Faraday's law -- circular loop with a triangle

[timetraveller123]

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

Homework Equations

3. The Attempt at a Solution
i transformed the given circular circuit into such
is it correct starting from point a i went a loop around the circuit clockwise by lenz law

i reasoned it as the magnetic field is decreasing through all part of the circle no matter which loop or triangle i pick the current must flow in clockwise hence i came up with this
is there any thing wrong with my reasoning[/B]

 

[cnh1995]

The induced electric fields are non-conservative. The line integral of E.dl from A to B along the circumference will not be same as the line integral along the triangle's side AB. Plus, the reading of the voltemeter will depend on how you connect the voltmeter between A and B. So I don't think your transformation of the circuit is valid. (Layout of the circuit is important here.)

 

[timetraveller123]

so how can i go about doing it

is it true that the current generally will flow from A to b then c?
i really don't see any other way to do this problem

 

[cnh1995]

so how can i go about doing it

is it true that the current generally will flow from A to b then c?
i really don't see any other way to do this problem

What is the flux through the triangle? What is the emf induced in the triangle? What is the emf induced along its each side? Make use of the symmetry.

 

[timetraveller123]

what is confusing me is that if i take the loop around the triangle will the outer wires play a role in the calculations?
what about the dot product in the electric field and the length how do i compute that
because all along i have only used simple applications of it in circles how will it change with triangles ?
it would be great if you could help with my questions i am having some conceptual problems with faraday law thanks

 

[cnh1995]

what is confusing me is that if i take the loop around the triangle will the outer wires play a role in the calculations?

No, not for induced emf.
What is the area of the triangle in terms of radius of the circle (a)? Use simple geometry. What is the induced emf in the triangular loop according to Faraday's law?

 

[timetraveller123]

1
What does this mean?

Homework Equations

The Attempt at a Solution

i transformed the given circular circuit into such
is it correct starting from point a i went a loop around the circuit clockwise by lenz law
View attachment 211466
i reasoned it as the magnetic field is decreasing through all part of the circle no matter which loop or triangle i pick the current must flow in clockwise hence i came up with this
is there any thing wrong with my reasoning[/B]

[/QUOTE]
very sorry i meant 2R₁ = 3R₂

 

[timetraveller123]

No, not for induced emf.
What is the area of the triangle in terms of radius of the circle (a)? Use simple geometry. What is the induced emf in the triangular loop according to Faraday's law?

ok then
$V_{ab} = V \frac{1}{4}\\ V= A k\\ A = \frac{3}{2} a^2 sin 120\\ V_{ab} = \frac{3 k a^2 \sqrt 3}{16}$
is this correct?

 

[cnh1995]

Vab=V14

Why divide by 4? The triangle is equilateral.

 

[cnh1995]

Something rotten here!

No, what you are getting is actually correct. The induced electric fields are non-conservative.
Hence, the emf between A and B will depend on the path you follow to go from A to B.
As I said in #2, this is exactly why the layout of the circuit is important and OP's transformation of the original circuit in #1 is not valid.

Here's a recent thread having a similar question.
https://www.physicsforums.com/threads/induced-emf-between-two-points.926298/

 

[cnh1995]

You mean, if I put a voltmeter at A and B I get two different voltages?
(Assume the voltmeter leads are outside the B field of course).

If the voltmeter loop is in the B field, you'd get different readings depending on where the voltmeter is. If the leads are outside the field, you'd get the same reading regardless of its position, but it will be the electrostatic voltage, which is not same as induced emf. That electrostatic voltage will depend on the given resistances.

BTW I assigned 6 separate currents to the figure & solved for them. The interesting result was that the sum of voltage drops from A to B over the two paths = k1 + k2 -2k3
where k1= emf around circle
k2 = emf around triangle
k3 = emf around one of the three loops between the circle and triangle

There are total three paths to go from A to B. Which two have you used? Side AB and smaller arc or side AB and larger arc?
If it's the former, the sum of the emfs should be simply k3.
If it's the latter, sum of the emfs should be k1-k3.

Whatever the resultant fields are in any part of the circuit, the closed loop integral of E.dl will be equal to the rate of change of flux through that loop.

