## College Physics (4th Edition)

We can rank the wires in order of decreasing drift velocity: $b \gt d \gt a = e \gt f \gt c$
We can write an expression for the drift velocity: $v_d = \frac{I}{n~q~A} = \frac{I}{n~q~\pi~r^2}$ $I$ is the current $n$ is the number of electrons per unit of volume $q$ is the charge of one electron $A$ is the cross-sectional area $r$ is the cross-sectional radius We can find an expression for the drift velocity in each case: (a) $v_d = \frac{80\times 10^{-3}}{n~q~\pi~(1\times 10^{-3})^2} = \frac{80\times 10^3}{n~q~\pi}$ (b) $v_d = \frac{80\times 10^{-3}}{n~q~\pi~(0.5\times 10^{-3})^2} = \frac{320\times 10^3}{n~q~\pi}$ (c) $v_d = \frac{40\times 10^{-3}}{n~q~\pi~(2\times 10^{-3})^2} = \frac{10\times 10^3}{n~q~\pi}$ (d) $v_d = \frac{160\times 10^{-3}}{n~q~\pi~(1\times 10^{-3})^2} = \frac{160\times 10^3}{n~q~\pi}$ (e) $v_d = \frac{20\times 10^{-3}}{n~q~\pi~(0.5\times 10^{-3})^2} = \frac{80\times 10^3}{n~q~\pi}$ (f) $v_d = \frac{40\times 10^{-3}}{n~q~\pi~(1\times 10^{-3})^2} = \frac{40\times 10^3}{n~q~\pi}$ We can rank the wires in order of decreasing drift velocity: $b \gt d \gt a = e \gt f \gt c$