In short

Two long parallel straight wires a distance d apart, carrying currents I_1 and I_2, exert a force per unit length on each other of magnitude

\boxed{\;\frac{F}{L} \;=\; \frac{\mu_0 I_1 I_2}{2\pi d}\;}.

The force is attractive if the currents flow the same way and repulsive if they flow opposite ways.

When a charge q moves through a region of crossed \vec E and \vec B fields (perpendicular to each other and perpendicular to the velocity), the electric and magnetic forces can cancel exactly for one particular speed:

\boxed{\;v \;=\; \frac{E}{B}\;}\qquad\text{(the velocity selector).}

Charges at other speeds are deflected; only charges at this speed pass straight through.

When current flows through a flat conductor immersed in a perpendicular magnetic field, a transverse voltage — the Hall voltage — appears across the conductor:

\boxed{\;V_H \;=\; \frac{IB}{nqt}\;},

where n is the carrier density, q the carrier charge, and t the conductor thickness. The sign of V_H reveals whether the carriers are positive or negative, and its magnitude reveals n. This is the experimental test that settled whether electric current is positive charge flowing one way or negative charge flowing the other (answer: usually negative in metals; positive in hole-doped semiconductors).

Parallel currents, velocity selectors, and the Hall effect all descend from the same microscopic law: \vec F = q\vec v\times\vec B on each moving charge.

A transmission tower near the Tilak Bridge in Delhi carries four huge bundled-conductor cables. In a normal summer afternoon, each cable flows around 1500 A of alternating current at 765 kilovolts. The cables are separated by 40 cm; each bundle hangs in the field of the other three; and every cable feels a steady attractive force — parallel currents attract — of half a newton per metre, pulling toward its neighbours. Over the 400-m span between two towers, that is 200 newtons — the weight of a 20-kg bag of rice — of magnetic tension laid on top of the mechanical tension the cables were already designed for. The engineers who built that tower knew this force existed because Ampère measured it in 1820.

This article is about three scenarios, all driven by the same one-line law \vec F = q\vec v\times\vec B: the force between two parallel current-carrying wires (an old friend from the conductor article but here given a deeper treatment), the velocity selector that uses crossed electric and magnetic fields to sift charges by speed, and the Hall effect — a transverse voltage that tells you what kind of charge carriers are flowing in a metal or a semiconductor and how many of them there are per cubic metre.

The force per unit length, re-derived with care

Two long, straight, parallel wires, a distance d apart. Wire 1 carries current I_1; wire 2 carries I_2. Take them to point in the \hat z direction; separate them along \hat x, with wire 1 at x = 0 and wire 2 at x = d.

Step 1. Compute the magnetic field from wire 1, at the location of wire 2. From the Biot–Savart law, an infinite straight wire carrying current I_1 produces a field that circles around the wire in the direction given by the right-hand rule (thumb along the current, fingers curl in the field direction) and has magnitude

B_1 \;=\; \frac{\mu_0 I_1}{2\pi d}.

At the location of wire 2, with wire 1's current in +\hat z and the separation in +\hat x, the right-hand rule (thumb along +\hat z, fingers curl from +\hat x toward +\hat y) gives the field direction as +\hat y. So \vec B_1 = B_1\hat y at wire 2's location.

Step 2. Compute the force on a length L of wire 2. From the wire-in-a-field formula, \vec F_2 = I_2\vec L_2 \times \vec B_1, where \vec L_2 = L\hat z (wire 2's current is in +\hat z, same as wire 1). Then

\vec F_2 \;=\; I_2 L \hat z \times B_1 \hat y \;=\; I_2 L B_1 (\hat z \times \hat y) \;=\; -I_2 L B_1 \hat x.

Why: \hat z \times \hat y = -\hat x in the standard right-handed orientation. The force on wire 2 points in -\hat x — toward wire 1 (wire 1 is at x=0, wire 2 at x = d > 0). That is attraction, which matches what experiment tells us for parallel currents.

Step 3. Write the magnitude.

\frac{F}{L} \;=\; I_2 B_1 \;=\; \frac{\mu_0 I_1 I_2}{2\pi d}.

Why: divide both sides by L. The factor I_2 B_1 is just the current in wire 2 times the field it sits in. Plugging in B_1 = \mu_0 I_1 / (2\pi d) gives the final form.

Step 4. Antiparallel case: flip the direction of I_2 (current in -\hat z). The cross product gains a minus sign, and the force reverses — the wires repel. Magnitude is the same.

