The electric field in a copper wire of cross-sectional area $4\, mm^2$ carrying a current of 1 A and having electrical conductivity of $6.25 × 10^7\, Sm^{-1}$, is

$4 × 10^{-3}\, Vm^{-1}$

The correct answer is Option (1) → $4 × 10^{-3}\, Vm^{-1}$

$\text{Given: } I = 1~\text{A},~ A = 4~\text{mm}^2 = 4 \times 10^{-6}~\text{m}^2,~ \sigma = 6.25 \times 10^7~\text{S/m}$

$\text{Current density: } J = \frac{I}{A} = \frac{1}{4 \times 10^{-6}} = 2.5 \times 10^5~\text{A/m}^2$

$\text{Ohm's law in microscopic form: } J = \sigma E \Rightarrow E = \frac{J}{\sigma}$

$E = \frac{2.5 \times 10^5}{6.25 \times 10^7} = 4 \times 10^{-3}~\text{V/m}$

$\text{Answer: } E = 4 \times 10^{-3}~\text{V/m}$