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

The electric fields of two plane electromagnetic plane waves in vacuum are given by \vec{E_1} = E_0\hat{j}\cos(\omega t - kx) and \vec{E_2} = E_0\hat{k}\cos(\omega t - ky) At t = 0, a particle of charge q is at origin with velocity \vec{v} = 0.8c\hat{j} (c is the speed of light in vacuum ). The instantaneous force experienced by the particle is:
Option: 1 E_0 q\left(0.8\hat{i} - \hat{j} + 0.4\hat{k} \right )
Option: 2 E_0 q\left(0.4\hat{i} - 3 \hat{j} + 0.8\hat{k} \right )
Option: 3  E_0 q\left(-0.8\hat{i} + \hat{j} + \hat{k} \right )
Option: 4 E_0 q\left(0.8\hat{i} + \hat{j} + 0.2\hat{k} \right )
  

Answers (1)

best_answer

 

 

 

 

 

Magnetic field vectors associated with this electromagnetic wave are given by

 

\overrightarrow{\mathrm{B}}_{1}=\frac{E_{0}}{\mathrm{c}} \hat{\mathrm{k}} \cos (\mathrm{k} \mathrm{x}-\omega \mathrm{t}) \ \ \& \ \ \overrightarrow{\mathrm{B}}_{2}=\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{i}} \cos (\mathrm{ky}-\omega \mathrm{t})

 

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

        = q(\vec{E_1}+ \vec{E_2}) + q\left (\vec{V}\times( \vec{B_1}+\vec{B_2})\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 particle is

\vec{F} = qE_0[0.8\cos(kx - \omega t )\hat{i} + \cos(kx -\omega t)\hat{j} + 0.2\cos(ky - \omega t)\hat{k}]

at t = 0 and at x = y = 0

\vec{F} = qE_0[0.8\hat{i} + \hat{j} + 0.2\hat{k}]

Hence the option correct option is (4).

Posted by

avinash.dongre

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