Question

A positive charge is passing through an electromagnetic field is which $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$ are directed toward $\mathrm{y}-$axis and $\mathrm{z}-$ axis respectively. If a charge particle passes through the region undeviated, then its velocity is/are represented by (here $\mathrm{a},\mathrm{b},$ and $\mathrm{c}$ are constant)

• A. $\overrightarrow{\mathrm{v}}=\frac{\mathrm{E}}{\mathrm{B}}\hat{\mathrm{i}}+\mathrm{a}\hat{\mathrm{j}}$
• B. $\overrightarrow{\mathrm{v}}=\frac{\mathrm{E}}{\mathrm{B}}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{k}}$
• C. $\overrightarrow{\mathrm{v}}=\frac{\mathrm{E}}{\mathrm{B}}\hat{\mathrm{i}}+\mathrm{c}\hat{\mathrm{i}}$
• D. $\overrightarrow{\mathrm{v}}=\frac{\mathrm{E}}{\mathrm{B}}\hat{\mathrm{i}}$

Answer 1

The correct option is B

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{E}}{\mathrm{B}}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{k}}$

Step 1: Given data

A charged particle is introduced at the origin, $(\mathrm{x}=0,\mathrm{y}=0,\mathrm{z}=0)$.

Initial velocity is $\overrightarrow{\mathrm{v}}$, uniform electric filed is $\overrightarrow{\mathrm{E}}$ and uniform magnetic field is $\overrightarrow{\mathrm{B}}$.

We have to find the cases where the particle will move in a straight line with constant velocity.

Step 2: Find the cases where the particle will move in a straight line with constant velocity.

To move a particle with constant velocity,

${\mathrm{F}}_{\mathrm{net}}=0$

So,

The electric force $=$ magnetic force

$qE=\mathrm{qvB}$

$\mathrm{E}=\mathrm{vB}$

$\mathrm{v}=\frac{\mathrm{E}}{\mathrm{B}}$

Only positive charged particles with speed $\frac{\mathrm{E}}{\mathrm{B}}$ can get through.

Hence, option $\mathrm{B}$ is correct answer.