Properties of Electromagnetic Waves Questions in English

(B) The given equation for the electric field is $E_y = E_0 \cos(\omega t - kx)$.
Comparing this with the standard form of a wave traveling in the positive $X$ direction,we identify the angular frequency $\omega = 2\pi \times 10^6 \ rad/s$ and the wave number $k = \pi \times 10^{-2} \ rad/m$.
The frequency $f$ is given by $\omega = 2\pi f$,so $2\pi f = 2\pi \times 10^6 \implies f = 10^6 \ Hz$.
The wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$,so $\frac{2\pi}{\lambda} = \pi \times 10^{-2} \implies \lambda = \frac{2}{\pi \times 10^{-2}} \times \pi = 200 \ m$.
Since the phase is $(\omega t - kx)$,the wave propagates in the positive $X$ direction.

(C) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by the equation $c = E/B$,where $c$ is the speed of light in vacuum.
Given:
$E = 6.3 \ V/m$
$c = 3 \times 10^8 \ m/s$
Rearranging the formula to solve for $B$:
$B = E/c$
$B = 6.3 / (3 \times 10^8)$
$B = 2.1 \times 10^{-8} \ T$ (or $Wb/m^2$)
Therefore,the magnitude of the magnetic field is $2.1 \times 10^{-8} \ Wb/m^2$.

(A) The average energy density $u$ of an electromagnetic wave is given by the sum of the electric and magnetic energy densities.
$u = u_E + u_B = \frac{1}{2} \varepsilon_0 E_0^2 = \varepsilon_0 E_{rms}^2$.
Given $E_{rms} = 720 \ N \ C^{-1}$ and $\varepsilon_0 = 8.854 \times 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$.
Substituting the values:
$u = 8.854 \times 10^{-12} \times (720)^2$
$u = 8.854 \times 10^{-12} \times 518400$
$u \approx 4.589 \times 10^{-6} \ J \ m^{-3}$.
Thus,the correct option is $A$.

(A) The speed of electromagnetic waves in vacuum is given by $c = 3 \times 10^8 \text{ m/s}$.
The frequency of the wave is given as $f = 8.2 \times 10^6 \text{ Hz}$.
The relationship between wavelength $\lambda$,speed $c$,and frequency $f$ is $\lambda = \frac{c}{f}$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{8.2 \times 10^6}$.
$\lambda = \frac{300}{8.2} \approx 36.585 \text{ m}$.
Rounding to one decimal place,we get $\lambda = 36.6 \text{ m}$.

(A) The intensity of an electromagnetic wave is given by the product of the electric field amplitude $(E_0)$ and the magnetic field amplitude $(H_0)$ in terms of the Poynting vector magnitude.
The intensity $I$ is given by $I = E_0 \times H_0$.
Given:
$E_0 = 100 \ Vm^{-1}$
$H_0 = 0.265 \ Am^{-1}$
Substituting the values:
$I = 100 \times 0.265$
$I = 26.5 \ Wm^{-2}$
Thus,the maximum intensity of radiation is $26.5 \ Wm^{-2}$.

(B) The relationship between the speed of light $(c)$,frequency $(f)$,and wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{c}{f}$.
Given: $c = 3 \times 10^8 \ m/s = 3 \times 10^{10} \ cm/s$ and $f = 8.196 \times 10^6 \ Hz$.
Substituting the values into the formula:
$\lambda = \frac{3 \times 10^{10}}{8.196 \times 10^6} \ cm$.
$\lambda = \frac{3}{8.196} \times 10^4 \ cm$.
$\lambda \approx 0.36603 \times 10^4 \ cm$.
$\lambda \approx 3660 \ cm$.
Thus,the wavelength is $3660 \ cm$.

(D) The intensity (flux) of electromagnetic radiation from the sun is given as $I = 10^3 \, W/m^2$.
The area of the roof is $A = \text{length} \times \text{width} = 8 \, m \times 20 \, m = 160 \, m^2$.
The power $P$ incident on the surface is given by the formula $P = I \times A$.
Substituting the values, we get $P = 10^3 \, W/m^2 \times 160 \, m^2 = 1.6 \times 10^5 \, W$.

(B) The average energy density of the electric field in an electromagnetic wave is given by the formula:
$\rho_E = \frac{1}{2} \epsilon_0 E_{rms}^2$
Since the root-mean-square electric field $E_{rms} = \frac{E_0}{\sqrt{2}}$,where $E_0$ is the amplitude of the electric field:
$\rho_E = \frac{1}{2} \epsilon_0 \left( \frac{E_0}{\sqrt{2}} \right)^2 = \frac{1}{4} \epsilon_0 E_0^2$
Given $E_0 = 1 \ V/m$ and $\epsilon_0 = 8.85 \times 10^{-12} \ F/m$:
$\rho_E = \frac{1}{4} \times 8.85 \times 10^{-12} \times (1)^2$
$\rho_E = 2.2125 \times 10^{-12} \approx 2.2 \times 10^{-12} \ J \ m^{-3}$.

