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Question about Faraday's law.

  1. Apr 14, 2012 #1
    When we talk about Faraday's law, we need a change in magnetic flux to create a voltage.
    So lets say I have a B field that exists in a square region and then a square loop that is a conductor and I push it into the B field. So as I originally push the square loop into the B field I have a change in magnetic flux, so there will be a voltage and the free electrons in the conductor will start to move. Or I could view it as the free electrons are approaching the B field with a speed v and they will experience a Lorentz force and this will start a current to flow in the loop. But now lets say the loop it completely in the B field, so the magnetic flux is constant. But I still would have a current in the loop because of the Lorentz pushing on the electrons. And I still would have a voltage because I have current in the loop. Am I thinking about this correctly?
     
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  3. Apr 14, 2012 #2

    jambaugh

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    Yes, provided the conducting loop has zero resistance. In fact, what will occur is that the current induced will create its own B field so that the total flux through the loop remains zero.

    Now usually there's some resistance so the current dissipates and the net B flux grows to the full amount you'd calculate. The heat from the dissipation of the current will tell you exactly how much work it took to push the loop into the B field and thus the force on the wire due to this current and the B field.

    The flux analysis is derivable by looking at the Lorentz forces on the charge carriers in the wire as it moves into the B field, (or the induced E field if you don't move the wire but rather change the B field over time).
     
  4. Apr 14, 2012 #3

    sophiecentaur

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    You need to remember that these 'free electrons' are in very rapid random motion and their paths in the metal are all affected by the magnetic field. The mean path is affected as you suggest, though. Also, the drift speed (RMS velocity) of conduction electrons is extremely low.
     
  5. Apr 14, 2012 #4

    jambaugh

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    I think you can ignore the random e.g. thermal fluctuations. The magnetic field will simply rotate the random motion which will in turn be symmetrically random as well. It is the net motion of the whole of the conductor which is relevant. Moving the wire across the B field, induces forces along the wire, in one direction for the positive charges, in the other direction for the negative. If either or both are mobile they will flow and you have your net current.
     
  6. Apr 14, 2012 #5

    sophiecentaur

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    I agree totally but the mean drift speed is very low and this is not always acknowledged. Electrons are so slow and low in mass that they move right down the potential slope - no curved paths because of their momentum and they accelerate ' instantaneously' to this speed.
     
  7. Apr 14, 2012 #6
    ok, interesting thanks for your responses
     
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