Drift velocity, mobility, conductivity: Exercises

This page is a self-test ladder. Every problem builds on the parent note. Try each one with the solution collapsed, then open it. The core toolbox you need is exactly four boxed relations from the parent:

Standard constants used throughout (unless a problem overrides them):

$e = 1.6 \times 1 0^{- 19} C$
$m = 9.1 \times 1 0^{- 31} kg$
Copper: $n = 8.5 \times 1 0^{28} m^{- 3}$

Figure — Drift velocity, mobility, conductivity

The figure above is your mental picture for every counting problem: a cylinder of length $v_{d} Δ t$ empties through the cross-section in time $Δ t$ . Keep glancing back at it.

Level 1 — Recognition

L1.1 — Read off the right formula

A wire carries current $I$ through area $A$ . Free electron density is $n$ , each electron carries charge magnitude $e$ , drift speed is $v_{d}$ . Which expression gives $I$ , and what does each symbol physically count?

Recall Solution

WHAT: We want the number of charges crossing per second, times the charge each carries. WHY this formula: Look at figure s01. In time $Δ t$ every electron within a distance $v_{d} Δ t$ of the line crosses it. That is a cylinder of volume $A \cdot v_{d} Δ t$ . Multiply by $n$ (electrons per m³) to get the count, by $e$ to get charge, divide by $Δ t$ to get charge per second = current. $I = n e A v_{d}$

$n$ = carriers per m³, $e$ = charge each, $A$ = "door" area, $v_{d}$ = how fast the average electron creeps.

L1.2 — Units of mobility

State the SI unit of mobility $μ$ from its definition $μ = v_{d} / E$ .

Recall Solution

WHAT: Divide the units of $v_{d}$ by the units of $E$ . $v_{d}$ is in $m/s$ . Field $E$ is in $V/m$ . $[μ] = \frac{m/s}{V/m} = \frac{m}{s} \cdot \frac{m}{V} = m^{2} V^{- 1} s^{- 1}$

Level 2 — Application

L2.1 — Drift speed in copper

A copper wire of cross-section $A = 2 mm^{2}$ carries $I = 3 A$ . Take $n = 8.5 \times 1 0^{28} m^{- 3}$ . Find the drift speed $v_{d}$ .

Recall Solution

WHAT: Invert the master relation $I = n e A v_{d}$ to isolate $v_{d}$ . WHY: We know everything except $v_{d}$ , so solve for it. Convert area first: $2 mm^{2} = 2 \times 1 0^{- 6} m^{2}$ (since $1 mm = 1 0^{- 3} m$ , and area squares that). $v_{d} = \frac{I}{n e A} = \frac{3}{( 8.5 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 2 \times 1 0 ^{- 6} )}$ Denominator $= 8.5 \times 1.6 \times 2 \times 1 0^{28 - 19 - 6} = 27.2 \times 1 0^{3} = 2.72 \times 1 0^{4}$ . $v_{d} = \frac{3}{2.72 \times 1 0 ^{4}} \approx 1.10 \times 1 0^{- 4} m/s$ About $0.11 mm/s$ — snail confirmed.

L2.2 — Mobility from field and drift

The wire in L2.1 sustains a field $E = 0.05 V/m$ along it. Find the mobility $μ$ .

Recall Solution

WHAT: Mobility is literally drift speed per unit field. WHY divide: $μ = v_{d} / E$ by definition — no other tool needed. $μ = \frac{v _{d}}{E} = \frac{1.10 \times 1 0 ^{- 4}}{0.05} = 2.21 \times 1 0^{- 3} m^{2} V^{- 1} s^{- 1}$

L2.3 — Relaxation time from mobility

Using the mobility from L2.2, find the relaxation time $τ$ . ( $m = 9.1 \times 1 0^{- 31} kg$ .)

Recall Solution

WHAT: Rearrange $μ = e τ / m$ to get $τ$ . WHY: $μ$ already packages $e$ , $τ$ , $m$ ; only $τ$ is unknown. $τ = \frac{μ m}{e} = \frac{( 2.21 \times 1 0 ^{- 3} ) ( 9.1 \times 1 0 ^{- 31} )}{1.6 \times 1 0 ^{- 19}}$ Numerator $= 2.21 \times 9.1 \times 1 0^{- 34} = 20.1 \times 1 0^{- 34} = 2.01 \times 1 0^{- 33}$ . $τ = \frac{2.01 \times 1 0 ^{- 33}}{1.6 \times 1 0 ^{- 19}} = 1.26 \times 1 0^{- 14} s$ About $12$ femtoseconds between collisions.

Level 3 — Analysis

L3.1 — Conductivity from measured quantities

For the copper wire above ( $n = 8.5 \times 1 0^{28}$ , $μ = 2.21 \times 1 0^{- 3}$ ), compute the conductivity $σ$ and hence the resistivity $ρ$ .

