Harvey,
YES to all the assumptions. I might add one other; the wire connecting the two resistors has a uniform resistance along its length totaling no more than 0.1 Ohms each leg.
Are you interested in offering an answer to the question I posed?
With the assumptions in place we know from Lenz's law the direction current will flow in the loop do the increasing flux denoted by Vector B which in most equations is called Phi ( Φ ). Therefore, the conventional current will flow clockwise through the loop such that if we start at A it will first reach the 100 Ohm resistor, then D then the 900 Ohm resistor and back to A. Consequently, we can expect a voltage (EMF) to be present directly at the leads of the 100 ohm resistor proportional to that current and polarized such that the A side is positive. That voltage would be 0.1V. Similarly, we can expect a voltage to be present directly at the leads of the 900 Ohm resistor such that the D side is positive. That voltage will be 0.9V. If both resistors are the same in size, and the leads between them are identical for the A side and the D side, then we are forced to realize that at that exact instant in time whereby the induced loop voltage is 1V that a differential exists across those leads 'totaling no more than 0.1 Ohms each leg' such that if we if we move our probe and its reference from the 900Ω resistor where it reads zero point nine volts (0.9V), over to the 100Ω resistor where it reads negative zero point one volt (-0.1V) we know that for the pair of leads (both the reference and the signal) we have 1V differential. And this is exactly that 1V that is induced. Therefore, when we position the reference at A and the signal probe at D we expect to see the midpoint voltage of zero point four volts (0.4V). We can illustrate this by dividing the length of the leg into 10 sections using 11 points where point 1 is the -0.1V position and point 11 is the 0.9V position. If the EMF is equally distributed, we can expect 0.1V differential at each point. Therefore, point 6 is exactly midpoint thus representing 0.4V. Why do we have such a high differential on a wire that measures only 0.1 Ohm? Because of inductive reactance. In order for the experiment to work as stated, the time-changing field must be done with such a time frame so as to allow the wire to act as a resistor to the 1 mA induced for the loop. Remember, this wire is functioning as a generator and thus appears as a supply to the circuit replacing the 1V battery previously referenced in Professor Lewin's video. Thus the reactance is part of the internal source impedance.  Edit to correct reference polarity 
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Topic: Professor Walter Lewin's Non-conservative Fields Experiment (Read 329987 times)
