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How is E.M.F induced?

  1. Dec 29, 2013 #1
    I know when the magnetic flux falling on a wire changes, emf is induced and a potential difference is created in the wire, I don't understand how can a change in the flux give electrons potential to create a current
     
  2. jcsd
  3. Dec 30, 2013 #2

    rude man

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    For a closed loop like a coil of wire,with time-varying magnetic field B inside the loop,

    del x E = - ∂B/∂t, found by experiment.
    where E is the electric field everywhere inside the loop.

    That plus Stokes' theorem gives you the emf induced around the loop.

    If an isolated wire of length L moves thru a B field with velocity v such that the B field and the wire direction are orthogonal (or partly orthogonal), then think of free electrons sitting along the wire. The force on an electron is F = q v x B so the electrons will be pushed to one end of the wire where they bunch up. That leaves the other end positively charged (lack of electrons). Equilibrilum is reached when the internal electric field = 0.

    So there is an emf generated by the moving wire = (B x L) * v.
     
  4. Apr 28, 2014 #3

    So induction actually happens due to Lorentz's force, doesn't it ?
     
  5. Apr 28, 2014 #4

    rude man

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    No. emf is induced around any closed path, per Stokes as mentioned. Even in thin air! Otherwise there'd be no e-m waves.

    The Lorentz force comes into play whenever there is a current density (moving charge) in a B field, providing the B field and current density vectors aren't aligned.

    If there is just an open moving wire, free charge does experience a Lorentz force when the velocity changes, but stops when the velocity is constant. I mentioned the F = qvB formula only to help you figure out which is the direction of the emf. I use that way myself.

    You can work this problem either by considering the change in loop flux as a function of time, or you can use the BLv law. I usually prefer the latter since it works for all moving medium situations I've ever encountered whereas the Maxwell equations do not. Be sure you look at each of the four loop segments individually to determine the magnitude and direction of the induced emf in each section.
     
    Last edited: Apr 28, 2014
  6. Apr 28, 2014 #5
    When a wire moves perpendicularly to a magnetic field B with velocity v
    The force acting on an electron is Bev
    That force is due to the magnetic field, it's very similar to Lorentz's force, the difference is that Lorentz's force happens when the electrons are moving in a wire due to an electric field so it's affected by a force perpendicular to B and to the direction of the current but in the case of inducing an emf the electrons are moving WITH THE WIRE not along it but along the v vector, so it's affected by a force along the wire which is perpendicular to v and B.
     
  7. Apr 28, 2014 #6

    rude man

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    Well, that's right, but as I said when the velocity is constant you get electrons bunched up at one end of the wire and 'holes' (positively charged ions) at the other end, and the induced emf is just enough to keep them separated. There is no flowing current in, nor net Lorentz force on, the wire when the velocity is constant.

    OK, a Lorentz force is still applied to the bunched-up charges, although there is no net force on the wire. (Actually, there are forces equal & opposite stretching the wire.) But come to think of it, you have a good point, the Lorentz force on the bunched-up charges is just enough to keep them separated from each other. In that way I suppose one can assert that the Lorentz force causes emf.

    Anyway, we digress ...
     
    Last edited: Apr 28, 2014
  8. Apr 28, 2014 #7
    But what causes Lorentz force and the force that makes the electrons accumulate at one side is almost the same.
     
  9. Apr 28, 2014 #8

    rude man

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    Exactly the same, so that equilibrium with zero steady-state current can be established.

    The charges are free to move and accumulate at the ends. So there would be an E field within the wire due to the separated charge. But this field is opposed by an exactly equal and opposite field due to induction. The net result is of course that the E field in the wire is zero.
     
  10. Apr 29, 2014 #9
    Thanks a lot I got that part, but I don't understand how a changing magnetic flux can force the electrons to do the same thing (accumulate at one side) inducing an emf
     
  11. Apr 29, 2014 #10

    rude man

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    It's not always flux, as I said. I can give you an example where emf = -dø/dt does not work.

