Home NCERT Solutions NCERT Exemplar CBSE Sample Papers NCERT Books Class 12 Class 10

Class 12th Physics 2018 Set1 Delhi Board Paper Solution

Question 12

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.

Answer

Consider a conductor of length L and cross sectional area A , when potental difference v is applied across its end the current produced is i. If n is the number of electrons per unit volume in conductor and Vd is the drift velocity then,

I = -nAeV          Equaltion 1.

Where e is charge on electron , n  is no of density

Electric field produced at each point, E = V/L           Equation 2

if t is relaxation time and e is the electric field strength then drift velocity

Vd  =  -etE / m                          Equation 3

Substituting value in equation 1:

I = - neA(-etE / m )

I = (n e2t / m ) AE                    Equation 4  

As e = V/L then,

I =   n e2t A / m  x V/L

Or

V/I = m/n e2t  x L/A

From other law R = V/I = m/n e2t  x L/A              Equation 5

 

 

 

Popular Questions of Class 12th physics

Recently Viewed Questions of Class 12th physics