How does property of material effect it's interaction in Faraday's Law?

Question

We all know that faraday law mathematically states:
$$ V = - frac{partial phi_B }{partial t}$$
But I think this law is a bit weird because it says that the voltage developed across any conductor due to changing flux is same.. which is a bit counter intuitive for me. I mean, is there no dependence of how the material is but I'm pretty sure different materials have different responses to coming in contact to magnetic field.
So, what exactly am I missing here? Taking this to the extreme we could even say take some sort of closed gaussian surface and talk about the flux change in it and hence the voltage generated on it's boundary is equal to magnetic flux through it... so is the voltage generated independent of material existing at all?
I bring up the previous point, because the explanation behind the law is that in when we have changing magnetic flux, the force pushes charged particles to different ends of a conductor and hence generates a potential difference which leads to a current.

Emmy · Accepted Answer

The Faraday law is no more than a different formulation of the Maxwell-Faraday equation: $$oint_{partialSigma}mathbf{E}cdot dmathbf{l} = -int_Sigmafrac{partialmathbf{B}}{partial t}cdot dmathbf{A}$$
where $Sigma$ is any closed surface, $dmathbf{A}$ a vector element of this surface; $partialSigma$ the contour of this surface and $dmathbf{l}$ an infinitesimal vector element of this contour. It leads to Faraday law if we define the electromotive force $V$ $$V = oint_{partialSigma}mathbf{E}cdot dmathbf{l}$$
and the flux : $$phi_B = int_Sigmamathbf{B}cdot dmathbf{A}$$
The important point of all this is that $Sigma$ can be any closed surface, so the electromotive force generated is indeed independent of material existing at all. But if material exists and if it is a conductor, this force will push its electrons and it generates a current.

How does property of material effect it's interaction in Faraday's Law?

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