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

Level 1 — Recognition

L1.1 — Name the magnetized species

A single radial field $B = 0.015 T$ threads the channel. At the same perpendicular speed, whose gyro-circle is smaller — the electron's or the xenon ion's — and by roughly what factor?

Recall Solution

What we use: $r_{L} = m v_{⊥} / (q B)$ . At equal $v_{⊥}$ , $q$ , $B$ , the only difference is mass, so $\frac{r _{L, i}}{r _{L, e}} = \frac{m _{i}}{m _{e}} = \frac{2.18 \times 1 0 ^{- 25}}{9.11 \times 1 0 ^{- 31}} \approx 2.39 \times 1 0^{5} .$ Answer: the electron's circle is smaller — about $2.4 \times 1 0^{5}$ times smaller. That huge gap is exactly why one field magnetizes electrons while ions barely notice it.

L1.2 — Read off the drift direction

$E$ points axially (toward the exit) and $B$ points radially (across the ring gap). Which way does the $E \times B$ drift point, and why does the ring shape matter?

Recall Solution

What we use: $v_{d} = (E \times B) / B^{2}$ . Axial $\times$ radial gives an azimuthal direction — around the ring. Why the ring: an azimuthal drift on a ring closes into a loop with no ends, so charge never piles up and the Hall current runs steadily. In a straight tube the drift would slam into a wall.

Level 2 — Application

L2.1 — Larmor radius of a thermal electron

$T_{e} = 15 eV$ corresponds to a perpendicular speed $v_{⊥} = 1.6 \times 1 0^{6} m/s$ . With $B = 0.018 T$ , find $r_{L, e}$ and compare it to a channel gap of $1.5 cm$ .

Recall Solution

$r_{L, e} = \frac{m _{e} v _{⊥}}{q B} = \frac{( 9.11 \times 1 0 ^{- 31} ) ( 1.6 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.018 )} \approx 5.06 \times 1 0^{- 4} m = 0.51 mm .$ Compare: $0.51 mm ≪ 15 mm$ , roughly $30 \times$ smaller. Electrons are strongly magnetized — they orbit many times before crossing the channel. See Larmor Radius and Cyclotron Motion.

L2.2 — Cyclotron frequency

For that same electron and $B = 0.018 T$ , find $ω_{c}$ and the orbital period $T = 2 π / ω_{c}$ .

Recall Solution

$ω_{c} = \frac{q B}{m _{e}} = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 0.018 )}{9.11 \times 1 0 ^{- 31}} \approx 3.16 \times 1 0^{9} rad/s .$ $T = \frac{2 π}{ω _{c}} = \frac{2 π}{3.16 \times 1 0 ^{9}} \approx 1.99 \times 1 0^{- 9} s \approx 2 ns .$ Meaning: each loop takes ~2 nanoseconds. In one microsecond an electron circles ~500 times — it is very well trapped between collisions.

L2.3 — Exhaust speed of xenon

Ions fall through $Δ V = 250 V$ . Find the exhaust speed $v_{e}$ .

Recall Solution

Why energy conservation: the ion is charge $q$ crossing a potential drop $Δ V$ , gaining kinetic energy $q Δ V = \frac{1}{2} m_{i} v_{e}^{2}$ . $v_{e} = \frac{2 q Δ V}{m _{i}} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 250 )}{2.18 \times 1 0 ^{- 25}} = 3.67 \times 1 0^{8} \approx 1.92 \times 1 0^{4} m/s .$

Level 3 — Analysis

L3.1 — Hall drift versus axial crawl

With $Δ V = 250 V$ across a channel length $L = 0.025 m$ and $B = 0.018 T$ , find the azimuthal drift speed $v_{d} = E / B$ . Then compare it to the electron's axial progress if it hops one gyroradius ( $r_{L, e} = 0.51 mm$ from L2.1) per collision at a collision rate $ν = 1 \times 1 0^{6} s^{- 1}$ .

Recall Solution

Step 1 — the axial field. $E = Δ V / L = 250/0.025 = 1.0 \times 1 0^{4} V/m$ . Step 2 — the drift. $v_{d} = \frac{E}{B} = \frac{1.0 \times 1 0 ^{4}}{0.018} \approx 5.56 \times 1 0^{5} m/s (azimuthal) .$ Step 3 — axial crawl. Each collision nudges the guiding centre one $r_{L, e}$ across $B$ (toward the anode), so the axial speed is roughly $v_{axial} \approx r_{L, e} ν = (5.06 \times 1 0^{- 4}) (1 \times 1 0^{6}) \approx 506 m/s .$ Interpretation: the electron whips around the ring at $\sim 5.6 \times 1 0^{5}$ m/s but crawls toward the anode at only ~500 m/s — a ratio of ~1100. The Hall (azimuthal) current dwarfs the leakage current, exactly as intended. This slow leak is the anomalous cross-field transport mentioned in the parent note.

