Question

An electron is released from the origin at a place where a uniform electric field $E$ and a uniform magnetic field $B$ exist along the negative $y$-axis and the negative $z$-axis respectively. Find the displacement of the electron along the $y$-axis when its velocity becomes perpendicular to the electric field for the first time. (Take $E=20\text{V/m}$ and $B=0.5\text{mT}$, charge of electron $e=1.6\times {10}^{-19}\text{C}$ , Mass of electron $m={10}^{-30}\text{kg}$)

• A. $0.5\text{mm}$
• B. $1\text{mm}$
• C. $2\text{mm}$
• D. $5\text{mm}$

Answer 1

The correct option is B $1\text{mm}$

Let us take axes as shown in the question figure . According to the right-handed system, the $z$-axis is upward in the figure and hence the magnetic field is in $-z$ direction. At any time, the velocity of the electron may be written as

$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$

Similarly the given electric and magnetic fields can be written as

$\overrightarrow{E}=-E\hat{j}$ and $\overrightarrow{B}=-B\hat{k}$

The net force acting on the electron is,

$\overrightarrow{F}=-e(\overrightarrow{E}+(\overrightarrow{u}\times \overrightarrow{B}\left)\right)$

$=eE\hat{j}+eB(u_y\hat{i}-u_x\hat{j})$

Thus, $F_x=e{u}_{y}B$

and $F_y=e(E-u_{x}B)$

The component of the acceleration are,

$a_x=\frac{d{u}_x}{dt}=\frac{eB}{m}{u}_y$ ......(i)

and $a_y=\frac{d{u}_y}{dt}=\frac{e}{m}(E-u_{x}B)$ ......(ii)

Differentiating the eq (i) w.r.t time we get,

$\frac{d^2{u}_y}{d{t}^2}=\frac{eB}{m}(-\frac{d{u}_x}{dt})$

$=-\frac{eB}{m}.\frac{eB}{m}{u}_y$

$=-{\omega }^2{u}_y$

where, $\omega =\frac{eB}{m}$ ........(iii)

This equations is similar to that of equation of a simple harmonic motion.

Thus,

$u_y=A\sin (\omega t+\delta )$ ....(iv)

and hence,

$\frac{d{u}_y}{dt}=A\omega \cos (\omega t+\delta )$ ..(v)

At $t=0,u_y=0$ and

$\frac{d{u}_y}{dt}=\frac{F_y}{m}=\frac{eE}{m}$

From (iv) and (v)

$\delta =0\text{and}A=\frac{eE}{m\omega }=\frac{E}{B}$

Thus, $u_y=\frac{E}{B}\sin \omega t$

The path of the electron will be perpendicular to the y-axis when $u_y=0$.

where , $\sin \omega t=0\text{(or)}$ $\omega t=\pi$

$\Rightarrow t=\frac{\pi }{\omega }=\frac{\pi m}{eB}$

Also, $u_y=\frac{dy}{dt}=\frac{E}{B}\sin \omega t$

Integrating both sides with proper limits we get,

${\int }_0^{y}dy=\frac{E}{B}{\int }_0^t\sin \omega dt$

$y=\frac{E}{B\omega }(1-\cos \omega t)$

At $t=\frac{\pi }{\omega }$

$y=\frac{E}{B\omega }(1-\cos \pi )=\frac{2E}{B\omega }$

Thus, the displacement along the y-axis is, $\frac{2E}{B\omega }$

$\Rightarrow y=\frac{2Em}{e{B}^2}$

Substituting the data given in the question we get,

$y=\frac{2\times 20\times {10}^{-30}}{1.6\times {10}^{-19}\times 0.25\times {10}^{-6}}\approx 1\text{mm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (b) is the correct answer.