(A) The relationship between the amplitude of the electric field $(E_0)$ and the amplitude of the magnetic field $(B_0)$ in an electromagnetic wave is given by $B_0 = \frac{E_0}{c}$,where $c$ is the speed of light in vacuum $(c \approx 3 \times 10^8 \, m/s)$.
Given: $E_0 = 48 \, V/m$.
Substituting the values: $B_0 = \frac{48}{3 \times 10^8} = 16 \times 10^{-8} \, T$ (or $Wb/m^2$).
Thus,the amplitude of the magnetic field is $16 \times 10^{-8} \, Wb/m^2$.

(C) The average energy density of the electric field is given by the formula: $u_E = \frac{1}{4} \varepsilon_0 E_0^2$.
Here,the amplitude of the electric field $E_0 = 1 \ V/m$.
The permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2/(N \cdot m^2)$.
Substituting the values:
$u_E = \frac{1}{4} \times (8.85 \times 10^{-12}) \times (1)^2$
$u_E = 2.2125 \times 10^{-12} \ J/m^3$.
Rounding to one decimal place,we get $u_E \approx 2.2 \times 10^{-12} \ J/m^3$.

(B) The intensity of the electromagnetic wave at a distance $r$ is given by $I = \frac{P}{4\pi r^2}$.
Also,the intensity is related to the electric field by $I = \frac{1}{2} c \epsilon_0 E_0^2$,where $E_0$ is the maximum (peak) electric field.
Equating the two expressions: $\frac{1}{2} c \epsilon_0 E_0^2 = \frac{P}{4\pi r^2}$.
Solving for $E_0$: $E_0 = \sqrt{\frac{2P}{4\pi r^2 c \epsilon_0}} = \sqrt{\frac{P}{2\pi r^2 c \epsilon_0}}$.
Given $P = 1500 \, W$,$r = 3 \, m$,$c = 3 \times 10^8 \, m/s$,and $\epsilon_0 = 8.854 \times 10^{-12} \, C^2 N^{-1} m^{-2}$.
$E_0 = \sqrt{\frac{1500}{2 \times 3.14 \times (3)^2 \times 3 \times 10^8 \times 8.854 \times 10^{-12}}}$.
$E_0 = \sqrt{\frac{1500}{50.24 \times 10^{-4}}} \approx \sqrt{298567} \approx 546 \, V/m$.
However,using the approximation $c \epsilon_0 = \frac{1}{120\pi}$:
$E_0 = \sqrt{\frac{2P}{4\pi r^2} \cdot 120\pi} = \sqrt{\frac{2P \cdot 30}{r^2}} = \frac{1}{r} \sqrt{60P} = \frac{1}{3} \sqrt{60 \times 1500} = \frac{1}{3} \sqrt{90000} = \frac{300}{3} = 100 \, V/m$.

(D) Comparing the given equation $\vec{E} = 10 \cos(10^7t + kx) \hat{j} \text{ V/m}$ with the standard form $\vec{E} = E_0 \cos(\omega t + kx) \hat{j}$,we get:
$E_0 = 10 \text{ V/m}$ and $\omega = 10^7 \text{ rad/s}$.
$(i)$ Wavelength $\lambda = \frac{2\pi c}{\omega} = \frac{2 \times 3.14 \times 3 \times 10^8}{10^7} = 188.4 \text{ m}$. This is correct.
$(ii)$ Wave vector $k = \frac{\omega}{c} = \frac{10^7}{3 \times 10^8} = 0.033 \text{ rad/m}$. This is incorrect (given $0.33 \text{ rad/m}$).
$(iii)$ Amplitude $E_0 = 10 \text{ V/m}$. This is correct.
$(iv)$ Since the argument is $(\omega t + kx)$,the wave propagates in the negative $X$-direction. This is incorrect.
Thus,statements $(i)$ and $(iii)$ are correct.

(B) The intensity $I$ of an electromagnetic wave is related to the energy density $u$ by the formula: $I = u \times c$,where $c$ is the speed of light.
Therefore,the energy density $u$ is given by: $u = \frac{I}{c}$.
Given: $I = 0.02 \ W/m^2$ and $c = 3 \times 10^8 \ m/s$.
Substituting the values: $u = \frac{0.02}{3 \times 10^8} = \frac{2 \times 10^{-2}}{3 \times 10^8} = 0.667 \times 10^{-10} = 6.67 \times 10^{-11} \ J/m^3$.

