What length of #22 B&S gauge copper wire is needed to make a resistance of 500 ohms?

This is what I came up with but don't think it is correct. looked up diameter for #22 gauge wire = .643 mm and converted to m 6.46 * 10^-4.
Used Eq. A = (pi)(d^2)/4 =>
A = (pi)(6.43 * 10^-4)^2/ = 3.24 * 10^-7 m
Then I used R=(p)l/A and rearranged so now its (R)(A)/p = l
p was given in a chart in my book for copper wire as p = 1.72*10^-8 so here is what I did and the answer I got for l
(500)(3.24*10^-7)/1.72*10^-8) =
9.42*10^3
to me this seems like a very long piece of wire so I think I did something wrong

To find the correct length of #22 B&S gauge copper wire needed to make a resistance of 500 ohms, you can follow these steps:

1. Determine the cross-sectional area (A) of the wire:
The diameter you mentioned for #22 gauge wire (0.643 mm) is correct. However, you converted it to meters incorrectly. The correct conversion is 6.43 * 10^-4 meters. Now, you can use the formula for the area of a circle (A = πr^2) to find the cross-sectional area:
A = π * (6.43 * 10^-4 / 2)^2
A ≈ 3.24 * 10^-7 square meters

2. Use the formula for resistance (R = ρ * l / A) to find l (the length of the wire):
The resistivity (ρ) for copper wire is indeed 1.72 * 10^-8 ohm-meters, as you mentioned. Rearranging the formula, we get:
l = (R * A) / ρ
l = (500 * 3.24 * 10^-7) / 1.72 * 10^-8
l ≈ 9.43 meters (rounded to two decimal places)

So, the correct length of #22 B&S gauge copper wire needed to make a resistance of 500 ohms is approximately 9.43 meters.