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
IIT 1991, Diffcult
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(d) 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 Similarly, according to equation $\overrightarrow {{F_m}} = q(\overrightarrow {v\,} \times \overrightarrow B )$
Force $\overrightarrow {{F_{\rm{m}}}} $ is also perpendicular to velocity vector .
Hence the rate of work done by the magnetic field = $0$
$K.E.$ 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 Similarly, according to equation $\overrightarrow {{F_m}} = q(\overrightarrow {v\,} \times \overrightarrow B )$
Force $\overrightarrow {{F_{\rm{m}}}} $ is also perpendicular to velocity vector .
Hence the rate of work done by the magnetic field = $0$
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