Force on a Moving Charge: qv×B

A hundred kilometres above the poles, streams of charged particles from the Sun slam into Earth's magnetic field — and instead of raining straight down, they are grabbed sideways, swung around the field lines, and funnelled toward the poles, where they set the sky on fire as the aurora. The force doing the grabbing is unlike any you have met. Gravity pulls whether you move or not. The electric force of Ch. E1 pushes a charge whether it moves or not. But the magnetic force is a conditional force: it acts only on moving charge, it grows with speed, and — strangest of all — it points sideways, perpendicular to the velocity it acts on, every instant, without exception. That sideways lock has a consequence so important it decides exam questions across every board: the magnetic force can bend a particle's path into circles and spirals, but it can never speed the particle up or slow it down. It steers; it never pays. A magnetic field a hundred times stronger than any on Earth still could not add one joule to a proton's kinetic energy. When the Large Hadron Collider wants its protons faster, it must use electric fields — its 8-tesla magnets are there purely to bend the beam around the ring. This module builds the law behind all of it: $F = qvB\sin\theta$ , the right-hand rule that gives its direction, the proof that it does no work, and its official union with the electric force — the Lorentz force, the single equation that tells charges what fields do to them. Master the force here; the next three modules spend it on circles, filters, and wires.

Key takeaways

A charge $q$ moving at speed $v$ in a magnetic field $B$ feels $F = qvB\sin\theta$ , where $\theta$ is the angle between the velocity and the field.
The direction is perpendicular to both $\vec{v}$ and $\vec{B}$ : flat right hand, thumb along $\vec{v}$ , fingers along $\vec{B}$ , palm push = force on a positive charge. Negative charge → exact opposite.
Moving along the field ( $\theta = 0$ ): zero force. Moving across it ( $\theta = 90^{\circ}$ ): maximum force $qvB$ . Stationary ( $v = 0$ ): nothing, always.
Because $\vec{F} \perp \vec{v}$ at every instant, the magnetic force does no work: it changes a particle's direction but never its speed or kinetic energy.
The full story when both fields act is the Lorentz force: $\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}$ — electric part pays energy, magnetic part steers.
The tesla, operationally: $1\ \mathrm{T}$ is the field in which $1\ \mathrm{C}$ crossing at $1\ \mathrm{m/s}$ feels $1\ \mathrm{N}$ ; equivalently $1\ \mathrm{T} = 1\ \mathrm{N/(A \cdot m)}$ .
Key idea: the magnetic force is velocity's perpendicular escort — proportional to charge, speed, field and $\sin\theta$ , forever sideways, able to redirect motion but never to change its speed.

The core insight

A moving charge in a magnetic field feels a force of magnitude $F = qvB\sin\theta$ , directed perpendicular to both the velocity and the field (right-hand rule for positive charge, reversed for negative) — so the force steers the motion without ever doing work on it, and vanishes entirely for charges at rest or moving along the field.

Where this lives today

Every particle accelerator on Earth is equation (3) with the terms deliberately separated. In the Large Hadron Collider, 1,232 superconducting dipoles at 8.3 T supply pure $q\vec{v}\times\vec{B}$ — steering 6.8-TeV protons around a 27-km circle while adding zero energy — and a few radio-frequency electric cavities supply pure $q\vec{E}$ , the only part of the machine that ever speeds a proton up. Proton-therapy hospitals run the same physics at bedside scale: magnets swing the beam through a rotating gantry to enter a patient from any angle, while the beam's energy — hence stopping depth — was set upstream by electric fields, untouched by all the steering. The no-work theorem is also why the sky glows: solar-wind particles, unable to cross Earth's field lines, are guided along them to the poles, arriving with their solar speed intact to splash green and red light from the upper atmosphere. The aurora is the Lorentz force painting with particles it never once slowed down.