(B) The wavelength $\lambda$ of an electromagnetic wave is given by the formula $\lambda = \frac{c}{f}$,where $c$ is the speed of light in vacuum $(3 \times 10^8 \, m/s)$ and $f$ is the frequency.
Given: $f = 2 \times 10^{10} \, Hz$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{2 \times 10^{10}} = 1.5 \times 10^{-2} \, m$.
Since $1 \, m = 100 \, cm$,we have $\lambda = 1.5 \times 10^{-2} \times 100 \, cm = 1.5 \, cm$.

(D) The intensity $I$ of an electromagnetic wave is given by the average power $P_{av}$ distributed over a spherical surface area $4\pi r^2$:
$I = \frac{P_{av}}{4\pi r^2}$
Also,the intensity in terms of the maximum electric field $E_m$ is:
$I = \frac{1}{2} c \epsilon_0 E_m^2 = \frac{E_m^2}{2 \mu_0 c}$
Equating the two expressions:
$\frac{P_{av}}{4\pi r^2} = \frac{E_m^2}{2 \mu_0 c}$
$E_m = \sqrt{\frac{2 \mu_0 c P_{av}}{4\pi r^2}} = \sqrt{\frac{\mu_0 c P_{av}}{2\pi r^2}}$
Substituting the given values ($P_{av} = 800 \, W$,$r = 3.5 \, m$,$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$,$c = 3 \times 10^8 \, m/s$):
$E_m = \sqrt{\frac{(4\pi \times 10^{-7}) \times (3 \times 10^8) \times 800}{2\pi \times (3.5)^2}}$
$E_m = \sqrt{\frac{2 \times 10^{-7} \times 3 \times 10^8 \times 800}{12.25}} = \sqrt{\frac{48000}{12.25}} \approx \sqrt{3918.36} \approx 62.6 \, V/m$

(B) The intensity of an electromagnetic wave is given by the formula $I_{av} = \frac{1}{2} c \epsilon_0 E_0^2$.
Given,$E_0 = 36 \ V/m$,$c = 3 \times 10^8 \ m/s$,and $\epsilon_0 = 8.854 \times 10^{-12} \ F/m$.
Substituting the values:
$I_{av} = \frac{1}{2} \times (3 \times 10^8) \times (8.854 \times 10^{-12}) \times (36)^2$
$I_{av} = 0.5 \times 3 \times 8.854 \times 10^{-4} \times 1296$
$I_{av} = 1.5 \times 8.854 \times 1296 \times 10^{-4}$
$I_{av} \approx 1.72 \ W/m^2$.

(B) The average intensity $I$ of an electromagnetic wave at a distance $r$ from a point source is given by $I = \frac{P_{av}}{4\pi r^2}$.
Also, the intensity in terms of the maximum electric field $E_m$ is $I = \frac{E_m^2}{2\mu_0 c}$.
Equating the two expressions: $\frac{P_{av}}{4\pi r^2} = \frac{E_m^2}{2\mu_0 c}$.
Solving for $E_m$: $E_m = \sqrt{\frac{\mu_0 c P_{av}}{2\pi r^2}}$.
Substituting the given values: $P_{av} = 800 \, W$, $r = 3.5 \, m$, $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$, and $c = 3 \times 10^8 \, m/s$.
$E_m = \sqrt{\frac{(4\pi \times 10^{-7}) \times (3 \times 10^8) \times 800}{2\pi \times (3.5)^2}} = \sqrt{\frac{2 \times 3 \times 10 \times 800}{12.25}} = \sqrt{\frac{48000}{12.25}} \approx 62.6 \, V/m$.

(C) The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$.
Substituting the values: $I = \frac{800}{4 \times 3.14 \times (3.5)^2} = \frac{800}{153.86} \approx 5.2 \, W/m^2$.
The relation between intensity and electric field amplitude $E_m$ is $I = \frac{1}{2} \epsilon_0 c E_m^2$.
$E_m = \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2 \times 5.2}{8.854 \times 10^{-12} \times 3 \times 10^8}} \approx \sqrt{3912} \approx 62.55 \, V/m$.
The maximum magnetic field $B_m$ is given by $B_m = \frac{E_m}{c}$.
$B_m = \frac{62.55}{3 \times 10^8} \approx 2.085 \times 10^{-7} \, T \approx 2.09 \times 10^{-7} \, T$.

