A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x,y) = (0,0) as shown. The magnetic field extends for a distance D = 0.79 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (4.4 X 105 m/s, 2.2 X 105 m/s).
What is the radius of curvature of the proton while it is in the region of the magnetic field?
V*M = Q*B*R
M-Mass Of Particle
Q-Charge Of Particle
R-Radius Of Circle
The Attempt at a Solution
I got what the velocity entering was since the particle does not speed up from the magnetic field. I know the particle moves in a circular path. I am just stuck on how to find the radius of the circle. From there everything else will fall into place.