Properties of Electromagnetic Waves Questions in English

(N/A) The total energy density $u$ of an electromagnetic wave is the sum of the energy density due to the electric field $(u_E)$ and the energy density due to the magnetic field $(u_B)$.
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 $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
Since in an electromagnetic wave,the energy is equally divided between the electric and magnetic fields $(u_E = u_B)$,the total energy density is:
$u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$.
Alternatively,it can be expressed as $u = \epsilon_0 E^2 = \frac{B^2}{\mu_0}$.

$(A)$ Microwaves are a part of the electromagnetic spectrum with frequencies ranging from $1 \text{ GHz}$ to $300 \text{ GHz}$.
They are produced by the oscillation of electrons in special vacuum tubes, such as klystrons, magnetrons, and Gunn diodes.
These devices facilitate the rapid oscillation of electrons, which in turn generates electromagnetic waves in the microwave frequency range.

(B) Electromagnetic waves emitted by a broadcasting station are plane-polarized.
This means the electric field vector of the wave oscillates in a specific plane.
For maximum reception,the antenna of the portable radio must be oriented parallel to the direction of the electric field vector of the incoming electromagnetic wave.
If the antenna is perpendicular to the electric field,the induced electromotive force $(EMF)$ will be minimal,resulting in poor or no signal reception.

(N/A) microwave oven heats food items containing water molecules most efficiently because the frequency of the microwaves produced by the oven matches the resonant frequency of the water molecules.
When these microwaves interact with the water molecules,they cause the molecules to oscillate and rotate rapidly due to resonance.
This rapid motion leads to internal friction between the molecules,which generates thermal energy (heat) throughout the food item,resulting in efficient heating.

(A) Comparing the given equation $B = 12 \times 10^{-8} \sin(1.20 \times 10^7 z - 3.60 \times 10^{15} t) \text{ T}$ with the standard wave equation,the amplitude of the magnetic field is $B_0 = 12 \times 10^{-8} \text{ T}$.
The average intensity $I_{\text{avg}}$ of an electromagnetic wave is given by the formula:
$I_{\text{avg}} = \frac{B_0^2}{2\mu_0} c$
Substituting the values $B_0 = 12 \times 10^{-8} \text{ T}$,$c = 3 \times 10^8 \text{ m/s}$,and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$:
$I_{\text{avg}} = \frac{(12 \times 10^{-8})^2 \times 3 \times 10^8}{2 \times 4\pi \times 10^{-7}}$
$I_{\text{avg}} = \frac{144 \times 10^{-16} \times 3 \times 10^8}{8\pi \times 10^{-7}}$
$I_{\text{avg}} = \frac{432 \times 10^{-8}}{25.12 \times 10^{-7}} \approx 1.72 \text{ W/m}^2$.

(N/A) In an electromagnetic wave,let $\vec E$ be in the $y$-direction,$\vec B$ be in the $z$-direction,and the electromagnetic wave propagate in the $x$-direction. The energy propagation will be in the direction of $\vec E \times \vec B$ (in the $x$-direction).
$\vec E = E_0 \sin(\omega t - kx) \hat j$
$\vec B = B_0 \sin(\omega t - kx) \hat k$
$\therefore \vec S = \frac{1}{\mu_0}(\vec E \times \vec B) = \frac{1}{\mu_0} E_0 B_0 \sin^2(\omega t - kx) (\hat j \times \hat k)$
$\therefore \vec S = \frac{E_0 B_0}{\mu_0} \sin^2(\omega t - kx) \hat i$ [since $\hat j \times \hat k = \hat i$]
The variation in the magnitude of $|\vec S|$ with time is shown in the graph below. The magnitude $|\vec S|$ varies as $\sin^2(\omega t - kx)$,meaning it is always non-negative and oscillates with a frequency double that of the electric or magnetic field,with a period of $T = \frac{\pi}{\omega}$.

(N/A) The radiant flux density (Poynting vector) is given by $S = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$.
Since $B = \frac{E}{c}$,the magnitude is $S = \frac{EB}{\mu_0} = \frac{E^2}{c\mu_0}$.
For an electromagnetic wave,$E = E_0 \cos(kx - \omega t)$.
Substituting this into the expression for $S$,we get $S = \frac{E_0^2 \cos^2(kx - \omega t)}{c\mu_0}$.
The average value of $\cos^2(\theta)$ over a full period $T$ is $\frac{1}{T} \int_0^T \cos^2(\omega t) dt = \frac{1}{2}$.
Therefore,the average radiant flux density $\langle S \rangle$ is $\langle S \rangle = \frac{E_0^2}{c\mu_0} \times \frac{1}{2} = \frac{E_0^2}{2c\mu_0}$.

