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Homework Help: Energy in a rectangular coil

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A rectangular coil with N = 2,000 turns that has a resistance of 7.90 Ohms is coplanar with a long wire which carries a current which depends on time according to I0 *exp(-t/tau), where I0 = 6.50 A and tau = 4.30 s. The rectangular loop has a width of W = 2.00 cm and length L = 7.60 cm. The near side of the loop is a distance D = 2.90 cm from the wire.

    What is the total energy dissipated in the entire coil from t = 0 to t = 2.60 s?


    2. Relevant equations
    I=I0*exp(-t/tau)
    Flux= [L*mu*I*ln(1+w/d)] / 2*pi
    V (or EMF) = d(flux)/dt = [ (L*mu*ln(1+w/d)) / 2*pi ]*(-I0*exp(-t/tau))
    where this, (-I0*exp(-t/tau)) , is the time derivative of I
    P=(V^2)/R
    Energy= integral(P)


    3. The attempt at a solution
    I already got the previous questions for this problem. The answers resulted in: Flux for one turn of the coil at 2.6 seconds= 2.83×10-8 T*m^2, EMF for entire coil at 2.6 seconds= 1.32×10-5 V, and the power dissipated from the coil at 2.6 seconds=2.19×10-11 W.

    I just don't know how to integrate the power equation.
    Thanks!
     
  2. jcsd
  3. Jun 21, 2011 #2

    ehild

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    You left out 1/tau while differentiating with respect to time.
    The dissipated energy is the integral of the power from t= to t=2.6 s. And the power is (V^2)/R as you have shown. The emf is of the form C exp(-t/tau) where C is a constant. What is its square?
    You certainly know what is the integral of an exponential function.

    ehild
     
  4. Jun 22, 2011 #3
    Well, I actually had the 1/tau in my work I just accidentally didn't type it. Well, the intergral of exp(-t/tau) would just be -tau*exp(-t/tau) ?? So the emf would just be -(tau^2)*exp(-(t^2)/(tau^2)) ? and divide by same R i was using ? It's that simple ? What would the constant be?
     
    Last edited: Jun 22, 2011
  5. Jun 22, 2011 #4

    ehild

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    Take care when you square an exponential function. (exp(x))^2=exp(2x)
    And the square of a negative quantity is positive.


    ehild
     
  6. Jun 22, 2011 #5
    Wait, I'm confused so you square it first then take the integral or what
     
  7. Jun 22, 2011 #6

    ehild

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    First square the emf as P=(V^2)/R , then integrate with respect to the time.

    You wrote in a previous post that

    "V (or EMF) = d(flux)/dt = [ (L*mu*ln(1+w/d)) / 2*pi ]*(-I0*exp(-t/tau))
    where this, (-I0*exp(-t/tau)) , is the time derivative of I"

    Here is a 1/tau missing. The time derivative of I is -I0*1/(tau)*exp(-t/tau), so the emf is

    V=[ (L*mu*ln(1+w/d)) / 2*pi ]*(-I0(1/tau)*exp(-t/tau)).

    You have to square this whole thing, divide by R and integrate.

    ehild
     
  8. Jun 22, 2011 #7
    Yea, I know I left that out by accident. So V^2= [L^2*mu^2*I0^2*(ln(1+w/d))^2*exp(-2t/tau)] / 4*pi^2*tau^2 ?

    So P= V^2/R= [(L^2*mu^2*I0^2*(ln(1+w/d))^2*exp(-2t/tau)) / 4*pi^2*tau^2] / R

    Take the integral of that, would you pull everything out except for exp(-2t/tau) ? So really you are just integrating that, the rest a constants ? So the integral of exp(-2t/tau) would equal -((tau*exp(-2t/tau))/2) ?



    Or no, it wouldn't be exp(-2t/tau) it would be exp(t^2/tau^2) ? but to integrate that I'm not sure of .. Ughh, I'm so confused for no reason =(
     
    Last edited: Jun 22, 2011
  9. Jun 22, 2011 #8

    ehild

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    Yes, integrate the exponential function between t=0 and t=2.6 s and multiply by everything else.

    Forget exp(t^2/tau^2). exp(x) is e on the x-th power ex. (ex)^2=(ex)*(ex)=e(x+x)=e2x.

    ehild
     
  10. Jun 22, 2011 #9
    Oh, okay. Thanks ^.^ I'll try it out and if there's a problem, I'll be back after I try it out a few times! Hopefully I won't have to.
     
  11. Jun 23, 2011 #10
    Okay, I have a problem. This is all of my work:


    V= -[itex]\frac{L \mu ln(1+w/d) (I0) exp(-t/\tau)}{2 \pi \tau}[/itex]

    V2 = [itex]\frac{L^2 \mu^2 (ln(1+w/d))^2 I0^2 exp(-2t/\tau)}{4 \pi^2 \tau^2}[/itex]


    P= [itex]\frac{V^2}{R}[/itex]

    [itex]\int P dt[/itex] = Energy = [itex]\frac {L^2 \mu^2 (I0^2) (ln(1+w/d))^2}{4 \pi^2 \tau^2 R}[/itex] * [itex]\int exp(-2t/\tau) dt[/itex]

    E= [itex]\frac {L^2 \mu^2 (I0^2) (ln(1+w/d))^2}{4 \pi^2 \tau^2 R^2}[/itex] * [ exp(-2t/tau)*(-tau/2) ]

    the boundaries on that is 0 to 2.6 ;
    my answer i got after many times of trying was 2.773358998E-17 J
    What is wrong ?
     
  12. Jun 23, 2011 #11

    ehild

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    I do not see N (the 2000 turns).

    ehild
     
  13. Jun 23, 2011 #12
    Ohhhhh ... *facepalm* I even thought of that but figured that wasn't right so I didn't even try it. Thanks! again!
     
  14. Jun 23, 2011 #13
    Can I just multiply by answer above by N^2 ?
     
  15. Jun 23, 2011 #14

    ehild

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    It would be enough, but there is some other computational error somewhere. It was not a good idea to square everything separately. Evaluate the constant of V first before squaring. And you can cancel pi as mu=4pi˙10-7.

    ehild
     
  16. Jun 24, 2011 #15
    I mean, how do you square it without squaring each thing separately ? ...and what do you mean evaluate the constant of V first?
     
  17. Jun 24, 2011 #16

    ehild

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    Here is a simple example: evaluate (2*3)2. You can do it by evaluating 2*3 first, then squaring the result: it is 62=36,
    or squaring separately: 22*32=4*9=36.

    And I found just now that you divided V^2 by R^2 instead of R.

    ehild
     
  18. Jun 24, 2011 #17
    No, I divided by R I dont know why I put R^2. But I got the answer. All I needed to do was multiply by N^2. thanks for the help
     
  19. Jun 24, 2011 #18

    ehild

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    Splendid!
    You are welcome.

    ehild
     
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