Question

A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2

. Calculate
the length of this wire required to make a 10 Ω coil.

Answer 1

the length of the wire is 134.62 m.

Step-by-step explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

`R = (\rho L)/(A) \\\\L = (RA)/(\rho) \\\\L= (10 * (35* 10^(-4)) * 10^(-4))/(2.6 * 10^(-8)) \\\\L = 134.62 \ m`

Therefore, the length of the wire is 134.62 m.