The eddy current demonstrator can teach a lot

Eddy Currents

The Eddy Current Demonstrator, prepared by The Magnet Source, and distributed by Edmund Scientific Co., makes the retarding effect of eddy currents very evident. The demonstrator consists of a piece of 14 mm copper tubing 332 mm long with plastic end caps, a magnet in the form of a cylinder, and an iron slug of the same appearance and weight. My magnet weighed 13.0 g, the slug 13.5 g.

A freely-falling body starting from rest will require 0.26 s to fall the length of the tube. If the iron slug is dropped from the top of the tube held vertically, it chatters around inside, so it takes longer and nothing is learned. It is best just to drop it beside the tube to see what 0.26 s looks like. When you drop the magnet through the tube, it takes much longer to come through, about 4 s, which is remarkable. Don't drop the magnet outside the tube -- it is fragile.

A magnet produces a pure magnetic field in its rest frame. Anything moving with respect to the magnet sees an electric field in addition to the magnetic field, that is roughly proportional to the relative velocity. An electron free to move, as in copper, will be set into motion by the electric field it sees. In the case of a length l metre of wire moving with speed v m/s in a magnetic field of B tesla, the electromotive force is e=Blv volt, so if the resistance of the wire is R ohm, the current is i=Blv/R ampere, if the circuit is closed. This current is called the eddy current, since it flows in closed loops in a conducting plate like eddying water.

In the eddy current demonstrator, we do not have a uniform magnetic field B, nor a definite length of wire l with resistance R. Indeed, what we have is a field problem. If we knew the magnetic field of the magnet as a function of position, and the resistivity of the copper, we could calculate the current density at every point in the wall of the tube. We do know, however, that this current, whatever it is, is proportional to the speed v.

When a wire in which a current i is flowing is moved at right angles to a magnetic field B, the force exerted on a length of wire l is F=Bli. Again, this is a field problem with the eddy current demonstrator, but knowing B and i as functions of position, we could again determine the total force F exerted on the currents in the wall of the tube, which must be equal and opposite to the force exerted on the magnet when we hold the tube still. We then find the equation F=mg=B2l2v/R. We do not know what B, l, and R to use in this equation, but we can reckon that doubling B will quadruple the force, but doubling the size of the apparatus will only double the force, since both l and R are proportional to size.

If the length of the tube is L, and t is the time of fall, then t=B2l2L/Rgm. The only parameter we can easily change in the demonstration is m, by dropping the magnet and slug together. There is a danger that adding the slug will change the magnetic field, but perhaps this will not change things greatly, only spreading the field out a little. On trying this, I find that the time of fall is halved, to 2 s rather than 4 s. This is consistent with the observation that the retarding effect of eddy currents is proportional to speed, an important result.

If the poles of the cylindrical magnet are at its ends (iron filings could show this, messily; a compass shows this to be the case, in fact), the magnetic field passes through the tube walls at top and bottom in opposite directions, producing eddy currents that are essentially rings about the tube, flowing in opposite directions at top and bottom, and moving with the falling magnet.

Let's make an approximate calculation, using reasonable assumptions for the distribution of the fields and currents. Suppose the eddy currents to flow in two rings of 5 mm height and 15 mm diameter in the copper of thickness 0.5 mm, with the magnetic field B normal to the walls of the tube. In this case, l = 0.047 m, L = 0.332 m, R = 0.0000325 ohm, m = 0.013 kg, g = 9.8 m/s/s, and t = 4 s seem reasonable. Solving the equation for B, we find B = 1.06 T, or 10,600 gauss, a reasonable figure for the Neodymium-Iron_Born magnet, for which the Edmund catalog quotes 12,000 gauss. The agreement is fortuitous, but whenever this happens, the scientific method demands that experiments be discontinued lest a worse result be obtained. It is the existence of such a strong small permanent magnet that makes the demonstration so vivid, since the time of fall is proportional to the square of the field.

[Eddy Currents] In the sketch, the magnet is falling with the N pole uppermost, and the tube has been imagined rolled out flat. The magnetic field is outward in the top rectangle, and inward in the lower. v is the velocity of the copper with respect to the magnet. The electric field E = v x B and the current density J = c E, where c is the conductivity, are shown at top and bottom. The force density F = J x B is seen to be the same at top and bottom. This is the force on the copper; the force on the magnet is -2F, opposite to its velocity. Of course, the force is assumed to have been integrated over the volume of the copper. The actual field is not constant over the rectangular areas, but varies smoothly, dropping to zero at the edges. The eddy currents flow circumferentially in the copper walls of the tube. The magnetic field produced by the eddy currents is at right angles to the magnet's field, distorting it to account for the forces between the tube and magnet, which we have accounted for in another way. Something similar occurs in the armatures of motors and generators when current flows through them.

Return to Physics Index

Composed by J. B. Calvert
Last revised 29 May 2004