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Average Induced Voltage

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data

    The magnetic field shown in Figure P20.61 has a uniform magnitude of 29.00 mT directed into the paper. The initial diameter of the kink is 0.30 cm.

    (a) The wire is quickly pulled taut, and the kink shrinks to a diameter of zero in 55.00 ms. Determine the average voltage induced between endpoints A and B. (Include the polarity, with + for positive A)

    (b) Suppose the kink is undisturbed, but the magnetic field increases to 96.00 mT in 3.65×10-3 s. Determine the average voltage across terminals A and B, including polarity, during this period.

    (a) B=.029 T d/2=r=.0015 m A=7.065E-6 m^2 Flux final = 0 t=.055

    (b) B=.096 T A=7.065E-6 m^2 Δt=3.65E-3 s

    2. Relevant equations

    I think I'm supposed to use the following:
    A=∏r^2 Flux=BAcos∅ Emf=-N ΔFlux / Δtime Emf=Flux(final)-Flux(initial) / Δt

    3. The attempt at a solution

    a) Flux =7.065E-6 X .029 = 2.04E-7 --> Emf= 2.03E-7 / .055 = 3.73E-6 Volts

    b)Flux2 =.096 X 7.065E-6 = 6.78E-7 6.78E-7 - 2.04E-7 = 4.74E-7
    Emf = 4.74E-7/3.65E-3 s = 1.30E-4 Volts
     
  2. jcsd
  3. Mar 2, 2013 #2

    Doc Al

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    Staff: Mentor

    Looks good to me. (Next time include the diagram.)
     
  4. Mar 2, 2013 #3

    PeterO

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    Homework Helper

    How did you go with the polarity? That's the part people most commonly get wrong. In (a) the flux reduced to zero, in (b) the flux increased so they can't both be positive, so I am guessing you have not attempted that part yet.

    By the way, the answers you gave are incorrect, they are merely copied from a calculator. part (a) may possibly be 3.73 x 10-6 Volts
     
  5. Mar 2, 2013 #4

    Doc Al

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    Why do you say the answers are incorrect? You gave the same answer.

    To deal with the polarity, we'll need the diagram.
     
  6. Mar 2, 2013 #5

    PeterO

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    I was commenting on the E-6 part of the answer.

    Any answer which includes such references is necessarily marked wrong in our school system. It gives no indication of understanding what the calculator is telling you.
    Same as quoting too many significant figures simply because you wrote down all the numbers the calculator was telling you.

    The answer I gave was in scientific notation, not calculator speak.
     
  7. Mar 2, 2013 #6

    Doc Al

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    OK, but that's fairly standard notation.
     
  8. Mar 2, 2013 #7

    PeterO

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    It may be a common way students write their answers but it is still wrong. In no way should it become accepted as a fairly standard notation.
    Students will only continue to write their answers in such a way if their answer is accepted as correct.
    Because of the way some calculators display, the answer 3.73-6 is sometimes offered up. Would you accept that as correct?

    Even to say 3.73 μV would be better.
     
  9. Mar 2, 2013 #8

    Doc Al

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    You are free to do whatever you like in your own classroom. Perhaps the OP just blindly copied an answer from the calculator and has no clue what it means. But I would not assume that.
    No.

    I think it's pretty unhelpful to make a blanket statement such as "Your answers are incorrect", when correct work was shown. A more helpful suggestion would be: You may want to express your answers in standard scientific notation.
     
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