A) \[\left( \frac{M{{v}^{2}}}{R} \right)2\pi R\]
B) zero
C) \[BQ.\,2\pi R\]
D) \[BQv\,.\,\,2\pi R\]
Correct Answer: B
Solution :
When particle describes circular path in a magnetic field, its velocity is always perpendicular to the magnetic force. Power \[P=\overset{\to }{\mathop{\mathbf{F}}}\,\cdot \overset{\to }{\mathop{\mathbf{v}}}\,=Fv\cos \theta \] Here, \[\theta ={{90}^{o}}\] \[P=0\] But \[P=\frac{W}{t}\] \[\Rightarrow \] \[W=P\cdot t\] Hence, work done \[W=0\] (everywhere)You need to login to perform this action.
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