Verified Solution • 2025–26 Session moderate

QJEEM24Ph279b: JEE Main 2024 Physics - Class 12 JEEM24Ph279b Solution: The electric field of an electromagnetic wave in free space is…

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} . The corresponding magnetic induction vector will be :

Multiple Choice Options

  • A B→=E0Ccos⁡(ωt+kz)j^\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}+\mathrm{k} z) \hat{j}
  • B B→=E0Ccos⁡(ωt−kz)j^\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{j}
  • C B→=E0Ccos⁡(ωt−kz)j^\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}-\mathrm{k} z) \hat{j}
  • D B→=E0Ccos⁡(ωt+kz)j^\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}+\mathrm{kz}) \hat{j}

Step-by-Step Verified Solution

amp; Given E⃗=E0cos⁡(ωt−kz)i^amp;B⃗=E0Ccos⁡(ωt−kz)j^amp;C^=E^×B^\begin{aligned}& \text { Given } \vec{E}=E_0 \cos (\omega t-k z) \hat{i} \\& \vec{B}=\frac{E_0}{C} \cos (\omega t-k z) \hat{j} \\& \hat{C}=\hat{E} \times \hat{B}\end{aligned}

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