[Physics] Spinning a bar magnet above a disc

electromagnetismhomework-and-exercisesinductionmagnetic fieldstorque

I'm preparing for a physics exam and this scenario was in one of the past papers:

If I have a horizontal bar magnet connected to a motor, and directly underneath there is a horizontal copper disc that is free to spin, when I turn the motor on and the magnet begins to rotate above the disc, which direction will the disc rotate?

I initially thought that as the magnet moves, eddy currents will be induced directly under where the magnet poles just moved away from, to make a magnetic field in the same direction as the pole that just moved, by Lenz's law. So induced N and S poles always lagging behind the moving N and S poles of the magnet, so the disc will always be repelled and hence rotate opposite to the magnet.

However, the answers say that the disc will rotate in the same direction as the magnet. The explanation is that "interaction of the magnetic fields produces a drag torque that tries to rotate the disc in the same direction as the spinning magnet" but this doesn't really explain why.

Could someone please help explain why the disc will chase the magnet in the same direction? What is wrong with my reasoning- how should I be properly approaching these induction questions? Thank you very much.

Best Answer

This experiment is called Arago's Disc, and is essentially no different from the one in which you drop a magnet into a copper tube and - surprisingly - it take much longer than expected to fall.

In both cases the explanation is that the changing magnetic field induces eddy currents in the copper disc, in a manner which opposes the change in flux, according to Lenz' Law. These currents produce magnetic fields in directions which opposes the change in the magnetic field from the magnet. Induced currents ahead of the magnet push against the magnet, while induced current behind the magnet pull on it. (I find it difficult to explain exactly how the opposing magnetic fields create forces between the magnet and the copper disc. All the websites I've read skip over the issue as I am doing here. I shall try to rectify this later.) Like mechanical friction, this magnetic friction opposes relative motion between the magnet and disc.

Lenz' Law is related to le Chatelier's Principle in chemistry, which can be stated as a much broader principle :

When a system is in dynamic equilibrium, any change in the status quo prompts an opposing reaction from the system.


Your reasoning is mostly correct, except that the induced poles to not lag, and are the same - not opposite - in polarity.

At the points immediately under the magnet, the magnetic field is approximately constant and parallel to the magnet and the disc. There is no change in magnetic flux through the disc in this region as the magnet moves, hence no eddy currents are induced and no force arises. It is the regions ahead and behind the magnet in which the magnetic flux has a component perpendicular to the disc. This component is changing as the magnet moves. That change in flux is what induces the eddy currents.

So a magnetic N pole is induced in the disc ahead of the magnet by the magnet's advancing N pole, and another magnetic N pole is induced in the disc behind the magnet by the magnet's receding S pole. In both cases the force between the magnetic and induced poles opposes the relative motion between the magnet and the disc.

Note that a S pole is not induced behind the receding S pole of the magnet. If this were so, the repulsions from ahead and behind would cancel out. I guess that is why you introduced the idea of a "lag" to explain why the two repulsions could add up to a net force on the disc. If in your explanation the induced N and S poles were directly under the like magnetic poles, the only force (repulsion) would be perpendicular to the disc. There would be no force parallel to it, either forward or backward.