Relation between Current and Drift Velocity

This article describes the relation between current and drift velocity. When an electric field is applied around a conductor, electrons start drifting with drift velocity and cause current to flow. The flow of electric current is proportional to the drift velocity and other factors.

What is Drift Velocity?

Drift refers to the slow movement of particles. When an external electric field is applied around a conductor, electrons move slowly toward the positive potential. The average velocity of these electrons is known as the drift velocity, and they move in the axial direction of the plane.

The drift velocity is the average velocity of electrons moving through a conductor in response to an electric field.

Relation between Current and Drift Velocity– Derivation

Let’s take a conductor whose length is L and the cross-section area is A. The volume of the conductor is;

Let the number of electrons be n in unit volume, then the total electrons in the conductor is;

Total number of free electrons in the conductor,

The charge of an electron is ‘e’; therefore, the total free charge in a conductor is,

The time taken by an electron to cross the conductor is;

Current flowing in the conductor is.

Putting the values of Q and T from equations 3 and 4 in equation 5, we can establish the relation between current and drift velocity.

Drift Velocity is,

Putting the value of Vd from equation 7 in equation 6, we get;

The equations, 7 and 8, show the relation between current and drift velocity.

Solved Examples

Problem 1: A conductor wire has 1032 free electrons/m3, and its cross-section area is 2 mm2. Calculate the drift velocity of electrons if the current 20 A flows through it. (e = 1.6 x 10-19 C)

Given data-
n=1032
A= 2 X 10-6 m2
I= 20 A
e= 1.6 x 10-19 C

Solution:

Problem 2: There are 8.4×1022 free electrons per cm3 in a conductor. If the drift velocity of electrons in a wire of 1mm2 size is 1.56 X 10-4 meters/second, then calculate the current flowing in the wire. (e=1.6 x 10-19 C).

Given data-
n=8.4 x 1022 X 106 per m3
A =10-6 m2
Vd=1.56 X 10-4 m/s
e= 1.6 x 10-19 C

Solution: