Hall-effect thruster — cross-field discharge, annular channel: Worked examples

This page is the problem gym for the Hall thruster. The parent note built the physics; here we drill every kind of number the topic can hand you — big fields, tiny fields, zero flow, the edge where an ion is just barely magnetized, a real mission problem, and an exam-style twist.

Before any algebra, one honest promise: we restate every symbol and formula the moment we use it, so you never have to flip back to the parent note.

The scenario matrix

Think of a Hall thruster problem as living in a grid. Each row is a quantity you might be asked for; each column is a regime (normal, an extreme, or a trap). The examples that follow fill in every cell.

Case class	What makes it tricky	Covered by
Magnetized? (electron)	tiny gyroradius, "normal" case	Ex 1
Magnetized? (ion, borderline)	ion radius $\approx$ channel — the degenerate edge	Ex 2
Limit: $B \to 0$	field off — trapping fails, drift diverges	Ex 3 (steps 1–2)
Limit: $B \to \infty$	huge field — radius and drift both collapse	Ex 3 (steps 3–4)
Exhaust speed / thrust	the standard delivered-performance sum	Ex 4
Zero / degenerate input	$\overset{m}{˙} = 0$ or $Δ V = 0$ — thrust vanishes	Ex 5
Hall drift direction (sign/quadrant)	which way does the loop spin? geometry	Ex 6
Real-world word problem	a station-keeping burn, uses Tsiolkovsky Rocket Equation	Ex 7
Exam twist	mix two ideas, hidden unit trap	Ex 8

The symbols and the three tools we will reuse

A few symbols to pin down first, because we lean on them repeatedly. Read them slowly — every one is earned here before it appears in a formula.

Definition What each symbol means

$q$ — the charge of one particle (electron or singly-charged ion), $q = 1.6 \times 1 0^{- 19} C$ (coulombs, the unit of electric charge).
$m$ — the mass of that particle ( $m_{e}$ for an electron, $m_{i}$ for an ion).
$v_{⊥}$ — the particle's speed across (perpendicular to) $B$ — the part of its motion that gets bent into a circle.
$B$ — the magnetic field strength, in teslas (T); it sets how hard the field bends a moving charge.
$E$ — the electric field strength, in volts per metre ( $V/m$ ). Picture it as the steepness of the voltage hill: if the accelerating voltage $Δ V$ drops over a channel length $L$ , then $E = Δ V / L$ . A large $E$ means the voltage falls off sharply over a short distance, so a charge placed there feels a strong push. This is the field that accelerates the ions and, crossed with $B$ , drives the electron drift.
$Δ V$ — the voltage drop (potential difference) from anode to channel exit, in volts (V).
$L$ — a length: the channel gap (for magnetization) or the channel length (for computing $E$ ).
$\overset{m}{˙}$ — the propellant mass flow rate, kilograms leaving per second ( $kg/s$ ).
$g_{0} = 9.81 m/s^{2}$ — standard gravity, used only to convert speed into a "seconds" fuel-economy figure.

Two derived symbols we lean on constantly:

$v_{i}$ — the ion speed after it falls through the accelerating voltage $Δ V$ .
$v_{e}$ — the exhaust speed the rocket "feels". Since the thrust is carried almost entirely by those ions, we set $v_{e} \approx v_{i}$ ; every time you see $v_{e}$ below, read it as "the ion exit speed $v_{i}$ ".

Constants used everywhere: $q = 1.6 \times 1 0^{- 19} C$ , $m_{e} = 9.11 \times 1 0^{- 31} kg$ , xenon ion mass $m_{i} = 2.18 \times 1 0^{- 25} kg$ , $g_{0} = 9.81 m/s^{2}$ .

Worked examples

Worked example Ex 1 — Electron gyroradius (normal magnetized case)

A $20 eV$ electron ( $v_{⊥} = 2 \times 1 0^{6} m/s$ ) sits in $B = 0.02 T$ . Channel gap $L = 1.5 cm$ . Is it magnetized?

Forecast: guess — will $r_{L}$ be micrometres, millimetres, or centimetres?

Write the tool: $r_{L} = \frac{m _{e} v _{⊥}}{q B}$ . Why this step? Magnetization is decided by comparing $r_{L}$ to the channel size $L$ ; "magnetized" means $r_{L} ≪ L$ (see the small red circle against the wide gap in the figure).
Plug in: $r_{L} = \frac{( 9.11 \times 1 0 ^{- 31} ) ( 2 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.02 )} \approx 5.7 \times 1 0^{- 4} m = 0.57 mm$ . Why this step? Numbers only after the formula, so units are traceable.
Compare: $0.57 mm ≪ 15 mm$ , ratio $\approx 26$ . Why this step? A factor $\sim 26$ smaller means the electron spins ~26 tight loops across the gap — firmly trapped.