(N/A) The radiation pressure $P$ is defined as the force per unit area,$P = \frac{F}{A}$.
According to Newton's second law,force is the rate of change of momentum,$F = \frac{dp}{dt}$.
For an electromagnetic wave,the relationship between energy $U$ and momentum $p$ is given by $p = \frac{U}{c}$.
Substituting this into the force equation,we get $F = \frac{d}{dt} \left( \frac{U}{c} \right) = \frac{1}{c} \frac{dU}{dt}$.
Intensity $I$ is defined as the energy incident per unit area per unit time,$I = \frac{1}{A} \frac{dU}{dt}$,which implies $\frac{dU}{dt} = I \cdot A$.
Substituting this into the expression for force,$F = \frac{1}{c} (I \cdot A)$.
Finally,the pressure $P = \frac{F}{A} = \frac{I \cdot A}{A \cdot c} = \frac{I}{c}$.

(N/A) The electric field of an electromagnetic wave is an oscillating field, and so is the electric force it exerts on a charged particle. The force is given by $F = qE_0 \sin(\omega t - kx)$. When averaged over an integral number of cycles, the net force is zero because the direction of the force reverses every half cycle. Since radiation pressure is defined as the time-averaged force per unit area, the oscillating electric field does not contribute to it. However, the electric field does perform work on the charge, thereby transferring energy.

(N/A) $(i)$ The line integral $\oint \vec E \cdot d\vec l$ over the loop $1234$ is $\int_1^2 \vec E \cdot d\vec l + \int_2^3 \vec E \cdot d\vec l + \int_3^4 \vec E \cdot d\vec l + \int_4^1 \vec E \cdot d\vec l$. Since $\vec E$ is along $\hat i$ and the segments $1-2$ and $3-4$ are along $\hat z$,the dot product is zero. For segments $2-3$ and $4-1$,$\vec E$ is parallel/anti-parallel to $d\vec l$. Thus,$\oint \vec E \cdot d\vec l = E(z_2, t)h - E(z_1, t)h = h E_0 [\sin(kz_2 - \omega t) - \sin(kz_1 - \omega t)]$.
$(ii)$ The magnetic flux $\phi_B = \int \vec B \cdot d\vec s$. With $d\vec s = h dz \hat j$,$\phi_B = \int_{z_1}^{z_2} B_0 \sin(kz - \omega t) h dz = \frac{B_0 h}{k} [\cos(kz_1 - \omega t) - \cos(kz_2 - \omega t)]$.
$(iii)$ Using Faraday's Law $\oint \vec E \cdot d\vec l = -\frac{d\phi_B}{dt}$,we differentiate the flux with respect to $t$ and equate it to the line integral. Comparing the amplitudes,we get $E_0 = c B_0$,hence $\frac{E_0}{B_0} = c$.
$(iv)$ Using Ampere-Maxwell Law $\oint \vec B \cdot d\vec l = \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E = \int E dA$. Following the same derivation steps as Faraday's law,we obtain $B_0 = \mu_0 \epsilon_0 c E_0$. Substituting $E_0 = c B_0$,we get $B_0 = \mu_0 \epsilon_0 c^2 B_0$,which simplifies to $c^2 = \frac{1}{\mu_0 \epsilon_0}$ or $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

(N/A) $(i)$ The energy density associated with an electric field $E$ is $u_E = \frac{1}{2} \epsilon_0 E^2$ and with a magnetic field $B$ is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
The total instantaneous energy density is $u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$.
For a plane wave,$E = E_0 \sin(kz - \omega t)$ and $B = B_0 \sin(kz - \omega t)$.
The time average of $\sin^2(kz - \omega t)$ over one cycle is $\frac{1}{2}$.
Thus,$\langle E^2 \rangle = \frac{E_0^2}{2}$ and $\langle B^2 \rangle = \frac{B_0^2}{2}$.
Substituting these into the average energy density expression:
$U_{av} = \langle u_E \rangle + \langle u_B \rangle = \frac{1}{2} \epsilon_0 \left( \frac{E_0^2}{2} \right) + \frac{1}{2 \mu_0} \left( \frac{B_0^2}{2} \right) = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0}$.
$(ii)$ Since $E_0 = c B_0$,we have $B_0 = \frac{E_0}{c}$. Also,$c^2 = \frac{1}{\mu_0 \epsilon_0}$,so $\frac{1}{\mu_0} = c^2 \epsilon_0$.
Substituting $B_0$ and $\frac{1}{\mu_0}$ into the energy density formula:
$U_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} (c^2 \epsilon_0) \left( \frac{E_0}{c} \right)^2 = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \epsilon_0 E_0^2 = \frac{1}{2} \epsilon_0 E_0^2$.
The intensity $I_{av}$ is the energy crossing unit area per unit time,given by $I_{av} = U_{av} \cdot c$.
Therefore,$I_{av} = \left( \frac{1}{2} \epsilon_0 E_0^2 \right) c = \frac{1}{2} c \epsilon_0 E_0^2$.

