Properties of Electromagnetic Waves Questions in English

(D) Given: Amplitude of electric field $E_0 = 66\, Vm^{-1}$,Wavelength $\lambda = 3\, mm = 3 \times 10^{-3}\, m$,Wave speed $c = 3 \times 10^8\, ms^{-1}$.
$1$. Calculate the amplitude of the magnetic field $B_0$:
$B_0 = E_0 / c = 66 / (3 \times 10^8) = 22 \times 10^{-8} = 2.2 \times 10^{-7}\, T$.
$2$. Calculate the angular frequency $\omega$:
$\omega = 2\pi f = 2\pi (c / \lambda) = 2\pi (3 \times 10^8 / 3 \times 10^{-3}) = 2\pi \times 10^{11}\, rad/s$.
$3$. Determine the directions:
The wave travels along $X$,and $E$ is along $Y$. Since the direction of propagation is given by $\vec{E} \times \vec{B}$,the magnetic field $B$ must be along the $Z-$ direction.
$4$. Write the equations:
$E_y = E_0 \cos(\omega(t - x/c)) = 66 \cos(2\pi \times 10^{11}(t - x/c))$.
$B_z = B_0 \cos(\omega(t - x/c)) = 2.2 \times 10^{-7} \cos(2\pi \times 10^{11}(t - x/c))$.
Comparing with the options,option $D$ matches these values.

(B) The average intensity $I_{av}$ of an electromagnetic wave is given by the formula:
$I_{av} = \frac{1}{2} c \epsilon_0 E_0^2$
Given:
$E_0 = 36 \, V/m$
$c = 3 \times 10^8 \, m/s$
$\epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)$
Substituting the values:
$I_{av} = \frac{1}{2} \times (3 \times 10^8) \times (8.85 \times 10^{-12}) \times (36)^2$
$I_{av} = 0.5 \times 3 \times 10^8 \times 8.85 \times 10^{-12} \times 1296$
$I_{av} = 1.5 \times 8.85 \times 1296 \times 10^{-4}$
$I_{av} = 17208.6 \times 10^{-4} \approx 1.72 \, W/m^2$

(B) The intensity $I$ of an electromagnetic wave 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 $E_0$ is $I = \frac{E_0^2}{2 \mu_0 c}$.
Equating the two expressions: $\frac{P_{av}}{4 \pi r^2} = \frac{E_0^2}{2 \mu_0 c}$.
Solving for $E_0$: $E_0 = \sqrt{\frac{2 \mu_0 c P_{av}}{4 \pi r^2}} = \sqrt{\frac{\mu_0 c P_{av}}{2 \pi r^2}}$.
Substituting the given values: $P_{av} = 800\,W$,$r = 3.5\,m$,$\mu_0 = 4 \pi \times 10^{-7}\,T\cdot m/A$,and $c = 3 \times 10^8\,m/s$.
$E_0 = \sqrt{\frac{(4 \pi \times 10^{-7}) \times (3 \times 10^8) \times 800}{2 \pi \times (3.5)^2}}$.
$E_0 = \sqrt{\frac{2 \times 10^{-7} \times 3 \times 10^8 \times 800}{12.25}} = \sqrt{\frac{48000}{12.25}} \approx \sqrt{3918.36} \approx 62.6\,V/m$.

(B) The energy density $u_E$ associated with the electric field in an electromagnetic wave is given by $u_E = \frac{1}{2} \epsilon_0 E^2$.
The energy density $u_B$ associated with the magnetic field in an electromagnetic wave is given by $u_B = \frac{B^2}{2 \mu_0}$.
In an electromagnetic wave,the total energy is shared equally between the electric and magnetic fields,such that $u_E = u_B$.

(C) The intensity of an electromagnetic wave is given by $I = U_{av} c$.
In terms of the electric field, the average energy density is $U_{av} = \frac{1}{2} \varepsilon_{0} E_{0}^{2}$.
In terms of the magnetic field, the average energy density is $U_{av} = \frac{1}{2} \frac{B_{0}^{2}}{\mu_{0}}$.
Since the relationship between the electric field amplitude $E_{0}$ and magnetic field amplitude $B_{0}$ is $E_{0} = c B_{0}$, we can substitute this into the electric field energy density expression:
$U_{av} (\text{electric}) = \frac{1}{2} \varepsilon_{0} (c B_{0})^{2} = \frac{1}{2} \varepsilon_{0} c^{2} B_{0}^{2}$.
Using the relation $c^{2} = \frac{1}{\mu_{0} \varepsilon_{0}}$, we get:
$U_{av} (\text{electric}) = \frac{1}{2} \varepsilon_{0} \left( \frac{1}{\mu_{0} \varepsilon_{0}} \right) B_{0}^{2} = \frac{1}{2} \frac{B_{0}^{2}}{\mu_{0}} = U_{av} (\text{magnetic})$.
Thus, the contributions from both fields are equal, and the ratio is $1 : 1$.

