Two particles, each having a mass m are placed at a separation d in a uniform magnetic field B as shown in they have opposite charges of equal magnitude q. at time t =0, the particles are projectd 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 = 2v_m` ? (d) Suppose `v =2v_m` and the collision between the particles is completely inelastic. Describe the motion after the collision.

Answer

(a)`V_(max)=(qBd)/(2m)` (b)`(pid)/(12 V_(max))=(pim)/(6qB)`
(c)`sqrt3d`

Explanation

Final mass `=3m`
Final charge `=-q`
`R_(f)=(3mV)/(qB)`
By conservation of linear momentum
`2 mV_(m)sin pi/6+4mV_(m)sin pi/6=3mV_(y)`
`4 mV_(m)cos pi/6-2mV_(m)cos pi/6=3mV_(x)`
`V_(x)=V_(m)/sqrt3`
`V=sqrt(V_(x)^(2)+V_(y)^(2))=(2Vm)/sqrt3, R_(f)=(6mV_(m))/(qBsqrt3)=(2sqrt3_(m))/(qB)xx(qBd)/(2m)=sqrt3 d`