Electromagnetic Waves

JEE Main level, single correct, medium

55 of 165

Question 55 of 165

The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by
Ex=E0sin⁡(kz−ωt){E}_{x}={E}_{0}\sin \left(kz-\omega t\right)
By=B0sin⁡(kz−ωt){B}_{y}={B}_{0}\sin \left(kz-\omega t\right)
Then the correct relation between E0{E}_{0} and B0{B}_{0} is given by

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Answer: (C) kE0=ωB0k{E}_{0}=\omega {B}_{0}

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\Rightarrow {E}_{0}=\dfrac{\omega }{k}{B}_{0}
⇒E0k=ωB0\Rightarrow {E}_{0}k=\omega {B}_{0}

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