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Conceptual difficulty with faradays law

  1. Dec 6, 2014 #1
    As i understand, the induced current in a wire equals -d(∫∫B⋅dS)/dt
    What is S? I know it has to be an open surface for the flux integral to be non-zero, but over what surface do i do the integral?
     
  2. jcsd
  3. Dec 6, 2014 #2

    Doc Al

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    That should be the induced EMF, not current.

    Consider the surface bounded by the wire.
     
  4. Dec 6, 2014 #3
    Oh thats right the voltage not current (i dont know why they say emf is a force if its really a potential)
    so the loop has to be closed then?
     
  5. Dec 6, 2014 #4

    jtbell

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    The loop has to be closed. In principle, you can use any open surface whose boundary is the loop in question. In practice, you use a surface that makes it easy to calculate the flux of ##\vec B##.
     
  6. Dec 6, 2014 #5
    Thanks that makes since I was just thinkin in terms of some general wire but I guess the circuit would have to be closed for you to attach a voltometer.

    So if the circuit is open there is no voltage generated?
     
  7. Dec 6, 2014 #6

    jtbell

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    A changing magnetic field is associated with an electric field, according to Maxwell's equations: $$\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$$ or $$\oint {\vec E \cdot d \vec l} = - \frac{d}{dt} \int {\vec B \cdot d \vec a}$$ The electric field in turn produces an emf along a specified path: $$\mathcal{E} = \int {\vec E \cdot d \vec l}$$ So I would say the emf exists because the electric field does, regardless of whether there is a wire along that path or not, or a voltmeter to measure it. In order to produce a steady-state current, you need a circuit, i.e. a closed loop of wire that electrons can travel along.
     
  8. Dec 6, 2014 #7
    interesting... i wish the book i read had put it that way. i never even knew the curl of the electric field could be non-zero this really gives me a whole new perspective of whats going on here
     
  9. Dec 6, 2014 #8

    jtbell

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    You started with electrostatics in which ##\vec E## does not change with time, and magnetostatics in which ##\vec B## does not change with time. In those situations, ##\vec \nabla \times \vec E = 0## so ##\vec E## is a conservative field and you can define the electric potential. Now you're starting on electrodynamics (time-varying ##\vec E## and ##\vec B## fields) in which this is no longer the case.
    Pretty soon you'll see Maxwell's Equations all together in their glory. :bow:

    (This stuff was what made me decide to be a physics major, by the way!)
     
  10. Dec 6, 2014 #9
    lol awesome i know relativity is up next so im getting pretty excited
     
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