Electric Current and Drift Velocity

In short

Electric current is the rate at which charge flows through a cross-section of a conductor:

I \;=\; \frac{dQ}{dt},

measured in amperes (1 A = 1 C/s). Conventional current points in the direction positive charges would move — the opposite of the actual motion of electrons in a metal wire.

Inside a metal, free electrons already move at about 10^6 m/s in random thermal motion. Applying an electric field biases this motion by a tiny amount, producing a net drift velocity \vec{v}_d parallel to the field — typically a fraction of a millimetre per second even for a brightly-lit household bulb. Current and drift velocity are linked by

\boxed{\;I \;=\; n A e v_d\;} \qquad\text{and}\qquad \boxed{\;J \;=\; \frac{I}{A} \;=\; n e v_d,\;}

where n is the free-electron number density, A is the cross-sectional area of the wire, and e is the magnitude of the electron charge. The bulb appears to light up instantly because the electric field inside the wire establishes itself at nearly the speed of light — even though the individual electrons themselves crawl.

Flip the light switch by the door of a hostel room in IIT Kanpur and the bulb above the desk lights up — no lag, no warm-up, just on. The copper wire between the switch and the socket is maybe five metres long. It seems obvious that the "electricity" has raced down the wire faster than you can blink. The question is: what actually moves?

If you could watch an individual free electron inside that copper wire under a microscope, you would see something surprising. Even with the bulb glowing brightly, the electron shuffles along the wire at roughly a tenth of a millimetre per second — slower than a monsoon ant walks on the hostel floor. At that pace, it would take over fifteen hours for any single electron to travel the five metres from switch to bulb. The bulb, meanwhile, lit up in a few tens of nanoseconds.

The resolution of this paradox is the subject of this chapter. You will define what electric current really is — the rate of flow of charge, carried by something, somewhere — and distinguish it from the much faster propagation of the signal that tells every electron along the wire to start drifting. You will connect the macroscopic current an ammeter reads, the current density, the drift velocity, and the huge background of random thermal motion that is all happening at once. By the end you will know why the bulb lights up instantly, why overhead tramlines in Kolkata carry kiloamps of current at a drift velocity you could outrun on a bicycle, and why the I = nAev_d formula is the single most important bridge in all of solid-state electromagnetism.

What electric current is

Charge, by itself, does not "flow". Only charge in motion flows. Electric current at a particular point in a conductor is defined as the rate at which positive charge crosses a cross-section at that point:

\boxed{\;I \;=\; \frac{dQ}{dt}.\;} \tag{1}

Why: a current is a flux of charge. The natural quantity is "how many coulombs per second pass through this cross-section?" That is exactly dQ/dt. The definition is a derivative — which means it is meaningful even when the current is changing with time, as in an AC circuit.

The SI unit of current is the ampere (A), named after André-Marie Ampère. One ampere is one coulomb per second:

1\ \text{A} \;=\; 1\ \text{C/s}.

A coulomb is an enormous unit on the scale of everyday charges: 1\ \text{C} = 6.24 \times 10^{18} electron charges. A single hairbrush-rub triboelectric transfer is a few nanocoulombs. But a current of 1 A through a wire means 1 coulomb every second passes the cross-section — 6.24 \times 10^{18} electrons per second. Household currents of a few amps are staggering in these units.

Average versus instantaneous current

When the current is constant — a steady DC circuit, like the torch-bulb-and-cell circuit you built in Class 9 — equation (1) simplifies to

I \;=\; \frac{Q}{t}.

If charge Q flows past in time t, the current is the ratio. For a time-varying current you need the derivative, and you can recover the total charge by integrating:

Q \;=\; \int_0^T I(t)\,dt.

Why: the fundamental theorem of calculus. If I = dQ/dt, then the total charge that passed is the integral of I over time. This is exactly how coulometers measure electroplated charge in an electrochemistry lab — they integrate the current.

