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Imagine a permanent magnet at rest in empty space. A proton initially travels along a straight path with a velocity $\vec{v}$ and enters the field of the magnet with $\vec{v}$ perpendicular to $\vec{B}$. The lorentz force deflects the proton, resulting in a circular motion. The lorentz force doesn't do work on the proton and the kinetic energy of the proton doesn't change. So far, so good.

It seems to me that the angular momentum $\vec{L}$ of the proton changes from zero to non-zero as it enters the $\vec{B}$-field. Without an external torque, the total angular momentum of the system 'proton plus magnet' is conserved.

a. Will the magnet start spinning because it acquires an angular momentum $-\vec{L}$?

b. If the magnet starts spinning, where does it get it's rotational energy from?

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  • $\begingroup$ Angular momentum with respect to what axis? Are you aware that angular momentum can come in the form of something moving on a linear path that does not pass through the center of the axis from which you define angular momentum? I suspect these considerations resolve the question. In the case that the proton gains angular momentum by being deflected, the magnet doesn't spin, it too is deflected. $\endgroup$
    – AXensen
    Commented May 23, 2023 at 23:07
  • $\begingroup$ Angular momentum with respect to an axis through the centre of the particle's circular path and perpendicular to the plane of the circular path. I apologise for not being specific. $\endgroup$
    – gamma1954
    Commented May 23, 2023 at 23:13

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"It seems to me that the angular momentum 𝐿⃗ of the proton changes from zero to non-zero as it enters the 𝐵⃗-field."

I don't agree. The angular momentum of a particle, P, about a point, O, is defined as $\vec L=\vec{OP}\times \vec p$, in which $\vec p=m \vec v$. So the magnitude of the angular momentum about O of a particle P moving in a straight line is $mv$ times the length of the perpendicular dropped from O on to that straight line (extended if necessary).

If we choose O to be the centre of the particle's circular path once it has entered the field (which has sharp edges, let us assume), $\vec L$ will stay the same, since the radius of the circle will be $r=|\vec{OP}|$ and the radius vector is always perpendicular to $m\vec v$. [About other points in the plane of the particle's motion it won't.]

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  • $\begingroup$ I wonder if the downvoter had the same reservation about this answer as I do: namely that I don't have anything to say about non-conservation of the proton's angular momentum about points in the plane of motion other than the circle centre. Any thoughts on the matter would be appreciated. $\endgroup$
    – Philip Wood
    Commented May 24, 2023 at 16:15
  • $\begingroup$ At this moment, I don't see that Philip Wood's answer has been downvoted. I upvoted and accepted his answer. Thank you, Philip. $\endgroup$
    – gamma1954
    Commented May 24, 2023 at 16:58
  • $\begingroup$ Thank you, gamma 1954. I'm pleased that the answer was helpful. It had received (by 18.10 British Summer time on 24th May 2023) 2 upvotes and a downvote. As I implied in my comment, I think that there's a loose end that needs attending to. $\endgroup$
    – Philip Wood
    Commented May 24, 2023 at 17:10
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The Lorentz force can be understood if one considers that the deflection of the charged particle is accompanied by a photon emission. The best example is the free-electron laser, where an intense laser light is generated between alternating magnetic poles. The electrons decelerate in the process.

The same happens in your example. Your proton ejects photons, is deflected and loses kinetic energy in the process. Its path is not a circle but a spiral, in the centre of which it comes to a standstill after exhausting all its kinetic energy.

On the question of the spinning magnet. The magnet interacts exclusively with the magnetic field of the proton. Calculably, the proton's magnetic field is aligned with that of the magnet. Although this also applies to the magnetic dipoles of the subatomic particles of the magnet, in practice it is only detectable for SQUID measuring devices and is also used in this way.

Short answer: No, the magnet as a whole does not rotate, not even theoretically.

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  • $\begingroup$ Consider a situation where a proton is deflected, but not captured by a localized magnetic field. The magnet exerts a force on the proton, and the proton exerts an equal and opposite force on the magnet. From the perspective of the magnet, this can be seen as the magnetic field produced by the moving point charge applying a force to induced surface current - and idk where you get this fact that a proton's magnetic field is generically parallel to any magnet it passes through. Note I didn't need larmor radiation at all there. $\endgroup$
    – AXensen
    Commented May 26, 2023 at 10:43

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