The electric field in a copper wire of cross-sectional area \( 2.0\,\mathrm{mm^2} \) carrying a current of 1 A is (Given: Resistivity of copper = \( 2 \times 10^{-8}\,\Omega\,\mathrm{m} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 10^{-2}\,\mathrm{V/m} \)
Ohm's law in vector form relates the field in a conductor to the current density: \( E = \rho J \), with \( J = I/A \).
\( \displaystyle A = 2.0\,\mathrm{mm^2} = 2.0\times10^{-6}\,\mathrm{m^2} \)
\( \displaystyle J = \frac{I}{A} = \frac{1\,\mathrm{A}}{2.0\times10^{-6}\,\mathrm{m^2}} = 5\times10^{5}\,\mathrm{A\,m^{-2}} \)
\( \displaystyle E = \rho J = 2\times10^{-8}\,\Omega\,\mathrm{m} \times 5\times10^{5}\,\mathrm{A\,m^{-2}} = 10^{-2}\,\mathrm{V/m} \)