CBSE Class 12 · 2009 · Current ElectricityShort Answer

Derive an expression for the drift velocity of free electrons in a conductor in terms of the relaxation time.

Answer: $v_{d} = - \frac{e E}{m} τ$ , so in magnitude $v_{d} = \frac{e E}{m} τ$ .

Solution

When a field

E

is applied, each electron feels a force

F = - e E

and hence an acceleration

a = - \frac{e E}{m}

. Between two successive collisions (mean time

τ

) an electron that emerged with thermal velocity

u_{i}

reaches

v_{i} = u_{i} + a τ_{i}

. Averaging over all electrons, the random thermal part averages to zero, so

v_{d} = ⟨ v_{i} ⟩ = a τ = - \frac{e E}{m} τ,

where

τ

is the average relaxation time. Thus

v_{d} = \frac{e E}{m} τ

CBSE Class 12 Board · 2009 · Subjective