(B) The relationship between the maximum electric field $(E_0)$ and the maximum magnetic field $(B_0)$ in an electromagnetic wave is given by the equation:
$c = \frac{E_0}{B_0}$
where $c$ is the speed of light in vacuum,approximately $3 \times 10^8 \, m/s$.
Rearranging the formula to solve for $B_0$:
$B_0 = \frac{E_0}{c}$
Substituting the given values:
$B_0 = \frac{18 \, V/m}{3 \times 10^8 \, m/s} = 6 \times 10^{-8} \, T$.
Thus,the maximum value of the magnetic field is $6 \times 10^{-8} \, T$.

(C) In an electromagnetic wave,the electric field vector $\overrightarrow{E}$ and the magnetic field vector $\overrightarrow{B}$ oscillate in phase and are perpendicular to each other.
According to Maxwell's equations and the properties of electromagnetic waves,the direction of energy propagation (the direction of the wave) is given by the direction of the Poynting vector $\overrightarrow{S}$.
The Poynting vector is defined as $\overrightarrow{S} = \frac{1}{\mu_0} (\overrightarrow{E} \times \overrightarrow{B})$.
Therefore,the direction of propagation of the electromagnetic wave is in the direction of the cross product $\overrightarrow{E} \times \overrightarrow{B}$.

(A) An electromagnetic wave propagates in the direction of the cross product of the electric field vector and the magnetic field vector,i.e.,$\vec{E} \times \vec{B}$.
Given that the direction of propagation is along the $z$-axis,the electric field and magnetic field must lie in the $xy$-plane.
Therefore,the electric field vector $\vec{E}$ must have components in the $x$ or $y$ direction,and the magnetic field vector $\vec{B}$ must have components in the $x$ or $y$ direction such that their cross product points in the $z$ direction.
Specifically,if $\vec{E}$ is along the $x$-axis $(E_x)$ and $\vec{B}$ is along the $y$-axis $(B_y)$,then $\hat{i} \times \hat{j} = \hat{k}$,which corresponds to the $z$-axis.
Thus,the pair $(E_x, B_y)$ is a valid configuration.

(D) In an electromagnetic wave,both the electric field $(E)$ and the magnetic field $(B)$ oscillate sinusoidally with time and space.
Mathematically,these fields are represented as $E = E_0 \sin(kx - \omega t)$ and $B = B_0 \sin(kx - \omega t)$.
The average value of a sinusoidal function over a complete cycle is zero.
Therefore,the average value of both the electric field and the magnetic field over a complete cycle is zero.

(A) The velocity $v$ of electromagnetic radiation in a medium is given by the relation $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu$ is the permeability and $\varepsilon$ is the permittivity of the medium.
For vacuum (or free space),the permeability is $\mu_0$ and the permittivity is $\varepsilon_0$.
Therefore,the velocity of electromagnetic radiation in vacuum is $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.

(B) The given electric field is $E_y = E_0 \cos(\omega t - kx)$.
Comparing this with the given equation $E_y = 2.5 \cos[(2\pi \times 10^6)t - (\pi \times 10^{-2})x]$,we identify the angular frequency $\omega = 2\pi \times 10^6 \text{ rad/s}$ and the wave number $k = \pi \times 10^{-2} \text{ rad/m}$.
Since the term inside the cosine is $(\omega t - kx)$,the wave is propagating in the positive $x$-direction.
The frequency $f$ is given by $\omega = 2\pi f$,so $f = \frac{2\pi \times 10^6}{2\pi} = 10^6 \text{ Hz}$.
The wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$,so $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi \times 10^{-2}} = 2 \times 10^2 = 200 \text{ m}$.
Thus,the wave is moving along the $x$-direction with a frequency of $10^6 \text{ Hz}$ and a wavelength of $200 \text{ m}$.

(D) Given equation: $\vec E = 10 \cos (10^7 t + kx) \hat j \, V/m$.
Comparing with the standard wave equation $\vec E = E_0 \cos (\omega t + kx) \hat j$,we get:
Amplitude $E_0 = 10 \, V/m$ (Statement $(3)$ is correct).
Angular frequency $\omega = 10^7 \, rad/s$.
Since the phase is $(\omega t + kx)$,the wave propagates in the $-x$ direction (Statement $(4)$ is incorrect).
For electromagnetic waves in free space,$c = \frac{\omega}{k} = 3 \times 10^8 \, m/s$.
Thus,$k = \frac{\omega}{c} = \frac{10^7}{3 \times 10^8} = 0.033 \, rad/m$ (Statement $(2)$ is incorrect).
Wavelength $\lambda = \frac{2\pi}{k} = \frac{2 \times 3.1416}{0.033} \approx 188.4 \, m$ (Statement $(1)$ is correct).
Therefore,statements $(1)$ and $(3)$ are correct.