The physical picture

The Magnetic Field & Field Lines left one question open: $\vec{B}$ is an instruction stationed at every point — but an instruction to do what? Here is the answer; read it slowly, because it defies everyday intuition about pushes. Take a region of uniform field — into the page, a rectangle of crosses in the notation you just learned. Fire a positive charge through it, rightward. The field does not push the charge along the field (into the page), nor along its motion (rightward). It pushes it upward — perpendicular to the velocity and the field, the one direction nothing else in mechanics would have suggested. Double the speed: the push doubles. Angle the velocity partly along the field: only the across-field part earns force. Stop the charge: the force vanishes as if the field were never there. So the magnetic force is a conditional, sideways force with four dials: charge $q$ , speed $v$ , field $B$ , geometry $\sin\theta$ . Each is an exam handle, and the sideways-ness is the physics. A force locked perpendicular to velocity is a pure steering force — never tangential. You have met one before: the tension of a string whirling a stone in Uniform Circular Motion changes the stone's direction constantly while its speed never budges. The magnetic force is nature's built-in version of that string, and Circular & Helical Paths in Magnetic Fields (next module) cashes the analogy in full.

Figure

The right-hand slap rule — thumb velocity, fingers field, palm pushes the positive charge

The law, anatomised

The magnitude first, in the form every board wants, with each dial read separately:

F = qvB\sin\theta \qquad (1)

where: $q$ — the magnitude of the charge, in coulombs; $v$ — its speed; $B$ — the field strength in tesla; $\theta$ — the angle between $\vec{v}$ and $\vec{B}$ . Read the structure: the force is a product of four factors, so it vanishes if any one of them is zero — no charge, no motion, no field, or motion parallel to the field. The $\sin\theta$ is doing honest geometry: only the component of velocity across the field, $v\sin\theta$ , generates force; the along-field component $v\cos\theta$ rides free. That free ride is precisely what will make helices in the next module.

Figure

The sin-theta dial — only the crossing component of velocity earns a force

The direction is the part the formula cannot say, and it is the exam's favourite half-mark. Use one rule, always, until it is reflex — this track uses the flat right hand: thumb along $\vec{v}$ , straightened fingers along $\vec{B}$ , and the palm pushes in the direction of the force on a positive charge. For an electron or any negative charge, read the rule, then reverse. (Fleming's left-hand rule, which you may know from Class 10, is the same physics dressed for currents in wires — it takes over officially in Force on a Current-Carrying Conductor, two modules on. Pick one rule per situation and never mix them mid-problem; a one-rule discipline for the whole chapter arrives as its own technique module at the end of E4.) Unfinished business: The Magnetic Field & Field Lines promised the tesla an operational definition, and equation (1) delivers it. Set $q = 1\ \mathrm{C}$ , $v = 1\ \mathrm{m/s}$ , $\theta = 90^{\circ}$ :

1\ \mathrm{T} = \frac{1\ \mathrm{N}}{1\ \mathrm{C} \cdot 1\ \mathrm{m/s}} = 1\ \frac{\mathrm{N}}{\mathrm{A \cdot m}} \qquad (2)

where: the last step uses $\mathrm{C \cdot m/s} = (\mathrm{C/s}) \cdot \mathrm{m} = \mathrm{A \cdot m}$ — a rearrangement worth doing once by hand, because "newton per ampere-metre" is exactly the form the wire-force module will need. A one-tesla field exerts one newton on each coulomb crossing it at one metre per second. Now the ladder from last module has teeth: Earth's $5 \times 10^{-5}\ \mathrm{T}$ sounds feeble, but a fast proton, as the worked example shows, still feels it ferociously.

Derivation AAssembling the force law from its experimental dials

We want equation (1) not as a decree but as the summary of what beams of charged particles actually do in magnetic fields — the way the law was built historically, and the way "justify the form of the force" questions expect it argued. The strategy: hold three dials fixed, vary the fourth, record the proportionality; then stack the four results into one product and fix the constant by choosing the unit of B. Watch where $\sin\theta$ enters — it is measured geometry, not decoration.

Step 1. Vary the charge. Send identical-velocity beams of particles with charge $e$ , $2e$ (helium ions), $3e$ through the same field: the deflecting force doubles and triples in step. $F \propto q$ .

Step 2. Vary the speed. Same particles, same field, faster beam: the force grows in exact proportion to $v$ — a beam twice as fast is bent by twice the force. $F \propto v$ . (Contrast this with the electric force $qE$ , which could not care less how fast the charge moves.)

Step 3. Vary the field. Doubling $B$ (two identical magnets stacked) doubles the force: $F \propto B$ . Combining: $F \propto qvB$ when velocity crosses the field at right angles.

