Faraday's law -- circular loop with a triangle

[rude man]

1. what is the "Lewin setup"? There are appareently several Lewin lectures and setups existent.
2.

Well, the total E-field can be conservative in some regions while non-conservative in other regions. For example, think about the Lewin setup, but only look at the right part of the circuit. Imagine a vertical wall between the inductor and the 900Ω resistor.

could you reference the "Lewin setup"? I think there are several Lewin setups existent.

Since the field is conservative in the whole region, there is a well-defined scalar potential, and you can easily measure voltages using a voltmeter, for example the voltage over the resistor is +0.9V, and the voltage over every part of the conductor is 0V (assuming ideal conductors).

The voltage in the wires is not zero. It is the net E field that is zero in the wires. There are equal and opposite E fields within the wires: E = E_s - E_m = 0.

Further, it is not easy to read voltages around the circuit with a voltmeter. A voltmeter loop outside the B field will read zero in a wire not because the voltage is zero but because the voltmeter introduces an E_s field on its own. Did you see the Mabilde video? Although his measurement setup only indirectly measures wire voltages, the data is correct. Lewin's statement that "Kirchhoff is wrong" is wrong. The circulation of E_s is always zero. The circulation of E_m = the applied emf. This can be a battery, a Faraday emf, a Seebeck effect device or any other source of emf.

The voltage across a pure inductor is the line integral of the axial E_s field which is why you can read the voltage across an inductor. An equal and opposite E_m field cancels the E_s field so the net E field is zero.

So, what I am trying to say is that as long as the total E field is conservative you always have a well-defined total potential, and thus a well-defined voltage

you are absolutely correct.

 

[rude man]

actually i am coming here from there

why does the voltmeter measure Es not the total E field as you said in the article

Afterthought: I think I have a better answer for you than just "it's by definition":
An emf-generated field has curl: ∇xE_m ≠ 0.
An electrostatic field does not: ∇xE_s = 0.
You will recall that a field with curl cannot have a potential V (a purely mathematical fact: ∇x (∇V) = 0).
Therefore, an E_m field cannot have a potential, i.e. a voltage V.

 

[timetraveller123]

no i know that
it is possible to know which is Em and which is Es when you look at one entire loop
but if you look at a open curve like,voltmeter connected between two points, how is it supposed to differentiate the two fields
or does this have to do with the inner workings of the voltmeter ?

 

[Stefan Gustafsson]

no i know that
it is possible to know which is Em and which is Es when you look at one entire loop
but if you look at a open curve like,voltmeter connected between two points, how is it supposed to differentiate the two fields
or does this have to do with the inner workings of the voltmeter ?

I disagree with @rude man .

A voltmeter really measures the current that passes through the voltmeter. Since the internal resistance of the voltmeter is known (very high) a voltage can be calculated using Ohms law, V = R*I

The voltmeter does not know anything about a separation between E_m and E_s.

To calculate the voltage displayed by a voltmeter you can always use Faraday's law.

Normally the voltmeter will display the voltage between the ends of the measuring probes, but if there is a changing magnetic flux in the loop formed by the measuring wires and the measured object there is also an induced emf that you need to account for.

 

[rude man]

no i know that
it is possible to know which is Em and which is Es when you look at one entire loop
but if you look at a open curve like,voltmeter connected between two points, how is it supposed to differentiate the two fields
or does this have to do with the inner workings of the voltmeter ?

∇xE doesn't apply just to a closed loop. The curl exists (or doesn't exist) at every point. Forget closed loops for a minute. The potential difference between any two points is the line integral of E_s between the two points. The curve connecting the two points is immaterial. That's why it's called a 'conservative' field. We would like the voltmeter to read the potential difference between any two points.

In the case of the battery-resistor circuit there is no problem: the voltmeter will read the potential difference between any two points along the circuit including across the battery because remember there are two E fields inside the battery and the voltmeter reads line-integrated E_s only. Check my paper if you need to.

When there is magnetic induction there's a problem because the voltmeter circuit forms a section of an alternate closed loop. Take a one-turn coil of resistance R and radius a. You get pure E_m field in the coil by Maxwell's ∇xE_m = -∂B/∂t. No E_s, no potential drop anywhere. Accordingly, we might expect a voltmeter, if connected, to read zero. But when the voltmeter is connected between two points along the coil the voltmeter, forming an alternate magnetic closed loop, will read a finite voltage. If the voltmeter circuit is outside the B field that voltage will be emf*s/2πa with s the length of the coil between the two points.

