Answer
$2.1~\times 10^{24}$ electrons flow through a cross-section of the wire each day.
Work Step by Step
We can find the electron current:
$i_e = n_e~A~v_d$
$i_e = n_e~\pi~r^2~v_d$
$i_e = (6.0\times 10^{28}~electrons/m^3)~(\pi)~(0.80\times 10^{-3}~m)^2~(2.0\times 10^{-4}~m/s)$
$i_e = 2.413\times 10^{19}~electrons/s$
We can find the number of electrons that flow through a cross-section each day:
$N_e = i_e~t$
$N_e = (2.413\times 10^{19}~electrons/s)(24\times 3600~s)$
$N_e = 2.1~\times 10^{24}~electrons$
$2.1~\times 10^{24}$ electrons flow through a cross-section of the wire each day.
