# Electronic conduction in a metal conductor

A metallic conductor moves with the speed $v_0= 10 m/s$ which is perpendicular to its section. At a certain moment of time the conductor slows constantly down until it stops. What number of conduction electrons go through the unit of its surface section when the ends of the conductor are connected by a wire having negligible resistance?. The resistivity of the metallic conductor is $\rho =2.5*10^{-9} \Omega*m$.

Answer

When the current I passes through the conductor having resistance R, in the time t, the lost energy is

$W =RI^{2}t$

The intensity of the current in this case in not constant. If the current I decreases uniformly from the value I to 0, then through the section S of the conductor will pass the charge:

$Q =\frac{1}{2}It$ so that $W =2QIR$

If N is the number of the free electrons from the conductor, and $l$ and S are its length and section, then:

$I =j S =e n v_0S =e v_0 \frac {(n l S)}{l} = e v_0 \frac {N}{l}$

and therefore

$W = (2ev_0QNR)/l$

If $\Delta W_k$ is the variation of the kinetic energy of an electron when its speed decreases from $v_0$ to 0 then for the total energy $W$ we can write:

$W =-N*\Delta W_c = -N*\Delta ((m_0*v)/2) =N*(m_0v_0)/2$

By comparing the two expressions for the energy W one gets:

$(2ev_0QNR)/l =N*(m_0v_0)/2$

or equivalent

$Q = (m_0v_0l)/(4eR)=(m_0v_0S)/(4e\rho)$

The number of conduction electrons that go through the unity of surface of its section is:

$n_0 = Q/(n S) = (m_0v_0)/(4e^{2}\rho) =3.55*10^{-15} m^{-3}$