Electrons crawl, the field runs

Pick a cross-section of a wire. Count the net charge that crosses it. Divide by the time. That is the current:

$$I_{\text{avg}} = \frac{\Delta Q}{\Delta t}, \qquad I = \frac{dQ}{dt}, \qquad 1 \ \text{A} = 1 \ \text{C/s}.$$

Three things in that sentence are load-bearing.

1. A surface. Current is always current through something. Asking for "the current at a point" means the current through the cross-section at that point.
2. Net charge. In a salt solution the positive ions go one way and the negative ions go the other, and both contribute to the same current in the same direction, because charge of one sign moving left is charge of the other sign moving right as far as the count is concerned.
3. Conventional direction. Current is drawn in the direction positive carriers would move. In a metal the actual carriers are electrons drifting the other way. Every circuit law in this unit is written in the conventional convention, so draw the arrow that way and stop translating.

When the charge is a function of time, differentiate. If $Q(t) = (3.0 \ \text{C/s}^2)t^2$, then $I(t) = 6.0t$ amperes, and the current at $t = 2$ s is $12$ A even though the average over the first two seconds is only $6$ A.

Build the current out of the carriers themselves. Let $n$ be the number of mobile carriers per unit volume, $q$ the charge each one carries, $A$ the cross-sectional area, and $v_d$ the average drift speed superimposed on their random thermal motion.

In a time $dt$, every carrier within a distance $v_d \, dt$ of the surface crosses it. That is a cylinder of volume $A v_d \, dt$, holding $nAv_d \, dt$ carriers and therefore $nqAv_d \, dt$ of charge. Divide by $dt$:

$$I = nqAv_d, \qquad v_d = \frac{I}{nqA}.$$

Read the second form when you want the scaling, because it puts the thing that is usually unknown on the left. For copper, $n \approx 8.5\times10^{28}$ carriers per cubic meter, one conduction electron per atom, and $q = e = 1.6\times10^{-19}$ C. A $3.4$ A current in a $1.0$ mm$^2$ wire gives

$$v_d = \frac{3.4}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})} = \frac{3.4}{1.36\times10^{4}} = 2.5\times10^{-4} \ \text{m/s},$$

a quarter of a millimeter per second. An electron leaving the switch would reach a lamp $1.5$ m away in about $6000$ seconds, close to two hours.

The related quantity is the current density $\vec{J}$, current per unit area, which points along the conventional current and satisfies

$$J = \frac{I}{A} = nqv_d, \qquad I = \int \vec{J} \cdot d\vec{A}.$$

This is the scaling students invert most reliably. Take one unbranched wire with a thick section and a thin section. Charge cannot pile up anywhere in steady state, so the same current crosses every section:

$$n q A_{\text{thick}} v_{\text{thick}} = I = n q A_{\text{thin}} v_{\text{thin}}.$$

Same material means the same $n$ and the same $q$, so $Av_d$ is what must stay constant. Bigger $A$ forces smaller $v_d$. The neck is where the drift speed peaks, exactly like a river running fast through a gorge and slow across a wide flat.

Run the geometry carefully, because the area carries the radius squared:

1. Double the current in a fixed wire: $v_d$ doubles.
2. Double the area at fixed current: $v_d$ halves.
3. Double the radius at fixed current: $A = \pi r^2$ quadruples, so $v_d$ falls to one quarter.
4. Change the metal to one with twice the carrier density: $v_d$ halves.

Note what is not on that list. The length of the wire never appears in $I = nqAv_d$. Neither does the resistance. Substituting a length where the area belongs is a common way to get an answer that is off by a factor of the wire's aspect ratio and looks plausible.

If the carriers crawl, why is there no two-hour delay between the switch and the light? Because the lamp does not wait for any particular electron to arrive. The wire is already full of mobile charge everywhere, including inside the filament. Closing the switch establishes an electric field along the conductor, and that field is set up by an electromagnetic disturbance that propagates at a speed comparable to $c$, roughly $2\times10^{8}$ m/s in a typical cable.

So for a lamp $3$ m away:

$$t_{\text{signal}} \sim \frac{3 \ \text{m}}{3\times10^{8} \ \text{m/s}} \approx 10^{-8} \ \text{s}, \qquad t_{\text{transit}} \sim \frac{3 \ \text{m}}{2.5\times10^{-4} \ \text{m/s}} \approx 10^{4} \ \text{s}.$$

Twelve orders of magnitude apart, and it is the small one that you observe. The analogy that carries the most weight is a garden hose that is already full of water: open the tap and water leaves the far end immediately, because the pressure change travels through the standing water at the speed of sound in water, not because the molecule at the tap sprinted the length of the hose.

Two conclusions worth committing:

1. The electrons in the filament were already in the filament. The battery does not ship them there; it drives the ones already present.
2. The delay you can measure is the field's, not the carriers'. Circuit response times in this course come from capacitance and inductance, never from drift transit.