A) \[\frac{qvR}{2}\]
B) \[qv{{R}^{2}}\]
C) \[\frac{qv{{R}^{2}}}{2}\]
D) \[qvR\]
Correct Answer: A
Solution :
As revolving charge is equivalent to a current, so \[I=q\,f\,=q\times \frac{\omega }{2\pi }\] But \[\omega =\frac{v}{R}\] where R is radius of circle and v is uniform speed of charged particle. Therefore, \[I=\frac{qv}{2\pi \,R}\] Now, magnetic moment associated with charged particle is given by \[\mu =I\,A=I\times \pi {{R}^{2}}\] \[or\mu =\frac{qv}{2\pi R}\times \pi {{R}^{2}}\] \[=\frac{1}{2}qvR\]
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