Faraday's law -- circular loop with a triangle

[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.

 

[rude man]

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

Hi timetraveller, welcome back to the melee!

I think cnh1995 probably has the best approach and I am going to direct my attention to that. I'm not surprised there are fewer than 18 parameters necessary but why worry about it when a computer can easily handle that. You are obviously right in pointing out that that would not be feasible in exam conditions.

There is one and only one correct answer to the problem and that is the line integral of the electrostatic field. If you integrate the total E field from A to B via the arc you get a different answer than if you integrate via the straight line.

 

[rude man]

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

Probably yes. In the absence of the triangle that is certainly the case. With it I'd have to look-see some more.

 

[rude man]

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.

@cnh1995, did you verify that the line integral of Es is the same for both the arc and the triangle side?

 

[Charles Link]

@cnh1995, did you verify that the line integral of Es is the same for both the arc and the triangle side?

This one has been an open-ended item for so long that I'm sure the staff will be ok with it. Just to be sure, let me ask.

 

[rude man]

This one has been an open-ended item for so long that I'm sure the staff will be ok with it. Just to be sure, let me ask.

NM I meant it to go to our private thread.

 

[Charles Link]

NM I meant it to go to our private thread.

@rude man The staff gave us permission to post a complete solution.

 

[timetraveller123]

Probably yes. In the absence of the triangle that is certainly the case. With it I'd have to look-see some more.

there would be no es even if there were a triangle provided they were all same resistance

 

[timetraveller123]

I might have goofed in posts 150-152, but surprisingly, I did not get the two I′1sI1′s I_1's or the two I′3sI3′s 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.

the reason why i said symetery is that if you look way back at post 27 tsny said
"By symmetry, two of the triangle branches will have the same current and two of the circular arcs will have the same current. "
but i am not entirely sure if the concept of balanced wheatstone bridge can be applied here because the concept of potential is not well defined here but i am leaning on the side that says symetery is correct.
furthermore he said
"
This can be used to reduce the number of unknowns. I did not use this symmetry. Instead, I let the software solve the six equations and then checked to see if the particular currents were equal.
"
so you that's why i mentioned that
if it is just six equations then it indeed is very solvable in exam

 

[Charles Link]

I finally got consistency and an answer that is consistent : $V_{AB} =-\frac{15}{32} \sqrt{3}Ca^2$.
I need to make a couple of changes to $I_4, I_5, I_6$ in post 152, which I will do momentarily.

 

[Charles Link]

For the $V_{AB}$ we have:
$I_5 r_2=\int\limits_{A}^{B \, straight \, line} \vec{E}_{induced} \cdot d \vec{l}+V_{AB}$.
The integral has value $\mathcal{E}_1=\frac{\sqrt{3}}{4} Ca^2$. Plugging in for $I_5$ from post 152, with $r_2=\frac{2}{3} r_1$, we get
$V_{AB}=-\frac{15}{32} \sqrt{3} Ca^2$.
====================================================================
Alternatively,
$I_2 r_1=\int\limits_{A}^{B \, arc \, path} \vec{E}_{induced} \cdot d \vec{l}+V_{AB}$.
The integral has the value $\mathcal{E}_2=\frac{\pi }{3} Ca^2$.
Plugging in for $I_2$ from post 152, we again get
$V_{AB}=-\frac{15}{32} Ca^2$.

 

[Charles Link]

The 6 KVL loop equations I used in post 152 were really quite straightforward. The hardest part was doing 6 equations and 6 unknowns by substitution by hand. Computer methods would be much easier and are not prone to algebraic and arithmetic errors.

 

[timetraveller123]

i don't think we even need 6 equation you only need 4 equations

so this is what i have come up with for why the corresponding currents branches in left and right must be same

my notation
$A_i$denotes area $I_i$is current $r$ is resistance$E_m , E_{m'}$ are the line integrals of electric field due to magnetism not the field itself and $V_1 , V_2$ are the line integrals of the electrostatic field
the net current through the left side is given by
$\frac{E_m + V_1}{1.5r_2} + \frac{E_{m'} + V_1}{r_2}$
and for the left side is
$\frac{E_m + V_2}{1.5r_2} + \frac{E_{m'} + V_2}{r_2}$
thus if
$V_1 \neq V_2$ then the current law cannot be satisfied and this implies the currents through the left triangle branch is the same the current through the right triangle branch and same for the arc

after this we only have three independent loops and one current law for the other junction thus four equations
i can't find any software where you can input variable constant like kA
someone help me to compute this
after getting i2 we can vab as
$i_2 r_2 - \frac{k A_0}{3}$

but if you don't note that the currents are not equal at the start then you get 2 more equations one from current law at the junction and another voltage loop equation which is 6

 

[Charles Link]

One correction to your diagram and your equations: By $I_4$ that needs to be $r_1$.
I should have a solution for you momentarily.

