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 \[=\frac{2}{\pi }M\]You need to login to perform this action.
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