A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some the the magnet :
(a) will stop in the tube.
(b) will move with almost constant speed.
(c) will move with an acceleration g.
(d) will oscillate.
Let the magnet be dropped with its north pole downward. We can consider long tube to be made of many circular loops joined together. At an instant $t=0$, the force mg acts on the magnet. $m$ is its mass. As it moves, by Lenz's law, each circular loop in the tube opposes its downward motion and hence upward force will act on it by many such loops. A time will come when the magnet experiences zero net force. Even eddy currents are produced in the magnet. The net force is $F-mg$ where at some instant will be nil, i.e. $F=mg$ as I already stated. Even if it has constant velocity at this instant magnetic flux by it linked with those coils is still varied and hence upward force still acts. Hence after this $F-mg<0$, and magnet accelerated upwards. Now again by Lenz' law coils oppose upward motion of magnet and this process continues till then again it move down.. and so on it just keep on oscillating...
So answer should be (d). But its wrong !! How ? Correct answer given in (b). Can anyone point out the flaw in my attempt ?