(D) The given equation for the electric field is:
$E = 3.1 \cos [ (1.8) y + (5.4 \times 10^8) t ] \hat{i}$ .......... $(i)$
Comparing this with the standard wave equation:
$E = E_0 \cos (ky + \omega t)$ .......... $(ii)$
We identify the wave number $k$ as:
$k = 1.8 \, rad \, m^{-1}$
The relationship between wavelength $\lambda$ and wave number $k$ is given by:
$\lambda = \frac{2 \pi}{k}$
Substituting the value of $k$ and using $\pi \approx 3.14159$:
$\lambda = \frac{2 \times 3.14159}{1.8} \approx 3.49 \, m$
Rounding to the nearest provided option,we get:
$\lambda \approx 3.5 \, m$

(A) The speed of light in vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
For an electromagnetic wave,the ratio of the magnitude of the electric field to the magnetic field is equal to the speed of light,i.e.,$\frac{E}{B} = c$.
Substituting this into the expression in option $A$:
$\sqrt{\mu_0 \varepsilon_0} \cdot \frac{E}{B} = \sqrt{\mu_0 \varepsilon_0} \cdot c = \sqrt{\mu_0 \varepsilon_0} \cdot \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 1$.
Since $1$ is a dimensionless constant,the expression $\sqrt{\mu_0 \varepsilon_0} \frac{E}{B}$ is dimensionless.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,$\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})$.
Since the velocity vector $\vec{v}$ of the electromagnetic wave is in the same direction as the energy flow (Poynting vector),the velocity is in the direction of $\vec{E} \times \vec{B}$.
Therefore,the correct option is $B$.

(D) The general equation for a wave is $E = E_0 \cos(\omega t - kx)$.
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 $\omega = 2\pi \times 10^6 \, rad/s$ and the wave number $k = \pi \times 10^{-2} \, rad/m$.
The wave propagates in the $+x$ direction because the term is $(\omega t - kx)$.
The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{2\pi \times 10^6}{2\pi} = 10^6 \, Hz$.
The wavelength $\lambda$ is given by $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi \times 10^{-2}} = 200 \, m$.

(B) The given electric field is of the form $E_y = E_0 \cos(\omega t - kx)$.
Comparing this with the standard wave equation, the wave is moving in the positive $X$-direction.
From the equation, the angular frequency is $\omega = 2\pi \times 10^6 \, \text{rad/s}$.
Since $\omega = 2\pi f$, we have $2\pi f = 2\pi \times 10^6 \, \text{Hz}$, which gives $f = 10^6 \, \text{Hz}$.
The wave number is $k = \pi \times 10^{-2} \, \text{rad/m}$.
Since $k = \frac{2\pi}{\lambda}$, we have $\frac{2\pi}{\lambda} = \pi \times 10^{-2}$.
Solving for wavelength, $\lambda = \frac{2\pi}{\pi \times 10^{-2}} = 2 \times 10^2 = 200 \, \text{m}$.
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) In an electromagnetic wave,the relationship between the amplitude of the electric field $E_0$ and the amplitude of the magnetic field $B_0$ is given by $E_0 = c B_0$,where $c$ is the speed of light in vacuum.
Since $B_0 = \mu_0 H_0$,we can substitute this into the equation: $E_0 = c \mu_0 H_0$.
We know that $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Substituting $c$ into the equation: $E_0 = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \cdot \mu_0 H_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} H_0$.
Rearranging for $H_0$,we get: $H_0 = E_0 \sqrt{\frac{\varepsilon_0}{\mu_0}}$.

(D) The relationship between the electric field magnitude $E$ and the magnetic field magnitude $B$ in an electromagnetic wave is given by the equation $E = B \cdot c$,where $c$ is the speed of light in vacuum.
Given: $E = 10^4 \, V/m$ and $c = 3 \times 10^8 \, m/s$.
Rearranging the formula to solve for $B$: $B = \frac{E}{c}$.
Substituting the values: $B = \frac{10^4}{3 \times 10^8} \, T$.
$B = \frac{1}{3} \times 10^{4-8} \, T = 0.333 \times 10^{-4} \, T = 3.33 \times 10^{-5} \, T$.
Therefore,the magnitude of the magnetic field strength is $3.3 \times 10^{-5} \, T$.

(C) In an electromagnetic $(EM)$ wave propagating through a vacuum,the electric field $(E)$ and the magnetic field $(B)$ oscillate in the same phase.
This means that both fields reach their maximum and minimum values at the same time and at the same position in space.
The relationship between the magnitudes is given by $E = cB$,where $c$ is the speed of light in a vacuum.

