Worked examples — Drift velocity, mobility, conductivity

This page is the drill room for the parent topic. The parent built the tools; here we throw every kind of question at them. Before any numbers, we map out what "every kind" even means — so you never meet a case we skipped.

Every symbol we use ($v_d$, $n$, $\tau$, $\mu$, $\sigma$, $\rho$, $J$, $E$, $A$, $e$, $m$) was defined in the parent. If you are shaky on any of them, re-read the parent's sections 1–3 first.

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The scenario matrix

Think of every problem on this topic as filling one cell of a grid. The two things that change from problem to problem are: which quantity is unknown, and what special situation the numbers describe (normal, zero, extreme, real-world, sign-tricky). The table below lists every cell. Each worked example is tagged with the cell it hits.

Cell	What's special about it	Example
C1 — solve for $v_d$	plain forward use of $I=neA{v}_d$	Ex 1
C2 — solve for $\mu$ then $\tau$	chain two definitions	Ex 2
C3 — solve for $\sigma ,\rho$	tie micro to macro	Ex 3
C4 — geometry twist (area/shape change)	$A$ changes, $I$ fixed	Ex 4 (figure)
C5 — sign & direction	electron flows against $E$	Ex 5 (figure)
C6 — zero / degenerate input	$E=0$, or $\tau \to 0$, or $n\to 0$	Ex 6
C7 — limiting / temperature behaviour	$T\uparrow \Rightarrow \tau \downarrow$	Ex 7
C8 — real-world word problem	full-circuit reasoning	Ex 8
C9 — exam-style twist	two wires, ratio, no plug-in numbers	Ex 9

Constants used throughout (memorise these): $e=1.6\times 10^{-19}\text{C},m=9.1\times 10^{-31}\text{kg}.$ For copper: $n=8.5\times 10^{28}{\text{m}}^{-3}$.

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Ex 1 — Forward: find the drift speed · Cell C1

Step 1 — Convert the area to SI. $A=2{\text{mm}}^2=2\times (10^{-3}\text{m}{)}^2=2\times 10^{-6}{\text{m}}^2.$ Why this step? $I=neA{v}_d$ only works if every quantity is in metres, coulombs, seconds. A raw "$2$" in mm² would be off by $10^6$.

Step 2 — Rearrange the master relation for the unknown. $v_d=\frac{I}{neA}.$ Why this step? $v_d$ is what we want, so isolate it. Everything else on the right is given.

Step 3 — Plug in. $v_d=\frac{3}{(8.5\times 10^{28})(1.6\times 10^{-19})(2\times 10^{-6})}.$ The denominator is $8.5\times 1.6\times 2\times 10^{28-19-6}=27.2\times 10^3=2.72\times 10^4$. $v_d=\frac{3}{2.72\times 10^4}\approx 1.1\times 10^{-4}\text{m/s}.$ Why this step? Just arithmetic — but track the powers of ten carefully; that's where marks are lost.

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Ex 2 — Sideways: mobility, then relaxation time · Cell C2

Step 1 — Use the definition of mobility. $\mu =\frac{v_d}{E}=\frac{1.1\times 10^{-4}}{8\times 10^{-3}}.$ Why this step? Mobility is literally drift speed per field (parent's definition). No new physics — just divide. $\mu =\frac{1.1}{8}\times 10^{-4+3}=0.1379\times 10^{-1}\approx 1.38\times 10^{-2}{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}.$

Step 2 — Get $\tau$ from $\mu =\frac{e\tau }{m}$. $\tau =\frac{\mu m}{e}=\frac{(1.38\times 10^{-2})(9.1\times 10^{-31})}{1.6\times 10^{-19}}.$ Why this step? The parent showed $\mu =e\tau /m$; rearranging is the only way to reach $\tau$ from measurable quantities ($\tau$ can't be measured directly — it's inferred).

