Integrate magnetic motion and recover a Hall sign

Write the coupled velocity equations

For B along +z and signed omega = (q/m)B, the Lorentz equations are dvx/dt = omega vy, dvy/dt = −omega vx and dvz/dt = 0. With initial velocity (3,0,2) m/s, vx = 3cos(omega t), vy = −3sin(omega t) and vz = 2. Integrating positions gives x = 3sin(omega t)/omega, y = 3[cos(omega t) − 1]/omega and z = 2t for nonzero omega.

Recover the neutral limit safely

If q/m = 0, use x = 3t, y = 0 and z = 2t directly. Do not divide by zero or draw a fictitious zero-radius orbit. For a charged body, transverse radius is 3/|omega| and period 2pi/|omega|. The zero placeholders in the neutral ledger are explicitly not physical radius or period. A field-parallel velocity creates a helix, not a closed three-dimensional circle.

Prove the energy statement

Magnetic power is q(v cross B) dot v = 0, so speed remains sqrt(13) m/s in this pure magnetic packet. The field changes direction of motion but not kinetic energy. An electric field, radiation or collision would change the account and is excluded here. This conclusion does not mean every macroscopic current-driven device has no source energy transfer.

Derive a Hall equilibrium with explicit polarity

Let conventional current density J point +x and applied B point +z in a single-carrier conductor. Mean carrier drift is vd = J/(nq). The transverse equilibrium is EHall = −vd cross B, so Ey = vd B. If the upper edge is at larger y by width w, Vupper − Vlower = −Ey w. Electron carriers have negative vd, giving negative Ey and a positive upper-minus-lower potential in this stated geometry. A reversal of either J or B reverses the Hall voltage.

Check a signed original Hall packet

Use the supplied carrier density n = 1×10²⁸ per m³, electron charge −e, J = 1×10⁶ A/m², B = 0.5 T and width 0.01 m. Then vd ≈ −6.24151×10⁻⁴ m/s, Ey ≈ −3.12075×10⁻⁴ V/m and Vupper − Vlower ≈ +3.12075 microvolt. A positive-carrier model with the same n and J predicts the opposite sign. Multiple carriers or uncertain contact polarity would require a richer model; this is not a material-identification guarantee.

Worked original case

At q/m = +1 C/kg and B = 2 T, omega = 2 rad/s, transverse radius is 1.5 m and period pi s. Initial acceleration is (0,−6,0) m/s². Reversing q/m reverses circulation while radius, period and speed stay unchanged.

Investigation

Compare negative, neutral and positive q/m. Derive the position functions by integration and check their initial values. Retain the no-work dot product and distinguish transverse from three-dimensional motion. Independently derive the Hall polarity under one reversal and explain the one-carrier assumption.

Changed-case practice

Sources and model boundary

Original teaching checked 31 October 2026 against the current official framework, Unit 12. The displayed magnetic calculations use the stipulated introductory approximation mu0 = 4pi×10⁻⁷ SI, not a claim that this is exact in modern SI. All apparatus and records are virtual, supplied or generated. No real magnet handling, electrical current setup, particle apparatus or material operation is requested. Public feedback and retained reports are practice evidence, not independent mastery or performed laboratory time.

Supplemental primary teaching reference: The Hall effect. The supplied cases, derivations and diagrams here are original; the source’s older exact-mu0 convention and classical spin language are not adopted literally.