Source: High school physics (Chinese)
Problem
A copper wire with cross-sectional area $S = 2.0\ \text{mm}^2$ carries a current $I = 30$ mA. Assume each copper atom contributes one free electron, and the random thermal speed of the electrons is $v_{\text{th}} = 10^5$ m/s. Useful constants: density of copper $\rho_{\text{Cu}} = 8.9 \times 10^3\ \text{kg/m}^3$, molar mass $\mu = 63.5$ g/mol, resistivity $\rho_e = 1.7 \times 10^{-8}\ \Omega\!\cdot\!\text{m}$, Avogadro's number $N_A = 6.02 \times 10^{23}$ /mol, electron charge $e = 1.6 \times 10^{-19}$ C, electron mass $m_e = 9.1 \times 10^{-31}$ kg.
- Find the current density in the wire.
- Find the drift velocity of the conduction electrons.
- Find the average time between two successive collisions of an electron with the lattice ions.
- Find the mean free path of the electrons.
(1) $j = 1.5 \times 10^4$ A/m$^2$; (2) $v_d \approx 1.1 \times 10^{-6}$ m/s; (3) $\tau \approx 2.5 \times 10^{-14}$ s; (4) $\lambda \approx 2.5 \times 10^{-9}$ m
\textbf{(1)} Current density:
$$j = \frac{I}{S} = \frac{30 \times 10^{-3}}{2.0 \times 10^{-6}} = 1.5 \times 10^4\ \text{A/m}^2.$$\textbf{(2)} Number density of free electrons (one per Cu atom):
$$n = \frac{N_A \, \rho_{\text{Cu}}}{\mu} = \frac{6.02 \times 10^{23} \times 8.9 \times 10^3}{63.5 \times 10^{-3}} \approx 8.44 \times 10^{28}\ \text{m}^{-3}.$$Drift velocity from $j = n e v_d$:
$$v_d = \frac{j}{n e} = \frac{1.5 \times 10^4}{8.44 \times 10^{28} \times 1.6 \times 10^{-19}} \approx 1.1 \times 10^{-6}\ \text{m/s}.$$\textbf{(3)} In the Drude model, conductivity $\sigma = n e^2 \tau / m_e$, so the mean time between collisions is
$$\tau = \frac{m_e}{n e^2 \rho_e} = \frac{9.1 \times 10^{-31}}{8.44 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 1.7 \times 10^{-8}} \approx 2.5 \times 10^{-14}\ \text{s}.$$\textbf{(4)} Since $v_{\text{th}} \gg v_d$, the mean free path is
$$\lambda = v_{\text{th}} \, \tau = 10^5 \times 2.5 \times 10^{-14} = 2.5 \times 10^{-9}\ \text{m} = 2.5\ \text{nm}.$$