Verify: units $\frac{kg \cdot m/s}{C \cdot T} = \frac{kg \cdot m/s}{C \cdot kg/(C\cdots)} = m$ ✔. Answer $0.57 mm < L$ ⇒ magnetized. Cell: electron magnetized. See Larmor Radius and Cyclotron Motion.

Worked example Ex 2 — The borderline ion (degenerate edge)

A slow just-born xenon ion, $v_{⊥} = 300 m/s$ , same $B = 0.02 T$ . Channel $L = 1.5 cm$ . Magnetized or not?

Forecast: ions are $1 0^{5} \times$ heavier — will $r_{L}$ land inside or outside the channel?

Same tool, ion values: $r_{L} = \frac{m _{i} v _{⊥}}{q B}$ . Why this step? $r_{L} \propto m$ , so at equal speed the ion radius is bigger by $m_{i} / m_{e}$ .
Plug in: $r_{L} = \frac{( 2.18 \times 1 0 ^{- 25} ) ( 300 )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.02 )} \approx 2.0 \times 1 0^{- 2} m = 2.0 cm$ . Why this step? This is the slowest ion (worst case for being trapped); faster ions have even larger $r_{L}$ .
Compare: $2.0 cm > 1.5 cm$ . The ion's would-be circle is wider than the channel — the wall interrupts it before it can complete an orbit. Why this step? This is the degenerate edge: $r_{L} ≳ L$ means "cannot circle" ⇒ effectively unmagnetized. Even the slowest ion clears the bar.

Verify: ratio $r_{L, i} / r_{L, e}$ at these speeds is $\frac{m _{i} \cdot 300}{m _{e} \cdot 2 \times 1 0 ^{6}}$ ; plug the two radii $0.02/5.7 \times 1 0^{- 4} \approx 35$ , and indeed the ion radius sits just above $L$ ✔. Cell: ion borderline, unmagnetized.

Worked example Ex 3 — The two limits:

$B \to 0$ and $B \to \infty$ Keep the $20 eV$ electron and $E = 1.2 \times 1 0^{4} V/m$ . What happens to $r_{L, e}$ and $v_{d}$ as $B$ shrinks to zero, then grows huge? Treat the two limits as separate physical worlds — steps 1–2 are the $B \to 0$ world, steps 3–4 the $B \to \infty$ world.

Forecast: which quantity blows up and which collapses in each limit?

World A — $B \to 0$ (field switched off):

From $r_{L} = m v_{⊥} / (q B)$ : as $B \to 0$ , $r_{L} \to \infty$ . Why this step? With no field there is no bending force, so the "circle" straightens into a line (left panel of the figure) — the electron is not magnetized. The trapping trick fails; there is no virtual grid.
From $v_{d} = E / B$ : as $B \to 0$ , $v_{d} = E / B \to \infty$ . Why this step? The drift formula literally diverges — physically this signals the drift picture itself breaks down: unmagnetized particles just free-accelerate along $E$ , they do not drift. A limit that "explodes" is your cue the model left its domain.

World B — $B \to \infty$ (over-strong field): 3. From $r_{L} = m v_{⊥} / (q B)$ : as $B \to \infty$ , $r_{L} \to 0$ . Why this step? The electron is pinned onto a single field line (right panel) — a completely different failure from World A. Here it is over-trapped, not under-trapped. 4. From $v_{d} = E / B$ : as $B \to \infty$ , $v_{d} \to 0$ . Why this step? The Hall loop crawls and cross-field leakage to the anode chokes off — the discharge starves. So the two extremes fail for opposite reasons, and real designs pick a middle $B$ .

Sanity number (finite $B$ ): at $B = 0.04 T$ (double the nominal $0.02$ ), $v_{d} = 1.2 \times 1 0^{4} /0.04 = 3 \times 1 0^{5} m/s$ , exactly half the $6 \times 1 0^{5}$ at $0.02 T$ — confirming $v_{d} \propto 1/ B$ between the extremes.

Verify: $v_{d} (0.02) = 6 \times 1 0^{5}$ , $v_{d} (0.04) = 3 \times 1 0^{5}$ , ratio exactly $2$ ✔. Cells: $B \to 0$ (World A) and $B \to \infty$ (World B), kept distinct.

Worked example Ex 4 — Exhaust speed, thrust,

$I_{s p}$ (delivered performance) Xenon, $Δ V = 300 V$ , $\overset{m}{˙} = 5 mg/s = 5 \times 1 0^{- 6} kg/s$ . Find $v_{e}$ , $F$ , $I_{s p}$ .

Forecast: will $v_{e}$ be near $1 0^{3}$ , $1 0^{4}$ , or $1 0^{5} m/s$ ?