So bottom line, in the magnetic induction case there is no potential difference anywhere but the voltmeter will nonetheless read a potential difference since it is included in the magnetic loop.

PS you will have to decide between myself and @Stefan Gustafsson.

 

[rude man]

I disagree with @rude man .

A voltmeter really measures the current that passes through the voltmeter. Since the internal resistance of the voltmeter is known (very high) a voltage can be calculated using Ohms law, V = R*I

correct

The voltmeter does not know anything about a separation between E_m and E_s.

Be careful where the resistor is located. If it's in series with the coil then you are right. V = iR and i = (E_m + E_s)/ρ where ρ is the resistance per unit length . But the resistor here is in parallel with a section of the coil so now the statement is wrong. There is no potential drop across the coil segment but the voltmeter reads a voltage because it forms an alternate magnetic closed loop.

To calculate the voltage displayed by a voltmeter you can always use Faraday's law.

This is true by coincidence. In reality there is no potential drop across the coil segment but because the voltmeter circuit forms an alternate magnetic closed loop that voltage is displayed. Coincidentally the voltmeter would read the same voltage as if the coil field were E_s without a dB/dt field, for example in the battery-resistor circuit.

Normally the voltmeter will display the voltage between the ends of the measuring probes, but if there is a changing magnetic flux in the loop formed by the measuring wires and the measured object there is also an induced emf that you need to account for.

That statement is correct but a voltmeter magnetic loop cannot be avoided. So you read what you think is the voltage but it's not, it's the line integral of the E_s field in the resistor which is generated by the induced emf which forces an E_s field in the voltmeter.
(The line integral of E_m over the length of the resistor is negligible since we assume a very short resistor.)

 

[Stefan Gustafsson]

Be careful where the resistor is located. If it's in series with the coil then you are right. V = iR and i = (E_m + E_s)/ρ where ρ is the resistance per unit length . But the resistor here is in parallel with a section of the coil so now the statement is wrong.

I am not sure exactly what circuit you are referring to here. Could you perhaps post a simple image of what you are talking about?

 

[rude man]

I am not sure exactly what circuit you are referring to here. Could you perhaps post a simple image of what you are talking about?

There is a simple one-turn coil of resistance r with a symmetric circular dB/dt field inside the coil, and you connect the voltmeter across a segment of this simple coil. You will measure a voltage only because the voltmeter wiring presents an alternative segment to the coil segment a-b. Look at my Insight paper figure 3 but let R1 = R2 = 0 and the ideal wire is replaced by a wire of finite resistance r.

 

[Stefan Gustafsson]

There is a simple one-turn coil of resistance r with a symmetric circular dB/dt field inside the coil, and you connect the voltmeter across a segment of this simple coil. You will measure a voltage only because the voltmeter wiring presents an alternative segment to the coil segment a-b. Look at my Insight paper figure 3 but let R1 = R2 = 0 and the ideal wire is replaced by a wire of finite resistance r.

Lets look at another problem for a second. Answering this should not require much effort.

The black line is a wire with resistance 1 ohm/cm
B is a linearly changing magnetic field that induces a current of 1A in the wire.
Outside the shaded region there is no changing magnetic field
L=2cm
r1=2cm
r2=4cm
V1 and V2 are voltmeters with very high resistance

1) What is the voltage between A and B?
2) What is the voltage between B and C?
3) What do the voltmeters V1 and V2 show?
4) What is ∇xE in the region X?
5) Is the E-field in region X conservative or non-nonservative?

 

[rude man]

Lets look at another problem for a second. Answering this should not require much effort.

I can answer two of your questions immediately:

4) What is ∇xE in the region X?

Obviously zero since there is no B field anywhere within X. This of course does not mean E = 0.

5) Is the E-field in region X conservative or non-nonservative?

Conservative, since ∇xE = 0.
LATE EDIT: sorry, this statement is not necessarily correct. The reason the field is conservative is that there is no net curl within any closed loop in the region.

BTW we assume self-generated B field << externally applied B field.

First, E_m + E_s = iR/L where R is the resistance of the segment A-B and i = current
.
Second, I think E_s = 0 throughout the ring since the ring assumes uniform resistance.

So E_m is uniform throughout the loop and the voltage between A and B is zero.

E_m = current x total loop resistance/total loop length. All E_m firelds are assumed > 0 and clockwise; all E_s fields are assumed counterclockwise and > 0. So current i is also clockwise.

However, the voltmeter does not read zero.

E_m is constant throughout the basic loop. However, we note that the loop can also be closed via the voltmeter circuit: A → voltmeter + wires → B → A instead of A → L → B → A.

