Question

Obtain the equation $\overrightarrow{J}=\sigma \overrightarrow{E}$ of Ohm’s on the basis of drift velocity. (Where symbols carry their usual meaning).

Answer 1

Hint: Calculate the expression for drift velocity of electron inside a metal in presence of external electric field. Then derive the relationship between electric current and drift velocity. From this relation write the drift velocity in terms of current density. And try to relate $\overrightarrow{J}\text{,}\sigma \text{ and }\overrightarrow{E}$to get the required relation.

Formulas used:
Average velocity of electrons inside a conductor in absence of electric field is
$\overrightarrow{u}=\dfrac{\overrightarrow{{{u}_{1}}}+\overrightarrow{{{u}_{2}}}+\overrightarrow{{{u}_{3}}}+......+\overrightarrow{{{u}_{n}}}}{n}=0$
Acceleration ,$\overrightarrow{a}=\dfrac{\text{Force}}{\text{Mass}}$
Electric force on an electron with charge $e$,$\overrightarrow{F}=-e\overrightarrow{E}$
Final velocity $\overrightarrow{v}$ of a particle with initial velocity $\overrightarrow{u}$ and acceleration $\overrightarrow{a}$
$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$
Current density
$J=\dfrac{I}{A}$

Complete step by step answer:
The electrons inside a metal were moving randomly in the absence of an external electric field. So their resultant velocity is zero. But when an electric field is applied the electron will be influenced by this electric field and will try to move in one direction. Thus the velocity of the electrons inside a metal in presence of an external electric field is called drift velocity$\left( {{v}_{d}} \right)$.
If \[\overrightarrow{{{u}_{1}}},\overrightarrow{{{u}_{2}}},\overrightarrow{{{u}_{3}}}......\overrightarrow{{{u}_{n}}}\]are the random velocity of n- electrons in absence of field then
$\overrightarrow{u}=\dfrac{\overrightarrow{{{u}_{1}}}+\overrightarrow{{{u}_{2}}}+\overrightarrow{{{u}_{3}}}+......+\overrightarrow{{{u}_{n}}}}{n}=0$
Thus there is no net flow of charge in any direction.
In the presence of an external electric field $\overrightarrow{E}$ each electron will experience a force $-e\overrightarrow{E}$in the opposite direction of field $\overrightarrow{E}$. And undergoes an acceleration given by
$\overrightarrow{a}=\dfrac{\text{Force}}{\text{Mass}}=-\dfrac{e\overrightarrow{E}}{{{m}_{e}}}$
As the electrons accelerate, they frequently collide with the positive metal ions or other electrons of the metal. Between two successive collisions, an electron gains a velocity component (in addition to its random velocity) in a direction opposite to $\overrightarrow{E}$. But the gain in velocity lasts for a short time and is lost in the next collision .At each collision, the electron starts afresh with a random thermal velocity. If an electron having random thermal velocity $\overrightarrow{{{u}_{1}}}$ accelerates for time $\overrightarrow{{{\tau }_{1}}}$, before it suffers next collision, then it will attain a velocity
$\overrightarrow{{{v}_{1}}}=\overrightarrow{{{u}_{1}}}+\overrightarrow{a}{{\tau }_{1}}$
Similarly the velocities of other electrons will be
$\begin{align}
  & \overrightarrow{{{v}_{2}}}=\overrightarrow{{{u}_{2}}}+\overrightarrow{a}{{\tau }_{2}} \\
 & \overrightarrow{{{v}_{3}}}=\overrightarrow{{{u}_{3}}}+\overrightarrow{a}{{\tau }_{3}} \\
 & .. \\
 & \overrightarrow{{{v}_{n}}}=\overrightarrow{{{u}_{n}}}+\overrightarrow{a}{{\tau }_{n}} \\
\end{align}$
Now the average velocity $\overrightarrow{{{v}_{d}}}$ of all n electrons will be
$\overrightarrow{{{v}_{d}}}=\dfrac{\overrightarrow{{{v}_{1}}}+\overrightarrow{{{v}_{2}}}+\overrightarrow{{{v}_{3}}}+......+\overrightarrow{{{v}_{n}}}}{n}$
Putting the values and calculating we get
$\overrightarrow{{{v}_{d}}}=\overrightarrow{a}\tau $
Where $\tau $average time of all electrons between two successive collisions.
But \[\overrightarrow{a}=-\dfrac{e\overrightarrow{E}}{{{m}_{e}}}\]
$\overrightarrow{{{v}_{d}}}=-\dfrac{e\overrightarrow{E}}{{{m}_{e}}}\tau $
The negative sign indicates that the direction of flow of electrons will be opposite to that of applied electric field $\overrightarrow{E}$.
For a conductor having length \[l\], potential difference between the two ends $V$, the electric field is given by.$E=\dfrac{V}{l}$
If the conductor has number of electrons per unit volume $n$then the total electrons inside the conductor lth length $l$ and area of cross-section A is
$\begin{align}
  & N=\text{ number of electrons per unit volume }\times \text{ volume of conductor} \\
 & N=n\times Al\left( \because \text{volume=Area }\!\!\times\!\!\text{ length} \right) \\
\end{align}$
Total charge contained in the conductor of length $l$ and area of cross-section $A$will be
$\begin{align}
  & q=\text{ number of electrons contained }\times \text{ electronic charge} \\
 & \Rightarrow q=enAl \\
\end{align}$
The electrons which enter the conductor at one end will exit it at other end in time
$t=\dfrac{\text{distance}}{\text{velocity}}=\dfrac{l}{{{v}_{d}}}$
So the current across the conductor will be
$I=\dfrac{q}{t}=\dfrac{enAl}{\left( \dfrac{l}{{{v}_{d}}} \right)}=neA{{v}_{d}}$
The current density $\overrightarrow{J}$ is defined as the current flowing per unit area. I.e.
$J=\dfrac{I}{A}=\dfrac{neA{{v}_{d}}}{A}=ne{{V}_{d}}$

