Question

What is drift velocity? Obtain the equation of Ohm’s law ($\overset{\to }{\mathop{J}}\,=\sigma \overset{\to }{\mathop{E}}\,$) on the basis of drift velocity where parameters are in usual meaning.

Answer 1

In the above given question to deduce the Ohm’s law with the help of drift velocity first we will discuss what is drift velocity then we will further use the formula of specific resistance, current density and specific conductivity so obtain the given equation.

Complete step-by-step solution:
Let us first discuss what is drift velocity?
The relaxation time of free electrons drifting in a conductor is defined as the average time elapsed between two successive collisions.
Now let us Deduce the Ohm's Law: Consider a conductor of length L and cross-sectional area A. When a potential difference of V is applied across the end of the conductor, the current produced is I. If n is the number of electrons per unit volume in the conductor and ${{V}_{d}}$ the drift velocity electrons, then the relation between current and drift velocity is
$I=-neAVd$ let this be equation (1)
In the above equation -e is the charge on the electron
Electric field produced at each point on the wire will be equal to
$E=\dfrac{V}{l}$Let this be equation (2)
Where E is the electric field produced.
Let $\Gamma$be the relaxation time and we know that E is the electric field so, drift velocity will be
${{V}_{d}}$=$\dfrac{erE}{m}$ let this be equation (3)
${{V}_{d}}$=$\dfrac{er}{m}\dfrac{V}{l}$ on substituting the value of equation (2)
Also, we know that
${{V}_{d}}=\dfrac{I}{Ane}$
So, we can say that
$\dfrac{erE}{m}$=$\dfrac{I}{Ane}$
V= $\dfrac{ml}{e\Gamma Ane}I$
V=$\dfrac{m}{n{{e}^{2}}\Gamma }\dfrac{l}{A}I$
$\dfrac{m}{n{{e}^{2}}\Gamma }$ is the specific resistance of the substance denoted by $\rho$
If the physical quantities do not change and the temperature remains constant then L and A will also be constant
So, we can say that
$\rho \dfrac{l}{A}$is also constant which is equal to R which is the resistance of the conductor.
Now, we can say that
$V=RI$
So, $V\alpha I$
“the potential difference developed across a conductor is directly proportional to current flowing through the conductor provided the physical conditions of the conductor remain unchanged.” This is Ohm’s law.
V=$\dfrac{m}{n{{e}^{2}}\Gamma }\dfrac{l}{A}I$
$\overset{\to }{\mathop{E}}\,=\dfrac{m}{n{{e}^{2}}\Gamma }\overset{\to }{\mathop{J}}\,$
As $\overset{\to }{\mathop{J}}\,=\dfrac{I}{A}$which is the current density
$\dfrac{m}{n{{e}^{2}}\Gamma }$ is the specific resistance of the substance denoted by $\rho$
So, we can say that
$\overset{\to }{\mathop{E}}\,=\rho \overset{\to }{\mathop{J}}\,$
$\dfrac{\overset{\to }{\mathop{E}}\,}{\rho }=\overset{\to }{\mathop{J}}\,$
$\overset{\to }{\mathop{J}}\,=\sigma \overset{\to }{\mathop{E}}\,$
As $\dfrac{1}{\rho }=\sigma$ which is the specific conductivity.

Note: To solve the above question and deduce the equation we used the fundamental relationship between the drift velocity and current. Further we reduced the equations to using specific resistance and also the current density. We also used the specific conductivity to deduce the above equation.