Recall Solution

WHAT: Use $σ = n e μ$ , then $ρ = 1/ σ$ . WHY $σ = n e μ$ (not the $τ$ form): We already have $μ$ , so this is one multiplication instead of re-deriving $τ$ . $σ = n e μ = (8.5 \times 1 0^{28}) (1.6 \times 1 0^{- 19}) (2.21 \times 1 0^{- 3})$ $= 8.5 \times 1.6 \times 2.21 \times 1 0^{28 - 19 - 3} = 30.1 \times 1 0^{6} = 3.01 \times 1 0^{7} Ω^{- 1} m^{- 1}$ . $ρ = \frac{1}{σ} = \frac{1}{3.01 \times 1 0 ^{7}} = 3.32 \times 1 0^{- 8} Ω m$ Real copper is $ρ \approx 1.7 \times 1 0^{- 8} Ω m$ ; our illustrative $E$ makes us a factor ~2 off, but the method is exact.

L3.2 — Two wires, same current, different thickness

Two copper wires carry the same current $I$ . Wire B has twice the cross-section area of wire A. Compare their drift speeds.

Recall Solution

WHAT: From $I = n e A v_{d}$ , at fixed $I$ , $n$ , $e$ : $v_{d} \propto 1/ A$ . WHY: Same current means the same charge-per-second must squeeze through. A bigger door lets each electron move slower. $\frac{v _{d, B}}{v _{d, A}} = \frac{A _{A}}{A _{B}} = \frac{A _{A}}{2 A _{A}} = \frac{1}{2}$ Wire B's electrons drift half as fast.

L3.3 — Effect of heating

A wire is heated so its relaxation time $τ$ falls to $60%$ of its original value, while $n$ , $e$ , $m$ stay fixed and the applied field $E$ is unchanged. By what factor does the drift speed change, and by what factor does the conductivity change?

Recall Solution

WHAT: Both $v_{d} = e E τ / m$ and $σ = n e^{2} τ / m$ are directly proportional to $τ$ . WHY: $E$ , $n$ , $e$ , $m$ are held constant, so only $τ$ moves the answer. $\frac{v _{d, new}}{v _{d, old}} = \frac{τ _{new}}{τ _{old}} = 0.60, \frac{σ _{new}}{σ _{old}} = 0.60$ Drift and conductivity both drop to 60%; resistivity rises to $1/0.60 \approx 1.67 \times$ . This is exactly why hot metals resist more — see Resistivity and temperature dependence.

Level 4 — Synthesis

L4.1 — Chain from current all the way to $τ$

A silver wire has $n = 5.86 \times 1 0^{28} m^{- 3}$ , area $A = 1 mm^{2}$ , and carries $I = 2 A$ under a field $E = 0.04 V/m$ . Find, in order: (a) $v_{d}$ , (b) $μ$ , (c) $σ$ , (d) $τ$ .

Recall Solution

WHAT & WHY: Each part feeds the next — this is the full logical chain of the topic.

(a) $v_{d} = \frac{I}{n e A}$ . Area $= 1 mm^{2} = 1 0^{- 6} m^{2}$ . Denominator $= (5.86 \times 1 0^{28}) (1.6 \times 1 0^{- 19}) (1 0^{- 6}) = 9.376 \times 1 0^{3}$ . $v_{d} = \frac{2}{9.376 \times 1 0 ^{3}} = 2.13 \times 1 0^{- 4} m/s$

(b) $μ = v_{d} / E = \frac{2.13 \times 1 0 ^{- 4}}{0.04} = 5.33 \times 1 0^{- 3} m^{2} V^{- 1} s^{- 1}$ .

(c) $σ = n e μ = (5.86 \times 1 0^{28}) (1.6 \times 1 0^{- 19}) (5.33 \times 1 0^{- 3}) = 5.00 \times 1 0^{7} Ω^{- 1} m^{- 1}$ .

(d) $τ = \frac{μ m}{e} = \frac{( 5.33 \times 1 0 ^{- 3} ) ( 9.1 \times 1 0 ^{- 31} )}{1.6 \times 1 0 ^{- 19}} = 3.03 \times 1 0^{- 14} s$ .

L4.2 — Reconciling $J = σ E$ with $J = n e v_{d}$

Show that the two microscopic forms of current density agree, i.e. that substituting $v_{d} = e E τ / m$ into $J = n e v_{d}$ reproduces $J = σ E$ with $σ = n e^{2} τ / m$ . Then verify numerically for the silver wire of L4.1 that $J$ computed both ways matches.

Recall Solution

WHAT (symbolic): Start from $J = n e v_{d}$ and swap in the drift expression. $J = n e v_{d} = n e \cdot \frac{e E τ}{m} = \frac{n e ^{2} τ}{m} E = σ E$ so the two forms are identical by construction — see Ohm's Law and Current density.

