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  • An electromagnetic wave going through vacuum is described by <img alt="\mathrm{E=E_{0} \sin (k x-w t)\: and \: B=B_{0} \sin (k x-w t)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathr

An electromagnetic wave going through vacuum is described by \mathrm{E=E_{0} \sin (k x-w t)\: and \: B=B_{0} \sin (k x-w t)} then ____

Option: 1

\mathrm{E_{0} k=B_{0} w}


Option: 2

\mathrm{E_{0} B_{0}=w k}


Option: 3

\mathrm{E_{0} w=B_{0} k}


Option: 4

None of these


Answers (1)

best_answer

Maxwell's equation is free space are-

\mathrm{\operatorname{curl} \bar{E}=-\mu_{0} \frac{\partial \bar{H}}{\partial t}\quad \ldots(1)}
\mathrm{and \: curl\: \bar{H}=\epsilon_{0} \frac{\partial \bar{E}}{\partial t}\quad \ldots(2)}

and\: \vec{E}=E_{0} \sin (k x-w t)
          \vec{B}=B_{0} \sin (k x-w t)
or \, \vec{H}=H_{0} \sin (k x-\omega t)

Equation (1) can be written as -

\vec{K} \times \vec{E} =\mu_{0} w \vec{H}
k(\hat{n} \times \hat{E}) =\mu_{0} w \vec{H}

In term of moduli

k|\vec{E}|=\mu_{0} \omega|\vec{H}|
k E_{0}=\mu_{0} \omega H_{0}
E_{0} k=w \mu_{0} H_{0}
E_{0} k=wB_{0} Ans


 

Posted by

Shailly goel

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