Current is a scalar (in this chapter)

Current itself is a scalar — a signed number — even though the underlying motion is directional. When you write "the current in the wire is 3 A", you mean 3 coulombs per second are crossing the cross-section in a specified direction. The direction is specified separately (by the arrow on the circuit diagram, or by the sign of the current relative to a chosen positive direction).

The finer-grained quantity that is a vector is the current density, which you will meet shortly.

Conventional current vs electron flow

Benjamin Franklin made a 50-50 guess about the sign of the charge carriers in electrolysis and lost — the charge carrier he called "positive" was actually what we today call the absence of an electron, and the one he called "negative" was the electron itself. The world has kept Franklin's sign convention ever since, which creates one of the most persistent confusions in introductory electricity:

Conventional current points in the direction positive charges would flow.

In a metal wire, the actual charge carriers are electrons — negatively charged, and they move in the direction opposite to the conventional current. When the diagram shows the current flowing from + to − through a copper wire, each electron is actually drifting from − to +.

Conventional current (red, top) runs from + to − through the wire, as if positive charges were flowing. The actual carriers in copper are electrons (black, bottom) and they move in the opposite direction — from − to +. Both pictures give the same current, because a negative charge moving left is electrically identical to a positive charge moving right.

Why both sign conventions give the same current

A negative charge moving left carries exactly the same current as a positive charge moving right. Quantitatively: if a charge \Delta q = -e crosses a cross-section while moving left over time \Delta t, the contribution to current (in the rightward direction) is \Delta q/\Delta t \cdot (-1) = +e/\Delta t, identical to a +e charge moving right.

Why: current is "positive charge per unit time crossing the cross-section in the chosen positive direction". A negative charge crossing in the negative direction contributes (-e)(-1) = +e to this count. The electron flowing left is indistinguishable from a proton flowing right, current-wise.

This is why, for virtually every circuit calculation you will ever do, you can ignore the sign subtlety and just use conventional current. The electron-level details matter for understanding what is physically going on inside the wire (Hall effect, thermoelectric effects, semiconductor physics), but they do not change the numbers your ammeter reads.

Inside the wire: free electrons, thermal chaos, and drift

A copper atom has 29 electrons. Of these, 28 are bound tightly to the nucleus. One — the single electron in the outermost 4s shell — is weakly bound and, in the metallic lattice, is free to wander through the whole crystal. Copper in bulk contains about

n \;\approx\; 8.5 \times 10^{28}\ \text{free electrons per cubic metre.}

Why: copper has a density of about 8.96 \times 10^3 kg/m³, a molar mass of 63.5 \times 10^{-3} kg/mol, and one free electron per atom. The number density is n = \rho N_A / M = (8.96 \times 10^3)(6.02 \times 10^{23})/(63.5 \times 10^{-3}) \approx 8.5 \times 10^{28} per cubic metre. That is the astronomical number of free electrons packed into every cubic metre of copper wire — far more than the number of stars in the observable universe.

Random thermal motion (no field)

These free electrons are not sitting still. At room temperature (300 K), each electron participates in the thermal motion of the lattice with an average kinetic energy of order k_B T \approx 4 \times 10^{-21} J. Equating this to \tfrac{1}{2}m_e v^2 gives a typical thermal speed

v_\text{th} \;=\; \sqrt{\frac{3 k_B T}{m_e}} \;\approx\; 10^5\ \text{m/s}\ \text{at room temperature.}

(The actual distribution is more nuanced — the Fermi-Dirac distribution places the relevant speed at about 1.6 \times 10^6 m/s, the Fermi velocity. For the present argument only the order of magnitude matters.)

But — and this is the crucial point — the thermal motion is isotropic. Each electron darts in a random direction, collides with an ion in the lattice after about 10^{-14} seconds (the mean collision time \tau), scatters off in a new random direction, and continues. Averaged over all free electrons, the net velocity is zero. No current.

With no field applied, free electrons move in random directions at enormous thermal speeds. The arrows point every which way; their vector average is zero. No net transport of charge — no current.

