Question

A positively charged particle having charge $q_1=1C$ and mass $m_1=40gm$ is revolving along a circle of radius $R=40cm$ with velocity $v_1=5m{s}^{-1}$ 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, 0.4 m, 0) and velocity was directed along positive x direction. Another particle having charge $q_2=1C$ and mass $m_2=10g$ moving uniformly parallel to positive z-direction will velocity $v_2=40/\pi m{s}^{-1}$ 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 $t=\pi /40s$

Answer 1

In three dimensional co-ordinates system, axes are assumed according to right hand screw rule. Consider such a system shown in fig. In this system positive z direction is normal to plane of paper and is directed toward the reader.
The particle is positively charged, centre of its circular path is at origin and when the particle is on positive y-axis, its velocity is directed along positive x-direction, it means that the particle is moving clockwise in fig.
Since the particle is positively charged, therefore, current associated with its motion is also clockwise or along positive x direction at (0, 0.4, 0).
Lorentz's force is toward origin. Therefore, according to Fleming's left hand rule, magnetic induction B must be along positive z-direction.
Radius of circular path followed by this particle is $r_1=0.4m$
But $r_1=\frac{m_1{v}_1}{q_{1}B}\Rightarrow B=\frac{m_1{v}_1}{q_1{r}_1}=0.5T$
Now the second particle collides and gets struck with it. Velocity of combined particle can be found out by applying law of conservation of momentum.
$(m_1+m_2)\overrightarrow{v}=m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2$
$(40\times {10}^{-3}+10\times {10}^{-3})\overrightarrow{v}$$=(40\times {10}^{-3)}\times 5\hat{i}+(10\times {10}^{-3})\frac{40}{\pi }\hat{k}$
$\therefore$ x-direction of velocity of combined particle is $v_x=4m{s}^{-1}$ and z-direction, $v_z=\frac{8}{\pi }m{s}^{-1}$
Due to $v_x$, combined particle tries to move clockwise along a circular path and due to $v_z$ it ties to move uniformly along z-axis.
Therefore, its ultimate path becomes helix.

Radius of this helix is: $R=\frac{(m_1+m2)v_x}{(q_1+q_2)B}=0.2m$

And period of revolution $T=\frac{2\pi (m_1+m_2)}{(q_1+q_2)B}=\frac{\pi }{10}s$

The particle turns by angle $\pi /2$ in time $\pi /4$ s.
$\therefore$ x-coordinate of new position of combined particle $=R=0.2m$
$\to y-coordinate=r_1-R=0.2m$
and $z-coordinate=v_{z}t=0.2m$