Electromagnetic Waves

$ \Rightarrow \quad E_x=E_0 \sin \left(k z-2 \pi \times 10^6 t\right)=E_0 \sin (k z-\omega t) $

⇒ Direction of propagation of wave is along $+z$ axis.

⇒ Speed of wave in medium

$ \begin{aligned} & v_m=\frac{v_0}{\sqrt{\varepsilon_r}}=\frac{c}{\sqrt{9}}=\frac{3 \times 10^8}{3} \\ & v_m=10^8 \mathrm{~m} / \mathrm{s} \end{aligned} $

⇒ Wavelength in medium $=\lambda_{\mathrm{m}}$

$ \lambda_m=\frac{v_m}{f_0}=\frac{10^8}{\frac{2 \pi \times 10^6}{2 \pi}}=100 \mathrm{~m}\left[\because f_0=\text { constant }\right] $

∴ Wavelength of the electromagnetic wave inside the medium is 100 m .

Note: Among the options, in the expression $B_y=\frac{B_0}{v} \sin \left(k z-2 \pi \times 10^6 t\right), B_0$ should be replaced by $E_0$.

In electromagnetic wave, $E$ and $B$ are in same phase and $B_0=\frac{E_0}{c}$; their planes are perpendicular to each other.

$\begin{aligned} & \therefore B_y=\frac{60}{c} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \\ & =\frac{60}{3 \times 10^8} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \\ & B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \end{aligned}$

The electromagnetic radiation with the smallest wavelength among the given options is Gamma rays.

Let's go through each option to understand more clearly:

Option A: X-rays

X-rays have very short wavelengths, typically in the range of 0.01 to 10 nanometers. They are used for medical imaging and other applications due to their ability to penetrate materials.

Option B: Gamma rays

Gamma rays have the smallest wavelengths of all electromagnetic radiation, often less than 0.01 nanometers (or 0.1 angstroms). They are highly energetic and are produced by nuclear reactions, radioactive decay, and other high-energy processes.

Option C: Ultraviolet rays

Ultraviolet (UV) rays have wavelengths ranging from about 10 to 400 nanometers. They are responsible for causing sunburns and are used in various scientific and industrial applications.

Option D: Microwaves

Microwaves have much longer wavelengths, typically ranging from 1 millimeter to 1 meter. They are commonly used in communication technologies and for heating food in microwave ovens.

Based on these explanations, the correct answer is:

Option B: Gamma rays

$\frac{\mu_r}{E_r}=\frac{1}{4}$

$\begin{aligned} \frac{E}{H}=\frac{E \mu}{B} & =v \mu \\ =\frac{1}{\sqrt{\mu \varepsilon}} \mu & =\sqrt{\frac{\mu}{\varepsilon}}=\sqrt{\frac{\mu_0 \mu_r}{\varepsilon_0 \varepsilon_r}} \\ & =\sqrt{\frac{\mu_0}{\varepsilon_0}} \sqrt{\frac{\mu_r}{\varepsilon_r}} \\ & =120 \pi\left(\frac{1}{2}\right) \\ & =\frac{60 \pi}{1} \end{aligned}$

Electromagnetic waves have several defining characteristics when they propagate through free space. Let's evaluate each of the options provided:

Option A: They are transverse in nature.

This is true. Electromagnetic waves are transverse waves, meaning the directions of the electric field and magnetic field oscillations are perpendicular to the direction of wave propagation. The electric field (E) and magnetic field (B) vectors are also perpendicular to each other and to the direction of propagation.

Option B: The energy density in the electric field is equal to the energy density in the magnetic field.

This is also true. In electromagnetic waves, the energy density stored in the electric field is equal to the energy density stored in the magnetic field. This is because the magnitudes of the electric and magnetic fields are related by $ c = \frac{E}{B} $ where c is the speed of light in vacuum. The energy density for each is given by $\frac{1}{2} \epsilon_0 E^2 $ for the electric field and $\frac{1}{2} \frac{B^2}{\mu_0}$ for the magnetic field. Given the relationship between E and B in a wave, these two energy densities are equal.

Option C: They travel with a speed equal to $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.

This statement is true. The speed of electromagnetic waves in vacuum is given by $ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} $, where $ \mu_0 $ is the magnetic permeability of free space and $ \varepsilon_0 $ is the electric permittivity of free space. This relationship derives from Maxwell’s equations in a vacuum.

Option D: They originate from charges moving with uniform speed.

This statement is false. Electromagnetic waves are not generally produced by charges moving with a uniform speed; rather, they are produced by charges that are accelerating. Uniform motion (where velocity is constant and acceleration is zero) does not result in radiation of electromagnetic waves. If a charge is accelerating – changing either the speed or direction of its motion – it emits electromagnetic radiation.

Therefore, the correct answer is Option D, as it is the property that is not true for electromagnetic waves traveling in free space.

$C=\frac{E_{0}}{B_{0}}$

$\mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}$

$=\frac{48}{3 \times 10^{8}}$

$=1.6 \times 10^{-7} \mathrm{~T}$

For electromagnetic wave,

$|\overrightarrow B | = {{|\overrightarrow E |} \over c}$

Here $\overrightarrow B $ is magnetic field associated with EM wave

$\overrightarrow E $ is electric field associated with EM wave

c is the speed of EM wave

$ \Rightarrow |\overrightarrow E | = c|\overrightarrow B |$

$ = 3 \times {10^8} \times 3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)$ V/m

Direction can be determined from

Poynting vector $ = {{\overrightarrow E \times \overrightarrow B } \over {{\mu _0}}}$

$\overrightarrow E = 9\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat k$ V/m

We know,

${{{E_0}} \over {{B_0}}}$ (in vacuum) = c

$\therefore$ ${{{B_0}} \over {{E_0}}} = {1 \over c}$

or, ${{{B_0}} \over {{E_0}}} = \sqrt {{\mu _0}{\varepsilon _0}} $

(a) - (ii)

(b) - (iii)

(c) - (iv)

(d) - (i)

$v = {1 \over {\sqrt {{\varepsilon _m}{\mu _m}} }}$

$v = {1 \over {\sqrt {{\varepsilon _0}{\varepsilon _r}{\mu _0}{\mu _r}} }}$

Since $c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }}$

$ \Rightarrow v = {c \over {\sqrt {{\varepsilon _r}{\mu _r}} }}$