Calculating turns in an inductor

[Kevin2341]

Homework Statement

Hey everyone, I got a question wrong on my homework that I'm trying to figure out. I'm given this:

"a 500uH solenoid(inductor) with an air core is 10cm long has an inside diameter of 1cm. How many turns of wire does it have?

Homework Equations

L = N²*u*A / l

N is the number of turns
u is the permeability of core material
A is the Area of the coil in square meters
l is the average length of the coil in meters.

The Attempt at a Solution

So first of all I rewrote the equation to be
N = sqrt([L*l]/[u*A]

L = 0.5H, l = 0.1m, u is u₀ * u_r, or 1 * (1.26x10^-6, and my area is going to ∏*(0.005)²

So,
N = √(0.5*0.1)/(1 * ([1.26x10^-6]*[∏*(0.005)²]

When I calculated this however, I'm getting an answer of 22477.9 turns, which I know is wrong. The answer actually is 712.

I'm thinking I either have my permeability of wrong (which I double checked, an air core is permeability 1 and the permeability of free space is 1.26x10^-6, so the permeability is 1.26x10^-6.

Do I possibly have my area wrong? The diameter is 1cm, so we know that the area of a circle is pi * r squared, so pi * (0.01/2), which I got 0.000079m².

Am I missing something obvious here?

When I solve for area, I get (0.5*0.1)/(712²*(1.26x10^-6)) which gives me 0.078278, which I'm guessing that's the area I SHOULD have gotten, but how is that?

 

[gneill]

The inductance is 500 μH, which is not 0.5H.