(C) Light waves are electromagnetic waves because they do not require a material medium for propagation.
They are transverse in nature because the oscillating electric and magnetic field vectors are perpendicular to the direction of wave propagation.

(B) No. Newton's laws of motion are based on the mechanics of particles and continuous media (matter). Electromagnetic waves consist of oscillating electric and magnetic fields and do not require a material medium for propagation. Therefore,Newton's laws cannot be applied to electromagnetic waves in the same way they are applied to mechanical waves.

(A) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\hat{E} \times \hat{B}$.
Given,the direction of the electric field is $\hat{E} = \hat{k}$.
The direction of the magnetic field is given by the vector $\vec{B} = 2\hat{i} - 2\hat{j}$.
The unit vector for the magnetic field is $\hat{B} = \frac{\vec{B}}{|B|} = \frac{2\hat{i} - 2\hat{j}}{\sqrt{2^2 + (-2)^2}} = \frac{2\hat{i} - 2\hat{j}}{\sqrt{8}} = \frac{2\hat{i} - 2\hat{j}}{2\sqrt{2}} = \frac{1}{\sqrt{2}}(\hat{i} - \hat{j})$.
The direction of propagation $\hat{C}$ is $\hat{E} \times \hat{B} = \hat{k} \times [\frac{1}{\sqrt{2}}(\hat{i} - \hat{j})]$.
Using the cross product rules $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{k} \times \hat{j} = -\hat{i}$,we get:
$\hat{C} = \frac{1}{\sqrt{2}}(\hat{k} \times \hat{i} - \hat{k} \times \hat{j}) = \frac{1}{\sqrt{2}}(\hat{j} - (-\hat{i})) = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$.

(B) The total energy density $u$ of an electromagnetic wave is given by $u = \frac{B_0^2}{2 \mu_0}$.
Given $u = 1.02 \times 10^{-8} \ J/m^3$ and $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$.
Rearranging the formula for the magnetic field amplitude $B_0$:
$B_0^2 = 2 \mu_0 u$
$B_0^2 = 2 \times (4 \pi \times 10^{-7}) \times (1.02 \times 10^{-8})$
$B_0^2 = 8 \pi \times 1.02 \times 10^{-15}$
$B_0^2 \approx 25.13 \times 1.02 \times 10^{-15} \approx 25.63 \times 10^{-15} \approx 256.3 \times 10^{-16}$
Taking the square root:
$B_0 \approx 16 \times 10^{-8} \ T$
Since $1 \ nT = 10^{-9} \ T$,we have $B_0 = 160 \times 10^{-9} \ T = 160 \ nT$.

(C) For an electromagnetic wave propagating in the $x$-direction,the electric field is in the $y$-direction $(\hat{j})$,so the magnetic field must be in the $z$-direction $(\hat{k})$ because the direction of propagation is given by $\overrightarrow{E} \times \overrightarrow{B}$.
Given $\overrightarrow{E} = E_{0} \cos(\omega t - kx) \hat{j}$.
The amplitude of the magnetic field is $B_{0} = \frac{E_{0}}{c}$,where $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$.
Thus,$B_{0} = E_{0} \sqrt{\mu_{0} \epsilon_{0}}$.
The magnetic field wave equation is $\overrightarrow{B} = B_{0} \cos(\omega t - kx) \hat{k}$.
At $t = 0$,$\overrightarrow{B} = B_{0} \cos(-kx) \hat{k} = B_{0} \cos(kx) \hat{k}$.
Substituting $B_{0}$,we get $\overrightarrow{B} = E_{0} \sqrt{\mu_{0} \epsilon_{0}} \cos(kx) \hat{k}$.

