Faraday's law -- circular loop with a triangle

[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.