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

3.3.42 · D3 · Physics › Rocket Propulsion › Hall-effect thruster — cross-field discharge, annular channe

Yeh page Hall thruster ka problem gym hai. Parent note ne physics build ki; yahan hum har tarah ke numbers drill karte hain jo yeh topic de sakta hai — bade fields, chhote fields, zero flow, woh edge jahan ek ion just barely magnetized hota hai, ek real mission problem, aur ek exam-style twist.

Koi bhi algebra shuru karne se pehle, ek seedha vaada: har symbol aur formula ko hum waheen restate karenge jab use karenge, taaki tumhe kabhi parent note pe wapas nahi jaana padega.

Scenario matrix

Ek Hall thruster problem ko ek grid mein sochte hain. Har row ek quantity hai jo poochi ja sakti hai; har column ek regime hai (normal, koi extreme, ya ek trap). Neeche ke examples har cell ko fill karte hain.

Case class	Kya cheez tricky banati hai	Covered by
Magnetized? (electron)	tiny gyroradius, "normal" case	Ex 1
Magnetized? (ion, borderline)	ion radius $\approx$ channel — degenerate edge	Ex 2
Limit: $B \to 0$	field off — trapping fail, drift diverge karta hai	Ex 3 (steps 1–2)
Limit: $B \to \infty$	huge field — radius aur drift dono collapse	Ex 3 (steps 3–4)
Exhaust speed / thrust	standard delivered-performance sum	Ex 4
Zero / degenerate input	$\overset{m}{˙} = 0$ ya $Δ V = 0$ — thrust vanish	Ex 5
Hall drift direction (sign/quadrant)	loop kis direction mein spin karta hai? geometry	Ex 6
Real-world word problem	ek station-keeping burn, uses Tsiolkovsky Rocket Equation	Ex 7
Exam twist	do ideas mix, hidden unit trap	Ex 8

Symbols aur teen tools jo hum baar baar use karenge

Pehle kuch symbols pin down karte hain, kyunki hum inhe baar baar use karenge. Inhe slowly padho — formulas mein appear hone se pehle har ek yahan earn kiya gaya hai.

Definition Har symbol ka matlab

$q$ — ek particle ki charge (electron ya singly-charged ion), $q = 1.6 \times 1 0^{- 19} C$ (coulombs, electric charge ki unit).
$m$ — us particle ki mass (electron ke liye $m_{e}$ , ion ke liye $m_{i}$ ).
$v_{⊥}$ — particle ki speed $B$ ke across (perpendicular) — uski motion ka woh hissa jo ek circle mein bend hota hai.
$B$ — magnetic field strength, teslas (T) mein; yeh set karta hai ki field ek moving charge ko kitna strongly bend karti hai.
$E$ — electric field strength, volts per metre ( $V/m$ ) mein. Ise voltage hill ki steepness samjho: agar accelerating voltage $Δ V$ channel length $L$ par drop hoti hai, toh $E = Δ V / L$ . Bada $E$ matlab voltage thodi si distance mein sharply girta hai, toh wahan rakha charge ek strong push feel karta hai. Yahi field ions ko accelerate karti hai aur, $B$ se crossed hokar, electron drift drive karti hai.
$Δ V$ — anode se channel exit tak voltage drop (potential difference), volts (V) mein.
$L$ — ek length: channel gap (magnetization ke liye) ya channel length ( $E$ compute karne ke liye).
$\overset{m}{˙}$ — propellant mass flow rate, har second nikalne wale kilograms ( $kg/s$ ).
$g_{0} = 9.81 m/s^{2}$ — standard gravity, sirf speed ko "seconds" fuel-economy figure mein convert karne ke liye use hoti hai.

