The electric field in a copper wire of cross-sectional area $2.0\, mm^2$ carrying a current of 1 A is

(Given: Resistivity of copper = $2 × 10^{-8} m$)

$10^{-2}\, V/m$

The correct answer is Option (1) → $10^{-2}\, V/m$

Given:

Current $I = 1 \, A$

Area $A = 2.0 \, mm^2 = 2.0 \times 10^{-6} \, m^2$

Resistivity of copper $\rho = 2 \times 10^{-8} \, \Omega m$

Current density: $J = \frac{I}{A} = \frac{1}{2.0 \times 10^{-6}} = 5.0 \times 10^{5} \, A/m^2$

Electric field: $E = \rho J = (2 \times 10^{-8})(5.0 \times 10^{5})$

$E = 1.0 \times 10^{-2} \, V/m$

Answer: The electric field in the wire is $1.0 \times 10^{-2} \, V/m$.