(D) For an electromagnetic wave propagating in 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 $E_0 = cB_0$,where $c$ is the speed of light in vacuum.
Therefore,the ratio of the amplitude of the magnetic field to the amplitude of the electric field is $\frac{B_0}{E_0} = \frac{1}{c}$.
This is equal to the reciprocal of the speed of light in vacuum.

(C) The energy density due to the electric field is given by $u_E = \frac{1}{2} \epsilon_0 E^2$.
The energy density due to the magnetic field is given by $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
In an electromagnetic wave,the relationship between the electric field $E$ and magnetic field $B$ is $E = cB$,where $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
Substituting $B = \frac{E}{c}$ into the magnetic energy density formula,we get $u_B = \frac{1}{2} \frac{(E/c)^2}{\mu_0} = \frac{1}{2} \frac{E^2}{c^2 \mu_0} = \frac{1}{2} \epsilon_0 E^2 = u_E$.
Thus,the average energy density is equally contributed to by both the electric and magnetic fields.

(C) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field.
In these waves,the electric field vector and the magnetic field vector oscillate perpendicular to each other and also perpendicular to the direction of wave propagation.
Since the oscillations of the fields are perpendicular to the direction of propagation,electromagnetic waves are classified as transverse waves.
Therefore,the correct option is $C$.

(A) Given: Frequency $f = 3 \times 10^{10} \, Hz$,Magnetic field amplitude $B_0 = 10^{-7} \, T$.
The amplitude of the electric field is $E_0 = c B_0 = (3 \times 10^8 \, m/s) \times (10^{-7} \, T) = 30 \, V/m$.
The wave propagates along the $X-$ direction and the magnetic field is along the $Y-$ direction,so the electric field must be along the $Z-$ direction $(E_z)$.
The wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi f}{c} = \frac{2\pi (3 \times 10^{10})}{3 \times 10^8} = 2\pi (100) \, rad/m$.
The angular frequency $\omega = 2\pi f = 2\pi (3 \times 10^{10}) \, rad/s$.
The expression for the electric field is $E_z = E_0 \sin(kx - \omega t) = 30 \sin(2\pi(100x - 3 \times 10^{10}t)) \, V/m$.

(B) 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 propagation of the wave (which is the direction of the velocity vector $\vec v$) is given by the cross product of the electric field vector and the magnetic field vector.
Therefore,the velocity vector $\vec v$ is parallel to $\vec E \times \vec B$.
This is also related to the Poynting vector $\vec S = \frac{1}{\mu_0} (\vec E \times \vec B)$,which represents the directional energy flux (the energy transfer per unit area per unit time) of an electromagnetic field.

(A) The given equation for the magnetic field is $B_y = B_m \sin(kz - \omega t)$.
In the wave equation $f(kz - \omega t)$,the term $(kz - \omega t)$ indicates that the wave is propagating in the positive $z-$ direction.
Electromagnetic waves are transverse in nature,meaning the electric field vector $\vec{E}$,the magnetic field vector $\vec{B}$,and the direction of propagation $\vec{k}$ are mutually perpendicular.
Since the magnetic field oscillates along the $y-$ axis and the wave propagates along the $z-$ axis,the electric field must oscillate along the $x-$ axis (as $\vec{E} \propto \vec{B} \times \vec{k}$).
Therefore,the direction of wave travel is the $+z-$ axis and the electric vector oscillates along the $x-$ axis.

(D) The total energy density $u$ of an electromagnetic wave is the sum of the electric energy density $u_E$ and the magnetic energy density $u_B$.
Electric energy density is given by $u_E = \frac{1}{2}\varepsilon_0 E^2$.
Magnetic energy density is given by $u_B = \frac{B^2}{2\mu_0}$.
Therefore,the total energy density is $u = u_E + u_B = \frac{1}{2}\varepsilon_0 E^2 + \frac{B^2}{2\mu_0}$.

(D) An electromagnetic wave consists of an oscillating electric field vector $\vec{E}$ and a magnetic field vector $\vec{B}$.
These vectors oscillate in mutually perpendicular planes,meaning the electric field is perpendicular to the magnetic field.
Furthermore,these fields oscillate in phase,which means they reach their maximum and minimum values at the same time and at the same position in space.
Therefore,the correct description is that they vibrate in mutually perpendicular planes but in phase.

(B) Radio waves are electromagnetic waves,which are transverse in nature. Only transverse waves can be polarized. Therefore,the Assertion is correct.
Sound waves in air are longitudinal waves,which cannot be polarized. Therefore,the Reason is also correct.
However,the fact that sound waves are longitudinal does not explain why radio waves can be polarized. Thus,the Reason is not the correct explanation of the Assertion.

