Question
A positron (q = +e) moves at 5.00 × 10^7 m/s in a magnetic field of magnitude 0.850 T. The magnetic force on the positron has a magnitude of 3.20 × 10^−12 N.
What is the component of the positron’s velocity perpendicular to the magnetic field?
What is the component of the positron’s velocity perpendicular to the magnetic field?
Answers
Madyson
I found the answer! I will post it here for anyone else who may also have issues figuring it out.
V(perpendicular) = F/qB
V(perpendicular) = (3.20 x 10^-12)/ [(1.602 x 10^-19)(0.85)]
V(perpendicular) = 2.35 x 10^7 m/s
V(perpendicular) = F/qB
V(perpendicular) = (3.20 x 10^-12)/ [(1.602 x 10^-19)(0.85)]
V(perpendicular) = 2.35 x 10^7 m/s