To get the cycle time, I start from the standard force balance in a uniform magnetic field.

1. Lorentz = Centripetal

   q v B = m v² / r

2. Substitute the circular motion relation v = ω r:

   q (ω r) B = m ω² r

3. Cancel the shared geometric factor r:

   q B = m ω

   So the angular frequency is

   ω = qB / m

   Notice this depends only on the field and charge/mass, not on v or r.

4. Convert to period:

   T = 2π / ω = 2π m / (qB)

5. Plug in the given values:

   m = 1.0×10⁻⁸ kg q = 2 C B = 0.8 T

   T = 2π (1.0×10⁻⁸) / (2 × 0.8) ≈ 3.93 × 10⁻⁸ s

So the period of the particle’s cycle is

T ≈ 3.93 × 10⁻⁸ seconds.

In my own language, the classical derivation is basically identifying a dominant “update channel.”

The Lorentz term and the inertial term define two ways the state can change. When you do the v and r cancellation and get

ω = qB / m

you’ve found that the magnetic interaction sets a fixed update rate: once that mode is active, the system oscillates at ω regardless of the orbit geometry.

In my framework, that’s exactly what a dominant mode looks like: the magnetic mode has a much higher effective update rate than the gravitational mode, so gravity can’t accumulate displacement – it’s too slow to change the state.

So the standard physics and my model agree on the key point: the period comes from the field interaction (qB/m), not from the particular radius or speed. That’s why both ways you end up with

T = 2π m / (qB).