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The magnetic field of a plane electromagnetic Wave is $\overrightarrow{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i }\, T$ Where $c=3 \times 10^{8} \,ms ^{-1}$ is the speed of light. The corresponding electric field is
$\overrightarrow{ E }=-10^{-6} \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$
$\overrightarrow{ E }=-9 \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$
$\overrightarrow{ E }=9 \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$
$\overrightarrow{ E }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m$
Solution
$\overrightarrow{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i } T$
$E _{0}= CB _{0} \Rightarrow E _{0}=3 \times 10^{8} \times 3 \times 10^{-8}$
$=9 V / m$
and direction of wave propagation is given as
$(\overrightarrow{ E } \times \overrightarrow{ B }) \| \overrightarrow{ C }$
$\hat{ B }=\hat{ i } \quad \& \quad \hat{ C }=-\hat{ j }$
so $\hat{ E }=-\hat{ k }$
$\therefore \overrightarrow{ E }= E _{0} \sin [200 \pi( y + ct )](-\hat{ k }) V / m$
