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The motion of a conductive wire in a uniform magnetic field causes electric field in the wire (thanks to the magnetic force that pushes free charges). From relativistic point of view, it should not matter whether the wire is moving and the field is stationary, or the field is moving and the wire is stationary. But how can you distinguish moving and stationary uniform field in equations?

Another example. Let us have two disc magnets and a conductive disc between them. The magnetic field passing through the disc is axially symmetric. If the disc spins, the electric field builds up within the disc (pointing radially). What happens if you spin the magnets? It should have the same effect, but the field is in fact still the same although the magnets are spinning, so how could it affect the disc?

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  • $\begingroup$ The Lorentz force on the lectrons is $\textbf{F}=-e(\textbf{E} + \textbf{v} \times \textbf{B})$, and here $\textbf{v}$ is the relative velocity between the electrons and the field $\textbf{B}$. When you move a metal wire or you move the field, does not matter, you move its free charges relative to the field, and they will be subjected to a force that makes them move, hence a current. $\endgroup$ – hyportnex May 7 '17 at 21:38
  • $\begingroup$ My question aims at the relative velocity. What is the reference frame? Does the voltage depend on the observer? If I move with the wire, is the voltage zero? $\endgroup$ – zemosh May 7 '17 at 21:57
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Yes, it does cause a current.

This is because when the coil is moved, the magnetic flux linked with the coil changes as $d \phi$, which in turn, induces an EMF = $-N \frac {d \phi }{dt}$, where $N $ is the number of turns in the coil.

The magnetic field may be uniform or variable, but the magnetic flux linked with the coil has to change, that's all.

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  • $\begingroup$ But if the magnetic field is same in all places, the magnetic flux is constant. I think. $\endgroup$ – zemosh May 7 '17 at 21:06
  • $\begingroup$ Moreover, for a wire, there is no flux. The voltage is caused by motional EMF. $\endgroup$ – zemosh May 7 '17 at 21:25

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