Question

Uniform electric and magnetic fields with strength E and induction B, respectively, are along $y$ -axis as shown in the figure. A particle with specific charge $q/m$ leaves the origin 0 in the direction of $x$ -axis with an initial non relativistic velocity $v_0.$ The coordinate $y_n$ of the particle when it crosses the $y$ -axis for the $n^{\text{th}}$ time is

• A. $\frac{2{\pi }^{2}m{n}^{2}E}{q{B}^2}$
• B. $\frac{{\pi }^{2}m{n}^{2}E}{q{B}^2}$
• C. $\frac{{\pi }^{2}m{n}^{2}E}{3q{B}^2}$
• D. $\frac{\sqrt{3}p{i}^{2}m{n}^{2}E}{q{B}^2}$

Answer 1

The correct option is A $\frac{2{\pi }^{2}m{n}^{2}E}{q{B}^2}$

We know, from given

$F_y=qE$

$a_y=\frac{qE}{m}$

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

given $t=nT=\frac{2\pi nm}{qB}$

Now,
$y=y_0+u_{y}t+\frac{1}{2}a{t}^2$

$y=\underset{0}{\underset{}{y_0+u_{y}t}}+\frac{1}{2}a{t}^2$

$y=\frac{1}{2}\times \frac{qE}{m}\cdot {\left(\frac{2\pi nm}{qB}\right)}^2$

$=\frac{2{\pi }^{2}m{n}^{2}E}{q{B}^2}$

Watch the next video for detailed solution.