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(D) magnetic wire of length $l$ has a magnetic moment $M = m l$,where $m$ is the pole strength of each end.When the wire is bent into an $L$-shape,the

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

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(D) magnetic wire of length $l$ has a magnetic moment $M = m l$,where $m$ is the pole strength of each end.When the wire is bent into an $L$-shape,the two segments of length $l/2$ are perpendicular to each other.The effective length (displacement between the poles) becomes the hypotenuse of a right-angled triangle with sides $l/2$ and $l/2$.Effective length $l' = \sqrt{(l/2)^2 + (l/2)^2} = \sqrt{l^2/4 + l^2/4} = \sqrt{l^2/2} = \frac{l}{\sqrt{2}}$.The new magnetic moment $M'$ is given by $M' = m l' = m \left( \frac{l}{\sqrt{2}} \right)$.Since $M = m l$,we get $M' = \frac{M}{\sqrt{2}}$.

$A$ metal wire of length $l$ has a magnetic moment $M$. What is the new magnetic moment if it is bent into an $L$-shape?

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