Describe the microscopic model of current and obtain general from of Ohm's law.

Explanation

(i) Consider a conductor with area of cross section A and an electric field `vecE` applied from right to left. Suppose there are n electrons per unit volumne in the conductor and assume that all the elecrons move with the same drift velocity `v_(d)` as shown in Figure. ,

(ii) The drift velocity of the electrons `=v_(d)`
The electrons move through a distance dx within a small interval of dt
`v_(d)=(dx)/(dt), dx=v_(d)dt " "...(1)`
(iii) Since A is the area of cross section of the conductor, the electrons available in the volume with time dt is
= voltmeter `xx` number per unit volume
`A dx xx n" "...(2)`
Substituting for dx from equation (1) in (2)
`=(A d_(v)dt)n`
Total charge in volume element dQ=
(charge) `xx` (number of elctrons in the volumne element).
`dQ=(e) (Av_(d)dt)n`
Hence the current `I=(dQ)/(dt)=("ne" Av_(d)dt)/(dt)`
`I=ne Av_(d) " "....(3)`
(iv) Current density (J) the current density (J) is defined as the current per unit area of cross section of the conductor.
`J=(I)/(A)`
The S.I unit of curent dencity is `(A)/(m^(2))(i.e.)`
`J=("ne" Av_(d))/(A)` (From equation 3)
`J='nev_(d)" ".....(4)`
(v) The above expression hold only when the direction of the current is perpendicular to the area A. In general the current density is a vector quantity and it is given by
`vecJ="ne" v_(d)`
Substituting `vecv_(d)` from equation `vec v_(d)=(e tau)/(m) vecE`
`vec J =(n. e^(2)tau)/(m)vecE" "...(5)`
`vec J =sigma vec E" "....(6)`
Where `sigma=-("ne"^(2)tau)/(m)` is called conductivity. the equation (5) is called micrsocopic form of ohm's law.