Step 3 — Crunch. Numerator $=1.38\times 9.1\times 10^{-2-31}=12.6\times 10^{-33}=1.26\times 10^{-32}$. $\tau =\frac{1.26\times 10^{-32}}{1.6\times 10^{-19}}\approx 7.9\times 10^{-14}\text{s}.$

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Ex 3 — Micro → macro: conductivity and resistivity · Cell C3

Step 1 — Use $\sigma =ne\mu$. $\sigma =(8.5\times 10^{28})(1.6\times 10^{-19})(1.38\times 10^{-2}).$ Why this step? This is the clean link the parent derived: $\sigma =ne\mu$ (no need to know $\tau$ separately). $\sigma =8.5\times 1.6\times 1.38\times 10^{28-19-2}=18.8\times 10^7\approx 1.9\times 10^8{\Omega }^{-1}{\text{m}}^{-1}.$

Step 2 — Invert for resistivity. $\rho =\frac{1}{\sigma }=\frac{1}{1.9\times 10^8}\approx 5.3\times 10^{-9}\Omega \text{m}.$ Why this step? Resistivity is defined as $1/\sigma$; they carry the same information, just flipped.

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Ex 4 — Geometry twist: a wire that gets thinner · Cell C4

Look at the figure: the same number of electrons per second must pass every cross-section (charge can't pile up). Where the pipe is narrow (teal section), each electron must move faster to keep the count up — like water in a garden hose when you pinch the end.

Step 1 — Write $I=neA{v}_d$ for each section. $I=ne{A}_1{v}_{d1}=ne{A}_2{v}_{d2}.$ Why this step? $I$, $n$, $e$ are identical in both sections (same wire, same current, same material). Only $A$ and $v_d$ differ.

Step 2 — Cancel the common factors and take the ratio. $A_1{v}_{d1}=A_2{v}_{d2}\Rightarrow \frac{v_{d2}}{v_{d1}}=\frac{A_1}{A_2}=\frac{4}{1}=4.$ Why this step? Ratios let all the messy constants ($n,e$) cancel — the fastest route when the question asks "compare."

Step 3 — Get an actual value in the thin part. $v_{d2}=\frac{I}{ne{A}_2}=\frac{2}{(8.5\times 10^{28})(1.6\times 10^{-19})(10^{-6})}\approx 1.47\times 10^{-4}\text{m/s},$ and $v_{d1}=v_{d2}/4\approx 3.68\times 10^{-5}\text{m/s}$.

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Ex 5 — Sign and direction: which way does an electron go? · Cell C5

Study the figure — three arrows, three colours.

Step 1 — Force direction. The electron charge is $-e$. Force is $\vec{F}=(-e)\vec{E}.$ Since $\vec{E}$ points $+x$ and the charge is negative, $\vec{F}$ points $-x$ (left, plum arrow). Why this step? Multiplying a rightward vector by a negative number flips it. Sign of charge is the direction rule.

Step 2 — Drift direction. ${\vec{v}}_d=\vec{a}\tau =\frac{-e\vec{E}}{m}\tau$ points the same way as $\vec{F}$: left ($-x$, orange arrow). Why this step? Drift is just accumulated acceleration; it inherits the force's direction.

Step 3 — Conventional current direction. Conventional current flows in the direction positive charge would move — i.e. along $\vec{E}$: right ($+x$, teal arrow). Why this step? By definition current follows positive-charge motion, which is opposite to the electrons.

Step 4 — Magnitudes are sign-blind. Speed and current use magnitudes: $v_d=\frac{eE\tau }{m},I=neA{v}_d,$ both positive. The minus sign only ever tells you direction, never size.

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Ex 6 — Degenerate inputs: what breaks and what doesn't · Cell C6

Case (a): $E=0$. $v_d=\frac{e(0)\tau }{m}=0\Rightarrow I=neA\cdot 0=0.$ Why this step? No field = no sideways nudge. Electrons still zoom thermally, but the average is zero (parent's swarm-of-bees picture). No net current — exactly the "no battery" baseline.

Case (b): $\tau \to \infty$. $v_d=\frac{eE\tau }{m}\to \infty .$ Why this step? If electrons never collide, each keeps accelerating forever — drift grows without bound, $\sigma =\frac{n{e}^2\tau }{m}\to \infty$, resistance $\to 0$. This is the idealised superconductor limit: collisions are the only thing that gives resistance, so removing them removes resistance.

Case (c): $n\to 0$. $I=neA{v}_d\to 0.$ Why this step? No free carriers means nothing to move, no matter how hard you push. That's an insulator. (Real Semiconductors sit between metals and insulators precisely because $n$ is small but not zero — and rises with temperature.)

