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If electric field intensity of a uniform plane electro magnetic wave is given as
$E =-301.6 \sin ( kz -\omega t ) \hat{a}_{ x }+452.4 \sin ( kz -\omega t )$ $\hat{a}_{y} \frac{V}{m}$
Then, magnetic intensity $H$ of this wave in $Am ^{-1}$ will be
[Given: Speed of light in vacuum $c =3 \times 10^{8} ms ^{-1}$, permeability of vacuum $\mu_{0}=4 \pi \times 10^{-7} NA ^{-2}$ ]
$+0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }+0.8 \sin ( kz -\omega t ) \hat{ a }_{ x }$
$+1.0 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ y }+1.5 \times 10^{-6}( kz -\omega t ) \hat{ a }_{ x }$
$-0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }-1.2 \sin ( kz -\omega t ) \hat{ a }_{ x }$
$-1.0 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ y }-1.5 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ x }$
Solution
$\overrightarrow{ E }=301.6 \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)+452.4 \sin ( kz -\omega t ) \hat{ a }_{ y }$
$\overrightarrow{ B }=\frac{301.6}{ C } \sin ( kz -\omega t )\left(-\hat{a}_{ y }\right)+\frac{452.4}{ C } \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)$
$\overrightarrow{ H }=\frac{\overrightarrow{ B }}{\mu_{0}}=\frac{301.6}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ y }\right)+\frac{452.4}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)$
$\overrightarrow{ H }=-0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }-1.2 \sin ( kz -\omega t ) \hat{ a }_{ x }$
For direction
$\overrightarrow{ E } \times \overrightarrow{ B }$ is direction of $\overrightarrow{ C }$
For first part $\hat{ E }=-\hat{ i }, \hat{ B }= ?$
$\hat{ E } \times \hat{ B }=\hat{ k } \Rightarrow \hat{ B }=-\hat{ j }$
Similarly for second
$\hat{ E }=\hat{ j }, \hat{ B }=?$
$\hat{ E } \times \hat{ B }=\hat{ k } \Rightarrow \hat{ B }=-\hat{ i }$
