Question

A charged particle of specific charge $1000\text{C/kg}$ is projected from the origin of the $x-y$ plane with velocity $2\hat{i}\text{m/s}$ at $t=0$. The plane has uniform magnetic field of $289\sqrt{19}\text{T}$ and electric field of magnitude ${10}^{-3}\text{N/C}$ along the positive $x-$ axis. Find the position of particle at $t=2\text{s}$.

• A. $(2,0,0)\text{m}$
• B. $(2,2,0)\text{m}$
• C. $(4,0,0)\text{m}$
• D. $(6,0,0)\text{m}$

Answer 1

The correct option is D $(6,0,0)\text{m}$

Given:
${\overrightarrow{v}}_0=2\hat{i}\text{m/s};\overrightarrow{E}={10}^{-3}\hat{i}\text{N/C}$
$\overrightarrow{B}=289\sqrt{19}\hat{i}\text{T};\frac{q}{m}=1000\text{C/kg}$

Since, velocity of charged particle and magnetic field are in same direction, so magnetic force on the particle will be zero.

$\therefore F_m=qvB\sin 0^{\circ }=0$

Therefore, charged particle will experience electric force only, which is

$\overrightarrow{F_e}=q\overrightarrow{E}$

So acceleration due to force along $+vex-$direction will be

$a_x=\frac{F_e}{m}=\frac{qE}{m}=1000\times {10}^{-3}=1{\text{m/s}}^2$

Now applying equation of motion along $x-$direction

$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Here, $u=2\text{m/s};t=2\text{s};a_x=1{\text{m/s}}^2$

$x=(2\times 2)+\frac{1}{2}\times 1\times 2^2$

$\therefore x=6\text{m}$

Since the particle is moving in a straight line along the direction of electric field with an acceleration, so the position of the particle at $t=2\text{s}$ will be $(6,0,0)\text{m}$.

Hence, option $\left(d\right)$ is the right choice.