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Another Electromagnetic Induction Question

  1. Apr 9, 2005 #1
    Again, this is a problem I got in my AP Physics Class can anyone help? I would really appreciate any help. My teacher didn't explain this concept too well. :surprised

    A metal bar with length [tex] L[/tex], mass [tex]m[/tex], and resistance [tex]R[/tex] is placed on frictionless, metal rails that are inclined at an angle [tex] \theta[/tex] above the horizontal. The rails have negligible resistance. A uniform magnetic field of magnitude [tex]B[/tex] is directed downward. The bar is released from rest and slides down the rails.

    a. What is the terminal speed of the bar?
    b. What is the induced current in the bar when the terminal speed has been reached?
    c. After the terminal speed of the bar has been reached, at what rate is electrical energy being converted to thermal energy in the resistance of the bar?
    d. After the terminal speed has been reached, at what rate is work being done on the bar by gravity?


    I attemped first to use the equations that
    [tex]\varepsilon=BLv[/tex]
    [tex] i=\frac{\varepsilon}{R}[/tex]
    [tex]F_B=iLB.[/tex]

    After some substitutions I find
    [tex]F= \frac{B^2L^2v}{R}[/tex]
    but what do I do next?
     
    Last edited: Apr 9, 2005
  2. jcsd
  3. Apr 9, 2005 #2
    I think I made a little progress. After drawing a force diagram, I think I could say that the magnetic force = the x component of the force of gravity(i split gravity into components rather than anything else). so then i said that:

    [tex]mg\sin{\theta}= \frac{B^2L^2v}{R}[/tex]

    and then solved for [tex] v[/tex]. Would this be right for the terminal velocity?
     
    Last edited: Apr 9, 2005
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