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  • An electromagnetic wave going through vacuum is described by <img alt="\mathrm{ E=E_0 \sin (k x-\omega t) }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20E%3DE_0%20%5Csi

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

Option: 1

\mathrm{E_0 K=B_0 \omega}


Option: 2

\mathrm{E_0 B_0=\omega k}


Option: 3

\mathrm{E_0 \omega=B_0 K}


Option: 4

None  of these


Answers (1)

best_answer

Maxwell's equation is free space are - 
 

\mathrm{ \text { curl } \bar{E}=-\mu_0 \frac{\partial \bar{H}}{\partial t} \quad-\text { (1) } }
\mathrm{ and cure \bar{H}=\epsilon_0 \frac{\partial \bar{E}}{\partial t}- (2). and }

\mathrm{ \text { and } \begin{aligned} \vec{E} & =E_0 \sin (k x-\omega t) \\ \vec{B} & =B_0 \sin (k x-\omega t) \\ \text { or } \quad \vec{H} & =H_0 \sin (k x-\omega t) \end{aligned} }

Equation (1) can be written as -

\mathrm{ \begin{aligned} \vec{k} \times \vec{E} & =\mu_0 \omega \vec{H} \\ k(\hat{h} \times \hat{E}) & =\mu_0 \omega \vec{H} \end{aligned} }

In term of moduli

\mathrm{ \begin{aligned} & k|\vec{E}|=\mu_0 \omega(\vec{H}) \\ & k E_0=\mu_0 \omega H_0 \\ & E_0 k=\omega \mu_0 H_0 \\ & E_0 k=\omega B_0 \end{aligned} }Ans
 

Posted by

rishi.raj

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