(A) The velocity of light in a medium is given by the formula $v = \frac{1}{\sqrt{\mu \epsilon}}$.
Substituting $\mu = \mu_{r} \mu_{0}$ and $\epsilon = \epsilon_{r} \epsilon_{0}$,we get $v = \frac{1}{\sqrt{\mu_{r} \epsilon_{r} \mu_{0} \epsilon_{0}}}$.
Since the speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}} = 3 \times 10^{8} \;m/s$,the formula becomes $v = \frac{c}{\sqrt{\mu_{r} \epsilon_{r}}}$.
Given $\mu_{r} = 1.0$ and $\epsilon_{r} = 1.44$,we have $v = \frac{3 \times 10^{8}}{\sqrt{1.0 \times 1.44}}$.
$v = \frac{3 \times 10^{8}}{\sqrt{1.44}} = \frac{3 \times 10^{8}}{1.2}$.
$v = 2.5 \times 10^{8} \;m/s$.

(A) The given magnetic field is $\overrightarrow{B} = B_0 \sin (kx + \omega t) \hat{j}$,where $B_0 = 3 \times 10^{-8} \; T$.
Since the wave propagates in the negative $x$-direction (as indicated by $+kx$),the direction of propagation is $-\hat{i}$.
The relationship between the amplitudes of electric and magnetic fields is $E_0 = c B_0$,where $c = 3 \times 10^{8} \; m/s$.
$E_0 = (3 \times 10^{8} \; m/s) \times (3 \times 10^{-8} \; T) = 9 \; V/m$.
In an electromagnetic wave,the direction of propagation is given by the direction of $\overrightarrow{E} \times \overrightarrow{B}$.
Here,$(-\hat{i}) = \hat{E} \times \hat{j}$.
Since $\hat{k} \times \hat{j} = -\hat{i}$,the electric field must be in the $\hat{k}$ direction.
Thus,$\overrightarrow{E} = 9 \sin (1.6 \times 10^{3} x + 48 \times 10^{10} t) \hat{k} \; V/m$.

(D) For an electromagnetic wave propagating in vacuum,the relationship between the electric field $\overrightarrow{E}$,the magnetic field $\overrightarrow{B}$,and the velocity vector $\overrightarrow{c}$ is given by $\overrightarrow{E} = \overrightarrow{c} \times \overrightarrow{B}$.
The wave propagates along the $z$-direction,so the velocity vector is $\overrightarrow{c} = c \hat{k} = (3 \times 10^{8}) \hat{k}\; m/s$.
The magnetic field is given as $\overrightarrow{B} = 5 \times 10^{-8} \hat{j}\; T$.
Substituting these values into the formula:
$\overrightarrow{E} = (3 \times 10^{8} \hat{k}) \times (5 \times 10^{-8} \hat{j})$
Using the cross product rules for unit vectors $(\hat{k} \times \hat{j} = -\hat{i})$:
$\overrightarrow{E} = (3 \times 5) \times (10^{8} \times 10^{-8}) \times (\hat{k} \times \hat{j})$
$\overrightarrow{E} = 15 \times 1 \times (-\hat{i})$
$\overrightarrow{E} = -15 \hat{i}\; V/m$.

(A) The direction of propagation is given by $\hat{n} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
The electric field polarization is given by $\hat{E} = \hat{k}$.
In an electromagnetic wave,the direction of propagation $\hat{n}$ is given by the cross product of the electric field direction $\hat{E}$ and the magnetic field direction $\hat{B}$,i.e.,$\hat{n} = \hat{E} \times \hat{B}$.
Substituting the known values: $\frac{\hat{i}+\hat{j}}{\sqrt{2}} = \hat{k} \times \hat{B}$.
We know that $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{k} \times \hat{j} = -\hat{i}$.
Therefore,$\hat{k} \times \left( \frac{\hat{i}-\hat{j}}{\sqrt{2}} \right) = \frac{\hat{j} - (-\hat{i})}{\sqrt{2}} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
Thus,the direction of the magnetic field is $\hat{B} = \frac{\hat{i}-\hat{j}}{\sqrt{2}}$.
Hence,the magnetic field vector is $\vec{B} = B_{0} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos \left(\omega t - \vec{k} \cdot \vec{r}\right)$.

