Current Density and Drift Velocity — Microscopic View of Current Flow

Question

How are current density ( $\vec{J}$ ) and drift velocity ( $v_d$ ) related, and what does the microscopic picture of current flow actually look like?

Solution — Step by Step

Inside a conductor, free electrons move randomly at high thermal speeds (around $10^5$ m/s). Without an external field, their net displacement is zero — random motion cancels out.

When we apply a potential difference, an electric field $\vec{E}$ develops inside the conductor. Each electron now experiences a small force $\vec{F} = -e\vec{E}$ and gains a tiny systematic drift superimposed on its random motion. This systematic component is the drift velocity $v_d$ — typically around $10^{-4}$ m/s. Incredibly slow, yet it produces measurable current.

Consider a conductor of cross-sectional area $A$ . In time $\Delta t$ , electrons drift a distance $v_d \Delta t$ . The volume of electrons passing through any cross-section is $A \cdot v_d \cdot \Delta t$ .

If $n$ is the number density of free electrons (electrons per unit volume), the total charge crossing the section:

\Delta q = n \cdot A \cdot v_d \cdot \Delta t \cdot e

Current:

I = \frac{\Delta q}{\Delta t} = neAv_d

Current density $\vec{J}$ is current per unit area:

J = \frac{I}{A} = nev_d

In vector form:

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

where $\sigma$ is the electrical conductivity. This connects the microscopic (drift velocity) to the macroscopic (conductivity, electric field).

graph LR
    A[Applied Voltage V] --> B[Electric Field E = V/L]
    B --> C[Force on electron F = eE]
    C --> D[Drift velocity vd = eE tau/m]
    D --> E[Current I = neAvd]
    E --> F[Current density J = I/A = nevd]
    F --> G[Ohms Law: J = sigma E]

    H[Relaxation time tau] --> D
    I[Number density n] --> E

Here $\tau$ is the relaxation time — the average time between successive collisions of an electron with the lattice ions.

Why This Works

The beauty of this microscopic model is that it derives Ohm’s law from first principles. The drift velocity $v_d = \frac{eE\tau}{m}$ is proportional to $E$ , which makes $J = nev_d$ proportional to $E$ — and that is precisely Ohm’s law ( $J = \sigma E$ ) with $\sigma = \frac{ne^2\tau}{m}$ .

This also explains why resistance increases with temperature: higher temperature means more lattice vibrations, shorter $\tau$ , smaller $\sigma$ , and hence higher resistance.

CBSE 12 boards frequently ask: “Derive the relation between current and drift velocity.” The derivation above (Step 2) is the standard 3-mark answer. Write it with a clear diagram showing the cylindrical volume element.

Alternative Method

Instead of the volume-element approach, we can start from the definition of current density $\vec{J} = nq\vec{v}_d$ directly (this is the fundamental definition in electrodynamics) and then integrate over the cross-section:

I = \int \vec{J} \cdot d\vec{A}

For uniform current density, this simplifies to $I = JA = neAv_d$ .

Common Mistake

Students confuse drift velocity with the speed of the electric signal. The drift velocity is about $10^{-4}$ m/s, but the electric signal (the field) propagates at nearly the speed of light ( $\sim 3 \times 10^8$ m/s). When you flip a switch, the field reaches the bulb almost instantly — the electrons themselves barely move. This distinction appears in CBSE and JEE conceptual questions regularly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next