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I am not sure if this a repeat. Also, I am a high school student, so I'm saying this from what I understood of my teacher's explanation of Maxwell's law.

Imagine a particle moving in a magnetic field. It will execute circular motion. But as it is accelerating, Maxwell says that it should radiate EM waves. If it is emitting EM waves, shouldn't it lose energy? If it does, it should lose velocity, right? And isn't the magnetic field applying a force $B$ on it?

So won't $W=B\cdot S$?

As $v$ isn't constant, $dx/dt$ isn't constant, so work should be done, right?

And so isn't the magnetic field here doing work?

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  • $\begingroup$ And isn't the magnetic field applying a force 𝐵 on it? $\vec B$ is the symbol for the magnetic field. A magnetic field is not a force. The force it exerts is $\vec F = q \vec v \times \vec B$. $\endgroup$
    – Ghoster
    Sep 17, 2022 at 17:23
  • $\begingroup$ Yea sorry my bad but consider qv×b as force B for this one $\endgroup$
    – Kugelblitz
    Sep 18, 2022 at 2:50
  • $\begingroup$ Since you specified that $\vec{B}$ (and $\vec{E}$) are constant, there should be no radiation. $\endgroup$
    – Bio
    Sep 18, 2022 at 4:49

1 Answer 1

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If it is emitting EM waves, shouldn't it lose energy? If it does, it should lose velocity, right?

Yes, a charged particle in circular motion will radiate energy and lose velocity. The fact that it loses velocity actually indicates that the B field is not doing work. All of the energy for the radiation comes from the particle itself, and not from the field. If the energy came from the field then there would be no need for the velocity of the particle to slow.

And isn't the magnetic field applying a force 𝐵 on it?

No, the force from the magnetic field is not equal to $\vec B$, it is $\vec F= q \vec v \times \vec B$. So $\vec F$ is perpendicular to $\vec v$, and a force that is perpendicular to the velocity does no work since the dot product $dW=\vec F \cdot d\vec x=0$.

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