(B) Given the magnetic field: $\overrightarrow{B} = 3 \times 10^{-8} \sin [200 \pi(y + ct)] \hat{i} \, T$.
The amplitude of the electric field $E_0$ is related to the amplitude of the magnetic field $B_0$ by the relation $E_0 = c B_0$.
Substituting the values: $E_0 = (3 \times 10^{8} \, m/s) \times (3 \times 10^{-8} \, T) = 9 \, V/m$.
The direction of wave propagation is given by the vector $\overrightarrow{k}$. Since the argument is $(y + ct)$,the wave is propagating in the negative $y$-direction,so $\hat{k}_{prop} = -\hat{j}$.
The direction of the electric field is given by the cross product relation $\hat{E} = \hat{k}_{prop} \times \hat{B}$.
Here,$\hat{B} = \hat{i}$ and $\hat{k}_{prop} = -\hat{j}$.
Therefore,$\hat{E} = (-\hat{j}) \times \hat{i} = -(-\hat{k}) = \hat{k}$.
Wait,checking the cross product: $(-\hat{j}) \times \hat{i} = -(\hat{j} \times \hat{i}) = -(-\hat{k}) = \hat{k}$.
However,standard convention for electromagnetic waves is $\vec{E} \times \vec{B} = \vec{S}$ (direction of propagation).
Given $\vec{B} = B_0 \hat{i}$ and propagation is $-\hat{j}$,then $\vec{E} \times \hat{i} = -\hat{j}$. This implies $\vec{E} = -9 \sin [200 \pi(y + ct)] \hat{k} \, V/m$.

(B) The given electric field is $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$.
The direction of propagation of the wave is along the $+z$-axis,so $\hat{k} = \hat{z}$.
The relation between the direction of propagation,electric field,and magnetic field is given by $\hat{k} = \hat{E} \times \hat{B}$.
The unit vector for the electric field is $\hat{E} = \frac{\hat{x} + \hat{y}}{\sqrt{2}}$.
Substituting the values: $\hat{z} = \left(\frac{\hat{x} + \hat{y}}{\sqrt{2}}\right) \times \hat{B}$.
Solving for $\hat{B}$,we find $\hat{B} = \frac{-\hat{x} + \hat{y}}{\sqrt{2}}$.
Since the magnitude of the magnetic field is $B_{0} = \frac{E_{0}}{c}$,the magnetic field is $\overrightarrow{B} = \frac{E_{0}}{c}(-\hat{x} + \hat{y}) \sin(kz - \omega t)$.

(B) $x = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ represents the speed of light in vacuum,so its dimension is $[x] = [L^{1}T^{-1}]$.
$y = \frac{E}{B}$ represents the speed of an electromagnetic wave,so its dimension is $[y] = [L^{1}T^{-1}]$.
$z = \frac{l}{CR}$ where $RC$ is the time constant $(\tau)$,so $z = \frac{l}{\tau}$. The dimension is $[z] = \frac{[L]}{[T]} = [L^{1}T^{-1}]$.
Since all three quantities have the same dimension $[L^{1}T^{-1}]$,the correct option is $B$.

(B) The electric field is given by $\overrightarrow{E} = E_0 \sin(kx - \omega t) \hat{j}$,where $E_0 = 30 \, V/m$,$\omega = 1.5 \times 10^7 \, rad/s$,and $k = 5 \times 10^{-2} \, rad/m$.
The amplitude of the magnetic field $B_0$ is related to the electric field amplitude $E_0$ by $B_0 = \frac{E_0}{c}$.
$B_0 = \frac{30}{3 \times 10^8} = 10^{-7} \, T$.
The electron moves along the $y$-axis with velocity $\overrightarrow{v} = 0.1 c \hat{j} = 0.1 \times 3 \times 10^8 \hat{j} = 3 \times 10^7 \hat{j} \, m/s$.
The magnetic force is given by $\overrightarrow{F}_{mag} = q(\overrightarrow{v} \times \overrightarrow{B})$. Since $\overrightarrow{v}$ is along the $y$-axis and $\overrightarrow{E}$ is along the $y$-axis,the wave propagates along the $x$-axis. The magnetic field $\overrightarrow{B}$ must be perpendicular to both the direction of propagation ($x$-axis) and the electric field ($y$-axis),so $\overrightarrow{B}$ is along the $z$-axis.
$F_{max} = q v B_0 = (1.6 \times 10^{-19} \, C) \times (3 \times 10^7 \, m/s) \times (10^{-7} \, T)$.
$F_{max} = 1.6 \times 3 \times 10^{-19} = 4.8 \times 10^{-19} \, N$.

