This question involves analyzing the motion of charged particles (electron and proton) entering a uniform magnetic field perpendicular to their velocity. Let's break down the concepts step by step.

Step 1: Force on a Charged Particle in Magnetic Field
When a charged particle moves in a magnetic field, it experiences a magnetic force given by: $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$ where $q$ is the charge, $\overrightarrow{v}$ is the velocity, and $\overrightarrow{B}$ is the magnetic field. Since the field is perpendicular to velocity, the force is perpendicular to both, causing circular motion.

Step 2: Direction of Force and Circular Path
The electron has negative charge ($q<0$), and the proton has positive charge ($q>0$). For the same velocity and magnetic field direction, the force on electron and proton will be in opposite directions due to opposite signs of charge. Thus, they will move in circular paths of opposite sense (e.g., one clockwise, other anticlockwise).

Step 3: Radius of Circular Path
The magnetic force provides the centripetal force for circular motion: $qvB=\frac{m{v}^2}{r}$ Solving for radius $r$: $r=\frac{mv}{qB}$ Mass of proton ($m_p$) is much larger than mass of electron ($m_e$), and charge magnitude is same ($\left|q\right|$). So, radius for proton is larger than for electron.

Step 4: Time to Come Out (if they do)
The time period for one full circle is: $T=\frac{2\pi m}{qB}$ Since the magnetic field region is semi-infinite, the particles will execute semicircular paths and exit. The time to exit is half the period: $t=\frac{T}{2}=\frac{\pi m}{qB}$ This time depends on mass and charge, but not on velocity or radius. Since charge magnitude is same and mass is different, the times are different: proton takes more time to exit than electron.

Step 5: Direction After Exiting
After executing a semicircular path, the velocity vector reverses its component perpendicular to the field but remains same along parallel (if any, but here it's purely perpendicular initially). So, upon exit, the velocity direction is reversed from initial for both particles. Since they entered with same velocity, they exit with same speed but opposite direction to entry. Their paths are parallel (antiparallel actually) to each other after exit.

Conclusion:
- They come out at different times (due to different masses).
- They come out travelling along parallel (actually antiparallel) paths.
- They do come out (since semicircular path in semi-infinite field).

Related Topics and Formulae

Magnetic Force on Moving Charge:
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$ This force is perpendicular to velocity, so it changes direction but not speed.

Circular Motion in Magnetic Field:
Radius: $r=\frac{mv}{qB}$
Time period: $T=\frac{2\pi m}{qB}$

Key Points:
- Opposite charges move in opposite circular directions.
- Time period depends on mass and charge, but not on speed or radius.
- In uniform perpendicular field, path is circular; if field region is limited, particles can exit after arc.