Calculate current in a cylinder with a hollow cavity

  • Thread starter Ylle
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  • #1
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Homework Statement



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:

[tex]\textbf{J}=J\hat{\textbf{z}}[/tex]

And that is pretty much it.

Now determine the total current I in the conductor.

Homework Equations



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

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.
 
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Answers and Replies

  • #2
288
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What is the definition for current density? This should answer your question.
 
  • #3
79
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Well, that is what I've written in the "Relevant equations" section.
 
  • #4
288
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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.
 
  • #5
79
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So instead of the above it's:

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

It gives the correct answer, but I don't know if that is what you meant ?
 
  • #6
288
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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.
 
  • #7
79
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Argh, ofc... That's much easier :)

Thank you :)
 

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