Calculate current in a cylinder with a hollow cavity

  • #1

Homework Statement

I got this problem (Sectional image of a cylinder): [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.

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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  • #2
What is the definition for current density? This should answer your question.
  • #3
Well, that is what I've written in the "Relevant equations" section.
  • #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.
  • #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 ?
  • #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.
  • #7
Argh, ofc... That's much easier :)

Thank you :)

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