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

Step 1 — Two arrows in a ring: naming the fields

WHAT. Before any formula, we need to agree on three directions inside the ring-shaped (annular) channel. Look at the figure: the channel is a hollow ring, like a doughnut sliced open.

The axial direction (blue arrow, call it $\overset{z}{^}$ ) points out of the thruster — the way we want ions to go.
The radial direction (yellow arrow, call it $\overset{r}{^}$ ) points across the gap, from inner wall to outer wall.
The azimuthal direction (green arrow, call it $\hat{θ}$ ) points around the ring, like walking a circular track.

WHY these three. A ring has exactly these three natural directions, and — crucially — they are mutually perpendicular at every point. That perpendicularity is the whole engine: it lets one field be "across" another. See Lorentz Force for where these directions plug in.

PICTURE.

Figure — Hall-effect thruster — cross-field discharge, annular channel

Step 2 — What a magnetic field does to a lone moving charge

WHAT. Put a single electron moving with velocity $v$ into the radial field $B$ . The force it feels is the Lorentz force:

$F = q v \times B .$

Read it term by term, right where each lives:

$q$ — the charge (negative for an electron); it sets the strength and sign of the push.
$v$ — how fast and which way the particle already moves.
$B$ — the radial field arrow.
$\times$ — the cross product: it hands back an arrow perpendicular to both $v$ and $B$ , with size $q v B sin ϕ$ where $ϕ$ is the angle between $v$ and $B$ .

WHY the cross product and not ordinary multiplication? Because experiment says the magnetic force is always sideways to the motion — it never lines up with $v$ . The cross product is the one operation that automatically returns a perpendicular arrow. That "always sideways" fact has a giant consequence, coming in Step 3.

PICTURE. The red force arrow sits at $9 0^{\circ}$ to the blue velocity.

Step 3 — Steering with constant speed makes a circle

WHAT. If the force is always perpendicular to $v$ , it does zero work (work needs force along motion). So the speed $∣ v ∣$ never changes — only the direction turns. A constant-speed, constantly-turning path is a circle. Balancing the magnetic force against the centripetal ("centre-seeking") force needed for that circle:

$magnetic force q v_{⊥} B = force to hold a circle of radius r_{L} \frac{m v _{⊥}^{2}}{r _{L}} .$

$v_{⊥}$ — the part of the speed across $B$ (the part that gets bent).
$r_{L}$ — the radius of the circle, called the Larmor radius.
$m$ — the particle's mass: heavier things are harder to turn.

Solve for $r_{L}$ :

$r_{L} = \frac{m v _{⊥}}{q B} .$

WHY this matters for the thruster. Notice $r_{L} \propto m$ . An electron and an ion at the same speed in the same field turn on circles whose sizes differ by their mass ratio $m_{i} / m_{e} \approx 2.4 \times 1 0^{5}$ for xenon. So one field can bend electrons into tiny loops while barely deflecting ions. That single fact is the Hall thruster. See Larmor Radius and Cyclotron Motion.

PICTURE. Tiny electron loop (green), giant ion arc (red), same $B$ .

Step 4 — Now switch on $E$ : the circle starts to march

WHAT. Keep $B$ radial, and turn on the axial $E$ . The electron now feels two forces: the electric push $q E$ (constant) plus the magnetic steering $q v \times B$ . Watch what happens along one loop:

On the half of the circle where the electron moves with $E$ 's push, it speeds up → bigger $r_{L}$ there.
On the half where it moves against the push, it slows down → smaller $r_{L}$ there.

A circle that is fat on one side and thin on the other does not close — its centre slides over a bit each lap. That slide is a net sideways march called the drift.

WHY it drifts sideways and not along $E$ . The magnetic bending turns every velocity by $9 0^{\circ}$ , including the extra velocity $E$ tries to add. So the response to a push along $\overset{z}{^}$ shows up as motion along $\hat{θ}$ — perpendicular to both. This is the seed of the boxed formula.

PICTURE. A cycloid — loops that creep sideways, fat-then-thin.

Step 5 — Pin down the drift: force balance in the moving frame

WHAT. Ride along with the marching centre at velocity $v_{d}$ . In that frame the loop closes again, meaning the average force is zero:

$q (E + v_{d} \times B) = 0.$

$q E$ — the steady electric push.
$q v_{d} \times B$ — the magnetic force caused by the drift itself.
Setting the sum to $0$ says: the drift is exactly the speed at which the magnetic reaction cancels the electric push.

WHY set it to zero. A steady drift means no net average acceleration — otherwise it would speed up forever. Zero net force is the definition of "steady."

