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Concentric Coils EMF from Graph

  1. Nov 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Two concentric circular coils of wire lie in a plane. The larger coil has 61 turns and a radius of a = 7.40 cm. The smaller coil has 58 turns and a radius of b = 0.95 cm. The current in the larger coil has a time dependence as shown in the figure.

    Picture and graph:
    http://imgur.com/fHB07bx

    Approximately what is the magnitude of the EMF induced in the small coil at t = 8 s?
    2. Relevant equations
    ξ=(d/dt)∫B⋅dA

    ∫B⋅dl=μ*I

    3. The attempt at a solution

    I tried taking the derivative of the slope given with respect to time
    So -15*e^(-3t)
    And multiplying that by the area of the small coil and the number of turns in the small coil.
    I don't really know where to go from there.


    I got 3.78E-12 V and thats not right
     
  2. jcsd
  3. Nov 6, 2016 #2

    TSny

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    The magnetic field produced by the large coil can be considered as uniform over the area of the small coil. This simplifies calculating the flux through the small coil.

    You'll need to recall how to find B at the center of a circular coil due to a current in the coil.
     
  4. Nov 6, 2016 #3
    So the Biot-Savart Law? So dB=(μIdl×r)/(4πR^2)
    Can dL be simplified to 2πR meaning the law could be written as (μI)/(2R) and then multiplied by N?
    But where does the smaller coil come into play? I guess I'm confused by the concept
     
  5. Nov 6, 2016 #4

    TSny

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    Yes
    The smaller coil comes into play when you are trying to find the induced emf in the smaller coil. What do you need to know in order to find the induced emf in the smaller coil?
     
  6. Nov 7, 2016 #5

    andrevdh

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    The large coil causes a flux change in the smaller one when the current changes in it.
    How are the flux change and the induced emf related?
     
  7. Nov 7, 2016 #6
    The derivative of the change in flux with respect to time is emf, but I guess I'm most confused about how to go about the induced part of this problem
     
  8. Nov 7, 2016 #7
    The flux with respect to time, is that where the graph comes into play?
     
  9. Nov 7, 2016 #8

    TSny

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    Yes. Note that the exponential decay part of the graph is proportional to ##e^{-(t-6)/2}##, not ##e^{-3t}##.
     
  10. Nov 7, 2016 #9
    Oh ok, I must have missed that.

    So it would be d/dt(μ*5e^(-(t-6)/2))/2Ra?

    Does that make sense? That would be the emf in the large coil, but how do I translate that into induced emf for the smaller one? Or is it all the same?
     
  11. Nov 7, 2016 #10

    TSny

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    What does this expression represent?
     
  12. Nov 7, 2016 #11
    The flux of the large coil, with current as a function
    When the derivative is taken with respect to time, its the emf in the large coil
     
  13. Nov 7, 2016 #12

    TSny

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    No, it is not the flux or the time derivative of the flux.

    In the expression d/dt(μ*5e^(-(t-6)/2))/2Ra, you are taking the time derivative of μ*5e^(-(t-6)/2))/2Ra. What does μ*5e^(-(t-6)/2))/2Ra represent? Refer back to post #3.
     
  14. Nov 7, 2016 #13
    Oh! The magnetic field created by the large coil. So I need to multiple in dA, which would be 2*pi*r to get flux. Then I can take the time derivative to get emf
     
  15. Nov 7, 2016 #14

    TSny

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    Is it the total magnetic field created by the large coil? Remember, the large coil has a certain number of turns.
    What does dA represent? The formula 2*pi*r gives you the circumference of a circle.

    Basically, yes. But don't forget that the small coil also has a certain number of turns.
     
  16. Nov 7, 2016 #15
    Ok...

    So would the correct equation be:
    Na*Nb*pi*Ra^2*μ*5e^(-(t-6)/2))/2Ra

    And then take the derivative with respect to time and plug in whatever value for t?

    I think I understand the concept.
     
  17. Nov 7, 2016 #16

    TSny

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    Sorry to be picky, but this is not an equation. It is an expression for something. Is it an expression for B? for flux? for emf?

    In the expression, you have an area pi*Ra^2. This would be the area of the large coil. Is that the appropriate area to use?

    You're getting close.
     
  18. Nov 7, 2016 #17
    That's the expression for flux.

    I would need to use the area of the small coil because that is what experiencing the induced emf.

    So Na*Nb*pi*Rb^2*μ*5e^(-(t-6)/2))/2Ra is the flux, and the time derivative will be the induced emf in the small coil
     
  19. Nov 7, 2016 #18

    TSny

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    That looks good.
     
  20. Nov 7, 2016 #19
    Thank you so much for all your help!
     
  21. Nov 7, 2016 #20

    TSny

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    OK, nice work. Welcome to PF.
     
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