So E_mL = E_mwl_w where E_mw is the emf field in the wires, E_sw is the static field in the wires, and l_w is the total length of the wires.
In the wires, E_mw = E_sw since E = 0 in the wires.

And E_swl_w = V = voltmeter reading.
Therefore, V = E_mL = iRL/L = iR, R = reistance of segment A-B. A is + with respect to B.

Now: you will say "that's what I said! It's iR!"
BUT - the voltage reading is NOT the potential difference between A and B. It is only by virtue of the fact that the voltmeter + wires offer an alternative path to your A-B segment in the magnetic closed loop and so induces an emf in the wires and also a potential in the wires and in the voltmeter resistor. The voltmeter reads ONLY E_swl_w.

This point is well described in the two K. McDonald papers I cited in my paper.

Sorry, I forgot to address V reading between B and C but the idea is the same and so is the result.

 

[Stefan Gustafsson]

I can answer two of your questions immediately:
Obviously zero since there is no B field anywhere within X. This of course does not mean E = 0.Conservative, since ∇xE = 0.

Great! At least we agree on this.

I will have more comments about this tomorrow - it is getting late here in Sweden now.

But in your answer you did not mention anything about the segment B-C.
Could you please tell me what voltage and voltmeter reading you expect there?

Also, one additional question:

6) Do you expect that connecting the voltmeters will change anything about the original circuit? Any change in voltages or currents?

 

[rude man]

Great! At least we agree on this.

I will have more comments about this tomorrow - it is getting late here in Sweden now.

Godnatt! (I was also born in Sweden! But never a Swedish citizen.)

But in your answer you did not mention anything about the segment B-C.
Could you please tell me what voltage and voltmeter reading you expect there?

I think it's the same thing but I will look at it more closely later.

Also, one additional question:

6) Do you expect that connecting the voltmeters will change anything about the original circuit? Any change in voltages or currents?

No. The voltmeter assumes infinite resistance.

 

[Stefan Gustafsson]

BUT - the voltage reading is NOT the potential difference between A and B. It is only by virtue of the fact that the voltmeter + wires offer an alternative path to your A-B segment in the magnetic closed loop and so induces an emf in the wires and also a potential in the wires and in the voltmeter resistor. The voltmeter reads ONLY E_swl_w.

I think we agree that both voltmeters will show 2V. (1A * 2 Ω)
A much easier explanation for this fact is to simply use Faraday's Law.
$$ \oint \vec E d \vec l = -{d \phi \over dt} $$
Note that $\vec E$ is the TOTAL field.
If we apply this for example to the loop A -> V1 -> B -> A , the right part is 0 because there is no flux change in the loop.
FL: R*I - V = 0, or V = R*I
Exactly the same calculation works for V2 as well.

There is no doubt that this is correct - it follows directly from Ohm's law and Faraday's law.

The thing we disagree on seems to be the definition of voltage and potential.
The normal definition of the voltage between two points a and b is the work per unit charge required to move a small test charge from a to b. Since the field in region X is conservative this is simply the line integral from a to b:
$$ V_{ab} = \oint_a^b \vec E d \vec l $$

You have your own definition of voltage, and I still don't really understand where you get it from.

This point is well described in the two K. McDonald papers I cited in my paper.

I disagree, I don't think it is well described at all. I am also not sure that McDonalds definition of voltage matches your definition.

In any case, as long as the electric field is conservative, I see no reason to use anything else than the original classic definition of voltage.

 

[rude man]

$$ \oint \vec E d \vec l = -{d \phi \over dt} $$

correct

Note that $\vec E$ is the TOTAL field.
In any case, as long as the electric field is conservative, I see no reason to use anything else than the original classic definition of voltage.

Me neither. But sometimes the field is non-conservative. Then there is no definition of voltage, not 'original classical' nor any other.
There is then in fact no voltage at all.
If a field consists of emf as well as conservative E fields then the voltage is the integral of just the conservative part. That is all I ever said.

 

[rude man]

Perhaps you may wish to respond to the attached pdf file.
The OP should also think about it. I have the answers but will withhold them until a response is received.

 

[timetraveller123]

Second, I think Es = 0 throughout the ring since the ring assumes uniform resistance.

i think i disagree with this
firstly if it were just the $E_m$ in action then there would be no electric field in the horizontal segments and furthermore the currents in each of the semicircles would be different given by $\frac{k r_i}{4 \rho}$ k is the rate of change of magnetic field
my guess for the configuration of the charge build up is A has small amount of positive charge and B has larger amount of negative charge while the other inner corner has small amount of negative charge and the other outer corner has large amount of positive charge. i am just guessing .

