What is the relationship between voltage and flux in Faraday's Law?

[alan123hk]

A former colleague asked me a simple question about Faraday's law. As shown below, the configuration is just a circular conductor and a changing B field, and then tried to measure the voltage across the diameter. But I am not sure what the answer is?

Any help would be greatly appreciated.

 

[jtbell]

I am not sure what the answer is?

Nevertheless, what are your thoughts, uncertainties, etc., about it? Then people here can address them directly, and try to clear up any errors in your thinking, instead of simply giving you the answer (which is frowned upon here).

 

[hutchphd]

You nowhere say exactly how the B field is changing in time (uniformly?).
When you have a changing field like this it is no longer a "circuits' problem and you need to consider the fields explicitly. The reading on the voltmeter will depend upon the exact path of the wires, not just their topology. Are you assuming perfect symmetry? So make an analysis.
This is of course Walter Lewin's famous headscratcher.

 

[alan123hk]

Nevertheless, what are your thoughts, uncertainties, etc., about it? Then people here can address them directly, and try to clear up any errors in your thinking, instead of simply giving you the answer (which is frowned upon here).

Thanks for your reply.

My idea is that from the front view, there are two loops connected to the voltmeter, the EMF induced on the right side seems to be +0.5V, and the EMF induced on the left side seems to be -0.5V. There seems to be a contradiction, maybe there is no contradiction at all. I think it is possible to add two values. So the answer should be that the maximum chance is 0V.

Is my inference correct?

 

[alan123hk]

You nowhere say exactly how the B field is changing in time (uniformly?).
When you have a changing field like this it is no longer a "circuits' problem and you need to consider the fields explicitly. The reading on the voltmeter will depend upon the exact path of the wires, not just their topology. Are you assuming perfect symmetry? So make an analysis.
This is of course Walter Lewin's famous headscratcher.

You make a lot of sense. Let's assume that the B field changes linearly with time, so the induced E field is a constant value. The path of the voltmeter connection is shown in the diagram, which is vertically to the plane of the circular conductor and located behind it. In this way relative to the plane of two parallel connecting lines of the voltmeter, the left and right are completely symmetrical. Therefore, the connection line of the voltmeter should be perpendicular to the induced electric field, which means that no EMF will be generated on them.

In addition, of course the resistivity of each point of the circular conductor is also the same, and the internal resistance of the voltmeter is infinite, etc.

Please let me know any hints about the answer, thanks.

 

[vanhees71]

Of course it depends on the geometry of the wires connected to the volt meter. You don't have a voltage here but an EMF, because the electric field is not conservative since $\vec{\nabla} \times \vec{E}=-\partial_t \vec{B} \neq 0$. The volt meter in fact measures a (usually very small) current going through it. So you measure the EMF as calculated along the wires connecting the volt meter.

 

[hutchphd]

So the easy way for me to consider the problem is to realize (as @vanhees71 points out ) that the voltmeter in fact measures the current through it. Current is absolutely conserved and the net current through the meter in the symmetric case is the sum of two equal and opposite currents if one looks at the two smaller loops. So zero is the reading. Any asymmetry will likely produce a nonzero reading, but it will not properly be called a "Voltage" in the usual Kirchhof sense.

 

[alan123hk]

Of course it depends on the geometry of the wires connected to the volt meter. You don't have a voltage here but an EMF, because the electric field is not conservative since ∇→×E→=−∂tB→≠0. The volt meter in fact measures a (usually very small) current going through it. So you measure the EMF as calculated along the wires connecting the volt meter

I totally agree with your explanation, your explanation is very accurate and helpful.

So the easy way for me to consider the problem is to realize (as @vanhees71 points out ) that the voltmeter in fact measures the current through it. Current is absolutely conserved and the net current through the meter in the symmetric case is the sum of two equal and opposite currents if one looks at the two smaller loops. So zero is the reading. Any asymmetry will likely produce a nonzero reading, but it will not properly be called a "Voltage" in the usual Kirchhof sense.

Your explanation is very logical and reasonable. I think you may not intend to explain some details for the time being, but your explanation has no flaws anyway.

