Answer:
[tex]E = \rho J = 3.13\times 10^{-5} ~{\rm N/m}[/tex]
Explanation:
The relationship between the current and the electric field inside a wire is given by the following equation:
[tex]\rho = \frac{E}{J}[/tex]
where ρ is the resistivity, and J is the current density. The current density is the current per area:
[tex]J = \frac{I}{A} = \frac{31.2 \times 10^{-3}}{\pi (\frac{4.57 \times 10^{-3}}{2})^2} = 1957[/tex]
The resistivity is the reciprocal of conductivity. Therefore,
[tex]\rho = \frac{1}{6.25 \times 10^7} = 1.6 \times 10^{-8}[/tex]
Finally, the electric field can be found:
[tex]E = \rho J = 3.13\times 10^{-5} ~{\rm N/m}[/tex]
