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Electromotive force in a moving bar

  1. Feb 1, 2016 #1
    Hi everyone,

    I've seen in some lecture notes that a moving conducting bar (length of L) in a uniform and constant magnetic field develops an electromotive force (emf) between the 2 edges of the bar.

    Lets assume that the bar moves to the right (with velocity v) and the magnetic field (B) points into the screen.

    So the explanation I've seen is that the magnetic field applies force on positive charges towards one side of the bar and negative charges towards the other side.
    The system comes to a steady state when the emf between the 2 edges of the bar is:
    emf=L*v*B

    So as long as the bar is moving there is emf.
    So the bar is now like a battery, and if we'll connect a wire between the 2 edges of the bar to create a closed circuit then we'll have current flowing in the loop.

    But since the magnetic field is constant and uniform and the circuit area does not change, then according to Faraday's law, the magnetic flux through the circuit area is constant, so there should be no emf....

    Can anybody help me solve this??

    Thanks a lot!
    Shai
     
  2. jcsd
  3. Feb 1, 2016 #2

    Simon Bridge

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    Make the loop square - there will be EMF in one "bar" of the loop - what about the opposite bar?
     
  4. Feb 1, 2016 #3
    Yeah I see now,
    So you mean that the opposite bar will develop the same emf, and will be considered as a second battery, so the 2 batteries facing each other, so no current will flow, right?

    Please confirm just to be sure...
    Thanks a lot...
     
  5. Feb 1, 2016 #4

    Simon Bridge

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    Yep - by the "generating emf" description.
    There's a bunch of puzzles like this... you can do it by rotating the loop so half moves one way and the other half moves the opposite way.
     
  6. Feb 1, 2016 #5
    If the bar is resting on rails and the rails are connected by a fixed connector then a current will flow from the moving wire as it is pulled, along the rails, to the right.
    If the field is into the screen as you quote then the top end of the moving wire will be +ve and the bottom end will be -ve
     
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