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Right answer for the Walter Lewin demo/paradox about electromagnetic induction in Lecture 16 ?
And greetings to you from sunny Essex UK.Walking clockwise through the loop, then the emf's have the right polarity, but the resistor drops are opposed to this direction. The sum will be ZERO
Using two voltmeters (in series) as in your "split" circuit, will give 0.2 Volts on each voltmeter.
For example, considering the upper left : 0.25 Vemf -0.05 Vdrop = 0.2V
and for upper right: - 0.25Vemf +0.45 Vdrop = 0.2 V etc...So the sum of both voltmeters will be 0.4 Volts and not 0.8 Volts!
Note that in the first (left) case your voltmeter Hi is connected to the positive side of the emf, but in the right case the voltmeter Hi is pointing to the negative side of the emf.
Hope this will help to understand also the rest of my video and thanks for watching it.
Best greetings from Belgium
Cyriel Mabilde
Right answer for the Walter Lewin demo/paradox about electromagnetic induction in Lecture 16 ?
Yes! I'll spare y'all the diatribe I wrote in a related post a while ago. But just point out that Kirchhoff's voltage and current laws are for LUMPED ELEMENT CIRCUITS. If you want to use them in the presence of induced emf, you must put some inductors in your schematic. This, of course, is why lumped element models don't work sometimes.Still, he was spot-on redressing Lewin for his Kirchhoff remark, synthesizing the actual voltages around the ring
Let's take a look at how the distinguished Professor Lewin explained the build up of charges in the circuit and the role of these charges in the overall electric field. I think this is a good explanation.Failure to identify the two electric fields of differing sources (Ec from charges only and Ei from induction is I feel a serious omission. For one thing, had he been asked if & where charges are produced...
Let's take a look at how the distinguished Professor Lewin explained the build up of charges in the circuit and the role of these charges in the overall electric field. I think this is a good explanation.
This video is relatively long, maybe we can watch it from 34 minutes.
or you might just ignore them.
signify potential difference, the E must be replaced by what you called Ec.
Hi @alan123hk,
good commentary, thank you.
My overall intention in the blog was to call attention to the existence of two E fields of separate provenance which, somewhat incredibly, many incl. Dr. Lewin seemed unaware of, and that that view provided an alternative explanation of the phenomena..
My criticism of Dr. Lewin's lecture was in essence his comment that "Kirchhoff was wrong". Kirchhoff's voltage law as applied to electric fields rather than potentials applies to what you call the ## E_c ## field only. Lewin implicitly defined voltage as ## \int (\bf {E_c + E_i}) \cdot \bf dl ##. He thus seemed unaware of what exactly "voltage" is, mistaking it for ## \int \bf E \cdot \bf dl ## where ## \bf E = \bf E_i + \bf E_c ##.
As you point out, the voltmeter readings nolens volens include the emf set up by the Faraday setup he has. He never explained that those readings are not actual voltages. Had he held up the voltmeter leads directly above the ring he would have read the correct voltage which is 0.4V and which would have canceled the emf of the ring itself with the emf induced by the flux cutting the voltmeter loop. But he didn't, alas.
Failure to identify the two electric fields of differing sources (## E_c ## from charges only and ## E_i ## from induction is I feel a serious omission. For one thing, had he been asked if & where charges are produced he would have had no answer, or so I believe.
What he did succeed in doing is that if your only goal is to analytically determine voltmeter readings for the different lead positions by considering the total E field alone then yes he did that. What he did not do was to point out that those readings are not the voltages between the two lead connections to the ring except if the leads are held directly across the resistors.
You might look at the references I provide, especially MacDonald's. He too addresses the Lewin "conundrum" & properly defines scalar potential aka voltage.
Also, did you see the infamous C. Mabilde video? That got me started on this whole odyssey. His data were correct but what he really showed was a simulation of the voltages around the ring (by equating the flux rate of his voltmeter lead setup to the actual voltage). He also, by not explicitly identifying the sepate but coincident E fields, came up with some hocus-pocus transformer-like coupling mechanism for the voltmeter readings. Still, he was spot-on redressing Lewin for his Kirchhoff remark, synthesizing the actual voltages around the ring, and although I never checked into it, Lewin's self-contradictory statements about the fields in the wire. Taken with that grain of salt the video is a real eye-opener.
