A straight steel wire of length ell has a mag | vedclass

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(C) The initial magnetic moment of the straight wire is $M = m \ell$,where $m$ is the pole strength of the wire.When the wire is bent into an $L$-shap

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$\frac{\sqrt{5} M}{3}$

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(C) The initial magnetic moment of the straight wire is $M = m \ell$,where $m$ is the pole strength of the wire.When the wire is bent into an $L$-shape with segments of length $\frac{\ell}{3}$ and $\frac{2\ell}{3}$ at an angle of $90^{\circ}$,the new magnetic moment $M^{\prime}$ is given by $M^{\prime} = m \ell^{\prime}$,where $\ell^{\prime}$ is the effective distance between the two poles.The effective length $\ell^{\prime}$ is the hypotenuse of the right-angled triangle formed by the two segments:$\ell^{\prime} = \sqrt{\left(\frac{\ell}{3}\right)^{2} + \left(\frac{2\ell}{3}\right)^{2}} = \sqrt{\frac{\ell^{2}}{9} + \frac{4\ell^{2}}{9}} = \sqrt{\frac{5\ell^{2}}{9}} = \frac{\sqrt{5}\ell}{3}$.Therefore,the new magnetic moment is $M^{\prime} = m \ell^{\prime} = m \left(\frac{\sqrt{5}\ell}{3}\right) = \frac{\sqrt{5}}{3} (m \ell) = \frac{\sqrt{5} M}{3}$.

$A$ straight steel wire of length $\ell$ has a magnetic moment $M$. It is bent into an $L$-shape as shown in the figure. The new magnetic moment is

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