ARCHETYPE: Charged-Particle Trajectories in E & B

Every collider detector is this archetype run in reverse. In CMS and ATLAS at the LHC, a charged particle's track curves in the solenoid's field; software fits a circle to the hits, reads $r$, and reports the momentum through $p = qBr$ — the radius as a momentum meter, millions of times per second, with the sign of the curvature tagging particle versus antiparticle: Anderson's 1932 trick industrialised. In hospitals, proton-therapy beamlines are a chain of "slab" problems — dipole magnets bending the beam through computed exit angles. And overhead the helix runs at planetary scale: solar-wind particles corkscrew along Earth's field lines toward the poles, where their arrival glows — Birkeland's aurora, the oldest trajectory diagram there is.

Geometry first, because in an archetype the geometry is the physics. A uniform field $\vec B$ fills a bounded region — a slab between two parallel faces in the standard version — and points into the page (drawn $\otimes$; out of the page is $\odot$). A positive charge $q$ enters the left face with speed $v$, perpendicular to the face.

The archetype rests on three facts, and the first two belong to other modules — we spend them here. Force on a Moving Charge: qv×B establishes that the force is always perpendicular to the velocity, so the speed never changes inside the region. Circular & Helical Paths in Magnetic Fields turns that into the two master numbers: $r = mv/qB$ and $T = 2\pi m/qB$, with the famous punchline that $T$ contains no $v$ at all. The third fact is the one this archetype owns, and it is pure geometry: the centre of the arc sits exactly one radius from the entry point, along the perpendicular to the entry velocity (on the side the force pushes — one right-hand-rule check at entry). Mark the centre and the particle's whole future is drawn: it rides the circle until the circle meets a boundary, and where the circle exits, the particle exits, along the tangent. That one construction converts every trajectory question into triangle work:

Which face does it leave? Draw the circle about the centre; see which boundary it cuts first.

At what angle? The velocity rotates by exactly the arc angle $\theta$; entry and exit speeds are equal, so only the direction turns.

How long inside? The clock rule. The particle sweeps $\theta$ out of a full $360^\circ$ lap that would take $T$, so

where: $\theta$ — the arc angle at the centre (equal to the angle by which the velocity direction turns); $T$ — the speed-independent period. Read it physically: a quarter-circle exit always costs $T/4$, a half-circle always $T/2$, no matter how fast the particle is moving — faster particles ride proportionally bigger arcs and finish on the same clock. This is the single most counterintuitive scoring fact in the archetype.

Perpendicular entry into a slab

The flagship geometry: field fills a slab of width $d$ between two parallel faces; the particle enters perpendicular to the first face. The centre sits on that entry face itself (perpendicular to $\vec v$ means along the face), so the deepest the circle can reach into the slab is exactly one radius. Two futures, decided by comparing $r$ with $d$:

Crossing ($r > d$). The circle pokes through the far face. The depth reached when the particle has turned by $\theta$ is $r\sin\theta$, so the exit condition $r\sin\theta = d$ gives

where: $\theta$ — both the arc angle and the angle between the exit velocity and the original direction (the deviation). Read it physically: a thin slab or a stiff (large-$r$) particle means a small nudge; as $r \to d$ the exit angle sweeps up toward $90^\circ$ and the particle barely escapes, grazing along the far face.

Turning back ($r \le d$). The circle never reaches the far face. The particle completes a half-circle inside the slab and exits through the entry face, moving exactly antiparallel to how it came in, displaced sideways by a full diameter $2r$. Time inside: $T/2$, by the clock rule — independent of speed.

Entry along the face, and the corner exit

Fire the particle along the boundary instead (tangential entry, a JEE favourite). Now the perpendicular-to-$\vec v$ direction points straight into the region, the centre sits a full radius deep, and the circle's far edge reaches a depth of one diameter: the crossing test becomes $2r > d$. The same construction runs the classic "enters one face of a square region, exits the adjacent face" problem: a quarter-circle arc, time $T/4$, entry and exit faces meeting at $90^\circ$.

Oblique to the field: the helix

Everything above lives in the plane perpendicular to $\vec B$. Launch at an angle $\theta$ to the field instead, and Circular & Helical Paths in Magnetic Fields says: the component along the field, $v_\parallel = v\cos\theta$, feels no force and coasts; the component across it, $v_\perp = v\sin\theta$, does all the circling. The path is a helix, and the archetype's job is the bookkeeping:

where: $r$ — the radius of the circular shadow (uses the perpendicular component); $p$ — the pitch, the distance advanced along $\vec B$ per full turn (uses the parallel component). Read the pair as a division of labour: $\sin\theta$ circles, $\cos\theta$ travels. The number of complete turns inside a region of length $L$ along the field is $N = L/p$ — count them like screw threads.

