KVL for non-conservative $E$-field

Question

Can we use KVL in a circuit having non-conservative field. I mean if its true then it denies the Maxwell equations which says that closed loop integral of E.dl is not zero in non-conservative fields.

Answer 1

Yes, we can use KVL in circuits with non conservative E fields without violating Maxwell’s equations. To see how this is possible, it is important to understand the basis of KVL. KVL is part of circuit theory, and circuit theory rests on three assumptions:

1) The circuit is small relative to the speed of light and the time scales of interest

2) Inside any circuit element the net charge is negligible

3) Outside any circuit element the magnetic flux is negligible

In particular the last assumption limits what types of non conservative E fields are compatible with KVL. Specifically, the assumption about magnetic flux requires that any nonconservative E field must be contained entirely within a circuit element. Then from the perspective of the circuit as a whole the EMF is simply treated as any other voltage across the terminals of a circuit element.

For a detailed justification of why this works see section 11.3 here: http://web.mit.edu/6.013_book/www/book.html But the basic idea is that as long as the fields go to zero outside of some region (the boundary of the circuit element) then regardless of the fact that the fields inside are non conservative, the energy crossing the boundary is equal to the current times the voltage and so it can be treated as a standard circuit element.

EDIT: I note that there is some comment about Dr Lewin’s famous anti-KVL lecture. The circuit he draws deliberately violates the third assumption, and so it is unsurprising that KVL fails since circuit theory is not justified in that case. KVL works any time the assumptions of circuit theory are satisfied

Answer 2

KVL works in every way possible. Its a nonconservative field that's agreed. When we apply KVL to Dr Lewin's circuit, one thing he has NOT considered is the inductance of the loop it has/appears when there is a changing magnetic field. He is only considering the resistors in the circuit loop while blatantly ignoring the mutual inductance between the circuit loop and the source from which the magnetic field is created.

If we target to model the mutual inductance by calculating the flux linking to the circuit loop due to current in the primary source, $\Phi/i_\text{source}$ will give you mutual inductance.

Now with this modeled inductance a change in current in the source will lead to EMF generation in the circuit loop, with $\text{EMF}=M\frac{di_\text{source}}{dt}=(R_1+R_2)i_\text{loop}$.

That's how you get current flowing in the circuit loop.

Secondly to further do advance analysis, this current $i_\text{loop}$ will create a magnetic field such that it opposes the change in magnetic field in the source.

Total flux linking the source coil is $Li_\text{source}+Mi_\text{loop}$. Therefore net voltage drop across the source coil is: $$Ri_\text{source}+L\frac{di_\text{source}}{dt}+M\frac{di_\text{loop}}{dt}$$ This is also equal to the source voltage feeding the coil. $R$ is the coil's ESR.

Therefore: $$V_\text{source}=Ri_\text{source}+L\frac{di_\text{source}}{dt}+M\frac{di_\text{loop}}{dt}$$ AND $$M\frac{di_\text{source}}{dt}=(R_1+R_2)i_\text{loop}$$

Solving the above two equations will give you the real-time values of current flowing in the loop and the variation of current in the source coil.

Lastly coming to the point where he says the voltage was measured across the same point but on different sides of the resistors. We will DEFINITELY get the SAME voltage. The mutual inductance that we derived counts in as a concentrated parameter. But actually it's a distributed parameter of the circuit loop. If somehow we were to model its distributed nature we would get the same voltage drop across the same points no matter how we measure it.