The formula is very simple each quarter of the ring (90 degree) will see a quarter of that 1V so 0.25V

Thus depending on the direction of the current you have +0.9V across the 0.9Ohm resistor but you subtract 0.25V thus +650mV

On the other side of the ring (again my assumption the other resistor is on the other half) you have 0.1V on the resistor and subtract 0.25V so -150mV

The copper wires are treated the same as the resistors so a 0.25V source and whatever resistance those wires have say is 1mOhm for each quarter segment in series

Ok, so you are Mabilde-like.

Hence you believe that the emf is located inside the wire in the form of a distributed voltage generator.

Can we summarize your KVL balance as

net voltage drop across 0.9 ohm + voltage gain across copper + net voltage drop across 0.1 ohm + voltage gain across copper = 0

(-0.9 +0.25) + 0.25 + (-0.1 + 0.25) + 0.25 = 0

Does that mean that there is an electric field inside the copper conductor? (either perfect or as you wish with 1mohm resistance per arc)?

If you disagree with this please provide your numbers and how you got to them.

I do not believe in a 'true' voltage (actually it's not a belief, the formulas tell me). In this case it all depends on how you measure it - I can get your numbers by suitably partitioning the disk with the probes, or many other values. But I tell you how I can balance Faraday here:

EMF = 1V, total loop resistance 1 ohm, current 1 amp

integral of E dl in 0.9 ohm + integral of E dl in 0.1 ohm + nearly nothing in copper = EMF

0.9 + 0.1 + 0 = 1

(signs come about when you consider the correct conventions)

My ohm's law still work. I had to give up uniqueness of voltage between two points, though.

Most importantly, there is practically no field inside the copper conductor, as predicted by Maxwell's equations.

In your case, well, what is the field inside the copper parts, if you have 0.25 V across each of them?