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Induced current in moving Mercury

  1. Feb 24, 2010 #1
    Hello everyone,
    This question is the third question from the 2001 IPhO:

    Problem text: http://www.jyu.fi/tdk/kastdk/olympiads/2001/IPhO2001-THEO.pdf
    Solution text: http://www.jyu.fi/tdk/kastdk/olympiads/2001/IPhO2001-THEO-SLN.pdf

    I'm having trouble understanding how they came to the conclusion that the current is in the [tex]-\hat y[/tex] direction. (Avoidance of a runaway process aside)
    I came to the conclusion that it's in the [tex]+\hat y[/tex] direction.

    My thought process is as follows:

    [tex]\vec F_{Lorentz}=0[/tex]

    [tex]\vec E = qvB \hat y[/tex]

    [tex]\vec J = \frac{\vec E}{\rho}[/tex]

    [tex]I=\vec J \cdot \vec A = \frac{E}{\rho}Lh[/tex]

    Now all this is still in the [tex]+\hat y[/tex] direction.

    I feel like I should somehow acknowledge the fact that the circuit is shorted out outside the material, and that that's the reason the current in the material goes in the opposite direction, but that doesn't seem right.

    Any advice on what I've missed would be greatly appreciated. :)
     
  2. jcsd
  3. Feb 24, 2010 #2

    rl.bhat

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    Homework Helper

    +y direction is the direction in which the electrons are forced to move. But the conventional current is in the direction opposite to the direction of motion of electrons.
    When you apply ohms law, you have to use the conventional current.
     
  4. Feb 24, 2010 #3
    I didn't think of Electrons at all.
     
  5. Feb 25, 2010 #4
    Speaking of electrons, an alternative analysis yields the following:

    Looking at this from a Hall-Effect point of view, let's look at the individual electrons. If we assume that the protons remain quasi-stationary throughout the buildup of the E-field, then we can look solely at the electrons.

    An electron of charge [tex]-e[/tex] will experience a magnetic force, [tex]-e\vec v\times\vec B=evB \hat y[/tex]

    After sufficient time has passed, there will be a "buildup" of electrons on the far plate (The one further along on the y axis) and an electron deficiency in the near plate. This will induce an electric field going from the positive plate to the negative plate, and thus, also produce a current in the same direction.

    Though I think I have an idea now.

    Throughout my analysis I've neglected the fact that the plates are shorted out. If by shorted out, we are to understand that current would rather go through the wire than through the Mercury, then a possible solution is the following:
    The electrons from the far plate travel along the wire to the positive plate. This in turn drives a motion of electrons through the Mercury from the positive plate to the negative plate. A motion of electrons in the [tex]+\hat y[/tex] direction is equivalent to a positive current in the [tex]-\hat y[/tex] direction.

    I think that's the answer. :) Any corrections would be appreciated.

    Thanks for making me think about the electrons, rl.bhat! Though I think you didn't quite understand what my mistake was.

    Another approach is to look at the closed loop integral of the E-Field around the loop "near plate (0 contribution) - Mercury (positive contribution) - far plate (0 contribution) - conducting wire (negative contribution)"

    And since we assume that the wire is of lower resistance than the mercury, and that the E-field in both is in the same direction, that means that the current in the [tex]+\hat y[/tex] direction must come in through the wire, leaving the return current to go through the Mercury.
     
    Last edited: Feb 25, 2010
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