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Charged particles in a magnetic field $\vec{B}$ usually perform some type of circular motion, unless they move parallel to the field lines, due to the Lorentz force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.

However, in the case where $\vec{E} = -\vec{v} \times \vec{B}$, a charged particle will translate uniformly. Given the link between special relativity and electrodynamics, is there some 'deeper' meaning behind this special case?

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  • $\begingroup$ Err, no. In that specific case, the forces just happen to cancel out, meaning there is no net for on the particle. This can be arranged any number of ways. (i could set up any arbitrary magnetic field, then specify an electric field that fulfills your equation. Whether or not one can actually create this setup, (when it non trivial) is another question. If in this case you know $ \vec{E}$ you can find the corresponding charge density via gauss's law, and then do the same for the current density with $\vec{B}$ $\endgroup$ – Haru Fujimura Sep 18 '16 at 14:05
  • $\begingroup$ But the link between special relativity (SR) and electrodynamics is as follows: Special relativity is already "built in" to ED. That is, it is consistent with SR. Trying to use simple newtonian mechanics to figure out the path of charged particles will not get you the correct answer. $\endgroup$ – Haru Fujimura Sep 18 '16 at 14:07
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Actually E is symmetric while B is anti symmetric.

So finding a E field that matches -vXB is highly improbable.

Either static or dynamic (by Maxwell laws any time variation of E will induce an additional B field).

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