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Electromagnetic Waves question

2020 · 4 Sep · Shift 2 · Q54
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  5. /2020 · 4 Sep · Shift 2 · Q54

Electromagnetic Waves question

2020 · 4 Sep · Shift 2 · Q54

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field of a plane electromagnetic wave is given by E→=E0(x^+y^)sin⁡(kz−ωt)\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)E=E0​(x+y​)sin(kz−ωt) Its magnetic field will be given by :
  1. A
    E0c(x^+y^)sin⁡(kz−ωt){{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)cE0​​(x+y​)sin(kz−ωt)
  2. B
    E0c(x^−y^)sin⁡(kz−ωt){{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)cE0​​(x−y​)sin(kz−ωt)
  3. C
    E0c(x^−y^)cos⁡(kz−ωt){{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)cE0​​(x−y​)cos(kz−ωt)
  4. D
    E0c(−x^+y^)sin⁡(kz−ωt){{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)cE0​​(−x+y​)sin(kz−ωt)
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