WHAT (numeric check): For L4.1, $J = I / A = 2/1 0^{- 6} = 2 \times 1 0^{6} A/m^{2}$ . Via $σ E$ : $J = (5.00 \times 1 0^{7}) (0.04) = 2 \times 1 0^{6} A/m^{2}$ . ✓ They match exactly.

Level 5 — Mastery

L5.1 — Why the bulb lights instantly (a quantitative estimate)

An electron drifts at $v_{d} = 1 \times 1 0^{- 4} m/s$ down a wire of length $L = 3 m$ from switch to bulb. (a) How long would a single electron take to travel the whole wire? (b) The electric field establishes itself along the wire at roughly $c = 3 \times 1 0^{8} m/s$ ; how long does that take? (c) Explain the enormous ratio.

Recall Solution

(a) WHAT: time = distance / speed for the crawling electron. $t_{electron} = \frac{L}{v _{d}} = \frac{3}{1 \times 1 0 ^{- 4}} = 3 \times 1 0^{4} s \approx 8.3 hours$

(b) WHAT: time for the field signal at speed $c$ . $t_{field} = \frac{L}{c} = \frac{3}{3 \times 1 0 ^{8}} = 1 \times 1 0^{- 8} s = 10 ns$

(c) WHY the bulb still lights instantly: The ratio $t_{electron} / t_{field} = 3 \times 1 0^{4} /1 0^{- 8} = 3 \times 1 0^{12}$ — twelve orders of magnitude. The wire is already packed with electrons (like a full pipe of marbles). The field arrives in nanoseconds and nudges every electron at once, so an electron already sitting at the bulb starts drifting immediately. No single electron needs to travel from switch to bulb. See Free electron model of metals and Electric field inside conductors.

L5.2 — Design a wire for a target drift speed

You need a copper conductor ( $n = 8.5 \times 1 0^{28} m^{- 3}$ ) carrying $I = 10 A$ but with drift speed capped at $v_{d} = 5 \times 1 0^{- 5} m/s$ (to keep it "gentle"). What minimum cross-section area $A$ is required?

Recall Solution

WHAT: Solve $I = n e A v_{d}$ for $A$ . WHY: All of $I$ , $n$ , $e$ , $v_{d}$ are fixed; $A$ is the free design variable, and lowering $v_{d}$ forces a bigger door. $A = \frac{I}{n e v _{d}} = \frac{10}{( 8.5 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 5 \times 1 0 ^{- 5} )}$ Denominator $= 8.5 \times 1.6 \times 5 \times 1 0^{28 - 19 - 5} = 68 \times 1 0^{4} = 6.8 \times 1 0^{5}$ . $A = \frac{10}{6.8 \times 1 0 ^{5}} = 1.47 \times 1 0^{- 5} m^{2} = 14.7 mm^{2}$ A fairly thick wire — halving the allowed drift would double the required area (since $A \propto 1/ v_{d}$ ).

L5.3 — Semiconductor vs metal (conceptual synthesis)

A semiconductor sample and a copper wire carry the same current density $J$ under the same field $E$ . The semiconductor has far fewer carriers, $n_{semi} ≪ n_{Cu}$ . What must be true of the carriers' mobility (or drift speed) in the semiconductor, and why?

Recall Solution

WHAT: Use $J = n e v_{d} = n e μ E$ . WHY: Same $J$ and same $E$ means $n e μ$ must be equal for both materials (since $J = σ E = n e μ E$ ). With far fewer carriers, the product $n μ$ can only stay equal if $μ$ (hence $v_{d} = μ E$ ) is much larger in the semiconductor. $n_{semi} μ_{semi} = n_{Cu} μ_{Cu} \Rightarrow μ_{semi} = μ_{Cu} \cdot \frac{n _{Cu}}{n _{semi}} ≫ μ_{Cu}$ Indeed real semiconductors have mobilities 10–100× that of metals; their conduction is limited by the small carrier count, not by sluggish carriers. See Semiconductors.

Recall wrap-up

Recall One-line reflexes (cover the right side)

Isolate $v_{d}$ from current ::: $v_{d} = I / (n e A)$ Mobility from drift and field ::: $μ = v_{d} / E$ $τ$ from mobility ::: $τ = μ m / e$ Conductivity from $n, e, μ$ ::: $σ = n e μ$ Same $I$ , double area → drift ::: halves ( $v_{d} \propto 1/ A$ ) Heat wire, $τ$ falls → $σ$ ::: falls in the same proportion Why bulb lights instantly ::: field travels at $\sim c$ , nudging all electrons at once

Connections

Ohm's Law — L4.2 is $J = σ E$ made concrete.
Current density — every problem here rests on $J = n e v_{d}$ .
Resistivity and temperature dependence — L3.3 is the heart of $ρ (T)$ .
Free electron model of metals — source of $n$ and the random motion (L5.1).
Electric field inside conductors — the driver in L5.1.
Semiconductors — L5.3 shows why mobility matters there.