Current Electricity

Concept:Drift velocity $v_d = I / (n e A)$, where $n$ is electron number density derived from material properties.Explanation:Step 1: Find electron number density $n$. Given: density $\rho = 6.35 \times 10^3\ \mathrm{kg\ m^{-3}}$, molar mass $M = 63.5\ \mathrm{g/mol} = 63.5 \times 10^{-3}\ \mathrm{kg/mol}$, Avogadro’s number $N_A = 6 \times 10^{23}\ \mathrm{atoms/mol}$, and one conduction electron per atom. $$n = \frac{\rho N_A}{M} = \frac{(6.35 \times 10^3)(6 \times 10^{23})}{63.5 \times 10^{-3}} = 6 \times 10^{28}\ \mathrm{m^{-3}}$$ Step 2: Find resistance of the wire. Conductivity $\sigma = 2 \times 10^8\ \mathrm{mho\ m^{-1}}$, length $L = 100\ \mathrm{m}$, area $A = 0.5\ \mathrm{mm^2} = 0.5 \times 10^{-6}\ \mathrm{m^2}$. $$R = \frac{L}{\sigma A} = \frac{100}{(2 \times 10^8)(0.5 \times 10^{-6})} = 1\ \Omega$$ Step 3: Find current in the circuit. Battery emf $V = 2\ \mathrm{V}$, internal resistance $r = 1\ \Omega$. $$I = \frac{V}{R + r} = \frac{2}{1+1} = 1\ \mathrm{A}$$ Step 4: Calculate drift velocity. Charge of electron $e = 1.6 \times 10^{-19}\ \mathrm{C}$, area $A = 0.5 \times 10^{-6}\ \mathrm{m^2}$. $$v_d = \frac{I}{n e A} = \frac{1}{(6 \times 10^{28})(1.6 \times 10^{-19})(0.5 \times 10^{-6})} = \frac{1}{4.8 \times 10^3}\ \mathrm{m/s}$$ Convert to $\mathrm{mm/s}$: $v_d = \frac{1000}{4800} = 0.2083\ \mathrm{mm/s}$. Answer:0.208 $\mathrm{mm\ s^{-1}}$ (option C).