Do derived symbols jin par hum constantly rely karte hain:

$v_{i}$ — ion speed accelerating voltage $Δ V$ se guzarne ke baad.
$v_{e}$ — exhaust speed jo rocket "feel" karta hai. Kyunki thrust almost entirely unhi ions ke zariye milta hai, hum $v_{e} \approx v_{i}$ rakhte hain; jab bhi neeche $v_{e}$ dekhein, ise "ion exit speed $v_{i}$ " padhein.

Har jagah use hone wale constants: $q = 1.6 \times 1 0^{- 19} C$ , $m_{e} = 9.11 \times 1 0^{- 31} kg$ , 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)

Ek $20 eV$ electron ( $v_{⊥} = 2 \times 1 0^{6} m/s$ ) $B = 0.02 T$ mein hai. Channel gap $L = 1.5 cm$ . Kya yeh magnetized hai?

Forecast: andaaza lagao — kya $r_{L}$ micrometres, millimetres, ya centimetres hoga?

Tool likho: $r_{L} = \frac{m _{e} v _{⊥}}{q B}$ . Yeh step kyun? Magnetization decide hoti hai $r_{L}$ ko channel size $L$ se compare karke; "magnetized" matlab $r_{L} ≪ L$ (figure mein wide gap ke against chhota red circle dekho).
Plug in karo: $r_{L} = \frac{( 9.11 \times 1 0 ^{- 31} ) ( 2 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.02 )} \approx 5.7 \times 1 0^{- 4} m = 0.57 mm$ . Yeh step kyun? Formula ke baad numbers, taaki units traceable rahein.
Compare karo: $0.57 mm ≪ 15 mm$ , ratio $\approx 26$ . Yeh step kyun? Factor $\sim 26$ chhota hona matlab electron gap mein ~26 tight loops spin karta hai — 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. Larmor Radius and Cyclotron Motion dekho.

Worked example Ex 2 — Borderline ion (degenerate edge)

Ek slow abhi-abhi bana xenon ion, $v_{⊥} = 300 m/s$ , same $B = 0.02 T$ . Channel $L = 1.5 cm$ . Magnetized hai ya nahi?

Forecast: ions $1 0^{5} \times$ zyada heavy hote hain — kya $r_{L}$ channel ke andar ya bahar padega?

Same tool, ion values: $r_{L} = \frac{m _{i} v _{⊥}}{q B}$ . Yeh step kyun? $r_{L} \propto m$ hai, toh equal speed par ion radius $m_{i} / m_{e}$ se bada hoga.
Plug in karo: $r_{L} = \frac{( 2.18 \times 1 0 ^{- 25} ) ( 300 )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.02 )} \approx 2.0 \times 1 0^{- 2} m = 2.0 cm$ . Yeh step kyun? Yeh slowest ion hai (trapped hone ke liye worst case); faster ions ka $r_{L}$ aur bhi bada hoga.
Compare karo: $2.0 cm > 1.5 cm$ . Ion ka would-be circle channel se wide hai — wall usse orbit complete karne se pehle rok deti hai. Yeh step kyun? Yahi degenerate edge hai: $r_{L} ≳ L$ matlab "circle nahi bana sakta" ⇒ effectively unmagnetized. Seedha slowest ion bhi isko paar kar leta hai.

Verify: in speeds par $r_{L, i} / r_{L, e}$ ka ratio $\frac{m _{i} \cdot 300}{m _{e} \cdot 2 \times 1 0 ^{6}}$ hai; dono radii plug karo $0.02/5.7 \times 1 0^{- 4} \approx 35$ , aur wakai ion radius $L$ ke just upar baithta hai ✔. Cell: ion borderline, unmagnetized.

Worked example Ex 3 — Do limits:

$B \to 0$ aur $B \to \infty$ $20 eV$ electron aur $E = 1.2 \times 1 0^{4} V/m$ rakho. Jab $B$ zero ki taraf shrink hota hai, phir huge banta hai toh $r_{L, e}$ aur $v_{d}$ ka kya hota hai? Dono limits ko alag physical worlds samjho — steps 1–2 $B \to 0$ world hain, steps 3–4 $B \to \infty$ world.

