new distance between the poles (the diameter of the semicircle) is $L' = 2R = \frac{2L}{\pi}$.The pole strength $m$ remains unchanged.Therefore,the new magnetic moment $M'$ is given by $M' = m 	imes L' = m 	imes \frac{2L}{\pi}$.Since $M = m 	imes L$,we can substitute $M$ into the equation:$M' = \frac{2M}{\pi}$.

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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