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 $P$ to $Q$ as shown in figure. The velocities at $P$ and $Q$ are $v\hat{i}$ and $-2v\hat{j}$. Among the following the correct statements is/are :

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

Answer 1

Answer: (B), (D)

Net work done $=$ Change in Kinetic Energy
Therefore, net work done from $P$ to $Q=\frac{3}{2}m{v}^2$
And all work is done by electric field therefore equating

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

Therefore, option A is correct.
Option B is wrong.
Using Dimensional Analysis of given option is of force not of power.
Option D is correct.
Because at Q, electric force is perpendicular to velocity vector.