Forecast: har limit mein kaun si quantity blow up hogi aur kaun si collapse?

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

$r_{L} = m v_{⊥} / (q B)$ se: jab $B \to 0$ , $r_{L} \to \infty$ . Yeh step kyun? Koi field nahi toh koi bending force nahi, toh "circle" straighten hokar ek line ban jaati hai (figure ka left panel) — electron not magnetized. Trapping trick fail; koi virtual grid nahi.
$v_{d} = E / B$ se: jab $B \to 0$ , $v_{d} = E / B \to \infty$ . Yeh step kyun? Drift formula literally diverge karta hai — physically yeh signal karta hai ki drift picture khud break down kar raha hai: unmagnetized particles sirf $E$ ke along freely accelerate karte hain, drift nahi karte. Jo limit "explode" kare woh tumhara cue hai ki model apni domain se bahar ho gaya.

World B — $B \to \infty$ (over-strong field): 3. $r_{L} = m v_{⊥} / (q B)$ se: jab $B \to \infty$ , $r_{L} \to 0$ . Yeh step kyun? Electron ek single field line par pin ho jaata hai (right panel) — World A se bilkul alag failure. Yahan woh over-trapped hai, under-trapped nahi. 4. $v_{d} = E / B$ se: jab $B \to \infty$ , $v_{d} \to 0$ . Yeh step kyun? Hall loop crawl karta hai aur anode ki taraf cross-field leakage choke off hoti hai — discharge starve ho jaata hai. Toh dono extremes opposite reasons se fail karte hain, aur real designs ek middle $B$ choose karte hain.

Sanity number (finite $B$ ): $B = 0.04 T$ par (nominal $0.02$ se double), $v_{d} = 1.2 \times 1 0^{4} /0.04 = 3 \times 1 0^{5} m/s$ , $0.02 T$ ke $6 \times 1 0^{5}$ ka exactly aadha — $v_{d} \propto 1/ B$ confirm karta hai extremes ke beech.

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) aur $B \to \infty$ (World B), alag rakhe gaye.

Worked example Ex 4 — Exhaust speed, thrust,

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

Forecast: kya $v_{e}$ $1 0^{3}$ , $1 0^{4}$ , ya $1 0^{5} m/s$ ke kareeb hoga?

Exhaust speed: $v_{e} \approx v_{i} = \frac{2 q Δ V}{m _{i}} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 300 )}{2.18 \times 1 0 ^{- 25}}$ . Yeh step kyun? Ion potential $Δ V$ se guzar kar girata hai; energy conservation $q Δ V = \frac{1}{2} m_{i} v_{e}^{2}$ speed deta hai. Quasineutral plasma ions ko poora $Δ V$ dekhne deta hai (koi space-charge cap nahi, Child–Langmuir ki tarah unlike).
Andar compute karo: $4.40 \times 1 0^{8} \approx 2.10 \times 1 0^{4} m/s$ . Yeh step kyun? Pehle fraction evaluate karo ( $2 \cdot 1.6 \times 1 0^{- 19} \cdot 300/2.18 \times 1 0^{- 25} = 4.40 \times 1 0^{8}$ ) aur phir square root lo, kyunki $v_{e}$ root ke neeche rehta hai — root lene tak speed read nahi kar sakte.
Thrust: $F = \overset{m}{˙} v_{e} = (5 \times 1 0^{- 6}) (2.10 \times 1 0^{4}) \approx 0.105 N \approx 105 mN$ . Yeh step kyun? Thrust momentum flux hai: mass per second nikalna times uski speed.
$I_{s p} = v_{e} / g_{0} = 2.10 \times 1 0^{4} /9.81 \approx 2.14 \times 1 0^{3} s$ . Yeh step kyun? $I_{s p}$ measure karta hai ki $1 N$ -worth propellant kitne seconds tak chalta hai — 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 — Zero inputs (degenerate: koi push nahi)

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

Forecast: har case mein thrust?