\boxed{\;\frac{F}{L} \;=\; \frac{\mu_0 I_1 I_2}{2\pi d},\qquad\text{attract if parallel, repel if antiparallel}\;}

Newton's third law — the forces are equal and opposite

A consistency check: the force on wire 1 due to wire 2's field must be the same magnitude as the force on wire 2 due to wire 1's field, and opposite in direction. You can verify this by redoing the calculation from wire 1's perspective — the field at wire 1's location from wire 2's current is \mu_0 I_2/(2\pi d), and the force per length on wire 1 is I_1 B_2 = \mu_0 I_1 I_2 / (2\pi d), directed toward wire 2 (for parallel currents). Same magnitude, opposite direction. Newton's third law holds — the magnetic interaction between two current-carrying wires is internally consistent.

(This is actually subtler than it looks. For curved wires, the force on wire 1 from wire 2 is not always equal and opposite to the force on wire 2 from wire 1 when computed piece-by-piece. The net forces on whole wires are equal and opposite, but the piece-by-piece forces can fail Newton's third law because the field propagates at the speed of light and the "action-at-a-distance" picture breaks down. For the infinite-straight-wire case we are doing here, symmetry guarantees Newton's third law works exactly.)

Crossed \vec E and \vec B — the velocity selector

Now a single charge, not a wire. Imagine a beam of charged particles with various speeds, moving in +\hat x. Apply an electric field \vec E = E\hat y (pointing up) and a magnetic field \vec B = B\hat z (out of the page). The forces on a positive charge q moving at velocity \vec v = v\hat x are:

The two forces are opposite. The total is

\vec F \;=\; q(E - vB)\hat y.

For the charge to pass straight through undeflected, the two must cancel: E = vB, or

\boxed{\;v \;=\; \frac{E}{B}\;}.

For any other speed, the net force is non-zero and the charge is deflected sideways. Only charges travelling at the special speed v = E/B continue in a straight line.

Velocity selector — crossed E and B fieldsA horizontal beam enters a region where E points up and B points out of the page. Only particles at speed v equal to E over B pass straight through. Faster particles are deflected down; slower particles are deflected up. − plate + plate $\vec E$ up $\vec B$ out of page selected: $v = E/B$ faster (deflected down) slower (deflected up)
A velocity selector. The electric field $\vec E$ points down (from + plate to − plate); on a positive charge moving right it produces a downward force. The magnetic field $\vec B$ comes out of the page; on the same charge it produces an upward force $qvB$. Only particles at $v = E/B$ have the two forces exactly cancel and pass straight through.

Why the cancellation is independent of the sign of the charge

Flip the sign of the charge: q \to -q. Both \vec F_E = q\vec E and \vec F_B = q\vec v\times\vec B pick up the same sign flip, so they stay antiparallel to each other and still cancel at v = E/B. A velocity selector selects the same speed for positive and negative charges alike — the physics is independent of sign.

Flip the direction of \vec v instead (charge moving backwards through the selector): \vec F_E does not change (it depends only on \vec E, not \vec v) but \vec F_B flips sign, so now both forces point the same way and the charge is deflected, not selected. A velocity selector only works in one direction — ions moving the wrong way are dumped out of the beam.

Flip \vec E and \vec B together: both forces flip. The cancellation survives. What matters is the geometric relation — \vec E, \vec B, and \vec v are mutually perpendicular — and the numerical balance E = vB, not the specific orientations of each vector individually.

Why it matters

Velocity selectors are used whenever you need a monoenergetic beam — a beam in which every particle has the same speed. They are the first stage of every mass spectrometer (covered in the motion-in-fields article), the clean-up step at the exit of accelerators, and the energy filter in electron microscopes. The principle is simple and geometric: a speed-dependent force (qvB) cancels a speed-independent force (qE) at exactly one speed.

The Hall effect — carriers with a side-push

Take a flat strip of conductor carrying current I in the +\hat x direction, and put it in a magnetic field \vec B pointing in +\hat z (out of the page). What happens to the charge carriers?

Hall effect geometryA rectangular slab of conductor. Current flows from left to right along the strip. Magnetic field B points out of the page. Positive carriers drift right and are deflected to the bottom edge; negative carriers drift left and are deflected to the bottom edge as well, giving an opposite Hall voltage sign. $I$ $\vec B$ out of page + $\vec v_d$ $q\vec v\times\vec B$ (down) top: depletes positives bottom: accumulates positives → $V_H > 0$ (top to bottom) thickness $t$
Current flows right through a rectangular conductor in a magnetic field out of the page. A positive carrier drifting in $+\hat x$ feels a force $q\vec v\times\vec B = qv_d B(\hat x\times\hat z) = -qv_dB\hat y$ — downward. Positive carriers pile up on the bottom edge, making it positive; a Hall voltage $V_H$ appears between top (negative) and bottom (positive). For electron-conductors the sign reverses.

