Question

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

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

Answer 1

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

Since the particle will accelerate along the axis in which $\overrightarrow{E}$ is applied, it will have a helical path. The circular path is due to the magnetic field.

$T=\frac{2\pi m}{qB}=2\pi$

After $\pi$ seconds, the particle will complete half the circle in the $z-x$ plane.

$r=\frac{mv}{qB}=\frac{1\times 1}{1\times 1}=1$

$X-$coordinate will be $x=0$.
$Z-$coordinate will be $z=2r=2$

Co-ordinate of the particle along $y-$axis: $y=ut+\frac{1}{2}a{t}^2=0t+\frac{1}{2}\frac{qE}{m}{t}^2=\frac{1}{2}\frac{qE}{m}{\pi }^2=\frac{{\pi }^2}{2}$

Hence, the co-ordinate of the particle will be $(0,\frac{{\pi }^2}{2},2)$.