A proton moves with a speed of \( 5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.4 \, \text{T}
**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. Radius r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 5 × 10⁶/1.6 × 10⁻¹⁹ × 0.4) = (8.35 × 10⁻²¹/6.4 × 10⁻²⁰) = 1.3047 × 10⁻¹ m = 13.05 cm . Using F = q v B sinθ,
Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion