JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Drift Velocity

Drift velocity is the average uniform velocity acquired by free electrons inside a metal by the application of an electric field which is responsible for current through it. Drift velocity is very small it is of the order of \[{{10}^{-4}}\,m/s\] as compared to thermal speed \[(\tilde{}\,{{10}^{5}}\,m/s)\] of electrons at room temperature.

If suppose for a conductor

n = Number of electron per unit volume of the conductor

A = Area of cross-section             

V = potential difference across the conductor  

E = electric field inside the conductor

i = current, J = current density, \[\rho =\] specific resistance, \[\sigma =\] conductivity \[\left( \sigma =\frac{1}{\rho } \right)\] then current relates with drift velocity as \[i=neA{{v}_{d}}\] we can also write

\[{{v}_{d}}=\frac{i}{neA}=\frac{J}{ne}=\frac{\sigma E}{ne}=\frac{E}{\rho ne}=\frac{V}{\rho \,l\,n\,e}\]. 

(1) The direction of drift velocity for electron in a metal is opposite to that of applied electric field (i.e. current density \[\vec{J}\]).

\[{{v}_{d}}\propto E\] i.e., greater the electric field, larger will be the drift velocity.

(2) When a steady current flows through a conductor of non-uniform cross-section drift velocity varies inversely with area of cross-section \[\left( {{v}_{d}}\propto \frac{1}{A} \right)\]

(3) If diameter (d) of a conductor is doubled, then drift velocity of electrons inside it will not change.

(1) Relaxation time \[\mathbf{(\tau )}\] : The time interval between two successive collisions of electrons with the positive ions in the metallic lattice is defined as relaxation time

\[\tau =\frac{\text{mean free path}}{\text{r}\text{.m}\text{.s}\text{. velocity of electrons }}=\frac{\lambda }{{{v}_{rms}}}\]. With rise in temperature \[{{v}_{rms}}\] increases consequently \[\tau \] decreases.

(2) Mobility : Drift velocity per unit electric field is called mobility of electron i.e. \[\mu =\frac{{{v}_{d}}}{E}\]. It?s unit is \[\frac{{{m}^{2}}}{volt-\sec }\].