Properties of Electromagnetic Waves Questions in English

(D) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$.
Substituting these,we get $v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$,where $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3 \times 10^8 \ m/s$.
Rearranging the formula,we have $\sqrt{\mu_r \varepsilon_r} = \frac{c}{v}$.
Squaring both sides,$\mu_r \varepsilon_r = \left(\frac{c}{v}\right)^2$.
Given $\mu_r = 2.0$ and $v = 1.5 \times 10^8 \ m/s$,we have $2.0 \times \varepsilon_r = \left(\frac{3 \times 10^8}{1.5 \times 10^8}\right)^2$.
$2.0 \times \varepsilon_r = (2)^2 = 4$.
Therefore,$\varepsilon_r = \frac{4}{2} = 2$.

(A) The intensity $I$ of an electromagnetic wave is given by the formula:
$I = \frac{1}{2} \epsilon_0 E_0^2 c$
Given:
$E_0 = 600 \ V/m$
$\epsilon_0 = 9 \times 10^{-12} \ C^2 N^{-1} m^{-2}$
$c = 3 \times 10^8 \ m/s$
Substituting the values:
$I = \frac{1}{2} \times (9 \times 10^{-12}) \times (600)^2 \times (3 \times 10^8)$
$I = \frac{1}{2} \times 9 \times 10^{-12} \times 360000 \times 3 \times 10^8$
$I = \frac{1}{2} \times 9 \times 36 \times 3 \times 10^{-12} \times 10^4 \times 10^8$
$I = \frac{1}{2} \times 972 \times 10^0$
$I = 486 \ W/m^2$

(B) The relationship between the maximum electric field $(E_0)$ and maximum magnetic field $(B_0)$ is given by $E_0 = B_0 c$.
Given $E_0 = 60 \text{ Vm}^{-1}$ and $c = 3 \times 10^8 \text{ ms}^{-1}$,we have $B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \text{ T}$.
The wave number $k$ is given by $k = \frac{2\pi}{\lambda}$. Given $\lambda = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}$,we have $k = \frac{2\pi}{4 \times 10^{-3}} = \frac{\pi}{2} \times 10^3 \text{ rad/m}$.
The angular frequency $\omega$ is given by $\omega = ck = (3 \times 10^8) \times (\frac{\pi}{2} \times 10^3) = \frac{3\pi}{2} \times 10^{11} \text{ rad/s}$.
The wave propagates in the $+x$ direction and the electric field is in the $+y$ direction. Since the direction of propagation is $\vec{E} \times \vec{B}$,the magnetic field must be in the $+z$ direction.
Thus,the equation for the magnetic field is $B_z = B_0 \sin(kx - \omega t) = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \times 10^3 (x - 3 \times 10^8 t) \right] \hat{k} \text{ T}$.

(D) The relationship between the amplitude of the electric field $E_0$ and the magnetic field $B_0$ is given by $E_0 = B_0 c$,where $c$ is the speed of light in vacuum $(c \approx 3 \times 10^8 \ ms^{-1})$.
Given $B_0 = 3.5 \times 10^{-7} \ T$,we calculate $E_0 = (3.5 \times 10^{-7}) \times (3 \times 10^8) = 105 \ Vm^{-1}$.
The wave propagates in the negative $x$-direction (indicated by the $+kx$ term). The magnetic field is in the $y$-direction $(B_y)$. Since the electric field,magnetic field,and direction of propagation are mutually perpendicular,the electric field must be in the $z$-direction $(E_z)$.
Thus,the electric field is $E_z = 105 \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \ Vm^{-1}$.

(C) Electromagnetic $(EM)$ waves are produced by accelerating charges.
According to the theory of electromagnetism,a charge moving with a uniform velocity produces a steady magnetic field and does not radiate energy in the form of $EM$ waves.
Therefore,the statement that they originate from charges moving with uniform speed is incorrect.
All other options are standard properties of $EM$ waves:
$1$. The energy density of the electric field $(u_E = \frac{1}{2} \varepsilon_0 E^2)$ is equal to the energy density of the magnetic field $(u_B = \frac{1}{2\mu_0} B^2)$.
$2$. The speed of $EM$ waves in free space is $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
$3$. $EM$ waves are transverse in nature,meaning the oscillations of electric and magnetic fields are perpendicular to the direction of wave propagation.

