Question

A copper wire is 2.0 mm in diameter and carries a current of 20 A.

Part A.
What is the electric field strength inside this wire?

Answer 1

Current density is, \(D=\frac{I}{A}=\frac{20 \mathrm{~A}}{\pi r^{2}}=\frac{20 \mathrm{~A}}{\pi\left(\frac{2 \times 10^{-3} \mathrm{~m}}{2}\right)^{2}}=6.36 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2}\)

Electric field is,

\(E=\frac{D}{C}=\frac{6.36 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2}}{5.96 \times 10^{7}}=0.1067 \mathrm{~V} / \mathrm{m}\)

Here, \(C\) is the conductivity of the copper.