Drift velocity and current — derive I = neAv_d for a conductor

Question

Derive the relation between current $I$ and drift velocity $v_d$ in a metallic conductor: $I = neAv_d$ , where $n$ is the number density of free electrons, $e$ is the electron charge, and $A$ is the cross-sectional area.

(NCERT Class 12, Chapter 3 — Current Electricity)

Solution — Step by Step

Consider a cylindrical conductor of cross-sectional area $A$ . Free electrons drift with velocity $v_d$ when an electric field is applied. We need to count how many electrons cross a given cross-section per second.

In time $\Delta t$ , each electron drifts a distance $v_d \cdot \Delta t$ . So all electrons within a cylindrical volume of length $v_d \Delta t$ and area $A$ will cross the chosen cross-section.

Volume of this cylinder:

V = A \cdot v_d \cdot \Delta t

If $n$ is the number of free electrons per unit volume, then the number of electrons in this volume is:

N = n \cdot A \cdot v_d \cdot \Delta t

Total charge crossing the cross-section:

Q = N \cdot e = neAv_d \cdot \Delta t

Current is charge per unit time:

I = \frac{Q}{\Delta t} = \mathbf{neAv_d}

This is the required relation. Current is directly proportional to drift velocity, electron density, and cross-sectional area.

Why This Works

The derivation is essentially a counting argument — we count how many charge carriers pass through a surface in a given time. The drift velocity $v_d$ is very small (of the order of $10^{-4}$ m/s), but since $n$ is enormous ( $\sim 10^{28}$ per m $^3$ for copper), the product $neAv_d$ gives a measurable current.

This also explains why current starts flowing almost instantly when you flip a switch — it is not the same electron travelling the whole length of the wire. The electric field propagates at nearly the speed of light, and all electrons throughout the wire start drifting simultaneously.

Alternative Method — Using Current Density

We can write the current density $\vec{J}$ (current per unit area) as:

\vec{J} = ne\vec{v}_d

Then the total current through area $A$ is:

I = J \cdot A = neAv_d

This relation is useful for quick numericals. For copper: $n \approx 8.5 \times 10^{28}$ m $^{-3}$ . If a 1 mm $^2$ wire carries 1 A, drift velocity works out to about $0.074$ mm/s. Examiners love asking this numerical in CBSE boards and NEET.

Common Mistake

Students sometimes confuse drift velocity with the speed of current (speed of the electric signal). Drift velocity is $\sim 10^{-4}$ m/s, but the electric signal travels at nearly $3 \times 10^8$ m/s. These are completely different quantities. In board exams, a common question is: “If drift velocity is so small, why does a bulb glow instantly?” The answer is the field propagation, not electron travel.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Current Density

Common Mistake

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