A plane electromagnetic wave travels in a medium of relative permeability $$1.61$$ and relative permittivity $$6.44$$. If magnitude of magnetic intensity is $$4.5 \times 10^{-2}$$ A m$$^{-1}$$ at a point, what will be the approximate magnitude of electric field intensity at that point?
(Given : Permeability of free space $$\mu_0 = 4\pi \times 10^{-7}$$ N A$$^{-2}$$, speed of light in vacuum $$c = 3 \times 10^{8}$$ m s$$^{-1}$$)

We are given: relative permeability $$\mu_r = 1.61$$, relative permittivity $$\varepsilon_r = 6.44$$, magnetic field intensity $$H = 4.5 \times 10^{-2}$$ A/m, $$\mu_0 = 4\pi \times 10^{-7}$$ N A$$^{-2}$$, and $$c = 3 \times 10^8$$ m/s.

The intrinsic impedance of a medium is:

$$ \eta = \frac{E}{H} = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{\mu_r \mu_0}{\varepsilon_r \varepsilon_0}} $$

Simplifying using the impedance of free space:

$$ \eta = \sqrt{\frac{\mu_r}{\varepsilon_r}} \times \sqrt{\frac{\mu_0}{\varepsilon_0}} $$

The impedance of free space is:

$$ \eta_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} = \mu_0 c = 4\pi \times 10^{-7} \times 3 \times 10^8 = 120\pi\ \Omega $$

The ratio is:

$$ \sqrt{\frac{\mu_r}{\varepsilon_r}} = \sqrt{\frac{1.61}{6.44}} = \sqrt{0.25} = 0.5 $$

Thus,

$$ E = \eta \times H = 0.5 \times 120\pi \times 4.5 \times 10^{-2} $$

$$ E = 0.5 \times 120 \times 3.1416 \times 4.5 \times 10^{-2} $$

$$ E = 60\pi \times 4.5 \times 10^{-2} $$

$$ E = 2.7\pi = 2.7 \times 3.1416 \approx 8.48 \text{ V/m} $$

Therefore, the correct answer is Option C.