Answer
(a) $E = 1.6\times 10^{-3}~V/m$
(b) $v = 1.1\times 10^{-5}~m/s$
Work Step by Step
(a) We can find the electric field:
$J = \sigma~E$
$E = \frac{J}{\sigma}$
$E = \frac{I/A}{\sigma}$
$E = \frac{I}{\sigma~A}$
$E = \frac{I}{\sigma~\pi~r^2}$
$E = \frac{20\times 10^{-3}~A}{(6.3\times 10^7~S/m)~(\pi)~(0.25\times 10^{-3}~m)^2}$
$E = 1.6\times 10^{-3}~V/m$
(b) We can find the drift speed:
$I = v~A~n~e$
$v = \frac{I}{A~n~e}$
$v = \frac{I}{\pi~r^2~n~e}$
$v = \frac{20\times 10^{-3}~A}{(\pi)(0.25\times 10^{-3}~m)^2(5.8\times 10^{28}~m^{-3})(1.6\times 10^{-19}~C)}$
$v = 1.1\times 10^{-5}~m/s$
