Question 2 of 6: Conduction in Copper — Carrier Fraction, Wire Sizing and Drift Velocity

Question 2: Conduction in Copper — Carrier Fraction, Wire Sizing and Drift Velocity (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. FCC copper, lattice parameter a = 3.6151 Å, one valence electron per atom; measured conductivity σ = 5.98×10⁵ (Ω·cm)^-1; electron mobility μ_n = 70 cm²/V·s; wire under J = 50,000 A/cm², V = 1000 V, R = 5 Ω.

Find. (a) The percentage of valence electrons actually carrying current; (b) the wire length; (c) the wire diameter; (d) the electron drift velocity.

Approach. Get the theoretical maximum free-electron density from the FCC unit cell (4 atoms/cell, 1 valence electron each), get the actual carrier density from the measured conductivity via σ = nqμ, then use Ohm's law and the resistivity form of R to size the wire, and finally get drift velocity from the current density.

1. Theoretical valence-electron density from the FCC cell. An FCC cell holds 4 atoms, each contributing 1 valence electron: $$V_{cell}=a^3=(3.6151\times10^{-8}\ \text{cm})^3=4.725\times10^{-23}\ \text{cm}^3$$ $$n_{total}=\dfrac{4\ \text{electrons}}{V_{cell}}=8.466\times10^{22}\ \text{cm}^{-3}$$
2. Actual carrier density from the measured conductivity. Using the Drude relation σ = nqμ: $$n_{actual}=\dfrac{\sigma}{q\mu_n}=\dfrac{5.98\times10^{5}}{(1.6\times10^{-19})(70)}=5.339\times10^{22}\ \text{cm}^{-3}$$ $$\%\ carrying=\dfrac{n_{actual}}{n_{total}}\times100=\boxed{63.1\%}$$
3. Wire length (part b). The applied current follows Ohm's law, and the cross-section follows from the stated current density: $$I=\dfrac{V}{R}=\dfrac{1000}{5}=200\ \text{A},\qquad A=\dfrac{I}{J}=\dfrac{200}{50{,}000}=4.0\times10^{-3}\ \text{cm}^2$$ The resistivity is ρ = 1/σ = 1.672×10^-6 Ω·cm, so from R = ρL/A, $$L=\dfrac{RA}{\rho}=\dfrac{(5)(4.0\times10^{-3})}{1.672\times10^{-6}}=11{,}960\ \text{cm}=\boxed{119.6\ \text{m}}$$
4. Wire diameter (part c). From the same cross-sectional area, treating the wire as circular: $$d=\sqrt{\dfrac{4A}{\pi}}=\sqrt{\dfrac{4(4.0\times10^{-3})}{\pi}}=0.0714\ \text{cm}=\boxed{0.714\ \text{mm}}$$
5. Drift velocity (part d), by two independent routes. Directly from J = nqv_d: $$v_d=\dfrac{J}{n_{actual}\,q}=\dfrac{50{,}000}{(5.339\times10^{22})(1.6\times10^{-19})}=\boxed{5.85\ \text{cm/s}}$$ As a check, the field along the wire is E = V/L = 1000/11,960 = 0.0836 V/cm, and v_d = μ_nE = (70)(0.0836) = 5.85 cm/s — the two routes agree exactly because n_actual and μ_n were tied together through σ = n_actualqμ_n in Step 2.