Learning Goal: To learn to apply the concept of current density and microscopic Ohm's law. A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric field E in the cable is 0.062 newtons per coulomb. Part A What is the material of the cable

Given Information:  

Current = I = 30 A

Diameter = d = 0.326 cm

Electric field = E =  0.062 N/C

Required Information:  

Material of jumper cable = ?

Copper material with resistivity ρ = 1.727x10⁻⁸ Ωm

Explanation

We can find out the material of the cable by finding the resistivity of the cable

ρ = E/j

The current density j is given by

j = I/A

Where I is the current and A is the area of conductor and is given by

A = πr²

r = d/2

r = 0.326/2

r = 0.163 cm

r = 0.00163 m

A = π(0.00163)²

A = 0.000008346 m²

j = 30/0.000008346

j = 3.59x10⁶ A/m²

The resistivity is given by

ρ = E/j

ρ = 0.062/3.59x10⁶

ρ = 1.727x10⁻⁸ Ωm

This is the resistivity of copper therefore, the material of this jumper cable is copper.

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