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Homework Help: Faraday's Law Help?

  1. Apr 12, 2006 #1
    The drawing shows a copper wire (negligible resistance) bent in a circular shape with a radius of .5 m. The radial section BC is fixed in place, while the copper bar AC sweeps around at an angular speed of 15 rad/s. The bar makes electrical contact with the wire at all times. The wire and bar have negligible resistance. A uniform magnetic field exists everywher, is perpendicular to the plane of the circle, and has a magnitude 3.8E-3T. Find the magnitude of the current induced in the loop ABC.

    http://edugen.wiley.com/edugen/courses/crs1000/art/qb/6e/ch22p_26.gif

    I thought I had done this right but my answer comes out incorrect. I found the equation E=-NB(deltaA/deltaT) using Faraday's Law. I then went about finding the ratio delta A/ Delta T. I took delta A/pi x r^2 = delta t/delta T. This gave delta A/delta t = pi x r^2/T, which I then got wr^2/2 from, so I found that delta A / delta t = wr^2/2.
    However, after plugging the numbers in, my answer was incorrect. I asked my instructor, and he said I was right there with the answer, but I'm not sure where else to go or if I need to use different figures since we're using rad/s for the angular velocity. Any help here at all is appreciated.
     
  2. jcsd
  3. Apr 12, 2006 #2

    berkeman

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    Staff: Mentor

    So what did you get for your answers for the loop voltage (what you call "E") and therefore the current? I get about 7mV and 2.4mA....
     
  4. Apr 12, 2006 #3
    I got the 7.13 mV but I wasn't sure how to get the current from this since it said the resistance was negligible?
     
  5. Apr 12, 2006 #4

    berkeman

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    Staff: Mentor

    There's an explicit 3 Ohm resistor in series with the wire and wiper. You probably just didn't notice it.
     
  6. Apr 12, 2006 #5
    Oooohhh now I see where the resistance comes from. I was using a number froma different problem somehow. Thanks a lot your answers actually helped me realize what I was doing!
     
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