Question

In a copper wire of diameter 2mm, 3×10²³ electrons flow in one second. Calculate the current density magnitude

Answer 1

The current by definition is:

$$I = \dfrac{ne}{t}$$

where $n = 3\times 10^{23}$ is the number of electrons, $e = 1.6\times 10^{-19}C$ is the charge of the electron, and $t = 1s$. Thus, obtain:

$$I = 4.8\times 10^4A$$

The charge density is:

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

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

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