A charged particle +q of mass m is placed at a distanced from another charged particle -2q of mass 2m in a uniform magnetic field B as shown in Fig. 1.28. If the particles are projected towards each other with same speed v,

a. find the maximum value of projected speed `v_m` so that the two particles do not collide.
b. find the time after which collision occurs between the particles if projection speed equals `2v_m.`
c. Assuming the collision to be perfectly inelastic, find the radius of the particle in subsequent motion. (Neglect the electric force between the charges.)

Answer

(a) `v_m<(Bqd)/(2m) l` (b) `t=(pim)/(6qB)` ; (c) `sqrt2d`

Explanation

(a) The particle will move in circular paths, as velocity vector is perpendicular to magnetic field. Time period of both the particle is same `(T=(2pim)/(Bq))`
So, for collision not to take place,
`r_1+r_2ltd`
`(mv)/(Bq)+(2mv)/(2Bq)ltd` or `vlt(Bqd)/(2m)`
Therefore, maximum speed should be `(Bqd)/(2m)`
i.e., `v_mlt(Bqd)/(2m)`
(b) From symmetry, it can be concluded that collision occurs at
`d//2` if
`v=2v_m=(qBd)/m`
`r=(mv)/(qB)=d, sin theta=(d//2)/d=1/2, theta=pi/6`
`t=T( theta/(2pi))=(2pim)/(qB)((pi//6)/(2pi))=(pim)/(6qB)`

(c) After collision, charge on the combined particle `=-q`,
Mass `=3m`
The combined particle will have velocity in y direction just after collision. Using conservation of linear momentum
`(mv cos thetai+mvsin thetaj)+(-2mvcos thetai+2mv sin thetaj)=3mvecv`
`3mvecv=-mv cos thetai+3mvsin thetaj`
`vecv=(-v/(2sqrt3)hati+v/2hatj)`
`|vecv|=vsqrt((1/(4xx3)+1/4))=vsqrt(1/4(4/3))=v/(sqrt3)`
`|v|=1/(sqrt3)xx2xx(qBd)/(2m)=(qBd)/(sqrt3m)`
`:. r=3mxx(qBd)/(sqrt3mxxqB)=sqrt3d`.