 

[Delta2]

So the easy way for me to consider the problem is to realize (as @vanhees71 points out ) that the voltmeter in fact measures the current through it. Current is absolutely conserved and the net current through the meter in the symmetric case is the sum of two equal and opposite currents if one looks at the two smaller loops. So zero is the reading. Any asymmetry will likely produce a nonzero reading, but it will not properly be called a "Voltage" in the usual Kirchhof sense.

I, on the other hand, believe that even in the case of asymmetry (that is even if we put the voltmeter leads anywhere along the circumference of the conductor) we will have again a zero reading, because there is no charge separation happening along the circumference of the conductor. A voltmeter measures indeed a current according to my opinion too, but this current must be a result of a scalar potential which is result of charge separation.

 

[hutchphd]

But the definition of a scalar potential assumes $\nabla \times \vec E =0$ so I don't know what scalar potential you mean. The internal workings of the Voltmeter in fact work by comparing magnetic force of an unknown current to a that of a spring

 

[vanhees71]

http://www.physics.princeton.edu/~mcdonald/examples/lewin.pdf

 

[alan123hk]

I personally tried to think more deeply about this problem, and believe that now I have found a very simple way to describe and predict the behavior of such circuits

I, on the other hand, believe that even in the case of asymmetry (that is even if we put the voltmeter leads anywhere along the circumference of the conductor) we will have again a zero reading, because there is no charge separation happening along the circumference of the conductor. A voltmeter measures indeed a current according to my opinion too, but this current must be a result of a scalar potential which is result of charge separation.

I think you made the very crucial point, that is, in fact, there is no charge separation happening along the circumference of the conductor, so from the perspective of the external circuit, there is no potential difference along the circumference of the circular conductor. This means that as long as the connecting wire of the voltmeter does not picked up the induced electric field generated by the changing magnetic field, regardless of the position of the connecting wire, the voltage measured by the voltmeter is still zero.

 

[hutchphd]

I would like to comment but I have no idea what you are saying:

This means that as long as the connecting wire of the voltmeter does not picked up the induced electric field generated by the changing magnetic field, regardless of the position of the connecting wire,

The Princeton citation from @vanhees71 is above is pretty comprehensive. Have at it.

 

[Delta2]

Well the paper states that
"The fact that the two meters give different readings when a = c and b = d (in sec. 2.1) indicates that the meter readings are not simply proportional to the electric scalar potential V at those points, as would be the case for a DC circuit"
I might be wrong in what i said in post #9 afterall.

 

[Delta2]

However the experiment setup is a bit different from what is discussed here so i am not completely sure.

 

[alan123hk]

I would like to comment but I have no idea what you are saying:

Sorry, my expression may not be clear enough, or I did not express my thoughts correctly.

Referring to the example I put forward on #1, the induced electric field generated by the changing B field will not only be concentrated on the circular conductor, but will also extend to all surrounding spaces, including the two connecting wires of the voltmeter.

What I mean is that since the circular conductor does not exert a potential difference on the two connecting wires, as long as the force of the induced electric field (generated by the changing B) acting on the charges on the two connecting wires does not generate a voltage across the voltmeter, the measured voltage on the voltmeter must be zero.

 

[Charles Link]

See https://www.physicsforums.com/threa...asure-a-voltage-across-inductor.880100/page-3 post 52 This is one of Professor Walter Lewin's videos.

 

[alan123hk]

After rethinking this issue, I think that there is no paradox.

We should not treat a and b are exactly same, even if there is no voltage drop due to Ohm's law, there does exist an induced electric field between them, so they are indeed two different points.

As long as we define the Electromotive force (EMF) and PD (potential difference) as follows,

Electromotive force, $~EMF_{ab} =−\int_a^b E \cdot dl$
Potential difference, $~PD_{ab} = \int_a^b E \cdot dl$

and put the EMF (created by the induced electric field) into the equations, the correct calculation result can be obtained.

For example, we can easily prove that if the resistance of each point of the circular conductor is the same, there is indeed no potential difference along the circumference of the conductor.