Drift: what an applied field does

Connect the two ends of the wire to a battery. An electric field \vec{E} is set up inside the wire. (Why a conductor can sustain a field when current is flowing — even though a conductor in electrostatic equilibrium cannot — is explained in Ohm's Law and Resistance.) Each electron now feels a force \vec{F} = -e\vec{E} that gives it a steady acceleration opposite to \vec{E}.

Between collisions, the electron picks up a small velocity increment \Delta \vec{v} = -e\vec{E}\,\tau/m_e, where \tau is the mean free time between collisions. A collision with a lattice ion randomises the direction again, wiping the drift increment (on average). Over many collision cycles, the electron's velocity decomposes into

\vec{v} \;=\; \vec{v}_\text{thermal} + \vec{v}_d,

where \langle \vec{v}_\text{thermal} \rangle = 0 is the isotropic thermal piece (unchanged by the field) and \vec{v}_d is the steady drift velocity the field has induced.

With an applied field, each electron's velocity arrow is biased slightly leftward (opposite to the rightward field). The vector average is now non-zero — a small drift velocity $\vec{v}_d$ pointing opposite to $\vec{E}$. That drift is the only part of the motion that transports charge.

The magnitude of the drift velocity is

v_d \;=\; \frac{e E \tau}{m_e},

a derivation covered in detail in Ohm's Law and Resistance. For now all you need is that v_d exists, it is parallel (or anti-parallel, for electrons) to \vec{E}, and it is typically a tiny fraction of the thermal speed.

The bridge: I = n A e v_d

This is the most important microscopic-to-macroscopic equation in conduction. It connects the current an ammeter reads (I) to the microscopic drift velocity of individual electrons (v_d). Here is the derivation.

Deriving I = n A e v_d

Assumptions: The conductor has a uniform cross-section A. The free carrier density n is the same everywhere in the conductor. Each free carrier has charge magnitude e (for electrons) or, more generally, q. All carriers have the same drift velocity v_d.

Step 1. Pick a cross-section of the wire perpendicular to the drift direction. Ask: in a small time dt, which electrons will cross this cross-section?

Exactly those electrons currently within a distance v_d\,dt of the cross-section (on the "upstream" side). Any electron farther away has not had time to reach the cross-section; any electron already past it has already been counted.

Step 2. The volume of that slab of wire is

dV \;=\; A \cdot (v_d\,dt).

Why: a cylinder of cross-section A and length v_d\,dt has volume A v_d dt. This is the geometric "scoop" of material that will flow through the cross-section in the time window.

Step 3. The number of free electrons inside that slab is

dN \;=\; n\,dV \;=\; n A v_d\,dt.

Why: n is number per unit volume, so the number in a volume dV is n\,dV.

Step 4. The total charge that crosses the cross-section in time dt is

dQ \;=\; e\,dN \;=\; n A e v_d\,dt,

where e = 1.6 \times 10^{-19} C is the magnitude of the electron charge.

Why: each electron carries charge e (the magnitude — we already absorbed the sign into "drift is opposite to conventional current"). The total charge is e times the number.

Step 5. Divide by dt to get current:

\boxed{\;I \;=\; \frac{dQ}{dt} \;=\; n A e v_d.\;} \tag{2}

Why: this is the definition of current applied to our carefully counted slab. The result is the cornerstone equation. Every quantity on the right is microscopic (density of charge carriers, area, charge per carrier, drift speed) — and their product is the macroscopic ammeter reading.

Reading equation (2)

Four quantities, one product. The equation tells you that to get a big current you can:

Make the cross-section A bigger — thicker wire carries more current at the same drift speed. This is why the aluminium overhead tram cables on a Kolkata tramline are thick aluminium tubes about 5 cm² in cross-section, not thin hair-like wires.
Use a material with large n — more carriers per cubic metre means more charge available to move. Metals (n ≈ 10^{28} to 10^{29} per m³) are far better conductors than semiconductors (n ≈ 10^{16} per m³ in pure silicon at room temperature) because of this single factor.
Increase v_d — which means increasing the applied field, which means increasing the voltage across the wire. This is the Ohm's law route, covered in the next chapter.
e is the same for every electron. Nothing to do here — except note that for charge carriers with charge 2e (like He²⁺ ions in a discharge tube) the current would double at the same density and speed.

