This will Blow your mind!

The way a homopolar generator works showed me you can't really trust your intuition.
Spin copper disk between magnets- current. Spin copper disk with magnets attached to the disk- current. Hold disk still and spin the magnets around the disk (exact converse of first case)- no current.
Dave

Dave__67 fired this volley in news:73a26711-3806- snipped-for-privacy@a32g2000yqm.googlegroups.com:
I think you entirely missed the point on that one. The pickup (the brushes) have to remain fixed relative to the magnets in order to see the potential in the disk. The potential is only developed under the magnet.
So, to remain stationary with respect to the magnets, the pickup would have to rotate with them.
So your case is NOT the "exact converse of the first case", because you didn't rotate the pickup.
'sides.... you can't have an "exact converse" of a three-variable arrangement by changing only two variables.
LLoyd

Can you even have the converse of a 3 variable system??
OK, how bout this non-intuition:
The Doppler effect:
You are standing still, an ambulance with sirens blasting approaches, then recedes. Observe frequency change. OK, now Ambulance is sitting still with sirens blasting (as they do in effing Yonkers, NY), and you approach, recede at same speed as the ambulance previously. Different frequency change!! goodgawd....
Lots of inneresting counter-intuitive stuff, just can't remember many. Marketers of course milk the shit out of our now topsy-turvy intuitive-less world. Like, Tobaccer companies telling us not to smoke.... WTF??????????????????????????

If you spin the magnets and the brushes, you will see no current, I believe.
With or without the brushes, there is a potential difference across the disk from center to edge when the disk rotates.
Dave

The prof is 'way overdue for a vacation.
He changed the resistors from being in series with the cell to being in parallel with the inductor.
V(d)-V(a) = V(r1)= V(r2) = 1.0 V
BTW I(r1)= V(r1)/r1=0.01 A I(r2)= V(r2)/r2=0.001111... A
--Winston

Nice one! What the 'prof' fails to acknowledge is that in a changing mag field the wires connecting the resistors have a voltage gradient along their length (think armature wires in a generator). He is playing on the normal assumption that the voltage along a wire is the same everywhere, given it's much lower resistance than the resistors. Art

There is no net l.dB/dt in the loop formed by the meter leads and the segment under test. So you only end up measuring the resistive voltage drop, not the induced emf. He's not stating/ignoring that point though :-)
I couldn't see what was so counter intuitive. Maybe EE students are more easily puzzled now than in days of yore.
Mark Rand RTFM

I finally got a chance to watch it...
I don't think that's it.
In the first case the voltage was the 'driving force', in the second case the current is forced.
The second case can be properly modeled with an ideal current source of zero internal resistance, I believe.
Dave

. But the Main point is, there are two scopes, both connected to the SAME points in the circuit, and yet they both have different readings. They are both connected to the same points!!
That is the mind blower.

... Well, as Artemus said (and I didn't mean to infer the wire resistance has nothing to do with it) the wires have resistance, and the voltmeters are not actually connected to the same point. *Electrically* with ideal wires they are at the same point in the circuit, but if they were connected *physically* to the same point they would read the same voltage.
Dave

om... I believe the leads are physically connected to the same points.. The leads themselves must be picking up voltage from the induction coil. But the voltages are not only different, but opposite polarities!
It takes several times of watching it.
I think it serves as a warning, as to what you think you are measuring may not be what you are really measuring. And that even a very simple circuit can really fool you.

I believe the leads are physically connected to the same points.. The leads themselves must be picking up voltage from the induction coil. But the voltages are not only different, but opposite polarities!
It takes several times of watching it.
I think it serves as a warning, as to what you think you are measuring may not be what you are really measuring. And that even a very simple circuit can really fool you.
===================================
Maybe I was too brief in my explanation. Think of a generator armature coil. The resistance of the wire is very small, usually < 1 ohm. With the generator off this can be measured with an ohmmeter. Now start the genny and, with no load, there is an output voltage. The wire resistance didn't change and no current is flowing. OK so far?
Back to the Prof and his bullshit explanation. If one were to open the loop there would be 1V at the open terminals. This induced voltage would be steadily and evenly increasing/decreasing as you went around the loop and measured the voltage with respect to one of the open terminals. As no current is flowing there is no IR drop across either resistor.
Now close the loop. There is now current flow and the induced voltage along a unit length of wire is still the same as in the open circuit case. The sum of the IR drops across each resistor is 1V.
Assume the top and bottom wires of his loop are of equal length and that measurement points A&D are in the mid points of each wire. Also assume the length of the resistors is much less than the wire. Using the top of R1 as the reference point and going around the loop clockwise the voltages are: R1 top = 0V D = +0.25V R2 top = +0.5V (half the total induced voltage) R2 bottom = -0.4V A = -.15V R1 bottom = +0.1V
The voltage across points A&D is +0.4V. The voltages across R1 = 0.1V and R2 = 0.9V.
If the leads to the meter form a second loop around the changing mag field then an additional voltage will be induced in it and a false reading of the D to A voltage will result.
HTH. Art

Experiencing that as a video is really frustrating! I want to ask Prof Lewin questions, I want to see the connections, I want to try changing them, etc. Is one scope connected across R1 & the other across R2, with lengths of wire between? I think. And that wire has voltage induced in it. (Am I able calculate the effect of that? No.)
As someone else said, the scopes cannot be literally connected to the same points. A mind experiment comes to mind :-) - a scope on the left connected to *points* A & D and one on the right also connected to A & D, as P. Lewin *implies*. Now I switch them by leaving them connected & simply moving them. Do the displays change? Obviously not. What makes the displays different? That the scopes are not literally connected to A & D, but across the resistors.
Bob

This professor is trying to confuse, bedazzle and impress rather than teach or elucidate. Nowhere does he explain that meters connected across the resistors are not connected to points A and D. Wires are assumed to be of zero resistance or they would be schematically shown as resistors, but the changing magnetic field induces a potential difference along each wire, with total loop induced voltage said to be 1 volt. He deceptively neglected to address the relevant matter of how this EMF was proportionally distributed in top wire, bottom wire and the two resistors, a matter which would depend on the physical geometry in a way left totally unaddressed in his performance. He very briefly mentioned path of integration and showed some (incorrect) definite integrals but nowhere did he show or speak of line integrals.
He should be taken to the academic woodshed. He should be reprimanded, and required to take and pass 9 credits from graduate student instructors (for which English is a third language) in each of political science, women's studies, elementary education for K thru 3 including a student teaching assignment, and boxing at 6 AM MWF. After completing that penance, he should then assigned to clean the latrines of all engineering fraternity houses for two years, and then be fired.

Yes.
(...)
Yes.
--Winston

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D Well, be careful, there *is no* open loop case.
Nothing more mysterious here than connecting 4 resistors with an idealized current source (no internal resistance).
Dave

Yes, not to mention the voltages induced in the loops connected to the meters.
What do you really think? ;^)
Best regards, Spehro Pefhany

"The Journey is the reward" The internal resistance of an oscilloscope is very high, high enough it will look like an open circuit (and hence no current), so I think the voltage would be negligible and ignorable.
Dave

No current means that there will be no voltage drop due to wire resistance.
An open loop of wire in the changing field such as the one that the good Dr. is discussing will have 1V induced in it, just like the secondary of a transformer.
So a voltmeter with the leads "shorted" together will read 1V (polarity depending on the orientation and voltmeter polarity).