So Em is uniform throughout the loop and the voltage between A and B is zero.

so i think voltage between A and B is not zero

i am not sure though but i think there might Es in the setup

 

[timetraveller123]

so let the static fields in upper circle be $E_{su}$ in segment AB be $E_{sh}$ and in lower circle be $E_{sl}$
right off the bat we can see that in segment ab the only field is $E_{s2}$ hence
$E_{sh}L = I \rho L$
if the left hand side is the voltage then the right hand side is the answer.

for voltage in bc

lower circle there are two opposing fields
$(\frac{k r_2}{4 \pi} - E_{sl}) \pi r_2 = I \pi r_2 \rho$
upper circle there are two supporting fields
$(\frac {k r_1}{4 \pi}+E_{su})\pi r_1 = I \pi r_1 \rho$
we need a third equation relating $E_{su}$ and $E_{sl}$
this can be obtained from the condition that the static fields are curless hence going in loop should give zero.
hence third equation

$2 E_{sh}L + E_{su}\pi r_1 - E_{sl}\pi r_2= 0$
we know $E_{sh}$ from previous part hence the three equations could be solved to get $E_{sl}$ from voltage bc can be gotten as $L E_{sl}$

 

[timetraveller123]

then as to the pdf file that rude man attached i think the solution follows in the same way as there will definitely be build up of charges at the junctions hence the solution follows in almost exact manner

$\frac{emf}{4r}+\frac{v}{2r} = I\\ \frac{emf}{2r} - \frac{v}{r}=I\\ v = \frac{emf}{6}$
as to what the voltmeters read in both the questions i am not sure voltmeters are screwing with my brain

 

[rude man]

[QUOTE="timetraveller123, post: 6104052, member: 611109"[/quote]

One thing though: the configurfation's poster and I did get the same value of voltmeter reading V_AB. Don't know if that proves anything ...

so i think voltage between A and B is not zero
i am not sure though but i think there might Es in the setup

Well, by my assumption of pure E_m in the configuration I determined that there is still E_s in the voltmeter circuit giving a finite voltage for V_AB which agreed with the poster's and I think yours.

 

[timetraveller123]

are you saying that you know the Em field at each point in the configuration by using Faraday's law as imaginary circular fields at each point in the configuration centered at the center of the B field,

yes the em has no regard for the actual setup
this also why the horizontal segments has no em
as it is pointing radially away(i am assuming)

 

[rude man]

yes the em has no regard for the actual setup
this also why the horizontal segments has no em
as it is pointing radially away(i am assuming)

Thanks, that's great, I suspected this might be a valid assumption, should have invoked it here.

 

[rude man]

then as to the pdf file that rude man attached i think the solution follows in the same way as there will definitely be build up of charges at the junctions hence the solution follows in almost exact manner

$\frac{emf}{4r}+\frac{v}{2r} = I\\ \frac{emf}{2r} - \frac{v}{r}=I\\ v = \frac{emf}{6}$
as to what the voltmeters read in both the questions i am not sure voltmeters are screwing with my brain

Exactly what I got for Vab. Your brain is working just fine.
V1 and V2 turn out to be just current times resistance. As I'm sure you'll discover also.

 

[Stefan Gustafsson]

Perhaps you may wish to respond to the attached pdf file.
The OP should also think about it. I have the answers but will withhold them until a response is received.

What do voltmeters VM1 and VM2 read?

VM1 = -i*r
VM2 = i*2r

Easy application of Faraday's Law and Ohm's Law.

What is Va – Vb? There can only be one answer!

Voltage and potential is not really defined in a non-conservative field so there is no one single correct answer to this question.

But I think the most sensible answer is that the voltage is path dependent.

Go back to the real definition of voltage: "The voltage between two points a and b is the work per unit charge required to move a small test charge from a to b"
The work going from b to a is different depending on if you go to the left or to the right of the magnetic field.
Note that since there is no B field outside the yellow region, the actual path does not matter - the only difference is if you go to the right or to the left.

So, the voltage is either -i*r if you go on the left side, or i*2r if you go on the right side.

Think of this like walking in a hurricane.

There is a wind blowing clockwise around the circuit. If you walk on the left you have the wind in your back so the work required is negative.
If you walk to the right you have the wind in your face so you have to work really hard to get to the same point.

