Question

Uniform electric and magnetic fields with strengths $E$ and $B$ are directed along the $y$-axis. A particle with specific charge $\left(\frac{q}{m}\right)$leaves the origin in the direction of $x$-axis with an initial velocity $V_0$. Find the $y$-coordinate of the particle when it crosses the $y$-axis for $n^{th}$ time.

• A. $\frac{n^{2}mE{\pi }^2}{3q{B}^2}$
• B. $\frac{n^{2}mE{\pi }^2}{2q{B}^2}$
• C. $\frac{n^{2}mE{\pi }^2}{q{B}^2}$
• D. $\frac{2n^{2}mE{\pi }^2}{q{B}^2}$

Answer 1

The correct option is D $\frac{2n^{2}mE{\pi }^2}{q{B}^2}$

According to the problem, the path of the particle will be a helix of increasing pitch.
The axis of the helix is parallel to $y$-axis (parallel to $E$) and the plane of circle of the helix is $xz$ (perpendicular to $B$).

The particle will cross the $y$-axis $nth$ times after time,

$t=nT=n\left(\frac{2\pi m}{Bq}\right)=\frac{2\pi mn}{Bq}$

The y-coordinates of particle at this instant is,

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

Where, $a_y=\frac{F_y}{m}=\frac{qE}{m}$

$\Rightarrow y=\frac{1}{2}\left(\frac{qE}{m}\right){\left(\frac{2\pi mn}{Bq}\right)}^2$

$\therefore y=\frac{2n^{2}mE{\pi }^2}{q{B}^2}$

Hence, option \(d) is the correct answer.