Current density: the vector that really lives inside the wire

A current of "3 A" says nothing about whether the wire is thin or thick. A thick wire carrying 3 A is quietly conducting; a hair-thin wire carrying 3 A is glowing red and about to fail.

The quantity that measures "how much current per unit area" — and is therefore a property of the material at a point, not of the whole wire — is the current density \vec{J}:

\boxed{\;\vec{J} \;=\; \frac{I}{A}\,\hat{n} \;=\; n e \vec{v}_d,\;} \tag{3}

where \hat{n} is the unit vector along the direction of positive carrier motion. \vec{J} is a vector, unlike I. Its SI unit is A/m² (amperes per square metre).

Why: dividing equation (2) by A gives the current per unit area, and attaching the direction of carrier motion (for conventional current) or opposite to it (for electron flow, picking up a sign) converts it to a vector. \vec{J} = n q \vec{v}_d is the microscopic definition: charge density (nq) times drift velocity. For electrons q = -e, so \vec{J} = -n e \vec{v}_{d,\text{electron}} — but since electrons drift opposite to the conventional direction, the minus sign cancels and \vec{J} points along the conventional current.

The current through any surface S is recovered from \vec{J} by integration:

I \;=\; \int_S \vec{J} \cdot d\vec{A}.

For a uniform current in a wire of cross-section A with \vec{J} parallel to the wire, this reduces to I = JA.

Typical current densities

A 15 A household wire in an Indian home has roughly a 2.5 mm² cross-section copper core. The current density is

J = \frac{15}{2.5 \times 10^{-6}}\ \text{A/m}^2 = 6 \times 10^6\ \text{A/m}^2.

That is six megaamperes per square metre — a huge number, but a wire this thick can carry it without burning because the heat dissipated per unit length (covered in Ohm's Law and Resistance) is still modest. If you tried to push the same 15 A through a 0.5 mm² wire, J would jump to 3 \times 10^7 A/m² — the wire would overheat and the insulation melt.

The BIS (Bureau of Indian Standards) wiring codes that electricians follow across India are, at heart, a table of maximum current densities for each wire gauge.

A drift-velocity calculation — brace for a surprise

Consider a copper wire of cross-section A = 1 mm² = 10^{-6} m² carrying a household current of I = 10 A — about what a 2300 W immersion heater draws in a Bengaluru flat. What is the drift velocity?

Use equation (2), solved for v_d:

v_d \;=\; \frac{I}{n A e} \;=\; \frac{10}{(8.5 \times 10^{28})(10^{-6})(1.6 \times 10^{-19})}.

Numerator: 10. Denominator: 8.5 \times 10^{28} \times 10^{-6} = 8.5 \times 10^{22}; times 1.6 \times 10^{-19} gives 8.5 \times 1.6 \times 10^3 = 13.6 \times 10^3.

v_d \;\approx\; \frac{10}{1.36 \times 10^4} \;\approx\; 7 \times 10^{-4}\ \text{m/s} \;=\; 0.7\ \text{mm/s}.

A fraction of a millimetre per second. At this rate, an electron that starts at the switch when you flip it on reaches the bulb five metres away after roughly two hours. And yet the bulb lights up instantly. How?

The resolution: the signal is not the electrons themselves. When you flip the switch, an electric field propagates down the wire at close to the speed of light (more precisely: the phase velocity of an electromagnetic wave in the copper-insulation composite, which is typically 60-80% of c). This field sets every electron along the length of the wire into drifting motion simultaneously. The electrons that were already next to the bulb when you flipped the switch were already there — they start drifting through the filament right away, delivering power. The ones that were at the switch will eventually arrive at the bulb hours later, but by then they have been shuffled by many other electrons pushing them along, and there is nothing special about any individual one.

Why: think of a long water pipe already full of water. When you open the tap at the far end, water comes out immediately — not the water that was at the source, but the water already at the tap. The tap at the source pushes the column along; the pressure wave (the "instruction to flow") propagates at the speed of sound in water, not at the speed of individual water molecules. Electrons in a wire are exactly the same: already there, ready to flow, just waiting for the field to tell them to drift.

