Crossed Fields & the Velocity Selector

In 1897, in a cramped room at the Cavendish Laboratory, J.J. Thomson pointed two invisible hands at a beam of cathode rays — an electric push one way, a magnetic push the other — and tuned them until the beam ran dead straight. That balancing act told him the beam's speed without ever catching a single particle, and it cracked open the atom: the "corpuscles" in the beam turned out to be lighter than anything chemistry had ever weighed. We call them electrons. The instrument he used is the velocity selector, and it is built from nothing but the two force laws you already own. Here is the puzzle it solves. Every real beam of charged particles comes out messy — ions boil off a source at a whole range of speeds, and no accelerator hands you a single clean velocity. Yet the best experiments need particles of exactly one speed. You cannot grab the particles, and you cannot see them. What you can do is exploit a loophole: the electric force $qE$ does not care how fast a charge moves, while the magnetic force $qvB$ is bought entirely by speed. Set the two in opposition and they cancel at exactly one speed, $v = E/B$ — every other particle gets swept into a wall. For your exams this module is a double-scorer: the balance condition is a one-line calculation loved by every board that tests it, and the idea — one force flat, one force climbing — is a reasoning pattern that JEE and NEET reuse in disguise year after year. Master the standoff once and you own the whole family.

Key takeaways

A velocity selector crosses a uniform electric field $E$ and a uniform magnetic field $B$ , both perpendicular to the beam, with their forces on the beam in opposition.
The electric force $qE$ is speed-blind; the magnetic force $qvB$ grows in proportion to speed — so they balance at exactly one speed.
The pass condition is $v = E/B$ . Both the charge and the mass cancel out: the selector sorts by speed and nothing else.
Too fast ( $v$ above $E/B$ ): the magnetic force wins. Too slow: the electric force wins. Either way the particle deflects into the slit wall.
Thomson's $e/m$ measurement is the selector plus one more step: balance the fields to find $v$ , then switch off $E$ and read the circle radius $r = mv/qB$ from Circular & Helical Paths in Magnetic Fields.
KEY IDEA: pit the speed-blind force $qE$ against the speed-proportional force $qvB$ and only one speed survives — $v = E/B$ , with charge and mass cancelling out entirely.

The core insight

In crossed fields arranged to oppose, the speed-blind electric force $qE$ and the speed-proportional magnetic force $qvB$ cancel at exactly one speed, $v = E/B$ — and because $q$ and $m$ both drop out of the condition, the device filters velocity and is blind to everything else.

Where this lives today

The velocity selector never retired; it moved into cleaner rooms. Electron microscopes use crossed-field monochromators to trim their beams to a whisker of energy spread, sharp enough to watch single atoms vibrate. Semiconductor fabs run the same physics at industrial scale: ion implanters filter dopant beams so that only the right species at the right energy reaches the silicon wafer — a selector feeding a mass analyser, exactly the two-stage pattern of the worked example, running around the clock in every chip plant on Earth. And the balance assembles itself inside ordinary matter. Put a current-carrying strip in a magnetic field: the drifting carriers get pushed sideways, pile up along one edge, and build a transverse electric field that grows until $qE$ exactly cancels $qv_dB$ — a velocity selector that constructs its own electric field and then holds itself in balance. The steady voltage that remains across the strip is the Hall voltage, and The Hall Effect (coming up in this chapter) turns it into a tool that measures magnetic fields and counts charge carriers. When you meet it, you will recognise the balance instantly.

The physical picture

Send a positive ion down the gap between two charged plates, with a magnetic field threading the gap at right angles to both the beam and the plate field. The ion now feels two hands on it at once. The electric hand pushes steadily toward the negative plate with force $qE$ — the same grip whether the ion crawls or streaks. The magnetic hand — Force on a Moving Charge: qv×B built this — pushes the opposite way with force $qvB$ : limp for a slow ion, ferocious for a fast one. Run the beam and watch the sorting happen. A slow ion barely wakes the magnetic force, so the electric hand wins and drags it into one plate. A fast ion whips up a magnetic force stronger than the electric one and is flung into the other plate. Only the ion moving at exactly the right speed feels the two hands cancel to nothing — and a particle with zero net force travels in a straight line, threads the exit slit, and joins a beam whose speed you now know without having measured it.

