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"Now we turn our attention to what happens in $S'$, in which the particle is at rest and the wire is running past (toward the left in the figure) with the speed $v$. The positive charges moving with the wire will make some magnetic field $\ B'$ at the particle. But the particle is now at $rest$, so there is no magnetic force on it! If there is no magnetic force on the particle, it must come from an electric field. It must be that the moving wire has produced an electric field. But it can do that only if it appears $charged$-it must be that a neutral wire with a current appears to be charged when set in motion."

So i got this from Feynman's lectures in physics volume 2 pg 13-7 and 13-8. I find it hard to believe that the particle experiences no magnetic force due to the moving wire although it's stationary. Since this is classical and not quantum electrodynamics if we assume the charged particle to have inertia, then by Newton's second law, the particle will remain at rest until acted upon by an external force, which i assume is the $magnetic$ $wind$ created by the moving wire past the stationary charged particle. Could someone please clarify??

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The whole point of Feynman's paragraph is to show that what we might believe is not what must happen by physical law. The full electromagnetic force on a particle is the Lorentz force, which is

$$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$$

Since the particle is stationary, the second summand is necessarily $0$ in the frame considered, and so, in this frame, there must be an electric field, since we know the total force is not zero.

I've just paraphrased Feynman above, this is exactly what he wants to tell you: As counterintuitive as it may seem, moving magnetic fields seem to be electric, and vice versa.

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  • $\begingroup$ What if the wire were circular and the charged particle was placed above and the wire rotated in such a way that the particle was always above it? $\endgroup$ – pkjag Jun 29 '14 at 12:14
  • $\begingroup$ Stationary particles cannot experience magnetic forces. There are no ifs and buts, experiencing a magnetic force is contingent on being already moving. The formula above holds always, and if $\vec{v}=0$, you can't outsmart it by constructing a somehow "special" situation where still $\vec{v} = 0$. $\endgroup$ – ACuriousMind Jun 29 '14 at 12:18
  • $\begingroup$ So the $v$ considered above is for the charged particle and not for the moving wire? $\endgroup$ – pkjag Jun 29 '14 at 12:23
  • $\begingroup$ Yes, indeed. Sorry if that was not clear. In the above, the $\vec{E}$ and $\vec{B}$ are the fields produced by the wire, and the $q$ and $\vec{v}$ are properties of the particle. $\endgroup$ – ACuriousMind Jun 29 '14 at 12:25

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