1. Magnetic force direction in each region

For a positive charge, $\vec F = q\,\vec v \times \vec B$ with $\vec v = v_0\hat i.$

• In the region $a<x<2a$, $\vec B = B_0\hat j.$ Hence, $\vec F = q(v_0\hat i)\times (B_0\hat j)=qv_0B_0\hat k.$ So the force is along $+\hat k$.

• In the region $2a<x<3a$, $\vec B = -B_0\hat j.$ Hence, $\vec F = q(v_0\hat i)\times (-B_0\hat j)=-qv_0B_0\hat k.$ So the force is along $-\hat k$.

Thus, the particle bends in opposite senses in the two regions.

2. Nature of motion in a uniform magnetic field

Since $\vec v \perp \vec B$, the speed remains constant and the particle moves in a circle of radius $R=\frac{mv_0}{qB_0}.$

So in each slab, the trajectory is a circular arc of the same radius $R$, but with opposite curvature because the magnetic field reverses direction.

3. Entry into the first region

The charge enters at $x=a$ with velocity along $+\hat i$.

Since the magnetic force in the first region is along $+\hat k$, the path initially curves toward $+z$. So from $x=a$ to $x=2a$, the trajectory must be an arc bending upward (toward $+z$).

4. Crossing into the second region

At $x=2a$, the magnetic field reverses. Therefore the curvature reverses as well. So after entering $2a<x<3a$, the path must bend toward $-z$.

Hence the overall path is made of two circular arcs of equal radius, joined smoothly at $x=2a$, with opposite curvatures:

• first arc concave upward,
• second arc concave downward.

This gives an S-like trajectory in the $x$-$z$ plane.

5. Checking the options

The correct figure must satisfy all of the following:

1. The motion stays in the plane perpendicular to $\hat j$, i.e. the $x$-$z$ plane.
2. In $a<x<2a$, it bends toward $+z$.
3. In $2a<x<3a$, it bends toward $-z$.
4. The radius of curvature is the same in both regions.

Among the given options, this corresponds to Option A.

6. Comparison with stored answer

Derived answer: A
Stored correct answer: A

They agree.