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Current density problem

  1. Sep 20, 2014 #1
    1. The problem statement, all variables and given/known data

    (a)The current density across a cylindrical region of radius R varies according to the equation: J=J0(1-r/R), where r is the distance from the axis of the cylinder. The current density is the maximum J0 at the axis r=0 and decreases linearly to zero at the surface r=R. Calculate the current in terms J0 and the region's cross sectional area A=pi*R^2.

    (b) Now suppose that a current density was a maximum Jo at the surface and decreased linearly to zero at the axis, so that: J=J0 r/R. Calculate the current. Why is the result different for these two cases?

    2. Relevant equations

    I=JA

    3. The attempt at a solution

    I uploaded a picture of the first part but I am not sure if it's correct. I also don't know how to get the area from this problem. I am thinking of plugging the value of R from I to get the area but I am pretty sure it is not right.
     

    Attached Files:

  2. jcsd
  3. Sep 20, 2014 #2

    ShayanJ

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    Gold Member

    You're integrating w.r.t. [itex] J_0 [/itex]???!!! That's strangely wrong!
    As you wrote, [itex] I=JA [/itex] but that's only when J is a constant through out A. Otherwise you have [itex] I=\int dI=\int J dA [/itex]. Here J is not constant and so you should integrate w.r.t. area. But what that area is, depends on the direction of [itex] J [/itex](remember current density is a vector!). You didn't give what's the direction of J but from the area formula you given, I take it that [itex] \vec J=J \hat z [/itex]. So the integral you should do is [itex] I=\int _0^R \int_0^{2\pi} J r d\varphi dr [/itex].
     
  4. Sep 20, 2014 #3
    Thanks for correcting [itex] J_0 [/itex]. Can you further explain this ∫2π0Jrdφ ??
     
  5. Sep 20, 2014 #4
  6. Sep 20, 2014 #5

    ZetaOfThree

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    Gold Member

    Since ##J## is changing throughout the area you are considering, you have to integrate it with respect to area to get the current. ##
    I=\int _0^R \int_0^{2\pi} J r d\varphi dr## is just ##
    \int J dA
    ## in polar coordinates (hopefully you've seen them before).
     
  7. Sep 21, 2014 #6
    I haven't but that makes sense. Thanks.
     
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