(C) The electric fields are $\overrightarrow{E}_{1}=E_{0} \hat{j} \cos (\omega t-kx)$ and $\overrightarrow{E}_{2}=E_{0} \hat{k} \cos (\omega t-ky)$.
At $t=0$ and origin $(0,0,0)$,$\overrightarrow{E}_{1} = E_{0} \hat{j}$ and $\overrightarrow{E}_{2} = E_{0} \hat{k}$.
The corresponding magnetic fields are given by $\overrightarrow{B} = \frac{1}{c} (\hat{n} \times \overrightarrow{E})$,where $\hat{n}$ is the direction of propagation.
For $\overrightarrow{E}_{1}$,$\hat{n} = \hat{i}$,so $\overrightarrow{B}_{1} = \frac{1}{c} (\hat{i} \times E_{0} \hat{j}) = \frac{E_{0}}{c} \hat{k}$.
For $\overrightarrow{E}_{2}$,$\hat{n} = \hat{j}$,so $\overrightarrow{B}_{2} = \frac{1}{c} (\hat{j} \times E_{0} \hat{k}) = \frac{E_{0}}{c} \hat{i}$.
The Lorentz force is $\overrightarrow{F} = q(\overrightarrow{E}_{1} + \overrightarrow{E}_{2}) + q(\overrightarrow{v} \times (\overrightarrow{B}_{1} + \overrightarrow{B}_{2}))$.
Substituting values: $\overrightarrow{F} = q(E_{0} \hat{j} + E_{0} \hat{k}) + q(0.8 c \hat{j} \times (\frac{E_{0}}{c} \hat{k} + \frac{E_{0}}{c} \hat{i}))$.
$\overrightarrow{F} = q E_{0} \hat{j} + q E_{0} \hat{k} + 0.8 q E_{0} (\hat{j} \times \hat{k}) + 0.8 q E_{0} (\hat{j} \times \hat{i})$.
Using cross products $\hat{j} \times \hat{k} = \hat{i}$ and $\hat{j} \times \hat{i} = -\hat{k}$:
$\overrightarrow{F} = q E_{0} \hat{j} + q E_{0} \hat{k} + 0.8 q E_{0} \hat{i} - 0.8 q E_{0} \hat{k} = q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$.

(N/A) Non-mechanical waves are waves that do not require a material medium for their propagation. These waves can travel through a vacuum.
They are also known as electromagnetic waves.
Examples of non-mechanical waves include:
$1$. Light waves
$2$. Radio waves
$3$. $X$-rays
$4$. Gamma rays
$5$. Microwaves

(B) The magnitude of the magnetic field $B$ is related to the electric field $E$ by the relation $B = \frac{E}{c}$,where $c = 3 \times 10^8 \; m/s$ is the speed of light in free space.
Substituting the given values: $B = \frac{6.3 \; V/m}{3 \times 10^8 \; m/s} = 2.1 \times 10^{-8} \; T$.
To find the direction,we know that the electromagnetic wave propagates in the direction of the vector $\vec{E} \times \vec{B}$.
Given that the wave propagates along the $x$-direction $(\hat{i})$ and the electric field is along the $y$-direction $(\hat{j})$,we have $\hat{j} \times \hat{B} = \hat{i}$.
Since $\hat{j} \times \hat{k} = \hat{i}$,the magnetic field must be in the $z$-direction $(\hat{k})$.
Therefore,$\vec{B} = 2.1 \times 10^{-8} \hat{k} \; T$.

(A) Comparing the given equation with the standard form $B_y = B_0 \sin(kx + \omega t)$:
Here,$k = 0.5 \times 10^3 \, rad/m$ and $\omega = 1.5 \times 10^{11} \, rad/s$.
Wavelength $\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{0.5 \times 10^3} \approx 1.26 \, m$.
Frequency $\nu = \frac{\omega}{2\pi} = \frac{1.5 \times 10^{11}}{2 \times 3.14} \approx 2.39 \times 10^{10} \, Hz = 23.9 \, GHz$.
$(b)$ The amplitude of the electric field is $E_0 = B_0 c = (2 \times 10^{-7} \, T) \times (3 \times 10^8 \, m/s) = 60 \, V/m$.
Since the wave propagates in the negative $x$-direction and the magnetic field is along the $y$-axis,the electric field must be along the $z$-axis. Thus,the expression is:
$E_z = 60 \sin (0.5 \times 10^3 x + 1.5 \times 10^{11} t) \, V/m$.

(A) The bulb,as a point source,radiates light in all directions uniformly. At a distance of $r = 3\; m$,the surface area of the surrounding sphere is $A = 4\pi r^2 = 4\pi(3)^2 = 113\; m^2$.
The power radiated by the bulb is $P = 100\; W \times 2.5\% = 2.5\; W$.
The intensity $I$ at this distance is $I = \frac{P}{A} = \frac{2.5\; W}{113\; m^2} \approx 0.022\; W/m^2$.
The average energy density is related to the electric field by $I = \varepsilon_0 E_{rms}^2 c$. Thus,$E_{rms} = \sqrt{\frac{I}{\varepsilon_0 c}}$.
Substituting the values: $E_{rms} = \sqrt{\frac{0.022}{(8.85 \times 10^{-12})(3 \times 10^8)}} \approx 2.87\; V/m \approx 2.9\; V/m$.
The peak electric field $E_0 = \sqrt{2} E_{rms} = \sqrt{2} \times 2.9 \approx 4.1\; V/m$.
The root mean square magnetic field is $B_{rms} = \frac{E_{rms}}{c} = \frac{2.9}{3 \times 10^8} \approx 9.7 \times 10^{-9}\; T$.
The peak magnetic field $B_0 = \sqrt{2} B_{rms} = \sqrt{2} \times 9.7 \times 10^{-9} \approx 1.37 \times 10^{-8}\; T$.