Explore it yourself

The interactive below lets you drag the current I from 0.1 A to 30 A in a copper wire of fixed cross-section 1 mm² and watch how the drift velocity changes. Notice that even at 30 A — a huge household current that would trip most circuit breakers — the drift velocity is still only about 2 mm/s.

Drag the red point to change the current in a 1 mm² copper wire. The drift velocity is tiny even for household currents: at 10 A (a typical heater), $v_d \approx 0.74$ mm/s. The readout shows how long it takes an individual electron to cross one metre of wire — it is on the order of tens of minutes for typical currents. The bulb, of course, lights up in nanoseconds, because the field propagates at nearly $c$, not at $v_d$.

Worked examples

Example 1: Drift velocity in an IIT-lab copper wire

A copper wire of cross-section 1.5\ \text{mm}^2 carries a current of 3\ \text{A} in a physics-lab bench setup at IIT Kanpur. Given the free-electron density of copper n = 8.5 \times 10^{28}\ \text{m}^{-3} and e = 1.6 \times 10^{-19}\ \text{C}, find the drift velocity of the electrons.

The cross-section of the copper wire is 1.5 mm² and the current is 3 A.

Step 1. Convert the cross-section to SI units.

A = 1.5\ \text{mm}^2 = 1.5 \times 10^{-6}\ \text{m}^2.

Why: every quantity must be in SI before the arithmetic. 1 mm² = 10^{-6} m² because 1 mm = 10^{-3} m and you square the factor.

Step 2. Apply I = n A e v_d and solve for v_d.

v_d = \frac{I}{n A e} = \frac{3}{(8.5 \times 10^{28})(1.5 \times 10^{-6})(1.6 \times 10^{-19})}\ \text{m/s}.

Step 3. Compute the denominator.

n A e = 8.5 \times 10^{28} \times 1.5 \times 10^{-6} \times 1.6 \times 10^{-19}.

Multiply the numbers: 8.5 \times 1.5 = 12.75; 12.75 \times 1.6 = 20.4.

Multiply the powers of 10: 10^{28} \times 10^{-6} \times 10^{-19} = 10^{28-6-19} = 10^{3}.

n A e \approx 20.4 \times 10^{3} = 2.04 \times 10^{4}\ \text{A·s/m}.

Why: breaking the product into mantissa and power of 10 keeps the arithmetic manageable. The units work out: m⁻³ · m² · C = C/m, and (C/m)·(m/s) = C/s = A, which is what I should be — check.

Step 4. Divide.

v_d = \frac{3}{2.04 \times 10^4}\ \text{m/s} \approx 1.47 \times 10^{-4}\ \text{m/s} \approx 0.15\ \text{mm/s}.

Step 5. Sanity check with thermal speed.

The drift velocity is 1.5 \times 10^{-4} m/s; the thermal speed is roughly 10^5 m/s. The ratio is 10^{-9} — the drift is one part in a billion of the random thermal motion.

Why: this is not just bookkeeping. It tells you that the drift is a tiny asymmetry superimposed on a gigantic random background. You cannot "see" the drift by following one electron — the thermal kicks dominate its motion completely. The drift shows up only when you average over 10^{28} electrons per cubic metre.

Result: v_d \approx 0.15 mm/s. An individual electron in this IIT-lab wire takes 6.8 seconds to cross a single millimetre — but the 3 A current is perfectly steady because 10^{22} electrons cross every square-millimetre cross-section every second.

Example 2: Kolkata tram overhead line

The overhead contact wire of a Kolkata tram carries a maximum current of about 600 A when the tram is accelerating. The copper-alloy wire has cross-section 60\ \text{mm}^2 = 6 \times 10^{-5}\ \text{m}^2. (a) Find the current density. (b) Find the drift velocity (assume n = 8.5 \times 10^{28} per m³). (c) How long does it take an individual electron to drift 1 km along the wire?

