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Induced EMF between two points

  1. Sep 22, 2017 #1
    1. The problem statement, all variables and given/known data
    ?temp_hash=bacb4187d6d47bb98dd0cfffab3b2f57.png

    2. Relevant equations


    3. The attempt at a solution

    First part is easy . Area of the triangular loop is A = (1/2)(2R)(R) = R2

    Induced EMF in loop ABC = Rate of change of flux = dΦ/dt = AdB/dt

    = R2(dB/dt) which is correct .

    Now the second part has stumped me .

    I would really appreciate if somebody could help me with this .

    Thank you .
     

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  3. Sep 22, 2017 #2

    TSny

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    My interpretation of part (ii) is that it is asking for the magnitude of the line integral ##\int \mathbf E \cdot \mathbf {dl}## along the straight line AB.

    If so, then use the result from (i) along with symmetry arguments.
     
  4. Sep 22, 2017 #3
    Thanks for replying .

    The induced electric field lines will be concentric circles . So I can't see how line integral can be calculated .

    How would we use symmetry in this ?
     
  5. Sep 22, 2017 #4

    TSny

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    What's the value of the line integral for the entire triangular circuit?

    What's the value of the line integral for the horizontal segment AC?
     
  6. Sep 22, 2017 #5
    You made it look so easy :) . I have been thinking about it for last couple of days .

    R2(dB/dt)

    As we move along from C to A along CA , direction of induced electric field would be perpendicular at every point of the path CA as CA is a a diameter and induced electric field lines would be concentric circles .The line integral would be zero along CA .

    Since field lines are circles , the paths AB and AC would look similar and their line integral would be equal .

    ##\int \mathbf E \cdot \mathbf {dl}## along the closed loop ABC =
    ##\int \mathbf E \cdot \mathbf {dl}## along the straight line CA + ##\int \mathbf E \cdot \mathbf {dl}## along the straight line AB +##\int \mathbf E \cdot \mathbf {dl}## along the straight line BC .

    ##\int \mathbf E \cdot \mathbf {dl}## along the straight line AB =
    R2(dB/2dt) .

    Is that alright ?
     
  7. Sep 22, 2017 #6

    TSny

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    Yes, that looks right to me.
     
  8. Sep 23, 2017 #7
    Thank you . Please clarify some points .

    1) Is KVL applicable in a closed circuit in the presence of induced EMF ? It does seem to apply in closed circuit ABC in this problem but one of the books I have suggests otherwise .It states that when electric field is non conservative Kirchoff's Voltage Law no longer holds true .

    2) In the branch AB what does the product iRAB signify ? It surely isn't the potential difference between points A and B . i2RAB is the power dissipated in wire AB .But what does iRAB tell us ?

    3) The line integral ##\int \mathbf E \cdot \mathbf {dl}## along the straight line AC is zero whereas the line integral along AB and BC is non zero . Is it something like having a battery in each of the branches AB and BC , but no battery and only resistance in branch AC ? Current is determined by the net EMF of the loop ?

    4) Is there an induced electric field line also on the circumference of radius R , i.e on the solenoid ?
     
  9. Sep 23, 2017 #8

    cnh1995

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    You'll have to modify the equation. In absence of a varying B-field, the closed loop integral of E.dl (in any loop) is zero. If there's a varying B-field, closed loop integral of E.dl is equal dΦ/dt, where Φ is the magnetic flux through that loop.
    It is the resultant emf in branch AB i.e.sum of electrostatic emf(conservative) and induced emf (non-conservative).
    Yes.
    Yes.
     
  10. Sep 23, 2017 #9
    Is there an electrostatic EMF in branch AB ? I think there is only non conservative EMF along the entire loop ABC .
     
  11. Sep 23, 2017 #10

    TSny

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    If the branches BC and CA are removed so that only AB is sitting there, then charge will be induced on the surface of AB to create a static E field that cancels the induced E field everywhere inside AB. This will happen almost instantaneously. So, Enet = 0 inside AB. Thus, no work would be required to move a test charge from A to B inside the branch AB. In that sense, there would be zero "net emf" between AB. (@cnh1995, please correct me if this is wrong.)

