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Magnetic field and current of solid wire

  1. Apr 6, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    This is also a problem on my Masteringphysics:

    A solid conducting wire of radius runs parallel to the axis and carries a current density given by J=J(0) * (1 - r/R)[tex]\hat{k}[/tex] , where J(0) is a constant and r is the radial distance from the wire axis.

    The parts are the same as in the textbook.


    2. Relevant equations
    biot savart law

    J=I/A


    3. The attempt at a solution

    I haven't answered part B, or C yet.

    ok, when i'm entering my answer for part A on masteringphysics, it keeps telling me "variables are case senstive, make sure that you have the right case on your variables." I've switched them around, and it keeps saying the same thing.

    Anyway, my simplified answer is J(0)pi*R[tex]^{2}[/tex](1-[tex]\frac{r}{R}[/tex])


    Since J is the current density of the wire, the current is just J*Area, right? That's basically what i'm doing, but something is off. Any help would be great.

    Thanks
     
  2. jcsd
  3. Apr 6, 2008 #2

    Doc Al

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    Staff: Mentor

    Why does your answer for total current have an r in it?

    You have to integrate to find the total current, since the current density is not constant.
     
  4. Apr 6, 2008 #3
    Since the density is not constant in the wire (depends on the distance from the center), I guess you will need to integrate over the area instead of just mupliplying. Btw, what is ^k?
     
  5. Apr 6, 2008 #4
    it's [tex]\hat{k}[/tex], the vector direction.
     
  6. Apr 6, 2008 #5
    ok, so i integrate from 0 to R in the given formula and i get

    J = J(0)*[tex]\frac{R}{2}[/tex] = [tex]\frac{I}{A}[/tex] -->

    I = J(0)*(piR^3/2)


    ??
     
  7. Apr 6, 2008 #6

    Doc Al

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    That's not correct. Show how you did the integration:
    [tex]I = \int J dA[/tex]
     
  8. Apr 6, 2008 #7
    ok, i actually integrated only J from 0 to R then multiplied by the area after the integration. So i'll integrate what you said.

    so DA = (pi*r)dr ??
     
  9. Apr 6, 2008 #8

    Doc Al

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    Almost. What's the circumference of a circle?
     
  10. Apr 6, 2008 #9
    circumference is 2pi*r

    Would dA be 2pi*r dr?

    why the need for the circumference?
     
  11. Apr 6, 2008 #10

    Doc Al

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    Yes.

    Because you dividing the disk into circular rings so you can integrate.
     
  12. Apr 6, 2008 #11
    ahhh indeed!

    i've got I = J(0)*((pi*R^2)/3))
     
  13. Apr 6, 2008 #12

    Doc Al

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    Looks good!
     
  14. Apr 6, 2008 #13
    it's correct, thank you..

    For part B, using ampere's Law i came up with ([tex]\mu[/tex](0)*J(0)*R^2)/6r which came out to be correct.

    Now for part C, do i need to integrate again since the current surrounded will not be from the whole wire? or i'm just replacing R with r for the total current now because the amperian line is inside the wire?
     
    Last edited: Apr 6, 2008
  15. Apr 6, 2008 #14

    Doc Al

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    Yes.
    Not sure what you mean... but don't do it! :wink:
     
  16. Apr 6, 2008 #15
    gotcha, haha. Thanks for the help, i'll let you know if i get it.
     
  17. Apr 6, 2008 #16
    SOLVED!

    thanks (again) Doc
     
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