Two particles, each having a mass `m` are placed at a separation `d` in a uniform magnetic field `B` as shown in the figure. They have opposite charges of equal magnitude `q`. At time `t = 0`, the particles are projected towards each other, each with a speed `v`. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value `v_(m)` of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if `v = v_(m) //2` ? ( c) At what instant will a collision occur between the particles if `v =2 v_(m)` ? (d) Suppose `v = 2 v_(m)` and the collision between the particles is completely inelastic. Describe the motion after the collision.

Explanation

(a) `R_(max) = d//2, v = v_(m)`
`(d)/(2) = (mv_(m))/(Bq) rArr v_(m) = (Bqd)/(2m)`
(b) `v = v_(m)//2` i.e. `R = d//4`

Minimum separation `= d//2`
Maximum separation `= 3d//2`
( c) `v = 2v_(m)` i.e. `R = d`. The particle will collide in middle

`sin theta = (d//2)/(R = d) = (1)/(2) rArr theta = pi//6`
The particles will collide after time
`t = (m theta)/(Bq) = (m pi)/(6 Bq)`
(d) Before collision :

After collision:

Momentum conservation:
`x: v_(x) = 0`
`y: m . 2v_(m) sin theta + m . 2v_(m) sin theta = (m + m) v_(y)`
`q` will cancel `-q`, net charge `= 0`
The combined mass will move in a straight line with speed `v_(m)`.