Current density and drift velocity — microscopic view of current flow

Question

A copper wire of cross-sectional area $2 \times 10^{-6}$ m $^2$ carries a current of 3 A. Given that copper has $8.5 \times 10^{28}$ free electrons per m $^3$ , find: (a) the current density, (b) the drift velocity of electrons, and (c) explain why the drift velocity is so small yet the bulb lights up instantly.

(CBSE 12 + JEE Main pattern)

Solution — Step by Step

Current density is current per unit cross-sectional area:

J = \frac{I}{A} = \frac{3}{2 \times 10^{-6}} = 1.5 \times 10^6 \text{ A/m}^2

This tells us how “concentrated” the current flow is. A thinner wire carrying the same current has higher current density — and therefore heats up more.

The key relation connecting macroscopic current to microscopic electron motion:

I = nAv_d e

where $n$ is free electron density, $A$ is cross-section area, $v_d$ is drift velocity, and $e = 1.6 \times 10^{-19}$ C.

v_d = \frac{I}{nAe} = \frac{3}{8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}

v_d = \frac{3}{2.72 \times 10^4} \approx 1.1 \times 10^{-4} \text{ m/s} \approx \mathbf{0.11 \text{ mm/s}}

The drift velocity is absurdly small — about 0.1 mm/s. At this speed, an electron would take hours to cross a 1 m wire. Yet the bulb lights up the instant we flip the switch. Why?

Because the electric field propagates at nearly the speed of light ( $\sim 3 \times 10^8$ m/s). When you close the switch, the electric field reaches every electron in the wire almost instantly. Every electron starts drifting simultaneously — like a long queue of people all taking one step forward at the same time. The signal travels fast; the electrons themselves crawl.

flowchart LR
    A["Apply voltage V"] --> B["Electric field E = V/L established instantly"]
    B --> C["All free electrons feel force F = eE"]
    C --> D["Electrons accelerate between collisions"]
    D --> E["Average drift velocity vd emerges"]
    E --> F["Current I = nAvd e"]
    F --> G["Current density J = I/A = nevd"]

Why This Works

At the microscopic level, free electrons in a conductor are constantly moving with random thermal velocities ( $\sim 10^5$ m/s). Without an electric field, their average displacement is zero — random motion in all directions cancels out.

When an electric field $\vec{E}$ is applied, a tiny systematic drift is superimposed on this random motion. The drift velocity $v_d$ is proportional to $E$ :

v_d = \frac{eE\tau}{m_e}

where $\tau$ is the mean relaxation time between collisions. Despite $v_d$ being tiny compared to thermal velocity, it produces measurable current because $n$ (the number of electrons) is enormous — of the order $10^{28}$ per m $^3$ .

Alternative Method — Using Current Density Vector

We can also write $\vec{J} = ne\vec{v_d} = \sigma \vec{E}$ , where $\sigma$ is conductivity. This links the microscopic picture (drift velocity) directly to Ohm’s law:

\vec{J} = \sigma \vec{E} \implies \frac{I}{A} = \frac{1}{\rho} \cdot \frac{V}{L} \implies V = I \cdot \frac{\rho L}{A} = IR

The relation $J = \sigma E$ is the microscopic form of Ohm’s law. Many JEE problems give you $J$ and $E$ and ask for resistivity: $\rho = E/J$ . This is faster than going through $V = IR$ and converting.

Common Mistake

Students often assume that since electrons drift slowly, there must be a delay before current starts flowing in a circuit. This confuses the drift velocity of individual electrons with the speed of the electric signal. The electric field propagates at nearly the speed of light — all electrons start moving together almost instantly. Think of it like a pipe full of water: push at one end and water comes out the other end immediately, even though individual water molecules move slowly.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Current Density Vector

Common Mistake

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