Question

Deduce Ohm’s law using the concept of drift velocity

Answer 1

Hint : We need to find the fundamental relation between the drift velocity and current. Also we need to find the relation between drift velocity and voltage across the conductor.

Formula used: In this solution we will be using the following formula;
 $ \Rightarrow {v_D} = \dfrac{{eVt}}{{ml}} $ where $ {v_D} $ is the drift velocity, $ e $ is the charge of an electron, $ V $ is the voltage across the wires, $ t $ is the relaxation time, $ m $ is the mass of an electron and finally, $ l $ is the length of the conductor. $ I = ne{v_D}A $ where $ I $ is the current, $ n $ is the free- electron density of the material, and $ A $ is the cross sectional area of the conductor.

Complete step by step answer
Generally, in a conductor, electrons start to move in a particular direction when a potential difference is applied across it. From the knowledge of drift velocity, it is related to the voltage of potential difference by
 $ \Rightarrow {v_D} = \dfrac{{eVt}}{{ml}} $ where $ {v_D} $ is the drift velocity, $ e $ is the charge of an electron, $ V $ is the voltage across the wires, $ t $ is the relaxation time, $ m $ is the mass of an electron and finally, $ l $ is the length of the conductor.
Hence, by making the voltage the subject of the formula, we have
 $ \Rightarrow V = \dfrac{{ml{v_D}}}{{et}} $
At the same time, the current is also related to the drift velocity by the expression
 $ \Rightarrow I = ne{v_D}A $ where $ I $ is the current, $ n $ is the free- electron density of the material, and $ A $ is the cross sectional area of the conductor.
Dividing the voltage by the current we have
 $ \Rightarrow \dfrac{V}{I} = \dfrac{{ml{v_D}}}{{et}} \div ne{v_D}A $
 $ \Rightarrow \dfrac{V}{I} = \dfrac{{ml{v_D}}}{{et}} \times \dfrac{1}{{ne{v_D}A}} $
Hence, by simplifying, we get
 $ \Rightarrow \dfrac{V}{I} = \dfrac{{ml}}{{n{e^2}At}} $
Here we can see that all the variables on the right hand side are constant for a particular temperature of a conductor. Hence,
 $ \Rightarrow \dfrac{V}{I} = K $
Hence, we can conclude that
 $ \Rightarrow \dfrac{V}{I} = R $
 $ \therefore V = IR $ which is Ohms law, where $ R = \dfrac{{ml}}{{n{e^2}At}} $ .

Note
For clarity we can derive the current and voltage relation with drift velocity from first principles.
We know that
 $ \Rightarrow I = \dfrac{{dq}}{{dt}} $ where $ q $ is charged, and $ t $ is time. For a conductor the amount of charge moving is given by the total amount of free electrons in the conductor, hence
 $ \Rightarrow I = \dfrac{{d(Ne)}}{{dt}} $ , we can write $ N $ as $ nV = nAl $ where $ n $ is number of electrons per unit volume and $ V $ is volume, $ l $ is the length of the conductor. Hence, for an infinitesimal length
 $ \Rightarrow I = \dfrac{{neAdl}}{{dt}} = neA{v_D} $ .
For the voltage, we know that
 $ \Rightarrow V = El $
the force experienced by the electrons is
 $ \Rightarrow F = qE = q\dfrac{V}{l} $ , hence, the acceleration is given by
 $ \Rightarrow a = \dfrac{F}{m} = \dfrac{{qV}}{{ml}} $
From equation of motion, the velocity after a time $ t $ is given by
 $ \Rightarrow v = u + at $ , hence
 $ \Rightarrow {v_D} = 0 + \dfrac{{qVt}}{{ml}} $
 $ \therefore {v_D} = \dfrac{{qVt}}{{ml}} $ where $ t $ has become the relaxation time.
The relaxation time is the time taken for electrons to come to rest (actually random motion where average velocity is zero) from its drift velocity if the voltage applied is suddenly removed.