A Kolkata tram drawing peak 600 A from the overhead contact wire at the moment of acceleration.

(a) Current density.

J = \frac{I}{A} = \frac{600}{6 \times 10^{-5}}\ \text{A/m}^2 = 10^{7}\ \text{A/m}^2.

Why: current per area, in SI units. 10^7 A/m² is high but well within the limit of heavy-duty bare copper, which can tolerate roughly 10^7 A/m² continuous. That is why the overhead line is such a thick wire — the tram needs this much area to handle peak current without excessive heating.

(b) Drift velocity.

Using equation (2):

v_d = \frac{I}{n A e} = \frac{J}{n e} = \frac{10^{7}}{(8.5 \times 10^{28})(1.6 \times 10^{-19})}\ \text{m/s}.

Denominator: 8.5 \times 1.6 \times 10^{28-19} = 13.6 \times 10^9 = 1.36 \times 10^{10}.

v_d = \frac{10^{7}}{1.36 \times 10^{10}}\ \text{m/s} \approx 7.4 \times 10^{-4}\ \text{m/s} \approx 0.74\ \text{mm/s}.

Why: even at 600 A, the electrons crawl at under a millimetre per second. This is because the wire's cross-section is also huge — the density of current (J) matters more than the absolute current when it comes to drift speed. Doubling the wire thickness at fixed current halves v_d; doubling the current at fixed thickness doubles v_d.

(c) Time to drift 1 km.

t = \frac{d}{v_d} = \frac{1000\ \text{m}}{7.4 \times 10^{-4}\ \text{m/s}} \approx 1.35 \times 10^{6}\ \text{s}.

Convert to more familiar units:

t \approx 1.35 \times 10^{6}\ \text{s} \times \frac{1\ \text{hour}}{3600\ \text{s}} \approx 375\ \text{hours} \approx 15.6\ \text{days}.

Why: a single electron would need over two weeks of continuous tram operation to drift from one end of a kilometre-long overhead line to the other. Meanwhile, the electromagnetic signal that coordinates their motion travels that kilometre in about 3 microseconds.

Result: (a) J = 10^{7} A/m². (b) v_d \approx 0.74 mm/s. (c) 1 km takes an individual electron about 15.6 days.

What this shows: High-current applications (power distribution, traction, industrial) deal with thousands of amperes by using large cross-sections, keeping the current density and drift velocity manageable. The drift velocity barely changes between a household wire and a tram cable; what changes is the total cross-sectional area.

Example 3: Charge that passes a point — integrating $I(t)$

In a capacitor-discharge experiment, the current through a lab resistor varies with time as I(t) = I_0\,e^{-t/\tau}, with I_0 = 2 A and \tau = 0.05 s. Find the total charge that flows through the resistor in the first 0.1 seconds.

Step 1. Write the definition of total charge.

Q = \int_0^{T} I(t)\,dt, \qquad T = 0.1\ \text{s}.

Why: the current I = dQ/dt inverts to Q = \int I\,dt. This is how a coulometer (or any charge-measurement device integrating an ammeter) computes total charge.

Step 2. Substitute and integrate.

Q = \int_0^{0.1} 2\,e^{-t/0.05}\,dt.

The antiderivative of e^{-t/\tau} is -\tau e^{-t/\tau}. So

Q = 2 \cdot (-0.05)\,\left[e^{-t/0.05}\right]_0^{0.1} = -0.1\,\left(e^{-2} - e^{0}\right) = -0.1\,(0.1353 - 1).

Q = -0.1 \times (-0.865) = 0.0865\ \text{C}.

Why: factor out the constants, apply the antiderivative, evaluate at the limits. The numerical evaluation uses e^{-2} \approx 0.1353. The sign works out positive because the upper bound of the exponential decay is smaller than the lower bound.

Step 3. Sanity check.

If the current were constant at I_0 = 2 A for 0.1 s, it would deliver 0.2 C. The actual answer, 0.0865 C, is less than half — because the exponential decay has fallen to e^{-2} \approx 13.5\% of its initial value by the end of the time window. A rough rule of thumb: an exponentially decaying current delivers, in the first time constant \tau, about (1 - 1/e)\,I_0\,\tau = 0.632\,I_0\,\tau = 0.063 C — consistent with the more-than-one-\tau answer above.

