# Faraday Law: integral and differential forms

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?

• Please use MathJax rather than links to scans. Commented Nov 5, 2019 at 1:33
• Have you read en.m.wikipedia.org/wiki/Faraday%27s_law_of_induction ? It has a long discussion of this. The two forms of Faraday's law you quote are not equivalent if the area is changing. Commented Nov 5, 2019 at 7:30

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.