- #71
Averagesupernova
Science Advisor
Gold Member
- 4,440
- 1,214
[Sorry, what is the DUT?]alan123hk said:I think the test method used by mabilde is correct. When we measure voltage with a voltmeter, the DUT has to supply energy, so the Lewin circuit has to supply energy, so the voltmeter loop has to somehow let the changing magnetic flux through to get the energy, otherwise it won't be able to measure anything thing.
Of course, according to Faraday's law, the test result is the induced EMF in the pie-shaped area due to the change of the passing magnetic flux. But this is not enough to prove that it cannot represent the scalar potential difference generated between the ring arcs at the same time.
I think whether you agree or disagree, both parties should try to come up with some stronger arguments to support it. The method you said of pressing the leads together so that the loop becomes a closed T-shape with no area is problematic because the leads of the voltmeter will be affected by the induced electric field. When you're trying to measure the potential difference created by a pure charge, you can't subject the voltmeter and its leads to an induced electric field.
I reiterate my opinion again, I agree with mabilde's method is correct.
Device Under Test.tedward said:[Sorry, what is the DUT?]
Doesn't the fact that he reads zero volts with the pie shaped leads in a shorted loop condition imply he's not measuring flux? If he were measuring flux a reading would show up here. As I said before, you can't have it both ways.tedward said:My problem is that he doesn't mention scalar potential, discuss the difference in voltage conventions, or acknowledge that he's really just measuring the flux through his loop.
You seem to contradict yourself. You say in one place you cannot have a voltage across a wire. But in the above quote....tedward said:I certainly do NOT agree that the correct measured voltage (meaning path voltage, IR drop, etc) is zero between any two points, as the emf is now evenly distributed across the ring.
Again, your model of alternating batteries and resistors is a great thought experiment for understanding the DIFFERENCE between induced voltage and the DC/battery variety. You HAVE to think about electric field. In a resistor, electric field points from high potential to low potential, in the direction of current. In the standard model of a battery, electric field points from high potential to low potential INSIDE the battery, which is OPPOSITE the direction of current. So in a simple circuit with a battery and few resistors in series, the sum of electric field (read sum of voltage drops) is zero.Averagesupernova said:You cannot have a voltage measured across the ring anywhere as long as the conductivity all the way around is the same. The voltage is lost across the internal resistance of the source and the source is the ring. The source and the load occupy the same space. The double A batteries in a loop with resistors example I gave above is not meant to be just some puzzle I use to make me look smart. It is supposed to illustrate that the same thing is happening with AC induced into the ring.
But they can't in a closed loop being driven inductively by a solenoid as in the @mabilde experiment. They will heat but you will never measure a voltage across them. Yes it seems impossible but its true. You may measure a little since individual wires that connect them are not the same as a resistor. But if you use resistor wire you will never measure anything. If you do it is due to voltmeter lead routing.tedward said:Imagine a string of 1-ohm resistors - they're small but they all share the voltage equally.
Cool experiment. Use resistor wire, put a solenoid in the center. Set up your infrared camera / heat sensor and calculate the power dissipation in any section, and calculate the honest-to-goodness voltage fromAveragesupernova said:Maybe an experiment could be devised using resistor wire. Measure the voltage all you want from any point to any other. By watching the heat with an infrared camera you should be able to tell if the voltage you measure is real. I guarantee all of the resistor wire will be heating evenly and that implies any voltage you measure at all can't be real because unless you are measuring 180 apart on the ring, one side has to be heating more than the other if the voltage measured is real.
You can't. The positive and negative field occupy the same space. They cancel just like they do in the case of a AA battery next to a resistor. There is still a current but the field around the section you are trying to measure is canceled by the field of the opposite polarity caused by the induction.tedward said:I guarantee that you will measure the same voltage that you measure from heat loss.
Try this - let's start with where we absolutely agree. We see, in the video, mabilde measure a sector of a resistor ring, with voltmeter lead wires pointing radially outward from the center, directly and symmetrically over the solenoid in a pie-slice shape. He measures zero volts - right? Right.Averagesupernova said:You can't. The positive and negative field occupy the same space. They cancel just like they do in the case of a AA battery next to a resistor. There is still a current but the field around the section you are trying to measure is canceled by the field of the opposite polarity caused by the induction.
-
Imagine hooking many voltmeters around the outside. They would all have to add up and by the time you get to the last part of the ring you would have to have all the voltage across that space. You won't. If you get readings on all those voltmeters it is currents induced into the leads in that case.
-
The basis of our disagreement is the placement of the leads. I happen to believe the opposite of what you do concerning this.
Averagesupernova said:You believe that I am tossing Faraday away saying it is wrong when I am not. I am saying it is misapplied. I believe that Lewin and you are tossing Kirchoff away saying it is wrong while it is not. You are misapplying it.
Lol. I've been 'finished' in my ideas on this for a few years. I would like to find someplace on the net if not here, someone who has taken Lewin's side and had it explained to them and realized the problems with it. All I ever find is us vs them. No one ever says: 'Oh NOW I get it. Yes, I can now see it. It certainly has been a lesson for me!" I never see that, maybe I need to look harder.hutchphd said:Can we be finished?
I mean, I could say the same exact thing from the other side. We're both confident, and one of us is certainly wrong. I am trying to understand your arguments and 'get inside your head' as it were, as I find the discussion useful even if we don't agree (I've really learned from thinking about this intently). So if we don't agree, no problem. See what you think of my energy argument :)Averagesupernova said:Lol. I've been 'finished' in my ideas on this for a few years. I would like to find someplace on the net if not here, someone who has taken Lewin's side and had it explained to them and realized the problems with it. All I ever find is us vs them. No one ever says: 'Oh NOW I get it. Yes, I can now see it. It certainly has been a lesson for me!" I never see that, maybe I need to look harder.
You may want to think twice about that statement before you dig yourself in any deeper.tedward said:The answer is certainly NOT the ohmic drop through the windings, which is zero.
Well of course that's what I mean! Hasn't that been the central point of the discussion? Just because the secondary winding in the transformer example I gave a few posts back is more than one turn makes no difference. The whole thing is exposed to the flux and the whole I*R voltage is lost across the resistance of the copper. Just how difficult are you trying to make this?tedward said:you mean take a secondary coil, put a changing flux through it (from any source - primary, whatever), and short the output with no resistors / load?
Apologies, I misunderstood what you were describing. I was thinking transformer with a load, not the uniform ring. In the last part of #91 I was describing a typical LR circuit where the winding resistance is neglected.Averagesupernova said:Well of course that's what I mean! Hasn't that been the central point of the discussion? Just because the secondary winding in the transformer example I gave a few posts back is more than one turn makes no difference. The whole thing is exposed to the flux and the whole I*R voltage is lost across the resistance of the copper. Just how difficult are you trying to make this?
So I'm writing a post on the lead argument / voltage masking debate right now, because it's a question that keeps coming up. But to this point - when you say the specific orientation of test leads are you referring to Mabilde's pie-shaped leads over the solenoid?Averagesupernova said:The whole point of what I have introduced into this is to prove it is not possible to measure a voltage between points on a shorted ring.... An agreement with my position causes the requirement to use specific orientation of test leads on other configurations of the resistance on the ring.
ANY placement. It's critical since a pair of leads for a voltmeter have the exact same properties as the DUT. They are not an insignificant part of the setup.tedward said:But to this point - when you say the specific orientation of test leads are you referring to Mabilde's pie-shaped leads over the solenoid?