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Suppose a moving proton traverses a uniform magnetic field with constant velocity. Note that because the proton is moving, it is generating its own magnetic field.

Is the magnetic force experienced by the proton completely explained by the Lorentz force law, i.e., $\vec{F} = q\, (\vec{v} \times \vec{B})$ ?

Or, is the proton also experiencing an additional magnetic force arising from the interaction of its own magnetic field with the external field? If that is incorrect, is the Lorentz force law actually just describing the result of proton field-external field interactions, such that any and all magnetic force on the proton is explained by the Lorentz force?

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    $\begingroup$ A moving proton can’t traverse a uniform magnetic field with uniform velocity, unless the field is parallel or antiparallel to the proton’s direction of motion. Otherwise, there will be a force making it accelerate. $\endgroup$
    – G. Smith
    Commented Feb 9, 2019 at 0:39
  • $\begingroup$ I added some further info in my answer to this question: physics.stackexchange.com/questions/483973/… $\endgroup$
    – Andrew Steane
    Commented Jun 3, 2019 at 13:52

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Magnetic fields do not interact with magnetic fields; they simply superpose. Instead, magnetic fields interact with charged particles.

It is reasonable to ask whether a moving charged particle feels its own magnetic field, just as it is reasonable to ask whether a stationary (or moving) charged particle feels its own electric field.

The answer in the case of stationary or uniformly-moving charged particle is that these electric and magnetic self-forces are observed to be zero. For a point particle, its electrical and magnetic fields become infinitely large near the particle, but they can be understood as acting on the particle in a symmetric way such that there is no net force in any direction.

In the case of an accelerating charged particle, things are more complicated. Then the Lorentz force due to external fields is not the whole story. An accelerating charge radiates electromagnetic waves that carry energy, momentum, and angular momentum away to infinity. The energy in these waves must come from the kinetic energy of the particle, so there must be a backreaction force on the particle slowing it down, in order to conserve energy. This backreaction can be understood as the self-force, where the accelerating particle’s own field exerts a nonzero force on it.

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  • $\begingroup$ The idea of the diverging electric and magnetic field cancelling out seems too convenient to me. I first saw it when reading the Schwinger book (chapter 1 or 2 I think) but the argument felt rather informal. Is there any formal way to prove it or to work around it? $\endgroup$
    – user137661
    Commented Feb 9, 2019 at 2:02
  • $\begingroup$ I think you can assume a finite radius for the particle, calculate the self-force, and then take the limit as the radius goes to zero. $\endgroup$
    – G. Smith
    Commented Feb 9, 2019 at 2:54
  • $\begingroup$ Thanks a lot for the response. So, is it fair to say that it is futile to try and understand the magnetic part of the Lorentz force as a result of interacting magnetic field lines somehow "pushing" or exerting a force on a moving charged particle? If I understand correctly, what is happening is the field is interacting directly with the particle, and the particle's own field is of no consequence. I suppose the next step is to learn how this discussion applies to permanent magnets and the forces between them, but that is probably beyond my grasp! $\endgroup$
    – M. A.
    Commented Feb 9, 2019 at 2:56
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    $\begingroup$ If you are referring to the “catapult field” mentioned in your newer question, I consider that a misconception. I don’t think you’ll find the concept of a catapult field in physics textbooks. (Where did you see it?) The only way classical magnetic fields interact is to superpose. For a charge in uniform motion, the self-force has no net effect and superposing it with the external field does not change this. You should think of the charge as directly interacting with the external field, and, if accelerating, also with its own field. $\endgroup$
    – G. Smith
    Commented Feb 9, 2019 at 3:09
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    $\begingroup$ Regarding the answer that talked about the catapult field... That answer got 3 + votes and 2 - votes. It is not mainstream physics. (Also, that account has been suspended “because of low-quality contributions”, and its reputation has been reduced to 1.) I recommend that you pay more attention to the answer that got 6 + votes and 0 - votes, by a member whose reputation exceeds 75,000. $\endgroup$
    – G. Smith
    Commented Feb 9, 2019 at 5:59

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