Question

A particle of charge $+q$ and mass $m$ moving under the influence of a uniform electric field $E\hat{i}$ and uniform magnetic field $B\hat{k}$ follows a trajectory from $\text{P}$ to $\text{Q}$ as shown in figure. The velocities at $\text{P}$ and $\text{Q}$ are $v\hat{i}$ and $-2v\hat{j}$. Which of the following statement(s) is/are correct?

• A. $E=\frac{3}{4}\left[\frac{m{v}^2}{qa}\right]$
• B. Rate of work done by the electric field at $\text{P}$ is $\frac{3}{4}\left[\frac{m{v}^2}{a}\right]$
• C. Rate of work done by the electric field at $\text{P}$ is zero
• D. Rate of work done by both the fields at $\text{Q}$ is zero

Answer 1

Answer: (A), (B), (D)

Rate of work done by both the fields at $\text{Q}$ is zero

Magnetic force $\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})$

$\because \overrightarrow{F_B}\perp \overrightarrow{v}$ ( always)

$\therefore$ Workdone by $\overrightarrow{F_B}$ on charged particle is,

$W_B=\int F_{B}dS\cos {90}^{\circ }=0$

$\therefore$ Workdone by magnetic force on a charge particle is always zero.

Now, workdone by electric force on a charge particle is,

$W_E=$ Work done by electric force

From work - energy theorem,

Workdone by all forces $=\Delta KE$

$\therefore W_B+W_E=\frac{1}{2}m[v{{}_2}^2-v{{}_1}^2]=\frac{1}{2}m[4v^2-v^2]$

$\therefore W_E=\frac{3}{2}m{v}^2........\left(2\right)$

$\because W_E=\int \overrightarrow{F_E}\cdot \overrightarrow{dS}=\int qEdS=qE\left(2a\right)........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$qE\left(2a\right)=\frac{3}{2}m{v}^2$

$\Rightarrow E=\frac{3m{v}^2}{4\left(qa\right)}........\left(3\right)$

Now , rate of work done = power $\left(P\right)$

$P=\overrightarrow{F}.\overrightarrow{v}$ at any instant.

$\therefore$ Rate of work done by $\overrightarrow{E}$ at $\text{P}$
$=qEv\cos \theta =qEv\cos 0^{\circ }=qEv...........\left(4\right)$

Substituting $\left(3\right)$ in $\left(4\right)$ we get,

$P=q\left[\frac{3}{4}\frac{m{v}^2}{qa}\right]v=\frac{3}{4}\frac{m{v}^3}{a}$

$\therefore$ Option (a) and (b) are correct

Implies, option (c) is wrong.

Now rate of work done by $\overrightarrow{E}$ & $\overrightarrow{B}$ at $\text{Q}$ is,

Work done the electric field,

$P=\overrightarrow{F_E}\cdot \overrightarrow{v}=2qEv\cos {90}^{\circ }=0$

$\because$ Workdone by magnetic force on a charge particle is always zero.

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

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