Question

If electric field intensity of a uniform plane electro magnetic wave is given as
$E=-301.6\, \sin (k z-\omega t) \hat{a}_{x}+452.4 \sin (k z-\omega t)$
$\hat{a}_{y} \frac{V}{m}$
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 $\mu_{0}=4 \pi \times 10^{-7} NA ^{-2}$ ]

• A. $+0.8 \sin (k z-\omega t) \hat{a}_{y}+0.8 \sin (k z-\omega t) \hat{a}_{x}$
• B. $+1.0 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ y }+1.5 \times 10^{-6}( kz -\omega t ) \hat{ a }_{ x }$
• C. $-0.8 \sin (k z-\omega t) \hat{a}_{y}-1.2 \sin (k z-\omega t) \hat{a}_{x}$
• D. $-1.0 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ y }-1.5 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ x }$

Answer 1

Answer: (C)

$\vec{ E }=301.6 \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)+452.4 \sin ( kz -\omega t ) \hat{ a }_{ y }$
$\vec{ B }=\frac{301.6}{ C } \sin ( kz -\omega t )\left(-\hat{ a }_{ y }\right)$
$+\frac{452.4}{C} \sin (k z-\omega t)\left(-\hat{a}_{x}\right)$
$\vec{ H }=\frac{\vec{ B }}{\mu_{0}}=\frac{301.6}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ y }\right)$
$+\frac{452.4}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)$
$\vec{ H }=-0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }-1.2 \sin ( kz -\omega t ) \hat{ a }_{ x }$
For direction
$\vec{ E } \times \vec{ B }$ is direction of $\vec{ C }$
For first part $\hat{ E }=-\hat{ i }, \hat{ B }=$ ?
$\hat{ E } \times \hat{ B }=\hat{ k }$
$\Rightarrow \hat{ B }=-\hat{ j }$
Similarly for second $\hat{ E }=\hat{ j }, \hat{ B }=?$
$\hat{ E } \times \hat{ B }=\hat{ k }$
$\Rightarrow \hat{ B }=-\hat{ i }$