question_answer

A magnetised wire of moment M is bent into an arc of a circle subtending an angle of \[60{}^\circ \] at the centre; then the new magnetic moment is

A) \[\,\frac{2\,M}{\pi }\]                 

B) \[\,\frac{\,M}{\pi }\]

C) \[\frac{3\sqrt{3}\,M}{\pi }\]                    

D) \[\frac{3\,M}{\pi }\]

Correct Answer: D

Solution :

\[r\theta =l\] \[r=\frac{l}{\theta }\] From figure \[\sin \,\frac{\theta }{2}=\frac{x}{r}\,\,\,\,\,\Rightarrow \,\,\,x=r\,\sin \,\,\frac{\theta }{2}\] Hence new magnetic moment \[M'\,\,=\,\,m\left( 2x \right)\,\,=\,\,m.2r\,sin\,\frac{\theta }{2}\] \[=\,\,m.\frac{2l}{\theta }\sin \,\frac{\theta }{2}=\frac{2ml\sin \theta /2}{\theta }\] \[=\,\,\frac{2M\,\sin \,\,(\pi /6)}{\pi /3}\,\,=\,\,\frac{3\,M}{\pi }\]