A uniform magnetic field B exists in the region between x=0 and `x=(3R)/2` (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge + Q and momentum p directed along x-axis enters region 2 from region 1 at point `P_(1)(y=-R)`. Which of the following option(s) is/are correct?

Options

A. For `B=8/13, P/(QR)`, the particle will enter region 3 through the `P_(2)` point on x-axis
B. For `B gt 2/3 P/(QR)`, the particle will re-enter region 1
C. For a fixed particles of same charge Q and same velocity v, the distance between the point `P_(1)` and the point of re-entry into region 1 is inversely proportional to the mass of the particle
D. When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point `P_(1)` and the farthest point from y-axis is `P//sqrt(2)`

Answer

A, B

Explanation

For `B=(8p)/(13QR)`
`r=P/(QB)=(p13QR)/(Q8P)=(13R)/8`
Clearly `x=sqrt(((13r)/8)^(2)-((5R)/8)^(2))`
`x=(3R)/2`
Hence particle enters region 3 through point `P_(2)`
(r = radius of circle in which charge moves)
For (B) The particle will reenter region 1 if `r lt (3R)/2`
`p/(QB)lt (3R)/2impliesBgt(2p)/(3QR)`
For (C) Distance between point and point of re-entry into region 1 = 2r
`y=(2mv)/(qB) prop m`
For (D) When particle re-enters region 1 through longest possible path, it will re-enter horizontally only.
Hence at farthest point from y axis, if must be vertical with magnitude of momentum being p. hence change in its linear momentum must be `sqrt(2)p`.