With two different velocities v1 and v2 d(t)= (2m(v1-v2)/eB)sin(eBt/2m) is the simplest form of the distance function. Ah yes, so d(t)=0 when eBt/2m=π, t=2mπ/eB and x= 2πmu/eB
It's not intuitive at all that no matter what the initial relative y velocity component, the two electrons will always...
Well, producing the equations for the helix will not solve this, as the electrons will continue along the helix but not come to a focus.
Assuming magnetic field in the x direction, and v a (very small) initial velocity in the y direction (without loss of generality):
x=ut, y=(mv/eB)sin(eBt/m)...
I thought that a nearly parallel entry path would result in a helix of very small, but constant, radius. I would not expect the electrons to focus at a point, but continue along the infinite helix. What have I missed?