The triage table

Exams love to open with "state the nature of the path". Answer from this table before touching a formula:

Board level 1 — the U-turn in a wide field region (NEET/board pattern)

A proton ($m = 1.67\times10^{-27}$ kg) enters a wide region of uniform magnetic field $B = 0.10$ T perpendicular to the field and perpendicular to the region's boundary face, with speed $v = 2.0\times10^{6}$ m/s. The region is deep enough that the proton cannot cross it. Find (a) the radius of its path, (b) the time it spends inside the field, and (c) how far along the boundary it emerges from where it entered.

Step 1. Radius from the master formula (quoted, not derived — it belongs to Circular & Helical Paths in Magnetic Fields): $r = \dfrac{mv}{qB} = \dfrac{(1.67\times10^{-27})(2.0\times10^{6})}{(1.6\times10^{-19})(0.10)} = \boxed{0.21\ \mathrm{m}}$.

Step 2. Perpendicular entry into a region it cannot cross means a half-circle: arc angle $180^\circ$. Clock rule: $t = \tfrac{1}{2}T = \tfrac{1}{2}\cdot\dfrac{2\pi m}{qB} = \dfrac{\pi (1.67\times10^{-27})}{(1.6\times10^{-19})(0.10)} = \boxed{3.3\times10^{-7}\ \mathrm{s}}$.

Step 3. Exit displacement along the face = one diameter: $2r = \boxed{0.42\ \mathrm{m}}$, moving antiparallel to its entry velocity, at the same speed (magnetic forces never change speed). Sanity: doubling the speed doubles (a) and (c) but leaves (b) untouched — say so and earn the reasoning mark.

Board level 2 — exit angle from a thin slab

An electron ($m = 9.1\times10^{-31}$ kg) moving at $v = 3.2\times10^{7}$ m/s enters a slab of magnetic field $B = 2.0\times10^{-3}$ T of width $d = 4.55$ cm, perpendicular to both the field and the entry face. Show that it crosses the slab, and find the angle by which it is deflected and the time spent inside.

Step 1. Radius: $r = \dfrac{mv}{qB} = \dfrac{(9.1\times10^{-31})(3.2\times10^{7})}{(1.6\times10^{-19})(2.0\times10^{-3})} = 9.1\times10^{-2}$ m $= 9.1$ cm. Since $r = 9.1$ cm $> d = 4.55$ cm, the electron crosses.

Step 2. Exit angle from the chord geometry, equation (2): $\sin\theta = \dfrac{d}{r} = \dfrac{4.55}{9.1} = 0.50 \Rightarrow \theta = \boxed{30^\circ}$.

Step 3. Time inside by the clock rule (1). Period: $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(9.1\times10^{-31})}{(1.6\times10^{-19})(2.0\times10^{-3})} = 1.79\times10^{-8}$ s. Then $t = \dfrac{30}{360}\,T = \dfrac{T}{12} = \boxed{1.5\times10^{-9}\ \mathrm{s}}$. The deflection is $30^\circ$ regardless of the electron's charge sign — the sign only decides which way the path bends.

Competitive level 1 — the helix bookkeeping (JEE-Main pattern)

A proton is launched at $v = 4.0\times10^{5}$ m/s at $60^\circ$ to a uniform field $B = 0.30$ T that fills a region of length $35$ cm along the field direction. Find the radius and pitch of the helical path, and the number of complete turns the proton makes before leaving the region.

Step 1. Split the velocity with respect to the field: $v_\parallel = v\cos 60^\circ = 2.0\times10^{5}$ m/s (coasts along $\vec B$); $v_\perp = v\sin 60^\circ = 3.46\times10^{5}$ m/s (circles).

Step 2. Radius uses only the circling component: $r = \dfrac{m v_\perp}{qB} = \dfrac{(1.67\times10^{-27})(3.46\times10^{5})}{(1.6\times10^{-19})(0.30)} = \boxed{1.2\ \mathrm{cm}}$.

Step 3. Period is component-blind: $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.30)} = 2.19\times10^{-7}$ s.

Step 4. Pitch uses only the coasting component, equation (3): $p = v_\parallel T = (2.0\times10^{5})(2.19\times10^{-7}) = \boxed{4.4\ \mathrm{cm}}$.

Step 5. Turns before exit: $N = \dfrac{L}{p} = \dfrac{35}{4.4} = 8.0$ — the proton screws through $\boxed{8\ \text{complete turns}}$. Timing note: ~2.5 minutes. The only decision is which trig function goes where — $\sin$ to the radius, $\cos$ to the pitch, when the angle is measured from $\vec B$. Mixing them gives pitch $7.6$ cm, an answer the option-setters will have waiting.

Competitive level 2 — accelerate, inject, then double the field (JEE-Advanced stretch)

An electron is accelerated from rest through a potential difference of $910$ V, then enters a slab of magnetic field $B = 2.0\times10^{-3}$ T, width $d = 4.4$ cm, perpendicular to the field and to the entry face. (a) Show it crosses the slab and find the exit angle and time inside. (b) The field is now doubled to $4.0\times10^{-3}$ T. Show the electron turns back, and find the time inside and where it re-emerges.

