Define current (flow of charge) and the ampere: Visual walkthrough

Step 1 — What does "current" even count?

WHAT. Imagine a slice cut straight through a wire, like a coin-sized window. Current is the answer to one question: how much electric charge slides through that window each second?

WHY. Before any formula, we must pin down the thing being measured. A "flow" always needs two ingredients — a how much and a per how long. Charge is our "how much"; the second is our "how long". Dividing one by the other gives a rate, and a rate is exactly what "flow strength" means.

PICTURE. Look at the violet window plane. The magenta dots are bits of charge streaming toward it. We only care about who crosses that plane.

Everything below is really just one long, careful effort to compute that $Δ Q$ from what's inside the wire.

Step 2 — Meet the movers: carriers in the metal

WHAT. Zoom inside the wire. It is packed with tiny charge carriers (in copper, these are free electrons). We describe the crowd with two numbers.

WHY. We can't count charges one by one — there are trillions. Instead we describe the crowd statistically: how tightly packed they are, and how much charge each one carries. That converts "count the marbles" into "multiply a density by a volume".

PICTURE. The orange dots below are carriers. Notice how evenly spread they are — that even spread is what " $n$ per cubic metre" means.

See Electric charge and the coulomb for what one coulomb actually is, and Conductors and insulators for why metals have so many free carriers while rubber has almost none.

Step 3 — They drift (slowly!) in one direction

WHAT. Without a battery the carriers jiggle randomly and go nowhere on average. Connect a battery — apply a voltage — and the whole crowd gains a tiny, steady sideways nudge called the drift velocity.

WHY. Random jiggling gives zero net crossing (as many go left as right). Only the shared, one-way drift produces a real flow through our window. So the quantity that controls current is not the frantic random speed but the gentle average drift.

PICTURE. Grey squiggles = random jiggle. The bold magenta arrow = the small average shift they all share. That average is $v_{d}$ .

Step 4 — How far does the crowd move in a moment $Δ t$ ?

WHAT. In a short time $Δ t$ , each carrier drifts forward by a distance $ℓ = v_{d} Δ t .$

WHY. This is just "distance = speed × time", the same rule as a car: go $v_{d}$ metres every second for $Δ t$ seconds and you cover $v_{d} Δ t$ metres. We need this length because it tells us how deep into the wire a carrier can be and still reach the window in time.

PICTURE. The length $ℓ$ is drawn as a violet bar behind the window. Any carrier further back than $ℓ$ simply won't arrive in time.

$ℓ$ — reach length: how far the drift carries a charge in the moment $Δ t$ .
$v_{d} Δ t$ — that reach, written as speed × time.

Step 5 — Sweep out the "cylinder of charge that makes it"

WHAT. Only the carriers inside a cylinder — window face of area $A$ , length $ℓ = v_{d} Δ t$ — cross the window during $Δ t$ . Its volume is $V = A \cdot ℓ = A v_{d} Δ t .$

WHY. A carrier crosses only if it is (a) behind the window and (b) close enough to reach it in time $Δ t$ . Those two conditions carve out exactly this cylinder. Turning "who crosses?" into "who lives in this box?" is the key trick — geometry replaces guesswork.

PICTURE. The whole shaded cylinder is the group that crosses. Its base is the wire's cross-section $A$ ; its length is the reach $ℓ$ from Step 4.

Step 6 — Count the carriers, then their charge

WHAT. Multiply the volume by the density $n$ to get how many carriers are in the cylinder, then by $q$ to get their total charge: $Δ Q = q \cdot (n \cdot V) = q n A v_{d} Δ t .$

WHY. Number of carriers = (carriers per m³) × (m³) = $n \times V$ . Total charge = (number) × (charge each) = that count $\times q$ . This is the same logic as "6 eggs per box × 5 boxes × ₹8 per egg = total money": density × volume × price-each.

PICTURE. The cylinder is now filled with orange dots. The count is $nV$ ; each dot carries $q$ ; stacked up they make $Δ Q$ .

Step 7 — Divide by time: the formula appears

WHAT. Go back to Step 1's definition $I = Δ Q /Δ t$ and substitute: $I = \frac{Δ Q}{Δ t} = \frac{q n A v _{d} Δ t}{Δ t} = n q A v_{d} .$

WHY. The whole point of building $Δ Q$ was to feed it into the current definition. The $Δ t$ on top cancels the $Δ t$ on the bottom — which is why the answer is independent of how long you watch. A steady flow gives the same current whether you count for one second or ten.

PICTURE. The cancellation shown visually: the time-bar on top and bottom shrink away, leaving four surviving letters.

Rearranged, this also feeds straight into Ohm's Law and resistance and explains what changes in Series and parallel circuits.

Step 8 — Degenerate & sign cases (don't skip these!)

Every case the reader could meet:

PICTURE. Three mini-panels: still crowd ( $v_{d} = 0$ ), empty insulator ( $n = 0$ ), and electrons drifting left while conventional current points right.

Step 9 — Sanity check: the numbers are tiny

WHAT. Solve for $v_{d}$ in a real copper wire carrying $1 A$ through $A = 1 \times 1 0^{- 6} m^{2}$ : $v_{d} = \frac{I}{n q A} = \frac{1}{( 8.5 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{- 6} )} \approx 7.4 \times 1 0^{- 5} m/s .$

WHY. Plugging real numbers tests the formula and delivers a shock: about 0.074 mm/s — slower than a snail. Yet lights turn on instantly, because the field pushing all electrons at once travels near light speed, even though each electron barely crawls.

PICTURE. A speed ruler: electron drift as a dot near zero, a snail far ahead, light off the chart.

The one-picture summary

The entire derivation on one canvas: a window of area $A$ , a crowd of density $n$ each carrying $q$ , drifting at $v_{d}$ so that in $Δ t$ they sweep a cylinder of volume $A v_{d} Δ t$ ; count × charge gives $Δ Q$ ; divide by $Δ t$ and the time cancels, leaving $I = n q A v_{d}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Picture a doorway in a wire (Step 1). Inside are countless tiny charge-carriers, spread evenly — that's $n$ per cube — and each carries a fixed pinch of charge $q$ (Step 2). Switch on a battery and they all start creeping forward together at the same gentle pace $v_{d}$ (Step 3). In a small moment $Δ t$ each one advances a little distance $v_{d} Δ t$ (Step 4). So exactly the carriers sitting in a box — doorway-sized on the front, $v_{d} Δ t$ deep — get to cross (Step 5). Count them ( $n$ times the box volume) and add up their charge (times $q$ ) to get the chunk $Δ Q$ that crossed (Step 6). Finally, "current is charge per second," so divide by $Δ t$ — and the time cancels, leaving the four survivors $I = n q A v_{d}$ (Step 7). Watch the edge cases: no drift means no current even with charge around, and no carriers (rubber) means no current even with a battery (Step 8). And the punchline: those electrons crawl slower than a snail, yet your lamp lights instantly because the push races through at light speed (Step 9).

Active recall

Why cylinder length

= v_{d} Δ t

It's the drift distance in time

Δ t

; only carriers within this reach can cross the window.

Where does

Δ t

cancel?

Δ Q = q n A v_{d} Δ t

, divided by

Δ t

I = Δ Q /Δ t

— steady current is time-independent.

Current when

v_{d} = 0

Zero — charge present but no net drift, so no crossing.

Current when

n \approx 0

Zero — an insulator has almost no free carriers even with voltage applied.

Why is drift speed so tiny yet devices instant?

Electrons crawl (~

1 0^{- 4}

m/s) but the electric field nudging them travels near light speed.

Visual walkthrough — Define current (flow of charge) and the ampere