Total length $~ L=\sum L_i$
Total EMF $~ \mathcal E=\sum \mathcal E_i=\sum C_1L_i$
Total Resistance $~R= \sum R_i= \sum C_2L_i$

where $~C_1$ and $C_2 ~$ are constants

For any arc length $~ L_i ~$ on the circumference, the actual potential difference should be : -

$$ \mathcal E_i - \frac {\sum \mathcal E_i} {\sum R_i} R_i$$
$$ C_1 Li - \frac {\sum C_1L_i} { \sum C_2L_i } ~ C_2L_i ~~ = ~ C_1 Li - \frac {C_1} { C_2 } ~ C_2L_i =~0 $$

 

[vanhees71]

Well the paper states that
"The fact that the two meters give different readings when a = c and b = d (in sec. 2.1) indicates that the meter readings are not simply proportional to the electric scalar potential V at those points, as would be the case for a DC circuit"
I might be wrong in what i said in post #9 afterall.

Sigh, well, I forgot that this imprecision is in this text. Please bear in mind that there is NO POTENTIAL in this case, because (in SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B} \neq 0$$
and that's the reason why the meters in Lewin's example show different readings. What the (idealized) volt meter measures is the EMF (NOT VOLTAGE!) along the wires. Note that it's a CLOSED line integral:
$$\mathcal{E}=\int_{\text{wires}} \mathrm{d} \vec{r} \cdot \vec{E}=-\dot{\Phi}_{\vec{B}},$$
where $\Phi_{\vec{B}}$ is the magnetic flux through an arbitrary surface with the wires making its boundary (assuming that the wires are at rest).

 

[Delta2]

@vanhees71 the scalar potential is defined even in the case of $\nabla\times\vec{E}\neq\vec{0}$, simply by the solution to the wave equation (in the lorentz gauge)$$\nabla^2 V-\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}=-\frac{\rho}{\epsilon_0}$$ but of course as you notice the voltmeter doesn't measure that (it usually does but not in this weird setup).

 

[hutchphd]

What the (idealized) volt meter measures is the EMF (NOT VOLTAGE!) along the wires. Note that it's a CLOSED line integral:

I believe that part of the confusion in the OP is from the geometry of the Voltmeter as the nexus of two such closed curves of wire. For me it is easier to understand the measurement result as the sum of two opposite induced currents (but I have never been on really friendly terms with the EMF).

 

[vanhees71]

@vanhees71 the scalar potential is defined even in the case of $\nabla\times\vec{E}\neq\vec{0}$, simply by the solution to the wave equation (in the lorentz gauge)$$\nabla^2 V-\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}=-\frac{\rho}{\epsilon_0}$$ but of course as you notice the voltmeter doesn't measure that (it usually does but not in this weird setup).

That's of course true. I misunderstood the statement to mean a static potential such that $\vec{E}=-\vec{\nabla} V$, which is of course wrong in the general case. Here the relations of the em. field to the potentials is (in SI units)
$$\vec{E}=-\vec{\nabla} V-\partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
In the Lorenz gauge
$$\Box V=\frac{1}{\epsilon_0} \rho, \quad \Box \vec{A}=\mu_0 \vec{j},$$
where the d'Alembert operator is defined as
$$\Box=\frac{1}{c^2} \partial_t^2 -\Delta.$$

 

[Charles Link]

The voltmeter is perhaps best modeled as a large resistor, with the voltage reading given by the product of the current and the voltage, as the resistance approaches infinity. The solution is to write out the loop equations, how Professor Lewin does, making sure to include any EMF from the changing magnetic flux, if it occurs inside the loop.
The location of the resistor of the voltmeter, along with the connecting leads, will affect what the voltmeter reads, rather than being able to assign a reading by just considering the two points in the circuit where the leads are attached.

 

[alan123hk]

I would like to express my views on the issue of non-conservative electric fields and potential differences.

Although the induced electric field is non-conservative, but for the fixed path and direction in the electric field, such as R1 and R2 shown in the figure below, there can indeed be a potential difference, because their voltage will respectively vary with their resistance, and they consume energy. But, of course, the placement and direction of the two voltmeters and their leads may affect the measured value.

Consider again the transformer shown in the figure, the secondary output voltage is also produced by the non-conservative electric field induced by changing magnetic field, therefore, its corresponding line integral of the electric field should only be along the path of the secondary coiled wire. If the line integral follows other paths, of course there may be different results.

In addition, We can see that the placement and direction of the secondary external output circuit loop has no (or extremely little) impact on the output voltage. The reason is obvious, because for the same magnetic flux , it only has one turn, but the number of turns inside the transformer is usually much larger than one turn.

