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Electromagnetic Induction Problem 
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#1
Jul3006, 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 102 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
Jul3006, 11:47 PM

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P: 6,654

AM Edit: corrected multiplication error in last line. 


#3
Jul3106, 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 


#4
Jul3106, 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. 


#5
Jul3106, 09:57 AM

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P: 6,654

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#6
Jul3106, 10:01 AM

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#7
Feb1908, 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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