Question

Establish a relation between electric current and drift velocity.
OR
Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.

Answer 1

Hint : To solve this question, we need to use the basic formula of the current and equate it to the total charge flowing through a given cross section of a conductor in the relaxation time. Then we have to use the basic definition of current density to evaluate its expression from that of the current.

Formula Used: In this solution we will be using the following formula,
$F = eE$ where $F$ is the force, $e$ is the charge and $E$ is the electric field.
$v = a\tau$ where $v$ is the drift velocity, $a$ is the acceleration and $\tau$ is the average relaxation time between two successive collisions.
$N = nV$ where $N$ is the total number of electrons, $n$ is the number density of electrons and $V$ is the volume.

Complete step by step answer
Let us consider a conductor which is subjected to an electric field $E$ . The magnitude of the force on an electron inside the conductor is given by
$F = eE$ …………………………..(1)
If the mass of an electron is ${m_e}$ , then the acceleration of the electron is given by
$a = \dfrac{F}{{{m_e}}}$
From (1)
$a = \dfrac{{eE}}{{{m_e}}}$ …………………………..(2)
We know that the electrons keep on moving in random motion inside a conductor. This random motion is due to its collision with ions. Let $\tau$ be the average relaxation time between two successive collisions. Then the velocity of the electron after a time interval of $\tau$ is given by
$v = a\tau$
From (2)
$v = \dfrac{{eE}}{{{m_e}}}\tau$ …………………………..(3)
This is the drift velocity of the electrons, because of which the charge will be transported across any given cross section perpendicular to the applied electric field. Let us consider such a cross section of area $A$ perpendicular to the electric field. In a time interval of $\Delta t$ , the distance travelled by the electrons after crossing the cross section is given by
$d = v\Delta t$
From (3)
$d = \dfrac{{eE}}{{{m_e}}}\tau \Delta t$ …………………………..(4)
So the volume swept by the electrons is given by
$V = Ad$
From (4)
$V = \dfrac{{eE}}{{{m_e}}}\tau A\Delta t$ …………………………..(5)
Now, let the number density of the electrons be $n$ . So the total number of electrons crossing the cross section is
$N = nV$
From (5)
$N = \dfrac{{neE}}{{{m_e}}}\tau A\Delta t$ …………………………..(6)
We know that the charge on an electron is $e$ . So the total charge which crossed the cross section is
$q = Ne$
From (6)
$q = \dfrac{{n{e^2}E}}{{{m_e}}}\tau A\Delta t$ …………………………..(7)
Let the current in the conductor be $I$ . So the charge crossing a cross section in the same time $\Delta t$ is
$q = I\Delta t$
From (7)
$\dfrac{{n{e^2}E}}{{{m_e}}}\tau A\Delta t = I\Delta t$
Cancelling $\Delta t$ from both the sides, we get
$I = \dfrac{{n{e^2}E}}{{{m_e}}}\tau A$ …………………………..(8)
From (2) the drift speed is
$v = \dfrac{{eE}}{{{m_e}}}\tau$
Substituting this in (8) we get
$I = nevA$ …………………………..(9)
This is the required relation between the current and the drift speed.
We know that the current density is define as the current per unit area, that is,
$j = \dfrac{I}{A}$
From (9)
$j = nev$
We can clearly see that the drift speed is proportional to the number density of the electrons.

Note
While calculating the acceleration of the electrons, we could take their initial velocity as zero due to the fact that in the absence of any electric field, the electrons are in random thermal motion. So the direction of motion of electrons is completely random and hence the average of the velocity of the electrons could be assumed to be zero.