(C) In an electromagnetic wave,the electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are mutually perpendicular to each other and are also perpendicular to the direction of propagation of the wave. Therefore,the statement that they are parallel to each other is false.

(A) An electromagnetic wave propagates in the direction of the vector product $\vec{E} \times \vec{B}$.
Given that the wave propagates along the $+z$-axis,the direction of propagation is $\hat{k}$.
We need to check which pair of fields satisfies $\hat{E} \times \hat{B} = \hat{k}$.
For option $A$: $\hat{i} \times \hat{j} = \hat{k}$. This matches the direction of propagation.
Therefore,the electric field is along the $x$-axis and the magnetic field is along the $y$-axis.

(B) For an electromagnetic wave propagating in a vacuum,the relationship between the amplitude of the electric field $(E_{0})$ and the amplitude of the magnetic field $(B_{0})$ is given by the equation:
$E_{0} = B_{0} c$
where $c$ represents the speed of light in a vacuum.
To find the ratio of the amplitude of the magnetic field to the amplitude of the electric field,we rearrange the equation:
$\frac{B_{0}}{E_{0}} = \frac{1}{c}$
Thus,the ratio is equal to $\frac{1}{c}$.

(A) The standard equation for an electromagnetic wave traveling in the $z$-direction is given by $\vec{E} = E_{0} \cos (kz - \omega t) \hat{i}$.
Comparing the given equation $\vec{E} = \hat{i} 40 \cos (kz - 6 \times 10^{8} t)$ with the standard form,we identify the angular frequency $\omega = 6 \times 10^{8} \, rad/s$.
In vacuum,the relationship between the wave vector $k$,angular frequency $\omega$,and the speed of light $c$ is given by $k = \frac{\omega}{c}$.
Given $c = 3 \times 10^{8} \, m/s$,we calculate $k = \frac{6 \times 10^{8}}{3 \times 10^{8}} = 2 \, m^{-1}$.

(D) In a microwave oven,the frequency of the microwaves is tuned to match the resonant frequency of water molecules.
When these frequencies match,the water molecules absorb the energy from the electromagnetic waves through the process of resonance.
This energy absorption increases the rotational kinetic energy of the water molecules,which manifests as an increase in temperature,thereby heating the food efficiently.

(C) According to Maxwell's theory,an accelerating charge produces time-varying electric and magnetic fields. These oscillating fields propagate through space as an electromagnetic wave. $A$ stationary charge produces only a static electric field,and a charge moving at a constant velocity produces a steady magnetic field along with an electric field,but neither generates a propagating electromagnetic wave.

(D) Given: $E_{rms} = 6 \, V m^{-1}$.
The relationship between the root mean square values of electric and magnetic fields in free space is given by $\frac{E_{rms}}{B_{rms}} = c$,where $c = 3 \times 10^8 \, m/s$ is the speed of light.
Calculating $B_{rms}$:
$B_{rms} = \frac{E_{rms}}{c} = \frac{6}{3 \times 10^8} = 2 \times 10^{-8} \, T$.
The peak value of the magnetic field $B_0$ is related to $B_{rms}$ by the formula $B_{rms} = \frac{B_0}{\sqrt{2}}$.
Therefore,$B_0 = B_{rms} \times \sqrt{2}$.
Substituting the value of $B_{rms}$:
$B_0 = 2 \times 10^{-8} \times 1.414 = 2.828 \times 10^{-8} \, T \approx 2.83 \times 10^{-8} \, T$.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\vec{S} \propto \vec{E} \times \vec{B}$.
Given that the velocity $\vec{v}$ is along the $+x$ axis $(\hat{i})$ and the electric field $\vec{E}$ is along the $+y$ axis $(\hat{j})$.
We know that $\vec{v} \parallel (\vec{E} \times \vec{B})$.
Substituting the known directions: $\hat{i} = \hat{j} \times \vec{B}_{dir}$.
Using the properties of unit vectors in a Cartesian coordinate system,we know that $\hat{j} \times \hat{k} = \hat{i}$.
Therefore,the direction of the magnetic field $\vec{B}$ must be along the $+z$ axis $(\hat{k})$.
Thus,the oscillating magnetic field is along the $+z$ direction.

