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 ?


1 Answer 1


Initially, due to acceleration of the magnet the rate of changing of linked flux ($-d\phi /dt$ is itself changing and hence the induced emf, and the force due to it, is increasing (changing). Ultimately at a sufficiently high velocity, the force due to induced current will be equal to the weight. When $F=mg$, then the net force and hence acceleration is zero, but the flux is still changing if it has a non zero velocity. After this, since velocity is constant and no acceleration exists, the induced emf will be constant because $-d\phi /dt$ is constant due to constant velocity. This means the forces will no longer vary and the magnet will continue with same speed.
Your analysis assumed the force due to induced emf to still change after $F=mg$ but that is incorrect as constant velocity will cause constant induced force, in the perfectly long tube's case.


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