(C) All electromagnetic waves,regardless of their wavelength or frequency,travel at the same speed in a vacuum.
This speed is denoted by $c$ and is approximately $3 \times 10^{8} \; m/s$.
Since $X$-rays,red light,and radio waves are all parts of the electromagnetic spectrum,they share this fundamental property in a vacuum.

(A) The electromagnetic wave travels in a vacuum along the $z$-direction. The electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are in the $x-y$ plane and are mutually perpendicular to each other and to the direction of propagation ($z$-direction).
Given frequency of the wave,$\nu = 30 \; MHz = 30 \times 10^{6} \; s^{-1}$.
Speed of light in a vacuum,$c = 3 \times 10^{8} \; m/s$.
The wavelength $\lambda$ is given by the formula $\lambda = \frac{c}{\nu}$.
Substituting the values: $\lambda = \frac{3 \times 10^{8}}{30 \times 10^{6}} = \frac{300 \times 10^{6}}{30 \times 10^{6}} = 10 \; m$.

(N/A) The radio tunes to a minimum frequency of $v_{1} = 7.5\; MHz = 7.5 \times 10^{6}\; Hz$.
The maximum frequency is $v_{2} = 12\; MHz = 12 \times 10^{6}\; Hz$.
The speed of light is $c = 3 \times 10^{8}\; m/s$.
The wavelength $\lambda$ is related to frequency $v$ by the formula $\lambda = \frac{c}{v}$.
For the minimum frequency $v_{1}$,the maximum wavelength $\lambda_{1}$ is:
$\lambda_{1} = \frac{c}{v_{1}} = \frac{3 \times 10^{8}}{7.5 \times 10^{6}} = 40\; m$.
For the maximum frequency $v_{2}$,the minimum wavelength $\lambda_{2}$ is:
$\lambda_{2} = \frac{c}{v_{2}} = \frac{3 \times 10^{8}}{12 \times 10^{6}} = 25\; m$.
Thus,the corresponding wavelength band is $25\; m$ to $40\; m$.

(A) According to the theory of electromagnetic waves,an oscillating charge is a source of electromagnetic radiation. The frequency of the electromagnetic waves produced by an oscillating charge is equal to the frequency of oscillation of the charge itself. Therefore,the frequency of the electromagnetic waves is $10^9 \; Hz$.

(D) The amplitude of the magnetic field of an electromagnetic wave in a vacuum is given as $B_0 = 510 \; nT = 510 \times 10^{-9} \; T$.
The speed of light in a vacuum is $c = 3 \times 10^8 \; m/s$.
The amplitude of the electric field $(E_0)$ of an electromagnetic wave is related to the amplitude of the magnetic field $(B_0)$ by the relation $E_0 = c B_0$.
Substituting the given values:
$E_0 = (3 \times 10^8 \; m/s) \times (510 \times 10^{-9} \; T)$
$E_0 = 3 \times 510 \times 10^{-1} \; N/C$
$E_0 = 1530 \times 10^{-1} \; N/C = 153 \; N/C$.
Therefore,the amplitude of the electric field part of the wave is $153 \; N/C$.

(N/A) Electric field amplitude,$E_{0} = 120 \; N/C$.
Frequency of source,$\nu = 50.0 \; MHz = 50 \times 10^{6} \; Hz$.
Speed of light,$c = 3 \times 10^{8} \; m/s$.
$(a)$ Magnitude of magnetic field strength is given as:
$B_{0} = \frac{E_{0}}{c} = \frac{120}{3 \times 10^{8}} = 4 \times 10^{-7} \; T = 400 \; nT$.
Angular frequency of source is given as:
$\omega = 2 \pi \nu = 2 \pi \times 50 \times 10^{6} = 3.14 \times 10^{8} \; rad/s$.
Propagation constant is given as:
$k = \frac{\omega}{c} = \frac{3.14 \times 10^{8}}{3 \times 10^{8}} = 1.05 \; rad/m$.
Wavelength of wave is given as:
$\lambda = \frac{c}{\nu} = \frac{3 \times 10^{8}}{50 \times 10^{6}} = 6.0 \; m$.
$(b)$ Suppose the wave is propagating in the positive $x$-direction. Then,the electric field vector will be in the positive $y$-direction and the magnetic field vector will be in the positive $z$-direction. This is because all three vectors are mutually perpendicular. Equation of electric field vector is given as:
$\vec{E} = E_{0} \sin(kx - \omega t) \hat{j} = 120 \sin(1.05x - 3.14 \times 10^{8}t) \hat{j} \; V/m$.
And,magnetic field vector is given as:
$\vec{B} = B_{0} \sin(kx - \omega t) \hat{k} = (4 \times 10^{-7}) \sin(1.05x - 3.14 \times 10^{8}t) \hat{k} \; T$.

