1. Radius of motion inside the magnetic field

A charged particle entering a uniform magnetic field normally moves along a circular arc of radius

Here the field exists only in the strip

and the particle enters at

with momentum along $+x$.

Since $\vec B$ is into the plane and charge is $+Q$, the magnetic force is upward at entry, so the center of the circular path lies vertically above $P_1$ by distance $r$.

Hence the circle is

So the center is

2. Condition for re-entering region 1

The particle re-enters region 1 if it comes back to the line $x=0$ after entering region 2.

For the circle, intersections with $x=0$ are obtained from

which gives

Thus the second intersection always exists mathematically, and the particle returns to region 1 after tracing the right semicircle provided it does not leave through the boundary $x=\frac{3R}{2}$ before that.

The maximum $x$-coordinate on the circular path is $x=r$.

So:

• if $r<\frac{3R}{2}$, the whole relevant arc lies inside region 2 and it returns to region 1;
• if $r=\frac{3R}{2}$, it just touches the boundary;
• if $r>\frac{3R}{2}$, it exits into region 3 before returning to region 1.

Therefore re-entry into region 1 occurs when

Using $r=\frac{p}{QB}$,

So Option A is correct.

3. Check whether for $B=\frac{8}{13}\frac{p}{QR}$ the particle enters region 3 through the point $P_2$ on the $x$-axis

Given

so

Since

particle indeed reaches the boundary $x=\frac{3R}{2}$ and enters region 3.

Now check whether this happens at the $x$-axis, i.e. at $y=0$.

Equation of path:

At boundary $x=\frac{3R}{2}$,

If crossing point is on $x$-axis, set $y=0$:

This is not the correct sign because from geometry the relevant point on the circle should satisfy

So for $y=0$,

with $x=\frac{3R}{2}$,

Squaring,

This is exactly the obtained radius. Hence the particle crosses into region 3 at

So Option B is correct.

4. Check Option C

The longest possible path in region 2 while still re-entering region 1 occurs for the limiting case

(when the path just touches the boundary $x=\frac{3R}{2}$ before returning).

The farthest point from the $y$-axis is then the rightmost point of the circle:

At this rightmost point, the momentum has turned by $90^\circ$ from its initial direction (initially along $+x$, there it is along $-y$).

So the magnitude of change in momentum is

Not $\frac{p}{\sqrt2}$.

Hence Option C is false.

5. Check Option D

For fixed $B$, same charge $Q$, same velocity $v$:

So $r\propto m$.

If the particle re-enters region 1, the second intersection with $x=0$ is at

Thus distance between entry and re-entry points is

which is directly proportional to mass, not inversely proportional.

Hence Option D is false.

6. Final conclusion

Correct options are:

These match the stored correct answer.