Step 4. Vary the geometry. Tilt the beam from across-field toward along-field, and the force falls — not linearly in the angle, but as its sine: at $30^{\circ}$ the force is exactly half its perpendicular value, at $0^{\circ}$ it is gone. Only the crossing component $v\sin\theta$ counts: $F = kqvB\sin\theta$ .

Step 5. Fix the constant. Define the tesla so that $k = 1$ — that is what equation (2) is: the SI's choice to make the force law clean. The direction, perpendicular to both $\vec{v}$ and $\vec{B}$ by the right-hand rule, is the remaining experimental fact the magnitude cannot encode. Verified: dimensions check — $\mathrm{C \cdot (m/s) \cdot N/(A\cdot m)} = \mathrm{(C/s)\cdot N/A} = \mathrm{N}$ ✓; and the $\theta = 30^{\circ}$ half-force reading matches $\sin 30^{\circ} = 1/2$ exactly, as used in Worked Example 2.

The theorem that decides marks: no work, ever

The single most consequential fact in this module follows from geometry alone.

Go deeper · Derivation B · 1 minThe magnetic force does no work on the charge

Derivation BThe magnetic force does no work on the charge

We prove that the magnetic force never changes a particle's speed or kinetic energy, no matter how the particle moves or how strong and complicated the field is. The strategy is to examine one tiny step of the motion, apply the definition of work from the mechanics track, and let the perpendicularity of force and velocity do the rest. The subtle point to watch: the argument never assumes the field is uniform or the path is a circle — it is completely general.

Step 1. In any tiny interval, the particle's displacement $\Delta \vec{x}$ points along its velocity $\vec{v}$ — that is what velocity means.

Step 2. The work done by a force over that step is $W = F\,\Delta x \cos\phi$ , with $\phi$ the angle between force and displacement (Work-Energy Theorem, from the mechanics track).

Step 3. But the magnetic force is locked perpendicular to the velocity — $\phi = 90^{\circ}$ at every instant, by the force law's own direction rule. So each step contributes $W = F\,\Delta x \cos 90^{\circ} = 0$ .

Step 4. The total work is the sum of the steps' contributions: a sum of zeros. By the work–energy theorem, the kinetic energy never changes: the speed is the same at every point of the path, however bent. Why this does not make the force useless: a force that cannot change speed can still change direction — and controlling direction is half of physics. The tension in a whirling string does zero work too, and it is the only reason the stone travels in a circle. *Verified: consistent with the observed physics of storage rings — protons circulate in the LHC for hours held by magnetic fields alone, and their energy stays fixed until the electric accelerating cavities fire.*

The division of labour this theorem enforces is the organising principle of the whole chapter. When both fields act at once, the total is the Lorentz force,

\vec{F} = q\vec{E} + q\vec{v} \times \vec{B} \qquad (3)

where: the first term is Ch. E1's electric force — acts on stationary charges, points along the field, does work; the second is this module's magnetic force — needs motion, points sideways, does none. (The symbol $\vec{v}\times\vec{B}$ is compact notation for "magnitude $vB\sin\theta$ , direction by the right-hand rule" — exactly what you have been computing; the full vector machinery is UG territory.) Every device in the chapters ahead is a deployment of equation (3) with one term or the other in charge: the next module lets the magnetic term steer alone, and Crossed Fields & the Velocity Selector sets the two terms against each other in a tug-of-war with one victor speed.

behaviour	electric force $q\vec{E}$	magnetic force $q\vec{v}\times\vec{B}$
acts on a stationary charge	yes	never
direction	along $\vec{E}$ ( $-$ for negative $q$ )	perpendicular to both $\vec{v}$ and $\vec{B}$
depends on speed	no	proportional to $v$
does work / changes KE	yes	never
what it changes	speed and direction	direction only
typical exam path	parabola (Charged Particle in a Uniform Field)	circle or helix (next module)

Limiting cases

case	result	meaning
$v = 0$	$F = 0$	magnetic fields ignore stationary charge — always
$\theta = 0^{\circ}$ or $180^{\circ}$	$F = 0$	motion along the field rides free
$\theta = 90^{\circ}$	$F = qvB$ (maximum)	the standard exam configuration
$\theta = 30^{\circ}$	$F = qvB/2$	$\sin\theta$ , not $\theta/90$ — half force at a third of the way
$q \to -q$	direction exactly reverses	electrons: apply rule, then flip
$v$ doubles	$F$ doubles	faster particles are steered harder
$\vec{E}$ and $\vec{B}$ both present	$\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}$	superpose the two forces as vectors — the selector's arena

