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I know that if we move a rectangular wire from no magnetic field to through a magnetic field, there would be an induced voltage because there is change in flux (b∆x). However, if we moved a wire/rod in the same situation, it will also induce a voltage but is it due to the change in flux (b∆x) or charge separation?

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The induced voltage depends on the change in flux according to faraday's law $$volt=-N\frac{\mathrm{d}\phi}{\mathrm{d}t}=-N\mathop{\iint}_{S^{'}}\frac{\mathrm{d}\vec{B}}{\mathrm{d}t}\cdot\mathrm{d}\vec{S}=-N\mathop{\oint}_{C^{'}}\frac{\mathrm{d}}{\mathrm{d}t}\left\{\nabla\times\vec{B}\right\}\cdot\mathrm{d}\vec{l}$$

So as you can see that the flux or the math associated with it needs a defined closed surface for inspection, since the potential induced in your case is NOT electrostatic, but electrodynamic. Electrodynamic potentials are NOT absolute and need a defined closed circuit for their realization. Hence, in this case, the question doesn't make sense since you don't have a closed circuit and are asking for an analysis of electrodynamic potential.

What would make sense is to ask that if you have a multimeter attached to the rod and then continuously monitor the potential, would the potential change? The answer depends on the state of motion of the multimeter. If the multimeter is static and the rod moves and while the rod moves the wires from the multimeter to the rod ends unfurl, then yes, you would see a voltage.

If, on the other hand the multimeter is soldered to the rod with two other rods so that they form a static hoop, then you will not see any voltage since, in that case, you'd have $$\mathop{\oint}_{C^{'}}\nabla\times\vec{B}\cdot\mathrm{d}\vec{l}=const$$ and the rate of change of that quantity wrt time is zero.

Now, when you talk of a Lorentz force, then $\vec{F}=q\left(\vec{v}\times\vec{B}\right)$ and you see the perpendicularity of the velocity and the magnetic field hence creating a force on the electrons which sift towards a given side. This creates a charge seggregation and hence a potential difference on the ends of the rod.

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  • $\begingroup$ Hi, thanks for the reply. Are you saying here that we cannot apply the concept of magnetic flux to the rod/wire but instead we talk about separation of charges due to lorentz force? Your explanation was quite technical (especially the formula) for me since im still in high school. $\endgroup$ – Hilkjh Sep 14 '17 at 7:34
  • $\begingroup$ yes, that's what I'm saying, but if you decide to complete a conducting loop, then you can talk about magnetic flux. Also you can accept the answer if you like it :) $\endgroup$ – ubuntu_noob Sep 14 '17 at 12:04

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