Electromagnetic Induction Problem

  1. A piece of copper wire is formed into a single circular loop of radius 14 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.75 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3 x 10-2 ohm/m. What is the average electrical energy dissipated in the resistance of the wire.

    For some reason, I keep on getting the wrong answer. I'm using the formula, E=NBA/t, where B= 0.75 T, and A=0.0616 m^2. When calculating that I get 0.10 V, and then using that, I get a current of 21.65 C. Then I use, P=IV, and I get an answer of 2.16 Watts, which divided by 0.45 secods is equal to 4.8 J.

    This answer is wrong, so I'm not sure where I am going wrong. Any help will be appreciated. Thanks.
     
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,875
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    I don't see how you get 21.65 amp. (I assume you mean C/sec=Ampere). The emf is .1 volt. The resistance is .029 ohms. So the current would be 3.45 amp. The power, then, is VI = .345 watt. The total electrical energy (I am not sure what "average" means) dissipated in .45 seconds would be .345*.45 = .155 J.

    AM

    Edit: corrected multiplication error in last line.
     
    Last edited: Jul 31, 2006
  4. How did you figure out the resistance? I'm a little confused on that step (I think thats where I went wrong).

    Thanks
     
  5. I actually redid the calculations. Using the same steps where V= 0.1026 and resistance of 0.029 ohms, I get 0.36 W. I then multiplied by .45 seconds and I got 0.16 J.

    But, this is wrong, so I'm still confused.
     
  6. Andrew Mason

    Andrew Mason 6,875
    Science Advisor
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    The resistance is given in ohms/metre. Just multiply .033 by the length of the wire to get the resistance of this wire.

    AM
     
  7. Andrew Mason

    Andrew Mason 6,875
    Science Advisor
    Homework Helper

    Perhaps they want the average rate of dissipation of energy, which would be .36 J./sec. The term "average energy dissipated" does not make sense in this context.

    AM
     
  8. induced emf = change in flux/change in time = (pi*(.14^2)*.75)/.45 = .1026V
    Resistance = circumference*resistance/distance = (2*pi*.14).033 = .029
    V=IR, so I=V/R = .1026/.029 = 3.5379
    P=IV, so P = 3.5379*.1026 = .363
    power (watts) is in Joules per second, so p=j/t or j=p*t = .363*.45 = .1633 Joules
    (just to sum things up - I had a question worded exactly the same on a homework and this method worked)
     
    Last edited: Feb 19, 2008
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