 

[Charles Link]

@timetraveller123 Yes, I get $I_1=\frac{\pi Ca^2}{3r_1}-\frac{\sqrt{3} Ca^2}{32r_1}$, and it agrees with my final solution. By solving yours, I actually found an error in mine, where I had reversed the sign of $N$.

 

[Charles Link]

@rude man @cnh1995 See final corrections to post 152 where I reversed the sign on $N$.

 

[Charles Link]

Additional comment: This one was exhausting. I was on it for 3 hours this morning. First I found, (after about an hour), that "timetraveller123" incorrectly labeled an $r_2$ in his diagram. Then somewhere, there was a wrong sign on a term, and I finally located it in my 8 pages of algebra where I flipped the sign on $N$.
I think I finally succeeded!
"timeteaveller123" took a shortcut with the symmetry of the problem. His solution is simpler. I used 6 equations and 6 unknowns. I solved his 4 equations for him, and his answers agree with mine of post 152 !

 

[cnh1995]

Here's how you can do it using node voltage method at nodes A and C.

 

[Charles Link]

@cnh1995 's method uses that the current into a node is the current out of the node. In addition it uses the current in any resistor wire is
$I=\frac{\int\limits_{X}^{Y} \vec{E}_{total} \cdot d \vec{l}}{R}$,
where $\vec{E}_{total}=\vec{E}_{induced}+\vec{E}_{electrostatic }$, and where
$\int\limits_ {X}^{Y} \vec{E}_{induced} \cdot d \vec{l}=\mathcal{E}_{XY}$,
and
$\int\limits_{X}^{Y} \vec{E}_{electrostatic \, XY} \cdot d \vec{l}=V_{XY}$.
======================================================================
Alternatively, KVL uses $\\$
$\oint \vec{E}_{induced} \cdot d \vec{l}=I_1 R+I_2 R +...=\mathcal{E}_{loop}$,
since $\oint \vec{E}_{electrostatic} \cdot d \vec{l}=0$.
($\vec{E}_{electrostatic}$ is a conservative field, unlike $\vec{E}_{induced}$).

 

[Charles Link]

Wit the above method, knowing that $\mathcal{E}_{arc}=\frac{ \pi Ca^2}{3}$, and knowing that
$I_1=\frac{\pi Ca^2}{3 r_1}-\frac{\sqrt{3} C a^2}{32 r_1}=\frac{ \mathcal{E}+V_{AB}}{r_1}$,
we quickly get the answer that
$V_{AB}=-\frac{\sqrt{3} Ca^2}{32 }$.
============================================================
The EMF can also be computed over the straight line path over one of the sides of the triangle to be
$\mathcal{E}_{side}=\frac{\sqrt{3} C a^2}{4}$. The current on this side of the triangle is
$I_2=\frac{21 \sqrt{3} C a^2}{64 r_1}$.
This gives, with $R=\frac{2 r_1}{3}$ that
$I_2 R=\frac{7 \sqrt{3} Ca^2}{32 }=\mathcal{E}_{side}+V_{AB}$.
Note $V_{AB}$ is electrostatic and thereby path-independent.
Once again we get
$V_{AB}=-\frac{\sqrt{3} Ca^2}{32}$.

 

[Charles Link]

Here's how you can do it using node voltage method at nodes A and C.
View attachment 240928

View attachment 240929

Our answers are all in agreement ! @cnh1995 needs to put his fractions in lowest terms: $\frac{3}{96}=\frac{1}{32}$, etc. Also I'm using $C$ for his $K$.

 

[timetraveller123]

One correction to your diagram and your equations: By $I_4$ that needs to be $r_1$.
I should have a solution for you momentarily.

oh you my bad
nice to have a final closure to this problem i have learned a lot from this problem thanks to all
which software do you use to solve equations

 

[Charles Link]

oh you my bad
nice to have a final closure to this problem i have learned a lot from this problem thanks to all
which software do you use to solve equations

I did them all by hand, by the algebraic substitution method. Your 4 equations and 4 unknowns was kind of easy. My 6 equations and 6 unknowns was painstaking, and even more difficult was isolating the error which I found=I had put a minus on the constant $N$ instead of the plus sign when I did an algebraic step. (When I solved your 4 equations and I had a slightly different answer, I had to then double-check both the solution I got to your 4 equations as well as my 6 equations=and there I found the incorrect $r_2$ in your diagram, but they still didn't agree. After another hour of looking, I found where I had reversed the sign on the $N$ in the equations that I had. Finally, with those 2 corrections, everything was in agreement).
I also found this problem very educational. Thank you for posting it !