 

[timetraveller123]

Voltage and potential is not really defined in a non-conservative field so there is no one single correct answer to this question.

yes it is true but in this case i believe rudeman is talking about the voltage and potential associated with the developed static field

 

[rude man]

yes it is true but in this case i believe rudeman is talking about the voltage and potential associated with the developed static field

Yes, since this is the only sensible definition of potential. It is the line integral of E_s. Nothing else. This is also per K. McDonald (see my citation in same blog. He calls it the scalar potential.)

Another example would be fig 3 in my Lewin Insight blog. A voltmeter would read zero but there is a potential difference between a and b even though there's nothing but zero-resistance wire between those points.

Consider the absurdity of saying there is no definable potential difference between two points with both E_m and E_s fields. So now let's see - at what point does a tiny amount of E_m negate the existence of a unique potential difference? E_m < 10^-6 E_s? E_m < 10^-100 E_s?
See my point?

 

[Stefan Gustafsson]

Yes, since this is the only sensible definition of potential. It is the line integral of E_s. Nothing else.

This is just your opinion.

My opinion is that voltage and potential are really undefined in a non-conservative field, but the most reasonable definition of the voltage between two points is the line integral of E_total.

I also believe that this is the definition used by for example Walter Lewin and Robert Romer
http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf
You should study the Romer Paper - it is very clear on these topics.

This is also per K. McDonald (see my citation in same blog. He calls it the scalar potential.)

I am not so sure that you and McDonald are saying the same thing. He is not talking about E_s and E_m
In any case I do not agree.

Another example would be fig 3 in my Lewin Insight blog. A voltmeter would read zero but there is a potential difference between a and b even though there's nothing but zero-resistance wire between those points.

All this falls back to your own definition of potential and voltage.

With my definition the voltage between these points is really zero.

Consider the absurdity of saying there is no definable potential difference between two points with both E_m and E_s fields. So now let's see - at what point does a tiny amount of E_m negate the existence of a unique potential difference? E_m < 10^-6 E_s? E_m < 10^-100 E_s?
See my point?

No absurdity at all. If you have a tiny E_m the differences we are talking about is very small. It does not matter if we use your definition, the McDonald definition or my definition.

But think about my example. Assume that you are confined to region X - there is a wall separating yourself from the magnetic field. You have no way to know that there is a changing magnetic field nearby. All you know is that there is a conservative electric field in the whole region you can access.

The real absurdity here is why you would say that the voltage A-B and B-C would be anything else than the voltmeters show.

 

[timetraveller123]

My opinion is that voltage and potential are really undefined in a non-conservative field, but the most reasonable definition of the voltage between two points is the line integral of Etotal.

i don't quite get your point here
you do realize that in the problem that rudeman suggested(also in the one you suggested) there is conservative field and a non conservative one. Just because you can't associate a voltage and potential to a non conservative one doesn't mean you can do the same with a conservative(static) field. it is these values that he was asking
and most often

line integral of Etotal.

this refers to emf and not voltage
https://en.wikipedia.org/wiki/Electromotive_force

I also believe that this is the definition used by for example Walter Lewin and Robert Romer

i also don't see where such a definition was used

But think about my example. Assume that you are confined to region X - there is a wall separating yourself from the magnetic field. You have no way to know that there is a changing magnetic field nearby. All you know is that there is a conservative electric field in the whole region you can access.

for all you know you just experience a field you can even say it is conservative

 

[Stefan Gustafsson]

i don't quite get your point here
you do realize that in the problem that rudeman suggested(also in the one you suggested) there is conservative field and a non conservative one. Just because you can't associate a voltage and potential to a non conservative one doesn't mean you can do the same with a conservative(static) field. it is these values that he was asking

I am saying that there is no point in separating the E field in a E_m and E_s. The only thing that counts is the total field.
If the total field is conservative in a region, then there is a well-defined voltage and potential.
If the total field is non-conservative then the voltage and potential is not well-defined

this refers to emf and not voltage
https://en.wikipedia.org/wiki/Electromotive_force

https://en.wikipedia.org/wiki/Voltage

"The difference in electric potential between two points (i.e., voltage) in a static electric field is defined as the work needed per unit of charge to move a test charge between the two points"

And I know that this only really applies to conservative electric fields.

for all you know you just experience a field you can even say it is conservative

Exactly - if you stay in region X all you know is that the total field is conservative everywhere - easily proved by Faraday's law since there is no changing magnetic field in the region.

You do not know anything about how to split E_total in E_s and E_m

 

[timetraveller123]

I am saying that there is no point in separating the E field in a Em and Es. The only thing that counts is the total field.

the point i am trying to make is that the point of splitting the fields is to find the electrostatic voltage
so what was your method for the problem you suggested

 

[Stefan Gustafsson]

the point i am trying to make is that the point of splitting the fields is to find the electrostatic voltage
so what was your method for the problem you suggested

As long as you stay in a region with a conservative total field (Like region X in my problem), there is no question: You find the voltage between two points by using the line integral of E_total*dl between the two points.

