# Homework Help: Faraday's law -- circular loop with a triangle

1. Sep 21, 2017

### vishnu 73

1. The problem statement, all variables and given/known data
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 UAB, the potential difference between points A and B

2. Relevant 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

Last edited: Sep 21, 2017
2. Sep 21, 2017

### 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.)

3. Sep 22, 2017

### vishnu 73

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 dont see any other way to do this problem

4. Sep 24, 2017

### cnh1995

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.

5. Sep 24, 2017

### vishnu 73

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

6. Sep 24, 2017

### cnh1995

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?

7. Sep 24, 2017

### rude man

1
What does this mean?

2. Relevant 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
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
[/QUOTE]

8. Sep 25, 2017

### vishnu 73

[/QUOTE]
very sorry i meant 2R1 = 3R2

9. Sep 25, 2017

### vishnu 73

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?

10. Sep 25, 2017

### cnh1995

Why divide by 4? The triangle is equilateral.

11. Sep 25, 2017

### rude man

@OP and cnh1995: OK if you use the triangle.
But what's wrong with using the semicircle AB instead? For that we get a different VAB:
VAB = Ak/3, A = πa2 so VAB = πa2k/3 ≠ 3kA√3/12 !!??
Something rotten here!

12. Sep 25, 2017

### cnh1995

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.

13. Sep 25, 2017

### rude man

Hint: why are all those resistances given?
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).
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

14. Sep 25, 2017

### cnh1995

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.
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.

15. Sep 25, 2017

### rude man

There are only 2 paths connecting A to B directly (without passing thru C). So the answer is the former. I computed two values of VAB, derived by computing currents thru the two segments and multiplying by the respective resistances, including 2r1 = 3r2.

I got VAB via the straight line to be (2/3){[34k1 + 3(23K2- 34k3)]/80}

and via the arc VAB = [86k1+ 3(17k2- 46k3)]/120.

So again, what would be the reading on the voltmeter? I don't need numbers; symbols will do. And surely you don't mean k3 do you? Thank you.

16. Sep 25, 2017

### cnh1995

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.
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.

17. Sep 26, 2017

### rude man

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.

18. Sep 26, 2017

### cnh1995

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

19. Sep 26, 2017

### rude man

I did exactly that. I called that emf k3.
No argument! I never integrated along the longer arc.

What I meant in my post 15 was "surely the voltmeter reading is not k3".

20. Sep 26, 2017

### vishnu 73

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