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Finding current from current density of wire

  1. Nov 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A cylindrical wire of radius 3mm has current density, J=3s |φ-[tex]\pi[/tex]| z_hat. Find the total current in the wire.


    2. Relevant equations



    3. The attempt at a solution
    I believe all I have to do is integrate over the area, but for some reason I can't get it to work. Is the differential area going to be da=s ds dφ? In that case s goes from 0 to 3mm and φ goes from 0 to 2Pi? The 'z' direction is throwing me off a bit. Thanks.
     
  2. jcsd
  3. Nov 24, 2010 #2

    vela

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    Yes, that's right. Post your work if you still can't get it to work out so we can see where you're going wrong.
     
  4. Nov 24, 2010 #3
    I just tried to work it out on a napkin, because I am not near a scanner at the moment. I ended up with a net result of 0 though. If you don't think this is correct, I can try to rewrite it and take a picture with my phone and upload it. Thanks again.
     
  5. Nov 24, 2010 #4

    berkeman

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    Yes, please upload it. Or you could use the Latex editor in the Advanced Reply window to write out your equations. Click on the [tex]\Sigma[/tex] symbol to the right in the toolbar to see your Latex options.
     
  6. Nov 24, 2010 #5
  7. Nov 24, 2010 #6

    vela

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    You're not dealing with the absolute value correctly. Break the integral over the angle into two ranges, one from 0 to π and the other from π to 2π. For the first integral, |φ-π|=-(φ-π), and for the other, |φ-π|=φ-π.
     
  8. Nov 24, 2010 #7
  9. Nov 24, 2010 #8

    vela

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    Your integrals look fine, but you made a mistake somewhere evaluating them.
     
  10. Nov 24, 2010 #9
    Hmm I can't find it. I checked it a few times after I posted it. Did you work it out and get a different result?
     
  11. Nov 24, 2010 #10

    vela

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    I entered it into Mathematica and got a different result. It looks like you messed up the angular integrations in several spot. Every integral should be proportional to π2, but you have π, π2, and π3.

    You can simplify the algebra a bit by separating the s integral and φ integral:

    [tex]I=\int_0^{R} 3s^2ds \int_0^{2\pi} |\varphi-\pi|\,d\varphi[/tex]

    and using the substitution u=φ-π to do the angular integrals.
     
  12. Nov 25, 2010 #11
    I seem to have gotten R3 Pi2where R=3 mm. Am I close? And of course, the answer is in Amps. Thanks.
     
  13. Nov 25, 2010 #12

    vela

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    Yup, that matches what I got.
     
  14. Nov 26, 2010 #13
    Ok thanks.
     
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