Figure

Crossed fields sort a beam by speed — one speed threads the slit

Why must exactly one speed survive? Plot each force against speed. The electric force is a flat horizontal line — $qE$ , whatever $v$ is. The magnetic force is a straight line climbing from zero — $qvB$ . A flat line and a climbing line cross exactly once, and the crossing point is the selected speed. Everything the device does — including what happens to the too-fast and too-slow particles — is written into that one picture.

Figure

Why exactly one speed survives — the flat force meets the climbing force

One more feature deserves a slow look before the algebra, because it is the most examined sentence in this module: the balance condition contains neither the charge nor the mass. Double the charge and you double both forces — still balanced. A heavier ion feels the same two forces as a lighter one at the same speed. Even a negative charge, which flips both force directions at once, still passes at the same speed. The selector is a pure speed filter, blind to everything else about the particle.

The balance condition

Set the geometry the way every exam draws it: the beam moves horizontally, the electric field points from the positive plate to the negative plate across the gap, and the magnetic field points into the page, perpendicular to both. For a positive charge moving at speed $v$ , the electric force has magnitude $qE$ toward the negative plate, and the right-hand rule from Force on a Moving Charge: qv×B gives a magnetic force of magnitude $qvB$ toward the positive plate — the two pushes are head-on. The particle sails through undeflected when they are equal, and setting them equal is the whole calculation:

qE = qvB \qquad (1)

where: $q$ — the particle's charge (C), $E$ — electric field strength (V/m or N/C), $B$ — magnetic flux density (T), $v$ — the particle's speed (m/s). Read it as a tug of war between a constant pull and a speed-bought pull; the equation asks "at what speed does the motion tax exactly match the flat fee?" Cancel the charge from both sides and the selected speed appears:

v = \frac{E}{B} \qquad (2)

where: the result is a speed, in metres per second — and nothing about the particle survives on the right-hand side. Check the units earn their keep: $E$ in newtons per coulomb divided by $B$ in newtons per (coulomb × metre/second) leaves exactly metres per second. A field of $3.0\times10^{4}$ V/m crossed with $0.20$ T selects $1.5\times10^{5}$ m/s — for protons, electrons, or doubly charged uranium ions alike.

Derivation AThe selected speed and what cancels on the way

We derive the pass condition of the velocity selector: the unique speed at which a charged particle crosses the crossed-field region undeflected. The strategy is a force balance in the direction across the beam — write both transverse forces, set them equal, and solve. The real cargo of the derivation is what cancels: watch both the charge and the mass disappear, because that double cancellation is the entire personality of the instrument, and it is what examiners test.

Step 1. Write the electric force. A charge $q$ between the plates feels $F_E = qE$ , directed from the positive plate toward the negative plate (for positive $q$ ). This force is identical for a parked particle and a speeding one — the field does not ask about motion.

Step 2. Write the magnetic force. The same charge moving at speed $v$ perpendicular to $B$ feels $F_B = qvB$ (the angle between $v$ and $B$ is $90°$ , so $\sin\theta = 1$ ). The right-hand rule, applied to this geometry, points it opposite to the electric force — the fields are arranged so that this happens; that is what "crossed and opposed" means in the device's design. Why the two forces share one line: the beam, $E$ and $B$ are mutually perpendicular, so both pushes land on the same across-the-beam axis. Nothing pushes along the beam — the particle's forward speed is untouched throughout.

Step 3. Demand no deflection. Undeflected means zero net transverse force: $qE = qvB$ . The charge $q$ appears on both sides — cancel it. The mass never appeared at all: no acceleration is being computed, so $m$ has no way into the equation.

Step 4. Solve: $v = E/B$ . A particle at this speed feels zero net force, moves in a straight line by Newton's first law, and exits through the slit. Why a negative charge changes nothing: flipping the sign of $q$ reverses the electric force AND the magnetic force — both at once. Two reversed forces that used to cancel still cancel. Electrons pass at the same setting as protons. *Verified: units of $E/B$ = (N/C)/(N·s/(C·m)) = m/s ✓; with $E = 3.0\times10^{4}$ V/m and $B = 0.20$ T, $v = 1.5\times10^{5}$ m/s, and both $qE$ and $qvB$ evaluate to $4.8\times10^{-15}$ N for a proton — equal, as demanded.*

What about the rejects? A particle at the wrong speed feels a net transverse force of magnitude $q\,(vB - E)$ — positive (magnetic side wins) when it is too fast, negative (electric side wins) when it is too slow. While that net force acts, the particle behaves exactly like the projectile of ARCHETYPE: Charged Particle in a Uniform Field: steady speed along the axis, steady sideways acceleration, a parabolic drift into the slit wall. The selector does not destroy the off-speed particles; it just refuses them the exit.

