Source: High school physics (Chinese)
Problem
A copper wire and an iron wire each have length $l = 10$ m and diameter $d = 2.0$ mm. They are connected in series, and a voltage $U = 100$ V is applied across the two ends. Take the resistivities to be $\rho_{\text{Cu}} = 1.7 \times 10^{-8}\ \Omega\!\cdot\!\text{m}$ and $\rho_{\text{Fe}} = 1.0 \times 10^{-7}\ \Omega\!\cdot\!\text{m}$.
- Find the electric field strength inside each wire.
- Find the current density in each wire.
- Find the potential difference across each wire.
(1) $E_{\text{Cu}} \approx 1.45$ V/m, $E_{\text{Fe}} \approx 8.55$ V/m; (2) $j_{\text{Cu}} = j_{\text{Fe}} \approx 8.55 \times 10^{7}$ A/m$^2$; (3) $V_{\text{Cu}} \approx 14.5$ V, $V_{\text{Fe}} \approx 85.5$ V
The cross-section of each wire is $S = \pi (d/2)^2 = \pi \times (1.0 \times 10^{-3})^2 = \pi \times 10^{-6}\ \text{m}^2 \approx 3.14 \times 10^{-6}\ \text{m}^2.$
\textbf{Setup.} In series, the same current $I$ (and same current density $j = I/S$) flows through both wires. From $\vec{E} = \rho \vec{j}$ and the constraint $U = (E_{\text{Cu}} + E_{\text{Fe}}) \, l$:
$$j = \frac{U}{l \, (\rho_{\text{Cu}} + \rho_{\text{Fe}})} = \frac{100}{10 \times (1.7 \times 10^{-8} + 1.0 \times 10^{-7})} \approx 8.55 \times 10^{7}\ \text{A/m}^2.$$\textbf{(2)} Current density (same in both wires): $j_{\text{Cu}} = j_{\text{Fe}} = j \approx 8.55 \times 10^{7}\ \text{A/m}^2.$
\textbf{(1)} Electric field inside each wire, $E = \rho \, j$:
$$E_{\text{Cu}} = \rho_{\text{Cu}} \, j = 1.7 \times 10^{-8} \times 8.55 \times 10^{7} \approx 1.45\ \text{V/m},$$ $$E_{\text{Fe}} = \rho_{\text{Fe}} \, j = 1.0 \times 10^{-7} \times 8.55 \times 10^{7} \approx 8.55\ \text{V/m}.$$\textbf{(3)} Potential difference $V = E \, l$:
$$V_{\text{Cu}} = E_{\text{Cu}} \, l \approx 14.5\ \text{V}, \qquad V_{\text{Fe}} = E_{\text{Fe}} \, l \approx 85.5\ \text{V}.$$Check: $V_{\text{Cu}} + V_{\text{Fe}} \approx 100$ V $ = U$. \checkmark