Question

A 51.0-g sample of a conducting material is all that is available. The resistivity of the material is measured to be 13 ✕ 10−8 Ω · m, and the density is 7.50 g/cm3. The material is to be shaped into a solid cylindrical wire that has a total resistance of 1.5 Ω. (a) What length of wire is required? m (b) What must be the diameter of the wire? mm

Answer 1

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

Cylinder V = πr²h

Several equations we need to combine
1.5 = (13*10^-8)L/A

we can get the volume from the mass and density
51 g / 7.05 g/cm³ = 7.23 cm³
7.23 cm³ x (1 m/100cm)³   =7.23 cm³ = 7.23e-6 m³

V = πr²h = = AL = 0.0723m³
A = 7.23e-6 m³ / L

substituting
1.5 = (13*10^-8)L/A
1.5 = (13*10^-8)L/(7.23e-6 m³ /L)
1.5 = (13*10^-8)L(L/7.23e-6 m³ )
1.5 = (13e-8)L^2/7.23e-6 m³
L^2 = 7.23e-6 m³ (1.5) / (13*10^-8)
L = 9.133 m

A = 7.23e-6 m³ / L
A = 7.23e-6 m³ / 9.133 m = 7.93e-7m² = πr²
r = 0.000502 m = 0.502*2 = 1.004 mm
d = 0.946 mm