A copper wire of length \( 10\text{ m} \) and radius \( (10^{-2}/\pi)\text{ m} \) has electrical resistance of \( 10\ \Omega \). The current density in the wire for an electric field strength of \( 10\text{ (V/m)} \) is:

Resistivity \( \rho = R \frac{A}{l} = 10 \frac{\pi (10^{-2}/\pi)^2}{10} = \frac{10^{-4}}{\pi}\ \Omega\text{m} \). Conductivity \( \sigma = \frac{1}{\rho} \). Current density \( J = \sigma E = \frac{E}{\rho} = \frac{10}{\frac{10^{-4}}{\pi}} = 10 \pi \times 10^4 \approx 3.14 \times 10^5 \). Wait, let's check \( J = \sigma E = \frac{l}{RA} E = \frac{10}{10 \times \pi (10^{-4}/\pi^2)} \times 10 = \frac{10}{\frac{10^{-4}}{\pi}} \times 10 \) - actually using \( J = \frac{E}{\rho} \) where \( \rho = R \frac{\pi r^2}{l} = 10 \frac{\pi \times 10^{-4}/\pi^2}{10} = \frac{10^{-4}}{\pi} \), then \( J = \sigma E = \frac{\pi}{10^{-4}} \times 10 = \pi \times 10^5 \approx 3.14 \times 10^5 \). Let's re-verify options. Oh wait, if \( \pi \) cancels or is approximated as \( 3.14 \), but options are powers of 10. Let's check \( J = \frac{I}{A} = \frac{V/R}{A} = \frac{E \cdot l / R}{\pi r^2} = \frac{10 \times 10 / 10}{\pi (10^{-2}/\pi)^2} = \frac{10}{\pi^{-1} 10^{-4}} = 10 \pi \times 10^4 \). If \( \pi \approx 3.14 \), maybe there's a simplified form. Let's check option 2: \( 10^5\text{ A/m}^2 \).