Question

A charged particle goes undeflected in a region containing electric and magnetic fields. It is possible that

• A. $\overrightarrow{E}||\overrightarrow{B},\overrightarrow{V}||\overrightarrow{E}$
• B. $\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$
• C. $\overrightarrow{V}||\overrightarrow{B},\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$
• D. $\overrightarrow{E}||\overrightarrow{B}$, but $\overrightarrow{V}$ is not parallel to $\overrightarrow{B}$

Answer 1

Answer: (A), (B)

$\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$


Case-I:

$\overrightarrow{E}||\overrightarrow{B}$ and $\overrightarrow{V}||\overrightarrow{E}$

Now $\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})=qvB\sin \theta =0$ (as $\sin 0^{\circ }=0$)

only electric force acts on it and which accelerates the charge particle along the direction of the force.

$\overrightarrow{F_E}=qE$ and $a_y=\frac{qE}{m}$

$\therefore$ charge particle can go undeflected. Only velocity will increase with time

$\therefore$ Option (a) is correct.

Case-II:

$\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$

$\overrightarrow{F_B}=q(v\hat{i}\times B\hat{k})=qvB(-\hat{j})$ and

$\overrightarrow{F_E}=q\overrightarrow{E}\hat{j}$

Hence, net force on the charge is zero and it will move with the same velocity in the same direction.


$\therefore$ Option (b) can also be correct.

Case-III:

$\overrightarrow{v}||\overrightarrow{B}$ but $\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$

$\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})=qvB\sin \theta =0$

only electric force acts on it and which accelerates the charge particle along the direction of the force.

$\overrightarrow{F_E}=qE$ and

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

$\therefore$ deflection will occur, hence option (c) is wrong

Case-IV:


$\overrightarrow{E}||\overrightarrow{B}$ but $\overrightarrow{v}$ is not parallel to $\overrightarrow{B}$

Now $\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})=q(v\sin \theta \hat{i}\times B\hat{j})=qvB\sin \theta \hat{k}$

$\overrightarrow{F_E}=qE\hat{j}$

Motion of the charge particle will be resultant of $\overrightarrow{F_E}\&\overrightarrow{F_B}$

$\therefore$ Deviaton will occur.

$\therefore$ Option (d) is also wrong.

Hence, options (a) and (b) are the correct answers.