(A) Given:
Frequency $v = 2.0 \times 10^{10} \; Hz$
Electric field amplitude $E_{0} = 48 \; V m^{-1}$
Speed of light $c = 3 \times 10^{8} \; m s^{-1}$
$(a)$ Wavelength $\lambda = \frac{c}{v} = \frac{3 \times 10^{8}}{2.0 \times 10^{10}} = 0.015 \; m$.
$(b)$ Magnetic field amplitude $B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{-7} \; T$.
$(c)$ Energy density of the electric field is $U_{E} = \frac{1}{2} \epsilon_{0} E^{2}$ and for the magnetic field is $U_{B} = \frac{B^{2}}{2 \mu_{0}}$.
Using $E = cB$ and $c = \frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}$,we have $E^{2} = c^{2} B^{2} = \frac{B^{2}}{\epsilon_{0} \mu_{0}}$.
Thus,$\epsilon_{0} E^{2} = \frac{B^{2}}{\mu_{0}}$.
Dividing by $2$,we get $\frac{1}{2} \epsilon_{0} E^{2} = \frac{B^{2}}{2 \mu_{0}}$,which implies $U_{E} = U_{B}$.

(A) The electric field is oscillating along the $x$-axis and varies with $y$ and $t$. Since the argument of the cosine function is $(ky + \omega t)$,the wave propagates in the negative $y$-direction,i.e.,$-\hat{j}$.
$(b)$ Comparing the given equation with the standard form $E = E_0 \cos(ky + \omega t)$,we have $k = 1.8 \; rad/m$. The wavelength $\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{1.8} \approx 3.49 \; m$.
$(c)$ The angular frequency $\omega = 5.4 \times 10^{6} \; rad/s$. The frequency $\nu = \frac{\omega}{2\pi} = \frac{5.4 \times 10^{6}}{2 \times 3.14} \approx 8.6 \times 10^{5} \; Hz$.
$(d)$ The amplitude of the magnetic field $B_0 = \frac{E_0}{c}$,where $E_0 = 3.1 \; N/C$ and $c = 3 \times 10^{8} \; m/s$. Thus,$B_0 = \frac{3.1}{3 \times 10^{8}} \approx 1.03 \times 10^{-8} \; T$.
$(e)$ Since $\vec{E}$ is along $\hat{i}$ and propagation is along $-\hat{j}$,the magnetic field $\vec{B}$ must be along $\hat{k}$ (as $\vec{E} \times \vec{B}$ gives the direction of propagation). The expression is $\vec{B} = \{(1.03 \times 10^{-8} \; T) \cos [(1.8 \; rad/m) y + (5.4 \times 10^{6} \; rad/s) t] \} \hat{k}$.

(A) Power rating of the bulb,$P = 100 \; W$.
It is given that about $5 \%$ of its power is converted into visible radiation.
$\therefore$ Power of visible radiation,$P' = \frac{5}{100} \times 100 = 5 \; W$.
$(a)$ At a distance of $d = 1 \; m$ from the bulb,the intensity $I$ is given by the formula $I = \frac{P'}{4 \pi d^2}$.
$I = \frac{5}{4 \times 3.14 \times (1)^2} \approx 0.398 \; W/m^2$.
$(b)$ At a distance of $d = 10 \; m$ from the bulb,the intensity $I$ is given by $I = \frac{P'}{4 \pi d^2}$.
$I = \frac{5}{4 \times 3.14 \times (10)^2} = \frac{5}{1256} \approx 0.00398 \; W/m^2$.

(N/A) Long distance radio broadcasts use short-wave bands because these waves are refracted by the ionosphere,allowing them to travel long distances around the curvature of the Earth.
$(b)$ $TV$ signals have very high frequencies and high energies. They are not reflected by the ionosphere; therefore,satellites are required to receive and retransmit these signals for long-distance communication.
$(c)$ The Earth's atmosphere absorbs $X$-rays,preventing them from reaching the ground. However,visible light and radio waves can penetrate the atmosphere,allowing optical and radio telescopes to operate on the ground.
$(d)$ The ozone layer absorbs harmful ultraviolet $(UV)$ radiation from the Sun,which would otherwise cause severe damage to living organisms,including skin cancer and genetic mutations.
$(e)$ Without an atmosphere,there would be no greenhouse effect. The Earth's average surface temperature would be significantly lower than it is now because the atmosphere traps heat radiated from the surface.
$(f)$ $A$ global nuclear war would release massive amounts of smoke and dust into the atmosphere. These particles would block sunlight from reaching the Earth's surface,leading to a drastic drop in temperature,known as a 'nuclear winter',and the depletion of the ozone layer.

