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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%7BE%3DE_0%20%5Csin%20%28k%2

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}}   .....(1)

\mathrm{ \text { curl } \bar{H}=\epsilon_0 \frac{\partial \bar{E}}{\partial t} }       .....(2)

and,  \mathrm{\overrightarrow{E}=E_0 \sin (k x-\omega t)}

         \mathrm{\overrightarrow{B}=B_0 \sin (k x-\omega t)}

or      \mathrm{\overrightarrow{H}=H_0 \sin (k x-\omega t)}
Equation (1) can be written as -
         \mathrm{\overrightarrow{K} \times \overrightarrow{E} =\mu_0 \omega \overrightarrow{H}}

     \mathrm{K(\hat{h} \times \hat{E}) =\mu_0 \omega \overrightarrow{H}}

In term of moduli
      \mathrm{ k|\overrightarrow{E}|=\mu_0 \omega|\overrightarrow{H}|}

         \mathrm{k E_0=\mu_0 \omega \mathrm{H}}

         \mathrm{E_0 k=\omega \mu_0 H}

         \mathrm{E_0 k=\omega B_0 \quad}   Ans.

Posted by

himanshu.meshram

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