As for my analysis of the battery circuit, that has generated enough controversy that I recommend ignoring it. What is certain is that the sum of line-integrated ## E_c ## fields around the circuit including the battery = 0 a la Kirchhoff.
I am curious about which consensus you think that would be. I'm also wondering if you are saying that it is impossible to position voltmeter leads in a way to obtain a measurement without having the emf affect the reading.I read this whole thread - there are several points that seem to have reached a consensus but are just plain wrong or misleading.
There is a simple resolution to whether or not a transformer has a 'voltage' across it. Let's just assume it's a coil of wire (solenoid) with an external B-field changing through it. If you want to measure the voltage between the terminals, there are two different paths you can consider. One path extends between the terminals on the outside of the solenoid, through free space. If you attach a voltmeter to these points, the outside path your are measuring IS the path of the voltmeter leads. There is definitely an electric field along this path, and the integral of this path is non-zero. Indeed, it is equal to the induced emf. But as it is a real electric field, it counts as a voltage ALONG that path. This is exactly how Feynman defines the voltage of an inductor in his E/M lectures.In an electromagnetic induction system, when current flows through a impedance, a voltage and corresponding electric field is generated. As for this voltage, or the approximate potential difference when the frequency is very low and tends to zero, it is difficult to say whether it can be measured with a voltmeter. I think it depends on the situation. In some cases, it can also be said that everyone's point of view may be different.
For example, when using a voltmeter to measure the voltage at the output load of an actual common transformer, one would think that it is measuring the transformer's EMF output, which is of course true. But some people may assumed that this is the voltage drop in the load impedance, because the measured voltage value is constant regardless of the position of the voltmeter itself and its leads, so this voltage is due to the electric field without curl, and this electric field should be generated by the electric charge of the system.
Another example, in real life the power company will say what is the voltage of the electricity it is supplying to the customer, not what the EMF is because from the consumer's point of view it is just a voltage. Regarding the work done when a charge is moved, of course all the energy comes from electromagnetic induction. But for consumers, I think it's acceptable in the context of their application to think of energy transfer as a voltage mechanism or even roughly a potential difference, since the difference is so small that it can be ignored.
There is no magnetic flux passing through the pie shaped conductors that radiate out from the center of the solenoid and make contact with the single loop conductor in the @mabilde video. The flux lines cut lengthwise along those conductors.If there is no magnetic flux passing through this loop, then your voltmeter is exactly in parallel with whatever you are measuring, and will show the intended voltage.
Lewin demo/paradox and general ordinary transformers are electromagnetic induction systems, so the working principle is basically the same.Your issue seems to be a different meaning of 'voltage' as the sum of ONLY the electrostatic E-field, as McDonald does in his paper. The sum of these static produced fields indeed is zero around any loop, but what use does it serve? The consequence is that you can define different 'potentials' at different points in the same conducting wire. This leaves you with potentials that 1) are not measurable by a voltmeter
I guess you mean that the electric field inside a perfect wire is 0. If the electric field in a region is zero, then of course there is no change in potential in that region. But note that this scale potential still acts on the wire surface and outer space2) vary in a region with ZERO net E- field
So isn't the voltage output by the transformer doing work on the load resistance?3) have no connection to the work done on a test charge as it moves from one point to another.
I completely disagree. The pie-wedge shaped surface of the voltmeter loops are parallel to the ground. The magnetic field of the solenoid points straight up, at least in the interior of the cylinder he is measuring. The magnetic field therefore penetrates this surface at a perfect 90 degree angle, straight through. All he's measuring is the rate-of-change of this flux on his voltmeter.There is no magnetic flux passing through the pie shaped conductors that radiate out from the center of the solenoid and make contact with the single loop conductor in the @mabilde video. The flux lines cut lengthwise along those conductors.
