Electron Drift Velocity

Suppose that in a conductor, the number of free electrons available per${\mathrm{<span\; style="font-family:\; "Times\; New\; Roman","serif";"> \; </span>m}}^3$ of the conductor material is n and let their axial drift velocity be v metres/second. In time dt, distance travelled would be v * dt. If A.is area of cross-section of the conductor, then the volume is vAdt and the number of electrons contained in this volume is vA dt. Obviously, all these electrons will cross the conductor cross-section in time dt. If e is the charge of each electron, then total charge which crosses the section in time dt is dq = nAev dt.

Since current is the rate of flow of charge, it is given as

$\mathrm{ \; i}=\frac{dq}{dt}=\frac{nAev dt}{dt}$   so, i=nAev

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Current density $j=\raisebox{1ex}{$i$}\!\left/ \!\raisebox{-1ex}{$A$}\right.=ne*v ampere/mete{r}^2$$\mathrm{}$

Assuming a normal current density  $\mathrm{J\; =\; 1.55}\times {10}^{6}A/m^2$;">, n = ${10}^{29}A/m^2$ for a copper conductor and e $\mathrm{=\; 1.6}\times {10}^{-19}$ coulomb, we get,

$\mathrm{ \; \; J\; =\; 1.55}\times {10}^6=$$\mathrm{<span\; style="font-family:\; inherit;"><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mo></mo><mo><span\; style="font-family:\; inherit;">*</span></mo></math></span>}$
$\mathrm{J\; =\; <span\; style="font-family:\; inherit;">9.7</span>}\times {10}^{-5}\mathrm{<span\; style="font-family:\; inherit;">m</span>}/$s= 0.58 cm/min

It is seen that contrary to the common but mistaken view, the electron drift velocity is rather very slow and is independent of the current flowing and the area of the conductor.

Note :-   Current density i.e., the current per unit area. is a vector quantity. It is denoted by the symbol$\overrightarrow{\mathrm{ \; J}}$.

Therefore. in vector notation, the relationship between current I and is:

$\mathrm{ \; I}=\overrightarrow{J }*\overrightarrow{a }$;">     [ where $\overrightarrow{a }$ is a vector notation for area a ]$\stackrel{}{}\stackrel{}{}$
For extending the scope of the above relationship. so that it becomes applicable for area of any shape, we write:

$\mathrm{<span\; style="font-family:\; "Times\; New\; Roman","serif";"> \; </span>i}=\int \overrightarrow{J.}d \overrightarrow{a }$  

The magnitude of the current density can. therefore be written as J*α  

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Example 1

A conductor material has a free-electron density of ${\mathrm{10}}^{24}$ electrons per ${\mathrm{meter}}^3$. When a voltage is applied, a constant drift velocity of  $\mathrm{1.5}\times {10}^{-2}$metre/second is attained by the electrons. If the cross-sectional area of the material is 1 ${\mathrm{cm}}^2$, calculate the magnitude of the current. Electronic charge is $\mathrm{1.6}\times {10}^{-19}{}^{\mathrm{}}$coulomb.

Solution:

The magnitude of the current is i = nAve amperes

Here         n =  ${\mathrm{10}}^{24}$ ;  A = ${\mathrm{1cm}}^2$ = ${\mathrm{10}}^{-4}$${\mathrm{ \; m}}^2$
                 e =$\mathrm{ \; 1.6}\times {10}^{-19}C{}^{\mathrm{}}$  
                 v =$\mathrm{ \; 1.5}\times {10}^{-2}{}^{\mathrm{}}$m/s

         Ans  i = 0.24 A

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