So, after all this posts you still have to answer what is the value of the resulting E field in V/m in copper.

You always ask silly questions that have no relevance to the problem. Showing that you do not understand the mechanisms behind all this.

Original problem had this as data

A closed loop made of two different resistors connected with some pieces of wire (nit that relevant as wire is also a resistor with lower value).

EMF = 1V (I think this was the same in my example and in Lewin example).

Resistor values 0.9Ohm and 0.1Ohm in my example not sure but I think it was 1000x more in Lewin example (not relevant).

Noting else was given and there is no need for anything else to find out what the voltage on those two symmetrically opposed points will be.

So let me explain to you what a 1V EMF means as you do not seem to understand (likely my fault as my examples should have been very clear).

You can have a wire open circuit no matter what shape how long or what type of conductor as long as you mention that EMF is 1V then you will measure 1V across that wire (this is a snapshot in time as of course there will be a variable magnetic field that generates this). There is also the assumption that magnetic field is uniform across the loop meaning that you will read 0.5V if you measure two points at half the wire length between them anywhere on the wire.

Say you close this wire in a loop (again not relevant what shape the loop has and is the exact same moment in time thus EMF=1V) then there will be a current generated in this loop that will be result of EMF divided by loop resistance.

Keep in mind that this is the case because the EMF = 1V was given as input for the closed loop meaning that the magnetic field was strong enough to generate that 1V EMF no matter how much current went trough the closed loop.

So in my example with a total loop resistance of around 1Ohm current was 1A and I think Lewin experiment had 1mA but there is no difference for the explanation.

Now if the loop was all made of copper wire with 1Ohm resistance it seems you do not have a problem with the fact that 1A will travel trough this wire as EMF is defined as 1V (am I correct in assuming that).

Your problem seems to be understanding why if say a quarter of this loop has 0.999Ohm and the remaining 3 quarters have 0.001Ohm

a) If this new loop is open circuit you will agree that you will read 1V between ends and not only that but you will read 0.5V on any half part of the loop.

b) If I cut this loop in two open parts one quarter size 0.999Ohm it will read 0.25V and the other 3 quarter size 0.001Ohm will read 0.75V

c) Each of this two separately can be equivalent with a battery with the respective internal resistance

Now can you understand why is simple to calculate the voltage in any two points on this close loop made of basically two resistors (you can add 10 different resistors in the loop if you want).

And you do not need to know anything else to calculate the voltage between any two points.

All your unrelated questions are useless and just show you do not understand what EMF is.

Just hope this made it clear to you and others as I love sharing and acquiring knowledge. If I'm not able to explain it so anyone can understand it means I do not understand it.