A positively charged particle having charge `q_1 = 1 C` and mass `m_1 =40 gm` is revolving along a circle of radius `R=40 cm` with velocity `v_1 = 5 ms^-1` in a uniform magnetic field with center of circle at origin `O` of a three-dimensional system. At `t=0`, the particle was at (0, 0.4m, 0) and velocity was directed along positive x direction. Another particle having charge `q_2 = 1 C` and mass `m_2 = 10g` moving uniformly parallel to positive z-direction with velocity `v_2 = 40//pi ms^-1` collides with revolving particle at `t=0` and gets stuck to it. Neglecting gravitational force and coulomb force, calculate x-, y- and z-coordinates of the combined particle at `t= pi//40 s`.

Answer

B

Explanation

xy-plane `T=(2pir)/v=(2pi(0.4))/5=8/50pi`
`since T=(2pim)/(Bq) or Tprop m/q`

After collision mass has become `5/4` times and charge two times.
`:. T'=(5/4xx1/2)T=5/8xx8/50i=pi/10s`
Given timet `t=T'/4` i.e. combined mass will complete one quarter circle.
Further `r=P/(Bq)`
or `rprop1/q(asP=constant)`
since, charge has become two times
`:. r'=r/2=0.2m`
At `t=(pi/40)` second, paticle will be at P in xy plane
`:. x=r'=0.2m` ltbrrgt `y=r'=0.2m`
z-coordinate mass of combined body has become 5 times of the colliding particle. Therefore, from conservation of inear momentum, velocity component in z-directini will become `1/5 ` times. Or
`v_z=1/5xx40/pim/s=8/pim/s`
`:. z=v-zt=8/pixxpi/40=0.2m`