Final answer:
To find the maximum length of a plain carbon steel wire with a diameter of 3 mm and conductivity of 6.0 x 10^6/ Ωm that has a resistance of 20 Ω or less, use the resistance formula R = ρL/A, recalculate for L, and plug in the known values of R, ρ, and A.
Step-by-step explanation:
The question is asking to find the maximum length of a plain carbon steel wire that can have a resistance of 20 Ω or less, given its diameter (3 mm) and electrical conductivity (6.0 x 10^6 / Ωm). To calculate the maximum length, we use the formula for resistance:
R = ρL/A
Where R is resistance, ρ (rho) is the resistivity (which is the reciprocal of conductivity), L is the length of the wire, and A is the cross-sectional area. Since we know the conductivity (σ), we can find resistivity:
ρ = 1/σ = 1/(6.0 x 10^6 / Ωm)
The cross-sectional area A for a wire of diameter d is calculated using the formula for the area of a circle, A = π(d/2)^2:
A = π(3 x 10^-3 m / 2)^2
Now, we can rearrange the resistance formula to solve for L:
L = R × A / ρ
Plug in the values to calculate L:
L = 20 Ω × (π(3 x 10^-3 m / 2)^2) / (1/(6.0 x 10^6 / Ωm))
By calculating L, we will get the maximum length that this wire can be to have a 20 Ω or lower resistance. Remember to ensure that the final answer is within an acceptable error margin for the calculation.