Faraday Law: integral and differential forms

Question

Let's consider both the integral and differential equations which express the Faraday Law (3rd Maxwell Equation):

$$\oint_{\partial \Sigma} \mathbf E \cdot d\mathbf l = -\frac{d}{dt}\iint_{\Sigma} \mathbf B \cdot d\mathbf S $$

And

$$\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}$$

They seem to me a bit in contrast. If we look at the first one, we see that a time variation of the flux of the magnetic field generates an induced voltage: let's suppose to do this variation by changing the surface S in time.

Correspondingly, there will not be any rotor of E since there is not time variation of magnetic field (but only of its flux).

Another situation is this: we change in time S and B, but in a way that the flux keeps constant (for instance we decrease S and increase B correspondingly). In this case there will be a rotor of E but not an induced voltage.

Where is the solution?

Answer 1

I think it’s important to know that the two forms of Faraday’s law is equivalent mathematically only when the surface remains constant (if the surface is time-dependent, see https://en.m.wikipedia.org/wiki/Faraday%27s_law_of_induction#Proof).

Starting with the differential form of Faraday’s law $$\nabla\times E=-\frac{\partial B}{\partial t}$$ It is a local statement. We first integrate on both sides about an arbitrary surface $\Sigma$, $$\int_{\Sigma}\nabla\times E \cdot da=-\int_{\Sigma}\frac{\partial B}{\partial t} \cdot da$$ On the left hand side of the above equation, we use Stokes theorem, $\int_{\Sigma}\nabla\times E \cdot da=\oint_{\partial \Sigma} E \cdot dl$, where $\partial \Sigma$ is the boundary of the surface. On the right hand side, we argue that the surface doesn’t change with time, therefore the derivative sign can be moved outside of the integral sign; in addition, the integral is now only a function of time, therefore it is justified to use the total derivative symbol. So we obtain $$\oint_{\partial \Sigma} E \cdot dl=-\frac{d}{dt}\int_{\Sigma}B \cdot da$$ which is the integral form of the Faraday’s law.