(D) The given electric field is $\vec{E} = 30(2 \hat{x} + \hat{y}) \sin \left[2 \pi \left(5 \times 10^{14} t - \frac{10^7}{3} z\right)\right]$.
Comparing this with the standard wave equation $\vec{E} = \vec{E}_0 \sin(\omega t - kz)$,we get $\omega = 2 \pi \times 5 \times 10^{14} \text{ rad s}^{-1}$ and $k = 2 \pi \times \frac{10^7}{3} \text{ m}^{-1}$.
The speed of the wave in the medium is $v = \frac{\omega}{k} = \frac{5 \times 10^{14}}{10^7 / 3} = 1.5 \times 10^8 \text{ m s}^{-1}$.
The refractive index $\mu = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$. Thus,option $(D)$ is correct.
The magnetic field is given by $\vec{B} = \frac{1}{v} (\hat{k} \times \vec{E})$.
$\vec{B} = \frac{1}{1.5 \times 10^8} \left[ \hat{k} \times 30(2 \hat{x} + \hat{y}) \right] \sin \left[2 \pi \left(5 \times 10^{14} t - \frac{10^7}{3} z\right)\right]$.
$\vec{B} = \frac{30}{1.5 \times 10^8} (2 \hat{y} - \hat{x}) \sin \left[2 \pi \left(5 \times 10^{14} t - \frac{10^7}{3} z\right)\right] = 2 \times 10^{-7} (- \hat{x} + 2 \hat{y}) \sin \left[2 \pi \left(5 \times 10^{14} t - \frac{10^7}{3} z\right)\right]$.
So,$B_x = -2 \times 10^{-7} \sin(\dots)$ and $B_y = 4 \times 10^{-7} \sin(\dots)$. Thus,option $(A)$ is correct and $(B)$ is incorrect.
The polarization direction is along $(2 \hat{x} + \hat{y})$,so $\tan \theta = \frac{E_y}{E_x} = \frac{1}{2} = 0.5$. Thus,$\theta = \tan^{-1}(0.5) \approx 26.57^{\circ}$,not $30^{\circ}$. Thus,option $(C)$ is incorrect.
Therefore,the correct options are $(A)$ and $(D)$.

(D) The given electric field is $\overrightarrow{E} = E_0 \cos(\omega t - \vec{k} \cdot \vec{r}) \hat{n}_E$,where $\hat{n}_E = \frac{4\hat{i} - 3\hat{j}}{5}$ is the unit vector in the direction of the electric field.
The wave vector is $\vec{k} = 5 \times 10^{-3} (3\hat{i} + 4\hat{j}) = 1.5 \times 10^{-2} \hat{i} + 2 \times 10^{-2} \hat{j}$.
The unit vector in the direction of propagation is $\hat{k} = \frac{3\hat{i} + 4\hat{j}}{5}$.
The magnetic field direction is given by $\hat{n}_B = \hat{k} \times \hat{n}_E$.
$\hat{n}_B = \left( \frac{3\hat{i} + 4\hat{j}}{5} \right) \times \left( \frac{4\hat{i} - 3\hat{j}}{5} \right) = \frac{1}{25} [3\hat{i} \times (-3\hat{j}) + 4\hat{j} \times 4\hat{i}] = \frac{1}{25} [-9\hat{k} - 16\hat{k}] = -\frac{25}{25} \hat{k} = -\hat{k}$.
The magnitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{57}{3 \times 10^8}$.
Thus,$\overrightarrow{B} = -\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3}(3 x + 4 y)\right] \hat{k}$.

(D) For an electromagnetic wave in free space, the relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ is given by $E = Bc$, where $c$ is the speed of light in vacuum $(c = 3 \times 10^8 \ ms^{-1})$.
Given $E_y = 9.3 \ Vm^{-1}$.
Using the relation $B_z = \frac{E_y}{c}$, we get:
$B_z = \frac{9.3}{3 \times 10^8} \ T$.
$B_z = 3.1 \times 10^{-8} \ T$.
Since the wave travels in the $+x$ direction and the electric field is in the $+y$ direction, the magnetic field must be in the $+z$ direction to satisfy the direction of propagation ($\vec{E} \times \vec{B}$ direction).