Case (a): $F = \overset{m}{˙} v_{e} = 0 \cdot v_{e} = 0 N$ . Yeh step kyun? Koi propellant nahi nikalta matlab koi momentum flux nahi, chahe voltage kitni bhi badi ho. Ek hot idle field kuch nahi pe push karta hai.
Case (b): $v_{e} = 2 q (0) / m_{i} = 0$ , toh $F = \overset{m}{˙} \cdot 0 = 0 N$ . Yeh step kyun? Zero accelerating voltage matlab ions zero speed se exit karte hain — mass zero thrust ke saath bahar tapakta hai. Dono degenerate inputs $F$ ko kill karte hain, par alag physical reasons se (koi mass nahi vs. koi speed nahi).
Product form $F = \overset{m}{˙} 2 q Δ V / m_{i}$ ise obvious banata hai: $F = 0$ jab bhi koi bhi factor zero ho. Yeh step kyun? Multiplicative structure recognize karne se recompute kiye bina edge cases predict ho jaate hain.

Verify: dono cases exactly $0$ dete hain ✔. Cell: zero/degenerate.

Worked example Ex 6 — Hall loop kis direction mein spin karta hai? (sign/geometry)

$E$ axially outward point karta hai (exit ki taraf, $+ \overset{z}{^}$ ). $B$ radially outward point karta hai ( $+ \overset{r}{^}$ ). Azimuthal Hall drift $v_{d} = (E \times B) / B^{2}$ kis direction mein hai?

Forecast: downstream se dekhne par clockwise ya anticlockwise?

Drift formula use karo $v_{d} = \frac{E \times B}{B ^{2}}$ ; direction cross product $\overset{z}{^} \times \overset{r}{^}$ se set hoti hai. Yeh step kyun? Magnitude $E / B$ hum jaante hain; sign poori tarah cross-product direction mein hai.
Cylindrical unit vectors mein, $\overset{z}{^} \times \overset{r}{^} = \hat{θ}$ (azimuthal direction). Toh $v_{d}$ $+ \hat{θ}$ ke along point karta hai. Yeh step kyun? $(\overset{r}{^}, \hat{θ}, \overset{z}{^})$ ek right-handed set hai, bilkul $(\overset{x}{^}, \overset{y}{^}, \overset{z}{^})$ ki tarah, toh unke cross products same right-hand rule follow karte hain (figure mein red arrow dekho).
Result: electron drift ring ke around azimuthally circulate karta hai, ek consistent sense mein — ek closed loop jiske koi ends nahi. Koi ek $E$ ya $B$ flip karo aur loop reverse ho jaata hai; dono flip karo aur woh stay karta hai. Yeh step kyun? Yahi sign-tracking hai jis wajah se parent note annular channel par insist karta hai: sirf ek ring $+ \hat{θ}$ ko khud par close karne deta hai. Ek straight tube yeh current ek wall mein slam kar deta.

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

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

Ek satellite jiska dry mass $500 kg$ hai, $12 kg$ xenon carry karta hai. Uska Hall thruster Ex 4 ke numbers run karta hai ( $v_{e} = 2.10 \times 1 0^{4} m/s$ , $F = 0.105 N$ ). (a) Total available $Δ v$ ? (b) Thruster ko saara xenon burn karne mein kitna time fire karna hoga?

Forecast: $500 m/s$ se zyada ya kam $Δ v$ ? Hours ya days ki firing?

