The fate of a bar magnet released in an infinitely long copper tube A bar magnet is released under gravity inside a long copper tube along the axis of the tube. 
The answer is given (edit :) it moves with constant velocity. But I don't understand how.
My research : 
The bar magnet is falling with an acceleration of $g$. So its speed increases. Thus the flux change also increases with time. And thus ultimately, the induced emf also changes as induced emf is directly proportional to the change of magnetic flux through the loop (the copper tube consists of infinite copper loops).
Now according to Lenz's Law, the induced emf produces a magnetic flux as to oppose the change in the magnetic flux through the loop. Consider, the magnet to fall vertically downward, thus a magnetic flux is induced vertically upwards. Thus a vertical force is applied to the magnet in the upwards direction.
What I do not understand is that won't the 'induced' force oppose the acceleration due to gravity. And if it opposes, the velocity of the magnet becomes less, thus the emf and ultimately the induced force becomes less. And the gravitational force dominates again. Will the motion be oscillatory, or will it stop? If it will stop, can you provide an elaborated reason? Thank You!
 A: TL;DR:  The magnet will asymptotically approach a "terminal velocity", at which the magnetic force from the currents in the walls of the tube is exactly balanced by the force of gravity.
To see this, let's denote $F_\text{Lenz}$ as the force from the currents in the wall.  This force will obviously depend on the velocity $v$ of the magnet.  Its magnitude will increase monotonically with the speed of the magnet (a faster magnet means the flux through a loop of the pipe is changing faster, which results in a larger EMF, which results in more currents in the pipe).  This implies that we have 
$$
F_\text{tot} = ma = m \frac{dv}{dt} = mg - F_\text{Lenz}(v).
$$
Here, we have defined $v$ to be positive in the downwards direction, while positive values of $F_\text{Lenz}(v)$ correspond to a force in the upwards direction.  
To carry this further, we would need to know the precise form of $F_\text{Lenz}(v)$.  We can, however, see that the only "stable" solution for a long period of time is $v(t) = v_\text{term}$, where $v_\text{term}$ is the value of the speed which satisfies
$$
F_\text{Lenz}(v_\text{term}) = mg,
$$
i.e., the weight of the magnet is counterbalanced by the force from the currents in the pipe.  We know that such a value will exist, since we argued above that $F_\text{Lenz}$ increases with $v$;  so for some value of $v$, the Lenz force will be large enough to counter-balance gravity.  Moreover, we note that if at any time $v < v_\text{term}$, the velocity will increase (since $mg > F_\text{Lenz}$), driving it towards $v_\text{term}$.  Similarly, if at any time $v > v_\text{term}$, the velocity will decrease back towards $v_\text{term}$.  Taken all together, this means that after a long period of time, the magnet will approach some terminal velocity.
Finally, note that the magnet cannot stop in the tube:  if $v = 0$, then the Lenz force will vanish, and so the only force on the magnet at such a moment will be gravity.  It will then be accelerated downwards by gravity and no longer be at rest.

As an aside: an exercise in Zangwill's Modern Electrodynamics (2013) actually presents the calculation of this terminal velocity as a (relatively involved) exercise.  Several simplifying assumptions must to be made to solve this problem, most notably that the magnet is a perfect dipole, the walls of the pipe are much thinner than the radius of the pipe, and the self-inductance of the pipe can be ignored.  In the end, the result is that
$$
v_\text{term} \propto \frac{a^4 M g}{\mu_0^2 m^2 \sigma d}
$$
where $M$ is the mass of the magnet, $m$ is its dipole moment, $a$ is the radius of the pipe, $d$ is the thickness of the pipe's walls, and $\sigma$ is the conductivity of the pipe.  (Finding the precise proportionality factor is left as an exercise for the reader.)