(B) The energy density of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E^2$.
The total energy $U$ contained in a cylinder of length $L$ and radius $R$ is $U = u \times V = \frac{1}{2} \epsilon_0 E^2 \times (\pi R^2 L)$.
Given that the energy $U$ remains the same for both cylinders,we have $U_1 = U_2$.
$\frac{1}{2} \epsilon_0 E_1^2 \pi R_1^2 L_1 = \frac{1}{2} \epsilon_0 E_2^2 \pi R_2^2 L_2$.
Since $L_1 = L_2$ and $R_2 = \frac{R_1}{2}$,the equation simplifies to $E_1^2 R_1^2 = E_2^2 (\frac{R_1}{2})^2$.
$E_1^2 R_1^2 = E_2^2 \frac{R_1^2}{4}$.
$E_2^2 = 4 E_1^2$,which implies $E_2 = 2 E_1$.
Given $E_1 = 100 \ NC^{-1}$,we get $E_2 = 2 \times 100 = 200 \ NC^{-1}$.
Thus,the electric field is $200 \sin(\omega t - kx) \ NC^{-1}$.

(A) The given magnetic field is $\overrightarrow{ B } = B_0 \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right] \hat{n}$, where $\hat{n} = \frac{\sqrt{3}}{2} \hat{ i } + \frac{1}{2} \hat{ j }$ and $B_0 = 30 \text{ T}$.
Since the wave propagates in the $+z$ direction $(\hat{k})$, the electric field $\overrightarrow{ E }$ is related to the magnetic field $\overrightarrow{ B }$ by the relation $\overrightarrow{ E } = c (\overrightarrow{ B } \times \hat{k})$.
Substituting the values: $\overrightarrow{ E } = c \left[ \left( \frac{\sqrt{3}}{2} \hat{ i } + \frac{1}{2} \hat{ j } \right) 30 \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right] \times \hat{k} \right]$.
Using the cross products $\hat{ i } \times \hat{k} = -\hat{ j }$ and $\hat{ j } \times \hat{k} = \hat{ i }$, we get:
$\overrightarrow{ E } = 30 c \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right] \left( \frac{\sqrt{3}}{2} (-\hat{ j }) + \frac{1}{2} \hat{ i } \right)$.
Rearranging the terms, we get $\overrightarrow{ E } = \left( \frac{1}{2} \hat{ i } - \frac{\sqrt{3}}{2} \hat{ j } \right) 30 c \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right]$.

(B) Electromagnetic waves carry both energy and momentum. The momentum $p$ of a photon is given by $p = E/c = h/\lambda$,where $E$ is the energy,$c$ is the speed of light,$h$ is Planck's constant,and $\lambda$ is the wavelength. Therefore,Assertion $(A)$ is false.
The rest mass of a photon is indeed zero,which is a correct physical fact. Therefore,Reason $(R)$ is true.
Thus,$(A)$ is false but $(R)$ is true.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\overrightarrow{S} = \frac{1}{\mu_0} (\overrightarrow{E} \times \overrightarrow{B})$.
Thus,the direction of propagation is along $\overrightarrow{E} \times \overrightarrow{B}$.
Given that the wave propagates along the $+x$ direction,we have $\hat{i} = \hat{E} \times \hat{B}$.
If $\overrightarrow{E}$ is along the $y$-axis $(\hat{j})$ and $\overrightarrow{B}$ is along the $z$-axis $(\hat{k})$,then $\hat{j} \times \hat{k} = \hat{i}$.
Therefore,the components of the electric field and magnetic field are $E_y$ and $B_z$ respectively.

(C) The velocity of light in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$.
Since $\mu = \mu_0 \mu_r$ and $\epsilon = \epsilon_0 \epsilon_r$,we have $v = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}} = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \cdot \frac{1}{\sqrt{\mu_r \epsilon_r}}$.
Given that the velocity of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we can write $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$.
Given $\mu_r = \frac{10}{\pi}$ and $\epsilon_r = \frac{1}{0.0885 \times 10^{-10}}$ (Note: $8.85 \times 10^{-12} = 0.0885 \times 10^{-10}$),let's calculate the product $\mu_r \epsilon_r$.
Actually,using the provided values: $\mu_r \epsilon_r = \frac{10}{\pi} \times \frac{1}{0.0885} \approx 3.183 \times 11.299 \approx 36$.
Thus,$v = \frac{c}{\sqrt{36}} = \frac{c}{6}$.
Therefore,$c = 6v$. The velocity of light in vacuum is $6$ times greater than the velocity in the medium.

