Question

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

• A.
• B.
• C.
• D. None of the above

Answer 1

The correct option is A

${\overrightarrow{F}}_{net}=q\overrightarrow{E}+q\overrightarrow{V}\times \overrightarrow{B}$

Given, $\overrightarrow{E}=$ E${}_0$$\hat{j}$ and $\overrightarrow{B}=$ B${}_0$$\hat{k}$

Since initially electron is at rest $\therefore \overrightarrow{v_0}=0$

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

only $\overrightarrow{F_E}=q\overrightarrow{E}=-eE\hat{j}$

$\therefore$ electron starts moving along negative Y- axis (initially)

Now it has velocity along negative Y-axis
$\therefore$ $\overrightarrow{F_B}=q(\overrightarrow{V}\times \overrightarrow{B})$

Using Right-hand thumb rule, the direction of magnetic force is along $-ve$ x-axis.

Now it will move under influence of $\overrightarrow{F_E}$ and $\overrightarrow{F_B}$ both.

$\therefore$ eletron start by moving along negative Y- axis but then move in $3$rd quadrant because of both forces.
So option $\left(A\right)$ is correct.