Faraday's law -- circular loop with a triangle

[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.

 

[Charles Link]

Letting $C=\frac{dB}{dt}$, I get the EMF for the 3 outer circular arcs is $\mathcal{E}_o=\frac{C \pi a^2}{3}$, and (assuming I computed it correctly), I get the EMF for each of the three straight line segments is $\mathcal{E}_i= \frac{C \sqrt{3} a^2}{4}$. It should be a simple matter to then compute all of the currents, using Kirchhoff's laws. It requires 6 loop equations. There are 6 currents to solve for. $\\$ @vanhees71 I will try to double-check my calculations, but might you concur?

 

[Charles Link]

Letting $C=\frac{dB}{dt}$, I get the EMF for the 3 outer circular arcs is $\mathcal{E}_o=\frac{C \pi a^2}{3}$, and (assuming I computed it correctly), I get the EMF for each of the three straight line segments is $\mathcal{E}_i= \frac{C \sqrt{3} a^2}{4}$. It should be a simple matter to then compute all of the currents, using Kirchhoff's laws. It requires 6 loop equations. There are 6 currents to solve for. $\\$ @vanhees71 I will try to double-check my calculations, but might you concur?

I think I have complete solution, but it would be easy to have mistake. Going clockwise around the outer loop, starting with CA, and calling currents $i_1,i_2,i_3$, and similarly around the triangle, $i_4,i_5, i_6$, I get $\\$ $i_1=\frac{9}{2}M-\frac{7}{8}N$ $\\$ $i_2=\frac{45}{2}M-\frac{35}{8}N$
$i_3=6M-\frac{5}{4}N$
$i_4=\frac{21}{4}M-\frac{21}{16}N$
$i_5=\frac{-51}{4}M+\frac{35}{16}N$
$i_6=\frac{15}{4}M-\frac{15}{16}N$
where
$M=\frac{D}{r_1}$ and
$N=\frac{9E}{2 r_1}$
where
$D=Ca^2(\frac{\pi}{3}-\frac{\sqrt{3}}{4})$ and
$E=Ca^2 (\frac{\sqrt{3}}{4})$.
My solution might contain errors, but this is what I got in solving the 6 loop equations. ===============================================================================
Corrections (3-27-19):
================================================================================
$i_1=i_2=M+\frac{7}{36}N$
$i_3=M+\frac{5}{18}N$
$i_4=i_5=+\frac{7}{24}N$
$i_6=+\frac{5}{24}N$ (Note: I corrected $I_4,I_5,I_6$ a second time at 7:30 PM 3-27-19 and corrected again 3-28-19 at 10:45 AM where I had the sign on N reversed.).

 

[Charles Link]

@rude man See my solution in posts 151 and 152.

 

[rude man]

@rude man 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 ##.

That's the wrong integral. The integral is with Es, not Em. And as I pointed out, the Es integral from A to B will be the same whether you go via the arc AB or the triangle side AB. The voltage can be computed as asked for.

I didn't complete the analysis but I did a somewhat similar one (see attached pdf, sorry it's not very good). In that one there were three separate Es paths and all three integrated to the same answer.

For the OP's problem I assumed the B field to be symmetrical, in fact circular & centered at the center of the circle. That's how I obtained the values of the Em fields.

I will reply to your newest 2 posts separately.

 

[rude man]

@Charles, I agree with your emf's for the three arcs; of course that's pretty straightforward. I never calculated the other Em's since that involves tedious geometry (weird areas and triangle lengths); congrats for doing it.

As for the rest, I think we have a fundamental disagreement as to what constitutes "voltage", and I reiterate that the problem must be attacked with separate Em and Es fields as I have blathered on for a long time.

 

[cnh1995]

I have sent you a PM with the answer I am getting for the voltmeter reading.

Any chance you have that PM @rude man?
I remember writing the whole solution in a notebook at that time, but I'm afraid I have lost that notebook.
Will try again and post my results.