(B) The intensity $I$ of an electromagnetic wave is given by the relation $I = \frac{1}{2} \varepsilon_0 E_0^2 c$,where $E_0$ is the amplitude of the electric field.
Rearranging the formula for $E_0$,we get $E_0^2 = \frac{2 I}{\varepsilon_0 c}$,which implies $E_0 = \sqrt{\frac{2 I}{\varepsilon_0 c}}$.
Since $E_0$ represents the electric field,its $SI$ unit is the same as the electric field,which is $\text{Volt per meter}$ $(V/m)$ or $\text{Newton per Coulomb}$ $(N/C)$.
Comparing this with the given options,the correct unit is $N/C$.

(A) The wave equation is given by $E_z = E_0 \cos(kx + \omega t)$. Comparing this with the given equation,we have $E_0 = 60 \text{ V/m}$,$k = 5 \text{ rad/m}$,and $\omega = 1.5 \times 10^9 \text{ rad/s}$.
The speed of the wave is $v = \frac{\omega}{k} = \frac{1.5 \times 10^9}{5} = 3 \times 10^8 \text{ m/s}$.
The amplitude of the magnetic field is $B_0 = \frac{E_0}{v} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \text{ T}$.
The direction of propagation is given by the vector $\vec{E} \times \vec{B}$. Since the wave propagates in the $-x$ direction and $\vec{E}$ is along the $z$-axis,we have $(-\hat{i}) = \hat{k} \times \vec{B}$. This implies $\vec{B}$ must be along the $y$-axis (since $\hat{k} \times \hat{j} = -\hat{i}$). Thus,the magnetic field is $B_y = B_0 \cos(kx + \omega t) = 2 \times 10^{-7} \cos(5x + 1.5 \times 10^9 t) \text{ T}$.

(C) The given equation for the magnetic field is $B_y = 3 \times 10^{-7} \sin(10^3 x + 3 \times 10^{11} t)$.
Comparing this with the standard wave equation $B_y = B_0 \sin(kx + \omega t)$,we identify the wave number $k$ as $k = 10^3 \ m^{-1}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$,we get $10^3 = \frac{2\pi}{\lambda}$.
Therefore,$\lambda = \frac{2\pi}{10^3} = 2 \times 3.14 \times 10^{-3} \ m = 6.28 \times 10^{-3} \ m$.
Converting meters to centimeters,$\lambda = 6.28 \times 10^{-3} \times 10^2 \ cm = 0.628 \ cm \approx 0.63 \ cm$.

(A) Statement-$I$ is true: In an electromagnetic wave,the electric field vector $\vec{E}$,the magnetic field vector $\vec{B}$,and the direction of wave propagation are mutually perpendicular to each other.
Statement-$II$ is true: For an electromagnetic wave in vacuum,the average energy density associated with the electric field is $u_E = \frac{1}{2} \epsilon_0 E_{rms}^2$ and the average energy density associated with the magnetic field is $u_B = \frac{1}{2} \frac{B_{rms}^2}{\mu_0}$. Since $E_{rms} = c B_{rms}$ and $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,it follows that $u_E = u_B$. Thus,the energy is equally shared between the electric and magnetic fields.

(B) The average total energy density $\mu$ of an electromagnetic wave is the sum of the electric energy density $\mu_E$ and the magnetic energy density $\mu_B$.
$\mu = \mu_E + \mu_B = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2$
Since $B_{rms} = \frac{E_{rms}}{c}$ and $c^2 = \frac{1}{\mu_0 \varepsilon_0}$,we have $\mu_B = \frac{1}{2\mu_0} \left(\frac{E_{rms}^2}{c^2}\right) = \frac{1}{2\mu_0} (E_{rms}^2 \mu_0 \varepsilon_0) = \frac{1}{2} \varepsilon_0 E_{rms}^2$.
Therefore,$\mu = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2} \varepsilon_0 E_{rms}^2 = \varepsilon_0 E_{rms}^2$.
Given $E_{rms} = 720 \ N/C$ and $\varepsilon_0 = 8.85 \times 10^{-12} \ F/m$.
$\mu = (8.85 \times 10^{-12}) \times (720)^2$
$\mu = 8.85 \times 10^{-12} \times 518400$
$\mu \approx 4.587 \times 10^{-6} \ J/m^3$.