Step 4. Number of electrons.

N = \frac{Q}{e} = \frac{0.0865}{1.6 \times 10^{-19}} \approx 5.4 \times 10^{17}\ \text{electrons.}

Result: Q \approx 0.0865 C pass through the resistor in the first 0.1 s — about 5.4 \times 10^{17} electrons.

What this shows: When the current varies with time, you cannot just multiply I \times t — you must integrate. The integral picks up the changing rate of charge flow, and the result is always less than (or more than, for increasing currents) the naïve product. This is how capacitor charges and discharges are analysed in RC-circuit problems.

Common confusions

"Conventional current is just a wrong guess we keep for historical reasons." Not quite — it is a choice of sign convention that happens to be opposite to the actual carrier motion in metals, but it gives identical predictions and is the universally used convention in physics and engineering. Sticking with it also makes the formulas work cleanly for positive carriers (protons in an accelerator, positive ions in a solution) without extra minus signs.
"Drift velocity means the electrons only move in one direction." No — drift is an average on top of huge random motion. Each individual electron is zigzagging at 10^5–10^6 m/s in all directions; the drift is only the tiny bias in the average direction. At any instant, most electrons near you are moving against the drift.
"In AC circuits, current doesn't flow — it just wiggles." Charge does not net-move in AC, but at every instant it is flowing. The current I(t) is meaningful instant by instant, and all the definitions here apply with I replaced by I(t). The drift velocity also oscillates: v_d(t) = I(t)/(nAe).
"The drift velocity of electrons equals the speed of the electrical signal." No — they differ by about a factor of 10^{11}. Signal speed (speed of electromagnetic field propagation) is close to c; drift speed is sub-millimetre per second. They are different physical quantities, and the confusion between them is the single biggest source of misconception about electricity.
"Current depends on the shape of the wire along its length." No — for a steady DC current in a single wire, the current I is the same through every cross-section. This is charge conservation: whatever flows in at one end must flow out at the other. What changes with the wire's cross-section is the current density J = I/A, which is higher where the wire is thinner.
"I = nAev_d needs electrons; it does not apply to semiconductors or ionic solutions." The same equation works for any charge carrier — just replace n with the appropriate carrier density and e with the magnitude of the charge on each carrier. In an NaCl solution, two kinds of carriers (Na⁺ and Cl⁻) contribute independently, so you sum the two contributions: J = n_+ q_+ v_{d+} + n_- q_- v_{d-} (with the signs worked out carefully).

If you have the definition of current, the conventional-vs-actual-flow distinction, and the I = nAev_d bridge, you have what you need for JEE Main. What follows is JEE Advanced content: the continuity equation, vector current density, the role of the Fermi velocity, and why metals have roughly constant n but semiconductors do not.

The continuity equation

Charge conservation, written locally, gives the continuity equation:

\nabla \cdot \vec{J} \;+\; \frac{\partial \rho}{\partial t} \;=\; 0,

where \rho is the local charge density. This is the differential form of "charge is neither created nor destroyed"; in a steady state \partial \rho/\partial t = 0, so \nabla \cdot \vec{J} = 0 — the current density is divergence-free, which is why the same I flows into and out of every cross-section of a series wire. In a capacitor being charged, \partial \rho/\partial t \neq 0 on the plates, and the continuity equation captures the accumulation of charge there.

Fermi velocity vs classical thermal velocity

The classical estimate v_\text{th} = \sqrt{3 k_B T / m_e} \approx 10^5 m/s is wrong. Electrons in a metal obey Fermi-Dirac statistics, not classical Maxwell-Boltzmann. At T = 0 K they fill all states up to the Fermi energy E_F and sit in those states — they do not "freeze" because the Pauli exclusion principle forbids them from going to lower states. The typical electron speed is therefore the Fermi velocity:

v_F \;=\; \sqrt{\frac{2 E_F}{m_e}}.