Thomson's move: from selector to electron

The selector's punchline is that it hands you $v$ — and once you own $v$ , one more measurement buys you the particle's identity. Thomson's 1897 sequence, still reproduced in school CRT experiments, runs in two acts.

Act one — balance. Cross the fields, tune until the beam is straight, and record $v = E/B$ . No mass, no charge needed.

Act two — bend. Switch off the electric field. Now the magnetic force acts alone, and Circular & Helical Paths in Magnetic Fields tells you what happens: the beam curves into a circle of radius $r = mv/qB$ . The radius is visible — a glowing arc on a screen — and measurable. Combine the two acts and the particle confesses its charge-to-mass ratio:

\frac{q}{m} = \frac{v}{Br} = \frac{E}{B^{2}r} \qquad (3)

where: $r$ — the measured radius of the circular arc with only $B$ on; everything on the right is a dial reading or a ruler measurement. Notice the division of labour: the selector pinned down $v$ , the circle converted $v$ into $q/m$ . With cathode rays Thomson found $q/m \approx 1.8\times10^{11}$ C/kg — over a thousand times larger than hydrogen's, meaning either an absurd charge or a mass far below any atom's. He bet on the tiny mass, and the electron entered physics.

Go deeper · Derivation B · 1 minCharge-to-mass ratio from a balance and a bend

Derivation BCharge-to-mass ratio from a balance and a bend

We derive equation (3), the two-step measurement that identified the electron. The goal is to express $q/m$ purely in terms of laboratory dials ( $E$ , $B$ ) and one ruler measurement ( $r$ ). The strategy is to chain the selector's result into the circular-motion result — each step is one line, and the power is in the chaining. Watch how $v$ , which no instrument can read directly, is eliminated between the two equations.

Step 1. Balance act: with both fields on and the beam straight, the selector condition gives $v = E/B$ . This is equation (2), used as a measured value of $v$ .

Step 2. Bend act: with $E$ off, the magnetic force alone steers the beam into a circle. From Circular & Helical Paths in Magnetic Fields (result recalled by name, derived there): $r = \dfrac{mv}{qB}$ , so $\dfrac{q}{m} = \dfrac{v}{Br}$ .

Step 3. Eliminate $v$ : substitute $v = E/B$ into the ratio: $\dfrac{q}{m} = \dfrac{E}{B^{2}r}$ . Every symbol on the right-hand side is now something you can set or measure: plate field, magnet current, arc radius. Why this was genius rather than routine: separately, neither act determines the particle — the balance gives speed but not identity, the bend gives only the combination $mv/q$ . Chained, they corner $q/m$ exactly. What they can never give is $q$ and $m$ separately; that took Millikan's oil drops twelve years later. *Verified: with Thomson-style values $E = 1.5\times10^{4}$ V/m, $B = 5.0\times10^{-4}$ T, the selected speed is $v = 3.0\times10^{7}$ m/s; a measured radius $r = 0.34$ m then gives $q/m = E/(B^{2}r) = 1.76\times10^{11}$ C/kg — the accepted electron value ( $1.76\times10^{11}$ C/kg) on the nose.*

This two-stage pattern — select the speed, then bend to identify — is the front end of every mass spectrometer. The selected beam enters a pure magnetic field, and since every particle arrives with the same $v$ , the radius $r = mv/qB$ now sorts purely by mass-to-charge: heavier ions sweep wider arcs and land farther along the detector. The machine anatomy and its numericals are coming up in ARCHETYPE: Cyclotron and Mass-Spectrometer Numericals; here, own the principle: the selector makes the speed known, so that the bend can make the mass known.

Limiting cases

case	result	meaning
$v = E/B$	zero net force, straight line	the selected particle — threads the slit
$v$ above $E/B$	net force $q(vB - E)$ toward the magnetic side	too fast — magnetic hand wins, particle rejected
$v$ below $E/B$	net force toward the electric side	too slow — electric hand wins, particle rejected
$q \to -q$	both forces flip; balance survives	the selector is charge-blind — electrons pass at the same setting
$B \to 0$	every particle deflects, no speed passes	just a pair of deflecting plates — the CRT geometry of ARCHETYPE: Charged Particle in a Uniform Field
$E \to 0$	pure magnetic steering, circle $r = mv/qB$	Thomson's act two — the identify stage
neutral particle ( $q = 0$ )	no force from either field	sails through at ANY speed — the selector only filters what it can touch

