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According to newtons third law, all forces occur in pairs. What is the reaction force that the third law predicts when a magnetic force acts on a charged particle moving in a magnetic field?

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Why do you ask about a magnetic field specifically? Wouldn't we have the same problem with a eletric field? Or a gravitational field?

The answers are the same for all of those fields: the particles that generates the field "feels" the other half of the pair force.

In your case, we have, for nonrelativistic $(v<<c)$ velocities, that the magnetic field generated by a charged particle is

$$ B(\vec{r}) = \frac{\mu_{0}}{4\pi} \frac{q \vec{v} \times (\vec{r}-\vec{x}) }{||\vec{r}-\vec{x}||^{3}} $$

In the equation, $ \vec{v} $ is the velocity of your particle, $q$ is it's eletric charge; $ \vec{r} $ is the position of the particle on your question, the one that is going to "feel" this field $ \vec{B} $; $\vec{x}$ is the position of the particle that generates the magnetic field $\vec{B}$; $\mu_{0}$ is a constant.

It looks just like the electromagnetic field generated by a point charge, with small adjustments. The denominator is powered to $3$, but it is multiplied by $||\vec{r}-\vec{x}||$ in the numerator, so in practice we have a power of $2$ in the denominator (it's just a matter of notation, we could have used the unit vector of $\vec{r}-\vec{x}$ in the numerator, so the denominator would by powered by $2$).

This equation for magnetic fields generated by point charges and the Lorentz Force: $$ F = q(\vec{E} + \vec{v}\times \vec{B}) $$ tell us who is the pair of forces that act in the two particles, the one that generates the field and the one the feels it.

So, after the mathematical description, I think it is nice to have an exemple. Consider two positive particles with equal charges $q$ going on a parallel trajectory with each other; they are side by side and both have velocity of magnitude $v$. We have the field generated by the particles, according to our first equation. Therefore, the magnitude of the magnetic force between the particles is $$ 2qv (\frac{\mu_{0}}{4\pi} \frac{qv}{d^{2}} ) = 2qvB $$ (the factor 2 is because we have the magnetic fields generated by both particles).

The magnetic force, in this case, is against the electrical force, and try to keep the particles close to each other.

Back to your question: when we write the magnetic force as B (and not explicitly with $q$, $v$, etc), we must remember that behind $B$ we have a particle that feels the other half of the pair force.

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  • $\begingroup$ My teacher asked me that question in an exam exactly the way i have written it. what is the answer to the question ? $\endgroup$ Jul 3, 2018 at 3:03
  • $\begingroup$ @unusable_username I think that question is unsuitable for an exam. The reason is that the Lorentz magnetic force does not conserve momentum. $\endgroup$
    – my2cts
    Jul 3, 2018 at 8:34
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I think that question is unsuitable for an exam. The reason is that the Lorentz magnetic force does not conserve momentum.

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It is induced emf. According to Lenz's law, in a magnetic field as flux linked with the object changes ,an emf is induced in it such that it opposes its very own cause. Mag. Flux = B.A

emf=-change in flux/time

Example: Suppose you move a magnet with its north pole ahead through a single turn coil from left to right. Then on the left face of coil north pole will develop when magnet enters the coil and north pole will develop on right face when it leaves. So it always opposes its motion

  1. Or think you drop a magnet vertically through a coil. Then its acceleration WILL ALWAYS BE LESS THAN g.
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In general, Newton's third law is not valid when EM forces due to distant bodies are involved. Two charged moving particles act on each other via electromagnetic forces, but these are not related as action and reaction in Newton's 3rd law; they may even have different directions and magnitudes.

Instead, there is a law of local conservation of momentum, but one has to include momentum of EM field.

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