Question

A current of 32.1 mA flows in a silver wire of diameter 4.57 mm. Find the electric field strength inside the wire. The conductivity of silver is 6.25 × 10 7 Ω − 1 m − 1 .

Answer 1

Answer:

$E = \rho J = 3.13\times 10^{-5} ~{\rm N/m}$

Explanation:

The relationship between the current and the electric field inside a wire is given by the following equation:

$\rho = \frac{E}{J}$

where ρ is the resistivity, and J is the current density. The current density is the current per area:

$J = \frac{I}{A} = \frac{31.2 \times 10^{-3}}{\pi (\frac{4.57 \times 10^{-3}}{2})^2} = 1957$

The resistivity is the reciprocal of conductivity. Therefore,

$\rho = \frac{1}{6.25 \times 10^7} = 1.6 \times 10^{-8}$

Finally, the electric field can be found:

$E = \rho J = 3.13\times 10^{-5} ~{\rm N/m}$