(D) The energy density $u$ of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E_0^2$,where $E_0$ is the amplitude of the electric field.
Given $E_0 = 50 \ N/C$,$\epsilon_0 = 8.85 \times 10^{-12} \ F/m$.
The volume $V$ of the cylindrical region is $A \times l = 10 \ cm^2 \times 50 \ cm = 10 \times 10^{-4} \ m^2 \times 0.5 \ m = 5 \times 10^{-4} \ m^3$.
The total energy $U$ is $u \times V = \frac{1}{2} \epsilon_0 E_0^2 V$.
$U = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 \times (5 \times 10^{-4})$.
$U = 0.5 \times 8.85 \times 10^{-12} \times 2500 \times 5 \times 10^{-4}$.
$U = 5.53 \times 10^{-12} \ J \approx 5.5 \times 10^{-12} \ J$.

(A) The intensity $I$ (average power per unit area) of an electromagnetic wave is given by $I = \frac{1}{2} \varepsilon v E_p^2$,where $E_p$ is the peak electric field amplitude and $v$ is the speed of the wave in the medium.
Given $E_p = 50 \ V/m$,$\mu = 4\mu_0$,and $\varepsilon = \varepsilon_0$.
The speed of the wave in the medium is $v = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{1}{\sqrt{4\mu_0\varepsilon_0}} = \frac{1}{2\sqrt{\mu_0\varepsilon_0}} = \frac{c}{2}$.
Substituting $c = 3 \times 10^8 \ m/s$,we get $v = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \ m/s$.
Now,calculating the intensity $I = \frac{1}{2} \varepsilon_0 v E_p^2 = \frac{1}{2} \times (8.854 \times 10^{-12}) \times (1.5 \times 10^8) \times (50)^2$.
$I = 0.5 \times 8.854 \times 10^{-12} \times 1.5 \times 10^8 \times 2500$.
$I \approx 1.66 \ W/m^2$. Using the approximation $\varepsilon_0 \approx 8.8 \times 10^{-12}$,we get $I = 0.5 \times 8.8 \times 10^{-12} \times 1.5 \times 10^8 \times 2500 = 1.65 \ W/m^2$.

(C) The intensity $I$ of an electromagnetic wave at a distance $d$ from a point source of power $P$ is given by $I = \frac{P}{4\pi d^2}$.
Also,the intensity is related to the peak electric field $E_m$ by the relation $I = \frac{1}{2} \epsilon_0 E_m^2 c$.
Equating these,we get $\frac{P}{4\pi d^2} = \frac{1}{2} \epsilon_0 E_m^2 c$.
Since $4\pi, d^2, \epsilon_0,$ and $c$ are constant for a fixed distance,we have $E_m^2 \propto P$,which implies $E_m \propto \sqrt{P}$.
Given $P_1 = 100\ W$ and $P_2 = 50\ W$,and the corresponding electric fields are $E_1 = E$ and $E_2 = E'$,we have $\frac{E'}{E} = \sqrt{\frac{P_2}{P_1}}$.
Substituting the values,$\frac{E'}{E} = \sqrt{\frac{50}{100}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$E' = \frac{E}{\sqrt{2}}$.

(A) The total energy $E$ of the photons in the unit volume is given by $E = N \times hf$,where $N = 35 \times 10^7$,$h = 6 \times 10^{-34} \text{ Js}$,and $f = 10^{15} \text{ Hz}$.
$E = (35 \times 10^7) \times (6 \times 10^{-34}) \times 10^{15} = 210 \times 10^{-12} = 2.1 \times 10^{-10} \text{ J}$.
The energy density $u$ of an electromagnetic wave is given by $u = \frac{B_0^2}{2\mu_0}$,where $B_0$ is the amplitude of the magnetic field.
Since the volume $V = 1 \text{ m}^3$,the total energy $E = u \times V = \frac{B_0^2}{2\mu_0} \times 1$.
Equating the energies: $2.1 \times 10^{-10} = \frac{B_0^2}{2 \times 4\pi \times 10^{-7}}$.
$B_0^2 = 2.1 \times 10^{-10} \times 8\pi \times 10^{-7} = 16.8\pi \times 10^{-17}$.
Using $\pi = \frac{22}{7}$,$B_0^2 = 16.8 \times \frac{22}{7} \times 10^{-17} = 2.4 \times 22 \times 10^{-17} = 52.8 \times 10^{-17} = 528 \times 10^{-18}$.
$B_0 = \sqrt{528} \times 10^{-9} \approx 22.978 \times 10^{-9} \text{ T}$.
Thus,$\alpha \approx 22.98$.

