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Electromagnetic Waves question

2020 · 9 Jan · Shift 1 · Q62
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  5. /2020 · 9 Jan · Shift 1 · Q62

Electromagnetic Waves question

2020 · 9 Jan · Shift 1 · Q62

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric fields of two plane electromagnetic plane waves in vacuum are given by E1→=E0j^cos⁡(ωt−kx)\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)E1​​=E0​j​cos(ωt−kx) and E2→=E0k^cos⁡(ωt−ky)\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)E2​​=E0​kcos(ωt−ky) At t = 0, a particle of charge q is at origin with a velocity v→=0.8cj^\overrightarrow v = 0.8c\widehat jv=0.8cj​ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
  1. A
    E0q(0.8i^−j^+0.4k^){E_0}q\left( {0.8\widehat i - \widehat j + 0.4\widehat k} \right)E0​q(0.8i−j​+0.4k)
  2. B
    E0q(−0.8i^+j^+k^){E_0}q\left( { - 0.8\widehat i + \widehat j + \widehat k} \right)E0​q(−0.8i+j​+k)
  3. C
    E0q(0.8i^+j^+0.2k^){E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)E0​q(0.8i+j​+0.2k)
  4. D
    E0q(0.4i^−3j^+0.8k^){E_0}q\left( {0.4\widehat i - 3\widehat j + 0.8\widehat k} \right)E0​q(0.4i−3j​+0.8k)
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