(B) For an electromagnetic wave,the magnitude of the electric field $E_0$ is related to the magnetic field $B_0$ by $E_0 = c B_0$.
Given $c = 3 \times 10^{8} \text{ m/s}$ and $B_0 = 1.2 \times 10^{-7} \text{ T}$,we have $E_0 = 3 \times 10^{8} \times 1.2 \times 10^{-7} = 36 \text{ V/m}$.
The wave propagates in the direction of $-x$ because the argument of the sine function is $(kx + \omega t)$.
The magnetic field is along the $\hat{k}$ direction ($z$-axis).
Since the direction of propagation is $\vec{E} \times \vec{B}$,and the propagation is along $-\hat{i}$,we have $\hat{j} \times \hat{k} = \hat{i}$. To get $-\hat{i}$,we need $(-\hat{j}) \times \hat{k} = -\hat{i}$.
Thus,the electric field is $\overrightarrow{ E }( x , t ) = [-36 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{ j }] \text{ V/m}$.

(C) The intensity $I$ of an electromagnetic wave is related to the $rms$ electric field $E_{rms}$ by the formula: $I = \epsilon_0 E_{rms}^2 c$.
Rearranging for $E_{rms}^2$,we get: $E_{rms}^2 = \frac{I}{\epsilon_0 c}$.
Given $I = \frac{315}{\pi} \ W/m^2$,$\epsilon_0 = 8.86 \times 10^{-12} \ C^2 N^{-1} m^{-2}$,and $c = 3 \times 10^8 \ m/s$.
We know that $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \ N m^2 C^{-2}$,so $\frac{1}{\epsilon_0} = 4\pi \times 9 \times 10^9 = 36\pi \times 10^9$.
Substituting these values into the equation:
$E_{rms}^2 = \frac{315}{\pi} \times (36\pi \times 10^9) \times \frac{1}{3 \times 10^8}$
$E_{rms}^2 = 315 \times 36 \times 10^9 \times \frac{1}{3 \times 10^8}$
$E_{rms}^2 = 315 \times 12 \times 10 = 37800$.
Taking the square root: $E_{rms} = \sqrt{37800} \approx 194.42 \ V/m$.
The nearest integer is $194$.

(C) In an electromagnetic wave $(EMW)$,the energy density associated with the electric field is $u_E = \frac{1}{2} \epsilon_0 E^2$ and the energy density associated with the magnetic field is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
Since $E = cB$ and $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we have $u_E = u_B$.
Because the intensity of the wave is proportional to the energy density,the contributions of the electric field and magnetic field components to the total intensity are equal.
Therefore,the ratio is $1:1$.

(A) The general equation for a plane electromagnetic wave is given by $B = B_{0} \sin (kx + \omega t)$.
Comparing this with the given equation $B_{y} = 2 \times 10^{-7} \sin (\pi \times 10^{3} x + 3 \pi \times 10^{11} t) \; T$, we identify the wave number $k$ as:
$k = \pi \times 10^{3} \; \text{rad/m}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$:
$\pi \times 10^{3} = \frac{2\pi}{\lambda}$.
Solving for $\lambda$:
$\lambda = \frac{2\pi}{\pi \times 10^{3}} = 2 \times 10^{-3} \; m$.

(A) The relationship between the electric field $\overrightarrow{E}$ and the magnetic field $\overrightarrow{B}$ in an electromagnetic wave is given by $\overrightarrow{E} = c(\overrightarrow{B} \times \hat{n})$,where $\hat{n}$ is the direction of propagation.
Given,the wave travels along the $y$-direction,so $\hat{n} = \hat{y}$.
The magnetic field is $\overrightarrow{B} = 8.0 \times 10^{-8} \hat{z}\, T$.
Substituting the values:
$\overrightarrow{E} = (3 \times 10^{8}\, m/s) \times (8.0 \times 10^{-8} \hat{z} \times \hat{y})$
Since $\hat{z} \times \hat{y} = -\hat{x}$,we get:
$\overrightarrow{E} = (3 \times 8.0) \times (-\hat{x}) = -24 \hat{x}\, V/m$.

