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We know that in a closed circuit connected to a battery, $\oint E.dl\ne0$, due to the non conservative nature fo the EMF generated by the battery. But, according to Faraday's law, then $\int_{\Sigma}\frac{\partial B}{\partial t}. dA\neq0$.

Where this varaiable B comes from? does it really exists? Assume all currents are stationary, how can a variable B be generated? or Faraday's law is not valid in this case for some reason I cannot figure it out? Thanks!

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The field inside a circuit is conservative. The thing is a little tricky: The battery does work on the charges via a separate chemical force or something. This force is the one that does the work, not the electric force. So, the electric field in the circuit is a conservative one and so no $ \boldsymbol{B} $ field is induced. Note that the field generated inside the battery though is not conservative, but the field around the circuit is.

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  • $\begingroup$ That disagree with what is in the books, inside the battery there are two contributions to the electric field, a conservative one, related to the separation of charges which is opposite to the current, and another E generated by the chemical reactions. Otherwise, what kind of force would make the electrons move against the conservative electric field inside the battery? $\endgroup$
    – user65081
    Dec 15, 2018 at 19:24
  • $\begingroup$ That's exactly what I said, but the fact that there is a non-conservative electric field inside the battery doesn't mean that the field that goes around the circuit is non-conservative. $\endgroup$
    – Kirtpole
    Dec 15, 2018 at 19:26
  • $\begingroup$ but the integral is different than zero around the circuit (how would the electrons otherwise move in a closed loop?, so the total E is non conservative. The nonconservative part shoudl generate a variable B flux, right? $\endgroup$
    – user65081
    Dec 15, 2018 at 19:28
  • $\begingroup$ The magnetic field will be induced inside the battery but not in the loop. The fact that the field around the loop is conservative is why Kirchoff's law holds. $\endgroup$
    – Kirtpole
    Dec 15, 2018 at 19:34
  • $\begingroup$ I found teh correct answer in an article, thanks, the point of my question was about the EMF, that happens not to be electrical as I wrongly assumed) see commment on Skorka answer $\endgroup$
    – user65081
    Dec 17, 2018 at 13:25
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The variable $B$ comes from the changing current, as magnetic field is directly proportional to current by the Biot-Savart Law.

If the current in the circuit were constant, as your are assuming, then the magnetic field would be constant, and by Faraday's Law and Kirchhoff's Law, this implies that $\oint \vec E \cdot d\vec{l} = 0$, which contradicts your earlier statement, so the magnetic flux is actually constant when the current is constant.

In the case that $\oint \vec E \cdot d\vec l \ne 0$, this is the result of not having circuit components to counter the emf of the battery. In this case, we consider the inductance of the circuit (caused by the shape of the wire) to model its behaviour as an LR circuit (accounting for the internal resistance of the battery).

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  • $\begingroup$ thanks, but the point of my question was about the EMF, that happens not to be electrical as I wrongly assumed) see commment on Skorka answer $\endgroup$
    – user65081
    Dec 17, 2018 at 13:24
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According to Ampere's law $\textrm{curl}\textbf{B}=\mu_0 \textbf{J}$ so for stationary steady state current the curl of $B$ is $J$. This is usually expressed as that the current generates the magnetic field, and an essentially equivalent expression is that o fthe Biot-Savart formula. If $B$ is time independent then from Faraday's law $\textrm{curl}\textbf{E}=0$.

This is true, that is the E field is conservative throughout, even if the source of the current is a battery. If you consider the battery as an electrostatic boundary condition, ie., something by which the potential between two fixed points (the "terminals") the voltage difference is a given constant then you still have in the rest of the space over any contour that $\oint \textbf{E} d\textbf{l}=0$, in other words a conservative field.

It is true that electrochemistry is not electrostatics, in other words no pure electric or magnetic reasoning from the equations of Maxwell/Ampere/Gauss/Faraday etc., can explain why a battery acts as an almost ideal voltage source. But this does not mean that once its behavior is understood it cannot be described/modeled phenomenologically as an ideal voltage source with a source impedance in an electric circuit, and this is what we are doing in circuit analysis.

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  • $\begingroup$ thanks, but the point of my question was about the EMF, that happens not to be electrical as I wrongly assumed) see commment on Skorka answer $\endgroup$
    – user65081
    Dec 17, 2018 at 13:23
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There are two kinds of force fields in your circuit:

  • The electrostatic field, due to the static charges accumulated in the circuit, maintains and smooths out the currents in all parts of the circuit. This field is conservative, i.e. if we move a unit charge around the circuit, the net work done by the electrostatic field is equal to zero.
  • The chemical force field, that exists only inside the battery and supplies the energy to the charges moving in the circuit. The chemical forces are a result od complex molecular interactions, that take place inside the battery. This field is not conservative.

If we consider a perfect battery, with zero internal resistance, the net force acting on the charges inside it must be equal to zero (otherwise, we would get infinite current). Thus, inside the battery, the electric field acts in the direction opposite to the current.

Going back to your question, in the integral we account only for the electic field inside the circuit and in this case, the electric field is conservative and the integral is equal to zero.

On the other hand, if you used a generator, instead of the battery, the electric field inside the circuit would no longer be conservative. In this case, you get two contributions to the electric field:

  • Electrostatic contribution from the static charges accumulated in the circuit, as mentioned above.
  • Electrodynamic contribution, induced by the change of flux in the generator. We need to include both contributions when calculating the integral. In this case, the integral is not equal to zero, but this is what we expect, because there is a changing flux, inside the generator.

There is a small subtlety in the above argument. The molecular interactions, responsible for the chemical forces are in fact electromagnetic in nature. Hence, one may ask, why they are not included in the integral. The answer is: The equations that we use to describe electrical circuits contain macroscopic fields i.e. fields that are averaged over small regions of space. The molecular fields, responsible for the chemical forces are averaged to zero by this procedure and hence need not be included in the integral.

This answer is based on the 7.1.2 from Gryffiths, 'Introduction to Electrodynamics.' You may find it helpful, to consult this textbook for more detailed description.

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  • $\begingroup$ Thanks,just I found an article from Am. J. Phys. 51 (9) 1983, how batteries work, a gravitational analog, explaining how the textbooks explanations are wrong. If it were true that the emf is electrical but this forces average to zero, then there woud be no net EMF $\endgroup$
    – user65081
    Dec 17, 2018 at 13:21

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