 

[Charles Link]

For an inductor or transformer coil, there are a couple of ways to look at it. The method that I prefer is to observe that the electric field in the conductor must be nearly zero, so there must necessarily be a conservative field $E_s$ (electrostatic) in the coil that is equal and opposite the $E_{induced}$. The $E_{induced}$ is zero outside the component, but we can write $\oint E_s \cdot ds=0$ over a complete loop, so that if outside the inductor there is just a resistor, the voltage drop across the resistor must be equal to $\int E_{induced} \cdot ds$.
It perhaps is left to the reader to decide how they like to treat EMF's, e.g. $\mathcal{E}=-L di/dt$. The $E_{induced}$ behaves very differently from $E_s$ in that it is non-conservative. One method is to simply say the sum of the EMF's in a circuit loop is equal to the sum of the voltage drops. Alternatively one can write $\oint E_s \cdot ds=0$, and compute the $E_s$ in the inductor as just mentioned.

 

[hutchphd]

Thank you for sharing. If it gives you any solace, according to Prof Lewin's discussion of his experiences, there are MIT professors who are also similarly confused. The physics has been explained. A potential difference between points cannot depend upon the path.

 

[Delta2]

E hehe its good to know that not only us, the common mortals, but also professors at big universities get confused by this...

 

[alan123hk]

I want to further describe the secondary output of the transformer I mentioned.

Electromotive force = The line integral of the electric field along the coil including the final loop of the load.

Total resistance =coil resistance + load resistance

Current = Electromotive force divided by total resistance

When the coil resistance is much less than the load resistance, because the voltage drop of the coil resistance is very small, the output is almost equal to the EMF. In this situation we may say that the output is only EMF, although from the point of view of load resistance, there is still a voltage drop across it.

On the other hand, when the coil resistance is much larger than the load resistance, the output becomes almost equal to zero. In this case, we have to admit that the coil EMF driven by the non-conservative electric field is subtracted by the voltage drop caused by its own resistance and therefore the final output voltage becomes almost zero.

Therefore, my own conclusion is that although this may seem confusing, it may simply be because everyone in different situations may have different perspectives and priorities when dealing with the same issue.

Difference Between Electromotive Force & Potential Difference : -
https://circuitglobe.com/difference-between-electromotive-force-and-potential-difference.html

 

[hutchphd]

Their is a certain schadenfreude inherent but it also slightly disconcerting. This is a fundamental albeit subtle mistake for EE professors to screw up. I cannot find the source where Lewin discusses this but I remember it distinctly and will post it if I do.

...Its in his lecture see below in post #36

 

[Charles Link]

See https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/
for a very detailed discussion on transformers and other related topics. It's a rather lengthy thread, but you might find it a good source for some of the basic concepts as well as some of the finer details.

 

[hutchphd]

Therefore, my own conclusion is that although this may seem confusing, it may simply be because everyone in different situations may have different perspectives and priorities when dealing with the same issue.

It is confusing, and that is why scientists agree on certain definitions. You can always think about electric fields as made of cotton candy...I could not care less. But you cannot define "the potential difference" in a non-conservative field. That definition is already taken for conservative fields. So when $\partial_t \vec B \neq 0$ we do not call it a voltage difference nor should you, regardless of your perspective.

 

[alan123hk]

It is confusing, and that is why scientists agree on certain definitions. You can always think about electric fields as made of cotton candy...I could not care less. But you cannot define "the potential difference" in a non-conservative field. That definition is already taken for conservative fields. So when ∂tB→≠0 we do not call it a voltage difference nor should you, regardless of your perspective.

OK, I understand and agree that I should not define "potential differences" in non-conservative fields. By definition there is no potential difference in non-conservative field.

But in actual transformer circuit engineering analysis, people usually call it "voltage drop". Of course, it does not mean that it is intended to say that there will be a potential difference in a non-conservative electric field. This may just be the name they use in circuit analysis, or just think of it as the voltage drop in the lumped circuit analysis equation.

Quote from https://www.electronics-tutorials.ws/transformer/voltage-regulation.html
" Thus voltage drops due to the windings internal resistance and its leakage reactance causes the output terminal voltage to change."

