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An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5\, R)$ respectively, in a uniform magnetic field ${\rm{\vec B = }}{{\rm{B}}_0}{\rm{\hat i}}$ , each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles ?
Solution
If radius of the circle made from combination of electron and positron pair then if this is more than $2 \mathrm{R}$, than these two circle does not intersect with each other.
Magnetic field $\overrightarrow{\mathrm{B}}$ is in the direction of $x$, so momentum of circular system is in plane $y z$. Both particles are opposite to each other from orbital radius $\mathrm{R}$.
Let $\mathrm{Pl}$ and $\mathrm{P}_{2}$ are momentum of electron and positron.
Suppose $\mathrm{P}_{1}$ form $\theta$ angle with $y$-axis and $\mathrm{P}_{2}$ also form same angle.
Centre of respective circle are at origin position and perpendicular to $\mathrm{R}$ distance.
Centre point of path of electron-positron circle as shown by $C_{e}$ and $C_{p}$.
Coordinate of $C_{e}(0,-R \sin \theta, R \cos \theta)$
co-ordinate of $C_{p}\left(0,-R \sin \theta, \frac{3}{2} R-R \cos \theta\right)$
If distance between centre point to radius for circle is greater than $2 \mathrm{R}$ than circle doesn't inter sect with each other.
Suppose, distance between $\mathrm{C}_{\mathrm{p}}$ and $\mathrm{C}_{\mathrm{e}}$ be $d$.
$\therefore d^{2}=(2 \mathrm{R} \sin \theta)^{2}+\left(\frac{3}{2} \mathrm{R}-2 \mathrm{R} \cos 0^{\circ}\right)^{2}$
$d^{2}=4 \mathrm{R}^{2} \sin ^{2} \theta+\frac{9}{4} \mathrm{R}^{2}-6 \mathrm{R}^{2} \cos \theta+4 \mathrm{R}^{2} \cos ^{2} \theta$
$=4 \mathrm{R}^{2}+\frac{9}{4} \mathrm{R}^{2}-6 \mathrm{R}^{2} \cos \theta$
