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Homework Help: Need help with an electromagnetic induction problem concerning two rods

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Two metal rods are placed on two parallel smooth conducting rails , under uniform B field lines . The metal rod on right-hand side is initially moving at a constant velocity of 3 m/s , while the other one is initially at rest . What would be the subsequent motions of the two rods and what are their final velocities ?

    Assume resistance of all conducting material is negligible .

    2. Relevant equations

    3. The attempt at a solution
    I guess they will eventually go at the same velocity , with their kinetic energy conserved ?then their velocities will be 3 divided by root 2 , rightward.

    Attached Files:

  2. jcsd
  3. Jun 6, 2012 #2
    You need to make a better attempt at that solution before we can further assist you.

    Why do you think they will eventually go at the same velocity?

    Seeing this problem makes me think about motional emf.

    [tex]\vec{F} = q ( \vec{v} \times \vec{B} )[/tex]
    Last edited: Jun 6, 2012
  4. Jun 6, 2012 #3
    As the moving rod (named rod B) is moving , by fleming's LHR , there will be an induced emf , and so current flows upward through rod B , now that current in rod B interact to push rod B leftward , the other rod (rod A) is pushed rightward by the same mechanism as well . So the two rods approaches each other , both rod A and B will produce an emf to oppose their current motion ( by lenz's law) and so a 'cycle' begins . I think i need to think deeper...but i barely can . i have a thought that each time they change their motion , they share their kinetic energy , and eventually they move together at the same velocity , where there is no current to change their motions.
  5. Jun 6, 2012 #4
    How is this?

    If you are implying rod A and rod B travel at the same velocity then there is no change in flux, correct? If there's no change in flux there's no induced current correct?
  6. Jun 6, 2012 #5
    current generated by the rod B flows downward through rod A while it flows upward through rod B . Current in rod A interacts with the external B field , causing rod A to be pushed rightward . now that Rod A moves rightward and rod B moves leftward , so there'll still be induced emf.
  7. Jun 6, 2012 #6
    Can you show some work justifying some of your claims?

    I think once you start poking around with some of the formulae things will become clearer.
  8. Jun 6, 2012 #7
    unfortunately i have not been taught any formulae . the claims can be justified by fleming's left hand rule
  9. Jun 7, 2012 #8
    Are you supposed to be neglecting the magnetic field produced by the induced current or not?

    If not then what you have explained (if I am understanding it correctly) seems to be correct.

    Assume rod A is stationary and rod B is moving a constant velocity. Now instantaneously your B field appears, and there is an induced current in the loop.

    This induced current exerts a rightward force on rod A and a leftward force on rod B. As these rods change velocity eventually the net flux passing through the loop will be decreasing as a opposed to increasing and thus the direction of induced current will reverse itself, causing leftward force on rod A and a rightward force on rod B.
  10. Jun 8, 2012 #9
    That's correct . However , what has been confusing me is...what happens afterwards ? After a leftward force on rod A and a rightward force on rod B are produced , will the two rods repeat their motions as they do after a current is induced in the loop for the first time ?
  11. Jun 11, 2012 #10
    My belief is that the rods will oscillate back and forth until the oscillations become so small that the two bars have reached a steady state and are (for our purposes) stationary.
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