 

[cnh1995]

And surely you don't mean k3 do you?

I did, but I was assuming you went from A to B along the smaller arc and from B to A along side AB, and added the two emfs.

The interesting result was that the sum of voltage drops from A to B over the two paths = k1 + k2 -2k3

But if you went from A to B along both the paths, surely the answer is not k3. I'll work it out and post later.

 

[cnh1995]

Thanks. I admit I'm bamboozled.
I'm not at this point really interested in computing the different emf's associated with different contours. Rather, I Hope you'll come up with the one & only voltmeter reading! Assume the voltmeter is tiny or whatever so the leads are not in any way affected by the B field - so we get the true voltage reading. Thanks again.

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

 

[timetraveller123]

Why divide by 4? The triangle is equilateral.

the v was the potential difference across the entire circuit
the potential across ab is just a fourth of it as it has resistance r while the entire circuit has a resistance 4r so naturally the potential difference across ab is a fourth of the potential difference across the whole circuit

 

[cnh1995]

the v was the potential difference across the entire circuit
the potential across ab is just a fourth of it as it has resistance r while the entire circuit has a resistance 4r so naturally the potential difference across ab is a fourth of the potential difference across the whole circuit

No. Induced emf doesn't depend on resistance. And you can't use the term 'potential difference' in case of non-conservative emf.
So what is the induced emf along AB?

After you find that, you'll have to find the electrostatic potential difference between A and B. That voltage will be shown by the voltmeter and there you have to make use of the resistances.

 

[timetraveller123]

No. Induced emf doesn't depend on resistance. And you can't use the term 'potential difference' in case of non-conservative emf.
So what is the induced emf along AB?

After you find that, you'll have to find the electrostatic potential difference between A and B. That voltage will be shown by the voltmeter and there you have to make use of the resistances.

sir i am having a huge misconception in faradays law i have never learned maxwells equation formally it would be helpful if you could answer my questions in detail here

1) in faraday law it is written in E.dl what exactly is E(electric field ) and where exactly is it
2) you kept saying non conservative emf and electric fields while i understand what the term non conservative means how exactly does it apply here
3) what is difference between induced emf and potential difference

So what is the induced emf along AB?

this seems to suggest that induced emf is the same at all point of the circuit how is that so
sorry if these seem like dumb question but even till now my understanding of induced emf and potential difference all muddled up could help if you could help me make a clear distinction between these thanks for your understanding

 

[cnh1995]

in faraday law it is written in E.dl what exactly is E(electric field ) and where exactly is it

If you pick any loop in your original diagram, there is a changing flux in that loop. The induced electric field is in concentric circles. But to calculate E.dl, we take the electric field on the boundary of the loop. Even though you don't know the actual electric field at each and every point on the contour, the closed loop integral of E.dl along the loop is equal to the rate of change of flux in that loop.

while i understand what the term non conservative means how exactly does it apply here

The emf will depend on the path you choose to go from A to B. In circuits with no varying fields, the electric fields in all the parts are conservative (except for the battery). Hence, no matter what path you choose to go from one node to another, potential difference between the two remains the same. Here, you can assign relative electrostatic potentials to the nodes.

what is difference between induced emf and potential difference

Induced emf is the result of varying magnetic field. Potential difference is the result of conservative electric fields due to induced charges on the conductors.

this seems to suggest that induced emf is the same at all point of the circuit how is that so

By symmetry, induced emf along each side will be same i.e. 1/3rd of the dΦ/dt through the triangle.

 

[timetraveller123]

thanks for the reply cleared up some but i still have some doubts it would be kind of you could answer that

If you pick any loop in your original diagram, there is a changing flux in that loop. The induced electric field is in concentric circles. But to calculate E.dl, we take the electric field on the boundary of the loop. Even though you don't know the actual electric field at each and every point on the contour, the closed loop integral of E.dl along the loop is equal to the rate of change of flux in that loop.

that there if the electric field is in concentric circles then does it mean that points on the same circle are of same potential ?

