# Current Density

1. Oct 31, 2009

### exitwound

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

2. Relevant equations

J=i/a

3. The attempt at a solution

Studying for a test. This sample exam went pretty well. I am having trouble with 3 out of the 20. I can't figure this one out. I can't seem to find how to get a denominator value.

The answer is B.

I know that current density is equal to the current over the area.
$$J=\frac{i}{A}$$
$$J_o(r/R)=\frac{i}{\pi (\frac{r}{R}^2)}$$

I know I'm doing something wrong. Anyone help?

2. Oct 31, 2009

### ApexOfDE

You are right at this formula:

$$J = \frac{i}{A}$$

But it is just the special case of this:

i = $$\int$$ JdA
and dA = 2 * pi * r * dr.

3. Oct 31, 2009

### exitwound

Oh. Right.

I still don't know how to get a correct answer out of that.

$$i = \int J \cdot da$$

$$i = \int_0^r \frac{J_or}{R} \cdot 2\pi r dr$$

pulling out the constants:

$$i = \frac{J_o2\pi}{R} \int_0^r r^2 dr$$

$$i = \frac{J_o2\pi}{R}\frac{r^3}{3}|_0^r$$

$$i = \frac{J_o2\pi}{R}\frac{r^3}{3}$$

Where's the extra r in the numerator coming from?

4. Oct 31, 2009

### ApexOfDE

When you set r = R, you will get what you want (because problem asks you to find total current, not current at radius r, i think).

Last edited: Oct 31, 2009
5. Oct 31, 2009

### exitwound

Oh. because I'm integrating from a radius of 0 to a radius of R. That makes sense.

$$i = \frac{J_o2\pi}{R}\frac{r^3}{3}|_0^R$$

$$i = \frac{J_o2\pi}{R}\frac{R^3}{3}$$

$$i = \frac{J_o2\pi R^2}{3}$$

Thanks :)

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