Question

A particle of charge q--8.2 HC has a velocity of 555 m/s that lies in the x-y plane and makes an angle of 65° with respect to the x axis as shown in the figure below. If there is a constant magnetic field of magnitude 1.7 T parallel to +y, what are the magnitude and direction of the magnetic force on the particle? magnitude 003 direction direction 65
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Answer #1


q = -8.2 uC

V = 555 m/s

theta = 65 deg

B = 1.7 T

Force is F = q*v*B*cos(theta)

F = 8.2*10^-6*555*1.7*cos(65)

F = 3.26*10^-3 N

direction is in the -z-direction

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