# Current density problem

1. Sep 20, 2014

### Aviegaille

1. The problem statement, all variables and given/known data

(a)The current density across a cylindrical region of radius R varies according to the equation: J=J0(1-r/R), where r is the distance from the axis of the cylinder. The current density is the maximum J0 at the axis r=0 and decreases linearly to zero at the surface r=R. Calculate the current in terms J0 and the region's cross sectional area A=pi*R^2.

(b) Now suppose that a current density was a maximum Jo at the surface and decreased linearly to zero at the axis, so that: J=J0 r/R. Calculate the current. Why is the result different for these two cases?

2. Relevant equations

I=JA

3. The attempt at a solution

I uploaded a picture of the first part but I am not sure if it's correct. I also don't know how to get the area from this problem. I am thinking of plugging the value of R from I to get the area but I am pretty sure it is not right.

#### Attached Files:

• ###### 10711264_757361231009229_384913907_n.jpg
File size:
41.8 KB
Views:
157
2. Sep 20, 2014

### ShayanJ

You're integrating w.r.t. $J_0$???!!! That's strangely wrong!
As you wrote, $I=JA$ but that's only when J is a constant through out A. Otherwise you have $I=\int dI=\int J dA$. Here J is not constant and so you should integrate w.r.t. area. But what that area is, depends on the direction of $J$(remember current density is a vector!). You didn't give what's the direction of J but from the area formula you given, I take it that $\vec J=J \hat z$. So the integral you should do is $I=\int _0^R \int_0^{2\pi} J r d\varphi dr$.

3. Sep 20, 2014

### Aviegaille

Thanks for correcting $J_0$. Can you further explain this ∫2π0Jrdφ ??

4. Sep 20, 2014

### Aviegaille

.

5. Sep 20, 2014

### ZetaOfThree

Since $J$ is changing throughout the area you are considering, you have to integrate it with respect to area to get the current. $I=\int _0^R \int_0^{2\pi} J r d\varphi dr$ is just $\int J dA$ in polar coordinates (hopefully you've seen them before).

6. Sep 21, 2014

### Aviegaille

I haven't but that makes sense. Thanks.