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My textbook suggests that the integral form of the law is evident from experiments, while the differential form can be obtained by considering a closed curve, constant in time, so that it is legitimate to "plug" the derivative under the sign of integral, in the definition of flux. Then, Stokes is enforced.

But how come, on the grounds of what was said, the differential form can be valid also in situations where the flux changes because of a deformation in the circuit, so that it shouldn't make sense to consider the curve constant in time?

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  • $\begingroup$ The curve doesn't have to be the actual wires in the circuit. Any hypothetical curve will work. This is analogous to the concept of a Gaussian surface which you may have learned. The derivation in fact requires us to consider all possible closed curved which are constant in time. $\endgroup$ – Spencer Feb 26 '18 at 15:06
  • $\begingroup$ Great question. May I request any answer address the use of Leibniz-Reynold's transport theorem? $\endgroup$ – cms Feb 26 '18 at 15:39
  • $\begingroup$ Being at the beginnings of my physics study, I don't know the theorem, but I appreciate the fact that you recognized some interest in the question, that was got rid of elsewhere i asked! $\endgroup$ – Francesco Bilotta Feb 26 '18 at 16:45
  • $\begingroup$ See this question: physics.stackexchange.com/questions/61851/… $\endgroup$ – jobe Feb 26 '18 at 19:36
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As jobe pointed out, there is a thorough treatment of your question here. You should read it. However, I think I can supplement with a more qualitative discussion and an example. Amendment: This could use some community critiquing, so I wouldn't take it as perfectly presented just yet.

Answer:

Allow you to give my answers first, then the explanation:

  1. You can apply Faraday's law in it's differential form to a fixed point in space even for time-dependent boundaries. It is always valid for a fixed point in space.
  2. If you really want to apply the differential form to a moving point, the easiest way is to transform the E and B-fields to a moving frame.

Explanation:

The question was how to properly treat the general case of Faraday's Law:
$$\begin{align} \oint_C\vec E \cdot d\vec l & = - {d \over d t} \iint_S \vec B \cdot d \vec A \\ \iint_S (\vec \nabla \times \vec E)\cdot d\vec A & = - {d \over d t} \iint_S \vec B \cdot d \vec A \\ \end{align}$$
Where $S$ is a smooth surface with one boundary curve $C(t)$ that may have explicit time dependence. Only when $C$ does not depend on $t$ can you bring the total derivative inside the double integral, downgrade it to a partial derivative, and then peel away the integral: $$ \begin{align} \iint_S (\vec \nabla \times \vec E)\cdot d\vec A & = - \iint_S {\partial \vec B \over \partial t} \cdot d \vec A \\ \vec \nabla \times \vec E &= - {\partial \vec B \over \partial t } \\ \end{align} $$
How do we know it is safe to peel off the integral? That is equivalent to declaring $\vec \nabla \times \vec E$ must equal $- {\partial \vec B \over \partial t}$ at every point $P$ on $S$. To see why this must be true imagine shrinking the surface $S$ down around the point $P$. At some point the smoothness of $\vec E$ and $\partial \vec B/\partial t$ must kick in and make these constant over the integral, thereby removing its need: stokes
And this brings me to my first point I would like to make:

$\oint_C\vec E \cdot d\vec l = - {\partial \over \partial t} \iint_S \vec B \cdot d \vec A$ is over a surface, while $\vec \nabla \times \vec E = - {\partial \vec B \over \partial t }$ is at a point.

Allow me to now show how Faraday's law in differential form applies to loops of changing area with the classic sliding bar example: sliding bar
The answer is of course that voltage $\mathcal{E} = BLv$ generates clockwise current around the loop, an answer readily acquired from the integral form of Faraday's law. But what does the differential form tell us? Consider a fixed point $P$ in space about to be overtaken by the sliding bar: zoomed in
Since the bar is a (perfect) moving conductor, it (perfectly) suppresses the magnetic field in its interior. This is how $P$ witnesses a changing magnetic field, even when $P$ is stationary. Let's write out the differential form with $\vec B = +B \hat z$: $$\begin{align} (\vec \nabla \times \vec E)_z &= -{\partial B_z \over \partial t}\\ {dE_y \over dx} - {dE_x \over dy} &= -{\partial B \over \partial t} \\ {\Delta E_y \over \Delta x} &\approx -{ \Delta B \over \Delta t} \\ E_y &\approx {\Delta x \over \Delta t} B \\ E_y &\approx v B \\ \mathcal{E} &\approx v B L \\ \end{align}$$
This is the same result as found with Faraday's law in integral form. The approximation can be made exact with a bit more rigor, but it gets the point across of how having a stationary point $P$ is no limitation in principle.

What if you really want $P$ to move:

Sure, $P$ can move. One method is to apply the Leibniz integral rule for two dimensions, which guides you through differentiation of integral limits: $${d \over dt} \iint_{S(t)} \vec F(\vec r, t) \cdot d\vec A = \iint_{S(t)} ({\partial \vec F(\vec r,t) \over \partial t} + [\vec \nabla \cdot \vec F(\vec r,t)\vec v ])\cdot d\vec A - \oint_{C(t)} [\vec v \times \vec F(\vec r,t)] \cdot d \vec l$$
Where $\vec v$ is the velocity at each point on $C(t)$ and $\vec F$ is your vector field. I am going to quote the result of applying this to Faraday's law, since that was done expertly in the previously linked answer: $$\vec \nabla \times (\vec E + \vec v \times \vec B) = -{\partial \vec B \over \partial t}$$
This is correct, but I find it misleading. The second method I would like to show is much simpler: convert $\vec E$ and $\vec B$ to the moving frame: $$\begin{matrix} \text{E and B in a moving inertial frame} \\ \begin{align} \vec E' &= \vec E + \vec v \times \vec B \\ \vec B' &= \vec B - {1 \over c^2} \vec v \times \vec E \\ \end{align} \end{matrix}$$
These are the (non-relativistic) conversions of electric and magnetic fields to those in an inertial frame moving at velocity $\vec v$. They make lots of quasistatic problems trivial, and Faraday's law is no different. Applying it to the sliding bar example from above: $$\begin{array}{c|c} \text{The rest frame} & \text{The moving frame}\\ \hline \vec E = 0 & \vec E' = \vec v \times \vec B \\ \vec B = +B\hat z & \vec B' = \vec B \\ \end{array}$$
$\vec E = 0$ in the rest frame because there is no charge density $\rho$ in this problem and we require $\vec E' = 0$ when $\vec v = 0$. Thus from the moving bar's perspective, it sees a constant $\vec B$ field and $\vec E$ field everywhere; no Faraday's law required. Since $E'_y = vB$, we can recover the voltage $\mathcal{E} = vBL$ as previously found. Note from this calculation we are actually measuring $\vec E'$ in the moving frame. Lastly, the result from applying the Leibniz integral rule: $$\begin{align} \vec \nabla \times ( \vec E + \vec v \times \vec B) &= - {\partial \vec B \over \partial t} \\ \vec \nabla' \times ( \vec E' ) &= - {\partial \vec B' \over \partial t} \\ \end{align}$$

is the same as applying the field transformations for E and B.

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