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Coil Magnetic Field Question

  1. Apr 9, 2014 #1

    patrickmoloney

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

    A planar 50-turn coil of area ## 0.20m^2 ## is rotated in a uniform ## 0.15 ## Tesla magnetic field by a motor at a constant angular velocity ## \omega = 315 s^-1 ## The axis of rotation is in the plane of the coil and is perpendicular to the magnetic field. At time ## t = 0 ##, the magnetic field is perpendicular to the plane of the coil.

    B.
    (i) Plot the total magnetic flux through the coil as a function of time. [3 marks]
    (ii) Find the maximum emf generated from the coil. [3 marks]
    (iii) What is value of the root-mean-square emf? [3 marks]

    C.
    If the rotating coil in part (b) forms part of an electrical circuit with total resistance ##30 \Omega ##(including the coil),

    (i) What is the average power generated? [3 marks]
    (ii) Plot the instantaneous power as a function of time. [3 marks]


    2. Relevant equations

    ## \epsilon = -N \frac{d\Phi_B}{dt} ##

    ## P = I^2R ##

    ## \Phi_B = BAcos(\omega t) ##


    3. The attempt at a solution

    (i) I need help with this part, I know Flux goes on the y-axis and Time on the x-axis. I'm not sure how to go about plotting the points. I think ## \omega ## depends on the amplitude of the flux.

    (ii) Maximum emf ## (\epsilon) ## when ## \theta = 90° ## i.e when ## \omega t = 90° ##

    ## \Phi = NBA(cos\theta) ##

    ## \epsilon = -N \frac{d\Phi}{dt} = -NBA \frac{d}{dt}(cos \theta) = -NBA \frac{d}{dt}(cos (\omega t)) ##

    ## \epsilon = -NBA(-\omega sin(\omega t)) ## , where ## \omega t = 90° ##

    ## \epsilon = NBA\omega = (50)(0.15)(0.2)(314) = 471 V ##

    (iii) ## \epsilon_{rms} = \sqrt{\frac{NBA\omega}{2}} ##

    ## = \sqrt{\frac{(50)(0.15)(0.2)(314)}{2}} = 15.35 V ##

    C.

    (i) ## P = I^2R ## Ohm's Law states ## E=IR \Rightarrow I^2 = \frac{E^2}{R^2} ##

    Now, ## P = \frac{E^2}{R} = \frac{(471)^2}{30} = 7394.7 W ##

    (ii) As you can see I really struggle with graphs, how do I go about plotting the instantaneous powers as a function of time?
     
  2. jcsd
  3. Apr 9, 2014 #2

    BvU

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    Hello Patrick.
    Your "I think ω depends on the amplitude of the flux" surprises me. In the problem statement it says that ω = 315 /s, so why should it not be that value ?
    And in (ii) you use ##\Phi## correctly, as far as I can see.
    Then again, in (iii) you surprise me: no suspicion if the rms of an alternating voltage with an amplitude of 471 V comes out so extremely low ?

    Next, C (i): Is that the average power or the amplitude of the instantaneous power ?
     
  4. Apr 9, 2014 #3

    patrickmoloney

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    Sorry I meant, I think the flux is dependent on the amplitude of omega. In B (iii) I thought that was the formula for the root-mean square of emf. I obtained it online, since I could not find it in my text book. C (i) ## P_{avg} = \frac{V_{rms}^2}{R} ## Where ## V_{rms} = \frac{V}{\sqrt{2}} ## is this the average power?
     
  5. Apr 10, 2014 #4

    BvU

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    Still don't understand what is blocking you: you write ## \Phi_B = BAcos(\omega t) ##, they tell you B, they tell you A, they tell you ##\omega## and all you have to do is plot a cosine and put the right values at the tick marks. The amplitude of the cosine comes in, but not the "amplitude" of ##\omega##.

    B(iii) formula is fine (check what it means); my comment was more that you should be suspicious if you get 15 when dividing 471 by ##\sqrt{2}##

    C (i): yes.

    rms is a simple thing for a sine wave: squaring gives sin2 which averages out to 1/2. taking the root gives you ##{1\over 2}\sqrt{2}##.

    For e.g. average power the squaring indeed comes from ##P = V \cdot I = V^2/R = I^2 R ## for a simple resistance (where there is no phase difference between V and I).

    With which I am opening a can of worms, because here we have a coil, that not only has a resistance (a part of the 30 ##\Omega##) but also a self-inductance that is introducing a phase difference between V and I. If you haven't been introduced to that stuff already, you might get away with ignoring it. After all, they tell you that the circuit has a total resistance of 30 ##\Omega##.

    [edit] Disregard last paragraph. L ≠ 0 clearly not intended. Any expert's comment on this issue ?
     
    Last edited: Apr 10, 2014
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