(N/A) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field.
In other words,electromagnetic waves are composed of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation.
These waves do not require a material medium for their propagation and can travel through a vacuum at the speed of light,$c \approx 3 \times 10^8 \ m/s$.

(N/A) The characteristics of electromagnetic waves are as follows:
$(1)$ In electromagnetic waves,the electric field and magnetic field are perpendicular to each other as well as perpendicular to the direction of wave propagation.
$(2)$ Electromagnetic waves are transverse in nature. If a plane electromagnetic wave propagates in the $z$-direction,the electric field $E_{x}$ oscillates in the $x$-direction and the magnetic field $B_{y}$ oscillates in the $y$-direction. They vary according to a sine function.
$(3)$ The mathematical representation is:
$E_{x} = E_{0} \sin(kz - \omega t)$
$B_{y} = B_{0} \sin(kz - \omega t)$
Thus,$\vec{E} = E_{0} \sin(kz - \omega t) \hat{i}$ and $\vec{B} = B_{0} \sin(kz - \omega t) \hat{j}$.
$(4)$ The speed of electromagnetic waves in vacuum is given by $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$,where $c = \frac{\omega}{k}$.
$(5)$ Electromagnetic waves do not require any material medium for propagation; they can travel through a vacuum.

(A) Jagdish Chandra Bose was a pioneer in the study of radio waves. In $1895$,he successfully produced and detected electromagnetic waves in the millimeter range,specifically between $5 \ mm$ and $25 \ mm$. This work was a significant contribution to the development of modern wireless communication.

(A) For an electromagnetic wave propagating in the $+x$ direction,the electric field vector $\vec{E}$ oscillates in the $y$-direction and the magnetic field vector $\vec{B}$ oscillates in the $z$-direction.
The general form of a plane electromagnetic wave traveling in the $+x$ direction is given by the wave functions:
$E_y = E_0 \sin(kx - \omega t)$
$B_z = B_0 \sin(kx - \omega t)$
Here,$E_0$ and $B_0$ are the amplitudes of the electric and magnetic fields,$k$ is the wave number,and $\omega$ is the angular frequency. These fields are in phase and satisfy the relation $E_0 = cB_0$,where $c$ is the speed of light.

(C) In an electromagnetic wave,the relationship between the electric field $(E)$ and the magnetic field $(B)$ is given by the equation $E = cB$,where $c$ is the speed of light in vacuum.
Therefore,the ratio of the electric field to the magnetic field is $\frac{E}{B} = c$.
This ratio represents the speed of light in the medium (or vacuum).

(C) plane electromagnetic wave traveling in the positive $x$-direction consists of oscillating electric and magnetic fields.
The electric field vector oscillates along the $y$-axis and is given by $E_y = E_0 \sin(kx - \omega t)$.
The magnetic field vector oscillates along the $z$-axis and is given by $B_z = B_0 \sin(kx - \omega t)$.
Here,$E_0$ and $B_0$ are the amplitudes of the electric and magnetic fields,$k$ is the wave number,and $\omega$ is the angular frequency.
Therefore,both equations represent the standard form of a plane electromagnetic wave.

(A) The speed of light in free space (vacuum) is a universal physical constant denoted by $c$.
Its value is approximately $3 \times 10^8 \ m/s$ (more precisely $299,792,458 \ m/s$).
Therefore,the correct option is $A$.

(A) In a simple pendulum,the displacement is represented by the angular displacement $\theta$,which is the angle the string makes with the vertical axis.
In the propagation of light (an electromagnetic wave),the displacement variable is the oscillating electric field vector $\vec{E}$ (and the magnetic field vector $\vec{B}$),which varies periodically in space and time.

We know that the electrostatic constant is $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \text{ Nm}^{2}/\text{C}^{2}$ and the magnetic constant is $\frac{\mu_{0}}{4 \pi} = 10^{-7} \text{ Tm/A}$.
Multiplying these two constants:
$\mu_{0} \epsilon_{0} = \left( \frac{\mu_{0}}{4 \pi} \right) \left( 4 \pi \epsilon_{0} \right) = (10^{-7}) \left( \frac{1}{9 \times 10^{9}} \right) = \frac{1}{9 \times 10^{16}}$.
Since the speed of light in vacuum is $c = 3 \times 10^{8} \text{ m/s}$, we can write:
$\mu_{0} \epsilon_{0} = \frac{1}{(3 \times 10^{8})^{2}} = \frac{1}{c^{2}}$.
Therefore, the relation is $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$.