# Electron transport in Cu and Al

Through a copper wire with a section $S=0.3 mm$ is passing a current $I=2.5 A$. Knowing the resistivity $\rho=1.55*10^{-8} \Omega * m$ of the Copper and the elementary charge of electron force acting on each conduction electron.

From the Ohm law it results that

$I= U/R=(U*S)/(\rho*l)=(S/\rho)*E$

where E is the electric field from the Copper conductor. The force that acts on an electron is given by

$F=e E= e*(\rho*I)/S =3.1*10^-20 N$

The density of current from an Al is $j= 1A/mm^2$. By knowing the atomic mass $M_{Al}=26.98 kg/kmol$ and the density of mass $\rho=2.7*10^3 kg/m^3$, find the transport speed of the conduction electrons, knowing that the number of free electrons from a cubic centimeter of Al is equal to the number of Al atoms. One knows the electron charge and the Avogadro’s Number $N_A$.

The density of current is defined as

$j= I/S = (n e/t)*(1/S)$

Taking a cubic volume of $V=1 m^3$ having an area $S=1 m^2$ the length is $L=1m$ and the time necessary to travel this distance for the electrons is

$t = L/v$ (here $v$ lowercase is speed). Therefore

$j = (n e v)/(LS) =(n e v)/V$

The number of electrons in the volume $V=1 m^3$ is given by

$n=(m/M)*N_A= (\rho*V*N_A)/M$ so that

$j= [(\rho V*N_A)/M]*(e v/V)= (\rho e v N_A/M)$

From above the transport speed of electrons is

$v = (j M)/(e\rho N_A)$