Question

In a copper wire of diameter 2mm, 3×10^23 electrons flow in one second. Calculate the electric field magnitude

Answer 1

The electric field can be found from the Ohm's law in differential form:

$$E = \dfrac{j}{\sigma}$$

where $j$ is the current density, and $\sigma=5.87\times 10^7S/m$ is the conductivity of the copper. The current density is the following:

$$j = \dfrac{I}{\pi d^2/4}$$

where $d = 2\times 10^{-3}m$ is the diameter and $\pi d^2/4$ is the cross-sectional area. Thus, obtain:

$$j = \dfrac{4\cdot 4.8\times 10^4}{\pi \cdot (2\times 10^{-3})} \approx 153A/m^2$$

Thus, obtain:

$$E = \dfrac{153A/m^2}{5.87\times 10^7S/m} \approx 2.6\times10^{-6}V/m$$

Answer. $2.6\times10^{-6}V/m$