Solution: Crossing Coordinates in Alternating B-Fields
Figure: Schematic of the trajectory projected onto the plane. The particle shifts coordinate with every semi-circle.
Analysis of Motion:
The particle moves in semi-circles. When crossing the plane $z=0$, the magnetic field changes, altering the plane of rotation and the radius.
Step 1: First Interval ($z > 0$)
Initial velocity $\vec{v} = v\hat{k}$. Field $\vec{B} = B_y \hat{j}$.
Force $\vec{F} = q(\vec{v} \times \vec{B}) = q(v\hat{k} \times B_y\hat{j}) = -qvB_y \hat{i}$.
The particle performs a semi-circle in the $x-z$ plane.
Radius $R_1 = \frac{mv}{qB_y}$.
Displacement after semi-circle: $\Delta x = -2R_1$, $\Delta y = 0$.
2nd Crossing Point: $(-2R_1, 0, 0)$. Velocity is now $-v\hat{k}$.
Step 2: Second Interval ($z < 0$)
Velocity entry $\vec{v} = -v\hat{k}$. Field $\vec{B} = B_x \hat{i}$.
Force $\vec{F} = q(-v\hat{k} \times B_x\hat{i}) = -qvB_x \hat{j}$.
The particle performs a semi-circle in the $y-z$ plane.
Radius $R_2 = \frac{mv}{qB_x}$.
Displacement in this interval: $\Delta x = 0$, $\Delta y = -2R_2$.
3rd Crossing Point: $x = -2R_1$, $y = -2R_2$. Velocity is now $+v\hat{k}$.
Step 3: Third Interval ($z > 0$)
The particle enters $z>0$ again with $\vec{v} = v\hat{k}$. This repeats the first type of motion (semi-circle in $x-z$ plane).
Displacement: $\Delta x = -2R_1$, $\Delta y = 0$.
4th Intersection Point:
$$ x_{final} = (-2R_1) + (-2R_1) = -4R_1 $$
$$ y_{final} = -2R_2 $$
Substituting $R_1 = \frac{mv}{qB_y}$ and $R_2 = \frac{mv}{qB_x}$: