# Homework Help: Current through a hollow copper cylinder

1. May 10, 2007

### Ellesar

1. The problem statement, all variables and given/known data
A 85.0 g hollow copper cylinder is 70.0 cm long and has an inner diameter of 1.0 cm. The current density along the length of the cylinder is 1.20×10^5 A/m^2. What is the current in the cylinder?

2. Relevant equations
possibly J = I/A where J is the current density

3. The attempt at a solution
I tried finding the current by using the formula I = JA, but when I did that, I didn't get the correct answer. Other than doing that, I have no idea about how to do this problem. Any pointers helping me to get started would be greatly appreciated.

2. May 10, 2007

### mbrmbrg

What did you use for the area of the cylinder?

3. May 10, 2007

### Ellesar

surface area = 2*pi*r*h + 2*pi*r^2

4. Mar 5, 2008

### octahedron

I'm stuck on this problem too. We know the inner radius, how do you get the outer radius? I'm thinking I should subtract the surface area of the outer circle - the surface area of the inner circle to get the right area. Any hints? I also found the drift velocity -- not sure that'd help though. Simply multiplying J and A (with A = Pi * (5/1000)^2) gives the wrong answer (but not by a big margin).

5. Mar 6, 2008

### octahedron

Just for future reference -- the mass is put in there for you to use it with the density of copper to get the volume. Then you can get the outer radius by using the relation V = h*Pi*(R^2-r^2), and solving for R. The surface area, therefore, is just S.A. = Pi*(R^2-r^2) and you're done.