This gives exactly the same result as using a voltmeter and it is consistent with ohms Law: V = I*R

There is absolutely no need to split E_total in E_s and E_m

 

[timetraveller123]

ok then we clearly have different definition of voltage
ok then how would you find the electrostatic potential difference

 

[timetraveller123]

furthermore

As long as you stay in a region with a conservative total field (Like region X in my problem), there is no question: You find the voltage between two points by using the line integral of Etotal*dl between the two points.

just because in region x curl is zero doesn't make it conservative
infact this exact thing is stated in the very article you mentioned

 

[Stefan Gustafsson]

ok then we clearly have different definition of voltage
ok then how would you find the electrostatic potential difference

I actually don't care about the electrostatic potential difference. It is purely a theoretical concept - it is not something you can measure. You have to calculate it by splitting the total electric field in E_m and E_s which is often very complicated.

 

[timetraveller123]

yes but that's the one thing that has one unique value in fact the voltage you said is ir but if you took different paths between a and b or b and c in the very circuit you suggested you would get different answer yes i agree it is a very theoretical concept

 

[Stefan Gustafsson]

furthermore

just because in region x curl is zero doesn't make it conservative
infact this exact thing is stated in the very article you mentioned
View attachment 235710

Yes, but when he is talking about region II he is talking about the full region outside the magnetic field. This region has a hole in the middle. As he says in the paper this region is not simply connected.

My region X is simply connected - the total E field in region X is conservative everywhere. There is a well-defined potential inside region X. No matter how you measure the voltage between two points in region X you will always get the same value.

This is not true for Romer's region II - if you wrap your leads around the hole in the middle you can get infinitely many different voltages depending on the number of turns and the winding direction.

 

[Stefan Gustafsson]

yes but that's the one thing that has one unique value in fact the voltage you said is ir but if you took different paths between a and b or b and c in the very circuit you suggested you would get different answer

As long as you stay in region X the voltage Vab and Vbc is path independent.

 

[timetraveller123]

oh i am sorry all along the picture at the back of my mind was x is everything outside like region 2

but anyways i think then the answers are same

but the main question for which i opened this thread i believe was asking for the electrostatic potential albeit a very theoretical concept

 

[Stefan Gustafsson]

but the main question for which i opened this thread i believe was asking for the electrostatic potential albeit a very theoretical concept

Well the original question asked for "Find UAB, the potential difference between points A and B".
It never says that they are asking about the "electrostatic potential"

Since "potential difference" is not clearly defined in a non-conservative field (which you have in the original question) you need to add your own interpretation somehow.

I have no idea what the original question was looking for, but my approach to solving this would have been to calculate the current in the straight line from A to B and then multiplying by r2 to get a voltage.

This may or may not be what the original question was after. I don't know.

 

[timetraveller123]

but wouldn't different paths give different results

 

[Stefan Gustafsson]

but wouldn't different paths give different results

Yes, that is exactly why "potential difference" is not clearly defined in this situation.

 

[timetraveller123]

yes that's why i assumed they were asking about electrostatic potential because after all they could have only set one answer

 

[rude man]

yes that's why i assumed they were asking about electrostatic potential because after all they could have only set one answer

Thx again for pointing out that the Em field can readily be determined along the entire contour of post 109.
So for example the E field in the straight sections is purely Es and the potential difference between A and B is zero. And for the circular sections we have a mixture of Es and Em with Es derivable by solving Em + Es = i dr/dl, Em, i and dr/dl known (i = current and dr/dl = resistance per unit length.)

 

[rude man]

r yes i agree it is a very theoretical concept

Very theoretical? Re: my post 115 pdf slide:

Hang a voltmeter so that it is suspended above the top of the coil at the half-way point with the leads going straight down to points a and b. What is the reading now? emf/6 maybe? The "theoretical" potential difference as computed by line-integrating Es and Es only?

And: I may have mentioned this before, but:

Kirchhoff said the sum of potential drops around a circuit = 0. If we consider each drop as the integral of Es only then his statement also applies to any closed loop in the sense of line-integrated Es fields.. Therefore. the line integral of Es must be the potential difference along each segment of any circuit.

Dr. Lewin's statement that "Kirchhoff was wrong" in the video with the 100 and 900 ohm resistors was absolutely incorrect.

