Question

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

• A. $vB\text{+x axis}$
• B. $vB\text{+y axis}$
• C. $vB\text{-y axis}$
• D. $vB\text{-x axis}$

Answer 1

The correct option is C $vB\text{-y axis}$

Magnetic force ${\overrightarrow{F}}_B$ on charge $q$ is given by, $${\overrightarrow{F}}_B=q(\overrightarrow{V}\times \overrightarrow{B})$$From the given figure, $\overrightarrow{B}=-B\hat{k}$ and $\overrightarrow{V}=v\hat{i}$

$\Rightarrow F_B=qvB[\hat{i}\times (-\hat{k}\left)\right]=qvB\hat{j}$

Now $\overrightarrow{E}$ electric field applied, so that charge particle follows straight path$${\overrightarrow{F}}_E=q\overrightarrow{E}$$Since the electric force will be balanced by the equal and opposite magnetic force so,

${\overrightarrow{F}}_E=-{\overrightarrow{F}}_B$

$\Rightarrow q\overrightarrow{E}=-qvB\hat{j}$

$\Rightarrow \overrightarrow{E}=vB(-\hat{j})$

Hence, option (c) is the correct answer.

$$\begin{array}{c}\text{Why this question: To familiarize oneself how to use Lorentz force equation in different}\\ \text{scenarios in order to get the asked parameters on which it depends.}\end{array}$$