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Question

(C) Let the original length of the wire be $L$. The original magnetic moment is $M = m L$,where $m$ is the pole strength.The wire is bent into a quart

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the original length of the wire be $L$. The original magnetic moment is $M = m L$,where $m$ is the pole strength.The wire is bent into a quarter-circle arc of radius $R$. The length of this arc is $L = \frac{1}{4}(2 \pi R) = \frac{\pi R}{2}$.Thus,$m = \frac{M}{L} = \frac{M}{\pi R / 2} = \frac{2M}{\pi R}$.The new magnetic moment $M^{\prime}$ is the product of pole strength $m$ and the straight-line distance between the two ends of the arc (the chord length).The distance between the ends is $L^{\prime} = \sqrt{R^2 + R^2} = \sqrt{2} R$.Therefore,$M^{\prime} = m L^{\prime} = \left( \frac{2M}{\pi R} \right) (\sqrt{2} R) = \frac{2 \sqrt{2} M}{\pi}$.

$A$ wire of magnetic moment $M$ is moulded according to the figure. What is the new magnetic moment?

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