    That's one of the reasons I wasn't sure of the interpretation of part (ii). I interpreted it to mean: find the line integral of the nonconservative field between A and B for the straight path from A to B. I interpreted it that way because I thought it made an interesting question.
     
  12. Sep 23, 2017 #11
    Does the answer obtained in post 5 depend on whether BC and AC are present or not ?

    Why did the question asked to remove BC and AC ?
     
  13. Sep 23, 2017 #12

    TSny

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    If you interpret (ii) as asking for the line integral of the nonconservative field along path AB then the answer does not depend on whether or not BC and AC are removed. But, maybe my interpretation of what (ii) is asking for is incorrect.
     
  14. Sep 23, 2017 #13
    Your interpretation must be correct as the answer obtained matches with the answer given in the book :smile:

    Do you think KVL can be applied in a closed loop where induced EMF is present irrespective of whether DC battery is present or not ?
     
  15. Sep 23, 2017 #14

    TSny

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    OK, good.

    Yes. The induced EMF in the loop just gets added in with the DC battery emf. That's esssentially what you are doing when you use emf = -LdI/dt for an inductor in the loop.

    This is worth watching:

     
    Last edited: Sep 23, 2017
  16. Sep 23, 2017 #15

    cnh1995

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    Wouldn't that depend on the resistances of AB and the circumference?
    I have worked it out using some numerical values for dB/dt, radius and resistances of the segments and I am not getting zero net field inside AB.
    If you have time, I will post it here. (It's a bit lengthy to followo0)).
     
  17. Sep 23, 2017 #16

    TSny

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    @cnh1995, Here's how I'm thinking about it. With sections BC and CA removed, you just have an isolated piece of wire AB that is sitting in the induced E field of the solenoid. This E field will induce a complicated surface charge density on AB such that the net field is zero everywhere inside AB. It's like putting a chunk of metal between the charged plates of a capacitor. Induced surface charge on the chunk will create an E field such that the total field is zero at every point inside the chunk.
     
  18. Sep 23, 2017 #17

    cnh1995

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    Here's how I thought about it: The induced charges will create an electric field in all the three parts (two arcs and segment AB), such that the sum of currents through smaller arc and AB is equal to the current through the larger arc. The field due to induced charges will depend upon the resistances of the segments.
    Does that sound correct?
     
  19. Sep 23, 2017 #18

    TSny

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    I interpret the circular arcs as representing the current in the windings of the solenoid. The circuit that we are interested in is the triangle ABCA. In (ii) this is dismantled with BC and CA removed so that just AB is left.
     
  20. Sep 23, 2017 #19

    cnh1995

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    Here's my working with some actual numbers.

    Let's assume radius=1. So, the area of the circle= 3.1415 unit, area of the smaller loop=0.2854 unit (π/4-1/2) and area of the bigger loop=2.8561 unit.

    Assume dB/dt=5.
    So, emf induced in the entire circle=AdB/dt=3.1415×5=15.70V.

    Emf induced in the bigger loop=(2.8561/3.1415)*15.70=14.2736V

    Emf induced in AB =R2/2*dB/dt=5/2= 2.5V (from A to B).

    Emf induced along the bigger arc= 14.2736-2.5= 11.7732V. (B to A)

    Emf induced in the smaller loop=15.70- 14.2736=1.4264V

    Emf induced along smaller arc= 2.5+1.426= 3.9268V (A to B).

    Now for each loop, these values satisfy closed loop ∫E.dl= dΦ/dt.

    Is this correct so far?
     
  21. Sep 23, 2017 #20

    TSny

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    I agree with your numbers. I'm interested in where you are going with this :oldsmile:.

    Edit: Added figure for reference
    upload_2017-9-23_14-33-39.png
     
    Last edited: Sep 23, 2017
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