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Imagine two axially-magnetized ring-shaped permanent magnets in a repulsive configuration where one is levitating on top of the other (and somehow radially constrained so the PMs remain ~concentric). Once we start rotating the top PM, what is the coefficient of friction between the two magnets?

I imagine it's a non-zero number -- if so, what is the source of friction?

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  • $\begingroup$ I think ideally it is zero. Since the magnetic field is symmetrical about the axis of rotation there should be no eddy currents. $\endgroup$
    – Drew
    Jun 2, 2022 at 5:52
  • $\begingroup$ I think the two magnetic fields would also affect one another and there would be magnetic drag when there is relative movement, such as rotation, between the two. $\endgroup$
    – Fred
    Jun 2, 2022 at 7:40
  • $\begingroup$ @Fred can you elaborate on that? $\endgroup$
    – MehranJ
    Jun 2, 2022 at 15:11

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Air drag and possible eddy currents due conductivity and less than perfect symmetry are already said, but there's one more: The magnets are made of particles and are not 100% uniform. There repulsive force varies slightly during the rotation. At a moment the upper ring climbs and soon after that it falls. You may think that those microscopic rolls uphill and downhill compensate each other with no loss and the net effect to the torque is zero.

But it is not. The slight vibration causes slight electromagnetic radiation, because there's field changes which have non-zero 2nd degree time derivative (=accelerating changes). The effect is not like friction between solid surfaces, it's more like the mentioned air drag - depends on the rotation speed.

A physicist is needed to calculate the amount and the structure of the caused radiation in a practical system. Unfortunately I am not one. I guess the effect is not measurable when the rotation speed is any practically possible.

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The air between the surfaces will provide the friction.

Also known as drag. As air friction is significant for airplanes, it is also significant for cars and trucks - check out aerodynamic force or drag..

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  • $\begingroup$ Assume it's in a vacuum. What is the coefficient of friction between the magnets? $\endgroup$
    – MehranJ
    Jun 2, 2022 at 15:10
  • $\begingroup$ @MehranJ on Mars or Venus or Neptune? Not interested if you keep changing the goalposts. $\endgroup$
    – Solar Mike
    Jun 2, 2022 at 15:20
  • $\begingroup$ I asked for "coefficient of friction", not "source of kinetic energy loss if we did this on my tabletop". Doesn't matter where it happens, because I want to know this one isolated variable, not a systems analysis $\endgroup$
    – MehranJ
    Jun 2, 2022 at 15:25

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