Tuesday, January 1, 2013

Drift Velocity

In absence of Electric field, electrons in a conductor will be moving due to thermal motion during which they collide with the fixed ions. An electron colliding with an ion emerges with the same speed as before the collision. However, the direction of its velocity after the collision is completely random.

If we consider all the electrons, their average velocity will be zero since their directions are random. Thus, if there are N electrons and the velocity of the ith electron (i = 1, 2, 3, ... N) at a given time is vi, then

Where –e is the charge and m is the mass of an electron.

Consider again the ith electron at a given time t. This electron would have had its last collision some time before t, and let ti be the time elapsed after its last collision. If vi was its velocity immediately after the last collision, then its velocity Vi at time t is

The average of vi’s is zero since immediately after any collision, the direction of the velocity of an electron is completely random.

Let τ is the time between to collisions. The average value of ti then is τ (known as relaxation time).

Thus, averaging over the N-electrons at any given time t gives us for the average velocity vd
This is the phenomenon of drift and the velocity vd is called the drift velocity.

Because of the drift, there will be net transport of charges across any area perpendicular to E. Consider a planar area A, located inside the conductor such that the normal to the area is parallel to E amount of time ∆t, all electrons to the left of the area at distances upto |vd|∆t  would have crossed the area.

If n is the number of free electrons per unit volume in the metal, then there are n ∆t |vd|A such electrons.

Since each electron carries a charge –e, the total charge transported across this area A to the right in time ∆t is –ne A|vd|∆t. E is directed towards the left and hence the total charge transported along E across the area is negative of this.

1 comment:

  1. It's a nice post about Drift Velocity. It's really helpful. Thanks for sharing it.