 

[Stefan Gustafsson]

Very theoretical? Re: my post 115 pdf slide:

Hang a voltmeter so that it is suspended above the top of the coil at the half-way point with the leads going straight down to points a and b. What is the reading now? emf/6 maybe? The "theoretical" potential difference as computed by line-integrating Es and Es only?

The voltage you are measuring in this case is just the line integral of the total E-field in a straight line from a to b. There is nothing special about this voltage - voltage in a non-conservative field is path dependent.

And: I may have mentioned this before, but:

Kirchhoff said the sum of potential drops around a circuit = 0. If we consider each drop as the integral of Es only then his statement also applies to any closed loop in the sense of line-integrated Es fields.. Therefore. the line integral of Es must be the potential difference along each segment of any circuit.

Dr. Lewin's statement that "Kirchhoff was wrong" in the video with the 100 and 900 ohm resistors was absolutely incorrect.

The discussion about Kirchhoff being wrong or not is pointless. It all depends on which "law" you want to call Kirchhoffs Law. Also, Lewin does not say that Kirchhoff was wrong. There is nothing wrong with KVL as long as you apply it in situations where it is valid. The experiment is designed to demonstrate a situation where KVL does not apply and you have to use Faraday's law instead.

In any case all of this discussion revolves around the definition of "voltage" and "potential" in a non-conservative field.

As I have said before these concepts are not really defined in non-conservative fields, so to talk about voltages you need to extend the usual definition somehow.

You want to extend it in your own way by saying that voltage is the line integral of Es only. This has the advantage of making the voltage path independent, but the disadvantage of creating a definition of voltage that can not be measured with a voltmeter. It also has the disadvantage that you now have to define what Es means. How do you calculate Es in a practical case, where all you really know is the total E-field?

I want to extend it by saying that voltage is the line integral of the total E-field. This has the advantage that it matches the readings of a voltmeter, but the disadvantage that voltage becomes path dependent.

None of the extensions is clearly "correct". I suppose they can both have value, but I clearly prefer my definition.

By the way, did you ever come to a conclusion about the answer to the voltages A-B and B-C in my problem in post 109?

Also, do you have any comments to this part of my post 126?

But think about my example. Assume that you are confined to region X - there is a wall separating yourself from the magnetic field. You have no way to know that there is a changing magnetic field nearby. All you know is that there is a conservative electric field in the whole region you can access.

The real absurdity here is why you would say that the voltage A-B and B-C would be anything else than the voltmeters show.

 

[rude man]

For those still curious, let us see step-by-step why voltmeter V1 reads -emf/2 when the actual voltage is zero:

1. The closed loop around the B field can include the ring (left half a→b ) OR it can be via the voltmeter and its wires. Faraday holds in either path.
2. Therefore, an Em field must exist in the voltmeter wires such that ∫E_mw dl = emf/πa (the voltmeter itself is assumed to have negligible length), a = radius, and E_mw is the Em field in the wires.
3. But if there is an Em field in the wires there must be an equal and opposite Es field in those wires since no net field can exist in an ideal (zero impedance) wire.
4. But if an Es field exists in the wires then the voltmeter reading must be -∫E_sw dl since the Es circulation around the voltmeter loop is zero, with dl an element of length of the wires and E_sw the Es field in the wires.
5. Thus, the voltmeter will read -∫E_sw dl = -emf/2. It does not read -∫Em dl even though the two computations happen fortuitously to be equal.

In the main loop the Em field is clockwise while the Es field is counterclockwise. The voltmeter reads line-integrated Es only and does not register an Em line integral. The actual voltage a↔b = 0.

QED I hope.

 

[Stefan Gustafsson]

V1 + V2 = 0 But then V1+V2 = ∫E dl around loop = = 0? Which is false since ∫E dl around loop = emf. So V1 and V2 can’t be voltages since voltages must add to 0 around any loop Cf. Kirchhoff.

From your diagram: V1 = -emf/2, V2 = emf/2
∫E dl = V2-V1 = emf
(Walk around the circuit in the direction of the current, V2 is counted positive and V1 is counted negative because you enter from the negative side)

So, summing the voltages around the loop givs the emf, just as Faraday's Law (FL) says.

Are you seriously treating Kirchoffs Voltage Law (KVL) as an axiom?
Are you trying to prove that V1 and V2 cannot be voltages because that would be against KVL?

The whole point of this experiment is to point out a situation where KVL does not apply and you have to use FL instead.

There is nothing wrong with KVL as long as you use it in situations where it applies, which is lumped model circuit analysis. This circuit explicitly violates the assumptions in the lumped model which is why KVL does not work.

