Question 124: EM waves

Question

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

Accepted Answer

$$\overrightarrow {{E_1}}  = {E_0}\widehat j\cos \left( {\omega t - kx} ight)$$
Its corresponding magnetic field will be
$$\overrightarrow {{B_1}}  = {{{E_0}} \over c}\widehat k\cos \left( {\omega t - kx} ight)$$
$$\overrightarrow {{E_2}}  = {E_0}\widehat k\cos \left( {\omega t - ky} ight)$$
Also its corresponding magnetic field will be
$$\overrightarrow {{B_2}}  = {{{E_0}} \over c}\widehat i\cos \left( {\omega t - ky} ight)$$
Net force on charge particle
= $$q\overrightarrow {{E_1}}  + q\overrightarrow {{E_2}}  + q\overrightarrow v  	imes \overrightarrow {{B_1}} $$$$ + q\overrightarrow v  	imes \overrightarrow {{B_2}} $$
= $${q{E_0}\widehat j}$$ + $${q{E_0}\widehat k}$$ + $$q\left( {0.8c\widehat j} ight) 	imes \left( {{{{E_0}} \over c}\widehat i} ight)$$ + $$q\left( {0.8c\widehat j} ight) 	imes \left( {{{{E_0}} \over c}\widehat i} ight)$$
= $${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} ight)$$

Answer

$${E_0}q\left( {0.8\widehat i - \widehat j + 0.4\widehat k} \right)$$

Answer

$${E_0}q\left( { - 0.8\widehat i + \widehat j + \widehat k} \right)$$

Answer

$${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)$$

Answer

$${E_0}q\left( {0.4\widehat i - 3\widehat j + 0.8\widehat k} \right)$$

EM waves

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