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Homework Help: Calculate current in a cylinder with a hollow cavity

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data

    I got this problem (Sectional image of a cylinder):
    http://img715.imageshack.us/img715/3448/cylinder.jpg [Broken]

    Besides that I know that the cylindrical conductor is infinite long, and the same is the cavity.

    And through the conducting material there is a current density that is given by:


    And that is pretty much it.

    Now determine the total current I in the conductor.

    2. Relevant equations

    [tex]\[J=\frac{I}{A}\Leftrightarrow I=JA\][/tex]

    3. The attempt at a solution

    I really have no idea...
    First I thought of doing this:

    [tex]I=\int_{0}^{R}{J}\left( 2\pi R \right)dR-\int_{0}^{R}{J}\left( 2\pi \left( R/2 \right) \right)dR[/tex]

    But that kinda did not work. So now I'm quite lost :)

    A hint would be much appreciated :)

    Oh yes, the correct answer should be:

    [tex]I=\frac{3}{4}\pi {{R}^{2}}\cdot J,[/tex]
    that's what I know.

    Thanks in advance.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 29, 2010 #2
    What is the definition for current density? This should answer your question.
  4. Mar 29, 2010 #3
    Well, that is what I've written in the "Relevant equations" section.
  5. Mar 29, 2010 #4
    Calculate the cross section area of the cylinder without the hole then find the cross section area of the hole. This will give the actual cross section area.
  6. Mar 29, 2010 #5
    So instead of the above it's:

    I=\int_{0}^{R}{J}\left( 2\pi R \right)dR-\int_{0}^{R/2}{J}\left( 2\pi R \right)dR

    It gives the correct answer, but I don't know if that is what you meant ?
  7. Mar 29, 2010 #6
    Yes, what you did is correct but there is another way that is simpler. Subtract the cross section area of the hole from the cross section area of a solid cylinder to find the net cross section area. This result mulltiplied by the current density equals the current. You found the net cross section area by integration.
  8. Mar 29, 2010 #7
    Argh, ofc... That's much easier :)

    Thank you :)
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