(C) Light is an electromagnetic wave in nature.
Electromagnetic waves do not require any material medium for their propagation.
Therefore,the statement that 'Light requires a material medium' is incorrect and is not a property of light.

(C) Electromagnetic waves do not require a material medium for their propagation; they can travel through a vacuum at the speed of light $(c \approx 3 \times 10^8 \ m/s)$.
This is because electromagnetic waves consist of oscillating electric and magnetic fields that sustain each other,as described by Maxwell's equations.
Therefore,the statement that a material medium is necessary for the propagation of electromagnetic waves is incorrect.

(B) Electromagnetic waves travel in free space or vacuum with the velocity of light,which is approximately $3 \times 10^{8} \ m/s$.

(A) The relationship between the speed of light $(c)$, frequency $(\nu)$, and wavelength $(\lambda)$ is given by the formula: $c = \nu \lambda$.
Given:
Wavelength $\lambda = 10 \, m$.
Speed of light $c = 3 \times 10^{8} \, m/s$.
Rearranging the formula to solve for frequency: $\nu = \frac{c}{\lambda}$.
Substituting the values: $\nu = \frac{3 \times 10^{8} \, m/s}{10 \, m} = 3 \times 10^{7} \, Hz$.

(C) According to electromagnetic theory,light is an electromagnetic wave. It consists of time-varying electric and magnetic fields that oscillate in space and time. These electric and magnetic field vectors are mutually perpendicular to each other and also perpendicular to the direction of propagation of the wave.

(C) The velocity of electromagnetic radiation in free space is given by the speed of light $(c)$.
According to Maxwell's equations,the speed of electromagnetic waves in a vacuum is related to the permeability of free space $(\mu_{0})$ and the permittivity of free space $(\varepsilon_{0})$ by the formula:
$c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$
Thus,the correct expression for the velocity is $\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.

(D) The ionosphere is a region of the Earth's upper atmosphere that is ionized by solar radiation. This ionization process strips electrons from atoms and molecules,resulting in a plasma composed of free electrons and positively charged ions.

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

(A) The relationship between speed of light $(c)$, frequency $(\nu)$, and wavelength $(\lambda)$ is given by $c = \nu \lambda$.
Therefore, the wavelength is $\lambda = \frac{c}{\nu}$.
For the lower frequency limit $\nu_1 = 6 \text{ MHz} = 6 \times 10^6 \text{ Hz}$, the wavelength is:
$\lambda_1 = \frac{3 \times 10^8}{6 \times 10^6} = \frac{300}{6} = 50 \text{ m}$.
For the upper frequency limit $\nu_2 = 12 \text{ MHz} = 12 \times 10^6 \text{ Hz}$, the wavelength is:
$\lambda_2 = \frac{3 \times 10^8}{12 \times 10^6} = \frac{300}{12} = 25 \text{ m}$.
Thus, the corresponding wavelength band is $25 \text{ m}$ to $50 \text{ m}$.

(A) According to the theory of electromagnetic waves, an oscillating charge is a source of electromagnetic radiation.
When a charged particle oscillates with a frequency $f$, it produces electromagnetic waves of the same frequency $f$.
Given that the frequency of the oscillating particle is $10^{9} \,Hz$, the frequency of the electromagnetic waves produced will also be $10^{9} \,Hz$.
Therefore, the correct option is $A$.

(B) The relationship between the magnitudes of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by $E = cB$,where $c$ is the speed of light in free space $(c = 3 \times 10^8 \text{ ms}^{-1})$.
Substituting the given values: $E = (3 \times 10^8 \text{ ms}^{-1}) \times (2.1 \times 10^{-8} \text{ T}) = 6.3 \text{ Vm}^{-1}$.
The direction of propagation of the electromagnetic wave is given by the direction of the vector $\overrightarrow{E} \times \overrightarrow{B}$.
Given that the wave travels along the $X$-direction $(\hat{i})$ and $\overrightarrow{B}$ is along the $Z$-direction $(\hat{k})$,we have $\overrightarrow{E} \times \hat{k} = \hat{i}$.
Since $\hat{j} \times \hat{k} = \hat{i}$,the electric field $\overrightarrow{E}$ must be in the $Y$-direction $(\hat{j})$.
Therefore,$\overrightarrow{E} = 6.3 \hat{j} \text{ Vm}^{-1}$.