Step 1. The carriers start out drifting along +\hat x with drift velocity \vec v_d. The magnetic force on each carrier is

\vec F_B \;=\; q\vec v_d \times \vec B.

For a positive carrier in +\hat x with \vec B in +\hat z: \vec F_B points in -\hat y (downward).

For a negative carrier — same current means the carriers move in -\hat x. Force: (-e)(-\hat x\cdot v_d)\times B\hat z = +evB\hat x \times \hat z = -evB\hat y — downward again!

Why: both positive and negative carriers, despite drifting in opposite directions, feel a force in the same transverse direction. This is a key point: the Hall force pushes carriers to the same edge regardless of sign. What differs is which edge becomes positive.

Step 2. Carriers accumulate at the bottom edge. If the carriers are positive, the bottom edge becomes positively charged and the top edge becomes negatively charged. If the carriers are negative, the bottom edge becomes negatively charged and the top edge positively charged. The sign of the transverse voltage tells you what kind of carrier you have.

Step 3. The accumulated charge produces an electric field across the conductor that pushes back on further carriers. Equilibrium is reached when the electric force balances the magnetic force:

qE_H \;=\; qv_d B \;\Rightarrow\; E_H \;=\; v_d B.

Step 4. The transverse voltage (the Hall voltage) across a conductor of width w is

V_H \;=\; E_H w \;=\; v_d B w.

Step 5. Relate v_d to the current. The current is I = nAqv_d where A = wt is the cross-sectional area (t = thickness, w = width) and n is the carrier density. So

v_d \;=\; \frac{I}{nqwt}.

Step 6. Substitute.

V_H \;=\; \frac{I}{nqwt} \cdot Bw \;=\; \frac{IB}{nqt}.
\boxed{\;V_H \;=\; \frac{IB}{nqt}\;}

Why: the width w cancels — the Hall voltage depends on the thickness t but not on the width. For a thin strip (small t), V_H is large; for a thick bar, V_H is small. This is why commercial Hall sensors use very thin films of semiconductor — nanometres to micrometres — to boost sensitivity.

The ratio R_H = 1/(nq) is called the Hall coefficient. Measuring V_H, I, B, and t on a slab gives you nq directly. The sign of V_H gives you the sign of q; the magnitude gives you n.

What the Hall effect measured historically

Before 1879, nobody knew whether current in a metal was positive charges flowing one way or negative charges flowing the other. Edwin Hall's experiment showed: in most metals, carriers are negative. For some metals (beryllium, zinc, cadmium — called "hole conductors"), carriers behave as if positive. This was decades before J. J. Thomson discovered the electron; the Hall effect was the first hint that conduction might involve negative particles.

In semiconductors, the Hall effect distinguishes between:

This is why every solid-state-physics course teaches the Hall effect: it was the experiment that established that charge carriers in solids can be of either sign, foreshadowing the band-theory understanding of conduction.

Modern applications — the Hall sensor in your phone

Tiny Hall sensors (indium-antimonide or gallium-arsenide thin films, a few hundred micrometres across) are in almost every electronic device you own:

The physics of every one of these is V_H = IB/(nqt). The sensor is typically a n-doped semiconductor with n \sim 10^{22}\,\text{m}^{-3} (not 10^{29} as in copper — semiconductors have far fewer carriers). That low n makes the Hall voltage large and easily measurable.

Explore the Hall voltage

Interactive: Hall voltage versus magnetic field A plot of Hall voltage in millivolts as a function of magnetic field for a typical semiconductor Hall sensor. A draggable point selects a field value and shows the resulting Hall voltage. magnetic field $B$ (T) Hall voltage $V_H$ (mV) 0 5 10 0.2 0.4 0.6 0.8 1.0 $V_H = IB/(nqt)$ with $I = 10$ mA drag the red point
For a Hall sensor with carrier density $n = 10^{22}\,\text{m}^{-3}$, thickness $t = 10^{-5}$ m ($10\,\mu$m), and current $I = 10$ mA, the Hall voltage is $V_H = IB/(nqt) = (0.01)B/(10^{22}\cdot 1.6\times 10^{-19}\cdot 10^{-5}) = 6.25\,B$ V per tesla — that is, 6.25 mV per 10 mT. Earth's field ($\sim 50\,\mu$T) gives $V_H \approx 0.3$ mV — marginal but detectable with amplification. An MRI field of 1 T saturates the sensor at 6.25 V.

