The basic physcs here is the Lorentz force on a moving charge, $q$, with velocity, $\vec{v}$ due to a magnetic field $\vec{B}$.
$$ F = q \left( \vec{E} + \vec{v} \times \vec{B} \right) \quad ,$$
where we ignore the electrical field so we get
$$ \vec{F}_B = q \vec{v} \times \vec{B}\quad .$$
The cross-product implies that the force acts perpendicularly to both field and velocity which means that the general path is (locally) helix with it's axis along the direction of the magnetic field.
The accelration is $\vec{a}_B = \frac{\vec{F}_B}{m} = \frac{q v_\perp B}{m}$ which implies a radius of curvature of in the plane normal to the magnetic field. Noting that this is a centripital force we use $a = \frac{v^2_{\perp}}{r}$ to find the radius of curvature as $$r_\perp = \frac{v^2_{\perp}}{a} = \frac{m v_\perp}{qB} = \frac{p_\perp}{qB} \quad .$$
Despite the classical derivation the final form involving momentum, charge and field is relativisitcally correct.
Momentum parallel to the field is unaffected.
Now, saying that the path is a helix is only correct for uniform fields which is not true over scales a significant fraction of the Earth's radius, but it is a good approximation of scales of a hundred kilometers or less, which suggests a reasonable step size for a low precision simulation.
Moving beyond the physics you are asking about the the desired result, these particles never lose any energy due to the effects of magnetic fields and are just pointed in different direction. The result is to first order no change in the flux ariving at objects in orbit.
There are some places where that first order result is insuffient (the flux can actually get amplified near the magnetic poles where particles are revesed and could pass the target twice), but it is a place to start.
Note that I put the charge in the wrong place in the comment I dashed off earlier and have used moderator superpower to fix it post facto.