By symmetry, induced emf along each side will be same i.e. 1/3rd of the dΦ/dt through the triangle.

this is what you confusing me the most about this question is that it is triangle i don't really get what symetry you are talking about if the electric field are curling around then in a circle the electric field is always in line with the length but in a triangle that is not the case then what symetry are you talking about

 

[cnh1995]

this is what you confusing me the most about this question is that it is triangle i don't really get what symetry you are talking about if the electric field are curling around then in a circle the electric field is always in line with the length but in a triangle that is not the case then what symetry are you talking about

Forget about the circle, just consider the equilateral triangle. The induced emf in the entire triangle i.e. the closed loop integral of E.dl along the triangle (path ABCA) is equal to dΦ/dt through the area of the triangle. (Here, E is the electric field on the side of the triangle.)
Now, if you just want to find the line integral of E.dl along side AB, you can divide the above "closed loop" integral by 3, because the triangle is equilateral and hence the line integrals of E.dl along the three sides will be equal. They will add up to dΦ/dt through the triangular area. This is the symmetry I am talking about.
If it were a scalene triangle instead, the induced emfs along the three sides would be unequal and it would be more complicated to solve.

that there if the electric field is in concentric circles then does it mean that points on the same circle are of same potential ?

No, there is no potential. Electric fields at any two points on the same circle are equal (in magnitude), but that doesn't mean they are equipotential. The fields are non-conservative, hence, concept of potential is not applicable.

.

 

[timetraveller123]

No, there is no potential. Electric fields at any two points on the same circle are equal (in magnitude), but that doesn't mean they are equipotential. The fields are non-conservative, hence, concept of potential is not applicable.

ok your previous post cleared up a lot of misconceptions one last one
you say electric field at any two point on the same circle are equal but here it is not a circle it is a triangle then wouldn't the magnitude of electric fields change along the triangle

 

[cnh1995]

wouldn't the magnitude of electric fields change along the triangle

Yes, but the integral of E.dl along the three sides will be the same, thanks to the symmetry of equilateral triangle. It is the line integral that you should be interested in, and not the actual electric field.

 

[timetraveller123]

ooh okay now i see it when you mean it in relation to other sides thanks i will get back to problem shortly

 

[cnh1995]

The above can be simplified as follows: there is no externally applied electrostatic voltage V around any part of the circle, therefore V = 0. The emf is not read because the IR drop is equal & opposite to the emf anywhere along the circle.

That would be true in absence of the triangle, or if r_AB=r_BC=r_AC.
But since r_AB≠r_BC, there should be some electrostatic voltage present in the circuit (so that KCL is satisfied at every node, especially at B).

 

[timetraveller123]

so sir sorry for the late reply was quite busy national exams coming up so sorry

so to sum up what i know about what you have told me so far
as long as the polygon in question is a regular one the emf induced would be the same in each side due to symetry is that right ?
so in this question
$emf_{ab} = k \frac{1}{2}a^2 sin120$
this is the emf induced then what is the potential difference between a and b please help me out thanks

 

[cnh1995]

@vishnu 73, I am traveling right now and the network connectivity is very poor here. So I won't be able to reply at least for the next 18-20 hours.
Meanwhile, you can check out a similar discussion here.
https://www.physicsforums.com/threads/induced-emf-between-two-points.926298/.

 

[timetraveller123]

oh understood sir i have been looking at the thread too i will wait for your reply

 

[TSny]

@vishnu 73 , I think a good approach to the problem is to label the currents in each of the six branches in the circuit. There are three branches for the triangle and three branches for the circle. You'll need to set up six equations for the six unknown currents; say, 4 loop equations and 2 junction equations.

For any loop, you can set up dΦ/dt = Σ(±IR) where the sum is over the branches in the loop. The signs depend on your choice of current directions.

Once you set up the equations, you can use some sort of software to solve the six equations.

By symmetry, two of the triangle branches will have the same current and two of the circular arcs will have the same current. 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.

I think the problem statement is unclear on what they actually want. If you go from A to B along the leg of the triangle you will get a different IR drop than if you go from A to B along the upper right circular arc. This was already pointed out by @cnh1995 in post #2.

