1. The problem statement, all variables and given/known data 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. 2. Relevant equations F=qvBsin(theta), F=qvxB 3. 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 dont understand how I can get the exit angle now.