A copper connector of mass `m` slides down two smooth cooper bars, set at an angle `alpha` to the horizontal due to gravity (Fig). At the top the bars is equal to `l`. The system is located in a unifrom magnetic field of induction `B`, perpendicular to the plane in which the connector slides. The resistances of the bars, the connector and the sliding contacts, as well as the self-inductance of the loop, are assumed to be negligible. Find the steady-state velocity of the connector.

Explanation

From Lenz's law, the current through the connector is directed from `A` to `B`. Here `xi_(I n) = vBl` between `A` and `B`
where `v` is the velocity of the rod at any moment.
For steady state, acceleration of the rod must be equal to zero.
Hence, `mg sin alpha - i l B` ....(1)
But, `i = (xi_(i n))/(R) = (v B l)/(R)`
From (1) and (2) `v = (mg sin alpha R)/(B^(2) l^(2))`