Common trapA velocity selector separates particles by mass

It cannot — and the algebra shouts why. Both $q$ and $m$ cancel out of the pass condition $v = E/B$ : a proton, an electron and a doubly charged ion all sail through together if their speeds match. Exam questions bait this trap constantly ("which ion emerges from the selector?" — answer: every ion at the selected speed, regardless of mass or charge). Mass separation happens after the selector, in a pure magnetic field, where the radius $r = mv/qB$ finally lets mass into the story — with $v$ already pinned by the selector so the radius reads $m/q$ cleanly. Second bite of the same trap: the balance holds at one speed only, never "for the beam". The whole operating principle is that $qvB$ changes with speed while $qE$ does not — say "the forces balance" without saying at $v = E/B$ and you have missed the physics.

Worked examples

Board level — set up the selector (CBSE/ISC pattern, 3 marks)

A beam of protons is to pass undeflected through a velocity selector in which the magnetic field is $0.20\ \mathrm{T}$ , perpendicular to both the beam and the electric field. (a) If the electric field between the plates is $3.0\times10^{4}\ \mathrm{V/m}$ , find the speed of the protons that pass straight through. (b) The plates are $1.0\ \mathrm{cm}$ apart. What voltage is applied across them? (c) Would electrons moving at the same speed also pass undeflected? Justify.

Step 1. Part (a): the balance condition. At the pass speed, electric and magnetic forces are equal: $qE = qvB$ , so $v = \dfrac{E}{B} = \dfrac{3.0\times10^{4}}{0.20} = \boxed{1.5\times10^{5}\ \mathrm{m/s}}$ .

Step 2. Part (b): uniform field between parallel plates (from Equipotentials & the Field–Potential Link, $E = V/d$ ): $V = Ed = (3.0\times10^{4})(1.0\times10^{-2}) = \boxed{300\ \mathrm{V}}$ — an ordinary lab supply.

Step 3. Part (c): yes. Reversing the sign of the charge reverses both the electric force and the magnetic force, so they still cancel. The condition $v = E/B$ contains neither $q$ nor $m$ — the selector is blind to charge and mass. (State the cancellation explicitly; that sentence carries the mark.)

Step 4. Sense check: $1.5\times10^{5}$ m/s is far below light speed and typical of ion-beam work — the numbers are self-consistent.

Competitive level — through the selector, into the bend (JEE-Main pattern, about 3 minutes)

A velocity selector has $E = 6.0\times10^{4}\ \mathrm{V/m}$ and $B = 0.30\ \mathrm{T}$ . Protons that pass through it then enter a region of uniform magnetic field $B' = 0.50\ \mathrm{T}$ (no electric field), perpendicular to their velocity. (a) Find the speed of the transmitted protons. (b) Find the radius of their circular path in the second region. (c) An alpha particle ( $m \approx 6.6\times10^{-27}\ \mathrm{kg}$ , $q = +2e$ ) also passes the selector. What is the ratio of its radius to the proton's in the second region? ( $m_p = 1.67\times10^{-27}\ \mathrm{kg}$ , $e = 1.6\times10^{-19}\ \mathrm{C}$ )

Step 1. Part (a): $v = E/B = \dfrac{6.0\times10^{4}}{0.30} = \boxed{2.0\times10^{5}\ \mathrm{m/s}}$ . Mass and charge irrelevant — the alpha particle in part (c) exits at this same speed.

Step 2. Part (b): in the pure field, $r = \dfrac{mv}{qB'}$ (from Circular & Helical Paths in Magnetic Fields): $r = \dfrac{(1.67\times10^{-27})(2.0\times10^{5})}{(1.6\times10^{-19})(0.50)} = \boxed{4.2\times10^{-3}\ \mathrm{m} \approx 4.2\ \mathrm{mm}}$ .

Step 3. Part (c): same $v$ , same $B'$ , so $r \propto m/q$ . The ratio is $\dfrac{r_\alpha}{r_p} = \dfrac{m_\alpha}{m_p}\cdot\dfrac{q_p}{q_\alpha} = \dfrac{6.6\times10^{-27}}{1.67\times10^{-27}}\cdot\dfrac{1}{2} \approx \boxed{2.0}$ . The alpha sweeps a circle twice as wide — the selector equalized the speeds precisely so that the bend could read $m/q$ . Timing note: three minutes, and part (a) should take fifteen seconds — never re-derive $v = E/B$ in an exam, quote it. The mark-winning insight is spotting that part (c) needs no fresh calculation: at equal $v$ and $B'$ , radii scale as $m/q$ , so one fraction finishes it. If you find yourself computing the alpha's radius from scratch, you are burning a minute the paper gave you for free.