Tsiolkovsky Rocket Equation se $Δ v$ : $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ , jahan $m_{0} = 512 kg$ , $m_{f} = 500 kg$ . Yeh step kyun? Rocket equation exhaust speed aur mass ratio ko velocity change mein convert karta hai jo tum khareed sakte ho — orbital maneuvers ki currency.
$Δ v = 2.10 \times 1 0^{4} \cdot ln (512/500) = 2.10 \times 1 0^{4} \cdot ln (1.024) \approx 2.10 \times 1 0^{4} \cdot 0.02372 \approx 498 m/s$ . Yeh step kyun? Chhota mass ratio $\Rightarrow$ chhota $ln$ , par huge $v_{e}$ phir bhi ek useful $\sim 0.5 km/s$ deta hai — high- $I_{s p}$ electric propulsion ka yahi point hai.
$\overset{m}{˙}$ se burn time: $t = \frac{m _{prop}}{m ˙} = \frac{12}{5 \times 1 0 ^{- 6}} s = 2.4 \times 1 0^{6} s$ . Yeh step kyun? Constant flow rate matlab time = fuel ÷ rate. Yahan electric thrusters bite karte hain: gentle force, long burns.
Convert karo: $2.4 \times 1 0^{6} s \div 86400 \approx 27.8 days$ . Yeh step kyun? Human units mein feel karo answer — half km/s ke liye lagbhag ek mahine ka continuous thrust.

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

Worked example Ex 8 — Exam twist (ek hidden unit trap + do ideas)

"Same thruster $Δ V = 300 V$ par run karta hai. Ek student claim karta hai ki Hall drift electrons ko axially $v_{d} = E \cdot B = 1.2 \times 1 0^{4} \times 0.02 = 240 m/s$ par move karta hai, toh yahi exhaust speed hai." Teen errors dhundho aur sahi exhaust speed do.

Forecast: galat formula, galat direction, aur galat species spot karo — aage padhne se pehle.

Error 1 — formula: force balance deta hai $E = v_{d} B$ , toh $v_{d} = E / B$ , $E \cdot B$ nahi. Yeh step kyun? Units decide karte hain: $(V/m) / T = m/s$ ✔, jabki $(V/m) \cdot T$ deta hai $V\cdotT/m$ — yeh speed nahi hai. Correct $v_{d} = 1.2 \times 1 0^{4} /0.02 = 6 \times 1 0^{5} m/s$ .
Error 2 — direction: $v_{d}$ azimuthal hai (ring ke around), axial nahi. Yeh closed Hall current hai, exhaust nahi. Yeh step kyun? Direction $E \times B$ se aati hai (Ex 6) — yeh $E$ ke along kabhi point nahi kar sakti.
Error 3 — species: exhaust ions hain, $Δ V$ se guzar kar accelerate hote hain, drifting electrons nahi. Yeh step kyun? Electrons trapped hain (Ex 1); thrust ion momentum hai. Toh $v_{e} = v_{i} = 2 q Δ V / m_{i}$ .
Sahi exhaust speed: $v_{e} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 300 )}{2.18 \times 1 0 ^{- 25}} \approx 2.10 \times 1 0^{4} m/s$ . Yeh step kyun? Same computation jaise Ex 4 — exam twist sirf teen distractors hain jo sahi idea ke around wrap hain.

Verify: correct $v_{d} = 6 \times 1 0^{5} m/s \neq = 240 m/s$ ✔; correct $v_{e} \approx 2.10 \times 1 0^{4} m/s$ ✔; aur $v_{d} ≫ v_{e}$ (electrons ions ke nikalne se kaafi faster whip around karte hain) ✔. Cell: exam twist.

Recall

Recall Har question ka answer kaun sa formula deta hai?

"Kya koi species magnetized hai?" ::: $r_{L} = m v_{⊥} / (q B)$ , channel $L$ se compare karo. "Hall loop kitni fast hai?" ::: $v_{d} = E / B$ (azimuthal). "Exhaust speed kya hai?" ::: $v_{i} = 2 q Δ V / m_{i}$ . "Thrust kya hai?" ::: $F = \overset{m}{˙} v_{e}$ . "Fuel kitna $Δ v$ khareedta hai?" ::: Tsiolkovsky $Δ v = v_{e} ln (m_{0} / m_{f})$ .

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