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Find the change in flux

  1. Jun 25, 2007 #1

    danago

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    Gold Member

    A simple single-phase generator has coils of 200 turns. The coil is 14cm long and 9cm wide. The magnetic field in the generator is 0.15T. The generator coil is turned at a rate of 3000rpm.

    Calculate the emf produced by this generator.


    [tex]
    \begin{array}{c} \\
    \frac{{d\theta }}{{dt}} = 3000rpm = 100\pi {\rm{ rad/sec}} \\
    \varepsilon = - N\frac{d}{{dt}}(AB\cos \theta ) \\
    = \frac{{d\theta }}{{dt}}NAB\sin \theta \\
    = (100\pi )(200)(0.09)(0.14)(0.15)\sin (100\pi t) \\
    = 37.8\pi \sin (100\pi t) \\
    \end{array}
    [/tex]

    Now it asks for 'the emf', which is a bit hard to give as a numerical answer, since the emf varies with time. I looked at the answer, and they give an average emf.

    So i proceeded as follows:

    [tex]
    \varepsilon _{{\rm{av}}} = \frac{{\varepsilon _{\max } }}{{\sqrt 2 }} = \frac{{37.8\pi }}{{\sqrt 2 }} = 83.97V
    [/tex]

    Now the book does it differently. They find the change in flux over a 90 degree turn (and i think are assuming that the rate of change of flux is constant), and then find the change in time.

    [tex]
    \begin{array}{c}
    \Phi _B = AB\cos \theta \\
    \Phi _i = 0.00189\cos \frac{\pi }{2} = 0 \\
    \Phi _f = 0.00189\cos 0 = 0.00189 \\
    \Delta \Phi _B = 0.00189 \\
    \Delta t = 0.005 \\
    \varepsilon = - N\frac{{\Delta \Phi _B }}{{\Delta t}} = 75.6V \\
    \end{array}
    [/tex]


    Which way is the correct way?

    Thanks in advance,
    Dan.
     
  2. jcsd
  3. Jun 26, 2007 #2

    danago

    User Avatar
    Gold Member

    Ahh everything seems to be contradicting everything else.

    My teacher gave us the formula:
    [tex]
    \begin{array}{l}
    \varepsilon(t) = 2\pi fNBA\sin (2\pi ft) \\
    \therefore \varepsilon _{\max } = 2\pi fNBA \\
    \end{array}
    [/tex]

    For a coil of N turns with cross sectional area A rotating through a magnetic field of flux density B, with a rotational frequency of f. Now i completely agree with this formula, and am able to derive it using some basic calculus and faradays law. However, another text book uses another method, which gives a different answer.
     
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