# Current in a wire and wire density

nateja

## Homework Statement

If the current density in a wire or radius R is given by J = kr, 0 < r < R , what is the current in the wire?

## Homework Equations

I used j = I/A, the definition of current density: current per unit cross sectional area.
the formula for the area of a circle and for a cylinder

## The Attempt at a Solution

First I tried to do the cross sectional area (a circle) times the current density, but I got none of the answer that were displayed, and by just using a circle, I'm only finding the current for a small section of the wire. So I found the area of a cylinder.

A = 2*pi*R*L (L is the length of the cylinder, not given in the problem)
j = kr

so A*j = 2*pi*R*L*k*r
if you set L and r = to R (no idea why you'd do this), then you get 2*pi*R^3*k, but the correct answer is (2*pi*k*R^3)/3?

What am I missing here, because this should be a pretty straight forward question... i think.

## Answers and Replies

Mentor
The current density is not constant across the cross section, but is a function of r. Integrate!

nateja
The current density is not constant across the cross section, but is a function of r. Integrate!

Ok, I understand that it's not constant everywhere, it's an average (from my understanding). So how do you get 2 *(k*pi*R^3)/3??

I can only get (k*pi*R^3)/3. What i did this time was use the cross sectional area A = pi*R^2 and I multiplied it by current density j = k*r. I then set up the integral:
I = ∫ [0,R] k*pi*R^2*dr (I substituted r for dr because for the current density function, r is any radius between 0 and R)
I = (k*pi*R^3)/3

A circle is the right cross sectional shape, correct? And am I integrating the correct bounds? I'm really confused about this question.

Mentor
I can only get (k*pi*R^3)/3. What i did this time was use the cross sectional area A = pi*R^2 and I multiplied it by current density j = k*r.
You can't use a disk, since the current density is not constant over a disk. Instead, break the cross section into rings of thickness dr. What's the area of each ring? Use that to set up your integral.
A circle is the right cross sectional shape, correct?
See above.