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Non-constant Current Density of Magnetic Field

  1. Apr 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Screen shot 2015-04-17 at 11.51.51 PM.png

    2. Relevant equations
    Ienclosed = ∫ JdA
    ∫B*ds = m0*I

    3. The attempt at a solution
    This is my take of finding
    when r is greater than R--so, the magnetic field is m0*I/2pir, since
    B(2pir) = m0*I
    This is because the amperian circle covers the whole wire.

    Can somebody verify that this is correct? (when r>R) I mean it doesnt matter that the density is nonunfirom correct?
  2. jcsd
  3. Apr 18, 2015 #2
    Correct but still you got to calculate I which will not be the same as if the current density was uniform.
  4. Apr 18, 2015 #3
    Oh I understand that "I" will be different when r is INSIDE the circle. But "I" outside should just be "I" total without having to integrate right?
  5. Apr 18, 2015 #4
    For an amperian circular loop that has radius any r>R , the current I will be the same "I total", but how you gonna find it without integrating the current density? it will be

    [itex]I=\int\limits_{0}^{R}\int\limits_{0}^{2\pi}\vec{J}\cdot d\vec{S}[/itex]

    while for r<=R it will be
    [itex]I=\int\limits_{0}^{r}\int\limits_{0}^{2\pi}\vec{J}\cdot d\vec{S}[/itex]
    Last edited: Apr 18, 2015
  6. Apr 18, 2015 #5
    Well originally I integrated it from limits R to r (r>R) for J*dA....and got an answer of a current equaling to I(r^5-R^5)/(R^5). And so B equaled m0*I(r^5-R^5)/(2piR^5*r)...so then I realized integrating for the amper circle outside would not be reasonable.
  7. Apr 18, 2015 #6
    I mean sure finding r inside the circle is a different current, but my main concern is outside.
  8. Apr 18, 2015 #7
    Doing it at 0 to R is the complete wire. It doesn't relate to the r of the amp. circle that is outside it, right?
  9. Apr 18, 2015 #8
    You integrated current density from R to r??? But the current density is zero outside the wire isnt it?
    when the radius r is >R then you integrate from 0 to R cause the loop encloses the whole wire.
  10. Apr 18, 2015 #9
    Oh yeah you make a great point, I just thought back on the section of Gauss and remembered that. Thanks!
    But as I integrated from 0 to R...and plugged in the Jo for the current density...I got a result of just "I". (a good indication that it is correct).

    So it wouldnt really be necessary at all to integrate right?--I mean everything cancels in the end to get a final "Itot". Or is it safe to just integrate anyways?
  11. Apr 18, 2015 #10
    Well integrating from 0 to R i seem to get a total current [itex]I=\frac{2}{5}\pi J_0R[/itex]
  12. Apr 18, 2015 #11
    Yup same here--when you find j0 and Im sure you could--you get 5I/2PiR :)...it then equals to I haha
  13. Apr 18, 2015 #12
    I guess integrating is a good way to show it, and indeed you're correct about the integration, but it would seem a tad redundant in having to do that since I that is enc equals the current of the wire in its entirety.
  14. Apr 18, 2015 #13
    yes ok, it is just that the total current I is not given, instead you are given the current density equation, R and J_0.
  15. Apr 18, 2015 #14
    I appreciate your help Delta!
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