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Eddy-current braking system, find the new velocity

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the eddy-current braking. A square loop, with 10 cm side is shot with the velocity 10 m/s into the uniform magnetic field with magnitude 0.1 T. The field is perpendicular to the plane of the loop, and the loop starts entering magnetic field at t=0. The resistance of the loop is 1.00 Ohm and the mass is 1.0 g. Assume the loop is moving to the right along x-axis and that x(t=0)=0. Find the velocity of the loop 0.1 seconds later. Comment on assumptions and approximation, or venture into the realm of differential equations…


    2. Relevant equations
    A = 0.1m X 0.1 m = 0.01m2
    V = 10m/s
    B = 0.1T
    R = 1Ω
    m = 0.001kg
    t = 0.1s
    Flux = BA = 0.001Wb
    ε = Flux/t = 0.01v
    I = ε/R
    l = 0.1m

    3. The attempt at a solution
    My first reflex was to find the current: I = .01v/1Ω = 0.01A

    And then I wanted to find the new velocity with the formula above, problem is; IR/Bl = (0.01A X 1Ω)/(0.1T X .1m) = 10 m/s, so that doesn't work

    And now I'm stumped, I can't really figure out this problem -_-
     
  2. jcsd
  3. Nov 11, 2012 #2

    mfb

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    Does your setup depend on the 0.1 seconds given in the problem statement? If not, why do you expect that the voltage depends on that time?
    You have to use another value for the time here.

    You could try energy conservation to calculate the velocity change.
     
  4. Nov 11, 2012 #3
    If I take this approach, I could find Ek = 0.5mv2 = 0.05j

    And then I assume I'd have to find the potential energy of the system once it has been slowed down after the 0.1 second (by using the 0.05j I found with law of conservation of energy), but again I am stumped and do not know how to go about finding that information.

    Second method does make a lot more sense than the first one though.
     
  5. Nov 11, 2012 #4

    mfb

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    There is no relevant potential energy. You'll lose some energy due to the current flow in the curcuit.
     
  6. Nov 11, 2012 #5
    If I lose energy to the current flow, how can I use energy conservation to calculate the velocity change like you previously mentioned?

    Or can I simply go ahead right away and find the energy of the system at 0.1s, and then use that in the kinetic energy formula to find the velocity? (Don't know how to do that yet, but if I'm on the right track I'll work on that)
     
  7. Nov 11, 2012 #6

    mfb

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    kinetic energy before = kinetic energy afterwards + energy "lost" due to the resistance

    "lost" in a technical sense here - the coil got warmer by a tiny amount.
     
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