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Sliding Bar in a Uniform Magnetic Field

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data
    A metal bar with length L, mass m, and resistance R is placed on frictionless metal rails that are inclined at an angle [itex]\phi[/itex] above the horizontal. The rails have negligible resistance. A uniform magnetic field of magnitude is directed downward in the figure B. The bar is released from rest and slides down the rails.

    (In the diagram it can be seen that the bar divides a rectangle into two smaller rectangles. As the bar slides, one grows, and the other shrinks.)

    What is the terminal speed of the bar?

    2. Relevant equations
    [itex]\cal{E}=IR[/itex]
    [itex]\cal{E}=-\frac{d\Phi_m}{dt}[/itex]


    3. The attempt at a solution
    Effective gravity has magnitude

    [itex]mg\sin\phi[/itex]

    The two wire loops formed by the bar have respective emfs

    [itex]{\cal E}_{1}=-\frac{d\Phi_{1}}{dt}=-BLv[/itex]
    [itex]{\cal E}_{2}=-\frac{d\Phi_{2}}{dt}=BLv[/itex]

    Therefore, the current along the bar is

    [itex]I=\frac{-{\cal E}_{1}+{\cal E}_{2}}{R}=\frac{2BLv}{R}[/itex]

    The negative sign appears because one loop runs clockwise and the other counterclockwise.

    This creates a force

    [itex]F=ILB=\frac{2B^{2}L^{2}}{R}v[/itex]

    and an effective force

    [itex]\frac{2B^{2}L^{2}}{R}v\cos\phi[/itex]

    The two are in equilibrium when

    [itex]\frac{2B^{2}L^{2}}{R}v = mg\tan\phi[/itex]
    [itex]v = \frac{Rmg\tan\phi}{2B^{2}L^{2}}[/itex]

    The online homework system says this is incorrect, and I can't fiqure out why.
     
  2. jcsd
  3. Nov 28, 2011 #2
    Oddly enough, I used my incorrect answer in all of the three remaining parts of the problem and got them correct. They asked for the terminal current, the power in the resistor, and the power of gravity (the latter two of which were equal).
     
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