Common trapA magnetic field accelerates the particle, so it speeds up

Half right, and the wrong half fails questions on every board. In physics, acceleration means any change of velocity — including pure changes of direction. The magnetic force does accelerate the particle (it bends the path, $a = F/m \neq 0$ ), but it never changes the speed: the force is perpendicular to the motion at every instant, so it does zero work and the kinetic energy is frozen. When an option says "the field increases the particle's kinetic energy" or "the particle exits the field region faster," it is wrong regardless of the numbers. Speed changes are the electric field's department — equation (3) keeps the books separate. If a question reports a particle gaining energy in a purely magnetic region, something else (an electric field, a collision) is hiding in the setup.

Worked examples

Board level — force, acceleration, and a sense of scale (CBSE/ICSE pattern, 3 marks)

A proton ( $q = 1.6 \times 10^{-19}\ \mathrm{C}$ , $m = 1.67 \times 10^{-27}\ \mathrm{kg}$ ) moves at $2.0 \times 10^{6}\ \mathrm{m/s}$ at right angles to a uniform field $B = 0.50\ \mathrm{T}$ . Find (a) the magnetic force on it, (b) its acceleration, and (c) the work done by the force in one full second.

Step 1. Part (a): with $\theta = 90^{\circ}$ , $\sin\theta = 1$ : $F = qvB = (1.6 \times 10^{-19})(2.0 \times 10^{6})(0.50) = \boxed{1.6 \times 10^{-13}\ \mathrm{N}}$ .

Step 2. Part (b): Newton's second law, $a = F/m = \dfrac{1.6 \times 10^{-13}}{1.67 \times 10^{-27}} = \boxed{9.6 \times 10^{13}\ \mathrm{m/s^2}}$ — ten trillion g. For scale, the proton's weight is $mg \approx 1.6 \times 10^{-26}\ \mathrm{N}$ : the magnetic force outguns gravity here by a factor of $10^{13}$ , which is why exam problems ignore gravity on particles without apology.

Step 3. Part (c): zero. The force is perpendicular to the velocity at every instant, so it does no work — the enormous acceleration of part (b) is pure direction-changing. Writing " $W = 0$ because $F \perp v$ throughout" is the full-credit answer; any attempt to compute $W = F \times d$ is a conceptual error, not an arithmetic one.

Competitive level — angles, signs, and species (JEE-Main/NEET pattern, ~2.5 minutes)

An electron moves at $4.0 \times 10^{6}\ \mathrm{m/s}$ in a region of uniform field $B = 0.20\ \mathrm{T}$ . (a) Find the force on it when its velocity makes $30^{\circ}$ with the field. (b) An alpha particle ( $q = +2e$ ) crosses the same field at the same speed, at $90^{\circ}$ . Find the force on it. (c) The electron is now redirected to move exactly along the field. Force?

Step 1. Part (a): $F = qvB\sin 30^{\circ} = (1.6 \times 10^{-19})(4.0 \times 10^{6})(0.20)(0.5) = \boxed{6.4 \times 10^{-14}\ \mathrm{N}}$ . Direction: apply the flat right hand to the velocity and field — then reverse, because the electron is negative. (In an MCQ, do the reversal last, once, deliberately; double-flipping is the classic error.)

Step 2. Part (b): the alpha carries $q = 2e = 3.2 \times 10^{-19}\ \mathrm{C}$ and crosses at $\sin 90^{\circ} = 1$ : $F = (3.2 \times 10^{-19})(4.0 \times 10^{6})(0.20) = \boxed{2.6 \times 10^{-13}\ \mathrm{N}}$ . Note the two dials that moved: charge doubled ( $\times 2$ ) and the angle factor went from $\tfrac12$ to $1$ ( $\times 2$ ) — force four times part (a)'s, read straight off the product structure without recomputing from scratch.

Step 3. Part (c): $\theta = 0$ , so $F = 0$ exactly. The electron cruises along the field line in a straight line at constant speed, completely unbothered — the free-ride direction.

