A charged particle is launched with a velocity of 5.2 × 10⁴ m/s at an angle of 35° with respect to a 0.0045-T magnetic field. If the magnetic field exerts a force of 0.0026 N on the particle, determine the magnitude of the charge on the particle.

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle is given by F = qvBsin(θ). Solving for q gives q = F / (vBsin(θ)). Substituting the given values: q = 0.0026 N / (5.2 × 10^4 m/s * 0.0045 T * sin(35°)) ≈ 19 µC.

A Charged Particle Is Launched with a Velocity of 5