(A) The number of waves in a given length $L$ is given by the formula: $\text{Number of waves} = \frac{L}{\lambda}$.
Since the radiation is passing through air,its velocity $v$ is approximately equal to the speed of light $c = 3 \times 10^8 \text{ m/s}$.
Given frequency $\nu = 9 \text{ GHz} = 9 \times 10^9 \text{ Hz}$ and length $L = 1 \text{ m}$.
Using the relation $\lambda = \frac{c}{\nu}$,we substitute this into the formula:
$\text{Number of waves} = \frac{L \times \nu}{c}$.
Substituting the values:
$\text{Number of waves} = \frac{1 \times 9 \times 10^9}{3 \times 10^8} = 3 \times 10^1 = 30$.
Thus,the number of waves is $30$.

(B) The relationship between the speed of light $(c)$,frequency $(v)$,and wavelength $(\lambda)$ is given by the formula: $c = v \lambda$.
Given that the speed of light $c = 3 \times 10^{8} \ m/s$ and the frequency $v = 8.196 \times 10^{6} \ Hz$.
Rearranging the formula to solve for wavelength: $\lambda = \frac{c}{v}$.
Substituting the values: $\lambda = \frac{3 \times 10^{8}}{8.196 \times 10^{6}}$.
$\lambda = \frac{3}{8.196} \times 10^{2} \ m$.
$\lambda \approx 0.3660 \times 10^{2} \ m$.
$\lambda = 36.60 \ m$.
Converting meters to centimeters: $36.60 \ m = 3660 \ cm$.
Therefore,the correct option is $B$.

(C) The intensity of radiation $I$ is defined as the power incident per unit area,given by $I = \frac{P}{A}$.
Given:
Intensity $I = 10^{3} \,W m^{-2}$
Area $A = 6 \,m \times 30 \,m = 180 \,m^{2}$
To find the total power $P$ incident on the roof:
$P = I \times A$
$P = 10^{3} \,W m^{-2} \times 180 \,m^{2}$
$P = 180 \times 10^{3} \,W$
$P = 1.8 \times 10^{5} \,W$
Thus,the correct option is $C$.

(D) The relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by the equation: $E/B = c$,where $c$ is the speed of light in vacuum.
Given: $E = 6.6 \,V \,m^{-1}$ and $c = 3 \times 10^8 \,m \,s^{-1}$.
Rearranging the formula to solve for $B$: $B = E/c$.
Substituting the values: $B = 6.6 / (3 \times 10^8) \,T$.
Calculating the result: $B = 2.2 \times 10^{-8} \,T$.
Therefore,the correct option is $D$.

(D) According to Rayleigh scattering, the intensity of scattered light is inversely proportional to the fourth power of its wavelength $(I \propto 1/\lambda^4)$.
Since infrared radiations have a longer wavelength compared to visible light, they undergo significantly less scattering by the particles in fog.
Because of this reduced scattering, infrared radiations can penetrate through fog more effectively than visible light.
Therefore, photographs taken with infrared radiations in foggy conditions are much clearer than those taken with visible light.

(B) In a vacuum,all components of the electromagnetic spectrum travel with the same speed,which is the speed of light,denoted by $c$.
This value is approximately $3 \times 10^{8} \ m/s$.
While their frequencies and wavelengths differ,their velocity remains constant in a vacuum.

(A) The intensity $I$ is defined as the power per unit area,so the power $P = I A$.
The momentum $p$ of a photon is given by $p = E / c$,where $E$ is the energy.
For a perfectly reflecting surface,the change in momentum $\Delta p$ for a photon is $p_{final} - p_{initial} = (-p) - (p) = -2p = -2E / c$.
The force $F$ is the rate of change of momentum,$F = \frac{dp}{dt} = \frac{2}{c} \frac{dE}{dt}$.
Since the power $P = \frac{dE}{dt} = I A$,the force exerted on the surface is $F = \frac{2 I A}{c}$.

