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