 

[rude man]

For those still interested I have added further explanation of my post 145. I think this will be my last shot unless newer posters have comments or questions.

 

[rude man]

EDITED: OK, one more thing: go back to my post 143 with the voltmeter & leads suspended directly above the B field So why is the reading now the correct potential? Answer: in this case there is a B-dot loop within the meter loop = B-dot/2 which acts to cancel the effects described in post #145. This loop generates an extra emf = half of the unconnected-meter emf, relulting in an Es field truly represents the potential between a and b. Cf. my Insight paper.

The physical distinction between Es and Em fields is that only in the former do flux lines begin and end on static charges. It's not just a theoretical distinction.

 

[rude man]

Homework Statement

A circular coil with radius a is connected with an equilateral triangle on the inside as shown in the figure below. The resistance for each section of the wire is labeled. A uniform magnetic field B(t) is pointing into the paper, perpendicular to the plane of the coil. B(t) is decreasing over time at a constant rate k. Given 2r₁ “ 3r₂. Find U_AB, the potential difference between points A and B
View attachment 211464

It's been a long grind, 1 1/2 years old or so, yet no one has IMO solved the OP's problem. I hereby submit a method of solution but won't carry out the computations as they are laborious and probably inappropriate anyway.

We recognize that the voltage between A and B is the line integral of the static electric field Es between those points. Any emf-generating field Em is irrelevant. The procedure is to solve for all fields in the six separate segments 1-6.

There are 18 parameters to be found (we need only 1 of them but we have to set up for all 18). They are:

Currents i1 - i6 in the 6 segments (3 arcs of 120 deg each plus 3 triangle sides),
Em in the 6 segments,
Es in the 6 segments.

We first find the 6 Em fields. This is readily done by constructing three dashed radii orthogonal to the triangle sides as shown. This enables easy computation by Faraday of for example Em1 in AY since Em is known in arc AX and = zero in XY. emf is obtained from the area AY-AX-XY. Etc. So the six Em are found immediately.

Next, 4 equations for Es are available. They are, by Kirchhoff voltage law (yes, Kirchhoff was right!),

(2$\pi$a/3)Es1 - L Es4 = 0
(2$\pi$a/3)Es3 - L Es6 = 0
(2$\pi$a/3)Es2 -L Es5 = 0
Es4 + Es5 + Es6 = 0
with L = length of one side of triangle.

Then, we invoke Ohm's law for E fields to get 6 more equations of the form Em+Es= i dr/dl. These are thus

$Em1 + Es1 = i1(3r1)/2\pi a$
$Em2 + Es2 = i2(3r1)/2\pi a$
(r1 is resistance of 1/3 of circumference)
etc.
Em4 + Es4 = i4(r2)/L with L = triangle side length,
Em5 + Es5 = i5(2r2)/L (since side BC has 2r2 resistance).
etc.

The final 2 equations are of the type i1 + i4 - i3 - i6 = 0 etc. in other words Kirchhoff current sums at A and B.

So we have 18 unknowns and 18 independent equations, allowing for finding Es1 (and/or) Es4. So U_BA = $2\pi aEs1/3 = L Es4$.

 

[Charles Link]

@rude man See this thread posts 21-40 : https://www.physicsforums.com/threa...-walter-lewins-paradox-comments.950284/page-2 The problem as given by the OP needs to be a magnetic field that is created by a solenoid such that its center coincides with the center of the wires. If not, the problem lacks the necessary symmetry. $\\$ Edit: Scratch this last couple of sentences. The only thing necessary to compute the EMF's of loop is the area. I actually did it the harder way by assuming the symmetry, and then I computed $I=\int \vec{E}_{induced} \cdot d \vec{l}$ for the line segments on the triangle. It's much easier to just compute the EMF's from the area. I still got the right answer, if I'm not mistaken. $\\$ @vanhees71 I think you will agree. $\\$ Under this assumption, and assuming the currents are small, I think you my have a good solution, but I need to look it over in more detail. $\\$ Edit: Then it should be possible to find all of the currents, as you may have done in your solution. It is, in any case, impossible to specify $U_{AB}$ as the question in the OP asks, because the path integral $I=\int\limits_{A}^{B} \vec{E}_{induced} \cdot d \vec{l}$ will be a function of the path that is taken between ## $and$ B $. Edit 3-28-19: By the potential$ U_{AB} $is meant$ U_{AB}=\int\limits_{A}^{B} \vec{E}_{electrostatic} \cdot d \vec{l} ## which is path independent. The problem is thereby perfectly well formulated.