The problem with your question is that you state a false premise, i.e. that the loop follows KVL.

KVL and KCL are only valid for lumped circuits, i.e. for circuits made of components (whose dimensionsare irrelevant) connected with ideal wires. Step out of these premises and you will start experiencing "weird" contradictions in your reasoning.

The fact is that KVL derives from a more general law, which is part of Maxwell's equation, i.e. Faraday's law.

What Faraday's law states is that a current is induced in a closed loop if the linked magnetic flux changes with time. If the loop is made of non-superconducting wire, it will have some resistance and the integral of the E field along that loop will be non-zero. This means the E field will perform some work to make that current do its merry-go-round.

Since Ohm's law is still valid (microscopically), that work (per unit time) will be equal to the R^2*I power dissipated in the loop. However you can't get a "voltage across two points" that has the same meaning of "voltage across a resistor", since the resistance is not lumped in a single element, but it's distributed along the loop.

The problem with your question is that you state a false premise, i.e. that the loop follows KVL.

KVL and KCL are only valid for lumped circuits, i.e. for circuits made of components (whose dimensions are irrelevant) connected with ideal wires. Step out of these premises and you will start experiencing "weird" contradictions in your reasoning.

The fact is that KVL derives from a more general law, which is part of Maxwell's equations, i.e. Faraday's law (see this).

What Faraday's law states is that a current is induced in a closed loop if the linked magnetic flux changes with time. If the loop is made of non-superconducting wire, it will have some resistance and the integral of the E field along that loop will be non-zero. This means the E field will perform some work to make that current do its merry-go-round.

Since Ohm's law is still valid (microscopically), that work (per unit time) will be equal to the R^2*I power dissipated in the loop. However you can't get a "voltage across two points" that has the same meaning of "voltage across a resistor", since the resistance is not lumped in a single element, but it's distributed along the loop.

Bottom line: there is no terminals across which you could find the induced voltage, as you would expect with KVL. You only have the circulation of E field along the loop.