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Faraday's Law and rail

  1. Jul 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A conducting rail in contact with zero-resistance conducting wires is pushed with constant speed, causing an induced current of 0.69A. Using Faraday's Law, calculate the rail's speed if the wires are separated by 0.34m.

    B = 0.50T
    R = 2.0


    2. Relevant equations


    Magnetic Flux = BAcos()

    B = Change in Magnetic Flux / Change in Area

    Induced Emf = Change in Magnetic Flux / Change in time



    3. The attempt at a solution

    I am rather confused, because Faraday's Law involves using an AREA, and I do not have the appropriate dimensions in order to solve for the area. Also, I can't find an equation that relates to Faraday's law that includes velocity.
     
    Last edited: Jul 1, 2008
  2. jcsd
  3. Jul 1, 2008 #2

    G01

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    As the bar moves outward, the area of the loop is increasing, causing a change in magnetic flux. The velocity will determine how fast the area is increasing.

    HINT: Can you find a relation for the rate of change of the area in terms of the velocity?
     
  4. Jul 1, 2008 #3

    Defennder

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    I assume the setup consists of a conductor rod lying on a pair of metal rails and the rod is moving parallel to the rails.

    So in this case, the change in magnetic flux would be the area covered by the rod within the confines of the metal rail when it moves over a small time period dt. So, the distance covered would be vdt and the area would then be Lvdt, where L is the wire separation distance. I'm sure you can take it from here.
     
  5. Jul 2, 2008 #4
    I am still lost. I don't see how I can use "Lvdt" when there is no variable for distance and I am solving for the velocity. I have two unknown variables in that equation.

    G01 -- The bar is positioned perpendicular to the rods, and it is moving inward, causing a decrease in the area of the loop.

    By the way, R = 2.0 = a RESISTOR, not the radius. Just thought I'd clear that up. :)
     
    Last edited: Jul 2, 2008
  6. Jul 2, 2008 #5

    G01

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    OK. You know have figured out that this is a Faraday's law problem. So, let's start there:

    [tex]V_{induced}=-\frac{d\Phi_m}{dt}[/tex]

    Now what is phi? You said that in your first post, therefore:

    [tex]V_{induced}=-\frac{d\Phi_m}{dt}=-\frac{d(BA)}{dt}[/tex]

    Now, the problem is we don't have v involved in the equation. My question to you is how can we involve v in a formula for the area? If we could do that, we could then plug that formula in for A in the above equation and solve for v, right? Now, can you take it from here? How can we represent A, the area of the loop in terms of v?

    HINT: Start with the standard equation for the area of a loop of that shape.
     
  7. Jul 2, 2008 #6
    A = Lvdt

    Am I getting somewhere?

    So now I plug in:

    Vinduced = -d(Lvdt) / dt

    Now I'm stuck again. There is still no distance traveled by the rail involved in the problem.
     
    Last edited: Jul 2, 2008
  8. Jul 2, 2008 #7

    G01

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    That's it. (Yes, you are getting somewhere.:smile:)

    Now, what are you going to do next?
     
  9. Jul 2, 2008 #8
    So now I plug in:

    Vinduced = -dB(Lvdt) / dt

    Factoring out... Vinduced = -dBLv

    Now I'm stuck again. There is still no distance traveled by the rail involved in the problem and I don't know Vinduced.

    (by the way -- thank you for helping me -- I know I'm totally clueless on this problem!)
     
  10. Jul 2, 2008 #9

    G01

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    OK, well your calculus is not correct, but if you haven't had any calculus, then I guess I can't hold it against you.:smile: Your equation is almost correct. Your final equation for v induced is:

    [tex]V_{induced} = -BLv[/tex] (that d is a part of a derivative operator that should not appear in your final line. Have you had calculus? If not, then lot's of what I wrote must be confusing!)

    This problem can be done without calculus as well, basically what you are doing is:

    [tex]V_{induced} = -\frac{B\Delta A}{\Delta t} = -BLv[/tex]

    Correct?

    Anyway, now you should be able to solve for v! L is given in the problem, and V induced can be found. How do you find V induced? HINT: What haven't you used yet?
     
    Last edited: Jul 2, 2008
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