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

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


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

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?


1 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.

  • $\begingroup$ @ChiralAnomaly thanks for pointing the typo out, I have edited my answer. $\endgroup$
    – Leo L.
    Nov 5, 2019 at 4:08
  • 1
    $\begingroup$ The question specifically says the area may change with time. $\endgroup$
    – ProfRob
    Nov 5, 2019 at 7:33
  • $\begingroup$ @RobJeffries Thanks for pointing it out, I have modified my answer. $\endgroup$
    – Leo L.
    Nov 5, 2019 at 13:50
  • $\begingroup$ I am thinking again about this topic. What I have not understood is what physically happens. If the magnetic flux change because surface is changing (while B is constant in time), there will be an induced voltage. But will E be rotational or not? Maxwell equations seem to say not. But how can physically exist an induced voltage without an induced electric field? $\endgroup$
    – Kinka-Byo
    May 23, 2020 at 7:43
  • $\begingroup$ Holding the field constant and ONLY varying the shape of the surface won't change the flux because the divergence of the field would have to be zero and there's a theorem from vector calculus that says that for divergence-less fields the flux only depends on the boundary curve and not the actual surface. (From the divergence divergence theorem and some straightforward reasoning). So NO, changing only the surface won't effect the flux and the emf will be zero. Makes sense intuitively too - the surface is just a mathematical construct so ONLY changing it shouldn't effect anything. $\endgroup$ Sep 25, 2023 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.