## College Physics (4th Edition)

$I = 1.2\times 10^{-2}~A$
We can write an expression for the drift speed: $v_d = \frac{I}{n~q~A} = \frac{I}{n~q~\pi~r^2}$ $I$ 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 the current $I$: $v_d = \frac{I}{n~q~\pi~r^2}$ $I = v_d~n~q~\pi~r^2$ $I = (6.5\times 10^{-6}~m/s)(5.90\times 10^{28}~m^{-3})(1.6\times 10^{-19}~C)~(\pi)~(2.5\times 10^{-4}~m)^2$ $I = 1.2\times 10^{-2}~A$