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  • If Electric field intensity of a uniform plane electro magnetic wave is given as <img alt="\mathrm{E}=-301.6 \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{x}+452.4 \sin (\mathrm{k}

If Electric field intensity of a uniform plane electro magnetic wave is given as
\mathrm{E}=-301.6 \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{x}+452.4 \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{y} \frac{\mathrm{V}}{\mathrm{m}} .
Then, magnetic intensity ' \mathrm{H}^{\prime}$ of this wave in $\mathrm{Am}^{-1} will be :
[Given : Speed of light in vacuum \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$, Permeability of vacuum $\left.\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2}\right]
 

Option: 1

+0.8 \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{y}+0.8 \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{x}


Option: 2

+1.0 \times 10^{-6} \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{y}+1.5 \times 10^{-6}(\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{x}


Option: 3

-0.8 \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{y}-1.2 \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{x}


Option: 4

-1.0 \times 10^{-6} \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{y}-1.5 \times 10^{-6} \sin (\mathrm{k} z-\omega \mathrm{t}) \hat{\mathrm{a}}_{x}


Answers (1)

best_answer

\mathrm{\left ( E_{0} \right )_{x}= 301.6}
\mathrm{\left ( B_{0} \right )_{x}= \frac{\left ( E_{0} \right )x}{c}= 10^{-6}}

\mathrm{\left ( E_{0} \right )_{y}= 452.4}
\mathrm{\left ( B_{0} \right )_{y}= \frac{\left ( E_{0} \right )_{y}}{c}= 1.5\times 10^{-6}}

\left ( \left ( \hat{E}_{0} \right )\times \left ( \hat{B_{0}} \right ) \right )
Points in the direction of propagation which is along +z^{-} axis

The correct option is (4)

Posted by

Deependra Verma

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