A silver wire with resistivity 1.59 × 10^-8Ω ∙ m carries a current density of 4.0 A/mm². What is the magnitude of the electric field inside the wire?

Question

Quizplus · Accepted Answer

The electric field (E) inside a wire can be calculated using Ohm's law in terms of current density (J) and resistivity (ρ): E = ρJ. Given ρ = 1.59 × 10^-8 Ω ∙ m and J = 4.0 A/mm² = 4.0 × 10^6 A/m², E = 1.59 × 10^-8 Ω ∙ m × 4.0 × 10^6 A/m² = 0.064 V/m.

A Silver Wire with Resistivity 1