Calculate current in a cylinder with a hollow cavity

[Ylle]

Homework Statement

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

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:

\textbf{J}=J\hat{\textbf{z}}

And that is pretty much it.

Now determine the total current I in the conductor.

Homework Equations

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

The Attempt at a Solution

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

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

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:

I=\frac{3}{4}\pi {{R}^{2}}\cdot J,
that's what I know.Thanks in advance.

 

[chrisk]

What is the definition for current density? This should answer your question.

 

[Ylle]

Well, that is what I've written in the "Relevant equations" section.

 

[chrisk]

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.

 

[Ylle]

So instead of the above it's:

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

It gives the correct answer, but I don't know if that is what you meant ?

 

[chrisk]

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.

 

[Ylle]

Argh, ofc... That's much easier :)

Thank you :)