Exam traps

Balancing the wrong pair. The balance is between the electric force $qE$ and the magnetic force $qvB$ — not between " $E$ and $B$ " (dimensionally illegal) and not between the force and the weight ( $mg$ is negligible for beam particles and no board expects it here).

"The selector selects the charge." Both $q$ and $m$ cancel. If the question asks what emerges: every species at speed $E/B$ , including negatives. Any answer that filters by charge or mass loses the mark.

Using volts instead of volts per metre. $E$ in $v = E/B$ is the field (V/m), not the plate voltage. If the paper gives plate voltage $V$ and separation $d$ , convert first: $E = V/d$ . This single slip is the most common numerical error in selector problems.

Geometry not perpendicular. The clean condition $v = E/B$ assumes beam, $E$ and $B$ mutually perpendicular ( $\sin\theta = 1$ in the magnetic force). If a problem tilts anything, go back to force components — but on CBSE/NEET the geometry is always the standard crossed one.

Faster means electric side, slower means magnetic side — reversed. Get the verdicts straight: too fast → magnetic force ( $qvB$ ) exceeds the electric — deflection toward the magnetic-force side; too slow → electric wins. Reading the force-versus-speed graph (flat line, climbing line) reconstructs this in five seconds if you blank.

Scope notes. CBSE/ISC/JEE/NEET: full quantitative treatment, including the $e/m$ chain. IB (SL and HL): motion in perpendicular fields is standard-level content — know the balance and the verdicts. A-Level 9702: "explain how electric and magnetic fields can be used in velocity selection" is an explicit learning objective — a wordy answer (flat force versus climbing force) is expected, not just the formula. AP Physics 2: the named device is not in the CED, but the underlying balance of independent field forces is — treat it as a two-force equilibrium question.

Origins · 1 minJ.J. Thomson and how this was discovered

J.J. Thomson

1856–1940 · Cavendish Laboratory, Cambridge

Joseph John Thomson was a theorist by training who inherited Cambridge's great experimental laboratory — and turned clumsy hands (his students allegedly steered him away from his own apparatus) into the sharpest measurement programme of the age. In 1897 he crossed electric and magnetic fields over a cathode-ray beam: balancing the deflections gave the rays' speed, and unbalancing them gave the charge-to-mass ratio — a thousand times larger than any ion's. His conclusion, that the rays were universal sub-atomic "corpuscles" torn from every element alike, was the discovery of the electron. Sixteen years later the same crossed-field instinct split a beam of neon into two parabolas — the first sighting of stable isotopes. Seven of his research assistants went on to win Nobel Prizes; the man who found the particle that powers electronics never patented a thing.

History

1897

— J.J. Thomson balances electric against magnetic deflection of cathode rays, measures e/m, and announces the electron.

1898

— Wilhelm Wien runs positive ions through crossed fields — the "Wien filter", the selector's formal name to this day.

1909

— Millikan's oil-drop experiment pins down e separately, turning Thomson's e/m into a mass for the electron.

1913

— Thomson's crossed-field apparatus splits neon into mass-20 and mass-22 — stable isotopes revealed.

1919

— Francis Aston, Thomson's student, builds the mass spectrograph and weighs the elements one by one.

Going further

Texts. NCERT Physics Class 12, Ch. 4 (Moving Charges and Magnetism, crossed-fields section) · HC Verma, Concepts of Physics Vol. 2, Ch. 34 · CIE 9702 Topic 20.3 (velocity selection, LO 20.3.6) · IB Physics Guide, Theme D.3.

Provenance. All results are standard. A verifier should recompute: board example $v = 3.0\times10^{4}/0.20 = 1.5\times10^{5}$ m/s, $V = 300$ V; forces on that proton $qE = qvB = 4.8\times10^{-15}$ N; competitive example $v = 2.0\times10^{5}$ m/s, $r_p = 4.2$ mm, $r_\alpha/r_p = (6.6/1.67)/2 \approx 2.0$ ; Thomson chain $v = 1.5\times10^{4}/5.0\times10^{-4} = 3.0\times10^{7}$ m/s, $q/m = E/(B^2r) = 1.76\times10^{11}$ C/kg at $r = 0.34$ m; gravity comparison $mg \approx 1.6\times10^{-26}$ N.

