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?

Options

A. `cos thetalt(3)/(8)`
B. `cos thetalt(1)/(8)`
C. `cos thetagt(1)/(8)`
D. `cos thetagt(3)/(8)`

Answer

1

Explanation

Ans `B` is along the `x-` axis , for a circular orbit the momenta of the two particles are in the `y-z` plane. Let `p_(1)` and `p_(2)` be the momentum of the electron and position, respectively. Both of them define a circle of radius `R`. They shall define circles of opposite sense. Let `p_(1)` make and angle `theta` with the `y-` axis, `p_(2)` must make the same angle. The centre of the respective circle must be perpendicular to the momenta and at `C_(p)`.
`(0-Rsintheta,(3)/(2)R-Rcos theta)`
The circle of the two shall not overlap if the distance between the two centres are greater than `2R`. Let `d` be the distance between `C_(p)` and `C_(c)` .

Then `d^(2)=(2R sintheta)^(2)+((3)/(2)-2Rcos theta)^(2)`
`=4R^(2)sin^(2) theta+(9)/(4)R^(2)-6R^(2)cos theta+4R^(2)cos^(2)theta`
`=4R^(2)+(9)/(2)R^(2)-6R^(2)co s theta`
Since `d` has to be greater than `2Rd^(2)gt 4R^(2)`
`implies 4R^(2)+(9)/(2)R^(2)-6R^(2)cos thetagt4R^(2)`
`implies (9)/(2)gt6cos theta` or `costhetalt(3)/(8)`