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Homework Help: Flux between two current-carrying wires

  1. Jun 14, 2018 #1

    Gene Naden

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    1. The problem statement, all variables and given/known data
    Two long, parallel copper wires of diameter 2.5 mm carry currents of 10 A in opposite directions. Assuming that their central axes are 20 mm apart, calculate the magnetic flux per meter of wire that exists between those axes. What fraction of this flux lies inside the wires?


    2. Relevant equations
    Inside a wire, ##B=\frac{\mu_0 i}{i R^2}r##.
    Outside a wire, ##B=\frac{\mu_0i}{2\pi r}## where R is the radius of the wire and r is the distance from the center of the wire.

    3. The attempt at a solution
    ##\Phi_1=2\int _{0}^{R} \frac{\mu_0 i x}{2\pi R^2} dx=\frac{\mu_0 i}{2\pi}##
    where x is the vertical distance downward from the center of the wire.
    ##\Phi_2=2\int_{R}^{s}\frac{mu_0 i}{2\pi r} dr=\frac{\mu_0 i}{2\pi} ln(\frac{s}{R})##
    where r is the vertical distance downward from the center of the wire and s is the separation.
    ##\Phi=\Phi_1+\Phi_2=13.09\mu W/m##
    ##Frac=\frac{\Phi_1}{\Phi_1+\Phi_2}=15.3 \:percent##

    For the flux, I agree with the textbook (Halliday, Resnick & Walker Fundamentals of Physics 5th edition, chapter 31, problem 23). But I disagree for the fraction of flux inside the wire. They get 17%. Can anyone verify my answer or point out where I went wrong?
     
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  3. Jun 14, 2018 #2

    kuruman

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    Your expression for the ##\Phi_2## integral should be twice of what you have. Would that do it?
     
  4. Jun 14, 2018 #3

    Gene Naden

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    Thanks for looking at this. But actually, I mistyped the expression when I posted the problem. I actually used the correct integral, ##\frac{\mu_0 i}{\pi} ln(\frac{s}{R})##. So that is not the problem.
     
  5. Jun 14, 2018 #4

    kuruman

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    If you don't substitute numbers, what do you get for the fraction ##\frac{\Phi_1}{\Phi_1+\Phi_2}##? Note that the common factor ##\frac{\mu_0i}{\pi}## cancels out.
     
  6. Jun 14, 2018 #5

    Gene Naden

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    Same result: ##\frac{\frac{1}{4}}{\frac{1}{4}+\frac{ln(\frac{s}{R})}{2}}=.153##
     
  7. Jun 14, 2018 #6

    kuruman

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    OK, then that's the answer. If your answer to part (a) agrees with the answer in the book and doing part (b) without relying on the numbers from (a) gives the same answer, then the answer in the book for part (b) must be incorrect.
     
  8. Jun 14, 2018 #7

    Gene Naden

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    Thank you
     
  9. Jun 14, 2018 #8

    haruspex

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    Not an area I know much about, but I believe flux lines in the same direction effectively repel each other. Is it possible the presence of the current in one wire would displace some into the other wire?
    What happens if you sum the two fields before performing the integral?
     
    Last edited: Jun 14, 2018
  10. Jun 15, 2018 #9

    TSny

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    Hello.
    If ##\Phi_1## denotes the total flux inside the two wires, then this is not quite correct. It does not take into account the flux produced inside wire 1 by the field of wire 2, and vice versa.

    If ##\Phi_2## is the total flux outside the wires, then the upper limit of your integral is not correct.

    The sum of your expressions for ##\Phi_1## and ##\Phi_2## does, however, give the correct value for the total flux (inside and outside the wires).
     
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