The question title sums it up pretty well, but to elaborate, is Kirchhoff's Law valid for situations where the emf is due to non-conservative electric field, like one produced due to a varying magnetic field (with the circuit stationary)?
This question was born out of another question here.
Thank you!
The Short answer is no. Kirchhoff's Law is derived by the assumption that the electric field is conservative, while the electric field induced by the change of an magnetic field is not conservative.
The Long answer is, that you have to clarify what voltages you are looking on. Kirchhoffs Law is usualy used to make a statement about the voltage drops in a closed path inside an electrical circuit, like \begin{align} U_1 + U_2 + U_3 = 0 \end{align}
Here, by $U_1$, the user usualy doesn't mean a line-integral of the electrical field, but instead the terminal voltage. The terminal voltage is the the line-integral of the conservative part of the electric field, or the voltage as it would be if you would remove the non-conservative part of the electric field. It is also the voltage that you can measure when you plug a multimeter to the two points that belong to $U_1$.
Using it like that, you can use Kirchhoff's Law even in cases that feature time dependend magnetic fields:
The next section makes the approximation of slowly varying current and fields inside the circuit. Because of that, I neglect radiation that is induced in the capacitor and the inductor by changing fields. I also assume that the time the charge distribution inside the circuit needs to cancel out an external electric field is small compared to the change of the electric field and the currents.
Kirchhoff's Law is commonly used to describe for example an LC-circuit. Here we usualy just state that
\begin{align}
U_L + U_C = 0
\end{align}
and we do that altough we know that the electric field inside the inductor isn't conservative. Why are we allowed to do that?
Coming from Maxwell's Law: \begin{align} \int_{Path} \vec{E}d\vec{x} = \int_{Surface} - \frac{\partial B}{\partial t} \end{align}
In the capacitor there is a conservative electrical field (generated by charges), in the inductor there is no field at all, because the non-conservative field generated by the changing magnetic field is directly cancelled out inside a conductor (conductors are field free). That means, the lhs of the equation is just $U_C$, the terminal voltage in the capacitor. The rhs of the equation instead is the negative terminal voltage in an inductor: \begin{align} -U_L = \int_{Surface} - \frac{\partial B}{\partial t} \end{align}
And then you arrive at
\begin{align}
U_L + U_C = 0
\end{align}
Kirchoff's laws are the electric equivalents of more fundamental laws.
Kirchoff's Law of Current: Is a formulation of the Charge Conservation Law, in a node there is no net charge, so from the continuity equation the net current going through the node is 0.
Kirchoff's Law of voltage: Corresponds to the energy conservation, altough you need current's law to see it. In a closed loop the net power delivered and consumed has to be 0 (Conservation of energy), since power is $P = V\cdot I$, since the current is governed by the current's law, with a little algebraic manipulation you get that the sum of voltages is 0.
Please notice that so far there is no dependence at all on the nature of the source, loads, voltage, currents of the circuit. All it need is that the circuit can be modeled as a circuit (lump model)See Wikipedia's page on Kirchoff's Laws for the discussion on the limitations of the laws.
In the particular case you are asking, you don't have a circuit model, thus you cannot apply the kirchoff's laws.
