# 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 [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:

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

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.

Last edited by a moderator:
What is the definition for current density? This should answer your question.

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

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.

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 ?

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.

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

Thank you :)