Question 1 of 6: Minimum Conductivity, Carrier Fraction and Temperature Dependence of a Silver Wire

Question 1: Minimum Conductivity, Carrier Fraction and Temperature Dependence of a Silver Wire (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 silver, lattice parameter a = 4.0862×10^-8 cm, one valence electron per atom; wire diameter d = 0.2 cm, current I = 10 A, maximum allowed power loss P/L = 4 W/m; electron mobility μ = 10 cm²/V·s; temperature coefficient of resistivity α_R = 0.0041 °C^-1.

Find. (a) The minimum conductivity that keeps the power loss at or below 4 W/m; (b) the percentage of valence electrons actually carrying current at that conductivity; (c) the temperature at which the resistivity is half its room-temperature value.

Approach. Power loss per unit length is P/L = I²ρ/A, so the largest tolerable resistivity (smallest tolerable conductivity) follows directly from the wire's cross-section and the 4 W/m ceiling. Compare the resulting carrier density (from σ = nqμ) against the theoretical maximum set by the FCC unit cell to get the conducting fraction, then use the formula sheet's linear resistivity law to find the temperature for half the room-temperature resistivity.

1. (a) Minimum conductivity from the power-loss ceiling. The wire cross-section is $$A=\pi\left(\dfrac{d}{2}\right)^2=\pi(0.001\ \text{m})^2=3.1416\times10^{-6}\ \text{m}^2$$ Power loss per unit length is P/L = I²ρ/A = I²/(σA), so the SMALLEST σ that keeps P/L ≤ 4 W/m is $$\sigma_{min}=\dfrac{I^2}{(P/L)\,A}=\dfrac{10^2}{(4)(3.1416\times10^{-6})}=\boxed{7.958\times10^{6}\ \Omega^{-1}\text{m}^{-1}}$$ (equivalently 7.958×10⁴ Ω^-1cm^-1). A larger σ only reduces the loss further, so this is the minimum acceptable value.
2. (b) Theoretical valence-electron density from the FCC cell. An FCC cell holds 4 atoms, each contributing 1 valence electron: $$V_{cell}=a^3=(4.0862\times10^{-8})^3=6.823\times10^{-23}\ \text{cm}^3,\qquad n_{total}=\dfrac{4}{V_{cell}}=5.863\times10^{22}\ \text{cm}^{-3}$$ Using σ_min = 7.958×10⁴ Ω^-1cm^-1 from step 1, the ACTUAL carrier density needed is $$n_{actual}=\dfrac{\sigma_{min}}{q\mu}=\dfrac{7.958\times10^{4}}{(1.6\times10^{-19})(10)}=4.974\times10^{22}\ \text{cm}^{-3}$$ $$\%\ contributing=\dfrac{n_{actual}}{n_{total}}\times100=\boxed{84.8\%}$$
3. (c) Temperature for half the room-temperature resistivity. The formula sheet's linear resistivity law is ρ(T) = ρ_RT[1+α_R(T−25)]. Setting ρ(T)/ρ_RT = 0.5: $$0.5=1+\alpha_R(T-25)\ \Rightarrow\ T=25+\dfrac{0.5-1}{\alpha_R}=25+\dfrac{-0.5}{0.0041}=\boxed{-97.0\,{}^{\circ}\text{C}}$$