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Induced Current

  1. May 29, 2006 #1
    This problem is difficult to describe, so I'll post a picture.
    [​IMG]

    The figure above shows a rod of length L caused to move at a constant speed v along horizontal conducting rails. The magnetic field B (the magnitude and direction of which are qualitatively shown by the figure) is not constant, but is supplied by a long wire parallel to the conducting rails. This wire is a distance a from the rail and has a current i.

    L=3.13 cm, v=3.11 m/s, a=15.6 mm, and i=11 A.

    What is the induced emf (e) in the rod?

    ---
    B = (u_0 I)/(2pi y), and it is not uniform, so I integrated over y=a...L
    I got (u_0 I)/(2pi)*ln(L/a).

    emf = vBL = v * (u_0 I)/(2pi)*ln(L/a) * L = 3.11 * 1.544597242E-6 * 0.0313 = 1.503557293E-7 V

    That is not the right answer, however.
    I double-checked that my calculations are correct. So I'm guessing that my steps are incorrect. Can anyone point me to where I'm going wrong?

    Thanks.
     
  2. jcsd
  3. May 29, 2006 #2

    Hootenanny

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    I would try this;

    [tex]B = \frac{\mu_{0} I}{2\pi y}[/tex]

    a is a constant; [itex]y = a + L[/itex]

    [tex]\frac{dB}{dL} = \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL[/tex]

    [tex]B = \int^{0.0313}_{0} \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL[/tex]

    See if that works.

    ~H
     
    Last edited: May 29, 2006
  4. May 29, 2006 #3

    Päällikkö

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    Shouldn't the integration be done over y=a..a+L?

    EDIT: I'm always too slow :smile:.
     
  5. May 29, 2006 #4
    B was approximately 3.326E-4, and emf was 3.23762818E-5 V, which was still incorrect.
    Am I using the right L for emf = vBL? I'm using 3.13 cm (0.0313 m).
     
  6. May 29, 2006 #5

    Hootenanny

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    I think you may have integrated in correctly you should obtain;

    [tex]B = \int^{0.0313}_{0} \frac{\mu_{0} I}{2\pi a + 2\pi L} \; dL[/tex]

    [tex]B = \left[ \frac{1}{2}\mu_{0}I \log (a + L) \right]^{0.0313}_{0}[/tex]

    Also ensure that you are converting correctly, note that a is given in mm. You are using the correct equation here;

    ~H
     
    Last edited: May 29, 2006
  7. May 29, 2006 #6
    Thanks, that worked.
    Weird, though, because I had used my calculator to graph the equation and integrate graphically.
     
  8. May 29, 2006 #7

    Hootenanny

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    No problem. I prefer pen and paper :wink:

    ~H
     
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