The IR drop for the leg of the triangle is the same as ∫Edl along the leg of the triangle. Similarly for the the circular arc.

 

[timetraveller123]

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

 

[TSny]

Is this correct?

Yes, all four equations look correct. I assume that $A$ in equation (3) is the total area of the circular loop.

Does the direction of the current I assume matter?

No. Pick any direction, the sign of your answer will tell you if you picked the actual direction.

there is one loop I haven't used then there would 5 equations for 4 unknown making it overdetermined

Right.

and lastly what sign should I use for the change in magnetic field

The flux is into the page and decreasing. So, what should be the "direction" of the induced emf in any loop (clockwise or counterclockwise)?

Looks like your equations correspond to the correct direction of the induced emf's.

 

[timetraveller123]

No. Pick any direction, the sign of your answer will tell you if you picked the actual direction.

thats where i am having confusion how is it that in physics even in mechanics problems i can assume random direction but end up with a correct direction despite having used a wrong direction in calculations
generally is having a overdetermined system a problem?

 

[TSny]

thats where i am having confusion how is it that in physics even in mechanics problems i can assume random direction but end up with a correct direction despite having used a wrong direction in calculations

Here's a long-winded attempt at an explanation. I'm sure others can do better.

Consider a resistor $R_1$ in a circuit. Suppose Bob chooses the direction of the unknown current $i_1$ through $R_1$ to be to the right while Jan chooses the direction to be to the left.

Suppose they are setting up $\sum \Delta V$ around a loop that contains $R_1$. We can also assume Bob and Jan go around the loop in the same direction and that this direction is such that they go from $a$ to $b$ when crossing $R_1$.

Then Bob would write the potential change from $a$ to $b$ as $\Delta V_{a \, \mapsto b} = -i_1 R_1$. Jan would write $\Delta V_{a \, \mapsto b} = +i_1 R_1$. It should be clear that wherever Bob has $i_1$ in an equation, Jan would have $-i_1$. (Same for a junction equation involving $i_1$.)

So, they will get the same answers for $i_1$ except they will get opposite signs. But note that they will both get the same magnitude and sign for the potential change $\Delta V_{a \, \mapsto b}$. The person who gets the positive answer for the current will have also have chosen the correct direction of the current.

For example, suppose that Bob gets a positive answer for $i_1$ when he solves the equations. Jan will necessarily get the negative of Bob’s answer. Based on their answers for $i_1$, both Bob and Jan will agree that the potential at $a$ is higher than the potential at $b$. But current flows from higher to lower potential. So Jan, who got the negative result for $i_1$, will conclude that she chose the wrong direction for the current. Bob, who got the positive answer for $i_1$, chose the correct direction.

 

[cnh1995]

@vishnu 73, your equations look correct to me. You can solve them simultaneously, get the six currents and find the line integral along path AB (and all other paths if you wish). That would be the " net emf" along side AB, which is not same when you choose different paths to go from A to B. What remains same is the "electrostatic potential difference" between A and B or V_AB.
Net emf is the resultant of electromagnetically induced emf and electrostatic potential difference.
So, for side AB, you know the net emf and electromagnetically induced emf. Find V_AB from that (take care of the polarity). I believe this is what the question in your OP is asking.

 

[timetraveller123]

@cnh1995 i am confused once again
1)what is net emf my guess is that if there was one and only equilateral triangle ABC then the net emf along AB is the k(area)/3 is that right
but in this case there is many more paths which have AB as one of their sides hence the sum of all such emf is the net emf along AB is that what net emf is
2)then no what is electrostatic potential difference prior to this problem my understanding is that electrostatic potential difference is just the difference in potential between two point
i am all confused with the terms again what is difference between net emf , electrostatic potential difference and how does that relate to V_ab

 

[timetraveller123]

@TSny
i am getting these are these correct
$I_1 = \frac{k(6A_1 - 1.5A_0 - 2A)}{12r_2}\\ I_3 = \frac{k(6A_1 + 1.5 A_o + 2A)}{-18r_2}$

 

[timetraveller123]

My thinking here was as follows: without the triangle we all agree the voltage between any 2 points along the circle = 0.

wait why is that so