Obtain the equation of resistivity of metal .

Explanation

`rArr ` Due to electric field there is unidirectional flo of charge through cross-section normal t electric field`vec(E)` .

`rArr` Consider plance cross-section A inside conductor. Here `vec(A)` and `vec(E)` are parallel.
`rArr` Due to drift in `Delta` t time interval, electrons residing in`|v_(d) |Delta t` length will be passing through cross-section.
`rArr` Let number density of free electron be n (per `m^(3)` ).
`rArr` Number of electron passing through area A in `Delta`t time interval will be N = `nA|v_(d)|Delta`t.
`rArr` If charge of one electron be (- e), then in At time charge passing in area A,
q = - nAe `|v_(d)|Delta` t
`therefore` Charge passing in direction of electric field ,
`(E)/(Delta Q) = + ` nAe `|v_(d)| Delta t `
`I = (Delta Q)/(Delta t) = "ne" Av_(d) " " `...(1)
`rArr` (But `" " v_(d) = (e|E|tau)/(m) `
I ` = "neA" [ (e |E| tau)/(m) ] `
I = `("ne"^(2) A|vec(E)| tau)/(m)`
`|vec(j)| = (I)/(A) = ("ne"^(2) A|vec(E)|tau)/(m) `
Direction of J and `vec(E)` is same hence, it can be considered as scalar,
`j = ("ne"^(2))/(m) tau `E
Comparing with equation obtained by Ohm.s law, (j = `sigma` E)
`sigma E = ("ne"^(2) tau E)/(m)`
`sigma = (1)/(rho)`
`(1)/(rho) = ("ne"^(2) tau)/(m)`
`therefore rho = (m)/("ne"^(2) tau)`
Which represent equation of resistivity of metal (conductor).