Step 4. The concept being scored: equation (1) is a product of independent dials. Competitive papers rarely want one number; they want you to re-scale an answer as $q$ , $\theta$ , $v$ change. Practise reading "alpha vs electron, $30^{\circ}$ vs $90^{\circ}$ " as " $\times 2 \times 2$ " rather than as two separate plug-ins. Timing note: about 2.5 minutes. Parts (a) and (b) are one multiplication each if you carry powers of ten separately. The direction call in (a) deserves ten careful seconds — sign errors on electron questions are the single largest mark-leak in this topic. Part (c) should take three seconds: parallel means free.

Exam traps

Forgetting $\sin\theta$ . The formula is $qvB\sin\theta$ , and $\theta$ is measured between $\vec{v}$ and $\vec{B}$ — not from the normal, not from the page. If the problem says "at $30^{\circ}$ to the field," the factor is $\sin 30^{\circ} = 0.5$ , full stop.

The electron double-flip. One reversal for negative charge — applied once, at the end. Students who flip while placing the hand and flip again "for the electron" restore the wrong answer. Ritual: hand for a positive charge first, then one deliberate reversal.

"The field did work." Any option, in any disguise, that has a purely magnetic field changing speed or kinetic energy is wrong. This kills entire option sets in seconds — and it is why part (c) of the board example is free marks for the prepared.

Stationary charge questions. $v = 0 \Rightarrow F = 0$ . A charge at rest in any magnetic field feels nothing magnetic. (An electric field would grab it — that contrast is a favourite one-marker.)

Hand hygiene. Right hand for charges (this module), Fleming's left hand for wires (two modules ahead) — and Fleming's "current" finger means conventional current: an electron beam moving left is a current pointing right. Skipping that translation is the standard route to a reversed answer.

IGCSE scope note. 0625 (Supplement) asks only for the direction of the deflection of charged-particle beams — no $F = qvB$ computation. The right-hand rule (or left hand with the current translation) plus the negative-charge flip is the entire requirement; spend your revision there.

Constants and conventions. $e = 1.6 \times 10^{-19}\ \mathrm{C}$ , $m_e = 9.1 \times 10^{-31}\ \mathrm{kg}$ , $m_p = 1.67 \times 10^{-27}\ \mathrm{kg}$ on all Indian boards; AP-2 and IB provide constants on the sheet. Gravity on beam particles is ignored unless the question insists — the board example's $10^{13}$ ratio is the licence.

Origins · 1 minHendrik Lorentz and how this was discovered

Hendrik Lorentz

1853–1928 · Leiden

Hendrik Antoon Lorentz held Leiden's chair of theoretical physics from the age of 24 and spent it building the bridge between Maxwell's fields and actual matter. His electron theory pictured metals and molecules as swarms of tiny charges, and in 1895 he wrote the complete recipe for what fields do to each one: $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$ — the equation this module has been assembling. (Heaviside had reached the magnetic term in 1889; Lorentz supplied the full contract and the theory around it.) Two years later, cathode-ray experiments measured the deflections his law predicted, and the electron became real. His force law, and the velocity transformations he devised to keep his electrons consistent with light, handed Einstein the raw material for special relativity — the deepest "why" behind the sideways force wears Lorentz's name too. Einstein kept his portrait on the study wall, calling him the greatest and noblest man of the age.

History

1820

— Ampère shows parallel currents attract and antiparallel repel: moving charge feels magnetic force.

1881

— J.J. Thomson extracts an early, half-strength formula for the force on a moving charge from Maxwell's theory.

1889

— Heaviside corrects it: the magnetic force is qv×B, full strength, purely sideways.

1895

— Lorentz publishes the complete force law F = q(E + v×B) inside his electron theory.

1897

— Thomson bends cathode rays with fields and weighs the electron — the law's first great harvest.

Going further

Texts. NCERT Physics Class 12, Ch. 4 §4.2–4.3 · HC Verma, Concepts of Physics Vol. 2, Ch. 34 · CIE 9702 Topic 20.3 (recall and use F = BQv sin θ) · IB 2023 Theme D.3 · AP Physics 2 CED Topic 12.2.

Provenance. All results standard. A verifier should recompute: board example $F = 1.6\times10^{-19} \times 2.0\times10^{6} \times 0.50 = 1.6\times10^{-13}\ \mathrm{N}$ , $a = F/m_p = 9.6\times10^{13}\ \mathrm{m/s^2}$ , weight ratio $F/m_p g \approx 10^{13}$ , $W = 0$ ; competitive example $6.4\times10^{-14}\ \mathrm{N}$ (sin 30° = 0.5), alpha $3.2\times10^{-19} \times 4.0\times10^{6} \times 0.20 = 2.6\times10^{-13}\ \mathrm{N}$ (= 4× part a), parallel case 0; tesla identity $\mathrm{N/(C\cdot m/s)} = \mathrm{N/(A\cdot m)}$ .

