If Electric field intensity of a uniform plane electro magnetic wave is given as
$$E = -301.6\sin(kz-\omega t)\hat{a}_x + 452.4\sin(kz-\omega t)\hat{a}_y$$ V m$$^{-1}$$. 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$$^{-2}$$]

The electric field of a uniform plane electromagnetic wave is given by:

$$\vec{E} = -301.6\sin(kz-\omega t)\hat{a}_x + 452.4\sin(kz-\omega t)\hat{a}_y$$ V/m

To find the corresponding magnetic field intensity $$\vec{H}$$, one uses the fact that for an electromagnetic wave propagating in the $$\hat{a}_z$$ direction, $$\vec{E}$$ and $$\vec{H}$$ are related by

$$\vec{H} = \frac{1}{\mu_0 c}(\hat{a}_z \times \vec{E})$$

Here, the product $$\mu_0 c$$ represents the impedance of free space and is given by $$\mu_0 c = 4\pi \times 10^{-7} \times 3 \times 10^8 = 120\pi \approx 376.8$$ ohm.

Evaluating the cross products, one finds

$$\hat{a}_z \times \hat{a}_x = \hat{a}_y,\quad \hat{a}_z \times \hat{a}_y = -\hat{a}_x$$

Substituting these into $$\hat{a}_z \times \vec{E}$$ yields

$$\hat{a}_z \times \vec{E} = -301.6\sin(kz-\omega t)\hat{a}_y + 452.4\sin(kz-\omega t)(-\hat{a}_x)$$

$$= -452.4\sin(kz-\omega t)\hat{a}_x - 301.6\sin(kz-\omega t)\hat{a}_y$$

Dividing each component by the impedance $$\mu_0 c = 376.8$$ gives

$$H_x = \frac{-452.4}{376.8} \approx -1.2\ \mathrm{A/m},\quad H_y = \frac{-301.6}{376.8} \approx -0.8\ \mathrm{A/m}$$

Therefore, the magnetic field intensity is

$$\vec{H} = -1.2\sin(kz-\omega t)\hat{a}_x - 0.8\sin(kz-\omega t)\hat{a}_y$$ A/m

The correct answer is Option C.