A copper pipe has an inside diameter of 6.525 cm and an outside diameter of 7.500 cm. What length of this copper pipe will have a resistance of 4.65 ×10−3 Ω?
To find the length of the copper pipe that will have a resistance of 4.65 × 10^(-3) Ω, we need to use the formula for resistance in a wire. The resistance of a wire is given by the formula:
R = (ρ * L) / A
Where:
- R is the resistance of the wire
- ρ (rho) is the resistivity of the material (in this case, copper)
- L is the length of the wire
- A is the cross-sectional area of the wire
First, we need to find the cross-sectional area of the copper pipe. The cross-sectional area is the difference between the areas of the outer circle and the inner circle.
Area of the outer circle = π * (radius_outer)^2
Area of the inner circle = π * (radius_inner)^2
Then, subtract the area of the inner circle from the area of the outer circle:
A = π * (radius_outer^2 - radius_inner^2)
Next, we need to convert the diameters to radii. The radius is equal to half the diameter:
radius_outer = outer diameter / 2
radius_inner = inner diameter / 2
Now, we can substitute the values into the formula and solve for L:
4.65 × 10^(-3) Ω = (ρ * L) / A
We know the resistivity of copper is approximately 1.7 × 10^(-8) Ω·m (ohm-meter).
Substituting all the known values, we can solve for L:
4.65 × 10^(-3) Ω = (1.7 × 10^(-8) Ω·m * L) / (π * (radius_outer^2 - radius_inner^2))
Finally, rearrange the equation to solve for L:
L = (4.65 × 10^(-3) Ω) * (π * (radius_outer^2 - radius_inner^2)) / (1.7 × 10^(-8) Ω·m)