Worked examples

Example 1: Current between two wires defining the ampere

Two infinite parallel wires are 1 metre apart in vacuum. Each carries 1 ampere of current in the same direction. Find the force per metre between them, and check the consistency with the historical definition of the ampere.

Two parallel wires, one ampere each, one metre apartTwo wires one metre apart, both carrying one ampere into the page, attracting each other with 2e-7 newtons per metre. 1 A 1 A $F/L = 2\times10^{-7}$ N/m $d = 1$ m
Two wires one metre apart, each carrying one ampere. The mutual attraction is $2\times 10^{-7}$ N per metre of length — the calibration point for the pre-2019 SI ampere.

Step 1. Apply F/L = \mu_0 I_1 I_2 / (2\pi d).

\frac{F}{L} \;=\; \frac{(4\pi\times 10^{-7})(1)(1)}{2\pi(1)}.

Step 2. Simplify.

\frac{F}{L} \;=\; \frac{4\pi}{2\pi} \times 10^{-7} \;=\; 2\times 10^{-7}\;\text{N/m}.

Why: 4\pi / (2\pi) = 2. The \pis cancel perfectly. This is the reason \mu_0 = 4\pi\times 10^{-7} is chosen with the factor of 4\pi — so that the ampere's definition comes out as a clean 2\times 10^{-7} rather than some awkward decimal.

Step 3. Check against the historical definition. The pre-2019 SI definition says: "one ampere is that current which, between two infinite parallel wires one metre apart in vacuum, produces a force of 2\times 10^{-7} newtons per metre." Our calculation reproduces this exactly.

Result. F/L = 2\times 10^{-7} N/m, attractive (same direction of currents).

What this shows. The ampere — until 2019 — was defined by this very force. The number 2\times 10^{-7} was not measured; it was declared to be exact, and \mu_0 was fixed to 4\pi\times 10^{-7} T·m/A to make it so. After the 2019 SI redefinition, the elementary charge e = 1.602176634\times 10^{-19} C is fixed exactly instead, and \mu_0 is measured. But the force-between-wires calculation is still the right physical intuition for "what is an ampere?"

Example 2: Velocity selector for a potassium-ion beam

A velocity selector uses E = 2.0\times 10^4 V/m and B = 0.10 T, with the fields perpendicular to each other and to the beam direction. Potassium ions (^{39}\text{K}^+) are fired through. Find the speed at which the ions pass through undeflected, and the kinetic energy this corresponds to. Mass of K-39: m = 39u = 6.47\times 10^{-26} kg; q = e = 1.6\times 10^{-19} C.

Velocity selector for potassium ionsA horizontal ion beam passes through a selector with E = 20000 V/m up and B = 0.1 T out of page. Only ions with v = 2e5 m/s pass straight. $v = E/B = 2\times 10^5$ m/s $\vec E$ $\vec B$ out
The velocity selector fields: $E$ up at $2\times 10^4$ V/m; $B$ out of the page at 0.10 T. Only ions at $v = 2\times 10^5$ m/s have the two forces on them cancel.

Step 1. Apply v = E/B.

v \;=\; \frac{E}{B} \;=\; \frac{2.0\times 10^4}{0.10} \;=\; 2.0\times 10^5\;\text{m/s}.

Why: division. The selected speed depends only on E and B, not on the ion's mass or charge. A helium ion and a uranium ion would both pass through at the same speed (if their charges are the same sign) — the mass cancels.

Step 2. Compute the kinetic energy.

\text{KE} \;=\; \tfrac{1}{2} m v^2 \;=\; \tfrac{1}{2}(6.47\times 10^{-26})(2.0\times 10^5)^2.
\text{KE} \;=\; \tfrac{1}{2}(6.47\times 10^{-26})(4\times 10^{10}) \;=\; \tfrac{1}{2}(2.59\times 10^{-15}) \;=\; 1.29\times 10^{-15}\;\text{J}.

Step 3. Convert to electron-volts for intuition.

\text{KE} \;=\; \frac{1.29\times 10^{-15}}{1.6\times 10^{-19}} \;=\; 8.08\times 10^3\;\text{eV} \;=\; 8.1\;\text{keV}.

