question_answer

A copper rod of mass \[m\] slides under gravity on two smooth parallel rails, with separation 1 and set at an angle of \[\theta \] with the horizontal. At the bottom, rails are joined by a resistance\[R\]. .There is a uniform magnetic field \[B\] normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is:                                                                                                    [JEE Online 15-04-2018 (II)]

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 \[\Rightarrow mg\,\,\sin \theta =\frac{{{B}^{2}}lV}{R}\] \[\Rightarrow V=\frac{mgR\sin \theta }{{{B}^{2}}{{l}^{2}}}\]