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Is the induced electromagnetic field (EMF) always equal to negative rate of change of flux, where $f$ is the flux? I'm curious how it can be true in every case.

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  • $\begingroup$ What are the cases for which you are curious? $\endgroup$ – garyp Oct 31 '18 at 10:58
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It is a manifestation of the fact that the induced current produced as a result of the induced emf will be in such a direction as to produce a magnetic field and hence magnetic flux which opposes the changing flux - Lenz’s law.

If this were not the case the law of conservation of energy would be violated. In other words no work would have to be done to produce an induced current which in turn could produce a heating effect, or light, or sound, or motion etc.
So when you push a magnet into a coil you will have to do work to produce an induced current. If the negative sign were not there then the magnet would be “sucked” into the coil with no intervention from you and electrical energy would be produced “spontaneously”.

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  • $\begingroup$ Actually i was asking whether the equation is valid in all cases..not about the negative sign $\endgroup$ – Alfred Mathew Oct 31 '18 at 9:16
  • $\begingroup$ In some cases the emf is purely motional yet we use this equation..i was asking how it is true even in such cases $\endgroup$ – Alfred Mathew Oct 31 '18 at 9:17
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There are two different effects, Faraday's Law and the Lorentz force law.

Faraday's Law states that $\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$, or $\mathcal{E} = - \frac{\partial \Phi}{\partial t}$. A changing magnetic flux will induce an electric field.

The Lorentz force law states that $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$. This is motional emf. A charge will experience this force when placed in electric or magnetic field.

Consider a conducting rod of length $L$ parallel to the $z$-axis, moving in the $x$-direction with speed $v$ through a uniform magnetic field $B$ pointing in the direction of the $y$-axis. In the lab frame, applying the Lorentz force law, the potential difference across the ends is easily obtained as $BLv$. Now, we switch to the rod's frame. In this frame, the rod is stationary, but there is now an electric field in the $z$-direction, given by $E = \gamma vB$. So the potential difference is $\gamma BLv$. Transforming this result back to the lab frame, we obtain $(1/ \gamma)(\gamma BLv) = BLv$, in agreement.

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