Faraday's law -- circular loop with a triangle

[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 !