    Think emf = BLv instead. L = length of the wire and v = velocity. You yourself have pointed out that once the charges are bunched at the ends the Lorentz force keeps them there. That force, which is indeed the force due to induction, is equal and opposite to the E field set up by the charges' positions. The same Lorentz force drives free charges to their end positions in the first place.
     
  12. Apr 29, 2014 #11
    What we've been taking about is motional emf, what about if we move a wire towards a magnet so there's a time varying flux, according faraday' law we say that an emf is induced,
    How is that ??
     
    Last edited: Apr 29, 2014
  13. Apr 29, 2014 #12

    rude man

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    emf = BLv. In this case B is not a uniform field but that makes no difference. Moving a wire in a constant field also generates an emf.
     
  14. Apr 29, 2014 #13

    But when moving a wire towards a magnet the velocity vector is no longer perpendicular to the field how does Bev and consequently Blv still work?
     
    Last edited: Apr 29, 2014
  15. Apr 29, 2014 #14
    "In thin air" the "induced emf" is pretty ambiguous. Unless a specific path is defined, the emf across 2 points in thin air is undefined unless a path is specified. The definition of potential, voltage, or emf, across points a & b, is the work done moving a charge from a to b, divided by the charge. In thin air charge is zero, as is the work. Thus emf in thin air is 0/0, which can be anything. In order for emf to have any real meaning there must be some finite charge value as well as work, and a specific path defined.

    We've discussed this issue many times on this forum, and as soon as we get to the "0/0" issue, the thread seems to stall. I don't want to belabor this, but I think it deserves mention. In thin air emf is ambiguous. With a conductor that has a shape, and free charges, emf is easily computed. Just my thoughts. Best regards.

    Claude
     
  16. Apr 29, 2014 #15

    jtbell

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    You take the angle into account via a factor of cos θ.

    http://farside.ph.utexas.edu/teaching/302l/lectures/node88.html

    (see equation 205 and the text immediately before and after it)
     
  17. Apr 29, 2014 #16

    Thanks a lot, but that's not what I meant.

    What I know is that when a current flows in a wire it induces a magnetic field so when this wire is place in a field of flux B the external field and the field of the current interact producing a force perpendicular to I and to B vector it can be calculated from : BIL

    So when a wire (has no current) cutting a magnetic field, the wire carries electrons that are moving with the wire along the v vector so these moving charged electrons induce a magnetic field which interacts with the external magnetic field producing a force that makes the electron to move perpendicularly to the v vector and to B, so it moves along the wire creating a potential difference and creating a current if it is connected to a closed circuit.

    But when a magnet is moving toward a still wire creating a time varying flux an emf is also induced, I can't understand why does this happen I can't relate it to what I understand.
    There things I'm missing causing misunderstanding, so I need some clarification, thanks in advance.!!
     
    Last edited: Apr 29, 2014
  18. Apr 29, 2014 #17

    rude man

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    I don't get your point.

    I said the emf is about a closed path. Why are you bringing up an open path?

    The emf around a closed path in 'thin air' is not at all ambiguous. That's why we have radio and TV.
     
  19. Apr 29, 2014 #18
    I was just pointing out that in thin air the emf reduces to 0/0. If we examine a closed path in thin air, we can define the emf as if a conductor were present along the specified path. In thin air, terms like emf are not well defined since it reduces to 0/0.

    We can however say that along this specific closed path defined by this equation/contour, the work per unit charge moving a charge around 1 time is this value. But the definition involves charges in a conductor. In thin air we are placing a conductor on that path and defining how charges behave.

    Maybe I'm being too strict, but emf in thin air is ambiguous. But I agree that if we define emf as what would be the work per charge if conductor and charge were present, it makes sense.

    Claude
     
  20. Apr 29, 2014 #19
    Please guys don't change the topic
     
  21. Apr 30, 2014 #20
    Ok, fair enough.

    Claude
     
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