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Electromagnetic Waves

JEE Main level, single correct, medium

98 of 165

Question 98 of 165

The electric fields of two plane electromagnetic plane waves in vacuum are given by $\vec{{E}_{1}}={E}_{0}\hat{j}\cos ⁡\left(\omega t-kx\right)$ and $\vec{{E}_{2}}={E}_{0}\hat{k}\cos ⁡\left(\omega t-ky\right)$ At $t=0,$ a particle of charge $q$ is at origin with a velocity $\vec{v}=08c\hat{j}$ ( $c$ is the speed of light in vaccum). The instantaneous force experienced by the particle is:

Show answer and solution

Answer: (D) ${\text{E}}_{0}\text{q}\left(0.8\hat{i}+\hat{j}+0.2\hat{k}\right)$

Magnetic field vectors associated with this electromagnetic wave are given by

{\vec{B}}_{1}=\dfrac{{E}_{0}}{c}\hat{k}\cos ⁡\left(kx-\omega t\right)

{\vec{B}}_{2}=\dfrac{{E}_{0}}{c}\hat{i}\cos ⁡\left(ky-\omega t\right)

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

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

By putting the value of

{\vec{E}}_{1},{\vec{E}}_{2},{\vec{B}}_{1}&{\vec{B}}_{2}

The net Lorentz force on the charged particles is

\vec{F}=q{E}_{0}\left[0.8\cos ⁡\left(kx-\omega t\right)\hat{i}+cos\left(ky-\omega t\right)\hat{j}+0.2cos\left(ky-\omega t\right)\hat{k}\right]

t=0

and at

x=y=0

\vec{F}=q{E}_{0}\left[0.8\hat{i}+\hat{j}+0.2\hat{k}\right]

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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