A copper rod of mass $m$ slides under gravity on two smooth parallel rails, with separation $l$ 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. The terminal speed of the copper rod is:
Answer: (B) $\dfrac{mgR\sin\theta}{B^2l^2}$
At speed $v$ the induced emf is $Blv$ and the current is $I = \dfrac{Blv}{R}$. Since $\vec B$ is normal to the rails, the magnetic force $IlB = \dfrac{B^2l^2v}{R}$ acts along the rails, up the incline.
At terminal speed it balances the component of gravity along the rails:
$$\frac{B^2l^2v}{R} = mg\sin\theta \quad\Rightarrow\quad v = \frac{mgR\sin\theta}{B^2l^2}$$
Solution by Sreeraj P, M.Sc Physics