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  • An electromagnetic wave going through vacuum is described by <img alt="\mathrm{E=E_0 \sin (k x-\omega t) ; B=B_0 \sin (k x-\omega t)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

An electromagnetic wave going through vacuum is described by \mathrm{E=E_0 \sin (k x-\omega t) ; B=B_0 \sin (k x-\omega t)}.

Which of the following equations is true?
 

Option: 1

\mathrm{E_0 k=B_{0 \omega}}


Option: 2

\mathrm{E_{0 \omega}=B_0 k}


Option: 3

\mathrm{E}_0 \mathrm{~B}_0=\omega \mathrm{k}


Option: 4

None of these


Answers (1)

best_answer

We know \mathrm{E_0=c B_0}, where c is the velocity of light.

\mathrm{\therefore \quad c=v \lambda=\frac{\omega}{2 \pi} \lambda=\frac{\omega}{k}}

Thus, \mathrm{E_0=\frac{\omega}{k} B_0 \text { or } \quad E_0 k=B_0 \omega}

Posted by

Divya Prakash Singh

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