Answer:
[tex]v_d=0.1\ mm/s[/tex]
Explanation:
The current flowing through a conductor is given as:
[tex]I=neAv_d\\where\ A =area\ of\ conductor=\pi d^2/4=\pi(1*10^-3)^2/4=7.85*10^{-7}\\\\I=current\ flowing\ through\ the \ conductor=1.1A\\\\e=charge\ of \ electron=1.602*10^{-19}\\\\N_A=Avogadro \ constant=6.023*10^{23}\ mol^{-1}\\\\Density=9*10^3kg/m^3=9*10^6g/m^3\\\\n=\frac{6.023*10^{23}\ mol^{-1}*9*10^6\ g/m^3}{63\ g/mol} =8.604*10^{28}\ m^{3}\\\\v_d=\frac{I}{neA}=\frac{1.1}{8.604*10^{28}*1.602*10^{-19}*7.85*10^{-7}}=0.0001\ m/s\\\\v_d=0.1\ mm/s[/tex]