Apparent paradox when we use the Kelvin–Stokes theorem and there is a time dependency I am having trouble to understand what is going on with the Maxwell–Faraday equation:
$$\nabla \times E = - \frac{\partial B}{\partial t},$$
where $E$ is the electric firld and $B$ the magnetic field. The equation is local, in the sense that any change at point $x$ will not affect what happens at another point $x'$, at least not instantaneously. That is, if there is a change in $B$ only at position $x$, then the change will need time to propagate to $x'$.
But we can use the Kelvin–Stokes theorem and write the equation in integral form:
$$\int_{\partial \Sigma} E.d\ell =  - \frac{\partial}{\partial t}\int_\Sigma B \cdot dS,$$
which is basically telling you that a change in $B$ at the center of the surface will affect instantaneously $E$ at the edge.  
What is it wrong with my interpretation of these equations?  
 A: Good question!
The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $\Sigma$. This can be interpreted in two ways:


*

*For a given closed loop $\ell\in\mathbb{R}^3$, there are infinitely many smooth surfaces $\Sigma$ such that $\partial\Sigma=\ell$;

*The closed loop $\ell$ itself could also be arbitrarily specified.


Therefore, while the integral form appears non-local, it is actually local, as you may take a "small" closed loop $\ell$ (e.g., a circle with an infinitesimal radius).
Further, even if you take a "large" closed loop $\ell$, you may still choose different surface $\Sigma$, such that a local change of $\mathbf{B}$ in $\Sigma$ would not effect the value of $\mathbb{E}$ on $\ell=\partial\Sigma$.
With these arguments, your question could be interpreted as follows. Suppose you have chosen some $\ell$ and $\Sigma$ with $\ell=\partial\Sigma$. Suppose $\mathbf{B}$ observes a tiny change in the interior of $\Sigma$. Then according to
$$
\oint_{\ell}\mathbf{E}\cdot{\rm d}\mathbf{l}=-\frac{\partial}{\partial t}\int_{\Sigma}\mathbf{B}\cdot{\rm d}\mathbf{S},
$$
it seems as if $\mathbf{E}$ also yields some changes along $\ell$. But wait! Since the change in $\mathbf{B}$ is tiny, you may want to find some $\Sigma'$, such that (1) $\partial\Sigma'=\ell$, and that (2) $\mathbf{B}$ does not have any change on $\Sigma'$. In this sense, you will obtain, at least for the moment,
$$
\oint_{\ell}\mathbf{E}\cdot{\rm d}\mathbf{l}=-\frac{\partial}{\partial t}\int_{\Sigma'}\mathbf{B}\cdot{\rm d}\mathbf{S}=0,
$$
with which you would have no idea whether or not $\mathbf{E}$ changes along $\ell$. For tiny changes in $\mathbf{B}$, you may apply the integral form around each point on $\ell$ with small closed loops $\ell'$ and surfaces $\Sigma''$ with $\partial\Sigma''=\ell'$ on which $\mathbf{B}$ does not find any change, and the arbitrariness of the choice of $\ell'$ and $\Sigma''$ would imply the free of change in $\mathbf{E}$. This trick fails only if the change in $\mathbf{B}$ hits $\ell$, which exactly indicates the locality of its physics.
Hope this could be helpful for you.
A: Since we have also $\nabla\times B = 0$, you can only change $B$ by adding an entire loop.  In this case, it will either cross the surface $S$ once in each direction, so be 0, or it will actually go around the perimeter current, and induce a current, which will change $E$. 
