Drift velocity, mobility, conductivity: Question bank

Definition Every symbol used on this page (define before you trap yourself)

$E$ = electric field strength inside the wire (units V/m) — the "push" per unit charge the battery sets up.
$E$ = the same field written as an arrow, so we can talk about its direction.
$e$ = the magnitude of the electron's charge, a positive number ( $1.6 \times 1 0^{- 19}$ C). The electron itself carries $- e$ (negative).
$m$ = the mass of one charge carrier (an electron, $9.1 \times 1 0^{- 31}$ kg).
$v_{d}$ = drift velocity: the tiny average velocity of the whole electron swarm, riding on top of huge random thermal jiggle.
$τ$ = relaxation time: the average time between two collisions of an electron with the lattice.
$n$ = number density: how many free electrons sit in one cubic metre of metal (units m⁻³).
$A$ = cross-sectional area of the wire; $I$ = current (charge per second, amperes).
$J = I / A$ = current density (current per unit area, A/m²).
$σ$ = conductivity (how easily the material conducts); $ρ = 1/ σ$ = resistivity (how strongly it resists).
$μ = v_{d} / E$ = mobility (drift speed gained per unit field).

True or false — justify

Every claim is either a subtle truth or a plausible-sounding lie. Say why before revealing.

With no battery connected, the current in a copper wire is exactly zero even though electrons move at ~

1 0^{5}

m/s.

True — the fast motion is random, so for every electron heading one way another heads the opposite way; the vector average is zero, hence no net charge transport.

Drift velocity is the speed at which an individual electron actually moves through the wire.

False — an individual electron zigzags at thermal speed (

\sim 1 0^{5}

m/s);

v_{d}

(

\sim 1 0^{- 4}

m/s) is only the tiny average drift of the whole swarm, not any single electron's true speed.

Electrons drift in the same direction as the applied field

E

False — the force is

- e E

, so electrons drift opposite to

E

; conventional current

I

points along

E

because current direction is defined for positive charge.

When you switch on a torch, the electron that lights the bulb had to travel the whole length of the wire first.

False — the wire is already packed with electrons; the field propagates at nearly

c

and nudges every electron at once, so an electron already at the bulb starts moving immediately.

Doubling the wire's cross-sectional area

A

while keeping the same current halves the drift velocity.

True — from

I = n e A v_{d}

with

I

n

e

fixed,

v_{d} \propto 1/ A

, so twice the area means half the drift speed.

Raising the temperature of a metal increases its drift velocity for a fixed applied field.

False — higher

T

means ions vibrate more, so collisions come sooner,

τ

falls, and since

v_{d} = e E τ / m

, drift drops (resistance rises).

Mobility

μ

is a property of the material, not of how strong the field is.

True —

μ = e τ / m

depends only on charge, mass and

τ

(a material property at fixed

T

); it does not contain

E

, so it stays fixed as you change the field.

Conductivity

σ = n e^{2} τ / m

increases if you use a metal with more free electrons per cubic metre.

True —

σ \propto n

, so more carriers per unit volume means more conductivity, all else equal (this is one reason silver/copper conduct so well).

In the relation

σ = n e μ

the charge

e

appears only once, so a sign error would flip conductivity negative.

False —

σ

is intrinsically positive; the

e

here is the magnitude of charge, and the two sign-flips (negative charge, drift opposite to field) cancel to give a positive

σ

Spot the error

Each line contains a mistake in reasoning or algebra. Name it.

"Since

v_{d} = e E τ / m

and drift is slow, the field

E

must be tiny."

The error: drift is slow mainly because

τ

is only femtoseconds (and

e

is tiny); the mass

m

sits in the denominator, so a small

m

actually raises

v_{d}

. So smallness of

v_{d}

comes from

τ

and

e

, not

m

— and a modest

E

still gives a crawl, so

v_{d}

being small does not imply

E

is small.

I = n e A v_{d}

, so a superconductor with zero resistance must have infinite

v_{d}

The error: zero resistance does not force infinite drift; a finite current with finite

n, A

needs only a finite

v_{d}

. The formula sets

v_{d} = I / (n e A)

, a finite number.

"Because

μ = v_{d} / E

, mobility grows without bound as you crank up the field."

The error:

v_{d}

grows proportionally with

E

(in the linear regime), so their ratio

μ

stays constant — it does not grow with field.

"Resistivity

ρ = m / (n e^{2} τ)

, so cooling a metal to raise

τ

makes

ρ

larger."