In vector form
$\overrightarrow{J}=ne\overrightarrow{{{v}_{d}}}$
But the magnitude of drift velocity is given by

${{v}_{d}}=\dfrac{eE}{{{m}_{e}}}\tau =\dfrac{eV\tau }{{{m}_{e}}l}\left( \because E=\dfrac{V}{l} \right)$
Now current,
$\begin{align}
  & I=neA{{v}_{d}}=neA\dfrac{eV\tau }{{{m}_{e}}l} \\
 & \Rightarrow \dfrac{V}{I}=\dfrac{{{m}_{e}}l}{n{{e}^{2}}\tau A} \\
\end{align}$
At a fixed temperature all the quantities on right hand side is constant .therefore
$\dfrac{V}{I}=\text{ a constant=}R$
Which proves Ohm’ law.
So the resistance of a conductor is given by
$R=\dfrac{{{m}_{e}}l}{n{{e}^{2}}\tau A}$
Resistivity
$\rho =\dfrac{RA}{l}=\dfrac{{{m}_{e}}}{n{{e}^{2}}\tau }$
To get a relation between $\overrightarrow{J},\sigma \text{ and }\overrightarrow{E}$
Charge on electron,$q=-e$ and $\overrightarrow{{{v}_{d}}}=-\dfrac{e\overrightarrow{E}}{{{m}_{e}}}\tau $
Now,
\[\overrightarrow{J}=ne\overrightarrow{{{v}_{d}}}=n\left( -e \right)\left( -\dfrac{e\overrightarrow{E}}{{{m}_{e}}}\tau \right)=\dfrac{n{{e}^{2}}\tau }{{{m}_{e}}}\overrightarrow{E}\]
The quantity \[\dfrac{n{{e}^{2}}\tau }{{{m}_{e}}}=\dfrac{1}{\rho }=\sigma \] is called conductivity of material.
So $\overrightarrow{J}=\sigma \overrightarrow{E}$ or $\overrightarrow{E}=\rho \overrightarrow{J}$

Note:
Note that the conductivity or resistivity depends upon the number of free-electrons or electron density of the conductor. Also depends upon relaxation time $\tau $, the average time between two successive collisions of electrons.