Why: 1\text{ eV} = 1.6\times 10^{-19}\text{ J}. This lets you quote the energy in the units accelerator physicists use.

Result. Ions pass through the selector at v = 2\times 10^5 m/s, equivalent to a kinetic energy of about 8.1 keV for K-39.

What this shows. A velocity selector tuned to v = E/B passes any ion at that speed, regardless of its mass. The corresponding kinetic energy does depend on mass (and charge), so a mass-filter stage downstream separates the ions. This is the standard architecture of a time-of-flight mass spectrometer — first select on velocity, then measure on mass.

Example 3: Hall voltage in a copper strip

A thin copper strip of thickness t = 0.5 mm and width w = 1 cm carries a current of 50 A in a uniform magnetic field of B = 1.2 T perpendicular to the strip. The free-electron density in copper is n = 8.5\times 10^{28}\,\text{m}^{-3}. Find the Hall voltage across the strip, and comment on the sign.

Hall voltage in a copper stripA copper strip 1 cm wide, 0.5 mm thick, carrying 50 A in a 1.2 T field. Hall voltage is about 8.8 microvolts. $I = 50$ A $\vec B = 1.2$ T out top: $-$ bottom: $+$ $w = 1$ cm
Current flows right through the copper strip; field is out of the page. Electrons (the actual carriers) drift left; $\vec F = -e(-\hat v\times\vec B)$ pushes them toward the top, so the top edge becomes *negative* — Hall voltage is bottom-positive.

Step 1. Apply V_H = IB/(nqt).

V_H \;=\; \frac{(50)(1.2)}{(8.5\times 10^{28})(1.6\times 10^{-19})(0.5\times 10^{-3})}.

Step 2. Compute the denominator.

n q t \;=\; (8.5\times 10^{28})(1.6\times 10^{-19})(5\times 10^{-4}).
= 8.5 \times 1.6 \times 5 \times 10^{28-19-4} \;=\; 68 \times 10^{5} \;=\; 6.8\times 10^{6}.

Step 3. Compute the numerator.

IB \;=\; 50 \times 1.2 \;=\; 60.

Step 4. Divide.

V_H \;=\; \frac{60}{6.8\times 10^{6}} \;=\; 8.82\times 10^{-6}\;\text{V} \;=\; 8.8\;\mu\text{V}.

Why: the Hall voltage in copper at everyday currents and modest fields is tiny — microvolts. This is because copper has so many free electrons (n = 8.5\times 10^{28}\,\text{m}^{-3}) that each one only needs a very weak transverse field to balance the magnetic push. This is also why copper is a poor choice for a Hall sensor — semiconductors with n \sim 10^{22} give Hall voltages a million times larger.

Step 5. Sign. The carriers in copper are electrons (q = -e). For current in +\hat x, electrons drift in -\hat x. The magnetic force on an electron moving in -\hat x in a field +\hat z is (-e)(-\hat x v_d)\times B\hat z = ev_d B(\hat x\times\hat z) = -ev_dB\hat y — pointing in -\hat y, i.e., toward the bottom of the strip. So electrons pile up on the bottom edge, making the bottom negative and the top positive.

Result. V_H \approx 8.8\,\muV, with the top edge positive and the bottom edge negative (opposite to what a positive-carrier material would give).

What this shows. The Hall voltage is negative for electron-conductors (metals like copper, n-type semiconductors), positive for hole-conductors (p-type semiconductors). This is how you distinguish them experimentally. Also, the small size of V_H in copper explains why Hall sensors are always made of semiconductors, where n is ten million times smaller and V_H correspondingly larger.

Common confusions

If you came to understand the force between parallel currents, the velocity selector, and the Hall effect — you have a complete working knowledge of this chapter. What follows is the deeper structure: why parallel currents attract from a relativistic viewpoint, the quantum Hall effect and its precise quantisation, and why the Hall effect is the foundation of the modern resistance standard.

Parallel currents attract: the relativistic derivation

The force between two parallel current-carrying wires is "magnetic" in the lab frame, but relativity says that what one observer calls magnetism another can call electricity. Let me sketch the argument.

Take two parallel wires, both carrying current in +\hat x in the lab frame. Each wire contains immobile positive ions at linear density \lambda_+ and mobile negative electrons at linear density \lambda_- drifting at velocity v_d in +\hat x. In the lab frame, the wire is electrically neutral: \lambda_+ + \lambda_- = 0 (with \lambda_- negative).

