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JEE Main 202324 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2023 · 24 January · Evening Shift · Physics · Electromagnetic Waves

The electric field and magnetic field components of an...

The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by

Ex=E0sin⁡(kz−\omegat){E}_{x} = {E}_{0}\sin \left(kz - \omegat\right)

By=B0sin⁡(kz−\omegat){B}_{y} = {B}_{0}\sin \left(kz - \omegat\right)

Then the correct relation between E0{E}_{0} and B0{B}_{0} is given by

Options

  1. AE0B0=\omegak{E}_{0}{B}_{0} = \omegak
  2. BE0=kB0{E}_{0} = k{B}_{0}
  3. CkE0=ωB0k{E}_{0} = \omega{B}_{0}
  4. DωE0=kB0\omega{E}_{0} = k{B}_{0}

Correct answer

C. kE0=ωB0k{E}_{0} = \omega{B}_{0}

Step-by-step solution

As we know, electric field and magnetic field are related as

E0=cB0{E}_{0} = c{B}_{0}, where, cc is speed of light in vacuum

Alsoc=ωk\text{Also}c = \dfrac{\omega}{k}

⇒E0=ωkB0⇒ {E}_{0} = \dfrac{\omega}{k}{B}_{0}

⇒E0k=ωB0⇒ {E}_{0}k = \omega{B}_{0}

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