(A) The direction of propagation of an $EM$ wave is given by the direction of the Poynting vector $\overrightarrow{S}$,which is in the direction of $\overrightarrow{E} \times \overrightarrow{B}$.
Given,the direction of propagation is $-z$ direction,so $\overrightarrow{E} \times \overrightarrow{B} = -\widehat{k}$.
From the figure,the electric field $\overrightarrow{E}$ is in the $+y$ direction,so $\overrightarrow{E} = E_0 \widehat{j}$.
Substituting these into the relation: $(E_0 \widehat{j}) \times \overrightarrow{B} = -\widehat{k}$.
Let $\overrightarrow{B} = B_x \widehat{i} + B_y \widehat{j} + B_z \widehat{k}$.
Since $\overrightarrow{E} \perp \overrightarrow{B}$ and $\overrightarrow{B} \perp \text{propagation direction}$,$\overrightarrow{B}$ must be along the $x$-axis.
So,$\widehat{j} \times (B_x \widehat{i}) = -B_x \widehat{k}$.
Comparing this with $-\widehat{k}$,we get $-B_x = -1$,which means $B_x = 1$.
Therefore,the magnetic field $\overrightarrow{B}$ is in the $+x$ direction.

(A) The average total energy density $u$ of an electromagnetic wave is the sum of the electric energy density and the magnetic energy density.
$u = u_E + u_B = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2$
Since $B_{rms} = \frac{E_{rms}}{c}$,we have $u_B = \frac{1}{2\mu_0} \left(\frac{E_{rms}}{c}\right)^2 = \frac{1}{2\mu_0} (E_{rms}^2 \varepsilon_0 \mu_0) = \frac{1}{2} \varepsilon_0 E_{rms}^2$.
Thus,the total energy density is $u = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2} \varepsilon_0 E_{rms}^2 = \varepsilon_0 E_{rms}^2$.
Given $\varepsilon_0 = 8.85 \times 10^{-12} \; F/m$ and $E_{rms} = 720 \; N/C$:
$u = (8.85 \times 10^{-12}) \times (720)^2$
$u = 8.85 \times 10^{-12} \times 518400$
$u = 4.58784 \times 10^{-6} \; J/m^3 \approx 4.58 \times 10^{-6} \; J/m^3$.

(B) The direction of polarization of an electromagnetic wave is defined by the direction of the oscillating electric field vector $\vec{E}$. Therefore,$\vec{X} \parallel \vec{E}$.
The direction of wave propagation $\vec{k}$ is given by the direction of the Poynting vector $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})$.
Thus,the direction of wave propagation is parallel to $\vec{E} \times \vec{B}$.

(B) Given,the peak value of the magnetic field is $B_{0} = 20 \ nT = 20 \times 10^{-9} \ T$.
The speed of light in vacuum is $c = 3 \times 10^{8} \ ms^{-1}$.
The relationship between the peak electric field $E_{0}$ and the peak magnetic field $B_{0}$ in an electromagnetic wave is given by $E_{0} = B_{0} \times c$.
Substituting the values:
$E_{0} = (20 \times 10^{-9} \ T) \times (3 \times 10^{8} \ ms^{-1})$
$E_{0} = 60 \times 10^{-1} \ Vm^{-1}$
$E_{0} = 6 \ Vm^{-1}$.
Therefore,the peak value of the electric field strength is $6 \ Vm^{-1}$.

(B) For an electromagnetic wave,the relationship between the amplitudes of the electric field $(E_0)$ and magnetic field $(B_0)$ is given by $E_0 = cB_0$,where $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
The electric energy density is given by $u_E = \frac{1}{2} \varepsilon_0 E_0^2$.
The magnetic energy density is given by $u_B = \frac{1}{2} \frac{B_0^2}{\mu_0}$.
Substituting $E_0 = cB_0$ into the electric energy density formula:
$u_E = \frac{1}{2} \varepsilon_0 (cB_0)^2 = \frac{1}{2} \varepsilon_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right) B_0^2 = \frac{1}{2} \frac{B_0^2}{\mu_0}$.
Therefore,$u_E = u_B$. The energy is equally divided between the electric and magnetic fields.

(B) The velocity of an $EM$ wave is given by $v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$.
In air,the wave equation is $\cos[2\pi v(\frac{z}{c} - t)] = \cos[\frac{2\pi v}{c}z - 2\pi vt]$. The phase velocity is $v_1 = c$.
In the medium,the wave equation is $\cos[k(2z - ct)] = \cos[2kz - kct]$. The phase velocity is $v_2 = \frac{kct}{2kz} = \frac{c}{2}$.
Since the medium is nonmagnetic,$\mu_{r_1} = \mu_{r_2} = 1$.
The ratio of velocities is $\frac{v_1}{v_2} = \frac{c}{c/2} = 2$.
Since $v = \frac{c}{\sqrt{\varepsilon_r}}$,we have $\frac{v_1}{v_2} = \sqrt{\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}}} = 2$.
Squaring both sides,$\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}} = 4$,which implies $\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{4}$.

