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