Step 1. Speed from the accelerating gap (energy bookkeeping, no field yet): $eV = \tfrac12 mv^2 \Rightarrow v = \sqrt{\dfrac{2eV}{m}} = \sqrt{\dfrac{2(1.6\times10^{-19})(910)}{9.1\times10^{-31}}} = 1.79\times10^{7}$ m/s.

Step 2. Radius at $B = 2.0\times10^{-3}$ T: $r = \dfrac{mv}{qB} = \dfrac{(9.1\times10^{-31})(1.79\times10^{7})}{(1.6\times10^{-19})(2.0\times10^{-3})} = 5.09$ cm. Since $r = 5.09 > d = 4.4$ cm: crosses.

Step 3. Exit angle: $\sin\theta = \dfrac{4.4}{5.09} = 0.865 \Rightarrow \theta = \boxed{60^\circ}$ (to the original direction). Time: $T = \dfrac{2\pi m}{qB} = 1.79\times10^{-8}$ s, so $t = \dfrac{60}{360}T = \boxed{3.0\times10^{-9}\ \mathrm{s}}$.

Step 4. Double the field: $r' = r/2 = 2.54$ cm $< 4.4$ cm — the circle no longer reaches the far face: turns back. Half-circle clock at the new, faster period $T' = T/2 = 8.9\times10^{-9}$ s: $t' = \dfrac{T'}{2} = \boxed{4.5\times10^{-9}\ \mathrm{s}}$.

Step 5. It re-emerges from the entry face, displaced $2r' = \boxed{5.1\ \mathrm{cm}}$ along it, moving antiparallel to its injection velocity at the same $1.79\times10^{7}$ m/s — the field did no work throughout. Timing note: ~4 minutes. Two separate reflexes are being tested: the crossing test ($r$ vs $d$) and the fact that doubling $B$ halves both $r$ and $T$. Notice the trap dodged in Step 4 — the turn-back time uses the new period, not the old one.

Radius forgets nothing, the clock forgets speed. $r = mv/qB$ carries the speed; $T = 2\pi m/qB$ does not. Questions that change $v$ and ask for time-inside on a turn-back are testing exactly whether you'll drag $v$ into a formula it doesn't appear in.

Degrees vs radians in the clock rule. $t = (\theta/360^\circ)T$ with degrees, $t = (\theta/2\pi)T$ with radians. Mixing them (dividing degrees by $2\pi$) is a silent factor-of-$57$ disaster. Pick one and label it.

Pitch takes the cosine. When the launch angle $\theta$ is measured from the field direction, $r \propto \sin\theta$ and $p \propto \cos\theta$. Papers sometimes quote the angle from the plane perpendicular to $\vec B$ instead — then the roles swap. Read which angle is given before touching the calculator.

The sign of the charge bends the path, not the physics. Electrons and protons in the same field curve opposite ways (right-hand rule for positive charges; reverse for negative). Every geometric result here — crossing test, exit angle, clock rule — is sign-blind; only the side of the deflection changes.

Kinetic energy is untouchable. The magnetic force is perpendicular to velocity, so speed and kinetic energy are the same at exit as at entry — any option offering an energy change inside a pure magnetic region is wrong by principle, not by arithmetic.

Constants each board expects. $e = 1.6\times10^{-19}$ C, $m_e = 9.1\times10^{-31}$ kg, $m_p = 1.67\times10^{-27}$ kg. JEE/NEET papers state them; AP-2, IB and A-Level provide data booklets — use the paper's values.

"Deep enough" language. Phrases like "the region is sufficiently wide" or "extends far to the right" are the examiner licensing the half-circle case. Don't hunt for a width — it's the $t = T/2$, shift-$2r$ pattern by design.

Texts. NCERT Physics Class 12, Ch. 4 (Moving Charges and Magnetism) · HC Verma, Concepts of Physics Vol. 2, Ch. 34 · SL Arora, Class 12, magnetic-effects chapter (region-entry problem sets).

Provenance. All geometry standard exam lore; every number recomputed for this edition. A verifier should recheck: Board 1 — $r = 0.21$ m, $t = 3.3\times10^{-7}$ s, shift $0.42$ m; Board 2 — $r = 9.1$ cm, $\theta = 30^\circ$, $t = T/12 = 1.5\times10^{-9}$ s; Competitive 1 — $r = 1.2$ cm, $p = 4.4$ cm, $N = 8.0$; Competitive 2 — $v = 1.79\times10^{7}$ m/s, $r = 5.09$ cm, $\theta = 60^\circ$ ($\sin\theta = 0.865$), $t = 3.0$ ns; doubled field: $r' = 2.54$ cm, $t' = 4.5$ ns, shift $5.1$ cm.

Prerequisites. Circular & Helical Paths in Magnetic Fields · Crossed Fields & the Velocity Selector. Leads to. Cyclotron & Mass-Spectrometer Numericals (this gym's machine hall) · Direction Triage (technique, coming up in this chapter) · Cyclotron Motion (UG deep dive).