# Suppose that you wish to fabricate a uniform wire out of 1.05 g of copper. Assume the wire has a resistance R?

Suppose that you wish to fabricate a uniform wire out of 1.05 g of copper. Assume the wire has a resistance R = 0.300 , and all of the copper is used.

(a) What will be the length of the wire?

m

(b) What will be the diameter of the wire?

µm

i know mass density is 8.92e3 kg/m^3

resistivity of Cu = 1.7e-8

m = 1.05e-3 = Pm * V

V = A*L

R = Pr * L/A

A = pi*d^2 / 4

i got that V = 1.17 e -7 using those equations and i know Pr. but Don't know how to get L and A when they are both together in both equations..

Update:

thank you! would the diameter be 3.22e-7 micrometers?

Relevance
• 9 years ago

Is that R = 0.300Ω ?

Resistance of a wire in Ω

R = ρL/A

ρ is resistivity of the material in Ω-m

L is length in meters

A is cross-sectional area in m²

A = πr², r is radius of wire in m

resistivity Cu 17.2e-9 Ω-m

(17.2e-9)L/(πr²) = 0.3

volume of the wire is V = πr²L

density copper 8960 kg/m³

Volume = 0.00105 kg / 8960 kg/m³ = 1.172e-7m³

so we have AL = πr²L = 1.172e-7

and (17.2e-9)L/A = 0.3

two equations in 2 unknowns.

0.3A = (17.2e-9)L

A = 57.3e-9L

AL = 1.172e-7

(57.3e-9L)L = 1.172e-7

L = 1.43 meters

A = 57.3e-9L

A = 8.19e-8 m²

.