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Emf max for a guitar string.

  1. Oct 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Review problem. A flexible metallic wire with linear density 3.00 multiplied by 10-3 kg/m is stretched between two fixed clamps 64.0 cm apart and held under tension 301 N. A magnet is placed near the wire as shown in Figure P31.25. Assume that the magnet produces a uniform field of 5.00 mT over a 2.00 cm length at the center of the wire and a negligible field elsewhere. The wire is set vibrating at its fundamental (lowest) frequency. The section of the wire in the magnetic field moves with a uniform amplitude of 1.50 cm.


    (b) Find the amplitude of the electromotive force (emfmax) induced between the ends of the wire.

    2. Relevant equations

    Emf = BA

    3. The attempt at a solution

    I am not quite sure I am doing the problem correctly. I had the area being 3.00cm X 2.00cm but that's not correct nor do I actually believe the above equation is the correct equation. I assume the equation needs to have a sine or a cosine in it, but we are needing to find Emfmax so that's not too important right? Thanks for the help.
  2. jcsd
  3. Nov 1, 2008 #2


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    Homework Helper

    Correction to your equation:

    Emf = \frac{d}{dt} (BA)

    Since B is constant, the problem comes down to finding dA/dt.
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