Question

In a region , uniform electric field is present as $\overrightarrow{E}=E_0\hat{j}$ and a uniform magnetic field is present as $\overrightarrow{B}=-B_0\hat{k}$. An electron is released from rest at origin. Which of the following best represents the path followed by electron after release.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Total force acting on the paricle is
$${\overrightarrow{F}}_T={\overrightarrow{F}}_B+{\overrightarrow{F}}_E=q(\overrightarrow{V}\times \overrightarrow{B})+q\overrightarrow{E}$$Given,
$\overrightarrow{E}=E_0\hat{j}$ and $\overrightarrow{B}=-B_0\hat{k}$

Since, initially the electron $e$ is at rest $\therefore \overrightarrow{V}=0$

$\therefore {\overrightarrow{F}}_B=q(\overrightarrow{V}\times \overrightarrow{B})=0$

So, only ${\overrightarrow{F}}_E=q\overrightarrow{E}=-e{E}_0\hat{j}$ exists

$\therefore e$ starts moving along $-y$ axis (initially)

$\therefore$ Now it has velocity along $-y$ axis

$\therefore {\overrightarrow{F}}_B=q(\overrightarrow{V}\times \overrightarrow{B})=qV{B}_0\left[\right(-\hat{j})\times (-\hat{k}\left)\right]$

$\Rightarrow {\overrightarrow{F}}_B=-eV{B}_0\hat{i}=eV{B}_0(-\hat{i})$

Thus, ${\overrightarrow{F}}_B$ will be acting along $-ve$ $x-$ axis.

$\therefore$ $e$ initially moving in $-y$ axis but then move in $3^{rd}$ quadrant because of both forces.

Hence, option (A) is the correct answer.