A particle with a charge of 2.80×10 8 C is moving with an instantaneous velocity
ID: 1461954 • Letter: A
Question
A particle with a charge of 2.80×108 C is moving with an instantaneous velocity of magnitude 40.5 km/s in the xy -plane at an angle of 54.0 o counterclockwise from the +x axis.
What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +zdirection? Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force.
What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?
Explanation / Answer
Force,magnetic field and velocity are mutually perpendicular to each other
direction of the force is 180+54 = 234 degrees
F = q*v*B = 2.8*10^-8*2*40.5*1000= 2.26*10^-3 N
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