I was fascinated by this demonstration in my high school physics class. The teacher went further and dropped the same magnet through another copper pipe of the same diameter that had a slit cut along its length - the magnet dropped straight through the slit pipe without slowing down. This provided an important "counterexample" demonstrating that the induced currents were circular around the circumference of the solid pipe - breaking the circle eliminated the braking force.
That's not true. A slit will cause it to fall faster, but not as fast as with a non conductive pipe. I did this experiment a couple of years ago to answer that question. Unfortunately I can't find the exact timing data. If you do the thinking and figure out how the currents should flow due to the moving magnet, this makes sense.
I do remember the demonstration clearly, and I have done the thinking to figure out how the currents should flow, but I would be interested in your explanation of why there would still be significant braking force in a slit cylinder.
Essentially, the eddy currents have to be moving perpendicular to the direction of the magnet's travel, using the right hand rule as mentioned in other comments. At any point along the surface of the cylinder, the current moves along the circumference, creating a magnetic field that pushes upward. Currents in any other direction would spin the magnet, which doesn't happen, so the slit effectively limits the eddy currents, except for some negligible, localized loops along the length of the cylinder.
Alternatively, in numbers. I'm ignoring constant factors and only doing proportional calculations. Where you see an = sign you should think "proportional to".
Suppose we have a magnet at the origin with magnetic moment 1 in the z direction: m = [0,0,1]. The magnet is moving in vertically (in the z direction). Further we have the pipe vertically with radius 1. The problem now is to determine the current through the pipe. Since the problem is symmetrical around the z-axis we can consider just the current though the points r = [1,0,z].
The magnetic field due to the dipole is B(r) = 3r(m dot r)/r^5 - m/r^3. The flux through the pipe at r = [1,0,z] is just the x component of the magnetic field, since the pipe is vertical. Plugging in and simplifying we get Phi = 3z/sqrt(1+z^2)^5. Since the magnet is moving in the z direction, lets say z = t, to compute the change in Phi we need to compute dPhi/dz. If we put that into Wolfram Alpha we get a picture like this:
As you can see you get a strong current in one direction around the pipe at the middle of the magnets, and two weaker currents in the opposite direction above and below the magnet.
This doesn't yet say anything about the case of the slit, but it does at least show that the currents are indeed moving like I sketched for the no slit case in the picture of my other response.
Okay I see how the currents move with the slit - thank you for the visuals! I just found a video using a solid vs. slit pipe that has a good example of the timing difference - not nearly as dramatic as my memory of the demonstration, but still informative!
I was thinking that the currents are like this: http://dl.dropbox.com/u/388822/magnet%2Bpipe.png (apologies for the crappy picture). The green lines are the magnetic field lines. These lines go through the copper pipe at the top, and at the bottom in opposite directions. Since the magnet is falling, just above the point where a line penetrates, the flux is decreasing. Just below, the flux is increasing. Hence there should be two opposite currents. When there is a slit, the currents get rerouted like in the right picture. Another way to think about it is that every field line creates a circular current around it. When you connect all these circular currents you get what's in the picture (if I did it correctly).
