(a) What is the magnitude of the magnetic force acting on the particle? (b) Dete

Question

(a) What is the magnitude of the magnetic force acting on the particle?


(b) Determine the magnitude of its charge.

A positively charged particle of mass 6.20 X 10^-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge. x 10^-8 kg is traveling due east with a speed of 20 m/s and enters a 0.35-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 6.40

Explanation / Answer

a)

w = 2*pi/T = 2*pi*/4*6.40*10^-3 = 245.44

F = m*v*w = 6.20*10^-8*20*245.44 = 0.0003043 N = 3.043*10^-4 N

b)

F =q*v*B

q = F/v*B = 0.0003043/20*0.35 = 4.348*10^-5 C

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