question_answer

If a steel wire of length \[l\] and magnetic moment M is bent into a semicircular arc, then the new magnetic moment is

A) \[M\times l\]

B)  \[\frac{M}{l}\]

C)  \[\frac{2M}{\pi }\]

D)  \[M\]

Correct Answer: C

Solution :

 When wire is bent in the form of semicircular arc then, \[l=\pi r\] \[\therefore \] The radius of semicircular arc, \[r=l/\pi \] Distance between two end points of semicircular wire \[=2r=\frac{2l}{\pi }\] \[\therefore \]Magnetic moment of semicircular wire \[=m\times 2r\] \[=m\times \frac{2l}{\pi }=\frac{2}{\pi }ml\] But ml is the magnetic moment of straight wire i.e.,           \[ml=M\] \[\therefore \]New magnetic moment \[=\frac{2}{\pi }M\]