JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 20)
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 :
$$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$ 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 :
$${E_0}q\left( {0.8\widehat i - \widehat j + 0.4\widehat k} \right)$$
$${E_0}q\left( { - 0.8\widehat i + \widehat j + \widehat k} \right)$$
$${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)$$
$${E_0}q\left( {0.4\widehat i - 3\widehat j + 0.8\widehat k} \right)$$
Explanation
$$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$
Its corresponding magnetic field will be
$$\overrightarrow {{B_1}} = {{{E_0}} \over c}\widehat k\cos \left( {\omega t - kx} \right)$$
$$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$$
Also its corresponding magnetic field will be
$$\overrightarrow {{B_2}} = {{{E_0}} \over c}\widehat i\cos \left( {\omega t - ky} \right)$$
Net force on charge particle
= $$q\overrightarrow {{E_1}} + q\overrightarrow {{E_2}} + q\overrightarrow v \times \overrightarrow {{B_1}} $$$$ + q\overrightarrow v \times \overrightarrow {{B_2}} $$
= $${q{E_0}\widehat j}$$ + $${q{E_0}\widehat k}$$ + $$q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)$$ + $$q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)$$
= $${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)$$
Its corresponding magnetic field will be
$$\overrightarrow {{B_1}} = {{{E_0}} \over c}\widehat k\cos \left( {\omega t - kx} \right)$$
$$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$$
Also its corresponding magnetic field will be
$$\overrightarrow {{B_2}} = {{{E_0}} \over c}\widehat i\cos \left( {\omega t - ky} \right)$$
Net force on charge particle
= $$q\overrightarrow {{E_1}} + q\overrightarrow {{E_2}} + q\overrightarrow v \times \overrightarrow {{B_1}} $$$$ + q\overrightarrow v \times \overrightarrow {{B_2}} $$
= $${q{E_0}\widehat j}$$ + $${q{E_0}\widehat k}$$ + $$q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)$$ + $$q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)$$
= $${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)$$
Comments (0)