(D) For an electromagnetic wave propagating in a vacuum,the relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ is given by the equation:
$v = \frac{E}{B}$
where $v$ is the speed of the electromagnetic wave.
In a vacuum,the speed of an electromagnetic wave is equal to the speed of light,$c$.
Therefore,the ratio is:
$\frac{E}{B} = c = 3 \times 10^8 \,ms^{-1}$
Thus,the ratio of the magnitudes of the electric field to the magnetic field is of the order of $10^8 \,ms^{-1}$.

(C) Given: $E_{0}=120 \text{ NC}^{-1}$,$f=50 \text{ MHz} = 50 \times 10^{6} \text{ Hz}$.
$(a)$ Magnetic field amplitude: $B_{0} = \frac{E_{0}}{c} = \frac{120}{3 \times 10^{8}} = 40 \times 10^{-8} \text{ T} = 400 \text{ nT}$. (Correct)
$(b)$ Angular frequency: $\omega = 2\pi f = 2\pi \times 50 \times 10^{6} = \pi \times 10^{8} \text{ rad/s}$. (Correct)
$(c)$ Propagation constant: $k = \frac{\omega}{c} = \frac{\pi \times 10^{8}}{3 \times 10^{8}} = \frac{\pi}{3} \approx 1.047 \text{ rad/m}$. The given value $2.1 \text{ rad/m}$ is incorrect.
$(d)$ Wavelength: $\lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{50 \times 10^{6}} = 6 \text{ m}$. (Correct)

(C) The electric field is given by $\vec{E} = E_{0} \sin(kx - \omega t) \hat{j}$.
The wave travels in the $+x$-direction,so the direction of propagation is $\hat{i}$.
The relation between the magnitudes of the electric and magnetic fields is $E_{0} = C B_{0}$,which implies $B_{0} = \frac{E_{0}}{C}$.
The direction of the magnetic field $\vec{B}$ is given by the direction of $\vec{c} \times \vec{E}$,where $\vec{c}$ is the direction of wave propagation.
Here,$\hat{i} \times \hat{j} = \hat{k}$.
Therefore,the magnetic field vector is $\vec{B}_{z} = \frac{E_{0}}{C} \sin(kx - \omega t) \hat{k}$.

(D) $X$-rays,gamma rays,and microwaves are all types of electromagnetic waves.
In a vacuum,all electromagnetic waves travel at the same speed,which is the speed of light $(c \approx 3 \times 10^8 \ m/s)$.
However,they differ in their frequencies and wavelengths according to the relation $c = f \lambda$,where $f$ is the frequency and $\lambda$ is the wavelength.
Since their frequencies are different,their wavelengths must also be different to maintain the same constant velocity $c$.

(C) The speed of all electromagnetic waves in a vacuum is given by the formula $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
Here,$\mu_{0}$ is the permeability of free space and $\varepsilon_{0}$ is the permittivity of free space.
Since both $\mu_{0}$ and $\varepsilon_{0}$ are universal constants,the speed of electromagnetic waves in a vacuum is a constant value,approximately $3 \times 10^{8} \ m/s$,regardless of the frequency,wavelength,or the source of the radiation.
Therefore,the speed is the same for all electromagnetic waves in a vacuum.

(C) The speed of light in vacuum $(c)$ is related to the fundamental electromagnetic constants of free space,namely the permeability of free space $(\mu_{0})$ and the permittivity of free space $(\varepsilon_{0})$.
According to Maxwell's equations,the speed of electromagnetic waves in vacuum is given by the formula:
$c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$
Thus,the correct expression is $\sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}$.

(A) The relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by the equation:
$E = cB$,where $c$ is the speed of light in vacuum.
Given values are $E = 6 \text{ V m}^{-1}$ and $c = 3 \times 10^{8} \text{ m s}^{-1}$.
Rearranging the formula to solve for $B$:
$B = \frac{E}{c}$
Substituting the values:
$B = \frac{6}{3 \times 10^{8}}$
$B = 2 \times 10^{-8} \text{ T}$.
Therefore,the magnitude of the magnetic field vector at that point is $2 \times 10^{-8} \text{ T}$.

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

(B) In a microwave oven,the frequency of the microwaves is chosen to be equal to the resonant frequency of water molecules.
When the frequency of the electromagnetic waves matches the natural resonant frequency of the water molecules,the molecules absorb the energy efficiently through the process of dielectric heating.
This resonance causes the water molecules to rotate and vibrate rapidly,generating heat that cooks the food.