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Current Electricity

Question
CBSEENPH12039396

Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm's law.

Solution

Let an electric field E be applied the conductor. Acceleration of each electron is
straight a space equals negative eE over straight m
Velocity gained by the electron
straight v space equals space minus eE over straight m straight t
Let the conductor contain n electrons per unit volume. The average value of time't', between their successive collisions, is the relaxation time, 't'.
Hence average drift velocity straight v subscript straight d space equals space fraction numerator negative eE over denominator straight m end fraction straight tau
The amount of charge, crossing area A, in time Δt is
=neAvdΔt = IΔt

Substituting the value of vd, we get
straight I space increment straight t equals space neA space open parentheses eEτ over straight m close parentheses increment straight t
therefore space straight I space equals space open parentheses fraction numerator straight e squared straight A space τn over denominator straight m end fraction close parentheses space straight E space equals space σE
open parentheses straight sigma space equals space fraction numerator straight e squared τn over denominator straight m end fraction close parentheses
But I = JA, where J is the current density
rightwards double arrow straight J space equals open parentheses fraction numerator straight e squared τn over denominator straight m end fraction close parentheses straight E
rightwards double arrow space straight J space equals σE

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