A) \[\frac{qvR}{2}\]
B) \[qv{{R}^{2}}\]
C) \[\frac{qv{{R}^{2}}}{2}\]
D) \[qvR\]
Correct Answer: C
Solution :
| [c] 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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