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Alternating Current Generator (Magentic Field.)

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    An alternating current generator consisting of a rectangular loop of length a = 0.1xm and width b = 0.1ym with
    N= 1w0 turns is rotated at a frequency f = 1.z Hz in a uniform magnetic field B= 0.1 T which points into the page.
    A load resistance R = 10 Ω completes the generator circuit. At time t = 0, the loop is momentarily vertical (Parallel to B really).
    x,y and z are known constants.
    (a) What is the induced emf in the circuit at t =0?
    (b) Calculate the current in the loop when it has
    rotated through an angle of π/6 radians.
    (c) Calculate the time-averaged power developed in
    the circuit over a complete revolution.
    Hint: calculate the average value of sin(2θ)

    2. Relevant equations
    emf = NABωsin(ωt).
    ω = θ(in radians) x f.
    emf = IR.


    3. The attempt at a solution
    The emf is 0 at t = 0 due to sin(0) = 0.
    For part (b), how do you find t?
    Is t = 1/f?
    Finally, I calculated the average value of sin(2θ) for part (c) but have no clue where to go from there. Any suggestions would be great! Thanks.
     
  2. jcsd
  3. Mar 5, 2013 #2

    NascentOxygen

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    The equation emf = NABωsin(ωt) is a general description for this situation, but t=0 is not arbitrary.

    A better form is: emf = NABωsin(ωt+)

    You are going to have to think about what's happening, rather than treat this as merely a maths substitution exercise. The first step is to draw a large diagram.
     
  4. Mar 5, 2013 #3

    rude man

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    Just tracking the thread ....
     
  5. Mar 7, 2013 #4
    Ok, so I've been having a think about it and haven't made too much progress to be honest. 1) Could you explain what that blank box stands for in the eq. that you've given me?

    2) So, emf = the negative of the rate of change of magnetic flux i.e. emf = -N d(∅)/dt. Which is the general formula, NABωsin(ωt). Could you explain the significance of this t? I mean, is it the time that taken to complete one revolution or part of a revolution?

    ω = d(θ)/dt ⇔ θ = ωt. So, I take it that I can replace this ωt with θ?

    Finally, at t = 0, the loop is vertical in the B field so the area vector of the loop is parallel/ anti-parallel to B. Thus θ = 0 (or some multiple of pi). Thus emf = NABωsin(0) which is 0. So, at t= 0 emf = 0.

    For (b) I believe it's emf = IR, I = emf/R....where emf = NABωsin(θ). Also, ω = θ(in radians) x f, yes?

    Thanks for all the help. Could you, or anyone else for that matter, take a look at that and see if it makes sense? Thanks!
     
  6. Mar 7, 2013 #5

    NascentOxygen

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    Sorry, that is supposed to show as the Greek symbol, Phi, standing for a phase angle.

    Maybe it renders correctly here: emf = NABωsin(ωt+ɸ)
    The time in seconds to complete one revolution = 1/f = 2π
    where ω is in radians/sec.
    When the loop is parallel to the flux lines, a tiny change in angle will cause a large change in the number of magnetic field lines passing through the loop. There is no orientation where this will be more pronounced.
    Correct.
     
  7. Mar 7, 2013 #6

    rude man

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    This is correct if the normal to the loop points into/out of page at t=0. Which I think is the case here, although the problem is very poorly worded.
    No. ω = 2πf. f =1.2 Hz if I understand the problem wording. One cycle is 1/f = T seconds.
    You don't.
    Use your emf formula for emf with θ = ωt = π/6.
    Average power over 1 cycle is ∫{(emf)2/R}dt integrated over time T, then divided by T. You have all the data you need to proceed from here.

    I would ignore the hint. Can't imagine what they had in mind there.
     
  8. Mar 8, 2013 #7
    Thanks for all the help guys. Great stuff!
     
  9. Mar 8, 2013 #8
    Wait, are the units of time averaged power just going to be Watts or Watts/s?
     
  10. Mar 8, 2013 #9
    Ok, so for part (b) I've calculated e=NABw(sin(theta)) where w = 2pi x f. I then divided this answer by R (10 ohms) to get the current. Apparently this is the wrong answer. I've checked the calculation a number of times so it isn't an arithmetic error. Any ideas?
     
  11. Mar 8, 2013 #10

    rude man

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    Did you use π/6 for theta?
     
  12. Mar 8, 2013 #11
  13. Mar 8, 2013 #12

    rude man

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    Power is power, whether averaged or not. Power is in Watts.
     
  14. Mar 8, 2013 #13

    rude man

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    What was your answer? I want your N, A, B, w and sin(theta) too.
     
  15. Mar 8, 2013 #14
    2.293 x10^-3. The letters w,x,y and z correspond to the values 6,2,3,6 respectively.
     
  16. Mar 8, 2013 #15

    rude man

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    Not what I got. Repeat: N,A,B,w, sin(pi/6) please ....
     
  17. Mar 8, 2013 #16
    Well, this is embarrassing....my calculator wasn't actually set to radian mode. Sorry for wasting your time man!
     
  18. Mar 8, 2013 #17

    rude man

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    No prob, man, but for further action please send the parameters I asked for. :smile:
     
  19. Mar 8, 2013 #18
    Will do! Thanks again.
     
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