Exhaust speed: $v_{e} \approx v_{i} = \frac{2 q Δ V}{m _{i}} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 300 )}{2.18 \times 1 0 ^{- 25}}$ . Why this step? The ion falls through the potential $Δ V$ ; energy conservation $q Δ V = \frac{1}{2} m_{i} v_{e}^{2}$ gives the speed. The quasineutral plasma lets ions see the full $Δ V$ (no space-charge cap, unlike Child–Langmuir).
Compute inside: $4.40 \times 1 0^{8} \approx 2.10 \times 1 0^{4} m/s$ . Why this step? We evaluate the fraction ( $2 \cdot 1.6 \times 1 0^{- 19} \cdot 300/2.18 \times 1 0^{- 25} = 4.40 \times 1 0^{8}$ ) first and then take its square root, because $v_{e}$ lives under the root — you cannot read off the speed until the root is taken.
Thrust: $F = \overset{m}{˙} v_{e} = (5 \times 1 0^{- 6}) (2.10 \times 1 0^{4}) \approx 0.105 N \approx 105 mN$ . Why this step? Thrust is momentum flux: mass leaving per second times its speed.
$I_{s p} = v_{e} / g_{0} = 2.10 \times 1 0^{4} /9.81 \approx 2.14 \times 1 0^{3} s$ . Why this step? $I_{s p}$ measures how many seconds $1 N$ -worth of propellant lasts — the fuel-economy figure.

Verify: $F / \overset{m}{˙} = 0.105/5 \times 1 0^{- 6} = 2.1 \times 1 0^{4} = v_{e}$ ✔. $I_{s p} \cdot g_{0} = 2140 \times 9.81 \approx 2.1 \times 1 0^{4} = v_{e}$ ✔. Cell: performance.

Worked example Ex 5 — The zero inputs (degenerate: no push)

Same thruster. (a) $\overset{m}{˙} = 0$ with $Δ V = 300 V$ . (b) $\overset{m}{˙} = 5 \times 1 0^{- 6} kg/s$ but $Δ V = 0$ .

Forecast: thrust in each case?

Case (a): $F = \overset{m}{˙} v_{e} = 0 \cdot v_{e} = 0 N$ . Why this step? No propellant leaving means no momentum flux, no matter how big the voltage. A hot idle field pushes on nothing.
Case (b): $v_{e} = 2 q (0) / m_{i} = 0$ , so $F = \overset{m}{˙} \cdot 0 = 0 N$ . Why this step? Zero accelerating voltage means ions exit at zero speed — mass drools out with no thrust. Both degenerate inputs kill $F$ , but for different physical reasons (no mass vs. no speed).
The product form $F = \overset{m}{˙} 2 q Δ V / m_{i}$ makes it obvious: $F = 0$ whenever either factor is zero. Why this step? Recognising the multiplicative structure lets you predict edge cases without recomputing.

Verify: both cases give exactly $0$ ✔. Cell: zero/degenerate.

Worked example Ex 6 — Which way does the Hall loop spin? (sign/geometry)

$E$ points axially outward (toward the exit, $+ \overset{z}{^}$ ). $B$ points radially outward ( $+ \overset{r}{^}$ ). Which way is the azimuthal Hall drift $v_{d} = (E \times B) / B^{2}$ ?

Forecast: clockwise or anticlockwise looking from downstream?

Use the drift formula $v_{d} = \frac{E \times B}{B ^{2}}$ ; direction is set by the cross product $\overset{z}{^} \times \overset{r}{^}$ . Why this step? The magnitude $E / B$ we know; the sign lives entirely in the cross-product direction.
In cylindrical unit vectors, $\overset{z}{^} \times \overset{r}{^} = \hat{θ}$ (the azimuthal direction). So $v_{d}$ points along $+ \hat{θ}$ . Why this step? $(\overset{r}{^}, \hat{θ}, \overset{z}{^})$ is a right-handed set, exactly like $(\overset{x}{^}, \overset{y}{^}, \overset{z}{^})$ , so their cross products follow the same right-hand rule (see the red arrow in the figure).
Result: the electron drift circulates azimuthally around the ring, one consistent sense — a closed loop with no ends. Flip either $E$ or $B$ and the loop reverses; flip both and it stays. Why this step? This sign-tracking is the whole reason the parent note insists on the annular channel: only a ring lets $+ \hat{θ}$ close on itself. A straight tube would slam this current into a wall.

Verify: magnitude $v_{d} = E / B = 1.2 \times 1 0^{4} /0.02 = 6 \times 1 0^{5} m/s$ ✔; and $(\overset{z}{^} \times \overset{r}{^}) \cdot \hat{θ} = 1 > 0$ confirms the $+ \hat{θ}$ sense ✔. Cell: sign/quadrant geometry. See Magnetic Mirror & Guiding-Centre Drifts for the guiding-centre view and Lorentz Force for the origin.

