A proton of kinetic energy 2.0 eV enters a uniform magnetic field B = (π/2) × 10⁻³ T at 60° to the field. The pitch of the helical path is ______ cm. [mₚ = 1.6 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C]
Physics · Circular Motion
JEE Mains 6 April 2023 Shift 2 Evening Shift

Question

A proton of kinetic energy 2.0 eV enters a uniform magnetic field B = (π/2) × 10⁻³ T at 60° to the field. The pitch of the helical path is ______ cm. [mₚ = 1.6 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C]

Physics · Circular Motion

JEE Mains 6 April 2023 Shift 2 Evening Shift

G1 Physics · Accepted Answer

For a charged particle entering a magnetic field at angle θ, the component parallel to B remains unchanged and produces the pitch. Kinetic energy: K = (1/2)mv² Given K = 2 eV: v = √(2K/m) Parallel component: v∥ = v cos60° = v/2 Cyclotron period: T = 2πm/(qB) Pitch: p = v∥T Using m = 1.6×10⁻²⁷ kg q = 1.6×10⁻¹⁹ C B = (π/2)×10⁻³ T gives p = 0.40 m = 40 cm Final value: 40