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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.

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The Faraday law is no more than a different formulation of the Maxwell-Faraday equation: $$\oint_{\partial\Sigma}\mathbf{E}\cdot d\mathbf{l} = -\int_\Sigma\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{A}$$ where $\Sigma$ is any closed surface, $d\mathbf{A}$ a vector element of this surface; $\partial\Sigma$ the contour of this surface and $d\mathbf{l}$ an infinitesimal vector element of this contour. It leads to Faraday law if we define the electromotive force $V$ $$V = \oint_{\partial\Sigma}\mathbf{E}\cdot d\mathbf{l}$$ and the flux : $$\phi_B = \int_\Sigma\mathbf{B}\cdot d\mathbf{A}$$ 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.

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  • $\begingroup$ So the force pushed on each is same but the current flow is different hmmm $\endgroup$
    – Babu
    Sep 28, 2020 at 12:38
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    $\begingroup$ The current flow is different if the resistivity of the conductor is different. If you consider a loop of resistivity $R$, the current that flows through it is given by Ohm's law $I = V/R$ $\endgroup$ Sep 28, 2020 at 18:36
  • $\begingroup$ I'm trying to wrap my head around all of this... so, the maxwell equation doesn't directly tell us about resistance but we find this extra experimental result. I'm a bit confused on how exactly the e.m fields interact with matter. As in do they care for material properties or not? or are they just caring that it has charge? $\endgroup$
    – Babu
    Sep 28, 2020 at 20:59
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    $\begingroup$ @Buraian Ohm's law might be an empirical law in the limit of low-electric fields and assumptions of linearity. Even if not, then you can consider $I=f(V)$ where $f$ can be linear or highly non-linear, but dependent on the material, and which can be completely derived from Maxwell's equations given a specific model for the structure of the material concerned. Ohm's law says $f$ is linear with slope $1/R$, where $R$ is a property of the material and intercept $0$ and thus Emmy's argument above passes through without assuming anything outside Maxwell (except maybe a model definition). $\endgroup$ Sep 29, 2020 at 4:48
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    $\begingroup$ physics.stackexchange.com/questions/582890/… $\endgroup$
    – Babu
    Sep 30, 2020 at 10:50

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