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My mind is blown. ΔV problem from MIT lecture.

  1. Feb 13, 2013 #1
    The lecture in question...


    So what happens if I actually build this circuit and hook my fluke to it? How can moving my meter from the right side of the table to the left change my voltage reading if the circuit and my meter's connection to it is untouched? What happens if I put the meter directly on top of the solonoid?

    Can someone try to explain this in a way that is different than the professor did?
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Feb 13, 2013 #2
  4. Feb 14, 2013 #3

    sophiecentaur

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    Good question.
    I think the answer, in practical terms, must be to do with the way your meter leads would routed (there have to be connecting leads, of course). They will also experience an induced emf as the magnet moves in and out. If they are on the left of the loop, the emf will be in the opposite direction to the emf induced when they're on the right side - this will add or subtract the emf from the emf across the various components and give you different answers.
     
    Last edited by a moderator: Sep 25, 2014
  5. Feb 14, 2013 #4
  6. Feb 14, 2013 #5

    sophiecentaur

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    Great article. I love it when someone has time, inclination and resources to demonstrate such basic stuff. Lewin's problem is that he sounds a bit too much a a showman at times and that can lead to doubt in the minds of students.
    Lesson: use Kirchoff with care, same as a chain saw.
     
  7. Feb 14, 2013 #6
    Thanks for the insights. This is making much more sense now.

    Suppose though that we connect 2 metal objects to different points in the circuit. We pulse the solonoid. At the same instant, while the field is moving, we disconnect the 2 metal objects. Each one should now have a static charge representitive of the potential at the point it was connected to. We've taken the meter leads completely out of the picture. We measure the potential difference between the 2 objects after the circuit is powered down. What do we get?
     
  8. Feb 14, 2013 #7

    sophiecentaur

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    The PD between your two objects cannot be defined in that case without specifying the path you want to take. It's not a 'conservative field, as with a battery supply.
    (Nice try!! :wink:)
     
  9. Feb 14, 2013 #8
    Electrons in a wire do not behave like water. It's not the case that you push electrons into the wire and those then push along the other electrons.
    All the electrons in the wire are pushed by an electric field and that field acts on the entire wire and all resistors in the circuit. That includes the wires that lead to the meter and the meter itself. It's all inside the electric field that is induced by the solenoid.
    The same can be said about a battery connected to a circuit. The battery is not simply pushing electrons into the circuit at one end and pulling them out at the other. The battery produces an electric field that is felt by the entire circuit and can push electrons in the entire circuit simultaneously. Resistors will cause electrons to "pile up" so that they become a source of an electric field themselves. If you overlap the electric fields produced by all the resistors with the field produced by the battery and use that to calculate the potential differences inside the circuit you will get the same results you would get with the "normal" method i.e. kirchoff as long as no changing magnetic fields are involved.
    If you briefly connect two metal objects to point A and D while the magnetic field is changing the PD you get between them depends on how much charge is being pushed into them by the electric field. That in turn depends on how the objects are connected and in which direction the field is pointing.
     
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