Faraday's law -- circular loop with a triangle

[timetraveller123]

thanks for your encouragement i will take a look at the article
but i have one more doubt
what exactly happens in a battery i am given to understand it provides a non conservative emf then shouldn't it be in the other side of kirchhoff laws

 

[rude man]

thanks for your encouragement i will take a look at the article
but i have one more doubt
what exactly happens in a battery i am given to understand it provides a non conservative emf then shouldn't it be in the other side of kirchhoff laws

I answer this question in the referenced article, but there are two E fields in the battery: one points - to + and is emf-generated (I call it E_m), and the other points + to - and is electrostatic-generated, I call it E_s. The two fields are equal and oppsite inside the battery (assuming zero internal resistance) so the net E field in the battery is zero.

As I also said earlier, Kirchoff's laws really apply to voltage drops and rises, not E fields. However, the voltage law applies also to E_s fields but not to E_m fields. That's because voltage by definition is $V_b - V_a = \int_a^b \vec E_s \cdot d \vec l.$ So for the battery plus resistor circuit, $\oint \vec E_s \cdot d \vec l = 0$ but $\oint (\vec E_s + \vec E_m) \cdot d \vec l = iR,$ i = current, R = circuit resistance. The total E field = E_s + E_m, vectorially added.

 

[timetraveller123]

I answer this question in the referenced article

actually i am coming here from there

i believe the electrostatic field is due to the buildup of charge at the resistor
ok i see what you are saying now the Es field points in the opposite direction from positive terminal to negative hence giving rise to voltage rise thanks for that
one more question

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

 

[rude man]

actually i am coming here from there

i believe the electrostatic field is due to the buildup of charge at the resistor
ok i see what you are saying now the Es field points in the opposite direction from positive terminal to negative hence giving rise to voltage rise thanks for that
one more question

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

This is something you just have to accept. A voltmeter reads voltage and voltage is the line integral of E_s, not E_m.

Think of a ring, uniform resistance, in a time-changing B field. You know the emf = -∂φ/∂t = $\mathcal E = \oint \vec E \cdot d \vec l.$ There is no E_s field in the ring so ΔV = 0 between any two points along the ring.

BUT - the trouble is that when you attach a voltmeter by its leads to any two points along the ring θ radians apart, the meter loop GENERATES an E_s field in the meter loop. The result is the meter would not read zero but |ΔV| = $(θ/2π)\mathcal E$. I am assuming the meter loop is outside the B flux. To understand why you need to carefully read my Insight blog.

Another example is the battery. The voltmeter reads the emf because, and ONLY because, there is also an E_s field inside the battery. The E_m field does not contribute to the voltmeter reading.

 

[timetraveller123]

ok then thanks once again

 

[Stefan Gustafsson]

That's because voltage by definition is $V_b - V_a = \int_a^b \vec E_s \cdot d \vec l.$

Hello, I have found this very interesting thread from the Lewin Videos.

I wonder why you say that voltage by definition is $V_b - V_a = \int_a^b \vec E_s \cdot d \vec l.$ ?

I would say that as long as you have a conservative total field, the voltage between two points a and b is defined as the scalar potential difference between the two points, which can be calculated as $V_b - V_a = \int_a^b \vec E \cdot d \vec l.$ Where $\vec E$ is the total field.

Why would you exclude an E_m field when calculating the voltage?

 

[rude man]

Hello, I have found this very interesting thread from the Lewin Videos.

I wonder why you say that voltage by definition is $V_b - V_a = \int_a^b \vec E_s \cdot d \vec l.$ ?

I would say that as long as you have a conservative total field, the voltage between two points a and b is defined as the scalar potential difference between the two points, which can be calculated as $V_b - V_a = \int_a^b \vec E \cdot d \vec l.$ Where $\vec E$ is the total field.

Why would you exclude an E_m field when calculating the voltage?

Because the definition of voltage is the line integral of the electrostatic field E_s only.
Did you read my Insight article https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/ ? That might help you understand the difference between electrostatic and electromotive E fields.

 

[Stefan Gustafsson]

Because the definition of voltage is the line integral of the electrostatic field E_s only.
Did you read my Insight article https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/ ? That might help you understand the difference between electrostatic and electromotive E fields.

Yes, I read your Insight.

Yes, I hear what you are saying about the definition of voltage. I just wonder where you got that definition from?

In the references I have looked in, voltage is simply defined as difference in scalar potential, which is well-defined in any conservative field.
for example: [Field and wave Electromagnetics by David K. Cheng]

 

[rude man]

Yes, I read your Insight.

Yes, I hear what you are saying about the definition of voltage. I just wonder where you got that definition from?

In the references I have looked in, voltage is simply defined as difference in scalar potential, which is well-defined in any conservative field.

Exactly. It is the scalar potential difference. Such a difference exists only in a conservative, i.e. electrostatic field. If an emf field is also present it does not contribute to the scalar potential since it is non-conservative.

I suggest reading the two papers I cited in my Insight article which do a fine job in explaining the situation. Also the Skilling textbook which is I'm sure still available on ebay etc. Skilling is the only one I have encountered so far to emphasize the difference between E_s and E_m and why voltage is the line integral of E_s only.

 

[Stefan Gustafsson]

Exactly. It is the scalar potential difference. Such a difference exists only in a conservative, i.e. electrostatic field. If an emf field is also present it does not contribute to the scalar potential since it is non-conservative.

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.

You are on the right side of the wall. Imagine that you do not even know that the E field comes from a changing magnetic field.

In this region, the total E field is conservative everywhere because there is no changing magnetic field which means that rot E = 0

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)

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.

I really do not understand why you would want to exclude the "emf field" to get some "true voltage" which is not the same as what you can measure with a voltmeter.

 

[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)