L3.2 — Is a fast (accelerated) ion still unmagnetized?

Take a from-rest xenon ion accelerated through $Δ V = 250 V$ (so $v_{i} = v_{e}$ from L2.3), moving mostly axially but with a small perpendicular component $v_{⊥} = 0.1 v_{i}$ . With $B = 0.018 T$ , find $r_{L, i}$ and compare to the $1.5 cm$ gap.

Recall Solution

$v_{⊥} = 0.1 \times 1.92 \times 1 0^{4} = 1.92 \times 1 0^{3} m/s$ . $r_{L, i} = \frac{m _{i} v _{⊥}}{q B} = \frac{( 2.18 \times 1 0 ^{- 25} ) ( 1.92 \times 1 0 ^{3} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.018 )} \approx 0.145 m = 14.5 cm .$ Compare: $14.5 cm ≫ 1.5 cm$ — about 10× the whole channel. The ion's "circle" is so large that inside the channel its path is essentially straight. Ions stay unmagnetized even after acceleration. ✔

Level 4 — Synthesis

L4.1 — Thrust, specific impulse, and efficiency budget

A xenon Hall thruster runs at $Δ V = 300 V$ , propellant flow $\overset{m}{˙} = 4 mg/s$ , discharge current $I_{d} = 4.5 A$ . (a) Exhaust speed $v_{e}$ . (b) Thrust $F = \overset{m}{˙} v_{e}$ . (c) Specific impulse $I_{s p}$ . (d) Input electrical power $P = I_{d} Δ V$ and the "jet" power $P_{jet} = \frac{1}{2} \overset{m}{˙} v_{e}^{2}$ ; report the efficiency $η = P_{jet} / P$ .

Recall Solution

(a) $v_{e} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 300 )}{2.18 \times 1 0 ^{- 25}} = 4.40 \times 1 0^{8} \approx 2.10 \times 1 0^{4} m/s .$ (b) $F = \overset{m}{˙} v_{e} = (4 \times 1 0^{- 6}) (2.10 \times 1 0^{4}) \approx 0.0840 N = 84.0 mN .$ (c) $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{2.10 \times 1 0 ^{4}}{9.81} \approx 2140 s .$ (d) $P = I_{d} Δ V = (4.5) (300) = 1350 W .$ $P_{jet} = \frac{1}{2} \overset{m}{˙} v_{e}^{2} = \frac{1}{2} (4 \times 1 0^{- 6}) (2.10 \times 1 0^{4})^{2} \approx 882 W .$ $η = \frac{P _{jet}}{P} = \frac{882}{1350} \approx 0.653 = 65%.$ Reading it: two-thirds of the electrical power ends up as ordered exhaust kinetic energy; the rest goes to ionization, wall losses, and electron leakage — a realistic Hall-thruster number.

L4.2 — Design the magnetic field

You must magnetize electrons ( $r_{L, e} < 1 mm$ ) but keep ions free ( $r_{L, i} > 5 cm$ ) for the same particles as L2.1 ( $v_{⊥, e} = 1.6 \times 1 0^{6}$ m/s) and L3.2 ( $v_{⊥, i} = 1.92 \times 1 0^{3}$ m/s). Find the range of $B$ that satisfies both conditions.

Recall Solution

Electron condition ( $r_{L, e} < 1 0^{- 3}$ m): $B > \frac{m _{e} v _{⊥, e}}{q ( 1 0 ^{- 3} )} = \frac{( 9.11 \times 1 0 ^{- 31} ) ( 1.6 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{- 3} )} \approx 9.11 \times 1 0^{- 3} T .$ Ion condition ( $r_{L, i} > 0.05$ m): $B < \frac{m _{i} v _{⊥, i}}{q ( 0.05 )} = \frac{( 2.18 \times 1 0 ^{- 25} ) ( 1.92 \times 1 0 ^{3} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.05 )} \approx 5.23 \times 1 0^{- 2} T .$ Answer: any $B$ with $9.1 mT < B < 52 mT$ works. This window is why real Hall thrusters run around $15$ – $25 mT$ — comfortably inside the band.

