Walter Lewin demo and question about how Electric fields change

This question is related to Walter Lewin "Complete Breakdown of Intuition" question asked earlier. Imagine an area where the magnetics flux is changing and the electric field lines form circular loops with E-field uniform along the circular loop. Now I introduce a wire loop with a resistor in it. What are the Electric fields now around the same loop including the resistor and the ideal wires? The integral of the electric field is still the same = to rate of change of magnetics flux but at each location around the loop they are no longer uniform. How did the electric field across the resistor got generated? Is the E-field due to the magnetic flux causing the charge in the ideal wires to move and pile up against the resistor which generates the E fields necessary to generate the current across the resistor? To make the discussion simpler we can assume constant rate of change of magnetics flux so E-fields are stationary after the wire loop is introduced. Is there a way to simulate the electric fields before and after the current loop is introduced and also show where charges are accumulated if any?

As an answer to Walter Lewin demo question I wrote:

Instead of using voltage and voltmeters let's use just the fundamentals like Electric field, charges, and force. Get rid of the voltmeters and explain what the electric fields are inside resistor R1, R2 and the ideal wires. Before we inserted the current carrying wires, the electric field lines formed circular loops uniform across the loop. Once current carrying loop including R1 and R2 is introduced the electric fields change completely although the integral around the same circular loop is still equal to the rate of change of the magnetics flux through it. Explain why did the electric fields around the same circular loop changed? Is there a charge build up across the resistors that provide the field across the resistor? Is there any electric field across the ideal wire which will cause infinite current?

If you can explain the above, then it is simple to explain the voltage readings without reference to voltmeters. You can connect a capacitor across R1 and another across R2 and if they build up opposite and unequal charges I can see how the scope will show different voltages.

I know this is not really an answer and I should perhaps word it as a question. But it might help to answer this question as well.

• Ask yourself what a wire does. Once you understand what wires do, you got your answer. – CuriousOne Jun 18 '16 at 1:10
• Closely related: physics.stackexchange.com/questions/7936/… – dmckee Jun 18 '16 at 14:39
• Could we have a link to "Walter Lewin "Complete Breakdown of Intuition" question" and a figure please. – jim Jun 18 '16 at 16:41

Thirdly, regarding the previous question, it is not possible to define a potential function (voltage) for a field whose curl is not 0. Therefore to explain the readings of the voltmeter you should use the currents, not voltages. The voltmeters are nothing but galvanometers calibrated in terms of voltage. (to calculate the current use microscopic form of ohm's law, $\vec j= \sigma \vec E$)
• @InquiringMinds the thing you will measure is only $\int \vec E\cdot d\vec r$ but that is not the potential difference as it is path dependent. The potential as in E=-grad V is simply not defined. For more details you can read Feynman's lectures on physics Vol II. – Kartik Jun 20 '16 at 10:48