Electromagnetic Induction Problem


by tigerguy
Tags: electromagnetic, induction
tigerguy
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#1
Jul30-06, 11:29 PM
P: 32
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.
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Andrew Mason
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#2
Jul30-06, 11:47 PM
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Quote Quote by tigerguy
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.
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.
tigerguy
tigerguy is offline
#3
Jul31-06, 12:07 AM
P: 32
How did you figure out the resistance? I'm a little confused on that step (I think thats where I went wrong).

Thanks

tigerguy
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#4
Jul31-06, 12:45 AM
P: 32

Electromagnetic Induction Problem


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.
Andrew Mason
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#5
Jul31-06, 09:57 AM
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Quote Quote by tigerguy
How did you figure out the resistance? I'm a little confused on that step (I think thats where I went wrong).

Thanks
The resistance is given in ohms/metre. Just multiply .033 by the length of the wire to get the resistance of this wire.

AM
Andrew Mason
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#6
Jul31-06, 10:01 AM
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Quote Quote by tigerguy
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.
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
jesusburger
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#7
Feb19-08, 10:38 PM
P: 1
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)


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