Q:

Particle beam in infinite magnetic field

In a physics exam at university I had the following problem:

A proton-deuteron beam is accelerated by a $\Delta V = 10^7\,\mathrm{V}$ difference of potential.
  At some point, the particles enter a uniform, infinite, magnetic
  field with $B = 2 T$, perpendicular to the beam's direction. Calculate,
  at the exit of the magnetic field, the distance between protons and
  deuterons.

My approach was that, in the exam's short available time, since the magnetic field is so strong, the forces exerted by the particles were negligible; also that the beam was accelerated before entering the field, so moving with constant speed. So the only force I thought would be significant was the Lorentz force; since this force does no work on the particles, these may only change the direction of their speed, not the module, and so they start rotating in a cycloidal motion along the direction of the beam.
My result so was:
$x_A(t) = \frac{v_0}{\omega} [1-cos(\frac{\omega t}{A})]$
$y_A(t) = \frac{v_0}{\omega} [sin(\frac{\omega t}{A})]$
$z_A(t) = 0$
$\omega = \frac{qB}{m}$
$v_0 = \sqrt{\frac{2q\Delta V}{m}}$
With y being the direction of the beam, z the position along the direction of B, the x axis perpendicular to both, and A the atomic mass.
My professor didn't tell me what was wrong, but implied that it was all wrong, and I have no idea where. I don't want to make wrong assumptions that may bring me even further from the right solution.
What's wrong in my approach, and what is the right solution/way to solve this problem?

A:

It looks like you were on the right track but weren't careful to label the velocities and frequencies with the associated particle species. $v_0$ and $\omega$ have different values for each beam. Additionally, the atomic mass $A$ shouldn't be in your formula for $x_A(t)$ because it's already in $\omega$. These may be the problems your professor saw.
Presumably the infinite magnetic field only occupies the y>0 region. So the beams enter the field at the same point, follow two semicircles of different radii, and leave. You just need to find the distance between the these exit points, and you've already calculated their paths $x_A(t), y_A(t)$. The distance is
$$D = x_p(\pi/\omega_p) - x_d(\pi/\omega_d) = 2\left(\frac{v_p}{\omega_p} - \frac{v_d}{\omega_d}\right) = \frac{2(m_p v_p - m_d v_d)}{qB} $$
It's also possible your professor was expecting you to "just know" the gyroradius formula $R= |\frac{mv}{qB}|$ leading to $D = 2(R_p - R_d)$, and wasn't looking for more detailed answers.

