JEE MainPhysicsToughEM wave

The electric fields of two plane electromagnetic plane waves in
vacuum are given by:
${\overset{\rightarrow}{E}}_{1} = E_{0}\overset{\hat{}}{j}cos(\omega t - kx)\ and\ {\overset{\rightarrow}{E}}_{2} = E_{0}\overset{\hat{}}{k}cos(\omega t - ky)$
At t = 0, a particle of charge q is at origin with a velocity
j $\overset{\rightarrow}{v} = 0.8c\overset{\hat{}}{j}$ (c is the speed
of the light in vacuum). The instantaneous force experienced by the
particle is

Show answer and solution

Magnetic field vectors associated with this electromagnetic
wave are given by

{\overset{\rightarrow}{B}}_{1} = \frac{E_{0}}{c}\overset{\hat{}}{k}cos(kx - \omega t)\&{\overset{\rightarrow}{B}}_{2} = \frac{E_{0}}{c}\overset{\hat{}}{i}cos(ky - \omega t)

\overset{\rightarrow}{F} = q\overset{\rightarrow}{E} + q\left( \overset{\rightarrow}{V} \times \overset{\rightarrow}{B} \right)

= q\left( {\overset{\rightarrow}{E}}_{1} + {\overset{\rightarrow}{E}}_{2} \right) + q\left( \overset{\rightarrow}{V} \times \left( {\overset{\rightarrow}{B}}_{1} + {\overset{\rightarrow}{B}}_{2} \right) \right)

By putting the value of

{\overset{\rightarrow}{E}}_{1},{\overset{\rightarrow}{E}}_{2},{\overset{\rightarrow}{B}}_{1}\&{\overset{\rightarrow}{B}}_{2}

The net Lorentz force on the charged particle is

\overset{\rightarrow}{F} = {qE}_{0}\lbrack 0.8cos(kx - \omega t)\overset{\hat{}}{i} + cos(kx - \omega t)\overset{\hat{}}{j} + 0.2cos(ky - \omega t)\overset{\hat{}}{k}\rbrack

At t = 0 and at x = y = 0

\overset{\rightarrow}{F} = qE_{0}\lbrack 0.8\overset{\hat{}}{i} + \overset{\hat{}}{j} + 0.2\overset{\hat{}}{k}\rbrack

Source: Grass-G / P-13 Physics paper 4 Nov.docx

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