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Obtain the equation of resistivity of metal . |
Answer» Solution :`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 areaA 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). |
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