The radius of circular path of an electron when subjected to a perpendicular magnetic field is

An electron is projected with velocity $\vec v$ in a uniform magnetic field $\vec B$ . The angle $\theta$  between $\vec v$ and $\vec B$  lines between $0^o$ and $\frac{\pi}{2}$ . It velocity $\vec v$ vector returns to its initial  value in time interval of 

When a magnetic field is applied in a direction perpendicular to the direction of cathode rays, then their

A particle having charge of $1 \,\,C$, mass $1 \,\,kg$ and speed $1 \,\,m/s$ enters a uniform magnetic field, having magnetic induction of $1$ $T,$ at an angle $\theta = 30^o$ between velocity vector and magnetic induction. The pitch of its helical path is (in meters) 

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$

Assertion $(A)$ : In an uniform magnetic field, speed and energy remains the same for a moving charged particle.

Reason $(R)$ : Moving charged particle experiences magnetic force perpendicular to its direction of motion.

A proton of mass $1.67\times10^{-27}\, kg$ and charge $1.6\times10^{-19}\, C$ is projected with a speed of $2\times10^6\, m/s$ at an angle of $60^o$ to the $X-$ axis. If a uniform magnetic field of $0.104\, tesla$ is applied along the $Y-$ axis, the path of the proton is