When you measure the output voltage of a transformer (taking a voltmeter as the load that completes the circuit), you can analyze it two ways. If you draw a loop that passes through the interior of the conducting wire in the solenoid, The only voltage drop is across the load. The emf is provided by the changing magnetic field which penetrates the 'loop'. The loop is a bit hard to visualize as it consists of a corkscrew, but imagine a piece of rotini pasta. The field lines penetrate this surface once every turn, giving the net emf. In this analysis, there is no voltage across the transformer, just an induced emf, using Faraday's law to analyze the loop.Lewin demo/paradox and general ordinary transformers are electromagnetic induction systems, so the working principle is basically the same.
Haven't the common transformers we use every day already shown that a voltmeter can be used to measure scalar potential in such systems?
Even though it is difficult to measure with a voltmeter in some cases like the Lewin demo/paradox, is there really no other way?
I guess you mean that the electric field inside a perfect wire is 0. If the electric field in a region is zero, then of course there is no change in potential in that region. But note that this scale potential still acts on the wire surface and outer space
So isn't the voltage output by the transformer doing work on the load resistance?
I think no matter anywhere in the system, including in the solenoid winding or in the external load circuit, whenever there is current passing through the resistor, there will be a relative scalar potential difference and corresponding energy loss, so this scalar potential difference provides energy for the resistor (The scalar potential around the solenoid winding wire is more complicated, it may not be equal to the current times the resistance of the wire due to the induced electric field), which is one of its meanings in the system. It works even if the net electric field in the solenoid zero resistance winding is zero. when current flows through the solenoid winding, the actual work of this scalar field is to accept the energy provided by the induced electric field, how can it be said that its existence is meaningless?But as this 'potential' ignores the induced emf completely, it has no relation to work on a test charge as it moves around in the solenoid windings. It costs nothing to move a charge from higher to lower 'potential' or vice versa along the solenoid path, and thus renders the concept of potential meaningless
Please note that the output port of the solenoid is defined by ourselves. In fact, any two endpoints on any wire segment can define an output, and even though the solenoid has a zero-resistance winding, there is a corresponding scalar potential.When you measure the output voltage of a transformer (taking a voltmeter as the load that completes the circuit), you can analyze it two ways. If you draw a loop that passes through the interior of the conducting wire in the solenoid, The only voltage drop is across the load. The emf is provided by the changing magnetic field which penetrates the 'loop'. The loop is a bit hard to visualize as it consists of a corkscrew, but imagine a piece of rotini pasta. The field lines penetrate this surface once every turn, giving the net emf. In this analysis, there is no voltage across the transformer, just an induced emf, using Faraday's law to analyze the loop.
The other analysis loop consists of the voltmeter and a path that connects the terminals of the solenoid OUTSIDE the windings (hence outside the circuit). There is no flux which penetrates this surface, but there is an electric field that runs outside the solenoid, whose integral is equivalent to the induced emf. Using this loop, you could fairly refer to that electric field integral as a voltage, and use Kirchoff's law to analyze the loop.
I don't understand why you say that the definition of this scalar potential does not conform to Ohm's law.this definition of potential or voltage does not agree with Ohm's law unless the resistor in question has zero length
Relative to the ground is not relevant. Coil relative to the pie shaped is relevant.I completely disagree. The pie-wedge shaped surface of the voltmeter loops are parallel to the ground.
The magnetic field of the solenoid points straight up, at least in the interior of the cylinder he is measuring. The magnetic field therefore penetrates this surface at a perfect 90 degree angle, straight through.
Well, naturally that is reflected by the voltage generated but I'm not sure this is how you meant it.All he's measuring is the rate-of-change of this flux on his voltmeter.
Right - this is the new definition of Ohm's law that accommodates scalar potential. But the good old V = IR that everyone actually uses and is measured by a voltmeter is the path voltage, Int(E.dl).I would like to introduce the relevant situation further. Does not violate Ohm's law. It is precisely because the definition of the scalar potential must be obeyed, and Ohm's law must be obeyed at the same time, so the calculation equation of the scalar potential must be ## V_1-V_2=IR_{12}-\varepsilon _{12}~~~\Rightarrow ~V_1-V_2=IR_{12}-\int_1^2~dl\cdot E^{'}~~~~~~##![]()
The only law that must be obeyed in physics territory is Maxwell. Period.It is precisely because the definition of the scalar potential must be obeyed, and Ohm's law must be obeyed at the same tim