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Electromagnetic Waves question

2019 · 10 Apr · Shift 1 · Q54
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Electromagnetic Waves question

2019 · 10 Apr · Shift 1 · Q54

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field of a plane electromagnetic wave is given by E→=E0i^cos⁡(kz)cos(ωt)\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)E=E0​icos(kz)cos(ωt) The corresponding magnetic field B→\overrightarrow BB is then given by
  1. A
    B→=E0Cj^sin⁡(kz)sin⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\sin (\omega t)B=CE0​​j​sin(kz)sin(ωt)
  2. B
    B→=E0Cj^sin⁡(kz)cos⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\cos (\omega t)B=CE0​​j​sin(kz)cos(ωt)
  3. C
    B→=E0Cj^cos⁡(kz)sin⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\cos (kz)\sin (\omega t)B=CE0​​j​cos(kz)sin(ωt)
  4. D
    B→=E0Ck^sin⁡(kz)cos⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat k\sin (kz)\cos (\omega t)B=CE0​​ksin(kz)cos(ωt)
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