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Question

(C) Let the length of the wire be $L$. The magnetic moment of a straight wire is $M = m \cdot L$,where $m$ is the pole strength.When the wire is bent

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the length of the wire be $L$. The magnetic moment of a straight wire is $M = m \cdot L$,where $m$ is the pole strength.When the wire is bent into an arc of angle $	heta = 90^o = \frac{\pi}{2}$ radians,the arc length $L = R	heta = R(\frac{\pi}{2})$,so the radius $R = \frac{2L}{\pi}$.The new magnetic moment $M'$ is given by $M' = m \cdot d$,where $d$ is the straight-line distance (chord length) between the two ends of the arc.The chord length $d = 2R \sin(\frac{	heta}{2}) = 2R \sin(45^o) = 2R \cdot \frac{1}{\sqrt{2}} = R\sqrt{2}$.Substituting $R = \frac{2L}{\pi}$,we get $d = \frac{2L}{\pi} \cdot \sqrt{2} = \frac{2\sqrt{2}L}{\pi}$.Thus,$M' = m \cdot \frac{2\sqrt{2}L}{\pi} = \frac{2\sqrt{2}}{\pi} (mL) = \frac{2\sqrt{2}M}{\pi}$.

$A$ wire of magnetic moment $M$ is bent to form a circular arc which subtends an angle $90^o$ at the center. Then,its new magnetic moment is:

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