Q.(a) In Example 3.1, the electron drift speed is estimated to be only a few $\text{mm s}^{-1}$ for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed?

(a) The electric field set up in the conductor propagates at nearly the speed of light ($\approx 3\times10^8\ \text{m s}^{-1}$), so it starts all free electrons drifting almost simultaneously the instant the circuit is closed — the current is established by the field, not by electrons travelling from the switch to the load.

(b) Between collisions the field accelerates an electron ($a = eE/m$), but each collision with a lattice ion randomises its velocity. Averaged over many collisions the electron acquires only a small steady drift speed $v_d = \dfrac{eE\tau}{m}$ ($\tau$ = relaxation time) — the collisions act like a resistive drag that prevents unlimited acceleration.

(c) Current is $I = neAv_d$. The free-electron density $n \sim 10^{28}\text{–}10^{29}\ \text{m}^{-3}$ is enormous, so even a tiny $v_d$ and tiny $e$ give currents of several amperes.

(d) No. The electrons keep their large random thermal motion in all directions ($\sim 10^5\ \text{m s}^{-1}$); the field only superimposes a very small net drift ($\sim 10^{-4}\ \text{m s}^{-1}$) opposite to $\vec E$. Only the average velocity is directed, not each electron. …

(a) Why current is established almost instantly

When the circuit closes, the electric field is set up throughout the conductor at nearly the speed of light ($\approx 3\times10^8\ \text{m s}^{-1}$). This field acts on every free electron at essentially the same moment, so all of them begin their slow drift together. The near-instant current is due to the rapid propagation of the field, not to any single electron travelling from the source to the load — like a water-filled pipe where pushing in one end makes water leave the far end at once.

(b) Why a steady drift speed

Between collisions an electron accelerates, $a = eE/m$. But it collides with the vibrating positive ions every $\tau \sim 10^{-14}\ \text{s}$, and each collision randomises its velocity. Averaging, the electron gains a constant drift velocity

opposite to $\vec E$. The collisions behave like a viscous drag that balances the electric force, so the speed does not grow without bound.

(c) Large current from small drift

The current is

where $n$ is the free-electron density. For a metal $n \sim 10^{28}\text{–}10^{29}\ \text{m}^{-3}$ — astronomically large — so even with $v_d \sim 10^{-4}\ \text{m s}^{-1}$ and $e = 1.6\times10^{-19}\ \text{C}$ the product gives amperes. The sheer number of carriers, not their speed, delivers the current.

(d) Do all electrons move the same way? …

Method: Drift Velocity Model (Microscopic View of Current)

This model explains current in terms of the average motion of free electrons under an electric field, accounting for collisions with ions.

(a) How is current established almost instantly if drift speed is so small?

Key idea: The electric field inside the conductor propagates at nearly the speed of light ($\approx 3 \times 10^8 \, \text{m/s}$).

• When the circuit is closed, the electric field is set up along the entire conductor almost instantly.
• All free electrons in the conductor begin drifting simultaneously in response to this field.
• Even though each electron moves slowly ($\text{mm/s}$), the signal (field) travels fast, so current starts flowing everywhere at once.

Analogy: A long pipe full of water — when you open the tap, water comes out almost immediately because the pressure wave travels fast, even though individual water molecules move slowly.

(b) Why do electrons acquire a steady average drift speed, not accelerate continuously?

Key idea: Electrons undergo frequent collisions with the positive ions (lattice) of the metal.

• Between collisions, an electron accelerates due to the electric field $E$:

where $e$ is electron charge, $m$ is electron mass.

• But after each collision, the electron’s velocity is randomized (direction and magnitude).
• The average velocity over many collisions becomes constant — this is the drift velocity $v_d$.

Mathematically, if $\tau$ is the average time between collisions:

So the steady speed arises because collisions reset the motion repeatedly.

(c) How can we get large currents if drift speed and electron charge are small?

Key idea: Current depends on number density of free electrons, which is enormous in metals.