 

[Charles Link]

It might interest you that in measuring the voltage from an inductor, the result is $V=+\int E_{induced} \cdot ds$. If the source is purely electrostatic, the result is $V=-\int E_s \cdot ds$. (Notice the signs on these two integrals).
It is conceivable that what is being measured in most cases is the resulting electrostatic field that necessarily occurs which, as mentioned in post 25, is equal and opposite the $E_{induced}$ that is contained within the power source. The electrostatic field, being conservative, must have $\oint E_s \cdot ds=0$, and thereby we will get the same voltage for any path between the two terminals.
Professor Lewin's case (where you can get two different readings) is probably not applicable to most circuits that one encounters, but one lesson from it would be when taking a voltage reading (e.g. with an oscilloscope), be sure your leads aren't surrounding or near a current carrying line (which may be generating a changing magnetic field). To minimize stray signals, it can be helpful to have the two wires that are doing the measurement to be in the form of a coaxial cable, that has an outer conductive sleeve that is separated from the center conductor by an insulator. Coaxial cables are commonly used in the electronics lab when taking oscilloscope readings of a circuit, especially if the component being measured and the oscilloscope are not in close proximity.

 

[alan123hk]

I came up with another idea.
The situation is like this, the induced electric field will exert a force to the charge in the conductor, and then the current will flow through the circuit, and then according to Ohm’s law, a voltage will be generated across the resistor, so there are electric charges accumulation on both ends of the resistor. These charges accumulated in fixed positions will generate a conservative electric field, so actually there are two different electric fields superimposed. This statement seems more convincing and accurate, and doesn't contradict the definition. I personally think that it seems quite reasonable, at least for the time being.

 

[DaveE]

Their is a certain schadenfreude inherent but it also slightly disconcerting. This is a fundamental albeit subtle mistake for EE professors to screw up.

I can only speak for my experience with EE professors, but, I don't think they do screw this up, at least not at Caltech or Santa Clara Univ. What they do do is partition the problem into "circuit elements". Some like inductors and transformers have induced voltage, which is correctly explained, in my experience. Others have no significant flux linkage with changing fields. This is a nice way to model the world of circuits since, as a rule, this is how things are built (except, of course when they aren't). They know that a transformer winding can generate voltage in a magnitude and polarity different than you would expect from it's ohmic resistance, they know why, and they move on. Or they do detailed EM modeling and analysis of a distributed, perhaps complex structure. Any analog EE understands that to really model the workings of a transformer winding is a significant undertaking, that, frankly, you hope someone else did for you; or you just model the external salient features because that's all you needed.

Further, I think every good analog or RF EE knows about induced voltages in wires. [generic insult directed to digital EE's deleted for civility]. As others have said, you don't have to spend too much time using oscilloscope probes to realize this happens. They may say something about RFI or antennas, but, you know, "a rose by any other name...".

I think this thread is the dual case of "why can current flow through a capacitor when there aren't any conductors across it." Good EEs are comfortable with the physics of both idealized inductors and capacitors. Frankly I think Lewin does a disservice to Kirchhoff's laws which are only intended to be applied in lumped element circuits, and which I think all the good EEs understand. In the lumped element world, this problem is "what L, and C really are", not in the circuits, not in the induced voltage in loops. Yes, we know L⋅(di/dt) is an induced voltage, we know why and then we use it as a circuit element just like we use C⋅(dv/dt); works great, less filling.

Honestly, I don't think it's that complicated unless people can't keep their contexts straight.

 

[hutchphd]

Honestly, I don't think it's that complicated unless people can't keep their contexts straight.

Well said. I can say that one of my best lifelong friends who is a very good analog EE had to be brought kicking and screaming to this truth. He is not an RF guy though.
And Lewin very clearly talks about colleagues who accused him of chicanery in his lecture (see 48:25 et seq) which I, too, found surprising. I believe something akin to "politically correct speech" may actually be important here.

.

 

[Delta2]

I came up with another idea.
The situation is like this, the induced electric field will exert a force to the charge in the conductor, and then the current will flow through the circuit, and then according to Ohm’s law, a voltage will be generated across the resistor, so there are electric charges accumulation on both ends of the resistor. These charges accumulated in fixed positions will generate a conservative electric field, so actually there are two different electric fields superimposed. This statement seems more convincing and accurate, and doesn't contradict the definition. I personally think that it seems quite reasonable, at least for the time being.

