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If a charged particle is deflected in a magnetic field due to some magnet, is there any experiment that shows (or measures) the opposite reaction force (due to Newton's third law) on the magnet?

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3 Answers 3

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A moving charge interacts with the external magnet via its magnetic dipole. To understand this, some basic understandings are required:

  1. the electron (as a representative of charges) has an intrinsic (permanently present) magnetic dipole.
  2. the magnetic field of a magnet does not weaken with time ( a permanent magnet can be used as a magnet). The magnet does not give energy permanently to the charge, whose path of motion nevertheless curves.
  3. the energy necessary for the deflection of the particle comes from the movement of the part. As the duration of curvature increases, the charge slows down. Its trajectory of motion is a spiral and the charge came to standstill in its center.
  4. the deflection is associated with an observable emission of EM radiation.

The magnetic field is something like a catalyst in chemistry; the field influences the process but is not consumed in the process. How does this work?

The key lies in the temporary alignment of the magnetic dipole of the charge to this external magnetic field together with a simultaneous emission of electromagnetic radiation.
As is known, EM radiation has no mass, but it has a moment. First, this moment of the emitted radiation is exactly the reason of the lateral deflection. Secondly, the alignment of the dipole to the external field is disturbed again. Because of 1 the charge slows down, because of 2 the cycle can start again.

Since the EM emission is the reason for the deflection of the charge, there is no opposing "opposite reaction force (due to Newton's third law) on the magnet". And - to emphasize it again - also no weakening of the external magnetic field.

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  • $\begingroup$ Thank you for your answer! Do you have references where to read more about this, maybe how to caculate the loss of velocity / emitted energy? $\endgroup$
    – StefanH
    Commented Jan 9, 2022 at 16:55
  • $\begingroup$ An example: 14 kW output. Some calculation. $\endgroup$ Commented Jan 9, 2022 at 21:35
  • $\begingroup$ Ok. Thank you. I have the feeling that all answers differ somewhat, and yours seem to be the "deepest". $\endgroup$
    – StefanH
    Commented Jan 9, 2022 at 21:49
  • $\begingroup$ StefanH, keine Ursache. But, why did you ask this question? $\endgroup$ Commented Jan 10, 2022 at 4:20
  • $\begingroup$ I am a computer scientist who wants to understand the physics behind quantum computing. I came across this question while learning about the interaction of particles, but I know this is probably quite basic :D $\endgroup$
    – StefanH
    Commented Jan 10, 2022 at 18:14
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If you are talking about a laboratory macroscopic magnet as the source of the magnetic field, then for a single charged particle, probably not. The reaction force is microscopic. To detect the force of a single charged particle would be massively difficult. I won't say it's impossible. It might be possible using something like a superconductor magnet such that the entire magnet can be treated as a single quantum state. Then the single charged particle might produce a detectable change in the magnet. But it would be a quite challenging experiment.

For a stream of charged particles, it's quite easy. When an electric motor operates, the force the rotating part of the motor produces is exactly balanced by the force the rest of the motor produces. So when you use the shaft of an electric motor to turn something, the rest of the motor must be supported to keep it from turning the opposite direction.

For a single particle in the magnetic field of a microscopic magnet, the situation is very complicated. It is certainly possible to detect recoil of particles undergoing collision with other particles. However, this situation is quite complicated due to the interaction involving electric and probably nuclear forces. If a proton (the charged particle) scatters off a nucleus (the source of the magnetic field) there will be a recoil of the nucleus. And such recoil can be detected and measured. However, the scattering is not going to be purely a charge moving through a magnetic field. There will be electric and nuclear force effects. And the magnetic field of a nucleus is not a simple thing as you might hope for in small regions of an electric motor. So resolving the part due to the magnetic field is a challenge.

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  • $\begingroup$ For an electric motor, it think it is more due to the fact that two magnets interact with each other, and not so much the individual forces on the moving charges.... and I do not understand your last paragraph, why should the proton scatter of the nucleus? Why do they have to collide in the first place? $\endgroup$
    – StefanH
    Commented Jan 6, 2022 at 17:44
  • $\begingroup$ @StefanH "Two magnets interact" is exactly what the reaction force consists of. $\endgroup$
    – Dan
    Commented Jan 6, 2022 at 19:57
  • $\begingroup$ I am not sure... moving charges create a magnetic field by Oersted's law which interacts with the given magnetic field. I do not see the connection to the Lorentz force? Are you suggesting Oersted's law is linked to the Lorentz force and its reaction force? $\endgroup$
    – StefanH
    Commented Jan 6, 2022 at 21:33
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The reaction force is not on the magnet (some distance away) but on the electromagnetic field right at the location of the charged particle. If the charged particle gains momentum $\Delta {\bf p}$ then the local electromagnetic field loses exactly that amount of momentum. The rate of this momentum change is the reaction force.

Now if the electromagnetic field is static overall, then it must be giving up this same momentum to something else. What happens is that the momentum is transported through the field and eventually goes to whatever is acting as boundary to the field. The boundary could, for example, be further charged particles or magnetic dipoles in a solid material such as a permanent magnetic or an electromagnetic. So in this case the reaction force does eventually push on the magnet producing the field. This force is experienced whenever an electric generator is used to provide an electric current, for example. The current is having work done on it, and the generator contains things like rotating magnets. For a large generator a considerable force has to be applied to those magnets to make them go around! This is happening in most of the power stations of the world. The applied force is countering the reaction force asked about in the question. It is obtained from high-pressure steam in a gas- or coal-fired power station, or from wind pressure on a turbine, or from water pressure on a turbine, etc. It is easily measurable.

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  • $\begingroup$ Is the working principle of an electric motor not due to the magnetic field induced by the moving charges, and the resulting magnet interacts with the given magnet to produce motion? I mean, that is a different force than the force given by the Lorentz force, which does not requires the magnetic field around the moving charge? $\endgroup$
    – StefanH
    Commented Jan 6, 2022 at 17:48
  • $\begingroup$ @StefanH The interaction between magnetic field and charge is always the Lorentz force. An electric generator works by producing such a force on the charges inside a conducting wire. I don't quite follow the wording of your comment; the first part of it appears to describe a motor. $\endgroup$ Commented Jan 6, 2022 at 18:08
  • $\begingroup$ The explanations I have seen say that the current in the wire induces a magnetic field which interacts with the given magnetic field in such a way to induce motion. Hence it only uses Oersted's law, not the Lorentz force, right? $\endgroup$
    – StefanH
    Commented Jan 6, 2022 at 18:18

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