f = frequency, Ȧ = angular frequency
m = 2.5×10í^7 kg and B = 0.50 T.
What is the angular frequency of
the particle’s motion?
Ȧ = 2.0 rad/s
28.14 - Sample problem: proton in a magnetic field
Variables
What is the strategy?
- Find the speed of the circling proton in terms of the radius and angular velocity of its motion.
- State the equation for the radius of the circular motion of a charged particle in a magnetic field. For the particle’s speed, use the
expression from strategy step 1. - Solve for the magnetic field and evaluate.
This proton is moving in a circular
path in a magnetic field. Its angular
velocity is 2.00×10^7 rad/s.
What is the strength of the magnetic
field?
angular velocity Ȧ = 2.00×10^7 rad/s
speed v
radius of circular path r
magnetic field strength B
mass of proton mp = 1.67×10í^27 kg
charge of proton q = 1.60×10í (^19) C