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Metal bar on rails, induced emf

  1. Mar 18, 2014 #1
    A metal bar with length L, mass m, and resistance R is placed on frictionless metal rails that
    are inclined at an angle α above the horizontal. The top end of the rails are connected with a
    conducting wire. The resistance of the rails and wire are negligible. The rails are embedded in a
    uniform magnetic field B perpendicular to the plane in which the rails sit. The bar is released from
    rest and slides down the rails. Determine the magnitude of the induced emf on the loop after a generic time τ , shorter than the time required to reach the terminal velocity.




    I know that [itex]|\epsilon|=\frac{d\phi}{dt}[/itex], where in this case [itex]d\phi=BdA=BL\frac{g}{2}sin(\alpha)\tau^{2}[/itex] and [itex]dt=\tau[/itex], so [itex]|\epsilon|=BL\frac{g}{2}sin(\alpha)\tau[/itex]. But it doesn't seem right to me because at some point, the bar will reach terminal velocity, which means the bar must be decelerating and I haven't taken that into account because I don't know how to. Any help will be appreciated.
     
  2. jcsd
  3. Mar 18, 2014 #2

    tiny-tim

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    hi subzero0137! :smile:
    the acceleration will be less than gsinα

    because the constant gravitational component of force is opposed by the velocity-dependent electromagnetic force :wink:
     
  4. Mar 18, 2014 #3
    I see. But how would I account for that? Would the acceleration simply be gsin(α)-F(v)? I don't know what F(v) is...
     
  5. Mar 18, 2014 #4

    tiny-tim

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    you have mx'' equals a function of x' …

    write the equation out, and solve it! :smile:
     
  6. Mar 18, 2014 #5
    Sorry, I'm confused now. Is mx'' the overall force? If yes, shouldn't it be mx''=mgsin(α)-F(x')?
     
  7. Mar 18, 2014 #6

    tiny-tim

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    your RHS is a function of x' :smile:
     
  8. Mar 18, 2014 #7
    Ohh, of course :tongue:

    But when you say "solve it", do you mean solve the differential equation? Because my class hasn't solved diff equations of that form yet.
     
  9. Mar 18, 2014 #8

    tiny-tim

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    it's something like dv/dt = Av + B, or dv/(Av + B) = dt …

    can you solve that? :smile:
     
  10. Mar 18, 2014 #9
    It'll be (1/A)ln(Av+B)=t+C, but I don't know how to use that result in this question :confused:
     
  11. Mar 18, 2014 #10

    tiny-tim

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    (or Av+B = AoeAt)

    that gives you v as a function of t, from which you can get the induced emf :smile:
     
  12. Mar 18, 2014 #11
    But what are the constants A, Ao and B, and how does v as a function of t give me F(v)? Sorry if I'm being difficult, but I genuinely don't understand. Wouldn't F(v)=BIL, where I=BLv/R?
     
  13. Mar 18, 2014 #12

    tiny-tim

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    write out the equation properly, and solve it (using v(0) = 0), and all that should become clear :smile:

    (and now i'm off to bed :zzz:)​
     
  14. Mar 18, 2014 #13
    Which equation, F=(v(BL)^2)/R or Av+B=Ao*e^AT? Before you go, can you please tell me what F(v) should be? I can't do other parts of the question if I don't get past this point....
     
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