Prerequisites. Force on a Moving Charge: qv×B · Circular & Helical Paths in Magnetic Fields · Equipotentials & the Field–Potential Link (for E = V/d). Leads to. Charged-Particle Trajectories in E and B (coming up in the E4 gym) · Cyclotron and Mass-Spectrometer Numericals (coming up) · The Hall Effect (coming up) · Magnetic Force on Moving Charges (UG deep dive — carries the premium velocity-selector video).

Why does the magnetic force depend on speed when the electric force does not?

Because the magnetic force is fundamentally a force on moving charge — that is its defining property, built into $F = qvB\sin\theta$ from the start (Force on a Moving Charge: qv×B). A parked charge feels the full electric force and zero magnetic force; the faster it moves, the harder the magnetic field grips it. The selector is an engine built out of exactly this asymmetry: one force that ignores the speedometer, one that reads it. At deeper levels of physics the two turn out to be aspects of a single electromagnetic force seen from different frames — a story for the UG tier — but at exam level, "electric: speed-blind, magnetic: speed-proportional" is the complete and correct picture.

If the magnetic force does no work, how does the selector throw particles into the walls?

Deflection is not work. The magnetic force always pushes sideways to the motion, so it changes the direction of the velocity without changing the speed — a rejected ion hits the plate moving exactly as fast as it entered. What the magnetic force does is bend the path until the wall happens to be in the way. (In the rejected particle's flight the electric force also acts and does do a little work, but the point stands: you can be steered into a wall by a force that never speeds you up.) This is the same no-work doctrine that makes speed constant in Circular & Helical Paths in Magnetic Fields.

Do I have to worry about gravity on the particles?

No — and it is worth once seeing why. For the proton of the board example, the electric force is $qE = 4.8\times10^{-15}$ N while its weight is $mg \approx 1.6\times10^{-26}$ N: the field forces win by more than eleven orders of magnitude. Beam particles are so light that gravity is invisible on the length of any selector. No board expects a gravity term in the balance; if a question ever wants weight balanced against a field force, it will be about charged droplets or dust, and it will say so explicitly.

What exactly happens to the too-fast and too-slow particles inside the plates?

They become projectiles. An off-speed particle feels a constant net sideways force $q(vB - E)$ while its forward speed stays untouched, which is precisely the setup of ARCHETYPE: Charged Particle in a Uniform Field: uniform acceleration one way, uniform velocity the other, giving a parabolic path. The farther the speed is from $E/B$ , the harder the net force and the sharper the parabola — badly off-speed particles die early against a plate, near-misses drift out slowly and are caught by the exit slit's edges. That is why real selectors quote a narrow band of transmitted speeds rather than a mathematical point.

Why did Thomson need the selector at all — could he not just deflect the beam and calculate?

A single deflection measurement always contains the unknown speed tangled with the unknown $q/m$ — one equation, two unknowns. The crossed-field balance was Thomson's second, independent equation: it handed him $v$ with no assumptions about the particle. Balance for $v$ , bend for $q/m$ — two acts, two unknowns, problem closed. This "add an independent measurement to break the degeneracy" move is one of the great experimental patterns, and exam questions that feel unsolvable ("too many unknowns!") are often quietly offering you a selector condition as the missing equation.

Is the velocity selector the same thing as a mass spectrometer?

It is the front door, not the machine. The selector's output — a beam of known, uniform speed — is exactly what the mass-analysing stage needs, because the bending radius $r = mv/qB$ only reads mass cleanly if $v$ is the same for every particle. Selector first, analyser second: speed made known, then mass made visible. The full instrument, its dees and detectors and its numericals ladder, is coming up in ARCHETYPE: Cyclotron and Mass-Spectrometer Numericals; what you should carry there from here is the division of labour.

Does the selector work for a beam containing both positive and negative particles?

Yes — and they all pass, which surprises people. For a negative particle both forces reverse together, so the balance at $v = E/B$ is untouched. The two species are deflected to opposite sides only when they are off-speed. If you need to separate positives from negatives, the selector is the wrong tool; use a single field (either one alone deflects them oppositely). The selector's blindness — to charge sign, charge size and mass — is not a weakness: it is the design. It answers exactly one question, "is this particle at speed E/B?", and refuses to be distracted by anything else.