# Question 0a80a

Nov 29, 2016

$L \approx 2.356 \text{ meters}$

#### Explanation:

The resistance, R, of a wire resistor (in Ohms) is:

$R = \rho \frac{L}{A} \text{ [1]}$

where $\rho$ is the resistivity (in Ohm•meters), A is the cross-sectional area of the wire (in meters²) and L is the length of the wire:

Solve equation [1] for L:

$L = \frac{R A}{\rho} \text{ [2]}$

I believe that there is a minor error the unit for resistivity should be Omega•meter not $\frac{\Omega}{m e t e r}$

We are given: rho = 5xx10^-7Omega•m and R = 6 Omega#

Convert the diameter to a a radius (in meters)

$r = 2.5 \times {10}^{-} 4 m$

The cross-sectional area is that of a circle:

$A = \pi {r}^{2}$

$A = 6.25 \times {10}^{-} 8 \pi {m}^{2}$

Substitute into equation [2]:

$L = \frac{\left(6 \Omega\right) \left(6.25 \times {10}^{-} 8 \pi {m}^{2}\right)}{5 \times {10}^{-} 7 \Omega \cdot m}$

Please notice how the unit cancel, leaving only meters:

$L \approx 2.356 \text{ meters}$