 

[rude man]

Any chance you have that PM @rude man?
I remember writing the whole solution in a notebook at that time, but I'm afraid I have lost that notebook.
Will try again and post my results.

Hi again cnh, I will look for that PM, I think i can find it & look at it.
Do you agree with my recent posts, at least in principle?

 

[Charles Link]

That's the wrong integral. The integral is with Es, not Em. And as I pointed out, the Es integral from A to B will be the same whether you go via the arc AB or the triangle side AB. The voltage can be computed as asked for.

I didn't complete the analysis but I did a somewhat similar one (see attached pdf, sorry it's not very good). In that one there were three separate Es paths and all three integrated to the same answer.

For the OP's problem I assumed the B field to be symmetrical, in fact circular & centered at the center of the circle. That's how I obtained the values of the Em fields.

I will reply to your newest 2 posts separately.

This is the whole "crux" of Professor Walter Lewin's paradox. The EMF's/voltages are path dependent.

 

[cnh1995]

and I reiterate that the problem must be attacked with separate Em and Es fields as I have blathered on for a long time.

I agree. That's how I remember having done it.

 

[Charles Link]

Any chance you have that PM @rude man?
I remember writing the whole solution in a notebook at that time, but I'm afraid I have lost that notebook.
Will try again and post my results.

@cnh I look forward to your solution. Because I split the solution into two parts, I think my second solution may still be right, even if I goofed on the first part with the algebra. I do think there is chance that I got both parts correct.

 

[cnh1995]

I think i can find it & look at it.

I guess it is deleted from our inboxes as we didn't revisit that conversation for a year.Never mind! I will try to work it out again.

 

[Charles Link]

I agree. That's how I remember having done it.

It is simply Kirchhoff's voltage laws as spelled out by Professor Lewin.

 

[rude man]

This is the whole "crux" of Professor Walter Lewin's paradox. The EMF's/voltages are path dependent.

Which is why i said he was wrong. He didn't know what "voltage" means. It is not necessarily what a voltmeter reads.
The emf's are path-dependent, the voltages are not.

 

[rude man]

I guess it is deleted from our inboxes as we didn't revisit that conversation for a year.Never mind! I will try to work it out again.

Thanks cnh. I also could not find it. Looking forward to your re-work!

 

[Charles Link]

I did double-check my results, and surprisingly the currents that I got all satisfy current into the junction is the current out of the junction. $\\$ I didn't check the EMF's yet, but there are basically 3 equivalent EMF loops, (the 3rd with a different resistor on the one side), and then also the 4th EMF loop around the center triangle. There are only two independent current junction equations, accounting for the 6 KVL equations. $\\$ To make the problem much simpler, they could have made all three resistors the same on the triangle, and they could even have chosen $r_1=r_2$.

 

[rude man]

There are only two independent current junction equations, accounting for the 6 KVL equations. $\\$

Yes that's correct so I am missing one independent equation somewhere. 17 equations and 18 unknowns! Hope @cnh1995 has better luck!
EDIT: Es4 + Es5 + Es6 = 0 (the triangle) is a third independent Es equation.

 

[Charles Link]

@rude man @cnh1995 I made an Edit: to post 150.

 

[Charles Link]

@rude man @cnh1995
I did the arithmetic by hand, so I probably don't have two decimal place accuracy for each of the currents. Writing them as $J_i=\frac{i_i r_1}{Ca^2}$, I get
$J_1=1.05$
$J_2=5.28$
$J_3=1.24$
$J_4=0.66$
$J_5=-3.55$
$J_6=0.46$
These do satisfy, (s well as the currents in post 152), that
$J_1+J_4=J_2+J_5=J_3+J_6$. $\\$ This is our two current into the junction =current out of the junction equations.
If I correctly wrote down the voltage loop equations, and have the correct EMF's in post 151, I think I got the correct answer.
Edit: Scratch this. I had an error in my algebra that propagated. See the corrections in post 152 for the solution.

