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Magnetic flux rule for calculating motional EMF

  1. Jun 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Griffith ( Introduction to electrodynamics , 3 ed.)
    says in Problem 7.9:
    An infinite number of different surfaces can be fit to a given boundary line, and
    yet, in defining the magnetic flux through a loop, Φ = ∫B.da, I never specified the particular
    surface to be used. Justify this apparent oversight.


    2. Relevant equations


    3. The attempt at a solution
    I have always taken that surface which is perpendicular to the B and contains the loop as its boundary.
    I guess that Griffith doesn't define the surface because what we want to calculate is
    ε = - dΦ/dt and dΦ/dt doesn't change w.r.t. surface.

    But ,I don' t know in which direction should I think?

    In page no.296 , Griffith says
    Apart from its delightful simplicity, it has the virtue
    of applying to non- rectangular loops moving in arbitrary directions through non- uniform
    magnetic fields; in fact, the loop need not even maintain a fixed shape.

    But, won't calculating Φ for non- rectangular loops and non uniform magnetic fields itself be hard?
     

    Attached Files:

  2. jcsd
  3. Jun 4, 2017 #2

    cnh1995

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    Yes. Because the small surface in which the flux is changing is contained in the given boundary line.
     
  4. Jun 4, 2017 #3
    This I didn't get.
    Will you please illustrate it?
     
  5. Jun 4, 2017 #4

    cnh1995

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    If I got the question correctly:

    Consider a big metal ring S , w.r.t. which you want to calculate dΦ/dt.

    Suppose there are 100 different small rings contained in the ring S and in one of them, there is a changing flux. Call this ring S'. Now the emf induced in S' is equal to the emf induced in S, since S' is contained in S.

    But I'm not sure if I got the original problem statement correctly.
     
  6. Jun 4, 2017 #5
    I think the question means something else.
    Consider a ring shaped loop.
    Consider two surfaces:
    One could be a surface of the disk bounded by the ring.
    Another could be a hemispherical surface like a bowl.
    What the question says that while applying flux rule (or may be calculating flux), it doesn't matter whether I calculate flux through disk surface or hemispherical surface.
    This I have to prove.
     
  7. Jun 4, 2017 #6

    cnh1995

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    Well in that case, you need to consider the fact that all the field lines that pass through the circular base of the hemisphere also pass throgh the curved surface of the hemisphere. So it doesn't matter which surface you choose.
     
  8. Jun 4, 2017 #7
    I want to show it mathematically.
    How to do this ?
     
  9. Jun 4, 2017 #8

    cnh1995

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  10. Jun 5, 2017 #9
  11. Jun 5, 2017 #10

    cnh1995

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    There is no mistake. The minus sign shows up because you are considering a closed surface, where incoming flux (through the circular base) is equal to the outgoing flux (through the curved surface). If incoming flux is positive, outgoing flux has a negative sign, making their total zero.
     
  12. Jun 5, 2017 #11
    Thank you.
    Now I got it.
    While applying Green's theorem,
    the positive direction of da for curved surface is ##\hat r## .
    But in the second case , for calculating flux through the curved surface when the current is flowing through the ring, the positive direction of da for curved surface is -##\hat r## (as if I approximate the curved surface by a plane surface then the positive direction of da for this surface is ##\hat z## which is possible only if the positive direction of da for curved surface is ##-\hat r##) .

    Hence, flux through the curved surface in the first case i.e. Green's theorem is negative of the flux through the curved surface in the Second case.
    Hence, using the above fact and eq.(4), eq.(6) could be shown.

    Generalizing it, magnetic flux through all different surfaces bounded by the same boundary line is same.

    Is this correct?
     
  13. Jun 5, 2017 #12

    cnh1995

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    Yes. All that goes in, comes out.

    In Faraday's law, the flux is calculated w.r.t. open surface and hence, the integral of B.da is not zero.
     
  14. Jun 5, 2017 #13
    The first question is solved, but the second still has to be solved.
     
  15. Jun 5, 2017 #14

    cnh1995

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    It is hard.
    I am not sure if I got your question correctly this time as well.
    Could you elaborate?
     
  16. Jun 5, 2017 #15
    What I have understood from the Griffith's book is:
    Griffith is saying that emf can be calculated (using flux rule ) even when
    (1) the loop is non- rectangular
    (2)the magnetic field is varying
    (3)the loop is moving in any arbitrary direction and
    (4) the shape of the loop doesn't remain fixed

    But, I, too, feel that it is hard( can be almost impossible in some cases,too) to calculate flux in any of these four cases.
     
  17. Jun 5, 2017 #16

    cnh1995

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    Griffith is assuming that you have all the necessary information (otherwise what's the point of creating and solving such exercise?). For example, for a loop increasing in area in a magnetic field, emf can be calculated if you know the rate of increase in area and the variation of the magnetic field. It might be harder because of the higher math involved, but with sufficient information and correct math, none of the four cases is impossible.
     
  18. Jun 5, 2017 #17
    o.k. thank you.
     
  19. Jun 5, 2017 #18

    cnh1995

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    You're welcome!

    It would be better if you asked such conceptual questions in General physics or Classical physics forums. You could get way better and insightful answers there from our experts.
     
  20. Jun 5, 2017 #19

    rude man

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    Green's theorem is just the special case of Stokes's theorem applied to a two-dimensional figure. At least that's the consensus. So I have a problem with the wording in the graphic of post 9.
     
  21. Jun 5, 2017 #20
    pictures from page no.29 and 34, chapter 1 , Introduction to electrodynamics,3 ed.
    green's theorem.png
    stoke.png

    I have never read that Green's theorem is a special case of Stokes' theorem. Will you please give me a reference (to read)?
     
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