Question

A positively charged particle of charge 1 C and mass 40 g, is revolving along a circle of radius 40 cm with velocity 5m/s in a uniform magnetic field with centre at origin O in x-y plane .At t=0, the particle was at (0,0.4 m,0) and velocity was directed along positive x-direction.Another particle having charge 1 C and mass 10 g moving uniformly parallel to z-direction with velocity $\frac{40}{\pi }m/s$ collides with revolving particle at t=0 and gets stuck with it . Neglecting gravitational force and coulombian force,calculate x,y and z coordinates of the combined particle at $t=\frac{\pi }{40}sec$.

Answer 1

x-y plane : $T=\frac{2\pi r}{v}=\frac{2\pi \left(0.4\right)}{5}=\frac{8}{50}\pi$
Since $T=\frac{2\pi m}{Bq}$ or $T\propto \frac{m}{q}$
After collison mass has become $\frac{5}{4}$ times and charge two times.
$\therefore T^{{}^{\prime }}=(\frac{5}{4}\times \frac{1}{2})T=\frac{5}{8}\times \frac{8}{50}\pi =\frac{\pi }{10}s$
Given time $t=\frac{T^{{}^{\prime }}}{4}$ i.e.,combined mass will complete one quarter circle.
Further $r=\frac{P}{Bq}$ or $r\propto \frac{1}{q}$ (as P=constant)
Since charge has become two times
$\therefore r^{{}^{\prime }}=\frac{r}{2}=0.2m$
At $t=\left(\pi /40\right)$ second particle will be at P in xy-plane.
$\therefore x=r^{{}^{\prime }}=0.2m,y=r^{{}^{\prime }}=0.2m$
Z coordinate: Mass of combined body has become 5 times of the colliding particle .Therefore from conservation of linear momentum, velocity component in Z-direction will become $\frac{1}{5}$ times .Or,
$v_z=\frac{1}{5}\times \frac{40}{\pi }m/s=\frac{8}{\pi }m/s$
$\therefore z=v_{z}t=\frac{8}{\pi }\times \frac{\pi }{40}=0.2m$