Question

Obtain the equation $\overrightarrow J = \sigma \overrightarrow E$ of Ohm’s law on the basis of drift velocity. (where symbols carry their usual meanings).

Answer 1

You are asked to derive the above given equation of Ohm’s law. In order to derive it, first you need to know what these quantities are. $\overrightarrow J$ is the current density vector, meaning how much current will pass through an area, $\sigma$ is the electrical conductivity of the material and $\overrightarrow E$ is the electric field vector. Consider the drift velocity equation and find the current density in terms of drift velocity and then substitute the value of drift velocity which is in terms of electric field and then finally rearrange to obtain the above equation.

Complete step by step answer:
A conductor contains a large number of free electrons. When an electric field is present inside the conductor, the electrons in it will experience force due to the electric field. As a result, electrons start drifting in the direction of the electric field. Obviously, it is not a free space and hence, they will suffer collisions. For the time being, we only consider the motion of one electron. At each collision, the electron slows down and starts a fresh motion. The velocity increases and then suddenly becomes zero.

If the electron drifts a distance equal to $d$ in a long time $t$, we say that the drift velocity of the electron is ${v_d} = \dfrac{d}{t}$. If the time required for successive collisions is ${t_0}$, then we will have $d = \dfrac{1}{2}a{t_0}^2$. The acceleration of the electron will be given by$F = ma = eE \to a = \dfrac{{eE}}{m}$.
d = \dfrac{1}{2}\dfrac{{eE}}{m}{t_0}^2 \\\ \Rightarrow\dfrac{d}{{{t_0}}} = {v_d} = \dfrac{1}{2}\dfrac{{eE}}{m}{t_0} \\\
Consider a cylindrical cross section of area $A$ and length ${v_d}\Delta t$. The volume of this cylinder will be $A{v_d}\Delta t$. As it is a conductor, there will be free electrons. Let the number of free electrons per unit volume be $n$, so, the number of free electrons in that region will be $nA{v_d}\Delta t$ and charge crossing that area will be$\Delta Q = enA{v_d}\Delta t$. The current through it will be $i = \dfrac{{\Delta Q}}{{\Delta t}} = A{v_d}ne$. As defined, current density is charge passing through an area, the current density will be $J = \dfrac{i}{A} = \dfrac{{A{v_d}ne}}{A} = en{v_d}$
Substitute the value of drift velocity obtained previously.
$J = en\left( {\dfrac{1}{2}\dfrac{{eE}}{m}{t_0}} \right)$
\Rightarrow J = \left( {\dfrac{{n{e^2}{t_0}}}{{2m}}} \right)E \\\ \therefore J = \sigma E \\\
Here, $\sigma = \left( {\dfrac{{n{e^2}{t_0}}}{{2m}}} \right)$ is the electrical conductivity. The vector form can be written as $\overrightarrow J = \sigma \overrightarrow E$.

Hence, we have $\overrightarrow J = \sigma \overrightarrow E$.

Note: We have discussed in detail about the current density, the motion of electrons, its drift velocity, and electrical conductivity. You need to go through the steps carefully in order to remember the whole derivation, so that, whenever you are asked anything or any quantity which is discussed within the derivation, you can easily find it out. Try to remember all the formulae we used here.