Answer: a)1.27 *10^6 A/m^2; b) 57.9*10^6 Ω*m; c) 94.16*10^-6 m/s;
d) 4.29*10^-3 m^2/V*s
Explanation: In order to explain this problem we have to take into account the following expressions:
The current density is equal:
J=I/A where A is the area of the wire.
J=4A/(π*1*10^-6m^2)=1.27 *10^6 A/m^2
The resistence of the wire is given by:
R=L/(σ*A) where σ and L are the conductivity and the length of the wire, respectively.
Then we have:
σ= L/(R*A)=1/(5.5*10^-3*π*1*10^-6)=57.9 *10^6 Ω*m
The drift velocity of free electrons is given by:
Vd=i/(n*A*e) where n is the numer of electrons per volume. e is the electron charge.
then we have:
Vd=J/(n*e)= 1.27 *10^6/(8.43*10^28*1.6*10^-19)=94.16 *10^-6 m/s
Finally, the mobility of free electrons is equal to:
μ=Vd/E ; E=J/σ
μ=E*σ/J= 94.16*10^-6*57.9 *10^6/1.27 *10^6=4.29*10^-3 m^2/V*s
