So there's a rectangular region in the xy plane subjected to a magnetic field B with strength 0.06T pointing in the -z. A proton moving in the xy plane enters the region with velocity v such that the velocity components are initially vx = v*cos(24), vy = v*sin(24). It leaves the rectangular region at a point 0.4m above (as in yf = yi + 0.4m, xf = xi) its entry with the same vy and opposite vx. What is v?
F = qv x B, qvB = mv^2/r
The Attempt at a Solution
I figured I could just use F = qb x B to find the force in the y direction and then divided by the mass and integrated twice and set it equal to 0.4m, then solve for v, like
F/m = a = qvBcos(∅)/m
x = 0.4 = ∫∫vqBcos(∅)/m d∅^2 + ∫vsin(θ)d∅ for ∅ = 24 to 154
then solve for v. The first integration is due to the B field and the second is just integrating the initial vy.
But I'm off by more than a couple orders of magnitude....
Any help would be appreciated! I suspect that my way isn't the easiest way, but if anyone could explain the flaw in my approach anyway that would be awesome.
Thanks for the help!