What Drift Velocity Means
In a metal, the conduction electrons are in ceaseless random thermal motion, colliding repeatedly with the heavy fixed ions of the lattice. As NCERT states, after each collision an electron emerges with the same speed but in a random direction. If we consider all the electrons, their average velocity is zero, because their directions are random. With $N$ electrons whose individual velocities at a given instant are $\mathbf{v}_i$, this is written
$$ \frac{\sum_{i=1}^{N} \mathbf{v}_i}{N} = 0 $$
When an electric field $\mathbf{E}$ is switched on, a tiny systematic velocity is superposed on this random motion. That net average velocity, directed opposite to $\mathbf{E}$ (because the electron charge is negative), is the drift velocity $\mathbf{v}_d$. It is the engine of electric current, yet, as we will see, it is extraordinarily small.
Deriving the Drift Velocity
Follow the NCERT derivation closely. With a field present, every electron is accelerated by
$$ \mathbf{a} = \frac{-e\mathbf{E}}{m} $$
where $-e$ is the charge and $m$ the mass of an electron. Consider the $i$-th electron at time $t$. It had its last collision a time $t_i$ before, and if $\mathbf{v}_i$ was its velocity immediately after that collision, its velocity at time $t$ is
$$ \mathbf{V}_i = \mathbf{v}_i - \frac{e\mathbf{E}}{m}\,t_i $$
since it was accelerated for the interval $t_i$. Now we average over all $N$ electrons. The average of the $\mathbf{v}_i$ is zero, because immediately after any collision the direction of velocity is completely random. The collisions occur not at regular intervals but at random times; we denote by $\tau$ the average time between successive collisions, called the relaxation time. The average of $t_i$ over all electrons is then $\tau$. Averaging gives the drift velocity:
$$ \mathbf{v}_d \equiv (\mathbf{V}_i)_{\text{average}} = (\mathbf{v}_i)_{\text{average}} - \frac{e\mathbf{E}}{m}(t_i)_{\text{average}} = -\frac{e\mathbf{E}}{m}\,\tau $$
This result is, as NCERT remarks, surprising: the electrons move with an average velocity that is independent of time even though they are being accelerated. The acceleration between collisions is repeatedly cut short by the next collision, so a steady average is reached. This phenomenon is drift, and the magnitude is
$$ v_d = \frac{eE\tau}{m} $$
Current and Current Density
Because of the drift there is a net transport of charge across any area perpendicular to $\mathbf{E}$. Consider a planar area $A$ inside the conductor with its normal parallel to $\mathbf{E}$. In a small time $\Delta t$, every electron lying within a distance $|v_d|\Delta t$ to the left of the area crosses it. If $n$ is the number of free electrons per unit volume, the number of such electrons is $n\,\Delta t\,|v_d|\,A$. Each carries charge $-e$, so the magnitude of charge crossing in time $\Delta t$ equals $I\,\Delta t$, giving
$$ I\,\Delta t = neA\,|v_d|\,\Delta t \quad\Longrightarrow\quad I = neA\,v_d $$
Substituting $v_d = eE\tau/m$ into $I = neAv_d$ and using $I = |\mathbf{j}|A$, the current density becomes
$$ \mathbf{j} = \frac{ne^2\tau}{m}\,\mathbf{E} $$
The vector $\mathbf{j}$ is parallel to $\mathbf{E}$. Comparing with the microscopic Ohm's law $\mathbf{j} = \sigma \mathbf{E}$, we identify the conductivity. A useful intermediate form is $J = nev_d$, which links current density directly to drift.
The relation $\mathbf{j} = \sigma\mathbf{E}$ is the microscopic statement of Ohm's Law — see how it connects to the familiar $V = IR$.
Origin of Resistivity
The expression $\mathbf{j} = (ne^2\tau/m)\mathbf{E}$ is exactly Ohm's law in the form $\mathbf{j} = \sigma\mathbf{E}$ if we identify the conductivity as
$$ \sigma = \frac{ne^2\tau}{m} $$
and therefore the resistivity, its reciprocal, as
$$ \rho = \frac{1}{\sigma} = \frac{m}{ne^2\tau} $$
This is the heart of the subtopic. Resistivity is not an arbitrary tabulated number; it emerges from the collisions of drifting electrons with the lattice ions. The smaller the relaxation time $\tau$ — that is, the more frequently an electron is interrupted by a collision — the larger the resistivity. As NCERT notes, this simple picture reproduces Ohm's law, having assumed that $\tau$ and $n$ are constants independent of $E$.
Memorising the wrong power of e in resistivity
Conductivity is $\sigma = ne^2\tau/m$ and resistivity is $\rho = m/ne^2\tau$. The charge appears squared. A frequent slip is writing $\rho = m/ne\tau$. Note also that $\tau$ sits in the denominator of $\rho$ — more collisions (smaller $\tau$) means higher resistivity.
