Electromagnetic Waves Chapter-Wise Test 20

The document states that the sun is an important source of ultraviolet light, though much of it is absorbed by the ozone layer.

The document notes that Maxwell's equations predicted that light is an electromagnetic wave, confirmed by the fact that the speed of EM waves matched the speed of light.

The document explains that microwaves are suitable for radar systems due to their short wavelengths, which allow for precise detection and ranging.

The document mentions that the ozone layer absorbs most ultraviolet radiation from the sun at an altitude of \( 40-50 \, \text{km} \).

\( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \). Substituting values, \( \mu_0 \varepsilon_0 = (4 \pi \times 10^{-7}) \times (8.85 \times 10^{-12}) \approx 1.11 \times 10^{-17} \). Thus, \( c = \frac{1}{\sqrt{1.11 \times 10^{-17}}} \approx 3 \times 10^8 \, \text{m/s} \).

Using \( E_0 = B_0 c \), we have \( E_0 = (3 \times 10^{-8}) \times (3 \times 10^8) = 9 \, \text{V/m} \).

Gauss's Law for magnetism, \( \oint \mathbf{B} \cdot \mathrm{d} \mathbf{A} = 0 \), implies that the magnetic flux through a closed surface is zero, indicating no magnetic monopoles.

Comparing with \( B_y = B_0 \sin(kz - \omega t) \), we have \( k = 3 \times 10^3 \, \text{rad/m} \). Wavelength \( \lambda = \frac{2 \pi}{k} = \frac{2 \pi}{3 \times 10^3} \approx 2.09 \times 10^{-3} \, \text{m} \).

The document provides the form \( E_x = E_0 \sin(kz - \omega t) \), where \( E_x \) is along the \( x \)-axis for a wave propagating along the \( z \)-direction.

The Ampere-Maxwell Law includes both conduction current and displacement current (due to the time rate of change of electric flux) as sources of a magnetic field, as per the document.