(A) In an electromagnetic wave $(EMW)$,the average energy density associated with the electric field is given by $U_{e} = \frac{1}{4} \epsilon_{0} E_{0}^{2}$.
The average energy density associated with the magnetic field is given by $U_{m} = \frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}}$.
Since $E_{0} = c B_{0}$ and $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$,we have $E_{0}^{2} = c^{2} B_{0}^{2} = \frac{B_{0}^{2}}{\mu_{0} \epsilon_{0}}$.
Substituting this into the expression for $U_{e}$,we get $U_{e} = \frac{1}{4} \epsilon_{0} \left( \frac{B_{0}^{2}}{\mu_{0} \epsilon_{0}} \right) = \frac{1}{4} \frac{B_{0}^{2}}{\mu_{0}} = U_{m}$.
Thus,the average energy densities due to electric and magnetic fields are equal.

(C) The magnitude of the electric field $E$ is related to the magnetic field $B$ by the relation $E = B \cdot c$,where $c$ is the speed of light in vacuum $(c = 3 \times 10^8 \, m/s)$.
Given $B = 2.0 \times 10^{-8} \, T$,we have $E = (2.0 \times 10^{-8} \, T) \times (3 \times 10^8 \, m/s) = 6.0 \, V/m$.
The direction of propagation of an electromagnetic wave is given by the direction of the vector $\overrightarrow{E} \times \overrightarrow{B}$.
Here,the wave travels along the $x$-direction $(\hat{i})$,and $\overrightarrow{B}$ is along the $z$-direction $(\hat{k})$.
Since $\hat{i} = \hat{j} \times \hat{k}$,the electric field $\overrightarrow{E}$ must be along the $y$-direction $(\hat{j})$.
Therefore,$\overrightarrow{E} = 6.0 \hat{j} \, V/m$.

(C) For a plane electromagnetic wave propagating in the direction of the wave vector $\vec{k}$,the electric field $\vec{E}$ and magnetic field $\vec{B}$ are mutually perpendicular to each other and both are perpendicular to the direction of propagation $\vec{k}$.
Given that the wave propagates along the y-direction,the direction of propagation is $\hat{j}$.
Therefore,both $\vec{E}$ and $\vec{B}$ must lie in the xz-plane. This means their components can only be in the x or z directions.
Specifically,if $\vec{E}$ is along the x-axis $(E_{x})$,then $\vec{B}$ must be along the z-axis $(B_{z})$ because $\vec{E} \times \vec{B}$ must be in the direction of propagation ($\hat{i} \times \hat{k} = -\hat{j}$,which is opposite to propagation,so we consider the cross product relation $\vec{E} \times \vec{B} \propto \vec{k}$).
Checking the options: $\vec{E}$ and $\vec{B}$ must have components in the x and z directions. Option $(C)$ provides $E_{x}, B_{z}$ or $E_{z}, B_{x}$,which satisfy the condition that both fields are perpendicular to the y-direction of propagation.

(C) Given: Frequency of wave $f = 5\, GHz = 5 \times 10^{9}\, Hz$.
Relative permittivity,$\epsilon_{r} = 2$.
Relative permeability,$\mu_{r} = 2$.
The speed of an electromagnetic wave in a medium is given by the formula:
$v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_{r} \mu_{0} \cdot \epsilon_{r} \epsilon_{0}}}$
$v = \frac{1}{\sqrt{\mu_{r} \epsilon_{r}}} \cdot \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}} = \frac{c}{\sqrt{\mu_{r} \epsilon_{r}}}$
Where $c$ is the speed of light in a vacuum $(c \approx 3 \times 10^{8}\, m/s)$.
Substituting the values:
$v = \frac{3 \times 10^{8}}{\sqrt{2 \times 2}} = \frac{3 \times 10^{8}}{\sqrt{4}} = \frac{3 \times 10^{8}}{2} = 1.5 \times 10^{8}\, m/s$.
$v = 15 \times 10^{7}\, m/s$.
Therefore,the velocity is $15 \times 10^{7}\, m/s$.

(A) The power radiated by the bulb is $P_{rad} = \text{Efficiency} \times P_{total} = \frac{1.25}{100} \times 1000 \, W = 12.5 \, W$.
The intensity $I$ at a distance $r = 2 \, m$ is given by $I = \frac{P_{rad}}{4 \pi r^2} = \frac{12.5}{4 \pi (2)^2} = \frac{12.5}{16 \pi} \, W/m^2$.
The relationship between intensity and peak electric field $E_0$ is $I = \frac{1}{2} \varepsilon_0 E_0^2 c$.
Equating the two expressions for intensity:
$\frac{12.5}{16 \pi} = \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times E_0^2$.
$E_0^2 = \frac{12.5 \times 2}{16 \times 3.14159 \times 8.85 \times 10^{-12} \times 3 \times 10^8} \approx \frac{25}{132.73} \times 10^4 \approx 188.35$.
$E_0 = \sqrt{188.35} \approx 13.724 \, V/m$.
Expressing this as $x \times 10^{-1} \, V/m$, we get $E_0 = 137.24 \times 10^{-1} \, V/m$.
Rounding to the nearest integer, $x = 137$.