I totally agree with this view. The electric field at all times (including the case where we have time varying magnetic flux) is simply $$E=-\nabla V-\frac{\partial A}{\partial t}$$

with the conservative component being the $-\nabla V$ and the non-conservative component $-\frac{\partial A}{\partial t}$ (A the magnetic vector potential).

V is the scalar potential here which obeys the equation of post #20 (which is the time dependent generalization of Poisson's equation) , which @hutchphd insists not to call it potential difference but only call it so when the aforementioned non-conservative component $-\frac{\partial A}{\partial t}$is zero i.e when the magnetic field is constant in time.

 

[hutchphd]

You may call it what you like. I won't even object to calling it a voltage (it has the appropriate units). I do object to calling it a potential difference because the potential, in fact, does not exist and so the difference is similarly illusory.

 

[Delta2]

because the potential, in fact, does not exist

I don't seem to get it, are you saying that V exists only in the static case (and DC steady state perhaps?)
Then what is the V in my post #37 and #20?

 

[etotheipi]

I don't seem to get it, are you saying that V exists only in the static case (and DC steady state perhaps?)
Then what is the V in my post #37 and #20?

I think, whenever $\partial_t \boldsymbol{B} \neq \boldsymbol{0}$, it's probably fine to call the scalar field $V$ in $\boldsymbol{E} = -\nabla V - \partial_t \boldsymbol{A}$ the potential so long as you made this terminology clear to the reader, and that the potential on its own isn't a meaningful thing anymore.

 

[Delta2]

Ok I think I see now thanks.
EDIT: I now get the meaning of the image of post #38, the word potential is overloaded here...

 

[Charles Link]

I think in the equation $E=-\nabla V-\frac{\partial{A}}{\partial{t}}$, the $-\nabla V=E_s$ is the electrostatic component that I previously mentioned, (posts 25 and 33), and $E_{induced}=-\frac{\partial{A}}{\partial{t}}$.
To apply it to the case of the voltage that we measure from an inductor or secondary coil of a transformer=outside of the component the magnetic field is essentially/(ideally) zero, and we then measure $V$ when we attach a large resistor (voltmeter/oscilloscope) across the component.
It becomes a more difficult problem when the magnetic field is changing across the components of interest. See https://www.physicsforums.com/threa...op-with-a-triangle.926206/page-8#post-6154120 We(a bunch of us) finally were all in agreement on the solution=see posts 192 to 197.

 

[DaveE]

Well said. I can say that one of my best lifelong friends who is a very good analog EE had to be brought kicking and screaming to this truth. He is not an RF guy though.
And Lewin very clearly talks about colleagues who accused him of chicanery in his lecture (see 48:25 et seq) which I, too, found surprising. I believe something akin to "politically correct speech" may actually be important here.

I think the basis of much of this confusion is that EEs and physicists don't speak the same language when they start drawing schematics. Lewin drew "lumped element" resistors connected with "wires". Then explains how those wires have induced voltages.

This can be confusing to EEs who assume that schematics are highly abstract descriptions of what the creator wanted to communicate about salient features of the circuit. In the EE mode, wires are always superconductors with no induced voltage, i.e. a topological connection between things that they think are worth paying attention to. To most EEs, if you think induced voltages are significant, then you need to draw an inductor to tell people "pay attention to magnetic effects". In some sense a schematic, to us, is a way of communicating all of the stuff that you need to pay attention AS WELL AS all of the complex stuff you are allowed to ignore (we tend to call these parasitic effects). My complaint with Lewin in these videos is that he was using mixed metaphors, if you will. If he had drawn this as a wire loop (perhaps with regions of different resistivity), and left out the "resistors", I think many EEs would "shift gears" and understand this is an EM problem, where Maxwell is more useful than Kirchhoff.

In defense of physicists, this abstract approach can often lead to oversimplification of real world behavior, and faulty assumptions about how the real world works. It is often the case that schematics can't really describe the true behavior of some analog/RF circuits. Problems also arise because in industry, schematics serve multiple roles: functional description for other engineers, data input to CAD SW, navigation data for people working with HW, data input to ERP/MIS systems (purchasing, etc.). IMO, there has never been a perfectly drawn schematic, ever; it's impossible.

