Solution: Range of Velocity for Closed Orbits
Figure: Particle bounces elastically in curved arcs.
Analysis: The particle travels in circular arcs of radius $r = \frac{mv}{qB}$. It bounces elastically off the cylinder wall of radius $R$. For the particle to return to point P, the total angle subtended by the trajectory at the center of the cylinder must be an integer multiple of $2\pi$.
Let $\phi$ be the angle subtended by one “hop” (arc between two collisions). From geometry, the relation between the Larmor radius $r$ and cylinder radius $R$ is: $$ \frac{r}{R} = \tan\left(\frac{\phi}{2}\right) $$
For a closed loop after $n$ collisions and $k$ full rotations around the center: $$ n \phi = k (2\pi) \implies \frac{\phi}{2} = \frac{k}{n} \pi $$ where $k, n$ are integers.
Substituting this back into the radius equation: $$ r = R \tan\left( \frac{k}{n} \pi \right) $$ $$ \frac{mv}{qB} = R \tan\left( \frac{k}{n} \pi \right) $$ $$ v = \frac{qBR}{m} \tan\left( \frac{k}{n} \pi \right) $$
Conclusion: The fraction $\frac{k}{n}$ represents any rational number. Since the set of values $\tan(\mathbb{Q} \pi)$ covers the entire positive real line (rational numbers are dense), there is a valid closed orbit for almost any velocity. Therefore:
(Assuming we can always find integers $k, n$ to approximate the required angle arbitrarily well).