A copper rod of mass $m$ slides under gravity on two smooth parallel rails $l$ distance apart and set an angle $\theta$ to the horizontal. At the bottom, the rails are joined by a resistance $R$ in figure. There is a uniform magnetic field $B$ perpendicular to the plane of the rails. The terminal velocity of rod is

• A

  $\frac{mgR\tan \theta }{B^2{l}^2}$

• B

  $\frac{mgR\cot \theta }{B^2{l}^2}$

• Correct answer
  C

  $\frac{mgR\sin \theta }{B^2{l}^2}$
• D

  $\frac{mgR\cos \theta }{B^2{l}^2}$

Hence, the correct answer is option: 'C'.