 

[vanhees71]

I would like to express my views on the issue of non-conservative electric fields and potential differences.

Although the induced electric field is non-conservative, but for the fixed path and direction in the electric field, such as R1 and R2 shown in the figure below, there can indeed be a potential difference, because their voltage will respectively vary with their resistance, and they consume energy. But, of course, the placement and direction of the two voltmeters and their leads may affect the measured value.

View attachment 272957

Consider again the transformer shown in the figure, the secondary output voltage is also produced by the non-conservative electric field induced by changing magnetic field, therefore, its corresponding line integral of the electric field should only be along the path of the secondary coiled wire. If the line integral follows other paths, of course there may be different results.

In addition, We can see that the placement and direction of the secondary external output circuit loop has no (or extremely little) impact on the output voltage. The reason is obvious, because for the same magnetic flux , it only has one turn, but the number of turns inside the transformer is usually much larger than one turn.

This language is, from my experience, precisely what confuses students. If there is no potential, how can there be a potential difference? What we discuss here are precisely no potential differences but the non-vanishing "electromotive forces" (where "force" in this context is not force in the Newtonian sense but means an energy-like quantity). It's precisely NOT a potential difference, because it's the line integral of a vector field along a CLOSED loop. If the vector field were conservative, i.e., would have a potential, this EMF would be 0. Then and only then is the potential difference given by line integrals along an arbitrary open path connecting the two points where the potential difference is taken, and then and only then is this potential difference independent of the path chosen to connect the two points.

 

[Charles Link]

The inductor and the secondary coil of a transformer that delivers the "voltage" to a resistive load is a problem that involves a changing magnetic field, but upon looking at this problem a few times over the last couple of years, I'm starting to believe more and more that what is measured at the output terminals is indeed a potential difference which is just the integral of the electrostatic field. The induced electric field $E_{induced}$ is most often contained within the voltage source. Whether the EMF is generated by a changing magnetic field, or is the result of a battery, or solar cell, the net result is that it behaves as if it were an electrostatic source, e.g. like that of a Van de Graaff generator. $\\$

For the inductor and transformer,we must have that $E_{total} \approx 0$ inside the coil, so that an electrostatic $E_s=-E_{induced}$ must be present, where $E_{total}=E_{induced}+E_s$. Since the electrostatic field is conservative, i.e. $\oint E_s \cdot ds =0$ for any and all closed paths, the integral $V=\int E_s \cdot ds$ is well defined outside the device, (it's absolute value is equal to $\mathcal{E}=\int E_{induced} \cdot ds$ inside the device), and is the potential difference.

For the case of the battery, the EMF is not from a changing magnetic field, but rather a voltage created by a chemical reaction, but the same conservative electrostatic field rules apply, external to the device.
It is this electrostatic field, i.e. $\int E_s \cdot ds$ that is measured by a voltmeter or oscilloscope, which can be modeled as a large resistor, with a small current passing through it. The voltage is the product of the resistance and the current.

Taking Professor Lewin's "paradox" into account, it would appear that with an oscilloscope or voltmeter we could get slightly different readings from an inductor type voltage source, depending on where we put the lead wires, but for a toroidal type transformer or similar geometry, the lead wires generally wouldn't encompass a region of changing magnetic flux, and the readings would be unambiguous. These readings, with minimal induced electric field component, would be readings of the electrostatic field, i.e. $V=\int E_s \cdot ds$ which is also equal in absolute value to the EMF $\mathcal{E}= \int E_{induced} \cdot ds$ inside the component.

 

[alan123hk]

This language is, from my experience, precisely what confuses students. If there is no potential, how can there be a potential difference? What we discuss here are precisely no potential differences but the non-vanishing "electromotive forces" (where "force" in this context is not force in the Newtonian sense but means an energy-like quantity). It's precisely NOT a potential difference, because it's the line integral of a vector field along a CLOSED loop. If the vector field were conservative, i.e., would have a potential, this EMF would be 0. Then and only then is the potential difference given by line integrals along an arbitrary open path connecting the two points where the potential difference is taken, and then and only then is this potential difference independent of the path chosen to connect the two points

I understand what you mean, according to definition and theory, it is true.

