An electron and a position are released from (0, 0, 0) and `(0, 0, 1*5R)` respectively, in a uniform magnetic field `vecB=B_0hati`, each with an equal momentum of magnitude `p=eBR`. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

Explanation

Since, B is along the x-axis, for a circular orbit the momenta of the two particle are in the y-z plane. Let `rho_(1) and rho_(2)` be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let `rho_(1)` make and angle `theta` with the y-axis `rho_(2)` must make the same angle.
The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at `C_(e)` and of the positron at `C_(p)`. The coordinates of `C_(p)` is `C_(p)-=(0,-R sintheta,(3)/(2)R-Rcostheta)`
the circles of the two shall not overlap if the distance between the two centers are greater than `2R`.
Let d be the distance between `C_(p) and C_(e)`
Let d be the distance between `C_(p) and C_(e)`
Then, `d^(2)=(2Rsintheta)^(2)+((3)/(2)R-2Rcostheta)^(2)`
`=4R^(2)sin^(2)theta+(9^(2))/(4)R-6R^(2)costheta=4R^(2)cos^(2)theta` ltBrgt `=4R^(2)+(9)/(4)R^(2)-6R^(2)costheta`
Since, d has to be greater than 2 R
`d^(2) gt 4R^(2)`
`implies4R^(2)+(9)/(4)R^(2)-6R^(2)costheta gt 4R^(2)`
`implies(9)/(4) gt 6costheta`
or, `costheta lt (3)/(8)`