Question

If electric field intensity of a uniform plane electromagnetic wave is given as
$E = -301.6 \sin(kz - \omega t) \hat{a}_x + 452.4 \sin(k\omega - \omega t) \hat{a}_y \, \frac{V}{m}$
Then, magnetic intensity ‘H’ of this wave in Am–1 will be :
[Given : Speed of light in vacuum c = 3 × 108ms–1, Permeability of vacuum $\mu_0 = 4\pi \times 10^{-7}$ NA–2]

• A. $0.8\sin(kz - \omega t)\hat{a}_y + 0.8\sin(k\omega - \omega t)\hat{a}_x$
• B. $0.1 \times 10^{-6} \sin(kz - \omega t) \hat{a}_y + 1.5 \times 10^{-6} \sin(k\omega - \omega t) \hat{a}_x$
• C. $-0.8 \sin(kz - \omega t)\hat{a}_y - 1.2 \sin(k\omega - \omega t)\hat{a}_z$
• D. $-0.1 \times 10^{-6} \sin(kz - \omega t) \hat{a}_y - 1.5 \times 10^{-6} \sin(k\omega - \omega t) \hat{a}_x$

Answer 1

Answer: (C)

$-0.8 \sin(kz - \omega t)\hat{a}_y - 1.2 \sin(k\omega - \omega t)\hat{a}_z$

Answer 2

We know,$\mathbf{\overrightarrow{B}} \times \mathbf{\overrightarrow{C}} = \mathbf{\overrightarrow{E}}$
Taking cross product of vector C both the sides-
$\mathbf{\overrightarrow{C}} \times (\mathbf{\overrightarrow{B}} \times \mathbf{\overrightarrow{C}}) = \mathbf{\overrightarrow{C}} \times \mathbf{\overrightarrow{E}}$
So
$\overrightarrow{B} = \frac{\mathbf{\overrightarrow{C}} \times \mathbf{\overrightarrow{E}}}{C^2}$
$\mathbf{\overrightarrow{C}} = C \mathbf{\hat{k}}$
$\mathbf{\overrightarrow{E}} = -301.6 \sin(Kz - \omega t) \hat{a}_x + 452.4 \sin(kz - \omega t) \hat{a}_y$
and
$\mathbf{\overrightarrow{H}} = \frac{\mathbf{\overrightarrow{B}}}{\mu_0}$
On solving
$\mathbf{\overrightarrow{H}} = -0.8 \sin(kz - \omega t) \hat{a}_y - 1.2 \sin(k\omega - \omega t) \hat{a}_z$