A) \[\frac{mgR\cos \theta }{{{B}^{2}}{{l}^{2}}}\]
B) \[\frac{mgR\sin \theta }{{{B}^{2}}{{l}^{2}}}\]
C) \[\frac{mgR\tan \theta }{{{B}^{2}}{{l}^{2}}}\]
D) \[\frac{mgR\cot \theta }{{{B}^{2}}{{l}^{2}}}\]
Correct Answer: B
Solution :
\[\in =\frac{d\phi }{dt}=\frac{d(BA)}{lt}\] \[=\frac{d(Bl)}{dt}\] \[=\frac{Bdl}{dt}=BVl\] \[F=ilB=\left( \frac{BV}{R} \right)({{l}^{2}}B)=\frac{{{B}^{2}}{{l}^{2}}V}{R}\] At equilibrium \[\Rightarrow mg\,\,\sin \theta =\frac{{{B}^{2}}lV}{R}\] \[\Rightarrow V=\frac{mgR\sin \theta }{{{B}^{2}}{{l}^{2}}}\]You need to login to perform this action.
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