(C) The intensity $I$ of an electromagnetic wave is defined as the average energy per unit area per unit time,which is given by $I = \langle u \rangle c$,where $\langle u \rangle$ is the average energy density and $c$ is the speed of light.
The total energy density $u$ is the sum of electric and magnetic energy densities: $u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2\mu_0} B^2$.
Since $E = cB$ and $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,the average energy density is $\langle u \rangle = \frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2 = \frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} (\frac{E_{rms}}{c})^2 = \epsilon_0 E_{rms}^2$.
Given $E_{rms} = \frac{E_{max}}{\sqrt{2}}$,we have $\langle u \rangle = \epsilon_0 \frac{E_{max}^2}{2}$.
Thus,$I = \langle u \rangle c = \frac{1}{2} \epsilon_0 c E_{max}^2$.
Substituting $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we get $I = \frac{1}{2} \epsilon_0 \frac{1}{\sqrt{\mu_0 \epsilon_0}} E_{max}^2 = \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_{max}^2$.

(B) The propagation of radio waves into a tunnel depends on the phenomenon of diffraction. Diffraction is significant when the wavelength $(\lambda)$ of the wave is comparable to the size of the aperture (diameter $D = 10 \, m$).
The wavelength is given by $\lambda = c/f$,where $c \approx 3 \times 10^8 \, m/s$.
For $f_1 = 3 \times 10^6 \, Hz$,$\lambda_1 = (3 \times 10^8) / (3 \times 10^6) = 100 \, m$.
For $f_2 = 30 \times 10^6 \, Hz$,$\lambda_2 = (3 \times 10^8) / (30 \times 10^6) = 10 \, m$.
For $f_3 = 3 \times 10^9 \, Hz$,$\lambda_3 = (3 \times 10^8) / (3 \times 10^9) = 0.1 \, m$.
Since the diameter of the tunnel is $10 \, m$,the signal with $\lambda_2 = 10 \, m$ matches the aperture size,leading to effective diffraction and better penetration into the tunnel compared to the others.

(C) The ionosphere acts as a reflecting medium for radio waves within a specific frequency range.
This phenomenon is known as sky wave propagation.
The ionosphere reflects electromagnetic waves having frequencies roughly between $2\ MHz$ and $30\ MHz$.
Frequencies higher than this range typically penetrate the ionosphere and escape into space.
Therefore,among the given options,the range $2\ MHz$ to $10\ MHz$ is the correct range that is reflected back by the ionosphere.

(B) The instantaneous electric energy density is given by $u_E = \frac{1}{2} \varepsilon_0 E^2$.
For an electromagnetic wave,the electric field varies sinusoidally as $E = E_0 \sin(kx - \omega t)$.
The average value of $\sin^2(kx - \omega t)$ over a complete cycle is $\frac{1}{2}$.
Therefore,the average electric energy density is $\langle u_E \rangle = \frac{1}{2} \varepsilon_0 \langle E^2 \rangle = \frac{1}{2} \varepsilon_0 (E_0^2 \cdot \frac{1}{2}) = \frac{1}{4} \varepsilon_0 E_0^2$.

(C) The critical frequency $f_c$ for the reflection of radio waves from an ionospheric layer is given by the relation $f_c = 9 \sqrt{N_{max}}$,where $N_{max}$ is the maximum electron density.
This implies that $f_c \propto \sqrt{N}$.
Given the electron densities $N_E = 4 \times 10^{11} \, m^{-3}$,$N_{F_1} = 9 \times 10^{11} \, m^{-3}$,and $N_{F_2} = 16 \times 10^{11} \, m^{-3}$.
The ratio of critical frequencies is:
$(f_c)_E : (f_c)_{F_1} : (f_c)_{F_2} = \sqrt{4 \times 10^{11}} : \sqrt{9 \times 10^{11}} : \sqrt{16 \times 10^{11}}$
$= \sqrt{4} : \sqrt{9} : \sqrt{16}$
$= 2 : 3 : 4$.

(B) The electromagnetic wave propagates along the $+ve$ $X$-axis. The electric field $\vec{E}$ is in the $-ve$ $Z$-direction $(\vec{E} = -E_z \hat{k})$. According to Maxwell-Ampere's law,$\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$.
Since the magnitude of the electric field is increasing,the electric flux $\phi_E$ through the rectangle is increasing in the $-ve$ $Z$-direction. By Lenz's law,the induced magnetic field will oppose this change.
Using the right-hand rule,the magnetic field $\vec{B}$ must circulate in a way that creates a flux in the $+ve$ $Z$-direction to oppose the increasing $-ve$ $Z$-flux. This corresponds to a counter-clockwise circulation of the magnetic field around the rectangle.
Furthermore,since the wave is travelling in the $+X$ direction,the field values are higher at positions closer to the wave front. Thus,the magnetic field magnitude will be relatively larger at the edge further along the $X$-axis compared to the edge closer to the origin. Diagram $B$ correctly depicts this counter-clockwise circulation with a larger magnitude on the right edge.

