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

The electric fields of two plane electromagnetic plane waves in v…

MCQ

The electric fields of two plane electromagnetic plane waves in vacuum are given by

$\overrightarrow{\mathrm{E}}_{1}=\mathrm{E}_{0} \hat{\mathrm{j}} \cos (\omega \mathrm{t}-\mathrm{kx})$ and

$\overrightarrow{\mathrm{E}}_{2}=\mathrm{E}_{0} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{ky})$

At $t=0,$ a particle of charge $q$ is at origin with a velocity $\overrightarrow{\mathrm{v}}=0.8 \mathrm{c} \hat{\mathrm{j}}$ ($c$ is the speed of light in vacuum). The instantaneous force experienced by the particle is

A
$\mathrm{E}_{0} \mathrm{q}(-0.8 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
B
$\mathrm{E}_{0} \mathrm{q}(0.8 \hat{\mathrm{i}}-\hat{\mathrm{j}}+0.4 \hat{\mathrm{k}})$
✓
$\mathrm{E}_{0} \mathrm{q}(0.8 \hat{\mathrm{i}}+\hat{\mathrm{j}}+0.2 \hat{\mathrm{k}})$
D
$\mathrm{E}_{0} \mathrm{q}(0.4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+0.8 \hat{\mathrm{k}})$

✓

Answer

Correct option: C.

$\mathrm{E}_{0} \mathrm{q}(0.8 \hat{\mathrm{i}}+\hat{\mathrm{j}}+0.2 \hat{\mathrm{k}})$

c
$\overrightarrow{\mathrm{E}}_{1}=\mathrm{E}_{0} \mathrm{j} \cos (\omega \mathrm{t}-\mathrm{kx})$

Its corresponding magnetic field will be

$\overrightarrow{\mathrm{B}}_{1}=\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{kx})$

$\overrightarrow{\mathrm{E}}_{2}=\mathrm{E}_{0} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{ky})$

$\overrightarrow{\mathrm{B}}_{2}=\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{i}} \cos (\omega \mathrm{t}-\mathrm{ky})$

Net force on charge particle

$=\mathrm{q} \overrightarrow{\mathrm{E}}_{1}+\mathrm{q} \overrightarrow{\mathrm{E}}_{2}+\mathrm{q} \overrightarrow{\mathrm{v}} \times \mathrm{B}_{1}+\mathrm{q} \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}_{2}$

$=\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{j}}+\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{k}}+\mathrm{q}(0.8 \mathrm{c} \hat{\mathrm{j}}) \times\left(\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{k}}\right)+\mathrm{q}(0.8 \mathrm{c} \hat{\mathrm{j}}) \times\left(\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{i}}\right)$

$=\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{j}}+\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{k}}+0.8 \mathrm{q} \mathrm{E}_{0} \hat{\mathrm{i}}-0.8 \mathrm{q} \mathrm{E}_{0} \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{F}}=\mathrm{q} \mathrm{E}_{0}[0.8 \hat{\mathrm{i}}+1 \hat{\mathrm{j}}+0.2 \hat{\mathrm{k}}]$