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I am familiar with the principles of electromagnetic induction, but I am stuck on the scenario below: a long straight wire is translating in a uniform magnetic field. According to my book, this causes an EMF to be induced across it, and subsequently a current.

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I'm struggling to understand why, since the wire is just translating. Applying Faraday's law $$ V = -\frac{\partial \Phi }{\partial t} = -\frac{\partial BA }{\partial t}% $$ there needs to be a change in flux cutting the conductor somewhere. The flux density is uniform so this is not the cause. I also don't see how the wire's reference area could be changing since it is simply a long cylinder (not a wire loop or something where the angle could be varied).

Thanks in advance.

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    $\begingroup$ Consider the Lorentz force: $\vec{F} = q \vec{v} \times \vec{B}$. $\endgroup$ – ytlu Feb 20 at 16:44
  • $\begingroup$ Okay so, does this mean the protons feel a force in the opposite direction to the electrons in the wire, hence accumulate at opposite ends, inducing an EMF? If so, am I correct in saying the protons accumulate at end 2, and the electrons at end 1? $\endgroup$ – S H Feb 20 at 17:40
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    $\begingroup$ In principle, yes. But in fact, the proton in the wire cannot move. They are strongly binded inside the nucleu. Only, electron is moving. Therefore. in one end, the electron number is higher than average, and the other side, electron number is lower than average. $\endgroup$ – ytlu Feb 20 at 17:53
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You should know the Lorentz force acting on he charges in the wire, if you want to use Faraday, you should attach a Voltmeter, and have a loop with the the Voltmeter outside the magnetic field then you see the area de- or increase.

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