The error:

ρ \propto 1/ τ

, so a larger

τ

makes

ρ

smaller; cooling reduces vibration, lengthens

τ

, and lowers resistivity.

"Drift velocity

\propto T

just like the thermal (rms) speed."

The error: thermal speed scales as

T

, but drift scales as

τ

, which decreases with

T

; the two temperature dependences go opposite ways.

"Current density

J = n e v_{d}

has units of velocity times charge, so it's measured in m/s."

The error:

J = n e v_{d}

has units

(m^{- 3}) (C) (m s^{- 1}) = C s^{- 1} m^{- 2} = A m^{- 2}

, a current per area, not a speed.

"Since electrons collide constantly, they keep speeding up forever between collisions until they're relativistic."

The error: each collision randomises direction and wipes the gained drift; the electron only accelerates for one short

τ

before resetting, so it never runs away to high speed.

The electric field spreads through the wire at nearly light speed and pushes every electron simultaneously, like water already filling a full pipe exiting the far end the moment you push the near end.

Why does resistance increase with temperature in a metal?

Heat makes the lattice ions vibrate harder, so an electron hits something sooner;

τ

shrinks,

σ = n e^{2} τ / m

falls and

ρ

rises. See Resistivity and temperature dependence.

Why is the macroscopic law

V = I R

really the microscopic law

J = σ E

in disguise?

J = σ E

says current density is proportional to field at every point; integrating over a uniform wire's length and area turns

E

into

V

and

J

into

I

, giving

V = I R

. See Ohm's Law.

Why do we bother defining mobility

μ

separately when

σ

already exists?

μ = v_{d} / E

isolates how responsive one carrier is to a field, independent of how many carriers there are;

σ = n e μ

then cleanly separates "how many" (

n

) from "how mobile" (

μ

) — essential in Semiconductors.

Why does the sign of the electron charge not make conductivity negative?

The electron's negative charge makes it drift backward against

E

, but that backward motion of negative charge is a forward conventional current; the two minus signs cancel, leaving

σ > 0

Why is the electric field inside a current-carrying wire nonzero, unlike in electrostatic equilibrium?

A steady current needs a continuous push to keep replacing the drift lost at each collision; that push is a nonzero

E

maintained by the battery. See Electric field inside conductors.

Why does raising the number density

n

raise current even if each electron's

v_{d}

is unchanged?

Current counts charge crossing per second; more electrons per cubic metre means more carriers crossing the same area, so

I = n e A v_{d}

grows even at fixed drift.

Edge cases

Push the ideas to their limits and degenerate inputs.

What happens to

v_{d}

at the instant the field is switched off (

E \to 0

The driving force vanishes, so no fresh drift is added; within a few collision times the ordered drift decays to zero and only random thermal motion remains — current dies out.

What does

v_{d} = e E τ / m

predict as

τ \to 0

(an extremely "dirty" or hot metal)?

v_{d} \to 0

and

σ \to 0

: if collisions are instantaneous, no drift can build between them, so the material barely conducts.

n = 0

(no free carriers, like a perfect insulator), what does

I = n e A v_{d}

give?

I = 0

for any field — with no free electrons to move there is no conduction current no matter how large

E

is.

For a fixed current, does a thinner wire (smaller

A

) make electrons drift faster or slower?

Faster —

v_{d} = I / (n e A)

so smaller

A

raises

v_{d}

; the same charge per second must squeeze through less area, so it must move quicker.

What is the drift velocity in the limit of zero applied field, and why is that consistent with the electrons still moving?

v_{d} = 0

because the ordered average is zero; the electrons still move at full thermal speed, but those random velocities cancel in the average. See Free electron model of metals.

In a semiconductor both electrons and holes drift — does the current from holes subtract from the electron current?

No, they add: holes (positive) drift along

E

and electrons drift against it, but both transport charge in the same conventional-current direction, so their contributions sum. See Semiconductors.

Is there any real field strength at which

v_{d} = e E τ / m

eventually breaks down?

Yes — at very high fields

τ

itself starts depending on the drift energy, so drift no longer grows linearly with

E

and the simple constant-

μ

picture fails (velocity saturation).

Connections

Ohm's Law — the macroscopic face of these traps.
Current density — where $J = n e v_{d}$ is built.
Resistivity and temperature dependence — the $τ (T)$ traps live here.
Free electron model of metals — source of random motion and $n$ .
Electric field inside conductors — why $E \neq = 0$ during current.
Semiconductors — two-carrier extensions of these questions.

Question bank — Drift velocity, mobility, conductivity

True or false — justify

Spot the error

Why questions

Edge cases

Connections