A magnetised wire of magnetic moment 3.14 , A | vedclass

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(B) Let the original length of the wire be $L$ and the pole strength be $m$. The original magnetic moment is $M = m 	imes L = 3.14 \, A-m^2$.When the

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(B) Let the original length of the wire be $L$ and the pole strength be $m$. The original magnetic moment is $M = m 	imes L = 3.14 \, A-m^2$.When the wire is bent into a semi-circle of radius $R$,the length of the wire becomes the circumference of the semi-circle: $L = \pi R$,so $R = L / \pi$.The new magnetic moment $M'$ is the product of the pole strength and the distance between the poles (the diameter of the semi-circle).The distance between the poles is $2R = 2(L / \pi) = 2L / \pi$.Therefore,$M' = m 	imes (2L / \pi) = (m 	imes L) 	imes (2 / \pi)$.Substituting $M = 3.14$ and $\pi \approx 3.14$:$M' = 3.14 	imes (2 / 3.14) = 2 \, A-m^2$.

$A$ magnetised wire of magnetic moment $3.14 \, A-m^2$ is bent in the form of a semi-circle. The new magnetic moment will be.....$A-m^2$

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