I am fascinated with magnets, and specifically the idea of magnetic levitation. I recently purchased a Levitron toy maglev train and have enjoyed finding the perfect sweet spot at which it levitates, and I understand that the Levitron does not violate Earnshaw's theorem because it is in rotating motion.

I would like to build a toy maglev train that makes no contact with the track, or any guide rails, using only permanent magnets and their repulsive force. I understand that if I attempt to build a flat track with no guide rails, the magnets will not be able to stably levitate, and will shunt the train car to one side or the other. However, what if I built a trough, or a V-shaped track, with the train car sitting inside the track, and a corresponding v-shaped bottom, with repulsive magnets running the length of the track, inside the V? I realize that it wouldn't be stable while sitting still, but if the train were in motion, would that enable it to levitate stably? In my mind there should be a sweet spot where the train's motion overcomes any sideways shunting effects as it travels above the magnetic fields of the magnets in the track, while gravity's pull on the train car into the V would be negated by the repulsive force of the magnets. I haven't been able to find anyone who has tried this approach. Every toy maglev train that I've seen has either used guide rails or a superconductor. My gut tells me it's not possible, but I want to believe. :)

Is a configuration like this possible? Is it possible to overcome the instability through one-directional motion?


2 Answers 2


What you want to investigate is called a Halbach array. You will still need a guide rail for the train, but it will pick up off the track at very low speeds.

See http://en.wikipedia.org/wiki/Inductrack for an example.

  • $\begingroup$ I'm familiar with the concept of a halbach array. I've read about the Inductrac maglev proof of concept and think it would be a wonderful and lower-maintenance alternative to the superconducting maglev trains in current use. However, would I be able to translate that down to a small enough scale to employ it on a toy train? I understand that coils respond to a halbach by generating their own magnetic field as they pass over the array. How would I do this on such a small scale? $\endgroup$ Aug 15, 2013 at 0:21
  • $\begingroup$ From a practical standpoint, it depends on how small, but yes. The Halbach array is just an arrangement of magnets that focuses the field on one side. I believe you can find pre-packaged kits on the web to make small arrays. $\endgroup$
    – cassius
    Aug 29, 2013 at 4:45
  • $\begingroup$ Maybe you could have magnets on the guide rails repelling magnets on the train as well. $\endgroup$
    – Jitter
    Nov 13, 2013 at 3:40

Perhaps a U-shaped track, or even a W shaped track, using properly calibrated magnets, would get the effect you want? Then there would be upward force and an equilibrium forces towards the middle? In order to totally "lock" the train in, as in a superconductor, you would also need a force coming down, which I'm not sure is possible without guid rails. I would probably expect the train to constantly fall out of the track due to upwards force.

  • $\begingroup$ Wouldn't gravity count as the downward force? The idea of having a v-shaped track or a trough was specifically to stop the magnets from forcing the car upward and out of the track. What I'm trying to find out is if there is a spot of adequate balance between the upward force of the magnets and the downward force of gravity (similar to a Levitron), with the side to side shunting motion kept in check by the forward momentum. $\endgroup$ Aug 15, 2013 at 0:19
  • $\begingroup$ @DustinLovell You would just have to build it very precisely, which is pretty hard with permanent magnets to be honest. In fact, now that I think about it, a u shaped or w shaped track is almost the same thing as having guard rails, you've just effectively made them one with the track. Unfortunately, you're never going to get a "quantum locking" effect without a superconductor. $\endgroup$
    – Mike Flynn
    Aug 15, 2013 at 1:43

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