(B) The speed of light in vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
In a medium with permeability $\mu$ and permittivity $\varepsilon$,the speed is $v = \frac{1}{\sqrt{\mu \varepsilon}}$.
We know that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 K$,where $\mu_r$ is the relative permeability and $K$ is the dielectric constant (relative permittivity).
Substituting these into the expression for $v$,we get $v = \frac{1}{\sqrt{(\mu_0 \mu_r)(\varepsilon_0 K)}} = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \cdot \frac{1}{\sqrt{\mu_r K}}$.
Since $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,we have $v = \frac{c}{\sqrt{\mu_r K}}$.

(D) Given: Wavelength $\lambda = 3 \ mm = 3 \times 10^{-3} \ m$. Amplitude of electric field $E_0 = 66 \ Vm^{-1}$.
Frequency $f = c / \lambda = (3 \times 10^8) / (3 \times 10^{-3}) = 10^{11} \ Hz$.
Angular frequency $\omega = 2\pi f = 2\pi \times 10^{11} \ rad/s$.
The electric field equation is $E_y = E_0 \cos(\omega(t - x/c)) = 66 \cos(2\pi \times 10^{11}(t - x/c))$.
The amplitude of the magnetic field is $B_0 = E_0 / c = 66 / (3 \times 10^8) = 22 \times 10^{-8} = 2.2 \times 10^{-7} \ T$.
Since the wave travels in the $X$-direction and $E$ is in the $Y$-direction,$B$ must be in the $Z$-direction.
Thus,$B_z = B_0 \cos(\omega(t - x/c)) = 2.2 \times 10^{-7} \cos(2\pi \times 10^{11}(t - x/c))$.

(A) The wavelength $\lambda$ of a signal is given by $\lambda = c/f$,where $c$ is the speed of light $(3 \times 10^8 \ m/s)$ and $f$ is the frequency.
For $AM$ at $1020 \ kHz$ $(1.02 \times 10^6 \ Hz)$,the wavelength is $\lambda_{AM} = (3 \times 10^8) / (1.02 \times 10^6) \approx 294 \ m$.
For $FM$ at $89.5 \ MHz$ $(89.5 \times 10^6 \ Hz)$,the wavelength is $\lambda_{FM} = (3 \times 10^8) / (89.5 \times 10^6) \approx 3.35 \ m$.
For efficient radiation and reception,the height of the antenna should be comparable to a quarter wavelength $(\lambda/4)$ of the signal.
Since $\lambda_{AM} > \lambda_{FM}$,the $AM$ channel requires a much longer antenna than the $FM$ channel.

(D) The given electric field is $\vec E = E_z \hat z = 6.0\,\cos(kx - \omega t)\hat z$.
Comparing this with the standard form $\vec E = E_0 \cos(kx - \omega t)\hat z$,we see that the wave is propagating in the $+x$ direction (since the argument is $kx - \omega t$).
Thus,the direction of propagation $\hat k$ is $+x$ (or $\hat i$).
The relationship between the direction of the electric field $\vec E$,magnetic field $\vec B$,and the direction of propagation $\hat k$ is given by $\hat k = \hat E \times \hat B$.
Here,$\hat E = \hat z$ (or $\hat k_{unit}$) and $\hat k = \hat i$.
Substituting these,we get $\hat i = \hat k_{unit} \times \hat B$.
Using the cross product rules ($\hat k \times \hat i = \hat j$,$\hat k \times \hat j = -\hat i$,etc.),we find that $\hat B = \hat j$ (or $+y$).
Therefore,the direction of the magnetic field is $+y$.

(A) For an electromagnetic wave,the direction of propagation is given by the direction of the vector $\vec E \times \vec B$.
Given that the wave travels in the $+y$ direction,we have the direction of propagation $\hat{v} = \hat{j}$.
The electric field $\vec E$ is in the $+z$ direction,so $\vec E = E_0 \hat{k}$.
Using the relation $\vec E \times \vec B = \text{direction of propagation}$,we have $\hat{k} \times \vec B = \hat{j}$.
We know that $\hat{k} \times \hat{i} = \hat{j}$. Therefore,$\vec B$ must be in the $+x$ direction.
Also,in an electromagnetic wave,the electric and magnetic fields are always in phase.
Thus,the magnetic field $\vec B$ is in the $+x$ direction and in phase with the electric field $\vec E$.