F the velocity of the particle is perpendicular to the direction of the magnetic
ID: 1512398 • Letter: F
Question
F the velocity of the particle is perpendicular to the direction of the magnetic field, what is the force on the particle? (cl9p28) A particle having an electric charge q = 3.20xl0^-19 C is injected into a magnetic field B = 0.10 T with a speed of 1.37x 10^7 m/s. The velocity of the particle is perpendicular to the direction of the magnetic field. What is a magnitude of the magnetic force on the particle? How great an electric field would be required to exert an electrostatic force with the same magnitude as the magnetic force acting on the particle? (V/m)Explanation / Answer
Given :-
A)
q = 3.20 x 10^-19 C
B = 0.10 T
V = 1.37 x 10^7 m/s
F = Bqvsin(theta)
F = 0.10 T x 3.2 x 10^-19 C x 1.37 x 10^7 m/s
F = 0.4384 x 10^-12 N
B)
Electric field, E = F/Q
E = (0.4384 x 10^-12) / (3.2 x10^-19)
E = 1.37 x 10^6 V/m
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