Current formula:

where:

• $n$ = number of free electrons per unit volume ($\approx 10^{28} \, \text{m}^{-3}$ for copper)
• $e$ = charge of electron ($1.6 \times 10^{-19} \, \text{C}$)
• $A$ = cross-sectional area
• $v_d$ = drift velocity

Even if $v_d$ is tiny ($\text{mm/s}$), the product $n e A$ is huge, so $I$ can be several amperes.

Example: For copper wire of $1 \, \text{mm}^2$ area and $v_d = 0.1 \, \text{mm/s}$, $I \approx 1.6 \, \text{A}$.

(d) Does drift mean all free electrons move in the same direction?

No. Only the average motion is in the direction of the electric field (from lower to higher potential). …

Here are the common mistakes students make on these drift velocity questions, along with the correct conceptual understanding and how to avoid each error.

(a) Instantaneous Current vs. Slow Drift Speed

Common Mistake:

Students think that the electrons themselves must travel from the switch to the bulb at the drift speed (mm/s). They conclude that if electrons move that slowly, the bulb should take minutes or hours to light up.

Why this is wrong:

The current is established by the electric field, which travels at nearly the speed of light ($\approx 3 \times 10^8 \text{ m/s}$). The field pushes all free electrons in the wire simultaneously. The drift speed is the net speed of an individual electron, but the signal (the field) propagates almost instantly.

How to avoid this mistake:

• Think of a pipe full of water. When you turn on the tap, water comes out almost instantly, even though the water molecule at the tap didn't travel from the tank. The pressure wave (analogous to the electric field) travels fast.
• Memorise the distinction: Signal speed $\approx c$ (speed of light). Drift speed $\approx \text{mm/s}$. They are not the same thing.

(b) Steady Drift Speed Despite Constant Force

Common Mistake:

Students apply Newton's Second Law ($F = ma$) and assume that a constant electric force should cause constant acceleration, leading to ever-increasing speed.

Why this is wrong:

Electrons in a metal constantly collide with the vibrating positive ions (lattice). Between collisions, they do accelerate. However, each collision randomises the electron's velocity. The net effect is that the electron loses the extra kinetic energy gained from the field. The average velocity in the field direction becomes constant — this is the drift velocity.

How to avoid this mistake:

• Visualise a pinball machine. The flipper (electric field) gives the ball a push, but the bumpers (ions) keep knocking it off course. The ball doesn't keep speeding up; it reaches a steady average speed.
• Use the relaxation time ($\tau$) concept. The drift velocity is given by $v_d = \frac{eE}{m} \tau$, where $\tau$ is the average time between collisions. This formula directly shows a steady speed proportional to the field, not an accelerating one.

(c) Small Drift Speed + Small Charge = Large Current?

Common Mistake:

Students focus only on the small values ($v_d \approx 10^{-4} \text{ m/s}$, $e \approx 1.6 \times 10^{-19} \text{ C}$) and cannot see how this yields amperes of current.

Why this is wrong:

The current formula is $I = n e A v_d$. The key is the number density of free electrons ($n$). In a metal like copper, $n$ is enormous — on the order of $10^{28}$ electrons per cubic metre. This huge number compensates for the small charge and small speed.

How to avoid this mistake:

• Always write the full formula: $I = n e A v_d$.
• Do a quick order-of-magnitude check: For copper, $n \approx 8.5 \times 10^{28} \text{ m}^{-3}$, $e \approx 1.6 \times 10^{-19} \text{ C}$, $A \approx 10^{-6} \text{ m}^2$, $v_d \approx 10^{-4} \text{ m/s}$. Multiply: $I \approx (8.5 \times 10^{28})(1.6 \times 10^{-19})(10^{-6})(10^{-4}) \approx 1.36 \text{ A}$. It works because $n$ is astronomically large.

(d) Direction of Drift of All Free Electrons

Common Mistake:

Students think that all free electrons in the metal are moving in the same direction (from lower to higher potential) at the drift speed.

Why this is wrong: …