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Answer to Question #3822 in Mechanics | Relativity for mia

Question #3822
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?
Expert's answer
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

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