For copper E_F \approx 7 eV, giving v_F \approx 1.6 \times 10^6 m/s — an order of magnitude higher than the classical thermal estimate. Thermal fluctuations around the Fermi surface are small (about k_B T / E_F \approx 0.4\% at room temperature), so for most conduction questions the ambient speed of a free electron is effectively v_F, not v_\text{th}. This does not change I = nAev_d — the drift velocity is computed from the bias, not the ambient speed.

Why n changes enormously between materials

Metals have n \approx 10^{28}–10^{29} per m³ because every atom contributes a free electron. In a solid, there are about 5 \times 10^{28} atoms per m³ (governed by the atomic size ≈ 0.2 nm), and in a metal one or two of those electrons are mobile — giving the carrier density directly.

Semiconductors (silicon, germanium) at room temperature have n \sim 10^{16} per m³ in the pure state — a factor of 10^{13} smaller than copper — because thermal energy is barely enough to promote electrons across the band gap (≈ 1 eV for Si) into conducting states. Doping with phosphorus brings n up to around 10^{22} per m³, bringing the semiconductor's conductivity to within a factor of 10^6 of copper.

Ionic solutions have n depending on concentration: a 0.1 M NaCl solution has about 6 \times 10^{25} ions per m³ — comparable in order of magnitude to a metal, but the ionic mobilities are much smaller (ions are much heavier and interact with water strongly) so the effective v_d per unit applied field is far smaller.

This is why "conductor", "semiconductor", and "insulator" come in three very different orders of magnitude: they are set mainly by n, not by v_d.

Hall effect: finding the sign of the carrier

Put a current-carrying conductor in a perpendicular magnetic field \vec{B}. The moving charges experience a Lorentz force q \vec{v}_d \times \vec{B}, which pushes them to one side of the conductor. A voltage develops across the sample — the Hall voltage — whose sign tells you whether the carriers are positive or negative.

V_H \;=\; \frac{I B}{n q t},

where t is the thickness of the slab. For most metals (copper, aluminium, gold) the Hall voltage has the sign consistent with negative carriers (electrons). For some metals (beryllium, zinc, cadmium) and for p-type semiconductors, the sign is opposite, consistent with positive carriers ("holes" in the band-structure picture). The Hall effect was discovered by Edwin Hall in 1879 and became one of the fundamental experimental proofs that electric current in metals is carried by particles of a specific sign and density.

Indian hall-effect research continues today at laboratories like CSIR-NPL in Delhi, which provides calibrated Hall standards for current and magnetic field measurements used across industry.

Beyond the free-electron model

The picture of "independent free electrons colliding with lattice ions" is the Drude model, a classical 1900 proposal that gets the right qualitative answers and, surprisingly, the right magnitudes. The quantum-mechanical refinement (Sommerfeld, 1927) replaces the classical thermal distribution with the Fermi-Dirac one and adjusts the collision picture to account for the de Broglie wavelength of the electrons — but keeps the key equation I = nAev_d unchanged.

Modern condensed-matter physics goes further: the "free electrons" are actually quasiparticles in a band structure, the scattering is by phonons and defects rather than static ions, and the drift is described by a full Boltzmann transport equation. But for every one of the current-and-voltage calculations you will do in a JEE Advanced problem — or in an undergraduate electromagnetism course — equation (2) is exact, and every refinement just tells you why n, \tau, and the effective mass take the values they do.

Where this leads next

Ohm's Law and Resistance — what determines the drift velocity, the relation V = IR, and the origin of resistance in electron-lattice collisions.
Electric Field — the field that drives the drift in the first place, and whose propagation (not the electrons themselves) explains the instantaneous turn-on of a circuit.
Electric Potential and Potential Difference — the voltage that sets up the field along the wire when a battery is connected.
Capacitance and the Parallel Plate Capacitor — the analysis of capacitors where current is strictly the rate of change of stored charge, not a steady flow.
Electric Charge — Properties and Conservation — the conservation law that underlies both the definition of current and the continuity equation.