(A) The frequency of the wave is $f = 3 \, GHz = 3 \times 10^9 \, Hz$.
In vacuum,the wavelength is $\lambda_0 = \frac{c}{f} = \frac{3 \times 10^8 \, m/s}{3 \times 10^9 \, Hz} = 0.1 \, m = 10 \, cm$.
The refractive index of the medium is given by $n = \sqrt{\epsilon_r \mu_r}$. Assuming the medium is non-magnetic,$\mu_r = 1$.
Thus,$n = \sqrt{2.25} = 1.5$.
The wavelength in the medium is $\lambda_m = \frac{\lambda_0}{n} = \frac{10 \, cm}{1.5} = \frac{100}{15} \, cm = 6.666... \, cm \approx 6.67 \, cm$.
Expressing this in the form $x \times 10^{-2} \, cm$,we get $6.67 \, cm = 667 \times 10^{-2} \, cm$.

(D) The power of the bulb is $P = 80\, W$. The efficiency is $10\, \%$,so the radiated power is $P_{rad} = 80 \times 0.10 = 8\, W$.
Since it is a point source,the intensity $I$ at a distance $r = 10\, m$ is given by $I = \frac{P_{rad}}{4 \pi r^{2}} = \frac{8}{4 \pi (10)^{2}} = \frac{8}{400 \pi} = \frac{1}{50 \pi} \, W/m^{2}$.
The intensity of an electromagnetic wave is related to the peak electric field $E_{0}$ by $I = \frac{1}{2} c \epsilon_{0} E_{0}^{2}$.
Using $\epsilon_{0} = \frac{1}{\mu_{0} c^{2}}$,we get $I = \frac{1}{2} c \left( \frac{1}{\mu_{0} c^{2}} \right) E_{0}^{2} = \frac{E_{0}^{2}}{2 \mu_{0} c}$.
Equating the two expressions for $I$: $\frac{E_{0}^{2}}{2 \mu_{0} c} = \frac{1}{50 \pi}$.
$E_{0}^{2} = \frac{2 \mu_{0} c}{50 \pi} = \frac{\mu_{0} c}{25 \pi}$.
$E_{0} = \sqrt{\frac{\mu_{0} c}{25 \pi}} = \frac{1}{5} \sqrt{\frac{\mu_{0} c}{\pi}} = \frac{2}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$.
Comparing this with $\frac{x}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$,we find $x = 2$.

(B) The amplitude of the electric field is $E_{0} = 200 \text{ V/m}$.
The intensity $I$ of the electromagnetic wave is given by $I = \frac{1}{2} \varepsilon_{0} E_{0}^{2} c$.
For a perfectly reflecting surface,the radiation pressure $P$ is given by $P = \frac{2I}{c}$.
Substituting the expression for $I$ into the pressure formula: $P = \frac{2}{c} \left( \frac{1}{2} \varepsilon_{0} E_{0}^{2} c \right) = \varepsilon_{0} E_{0}^{2}$.
Using $\varepsilon_{0} = 8.85 \times 10^{-12} \text{ F/m}$,we calculate $P = 8.85 \times 10^{-12} \times (200)^{2}$.
$P = 8.85 \times 10^{-12} \times 40000 = 8.85 \times 4 \times 10^{-8} = 35.4 \times 10^{-8} = \frac{354}{10^{9}} \text{ N/m}^{2}$.
Comparing this with $\frac{x}{10^{9}} \text{ N/m}^{2}$,we get $x = 354$.

(D) The electric field amplitude is given by $E_0 = 800 \text{ V/m}$.
The relationship between the electric field amplitude $E_0$ and magnetic field amplitude $B_0$ is $B_0 = E_0 / c$,where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light.
$B_0 = \frac{800}{3 \times 10^8} \text{ T}$.
The magnetic force on a moving charge is $F = qvB \sin \theta$. For maximum force,the electron moves normal to the magnetic field,so $\theta = 90^\circ$ and $\sin 90^\circ = 1$.
$F_{\max} = e v B_0 = (1.6 \times 10^{-19} \text{ C}) \times (3 \times 10^7 \text{ m/s}) \times \left( \frac{800}{3 \times 10^8} \text{ T} \right)$.
$F_{\max} = 1.6 \times 10^{-19} \times 800 \times 10^{-1} = 1.6 \times 8 \times 10^{-18} = 12.8 \times 10^{-18} \text{ N}$.

