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Electromagnetic Induction Problem

  1. Jul 30, 2006 #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. Jul 30, 2006 #2

    Andrew Mason

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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. Jul 31, 2006 #3
    How did you figure out the resistance? I'm a little confused on that step (I think thats where I went wrong).

    Thanks
     
  5. Jul 31, 2006 #4
    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. Jul 31, 2006 #5

    Andrew Mason

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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. Jul 31, 2006 #6

    Andrew Mason

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    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. Feb 19, 2008 #7
    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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