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Magnatice field, moving rantangle conducting rod, incline plane

  1. Dec 26, 2005 #1
    Imagina an incline plane of angle @. A rod of length L, is at rest( becasuse you are holding it) at the very top of the incline. A magnatice field is pointing up in your mental picture. Find the velocity v of the rod as a function of time once your let go of the rod. Two things in the problem 1) there is no friction involved in this problem. 2) gravitation field exist. Find the velocity v of the rod as a function of time once your let go of the rod.
     
  2. jcsd
  3. Dec 26, 2005 #2
    Same here, show us what you have done so far.

    regards
    marlon

    EDIT : you will need to write down the equations of motion along the incline and perpendicular to the incline. Call those direction x and y, if you want. The forces acting here are gravity and the Lorentz force. Write down their components along each direction...Can you achieve this ?
     
  4. Dec 26, 2005 #3
    I got the answer, but i am curious how other people do the problem. it is not that hard, it is kind of visual. want to compare answers?
     
    Last edited: Dec 26, 2005
  5. Dec 26, 2005 #4
    What is the answer you got?
     
  6. Dec 26, 2005 #5
    v=MgRsin @/ B( square) L( square)cos@( square)

    R is the resistence.
     
  7. Dec 27, 2005 #6
    If this is an isolated rod (meaning no closed current loop) only force in the direction of the velocity is gravitational. In that case I don't see how you got 'B' in your final answer.

    If I assume that somehow this rod is part of a closed circuit, then induced emf,

    [itex]LvB cos(\theta) = IR [/itex]

    The induced current intern give rise to a force along the slope opposite to the velocity.

    [itex] F_B = L^2 B^2 v cos(\theta) / R [/itex]

    Equation of motion,

    [itex] m \frac{dv}{dt} = - L^2 B^2 cos(\theta)/R *v + mg sin \theta [/itex]

    Upon solving this, I get

    [itex]v(t) = \frac{mgR sin(\theta)}{L^2 B^2 cos (\theta)} ( 1- exp( \frac{-L^2 B^2 cos (\theta)*t}{mR})[/itex]


    Gamma
     
    Last edited: Dec 27, 2005
  8. Dec 27, 2005 #7
    Is a close loop. I forgot to put that in. Looking at it from the top view, the rod form one side of the loop. The rod can more up or down the incline.
     
  9. Dec 27, 2005 #8
    sounds good then.
     
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