In a vacuum chamber, a particle of charge 1 μC and mass 1 mg is projected with a velocity $(\hat{i}+2\hat{j}){\text{ms}}^{-1}$ from the XZ plane at time t = 0 in an electric field of $1\hat{i}{\text{Vm}}^{-1}$. At t = 0.2 s, the electric field is switched off and a magnetic field of $6\hat{j}\text{T}$ is switched on. The acceleration due to gravity is $-10\hat{j}{\text{ms}}^{-2}$. Correct option(s) is/are:

The correct options are:
1. The vertical distance of the particle from the XZ plane at t = 0.3 s is 15 cm.
2. The radius of the trajectory of the particle for t > 0.2 s is 20 cm.

Step-by-step Solution:

1. Given Data:
Charge of the particle, $q=1\mu \text{C}={10}^{-6}\text{C}$
Mass of the particle, $m=1\text{mg}={10}^{-6}\text{kg}$
Initial velocity at $t=0$, ${\overrightarrow{v}}_0=\hat{i}+2\hat{j}{\text{ms}}^{-1}$
Acceleration due to gravity, $\overrightarrow{g}=-10\hat{j}{\text{ms}}^{-2}$

2. Motion during Phase 1 ($0\le t\le 0.2\text{s}$):
Electric field applied: $\overrightarrow{E}=1\hat{i}{\text{Vm}}^{-1}$
The electrostatic force is along the x-direction:

${\overrightarrow{F}}_e=q\overrightarrow{E}={10}^{-6}\times 1\hat{i}={10}^{-6}\hat{i}\text{N}$

Thus, acceleration due to electric field is:

${\overrightarrow{a}}_e=\frac{{\overrightarrow{F}}_e}{m}=\frac{{10}^{-6}\hat{i}}{{10}^{-6}}=1\hat{i}{\text{ms}}^{-2}$

Total acceleration in Phase 1:

$\overrightarrow{a}={\overrightarrow{a}}_e+\overrightarrow{g}=1\hat{i}-10\hat{j}{\text{ms}}^{-2}$

Position along the Y-axis (vertical distance from XZ plane) at $t=0.2\text{s}$:
Initial y-velocity: $u_y=2{\text{ms}}^{-1}$
Acceleration along y-axis: $a_y=-10{\text{ms}}^{-2}$

$y\left(0.2\right)=u_{y}t+\frac{1}{2}{a}_y{t}^2=2\left(0.2\right)+\frac{1}{2}(-10){\left(0.2\right)}^2=0.4-0.2=0.2\text{m}=20\text{cm}$

Velocity at $t=0.2\text{s}$:

$v_x=u_x+a_{x}t=1+1\left(0.2\right)=1.2{\text{ms}}^{-1}$

$v_y=u_y+a_{y}t=2-10\left(0.2\right)=0{\text{ms}}^{-1}$

So at $t=0.2\text{s}$, the velocity of the particle is $\overrightarrow{v}=1.2\hat{i}{\text{ms}}^{-1}$.

3. Motion during Phase 2 ($t>0.2\text{s}$):
Electric field is turned off, and a magnetic field $\overrightarrow{B}=6\hat{j}\text{T}$ is switched on.
Magnetic Force on the particle:

${\overrightarrow{F}}_m=q(\overrightarrow{v}\times \overrightarrow{B})={10}^{-6}(1.2\hat{i}\times 6\hat{j})=7.2\times {10}^{-6}\hat{k}\text{N}$

This magnetic force acts perpendicular to the velocity ($1.2\hat{i}$) and magnetic field ($6\hat{j}$) in the XZ plane, causing circular motion in the XZ plane.
The radius of this circular trajectory is given by:

$R=\frac{m{v}_{\perp }}{qB}=\frac{{10}^{-6}\times 1.2}{{10}^{-6}\times 6}=0.2\text{m}=20\text{cm}$

Thus, the radius of the trajectory for $t>0.2\text{s}$ is 20 cm.

4. Vertical Distance at $t=0.3\text{s}$:
Since the magnetic force acts purely in the z-direction (in the XZ plane), the only force along the vertical y-axis for $t>0.2\text{s}$ is gravity ($a_y=-10{\text{ms}}^{-2}$).
For the interval $\Delta t=0.3-0.2=0.1\text{s}$:
Initial y-velocity at $t=0.2\text{s}$ is $v_y=0$.

$y\left(0.3\right)=y\left(0.2\right)+v_y\left(\Delta t\right)+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$

$y\left(0.3\right)=0.20+0+\frac{1}{2}(-10){\left(0.1\right)}^2=0.20-0.05=0.15\text{m}=15\text{cm}$

Thus, the vertical distance of the particle from the XZ plane at $t=0.3\text{s}$ is 15 cm.