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Energy in a rotating square loop

  • #1
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Homework Statement


A square circuit of resistance R=20Ω and side ℓ = 0,2 m spins 100 times per second around an horizontal axis that splits it in two. There is an uniform magnetic field B=1T perpendicular to the position ocupied by the circuit at t=0s.
Calculate (1) the magnetic flux, (2) the inducted current and (3) the dissipated energy by Joule heating in 2 minutes

Homework Equations


3. The Attempt at a Solution [/B]

So I had no trouble in the first 2 points, reaching:

$$\Phi_B=Bl^2 \cos(200 \pi t)$$
$$I=\frac{Bl^2 200 \pi \cos(200 \pi t)}{R}$$

Now I had trouble evaluating the energy.
I know that
$$P=RI^2$$

Taking that expression and integrating it between 0 and 120 seconds, and substituting values, I reach 1,9 kJ, which exactly half of what I was supposed to get (3.8 kJ). What am I doing wrong? Thanks!
 

Answers and Replies

  • #2
haruspex
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So what are your thoughts on the last part?
 
  • #3
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So what are your thoughts on the last part?
Well than again since they only asked for the energy dissipated by Joule heating, we don't have to count with other forces (e.g.. rotating the circuit).
I don't understand why I'm getting half of what I was supposed to though...
 
  • #4
haruspex
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I don't understand why I'm getting half of what I was supposed to though...
For the last part? I do not see any attempt on that.
I see your answers to the first two parts, as functions of t, and a generic formula for P. What were your steps from there to answer the question?
 
  • #5
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Substituting the values on I expression we obtain ##I=1.26 \sin(200 \pi t)##.
Substituting in P we get to ## P=31.752 \sin^2(200 \pi t)##.
Then ##\int_{0}^{120} 31.752 \sin^2(200 \pi t) dt = 1.9 kJ##
So that's it...
 
  • #6
haruspex
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Substituting the values on I expression we obtain ##I=1.26 \sin(200 \pi t)##.
Substituting in P we get to ## P=31.752 \sin^2(200 \pi t)##.
Then ##\int_{0}^{120} 31.752 \sin^2(200 \pi t) dt = 1.9 kJ##
So that's it...
That all looks right to me.
 

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