Answer to Question #3822 in Mechanics | Relativity for mia

Question

A stream of doubly-ionized particles (missing two electrons and thus carrying a net charge
of two elementary charges) moves at a velocity of 3.0 × 104 m/s perpendicular to a
magnetic field of 9.0 × 10–2 T. What is the magnitude of the force acting on each ion?

Answer 1

We know that the Lorentz force is
F = q [v × B], where q is the charge of the particle 2e = 2 * 1.6*10 ^-19 C = 3.2*10 ^-19 C , [v × B] - is the vector multiplication of the velocity and the magnetic field, v is perpendicular to B.
F = 3.2*10 ^-19 * 3.0*10⁴ * 9.0 *10^–2 = 8.64 x 10^–16