Prerequisites. The Magnetic Field & Field Lines · Charged Particle in a Uniform Field (the electric contrast). Leads to. Circular & Helical Paths in Magnetic Fields · Crossed Fields & the Velocity Selector · Force on a Current-Carrying Conductor · Direction Triage (E4's technique closer) · Magnetic Force on Moving Charges (UG deep dive).

Why is the force sideways? Every other force I know pushes along something.

At this level, take it as the experimentally certified shape of the law — beams in fields genuinely deflect perpendicular to both $\vec{v}$ and $\vec{B}$ , and the right-hand rule is the recorded pattern. But here is the trailer for the deep answer: magnetism is what the electric force looks like when its source charges move relative to you. Run alongside the charge and part of the magnetic force becomes plain electric push; the sideways twist is the geometric price of switching viewpoints. Making that exact is special relativity — Lorentz built the scaffolding, Einstein finished it. The sideways-ness is not a quirk; it is a clue.

How can the force be huge — 10¹³ g in the worked example — and still do zero work?

Because work is not force; work is force along the displacement. The magnetic force sits at $90^{\circ}$ to each tiny step of the path, and $\cos 90^{\circ} = 0$ kills every contribution. The enormous force is spent turning the velocity vector, never stretching it. Mechanics has an exact precedent: the normal force on a car sweeping a banked curve can exceed the car's weight and still does no work. Huge influence on the path, zero deposit in the energy account — the ledgers are independent.

Do I use the right-hand rule or Fleming's left-hand rule? My Class 10 book swears by the left.

Both are correct — they encode the same cross product. Fleming's left hand is tuned for wires: forefinger field, middle finger conventional current, thumb force; it stars in Force on a Current-Carrying Conductor. For free charges this track uses the flat right hand (thumb $\vec{v}$ , fingers $\vec{B}$ , palm push on positive) because it handles the electron flip most cleanly: rule first, one reversal after. What loses marks is mixing conventions mid-problem or forgetting an electron beam moving left is a current pointing right. One rule per situation, muscle memory — the E4 closer drills exactly this.

If both E and B act on a charge, which one wins?

Neither "wins" by rank — you add the two forces as vectors: $\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}$ . The electric piece is fixed by position; the magnetic piece grows with speed. So the balance can be engineered: oppose the two forces and there is exactly one speed at which they cancel and the charge sails straight. That device — the velocity selector — gets its own module two steps ahead; it is the trick Thomson used to weigh the electron.

Why does only the perpendicular part of the velocity matter?

Split $\vec{v}$ into parts along and across the field. The along-field part earns nothing — $\vec{v} \parallel \vec{B}$ gives $\sin\theta = 0$ . Only the crossing component $v\sin\theta$ "cuts across" the field and earns deflection. The consequence is lovely: the across part turns in circles while the along part cruises untouched, and "circle plus cruise" is a helix — the corkscrew the next module builds, and the reason aurora particles spiral along field lines instead of drifting across them.

Does a magnetic force act between two charges both sitting still?

No. Magnetic fields are made by moving charges and currents (that story starts with The Biot-Savart Law later in this chapter), and they act only on moving charges — this module. Two stationary charges interact purely electrically, by Coulomb's law. Set both moving and each makes a magnetic field the other responds to — genuinely there, but tiny at everyday speeds, which is why electrostatics never needed it. Currents in wires — enormous numbers of drifting charges — are how nature makes the effect big enough to build motors from.

The force changes direction of motion, the motion changes the force direction… doesn't this chase its own tail?

It does, and the tail-chase has a beautiful closed answer. At every instant the force is perpendicular to the current velocity; as the velocity turns, the force turns with it, staying exactly $90^{\circ}$ behind. A constant-magnitude force locked perpendicular to motion is the recipe for uniform circular motion — the across-field path closes into a perfect circle at constant speed. The full geometry (radius $r = mv/qB$ , and a period that spectacularly ignores the speed) is the next module's whole business.