An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5R)$ respectively,in a uniform magnetic field $\vec{B} = B_0 \hat{i}$,each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

(D) The radius of the circular path for both the electron and the positron is $R = \frac{P}{eB}$. Since the magnetic field is along the $x$-axis,the motion of both particles is confined to the $yz$-plane.
Let the momentum vectors of the electron and positron make an angle $\theta$ with the $y$-axis in the $yz$-plane. The centers of the circular orbits,$C_e$ and $C_p$,are located at a distance $R$ from the respective starting positions,perpendicular to the momentum vectors.
For the electron starting at $(0, 0, 0)$ with momentum $P_1$ at angle $\theta$ to the $y$-axis,the center $C_e$ is at $(0, -R \sin \theta, R \cos \theta)$.
For the positron starting at $(0, 0, 1.5R)$ with momentum $P_2$ at angle $\theta$ to the $y$-axis,the center $C_p$ is at $(0, -R \sin \theta, 1.5R - R \cos \theta)$.
The orbits will be non-intersecting if the distance $d$ between the centers $C_e$ and $C_p$ is greater than the sum of their radii,i.e.,$d > 2R$.
Calculating the distance squared $d^2$ between $C_e(0, -R \sin \theta, R \cos \theta)$ and $C_p(0, -R \sin \theta, 1.5R - R \cos \theta)$:
$d^2 = (0 - 0)^2 + (-R \sin \theta - (-R \sin \theta))^2 + (1.5R - R \cos \theta - R \cos \theta)^2$
$d^2 = 0 + 0 + (1.5R - 2R \cos \theta)^2$
$d^2 = (1.5R - 2R \cos \theta)^2$
For non-intersecting orbits,$d > 2R$,so $d^2 > 4R^2$:
$(1.5R - 2R \cos \theta)^2 > 4R^2$
$|1.5R - 2R \cos \theta| > 2R$
Case $1$: $1.5R - 2R \cos \theta > 2R \implies -2R \cos \theta > 0.5R \implies \cos \theta < -0.25$
Case $2$: $1.5R - 2R \cos \theta < -2R \implies -2R \cos \theta < -3.5R \implies \cos \theta > 1.75$ (Not possible as $\cos \theta \le 1$)
Thus,the condition for non-intersecting orbits is $\cos \theta < -0.25$.