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

  1. Sep 21, 2017 #1
    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
    upload_2017-9-21_22-45-10.png

    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
    upload_2017-9-21_22-48-54.png
    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. jcsd
  3. Sep 21, 2017 #2

    cnh1995

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    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.)
     
  4. Sep 22, 2017 #3
    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
     
  5. Sep 24, 2017 #4

    cnh1995

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    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.
     
  6. Sep 24, 2017 #5
    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
     
  7. Sep 24, 2017 #6

    cnh1995

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    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?
     
  8. Sep 25, 2017 #7
    [/QUOTE]
    very sorry i meant 2R1 = 3R2
     
  9. Sep 25, 2017 #8
    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 #9

    cnh1995

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    Why divide by 4? The triangle is equilateral.
     
  11. Sep 25, 2017 #10

    cnh1995

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    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/
     
  12. Sep 25, 2017 #11

    cnh1995

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    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.
     
  13. Sep 25, 2017 #12

    cnh1995

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    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.
     
  14. Sep 26, 2017 #13

    cnh1995

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    I have sent you a PM with the answer I am getting for the voltmeter reading.
     
  15. Sep 26, 2017 #14
    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
     
  16. Sep 26, 2017 #15

    cnh1995

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    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.
     
  17. Sep 26, 2017 #16
    sir i am having a huge misconception in faradays law i have never learnt 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
    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
     
  18. Sep 26, 2017 #17

    cnh1995

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    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.
    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.
    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.
    By symmetry, induced emf along each side will be same i.e. 1/3rd of the dΦ/dt through the triangle.
     
  19. Sep 26, 2017 #18
    thanks for the reply cleared up some but i still have some doubts it would be kind of you could answer that

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

    this is what you confusing me the most about this question is that it is triangle i dont 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
     
  20. Sep 26, 2017 #19

    cnh1995

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    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.
    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.

    .
     
  21. Sep 27, 2017 #20
    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
     
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