A magnetic field \(B^{-}=\) Bo \(j \wedge\) exists in the region \(a

The Lorentz force acting on a particle with charge q and velocity \(\textbf{v}\) in a magnetic field \(\textbf{B}\) is given by the vector formula: \(\textbf{F} = q(\textbf{v} \times \textbf{B})\)

In the first region, with \(a < x < 2a\) and \(\textbf{B} = B_0j\): \(\textbf{F}_1 = q(\textbf{V}_0 \times B_0j) = q(V_0 \times B_0)k\) In the second region, with \(2a < x < 3a\) and \(\textbf{B} = - B_0j\): \(\textbf{F}_2 = q(\textbf{V}_0 \times (-B_0j)) = -q(V_0 \times B_0)k\) The force is in the z-direction in both regions, and the charge is positive, thus causing a force in a circular motion.

Starting with the first region, we can write the equation of motion using Newton's second law: \(m\frac{d^2 \textbf{r}}{dt^2} = q(V_0 \times B_0)k\) Next, we integrate the equation to find the position vector as a function of time: \(\int \frac{d^2\textbf{r}}{dt^2} dt = \frac{d\textbf{r}}{dt} = q(V_0\times B_0)t + C_1\) Integrating once more: \(\int \frac{d\textbf{r}}{dt} dt = \textbf{r} = \frac{q(V_0\times B_0)}{2}t^2 + C_1t + C_2\) Then, do the same for the second region with the negative force: \(m\frac{d^2 \textbf{r}}{dt^2} = -q(V_0 \times B_0)k\) Repeating the integration steps, we get the relation for the position vector of the second region: \(\textbf{r} = -\frac{q(V_0\times B_0)}{2}t^2 + C_3t + C_4\)

From the above equations, we can see that the motion is primarily in a circular path since the force is perpendicular to the velocity. In the first region, the path is a clockwise circle due to the positive force. In the second region, the path is an anti-clockwise circle due to the negative force. Therefore, the trajectory of the charge in the regions will be circular paths with opposite directions, making a figure-eight pattern.