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Hw help please

  1. Feb 11, 2004 #1
    can anyone help me with this question from my homework?
    i have the system in the picture and i need to find out the type of movement of the bar.
    the box named A is an inductor, the 2 long parallel lines are infinite conductors. there's a magnetic field into the page. the black bolded line is a conductor with a mass of M and weight of Mg.
    there's no resistence in the circuit.
    and all i need to find is the type of the movement
    thank you very much!!!
     
  2. jcsd
  3. Feb 11, 2004 #2
    i forgot to attach...

    the circuit
     

    Attached Files:

  4. Feb 11, 2004 #3
    u forget to give ur try
     
  5. Feb 11, 2004 #4
    my try...

    as far as i know about inductors i think it'll make an harmonic type of move...
    but it seems to make no sense
    can anyone help?

    (i post it as a new thread by mistake, sorry...)
     
  6. Feb 11, 2004 #5

    NateTG

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    What is the net force on the system?
    Do you think there will be current flowing in the wires?
    If yes, which direction do you think the current will flow in?
     
  7. Feb 11, 2004 #6
    >>"What is the net force on the system?"
    i know that the force is the sum of gravity and lentz law
    i don't know if the'lly be equal or not

    >>"Do you think there will be current flowing in the wires?"
    i think there'll be current in the wires, there's also no resistence in the system, but the inductor has its induction which i didn't learn well enough

    >>"If yes, which direction do you think the current will flow in?"
    i know that the flow direction will be in the way that the lorentz force it'll cuase will be upside (resist to the change of the magnetic fkux)
     
  8. Feb 11, 2004 #7

    turin

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    They will not in general be equal. But, at least nature was nice enough to allow you to approximate gravity as a constant. If the rod starts from rest, then what is the Lentz force at this point in time? Better yet, what is the variable in this problem on which the Lentz force will depend (and I'll give you a hint, it isn't the area enclosed exactly, but it is related to that)?




    Do you know the value of the inductor? A voltage will develop across it to counteract a change in current (which is very much like the same principle as the Lentz force. In fact, the big rectangle is a big inductor). The formular is V = L(di/dt). You can relate this to a change in a magnetic field (and therefore flux).




    That's correct, and I'm pretty sure that is all you need to know about the direction. Both the inductor and the Lentz force will oppose change in the magnetic flux.

    EDIT:
    Hmm. Actually, I have to second guess myself regarding the effect of the inductor. I'll get back to this.
     
    Last edited: Feb 11, 2004
  9. Feb 11, 2004 #8

    NateTG

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    [tex]V=L\frac{dI}{dt}[/tex]
    where L is the inductance, and I is the current.

    It may be instructive to solve this problem ignoring the inductor first. (That's probably what I would do since I know almost nothing about AC circuits.)
     
  10. Feb 11, 2004 #9

    turin

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    I don't understand how one would do this. The more I work on this problem, the harder I think it is. Do you know the solution, NateTG?




    I got a result of the form:

    x(t) = x0 + A t sin(ω t + φ(v0))

    I neglected the energy term in the magnetic field produced by the current in the loop, and only used the magnetic reactive energy in the inductor and the kinetic energy of the rod. I assumed an ideal inductor and frictionless rails, so the sum of these two energies should be a constant. Then I used the diff. eq. for the inductor and Faraday's Law to get a 2nd order ODE in velocity. Anyone please verify this for ismael and me. I am utterly uncertain of it.




    If the form above is correct, then it looks like you were correct about the harmonic motion, ismael. Good intuition.
     
    Last edited: Feb 11, 2004
  11. Feb 11, 2004 #10

    NateTG

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    You should definitely not be getting any kind of occilations.

    Let me give it a whirl:

    The energy stored in the inductor is:
    [tex]E=\frac{1}{2}LI^2[/tex]
    where [tex]L[/tex] is the inductance.
    The kinetic energy of the bar is
    [tex]KE=\frac{1}{2}mv^2[/tex]
    so I have:
    [tex]mgh=\frac{1}{2}LI^2+\frac{1}{2}mv^2[/tex]

    Now all I need is an equation relating the velocity to the current which comes from Faraday's laws...
     
  12. Feb 11, 2004 #11

    turin

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    Oh crap! I forgot to include the grav. pot. E. I think that makes it orders of magnitude more complicated than even the way I did it, for which I had to look up an integral (though it was an inverse trig one, so it wasn't that bizarre, and I really should have it memorized).
     
  13. Feb 12, 2004 #12

    NateTG

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    Now, the rate of change of the flux is going to be proportional to the velocity, so let's (for now) say that
    [tex]I=kv[/tex] for some [tex]k[/tex].

    so we have:
    [tex]mgh=\frac{1}{2}Lk^2v^2+\frac{1}{2}mv^2=[/tex]
    simplify a little
    [tex]mgh=(\frac{1}Lk^2+\frac{1}{2}m)v^2[/tex]
    so
    [tex]\sqrt{\frac{2mgh}{Lk^2m}}=v[/tex]

    Of course this is somewhat less than ideal if you want to have velocity as a function of time, instead of velocity as a function of position.

    If you're looking for intuition, I would guess that you will end up with constant acceleration.
     
  14. Feb 12, 2004 #13

    turin

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    Whince did you get this? I thought that the changing flux would induce an emf directly, and only indirectly induce a current. The way I understand it is that the emf around the circuit is
    Vccw = -Blv.
    Then, this is the emf across the inductor (because the inductor is the only non-trivial electronic element), which gives an expression for the current as
    -Blv = L(diccw/dt)
    or
    Bl(dy/dt) = L(diccw/dt).
    That's one diff. eq. with two dependent variables, y(t) and i(t). Then, to get another equation, you use the enery relationship (the one that you corrected for me by including gravity). But this system is non-linear (and therefore scary to me).

    I think it would be more interesting (and/because more realistic) to add a resistance to the loop. If one must go through such a complicated computation, one might as well throw in a resistor for realistic good measure, IMO.
     
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