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 and distance d. 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.