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 :

[b] \[\in =\frac{d\phi }{dt}=\frac{d(BA)}{lt}\]

\[=\frac{d(Bl)}{dt}\]

\[=\frac{Bdl}{dt}=BVl\]

\[F=ilB=\left( \frac{BV}{R} \right)({{l}^{2}}B)=\frac{{{B}^{2}}{{l}^{2}}V}{R}\]

At equilibrium

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

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