MP Board · Class 12 · Physics · Current ElectricityWhat is drift velocity of free electrons in a conductor? Establish the relation between drift velocity and electric current. Hence, deduce Ohm's law.

Definition of Drift Velocity

\nIn the absence of an electric field, the free electrons in a metallic conductor move randomly in all possible directions due to thermal agitation, resulting in a net current of zero. When an external electric field $E$ is applied across the ends of the conductor, the free electrons experience an electrostatic force and are accelerated towards the positive terminal. However, due to continuous collisions with the positive metal ions, their acceleration is checked, and they acquire a constant average velocity opposite to the direction of the electric field. This constant average velocity with which free electrons get drifted towards the positive terminal of a conductor is called the drift velocity ($v_d$). Its magnitude is typically of the order of $10^{-4}\text{ m/s}$.

Relation Between Drift Velocity and Electric Current

\nConsider a uniform metallic conductor of length $L$ and cross-sectional area $A$. Let:

• $n = $ number density of free electrons (number of free electrons per unit volume)
• $e = $ magnitude of charge on an electron
• $v_d = $ drift velocity of electrons
• $I = $ electric current flowing through the conductor \nThe total volume of the conductor is $V = A \times L$. \nThe total number of free electrons in the conductor is: $$N = n \times A \times L$$ \nThe total charge $Q$ contained in the conductor is: $$Q = N \times e = nALe$$ \nIf the time taken by an electron to travel the entire length $L$ of the conductor with drift velocity $v_d$ is $t$, then: $$t = \frac{L}{v_d}$| \nThe electric current $I$ is defined as the rate of flow of charge through any cross-section of the conductor: $$I = \frac{Q}{t}$| \nSubstituting the values of $Q$ and $t$ into the equation: $$I = \frac{nALe}{L / v_d}$| $$I = nAeAv_d$$ (or simply $I = neAv_d$) \nThis is the fundamental relation between electric current and drift velocity.

Deduction of Ohm's Law

\nWhen a potential difference $V$ is applied across a conductor of length $L$, the magnitude of the electric field $E$ inside the conductor is given by: $$E = \frac{V}{L}$| \nThe electrostatic force acting on each electron of mass $m$ is $F = eE$. Therefore, the acceleration $a$ experienced by the electron is: $$a = \frac{F}{m} = \frac{eE}{m}$| \nDrift velocity is related to acceleration and relaxation time $\tau$ (the average time interval between two successive collisions) by the kinematic equation ($v = u + at$, where initial thermal velocity is zero): $$v_d = a \tau = \frac{eE \tau}{m}$| \nSubstituting $E = \frac{V}{L}$ into the expression for $v_d$: $$v_d = \frac{e \left(\frac{V}{L}\right) \tau}{m} = \frac{e V \tau}{m L}$| \nNow, substitute this expression for $v_d$ into our earlier current equation ($I = neAv_d$): $$I = neA \left( \frac{e V \tau}{m L} \right)$| $$I = \frac{n e^2 A \tau}{m L} V$$ \nRearranging the terms to find the ratio $\frac{V}{I}$: $$\frac{V}{I} = \frac{m L}{n e^2 A \tau}$| \nFor a given conductor at a constant temperature, mass $m$, length $L$, cross-sectional area $A$, electron density $n$, and relaxation time $\tau$ are all constant. Therefore, the term $\frac{m L}{n e^2 A \tau}$ is a constant, which is defined as the electrical resistance ($R$) of the conductor: $$R = \frac{m L}{n e^2 A \tau}$| \nThus, we get: $$\frac{V}{I} = R \implies V = IR$$ \nThis completes the deduction of Ohm's law from microscopic parameters.