Now Lorentz-transform into the frame of the electrons of wire 1. In this frame:

  • The electrons of wire 1 are at rest.
  • The ions of wire 1 are moving in -\hat x at speed v_d.
  • The electrons of wire 2 are also moving in... well, if both wires' electrons drift at the same v_d, then in the electron-1 frame, the electrons of wire 2 are at rest.
  • The ions of wire 2 are moving in -\hat x at speed v_d.

Because of relativistic length contraction, moving charge densities appear enhanced by \gamma = 1/\sqrt{1 - v_d^2/c^2}. In the electron-1 frame:

  • Wire 2's ions (moving at -v_d): linear density \gamma\lambda_+.
  • Wire 2's electrons (at rest): linear density \lambda_-.

Wire 2's total charge density in the electron-1 frame is \gamma\lambda_+ + \lambda_-. If \lambda_+ = -\lambda_- (neutral in lab), this is (\gamma - 1)\lambda_+ > 0 — wire 2 is slightly positively charged in the electron-1 frame.

A positively charged wire 2 exerts an electrostatic attraction on the negatively charged electrons of wire 1 (which are at rest in this frame). This is the "magnetic" attraction of the lab frame, re-expressed as electrostatics in a different frame.

When you work out the algebra carefully, the magnitude is exactly what you would compute from the magnetic formula F/L = \mu_0 I_1 I_2 / (2\pi d). The two frames agree. Magnetic force between currents is relativistic electrostatics — not a separate fundamental force, but a kinematic cousin of Coulomb's law.

This is the deepest reason why \mu_0 \epsilon_0 = 1/c^2: the magnetic coupling constant and the electric coupling constant are tied together through the Lorentz factor \gamma, and c sets the conversion between them.

The quantum Hall effect — von Klitzing's discovery

At ultra-low temperatures (milli-kelvin) and in very strong magnetic fields (\sim 10 T), the Hall voltage of a two-dimensional electron gas becomes quantised. The Hall resistance R_H = V_H/I takes the values

R_H \;=\; \frac{h}{e^2}\cdot \frac{1}{N},\qquad N = 1, 2, 3, \ldots

where h is Planck's constant. The constant h/e^2 = 25812.807\,\Omega is called the von Klitzing constant, denoted R_K. It is reproducible to one part in 10^9 — more precisely than any other electrical measurement — and is the basis for the modern SI ohm.

The quantisation is a consequence of Landau-level quantisation combined with the topology of the edges of the two-dimensional electron system. It is robust: the value of R_K is independent of the material, the exact sample geometry, or the disorder — an incredible fact that is only explained by the topological nature of the Hall conductance.

India's National Physical Laboratory in Delhi maintains a quantum Hall resistance standard for calibration; every multimeter in the country's scientific infrastructure is traceable back to this one experiment.

The anomalous Hall effect in magnetic materials

In ferromagnetic metals (iron, cobalt, nickel), even in zero applied magnetic field, a Hall voltage appears in response to the internal magnetisation. This is the anomalous Hall effect (AHE), and it is purely a solid-state-physics effect — the electron band structure in a magnetised material is not symmetric under time reversal, and this asymmetry produces a transverse current even in \vec B = 0.

Modern spintronics uses AHE (and its close cousin, the spin Hall effect) for reading information from magnetic hard drives. The read head of a terabyte hard drive contains a magnetic sensor that is, essentially, measuring an AHE voltage.

The Hall effect as carrier-concentration spectroscopy

Because V_H = IB/(nqt), measuring V_H at known I, B, t tells you n directly. For a semiconductor engineer, this is how you measure doping concentrations. A boron-doped silicon wafer at 0.01 mm thickness, carrying 1 mA of current in a 0.1 T field, gives V_H in millivolts — and V_H is directly proportional to the dopant concentration. Every fab process uses Hall measurements as a quality-control check on doping uniformity.

Why the selector selects velocity, not momentum

The velocity selector's condition v = E/B does not involve the charge or the mass. A 1 keV electron and a 100 keV uranium ion would both be selected at the same speed (if tuned to that speed) — despite having vastly different momenta.

This is in contrast to the mass analyser, which selects on r = mv/(qB) — and therefore on momentum per charge. A velocity selector followed by a mass analyser gives you a selection on (v, m/q) — equivalently on (v, m/q), or on energy and mass-to-charge ratio. That is enough information to do full mass spectrometry.

The reason the selector's condition is velocity (not momentum) is that the electric force is proportional to q (so it scales out of the balance condition) while the magnetic force is proportional to qv (so v remains). The q cancels; the mass never enters. Pure geometry.

Where this leads next