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Electromagnetic Waves question

2024 · 30 Jan · Shift 1 · Q70
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  5. /2024 · 30 Jan · Shift 1 · Q70

Electromagnetic Waves question

2024 · 30 Jan · Shift 1 · Q70

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field of an electromagnetic wave in free space is represented as E→=E0cos⁡(ωt−kz)i^\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{i}E=E0​cos(ωt−kz)i^. The corresponding magnetic induction vector will be :
  1. A
    B→=E0Ccos⁡(ωt+kz)j^\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}+\mathrm{k} z) \hat{j}B=E0​Ccos(ωt+kz)j^​
  2. B
    B→=E0Ccos⁡(ωt−kz)j^\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{j}B=CE0​​cos(ωt−kz)j^​
  3. C
    B→=E0Ccos⁡(ωt−kz)j^\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}-\mathrm{k} z) \hat{j}B=E0​Ccos(ωt−kz)j^​
  4. D
    B→=E0Ccos⁡(ωt+kz)j^\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}+\mathrm{kz}) \hat{j}B=CE0​​cos(ωt+kz)j^​
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