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Electromagnetic induction

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    We have a coil with cross-sectional area of 2.4m^2.
    Magnetic flux density is of 0.29T
    Number of turns are 1.

    Initially the coil is parallel to magnetic field lines. (Coil's normal vector is perpendicular to field vector)

    Calculate the change in magnetic flux when coil is moved through an angle of 1 degrees

    2. Relevant equations
    Δ
    Φ = ΔBAcosθ

    3. The attempt at a solution
    My attempt
    ΔΦ = ΔBAcosθ
    ΔΦ = BA Δcosθ
    ΔΦ = BA (cos 89 - cos 90)
    Which leads to a correct answer

    Examiner attempt
    ΔΦ = ΔBAcosθ
    ΔΦ = BA Δcosθ
    ΔΦ = BA cos (90-1) How do they do this?
    Which leads to a correct answer

    Another question inquired me to calculate the change of flux when the change in θ is from 0 to 90. I used the normal way of multiplying BA by (cos 90 - cos 0). However, the examiner stated that this is simplified treatment, a rigorous method would involve averaging of cos θ leading to a factor of 1/sqrt2.
    How is this factor achieved? I tried to calculate the mean value of cos θ using integration but the answer was different.
     
  2. jcsd
  3. Mar 21, 2017 #2

    BvU

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    To me both answers look the same: cos(89) because of the value of cos(90) :smile: You sure your calculator is on degrees and not on radians ?
    And how do they do it ? Probably they start with making a (top view) sketch !

    There must be a misunderstanding there. Do you have the full litteral text of thte question that was asked you ?
     
  4. Mar 21, 2017 #3
    No I meant, if I wanted to calculate the change in a value of a function at the interval ##a## and ##b##, I will calculate ##f(b)-f(a) \neq f(b-a)##. But in the provided example taking ##cos(90-1)## gave the answer. How was that possible?
     

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  5. Mar 21, 2017 #4

    BvU

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    In the 89 degree case the change in flux was asked for. That is a simple difference.
    In exercise 6 they explicitly ask for an average emf. Do you have a relationship (formula, expression) that links ##\Phi## to the emf ?
     
  6. Mar 21, 2017 #5
    ##Φ=\frac{-dV}{dt}=\frac{BANcos\theta}{\delta t} ##
     
  7. Mar 21, 2017 #6

    cnh1995

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    Are you sure?
    As per Faraday's law, what is the derivative of what?
     
  8. Mar 21, 2017 #7
    Oh sorry my bad
    ##V=N\frac{-dΦ}{dt}=\frac{\delta BANcos\theta}{\delta t} ##[/QUOTE]
     
  9. Mar 21, 2017 #8

    BvU

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    Don't know why you switch to a partial derivative ?
    Write ##\theta## as a function of ##t## and you can take the average of V for a quarter turn in 0.14 s: $$<V> = {1\over 0.14}\int_0^{0.14 s} V\,dt$$
     
  10. Mar 21, 2017 #9
    And how do I write theta as a function of t?
    I tried 0.14/90 t = theta but ended up getting the wrong answer
     
  11. Mar 21, 2017 #10

    BvU

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    You may assume it turns with a constant angular speed from ##\pi\over4## to ##0## in 0.14 seconds

    [edit] sorry,: yes, that should be ##\pi\over 2##.

    Check your dimensions: on the left you have s2/degree and on the right an angle in degrees!
     
    Last edited: Mar 21, 2017
  12. Mar 21, 2017 #11
    <Post deleted>
     
    Last edited: Mar 21, 2017
  13. Mar 21, 2017 #12
    Don't you mean pi/2? The rotation is of 0 to 90 degrees.

    Is this correct?
    $$ <V> = \frac{0.0123}{0.14}\int^{0.14}_0 \cos{\frac{90}{0.14}t}~ dt $$
     
    Last edited: Mar 21, 2017
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