Therefore, I think we should probably not say that the output of the transformer is a potential difference.
For example, It is not quite correct to say "..potential difference induced across the secondary winding.." in this education material https://phys.libretexts.org/Bookshelves/University_Physics/Book:_University_Physics_(OpenStax)/Map:_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/15:_Alternating-Current_Circuits/15.07:_Transformers

Even if the load is far away from the transformer and will not be interfered by the electromagnetic field of the transformer, it is always the one turn of the output winding of the transformer, and the close line integral of this turn along with the output winding is still not zero.

In addition, all AC circuit should not have a potential difference, because it involves an varying electromagnetic field. So many people like to say "AC potential difference" is not very correct.

I am not arguing anything about the definition and theory, I just think that they may be applying an equivalent model to solve practical problem under certain conditions, the model may not be perfect, but for them, this method is very useful and reasonable.

 

[Charles Link]

See https://www.physicsforums.com/threa...op-with-a-triangle.926206/page-8#post-6154120 especially posts 192-197 and the solution by @cnh1995 (post 192), and the subsequent posts explaining the solution.

When we do encounter a problem where the magnetic flux is changing inside the circuit loops, so that EMF's need to be included, it is apparent from this problem that there is still a (electrostatic) potential difference between A and B, given by $V_{AB}=\int\limits_{A}^{B} E_s \cdot ds$.

It is perhaps a very good thing that Professor Lewin has attracted some attention to this problem, but it looks as if we now have satisfactory solutions to it, and a good interpretation of it, along with suitable definitions to resolve the "paradox".

 

[vanhees71]

Yes, BUT $V_{AB}$ depends on the path you use to connect $A$ and $B$, i.e., you have to write
$$V_{AB}=\int_{C} \mathrm{d} \vec{r} \cdot \vec{E}.$$
It depends on the path, because if you have two different paths $C_1$ and $C_2$, both connecting $A$ with $B$ you get, for the closed path $C'=C_1-C_2$ (i.e., you go from $A$ to $B$ along path $C_1$ and then back from $B$ to $A$ along path $C_2$),
$$\int_{C'} \mathrm{d} \vec{r} \cdot \vec{E} = \int_{C_1} \mathrm{d} \vec{r} \cdot \vec{E} - \int_{C_2} \mathrm{d}\vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B} \neq 0.$$

 

[Charles Link]

Yes, BUT $V_{AB}$ depends on the path you use to connect $A$ and $B$, i.e., you have to write
$$V_{AB}=\int_{C} \mathrm{d} \vec{r} \cdot \vec{E}.$$
It depends on the path, because if you have two different paths $C_1$ and $C_2$, both connecting $A$ with $B$ you get, for the closed path $C'=C_1-C_2$ (i.e., you go from $A$ to $B$ along path $C_1$ and then back from $B$ to $A$ along path $C_2$),
$$\int_{C'} \mathrm{d} \vec{r} \cdot \vec{E} = \int_{C_1} \mathrm{d} \vec{r} \cdot \vec{E} - \int_{C_2} \mathrm{d}\vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B} \neq 0.$$

In the notation I used, $E_s$ is the electrostatic component of the electric field. This integral will be path independent. If you study the solution by @cnh1995 in detail,(post 192 of the "link"), I think you will find it to be a very good one. (To add more detail, I solved the problem for the currents using loop equations, essentially 6 equations and 6 unknowns. By introducing the electrostatic integrals, e.g $V_{AB}$, @cnh1995 came up with a much simpler solution, the details of which I explained in post 193).

See also post 42. I think we are justified in making a distinction between the electrostatic $E_s$ and the induced $E_{induced}$, with the result that $E_{total}=E_s+E_{induced}$.

 

[alan123hk]

In short, there is charge accumulation at both ends of the output port of the inductor or transformer, thereby forming a voltage. In the actual working environment, if there is no changing magnetic field in the external space, for example, there is no leakage flux from the inductor or transformer, the output voltage is basically the same no matter where the external connection wires and load are placed. In this case, as long as we know the reasons and the constraints behind it, then I guess the voltage can be approximated as a potential difference, but of course there are exceptions, that is, we cannot cross the two output connection wires and form another closed loop around the magnetic field of the inductor or transformer, because according to Faraday's Law, there will be an induced voltage, and then the output voltage will change.