Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

Solution

As we know that,

$R = \frac{ρ l}{A} V = IR V = \frac{Iρ l}{A} \frac{I}{A} = j = current density V = jρ l V = E . l E . l = jρ l E = jρ j = \frac{E}{ρ} \frac{1}{ρ} = σ j = σE V = u + at a = \frac{F}{m} a = \frac{- eE}{m} | V_{d} | = \frac{eE}{m} τ I . ∆ t = neA (V_{d}) ∆ t j = i / A j = neA \frac{eE}{m} τ j = \frac{{ne}^{2} τ}{m} E$

For Daily Free Study Material Join wiredfaculty

Current Electricity

Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

Some More Questions From Current Electricity Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction