Question

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 $A{m}^{-1}$ will be:'
[Given: Speed of light in vacuum $c=3\times 10^{8}m{s}^{-1}$, permeability of vacuum ${\mu }_0=4\pi \times 10^{-7}N{A}^{-2}$ ]

• A. $+0.8\sin (kz-\omega t){\hat{a}}_y+0.8\sin (kz-\omega t){\hat{a}}_x$
• 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. $-0.8\sin (kz-\omega t){\hat{a}}_y-1.2\sin (kz-\omega t){\hat{a}}_x$
• 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)

$\vec{E}=301.6\sin (kz-\omega t)\left(-{\hat{a}}_x\right)+452.4\sin (kz-\omega t){\hat{a}}_y$
$\vec{B}=\frac{301.6}{C}\sin (kz-\omega t)\left(-{\hat{a}}_y\right)$
$+\frac{452.4}{C}\sin (kz-\omega t)\left(-{\hat{a}}_x\right)$
$\vec{H}=\frac{\vec{B}}{{\mu }_0}=\frac{301.6}{\mu C}\sin (kz-\omega t)\left(-{\hat{a}}_y\right)$
$+\frac{452.4}{\mu C}\sin (kz-\omega t)\left(-{\hat{a}}_x\right)$
$\vec{H}=-0.8\sin (kz-\omega t){\hat{a}}_y-1.2\sin (kz-\omega t){\hat{a}}_x$
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 $\hat{E}=\hat{j},\hat{B}=?$
$\hat{E}\times \hat{B}=\hat{k}$
$\Rightarrow \hat{B}=-\hat{i}$