Question

There exist a uniform magnetic and electric fields of magnitudes $1\text{T}$ and $1\text{V/m}$ respectively along the positive $y-$axis. A charged particle of mass $1\text{kg}$ and charge $1\text{C}$ is having velocity $1\text{m/s}$ along $x-$axis and is at origin at $t=0$. Then, the coordinates of particle at $t=\pi \text{sec}$ will be:

• A. $(0,1,2)\text{m}$
• B. $(0,-{\pi }^2/2,-2)\text{m}$
• C. $(2,-{\pi }^2/2,2)\text{m}$
• D. $(0,{\pi }^2/2,2)\text{m}$

Answer 1

The correct option is D $(0,{\pi }^2/2,2)\text{m}$

Given:

$q=1\text{C};m=1\text{kg};B=1\text{T};E=1\text{V/m};v=1\text{m/s}$


Due to the presence of initial velocity, perpendicular to the magnetic field, the path of the particle will tend to be circular.

Moreover, due to the presence of electric field, it will accelerate along $y-$axis.

Thus, the particle will follow a helical path with increasing pitch, as shown in the above figure.

So, the time period of mentioned circular motion will be

$T=\frac{2\pi m}{qB}=\frac{2\pi \times 1}{1\times 1}=2\pi \text{sec}$

The particle will move in circular path in $(x-z)$ plane.

After $t=\pi \text{sec}=\frac{T}{2}$ time interval, the $x-$coordinate of particle $\left(\text{P}\right)$ will be zero as shown below.

$\therefore x=0$


For linear motion along $y-$direction, applying equation of motion

$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$y=(0\times t)+\frac{1}{2}(\frac{qE}{m})\left({\pi }^2\right)=\frac{1}{2}(\frac{1\times 1}{1})\left({\pi }^2\right)$

$\therefore y=\frac{{\pi }^2}{2}\text{m}$

The $z-$coordinate of the particle will be

$z=2R=\frac{2mv}{qB}=2\text{m}$

Thus, coordinates are $(0,\frac{{\pi }^2}{2},2)\text{m}$.

Therefore, option $\left(D\right)$ is the right choice.