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

The correct option is B Rate of work done by electric field at P is $\frac{3}{4}\frac{m{v}^3}{a}$
Kinetic energy of the particle at point $P=\frac{1}{2}m{v}^2$



K.E. of the particle at point $Q=\frac{1}{2}m(2v{)}^2$
Increase in K.E.$=\frac{3}{2}m{v}^2$
It comes from the work done by the electric force qE on the particle as it covers a distance 2a along the x-axis. Thus $\frac{3}{2}m{v}^2=qE\times 2a\Rightarrow E=\frac{3}{4}\frac{m{v}^2}{qa}$ .
The rate of work done by the electric field at P
$=F\times v=qE\times v=3\frac{m{v}^3}{4a}$
At Q, $\overrightarrow{F_e}=q\overrightarrow{E}$ is along x-axis while velocity is along negative y-axis. Hence rate of work done by electric field $=\overrightarrow{F_e}.\overrightarrow{v}=0$ $(\because \theta ={90}^{\circ })$ Similarly, according to equation $\overrightarrow{F_m}=q(\overrightarrow{v}\times \overrightarrow{B})$
Force $\overrightarrow{F_m}$ is also perpendicular to velocity vector v .
Hence the rate of work done by the magnetic field = 0