 

[cnh1995]

As per my calculations, the "electrostatic" voltage V_AB= V_A-V_B= -3sqrt(3)a²k/96.
(Here, k=dB/dt as mentioned in the OP).
Electrostatically, B comes out to be at a higher potential than A and C.(PS: How do I type mathematical signs and symbols here on PF5? I don't see any option in the editor.)

Edit: I verified the above result with actual numerical values as follows:
Radius a= 10m, k= 5T/s, r1= 15 ohm, r2= 10 ohm (2r1=3r2).
All the six currents are satisfying KCL at the 3 nodes (total incoming node current= total outgoing node current = 52.046 A).

Typing the entire solution here would be a tedious task. Let me know if you want to see the full solution. I will try to post an image of the whole solution.

 

[cnh1995]

and they could even have chosen r1=r2r1=r2 r_1=r_2 .

Wouldn't that eliminate the electrostatic field E_s from this circuit?
See #36.

 

[timetraveller123]

Yes that's correct so I am missing one independent equation somewhere. 17 equations and 18 unknowns! Hope @cnh1995 has better luck!
EDIT: Es4 + Es5 + Es6 = 0 (the triangle) is a third independent Es equation.

i don't think you need that many equation(please do correct me if i am wrong) there is some symetry involved by which you can say that i4=i6 and i1=i3
you also have that em1=em2=em3 and em2=em4=em6
thus you have es4 = es6 and es1 =es3

edit:
i am sorry i am wrong i am still thinking of this as a balanced wheatstone bridge which only holds for static cases
on a side note i think the setters of this question were not expecting this solution with 17 equations (i don't think it can be solved in exam conditions) they were just expecting potential difference between a and b as the line integral of total electric field along the straight path from a to b (ie i4*r2 ) although the official solution were not provided i am inferring this from his words

 

[Charles Link]

Is this correct? Does the direction of the current I assume matter? there is one loop I haven't used then there would 5 equations for 4 unknown making it overdetermined and lastly what sign should I use for the change in magnetic field thanks for the help
View attachment 213131

I might have goofed in posts 150-152, but surprisingly, I did not get the two $I_1's$ or the two $I_3's$ equal as in your diagram. The circulation in a given direction, (clockwise or counterclockwise) may, in fact, destroy what appears to be a symmetry that may be non-existent.=Edit: I goofed somewhere in the algebra. Hopefully I will have a correction shortly.

 

[Charles Link]

@rude man @cnh1995 @timetraveller123 See my post 152, with the corrections that I just posted. I believe I now have the correct answer for the currents, and it has the expected symmetry.

 

[rude man]

As per my calculations, the "electrostatic" voltage V_AB= V_A-V_B= -3sqrt(3)a²k/96.
(Here, k=dB/dt as mentioned in the OP).
Electrostatically, B comes out to be at a higher potential than A and C.(PS: How do I type mathematical signs and symbols here on PF5? I don't see any option in the editor.)

Edit: I verified the above result with actual numerical values as follows:
Radius a= 10m, k= 5T/s, r1= 15 ohm, r2= 10 ohm (2r1=3r2).
All the six currents are satisfying KCL at the 3 nodes (total incoming node current= total outgoing node current = 52.046 A).

Typing the entire solution here would be a tedious task. Let me know if you want to see the full solution. I will try to post an image of the whole solution.

Thank you @cnh1995. I would love to see a summary of your approach a la my post 149. Did you have 18 equations & 18 unknowns, what are they, etc. I'm not interested in quantitative results since I don't have any myself (I am too lazy to compute things like areas and triangle lengths ). The approach is what's interesting to me. Thanks.

 

[cnh1995]

Did you have 18 equations & 18 unknowns, what are they, etc

Nah, just 2 equations with 2 unknowns!
Will add you in an ongoing conversation as I am not sure if I can post the "complete" solution here in the HW thread.