Lewin's NCF Experiment and Lecture

[tinman]

OK Brad, no problem.

What is your prediction for the wire segments?

And sorry, your prediction AND measurement for the voltage across points D and A was/is correct.

I believe that it should be 0 as seen it the test with the wires running along side the resistor connector wire,but something is telling me that that will not be the case.

In all this(so far) dose this not make Lewins claim correct?,as we do not have 0 volts across the resistors as Kirchhoff would suggest.

Brad

[poynt99]

I'm not going to give you the answer (I know the answer), just curious what your thoughts were.

Perform the measurement and let's see what it is.

[tinman]

Well oddly enough,i get around 400mV across the bottom wire,and around 400mV across the top wire.

This was with the duel cable going to the scope probe vertical to the resistor loop,as in the other tests.

Why the two are different?-i have no idea,but i carried out the test many times on each wire,and the results were always the same.

Brad

[poynt99]

Well oddly enough,i get around 400mV across the bottom wire,and around 400mV across the top wire.

It is not 0V as most would seem to expect/guess.

Why the two are different?-i have no idea,but i carried out the test many times on each wire,and the results were always the same.

Brad

Do you mean why in the first case the measurement was 0V on each wire segment, and in this last case the sum is close to the calibration emf?

I was hoping you would have put it all together, as you've done quite well so far. Isn't this just a step-down transformer? We have 50 to 100 turns on the primary, and one turn on the secondary, loaded with a resistance of R1+R2.

You are measuring the actual emf induced in the secondary loop because the measurement device is decoupled from the primary (and secondary). That is why the measurements are different.

Now as a final step, use your vertical probe to go around measuring the 4 components (R1-->wire_seg-->R2-->wire_seg) in serial fashion following the current path and write the voltage down for each. If performed correctly, two measurements will be positive and two negative in polarity (the two resistors will be one polarity, and the two wire segments will be the other polarity). Add the 4 values together and you should be left with something close to 0V.

[tinman]

It is not 0V as most would seem to expect/guess.
Do you mean why in the first case the measurement was 0V on each wire segment, and in this last case the sum is close to the calibration emf?

I was hoping you would have put it all together, as you've done quite well so far. Isn't this just a step-down transformer? We have 50 to 100 turns on the primary, and one turn on the secondary, loaded with a resistance of R1+R2.

You are measuring the actual emf induced in the secondary loop because the measurement device is decoupled from the primary (and secondary). That is why the measurements are different.

Now as a final step, use your vertical probe to go around measuring the 4 components (R1-->wire_seg-->R2-->wire_seg) in serial fashion following the current path and write the voltage down for each. If performed correctly, two measurements will be positive and two negative in polarity (the two resistors will be one polarity, and the two wire segments will be the other polarity). Add the 4 values together and you should be left with something close to 0V.

Yes,the sum equals 0,but what has that to do with what Lewin claims?.
We can sum up voltages around any circuit to equal 0 if we measure from certain points using certain polarities.

It would seem to me that Lewins claim is correct,when measuring across the two resistors,as the sum is not 0,it is the induced 1volt EMF in this case,being the total across each resistor.
If we have a current flow,then we are not measuring just the E filed,but the EM field.
If it were just the E field,then Kirchhoff law may hold,in that the measuring was carried out incorrectly. But as we also have current flow,and a magnetic field,then it would seem to me that Kirchhoff's law dose not hold in this situation.


Brad