§ 15. Lorentz Force & 3D Cyclotron Helix (Grade 11)
Helical trajectory of a charged particle in a uniform magnetic field B. Because the magnetic Lorentz force is always perpendicular to a particle's velocity, it does no mechanical work, but curves the trajectory into a circle or 3D helix.
Theory Summary & Scientific Mechanism
A point charge q moving at velocity v through a uniform magnetic field B experiences the magnetic Lorentz force F_L = q[v × B]. Because the cross product is always orthogonal to v, the mechanical power delivered by the magnetic field is identically zero (P = F_L · v = 0), leaving the particle’s speed v and kinetic energy strictly constant!
Decomposing velocity relative to B at pitch angle α yields uniform drift along the field lines at v_∥ = vcos α = const and circular motion in the transverse plane at v_⊥ = vsin α. Equating centripetal force to Lorentz force gives (mv_⊥²)/(R) = |q|v_⊥ B, defining the Larmor (cyclotron) radius R = (mvsin α)/(|q|B).
Combining circular gyration with parallel drift traces a 3D helix. the cyclotron period T = (2π m)/(|q|B) and frequency ω_c = (|q|B)/(m) are independent of particle speed, with helical pitch h = v_∥ T = (2π m vcos α)/(|q|B). This governs cyclotrons, mass spectrometers, and magnetic plasma confinement.
Everyday Analogy
Earth’s magnetic field acts as an invisible shield: it twists charged solar-wind particles into spirals toward the poles, creating the glowing aurora! A magnetic field steers a particle sideways without ever changing its speed.
Interactive Question & Derivation
A charged particle enters a uniform magnetic field B at an acute pitch angle α to the field lines. How do the particle's kinetic energy Eₖ and trajectory shape change if the magnetic field strength B is doubled?
- A: Kinetic energy doubles and the particle flies along a straight line
- B: Kinetic energy Eₖ remains unchanged (since F_L ⊥ v and work A = 0), while the helix radius R decreases by half
- C: The particle rapidly stops due to magnetic braking
Study Hint: The Lorentz force F_L = q[v × B] is always perpendicular to velocity v. Can a force perpendicular to motion do work or change kinetic energy?
Proof & Explanation: The magnetic Lorentz force F_L = q[v×B] is everywhere perpendicular to velocity v, so its elementary work is zero (dA = F_L· dr = 0) and speed and kinetic energy Eₖ = const remain strictly constant. The transverse velocity component v_⊥ = vsin α drives circular rotation of radius R = (mvsin α)/(|q|B), while the longitudinal component v_∥ = vcos α carries the particle uniformly along B, tracing a 3D helix.
Key Equations & 3D Simulation Parameters
R = (m v sin α)/(|q| B), h = (2π m v cos α)/(|q| B), A_Lorentz = 0
- Magnetic field B: 0.6 … 3 T (default: 1.5 T)
- Initial particle speed v₀: 2 … 6.5 m/s (default: 4.2 m/s)
- Pitch angle α to B: 25 … 90 ° (default: 68 °)
- Specific charge |q|/m: 0.8 … 3.2 C/kg (default: 1.8 C/kg)