A neutral particle is initally at rest in a uniform magnetic field `B` as shown in the diagram. The particle then spontaneously decays into two fragments, one with a positive charge `+q` and mass `3m` and the other with negative charge `-q` and mass `m`. Neglecting the interaction between the two charged particles and assuming that the speeds are much less than speed of light, determine the time (in `mus`) after the decay at which the two fragments first meet. (use the following data `q=1 muC B=2pimuT, m=10^(-15) kg`). Both the charges have initial velocity in `x-y` plane.

Options

A. `250`
B. `500`
C. `600`
D. `750`

Answer

D

Explanation

Initial momentum `(p)` of them all be same but opposite direction.
`R=p/(qB)` = same for both.
Let they meet at angle `theta`, at time `t`. Then
`R theta=v_(0)t.....(i)`
and `R(2pi- theta)=3v_(0)t.....(ii)`
from (i) and (ii), `t=(piR)/(2v_(0))=(pi)/(2v_(0))=(pi)/(2v_(0)) (3mv_(0))/(qB)`
`=(3pim)/(2qB)=(3pixx10^(-15))/(2xx10^(-6)xx2pixx10^(-6))`

`=3/4xx10^(-3)s=750 mus`