Now solve. Cross the whole equation with $B$ on the right:

$q E \times B + q (v_{d} \times B) \times B = 0.$

Use the vector identity $(v_{d} \times B) \times B = (v_{d} \cdot B) B - B^{2} v_{d}$ . Since the drift is perpendicular to $B$ , the dot product $v_{d} \cdot B = 0$ , leaving:

$E \times B - B^{2} v_{d} = 0 ⟹ v_{d} = \frac{E \times B}{B ^{2}}, ∣ v_{d} ∣ = \frac{E}{B} .$

Numerator $E \times B$ — perpendicular to both fields, i.e. azimuthal.
Divide by $B^{2}$ — makes the units come out as m/s and shows stronger $B$ ⇒ slower drift (tighter loops march less each lap).

PICTURE. The two triangles cancelling; the surviving green azimuthal arrow.

Step 6 — Why the ring: the drift must close into a loop

WHAT. In the annular channel, $E$ is axial and $B$ is radial, so $E \times B$ points azimuthally — around the ring. Because the ring has no ends, this drift path closes on itself into a complete circle of current: the Hall current.

WHY a ring and not a straight tube. In a straight tube the azimuthal drift would run into a wall, dump charge there, build up an opposing field, and stall. Only a loop with no ends lets the current flow forever. This is why the geometry is annular, not incidental.

PICTURE. Top-down view of the ring with the green Hall-current loop.

Step 7 — The ions ignore all of this and get slung out

WHAT. Ions are unmagnetized (Step 3: their $r_{L}$ is bigger than the channel). They feel essentially only the axial $E$ , so they simply fall through the potential drop $Δ V$ . Energy conservation:

$q Δ V = \frac{1}{2} m_{i} v_{i}^{2} ⟹ v_{i} = \frac{2 q Δ V}{m _{i}} .$

$q Δ V$ — energy gained falling through the voltage (like a ball down a hill).
$\frac{1}{2} m_{i} v_{i}^{2}$ — where that energy goes: kinetic energy of the ion.

Then thrust follows from mass being thrown out (the Tsiolkovsky Rocket Equation world):

$v_{e} \approx v_{i}, F = \overset{m}{˙} v_{e} .$

WHY $B$ never appears here. The magnetic field does no work (Step 3); it only trapped the electrons so this axial field could exist inside the plasma. The push on the ions is purely electric.

PICTURE. Ion rolling down the voltage "hill" to exhaust speed.

Step 8 — Edge & degenerate cases (never hit a wall you weren't shown)

WHAT / WHY, each drawn in the summary of this step:

$E = 0$ : numerator $E \times B = 0$ , so $v_{d} = 0$ . The loop closes; electrons just gyrate in place. No Hall current, no discharge.
$B \to 0$ (unmagnetized): $r_{L} = \frac{m v _{⊥}}{q B} \to \infty$ ; the "loop" is infinitely big — a straight line. This is exactly the ion case, and why ions don't drift.
$B \to \infty$ : $v_{d} = E / B \to 0$ ; loops so tight they barely march. Electrons freeze azimuthally — too much field also kills the current.
$E ∥ B$ (not crossed): $E \times B = 0$ again — no drift. The design requires the $9 0^{\circ}$ crossing; that is why we insisted on it in Step 1.
$q$ sign flip (ion vs electron): $v_{d} = E \times B / B^{2}$ has no $q$ in it — so electrons and ions would drift the same direction and speed if both were magnetized. They differ only because ions aren't magnetized at all.

PICTURE.

The one-picture summary

Everything above, compressed: crossed fields → tiny electron loops that march azimuthally into a closed Hall current → quasineutral trap → strong axial field → ions flung out as thrust.

Recall Feynman retelling (say it to a friend)

Picture a round racetrack. A magnetic field points across the track (wall to wall); an electric field points along the way out. A heavy marble (ion) doesn't care about the magnet — it just rolls down the electric hill and shoots out the exit fast: that's the thrust, and it's electric, not magnetic. But a light bee (electron) gets bent by the magnet into tiny circles. When you also push the bee with the electric field, its circles don't close — they creep sideways, around the ring, forever, because the track has no ends. That circling swarm of bees is the Hall current, and it keeps the crowd of charges balanced so the electric hill stays steep. Fast marbles out one side = push. The magnet's whole job was just to hold the bees in place.

Recall Rebuild the key formula from scratch

Perpendicular magnetic force does no work ::: speed constant, path is a circle of radius $r_{L} = m v_{⊥} / (q B)$ Steady drift means net average force is zero ::: $q (E + v_{d} \times B) = 0$ Solve that by crossing with $B$ ::: $v_{d} = E \times B / B^{2}$ , size $E / B$ , pointing azimuthally Why the annulus ::: only a ring lets that azimuthal drift close into a steady loop What accelerates the ions ::: the axial $E$ -field alone (the magnet does no work)