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Ex 7 — Limiting behaviour: heat the wire · Cell C7

Step 1 — Drift velocity. $v_d=\frac{eE\tau }{m}\propto \tau$, so $v_d\to \frac{v_d}{3}.$ Why this step? Only $\tau$ changed and $v_d$ is directly proportional to it. Hotter ions vibrate more → collisions come sooner → less time to build up drift.

Step 2 — Conductivity. $\sigma =\frac{n{e}^2\tau }{m}\propto \tau$, so $\sigma \to \frac{\sigma }{3},\rho =\frac{1}{\sigma }\to 3\rho .$ Why this step? Same direct-proportionality on $\tau$. Resistivity triples — this is exactly the parent's "$\rho$ rises with $T$."

Step 3 — Current. At fixed voltage $V$, $I=\frac{V}{R}$ and $R\propto \rho$, so tripling $\rho$ gives $I\to \frac{I}{3}.$ Why this step? Higher resistivity chokes the current for the same push.

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Ex 8 — Real-world word problem: the light-bulb delay · Cell C8

Step 1 — Travel time of one electron. $t=\frac{\text{distance}}{\text{speed}}=\frac{30}{1\times 10^{-4}}=3\times 10^5\text{s}.$ Why this step? Plain time = distance/speed. $3\times 10^5\text{s}\approx 83$ hours $\approx 3.5$ days.

Step 2 — Compare with the signal time. The electric field spreads at nearly $c=3\times 10^8\text{m/s}$: $t_{\text{signal}}=\frac{30}{3\times 10^8}=1\times 10^{-7}\text{s}=0.1\mu \text{s}.$ Why this step? The field — not the electron — is what tells all the electrons along the wire to start drifting. It travels near light speed.

Step 3 — Resolve the paradox. The wire is already full of electrons everywhere (parent's water-pipe picture). The switch nudges the field, which reaches every electron in $\sim 0.1\mu \text{s}$; all of them start drifting at once, so an electron already sitting at the lamp moves immediately. Nobody had to make the $3.5$-day trip.

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Ex 9 — Exam twist: two wires, pure ratio · Cell C9

Step 1 — Express drift from the master relation. $v_d=\frac{I}{neA}\Rightarrow v_d\propto \frac{1}{A}(\text{since }I,n,e\text{ are equal for both}).$ Why this step? Same metal ($n$ equal), same current ($I$ equal), same charge $e$. The only difference is area, so everything else cancels — the hallmark of a ratio problem.

Step 2 — Relate area to radius. For a circular cross-section $A=\pi r^2$, so $A\propto r^2$. With $r_B=2r_A$: $\frac{A_B}{A_A}={\left(\frac{r_B}{r_A}\right)}^2=2^2=4.$ Why this step? Doubling a radius quadruples the area — squaring is where the factor 4 (not 2) comes from. A classic trap.

Step 3 — Combine. $\frac{v_{d,A}}{v_{d,B}}=\frac{A_B}{A_A}=4.$ Why this step? Since $v_d\propto 1/A$, the wire with the smaller area (A) has the larger drift. A is 4× faster.

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Recall drill

Recall Cover the answers — can you name the cell each hits?

• Q: Same current, wire narrows to ¼ area. Drift speed changes how? A: Rises ×4 (Ex 4, C4) — $A{v}_d$ conserved.
• Q: Field removed. Current? A: Zero — $v_d=0$ (Ex 6, C6).
• Q: Heating cuts $\tau$ to a third. Resistivity? A: Triples (Ex 7, C7) — $\rho \propto 1/\tau$.
• Q: Radius doubles, same $I$, same metal. Drift ratio old/new? A: 4 (Ex 9, C9) — $A\propto r^2$, $v_d\propto 1/A$.
• Q: Which is faster, the field or the electron, and by how much (Ex 8)? A: The field, by ~$10^{12}$ times.

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Connections

• ← Back to parent topic
• Current density — Ex 4 and Ex 9 are really statements about $J=ne{v}_d$ being fixed across a wire.
• Ohm's Law — Ex 8's $I=V/R$ closes the loop with the micro picture.
• Resistivity and temperature dependence — Ex 7 in full.
• Semiconductors — the $n\to 0$ limit of Ex 6 lives here.
• Free electron model of metals — where $n$ and random motion come from.
• Electric field inside conductors — the field of Ex 5 and Ex 8.