# Does a wire/rod moving through a magnetic field have any change in flux?

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