Worked example Ex 7 — Real-world word problem (station-keeping)

A satellite of dry mass $500 kg$ carries $12 kg$ of xenon. Its Hall thruster runs the numbers from Ex 4 ( $v_{e} = 2.10 \times 1 0^{4} m/s$ , $F = 0.105 N$ ). (a) Total $Δ v$ available? (b) How long must the thruster fire to burn all the xenon?

Forecast: more or less than $500 m/s$ of $Δ v$ ? Hours or days of firing?

$Δ v$ from the Tsiolkovsky Rocket Equation: $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ , with $m_{0} = 512 kg$ , $m_{f} = 500 kg$ . Why this step? The rocket equation converts exhaust speed and mass ratio into the velocity change you can buy — the currency of orbital maneuvers.
$Δ v = 2.10 \times 1 0^{4} \cdot ln (512/500) = 2.10 \times 1 0^{4} \cdot ln (1.024) \approx 2.10 \times 1 0^{4} \cdot 0.02372 \approx 498 m/s$ . Why this step? A small mass ratio $\Rightarrow$ small $ln$ , but the huge $v_{e}$ still yields a useful $\sim 0.5 km/s$ — the point of high- $I_{s p}$ electric propulsion.
Burn time from $\overset{m}{˙}$ : $t = \frac{m _{prop}}{m ˙} = \frac{12}{5 \times 1 0 ^{- 6}} s = 2.4 \times 1 0^{6} s$ . Why this step? Constant flow rate means time = fuel ÷ rate. This is where electric thrusters bite: gentle force, long burns.
Convert: $2.4 \times 1 0^{6} s \div 86400 \approx 27.8 days$ . Why this step? Human units to feel the answer — nearly a month of continuous thrust for half a km/s.

Verify: $Δ v \approx 498 m/s$ ✔; $t \approx 2.4 \times 1 0^{6} s \approx 27.8 d$ ✔. Cross-check impulse: $F \cdot t = 0.105 \times 2.4 \times 1 0^{6} = 2.52 \times 1 0^{5} N\cdots$ , and $\overset{m}{˙} v_{e} \cdot t = 0.105 \times 2.4 \times 1 0^{6}$ matches ✔. Cell: real-world.

Worked example Ex 8 — Exam twist (a hidden unit trap + two ideas)

"Same thruster runs at $Δ V = 300 V$ . A student claims the Hall drift makes electrons move axially at $v_{d} = E \cdot B = 1.2 \times 1 0^{4} \times 0.02 = 240 m/s$ , so that's the exhaust speed." Find three errors and give the right exhaust speed.

Forecast: spot the wrong formula, the wrong direction, and the wrong species — before reading on.

Error 1 — formula: force balance gives $E = v_{d} B$ , so $v_{d} = E / B$ , not $E \cdot B$ . Why this step? Units decide it: $(V/m) / T = m/s$ ✔, whereas $(V/m) \cdot T$ gives $V\cdotT/m$ — not a speed. Correct $v_{d} = 1.2 \times 1 0^{4} /0.02 = 6 \times 1 0^{5} m/s$ .
Error 2 — direction: $v_{d}$ is azimuthal (around the ring), not axial. It is the closed Hall current, not the exhaust. Why this step? Direction comes from $E \times B$ (Ex 6) — it can never point along $E$ itself.
Error 3 — species: the exhaust is ions, accelerated by falling through $Δ V$ , not the drifting electrons. Why this step? Electrons are trapped (Ex 1); the thrust is ion momentum. So $v_{e} = v_{i} = 2 q Δ V / m_{i}$ .
Right exhaust speed: $v_{e} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 300 )}{2.18 \times 1 0 ^{- 25}} \approx 2.10 \times 1 0^{4} m/s$ . Why this step? Same computation as Ex 4 — the exam twist is just three distractors wrapped around the right idea.

Verify: correct $v_{d} = 6 \times 1 0^{5} m/s \neq = 240 m/s$ ✔; correct $v_{e} \approx 2.10 \times 1 0^{4} m/s$ ✔; and $v_{d} ≫ v_{e}$ (electrons whip around far faster than ions leave) ✔. Cell: exam twist.

Recall

Recall Which formula answers each question?

"Is a species magnetized?" ::: $r_{L} = m v_{⊥} / (q B)$ , compare to channel $L$ . "How fast is the Hall loop?" ::: $v_{d} = E / B$ (azimuthal). "What's the exhaust speed?" ::: $v_{i} = 2 q Δ V / m_{i}$ . "What's the thrust?" ::: $F = \overset{m}{˙} v_{e}$ . "What $Δ v$ does the fuel buy?" ::: Tsiolkovsky $Δ v = v_{e} ln (m_{0} / m_{f})$ .

Related builds: Larmor Radius and Cyclotron Motion, Plasma Quasineutrality, Ion Thruster (Gridded).