Level 5 — Mastery

L5.1 — Mission burn with the rocket equation

The 84-mN thruster from L4.1 fires continuously to give a spacecraft a velocity change $Δ v = 2.0 km/s$ . The spacecraft starts at wet mass $m_{0} = 500 kg$ . (a) Using Tsiolkovsky Rocket Equation with $v_{e} = 2.10 \times 1 0^{4} m/s$ , find the propellant mass burned $m_{p} = m_{0} (1 - e^{- Δ v / v_{e}})$ . (b) At $\overset{m}{˙} = 4 mg/s$ , how long (in days) must the thruster fire?

Recall Solution

Why Tsiolkovsky: thrust ejects mass, so the ship's own mass shrinks as it accelerates — the exponential captures that shrinking-mass bookkeeping. (a) Exponent: $Δ v / v_{e} = 2000/21000 = 0.09524$ . $m_{p} = m_{0} (1 - e^{- 0.09524}) = 500 (1 - 0.9091) = 500 (0.09091) \approx 45.4 kg .$ (b) Burn time $= m_{p} / \overset{m}{˙} = 45.4/ (4 \times 1 0^{- 6}) \approx 1.136 \times 1 0^{7} s$ . $t = \frac{1.136 \times 1 0 ^{7}}{86400 s/day} \approx 131 days .$ Reading it: a tiny 84-mN thrust gets a big $Δ v$ — but takes months. That is the electric-propulsion bargain: low thrust, superb fuel economy (only 45 kg of xenon for 2 km/s).

L5.2 — Compare against a gridded ion engine's space-charge limit

A gridded ion engine (Ion Thruster (Gridded)) with grid gap $d = 1 mm$ and voltage $Δ V = 300 V$ is limited by the Child–Langmuir Law $J = \frac{4}{9} ε_{0} \frac{2 q}{m _{i}} \frac{Δ V ^{3/2}}{d ^{2}}, ε_{0} = 8.85 \times 1 0^{- 12} F/m .$ (a) Compute the maximum ion current density $J$ (A/m²). (b) The Hall thruster of L4.1 pushes $I_{d} = 4.5 A$ through an annular exit area $A = 5 cm^{2} = 5 \times 1 0^{- 4} m^{2}$ ; find its current density $J_{Hall}$ and the ratio $J_{Hall} / J$ . Comment on why the Hall device wins.

Recall Solution

(a) First the mobility factor: $\frac{2 q}{m _{i}} = \frac{2 ( 1.6 \times 1 0 ^{- 19} )}{2.18 \times 1 0 ^{- 25}} = 1.468 \times 1 0^{6} \approx 1.212 \times 1 0^{3} .$ $Δ V^{3/2} = 30 0^{1.5} = 5196$ . $d^{2} = (1 0^{- 3})^{2} = 1 0^{- 6}$ . $J = \frac{4}{9} (8.85 \times 1 0^{- 12}) (1.212 \times 1 0^{3}) \frac{5196}{1 0 ^{- 6}} \approx 24.8 A/m^{2} .$ (b) $J_{Hall} = \frac{I _{d}}{A} = \frac{4.5}{5 \times 1 0 ^{- 4}} = 9000 A/m^{2} .$ $\frac{J _{Hall}}{J} = \frac{9000}{24.8} \approx 363.$ Why it wins: the gridded engine accelerates ions across a vacuum gap, so its own ion cloud repels itself — the Child–Langmuir ceiling. The Hall thruster accelerates ions inside a quasineutral plasma where trapped electrons cancel the space charge, so no such ceiling applies. The result is a current density hundreds of times larger, hence far higher thrust per unit area — the entire reason the design exists.

Wrap-up recall

Recall One-line takeaways

$r_{L} \propto m$ — same field, electron circle $2.4 \times 1 0^{5} \times$ smaller than xenon's. ::: L1.1
$ω_{c} = q B / m$ has no velocity — orbit rate is speed-independent. ::: L2.2
Drift is azimuthal and does no work; axial leak is a slow collisional crawl. ::: L3
Magnetizing electrons yet freeing ions defines a window of $B$ , not a maximum. ::: L4.2
Quasineutral plasma dodges the Child–Langmuir ceiling → ~ $1 0^{2} \times$ current density. ::: L5.2