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Homework Statement
A proton with a speed 1.00x10^6 m/s enters a region with a uniform magnetic field of 0.800 T, and points into the page. The proton enters the region at an angle of 60º Find the exit angle and distance d.
Homework Equations
F=qvBsin(theta), F=qvxB
The Attempt at a Solution
I know that the charge of a proton is 1.6x10^-19C so I solved for the magnetic force.
F=(1.6x10^-19C)(1.00x10^6m/s)(.800T)sin(60) =1.109x10^-13N
I don't understand how I can get the exit angle now.