Question

A positively charged particle with charge $q$ is moving with speed $V$ in a region of uniform magnetic field $B$ at the instant shown in figure. An external electric field is to be applied such that the charged particle follows a straight line path. The magnitude and direction of electric field required are respectively.

• A. $\frac{VB}{q},(-ve)x-$ axis
• B. $VB,(-ve)y-$ axis
• C. $qVB,(-ve)y-$ axis
• D. $qVB,(+ve)y-$ axis

Answer 1

The correct option is B $VB,(-ve)y-$ axis

For the given instant force on the particle due to magnetic field is

$\overrightarrow{F_m}=qVB\hat{j}$

So, magnetic force is acting in the positive $y-$ axis and the particle is following a straight path, means net force on the charged particle in $y-$ direction is zero.


$\therefore \overrightarrow{F_m}+\overrightarrow{F_e}=0$

$\overrightarrow{F_m}=-\overrightarrow{F_e}$

$qVB\hat{j}=-q\overrightarrow{E}$

$\therefore \overrightarrow{E}=-VB\hat{j}=VB(-\hat{j})$

Therefore, magnitude of electric field is $VB$ and it is acting along $-vey-$axis.