1. Force on charged particle in magnetic field

A charged particle entering a uniform magnetic field \vec B perpendicular to its velocity $\vec v$ moves in a circular path because the magnetic force provides the centripetal force:

$qvB = \frac{mv^2}{r}$

So,

$r = \frac{mv}{|q|B}$

Also, the angular speed is

$\omega = \frac{|q|B}{m}$

and the time period is

$T = \frac{2\pi m}{|q|B}$

2. Electron and proton in the field

For the electron and proton:

• They have the same speed $v$.
• Their charges are equal in magnitude: $|q| = e$.
• Their masses are different: $m_p \gg m_e$.

Hence their radii are

$r_e = \frac{m_e v}{eB}, \qquad r_p = \frac{m_p v}{eB}$

Thus,

$r_p > r_e$

So both move on circular arcs, but with different radii.

3. Direction of curvature

Because the charges are opposite:

• electron bends in one direction,
• proton bends in the opposite direction.

Since they enter a semi-infinite region of magnetic field, the boundary is a straight line and the field exists only on one side of that line. A particle entering from the boundary and bending inside will trace a circular arc and will meet the boundary again after completing a semicircle inside the field region.

Therefore, each particle comes out of the field region after traversing a semicircular path.

So:

• Statement A: "They will never come out" is false.

4. Will they come out along parallel paths?

A particle entering normally to the boundary and moving in a circular path inside the field emerges after a semicircle. At the exit point, its velocity becomes opposite to the initial velocity direction.

Both electron and proton start with the same initial velocity direction. After semicircular motion, each emerges with velocity reversed relative to the initial direction.

Thus both emerge moving along straight lines parallel to each other.

So:

• Statement B is true.

5. Do they come out at the same time?

Time spent inside the field is the time for a semicircle:

$t = \frac{T}{2} = \frac{\pi m}{|q|B}$

So,

$t_e = \frac{\pi m_e}{eB}, \qquad t_p = \frac{\pi m_p}{eB}$

Since $m_p \ne m_e$ and in fact $m_p \gg m_e$,

$t_p > t_e$

Hence they come out at different times.

So:

• Statement C is false.
• Statement D is true.

6. Evaluate all options

• A: False
• B: True
• C: False
• D: True

7. Final answer

The correct statements are:

$\boxed{\text{B and D}}$

This matches the stored correct answer.