Lock in: $\;\rho = \dfrac{m}{ne^2\tau}, \quad \sigma = \dfrac{ne^2\tau}{m}.$
Confusing J = nevd with I = neAvd
Current density $J = nev_d$ has no area in it (it is current per unit area), whereas total current $I = neAv_d$ does. Examiners pair these to test whether you carry the cross-sectional area $A$. The 2022 NEET item used $J = \sigma E$ together with $\rho = L/(\sigma A)$ to back out the answer.
$J = nev_d = \sigma E$ (per unit area); $\;I = JA = neAv_d$ (total current).
Mobility
Conductivity arises from mobile charge carriers — electrons in metals, electrons and positive ions in an ionised gas, and positive or negative ions in an electrolyte. An important quantity is the mobility $\mu$, defined by NCERT as the magnitude of the drift velocity per unit electric field:
$$ \mu = \frac{|v_d|}{E} $$
Since $v_d = eE\tau/m$, this gives
$$ \mu = \frac{|v_d|}{E} = \frac{e\tau}{m} $$
where $\tau$ is the average collision time for electrons. The SI unit of mobility is $\text{m}^2/\text{V}\,\text{s}$, and it is $10^4$ times the mobility expressed in the practical unit $\text{cm}^2/\text{V}\,\text{s}$. Mobility is always positive. Unlike drift velocity, which depends on the applied field, mobility is a property of the material at a given temperature.
A carrier has drift velocity $7.5\times10^{-4}\ \text{m s}^{-1}$ in a field of $3\times10^{-10}\ \text{V m}^{-1}$. Find its mobility.
Using $\mu = |v_d|/E = (7.5\times10^{-4})/(3\times10^{-10}) = 2.5\times10^{6}\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$. This is exactly the structure of NEET 2020 Q.104.
How Small Is the Drift
NCERT Example 3.1 estimates the drift speed in copper. For a wire of cross-section $1.0\times10^{-7}\ \text{m}^2$ carrying $1.5\ \text{A}$, with free-electron density $n = 8.5\times10^{28}\ \text{m}^{-3}$ (one conduction electron per Cu atom), the relation $v_d = I/(neA)$ gives
$$ v_d = \frac{1.5}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-7})} = 1.1\times10^{-3}\ \text{m s}^{-1} = 1.1\ \text{mm s}^{-1} $$
This is about $10^{-5}$ times the thermal speed of the copper atoms (roughly $2\times10^{2}\ \text{m/s}$ at 300 K), and smaller by a factor of about $10^{-11}$ than the speed at which the electric field propagates along the conductor, $3.0\times10^{8}\ \text{m s}^{-1}$. The very large electron number density, of order $10^{29}\ \text{m}^{-3}$, is what allows a few amperes to flow despite this crawl.
"Slow drift means slow current" — false
The drift speed is only a few mm/s, yet a bulb lights the instant the switch closes. This is not a contradiction: the field is set up throughout the wire at nearly the speed of light, so every free electron starts drifting almost simultaneously. The current does not wait for an electron to travel from the switch to the bulb.
Drift speed $\sim$ mm/s; field/signal propagation $\sim 3\times10^{8}$ m/s. Also note $v_d \propto E$, so doubling the field doubles the drift.
Relations at a Glance
| Quantity | Symbol | Defining relation | SI unit |
|---|---|---|---|
| Drift velocity | v_d | v_d = eEt/m = I/(neA) | m s-1 |
| Relaxation time | t | avg. time between collisions; t = m/(ne^2 r) | s |
| Current density | J | J = nev_d = sE = I/A | A m-2 |
| Total current | I | I = neAv_d = JA | A |
| Conductivity | s | s = ne^2 t/m | S m-1 |
| Resistivity | r | r = m/(ne^2 t) = 1/s | Ω m |
| Mobility | m | m = |v_d|/E = e t/m | m2 V-1 s-1 |
Five lines to remember
- Drift velocity is the small net velocity ($\sim$ mm/s) of electrons opposite to $E$, superposed on random thermal motion: $v_d = eE\tau/m$.
- Relaxation time $\tau$ is the average time between collisions; it controls both drift and resistivity.
- Current and current density: $I = neAv_d$ and $J = nev_d = \sigma E$.
- Resistivity arises from electron-lattice collisions: $\rho = m/ne^2\tau$, conductivity $\sigma = ne^2\tau/m$.
- Mobility $\mu = |v_d|/E = e\tau/m$, in $\text{m}^2/\text{V}\,\text{s}$; it is a material property, always positive.