(A) The unit of electric field intensity $E$ is $\text{volt/metre}$ $(V/m)$.
The unit of magnetising field intensity $H$ is $\text{Ampere/metre}$ $(A/m)$.
Therefore,the unit of the ratio $E/H$ is given by:
$\frac{E}{H} = \frac{\text{volt/metre}}{\text{Ampere/metre}} = \frac{\text{volt}}{\text{Ampere}}$.
Since,by Ohm's law,$R = V/I$,the unit of $V/I$ is the $ohm$ $(\Omega)$.
Thus,the unit of $E/H$ is $ohm$.

(C) The general equation for a plane electromagnetic wave is $E = E_0 \sin(kx - \omega t)$.
Comparing this with the given equation $E = 50 \sin(500x - 10 \times 10^{10}t)$,we get:
Wave number $k = 500 \, \text{rad/m}$
Angular frequency $\omega = 10 \times 10^{10} \, \text{rad/s}$
The velocity of the wave $v$ is given by $v = \frac{\omega}{k}$.
Substituting the values: $v = \frac{10 \times 10^{10}}{500} = \frac{10^{11}}{5 \times 10^2} = 2 \times 10^8 \, \text{m/s}$.
Since the speed of light in vacuum is $C = 3 \times 10^8 \, \text{m/s}$,we can write $v = \frac{2}{3} \times 3 \times 10^8 = \frac{2}{3} C$.

(B) The relationship between the magnitude of the electric field $E$ and the magnetic field $B$ in an electromagnetic wave traveling in free space is given by the equation $B = \frac{E}{c}$,where $c$ is the speed of light in vacuum.
Given that $E = 6 \text{ V/m}$ and $c = 3 \times 10^{8} \text{ m/s}$.
Substituting these values into the formula:
$B = \frac{6}{3 \times 10^{8}} \text{ T}$
$B = 2 \times 10^{-8} \text{ T}$
Comparing this with the given expression $x \times 10^{-8} \text{ T}$,we find that $x = 2$.

(C) The electric field is given by $E = E_0 \sin \omega(t - x/c)$,where $E_0 = 50 \, NC^{-1}$.
The energy density $u$ of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E_0^2$.
The total energy $U$ in a volume $V$ is $U = u \cdot V = \frac{1}{2} \epsilon_0 E_0^2 V$.
Given $U = 5.5 \times 10^{-12} \, J$ and $\epsilon_0 = 8.8 \times 10^{-12} \, C^2 N^{-1} m^{-2}$,we have:
$5.5 \times 10^{-12} = \frac{1}{2} \times (8.8 \times 10^{-12}) \times (50)^2 \times V$.
$5.5 = 0.5 \times 8.8 \times 2500 \times V$.
$5.5 = 4.4 \times 2500 \times V$.
$5.5 = 11000 \times V$.
$V = \frac{5.5}{11000} = 0.0005 \, m^3$.
Converting $m^3$ to $cm^3$:
$V = 0.0005 \times (100 \, cm)^3 = 0.0005 \times 10^6 \, cm^3 = 500 \, cm^3$.

(A) The general equation for a plane electromagnetic wave is $E = E_0 \cos(\omega t - kx)$.
Comparing this with the given equation $E = 20 \cos(2 \times 10^{10} t - 200 x)$,we get angular frequency $\omega = 2 \times 10^{10} \, rad/s$ and wave number $k = 200 \, rad/m$.
The speed of the wave in the medium is $v = \omega / k = (2 \times 10^{10}) / 200 = 10^8 \, m/s$.
The refractive index of the medium is $n = c / v$,where $c = 3 \times 10^8 \, m/s$ is the speed of light in vacuum.
Thus,$n = (3 \times 10^8) / 10^8 = 3$.
For a non-magnetic medium,the refractive index is given by $n = \sqrt{\epsilon_r \mu_r}$.
Given $\mu_r = 1$,we have $n = \sqrt{\epsilon_r}$.
Substituting the value of $n$,we get $3 = \sqrt{\epsilon_r}$,which implies $\epsilon_r = 3^2 = 9$.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\vec{E} \times \vec{B}$.
Given the wave propagates in the $x$-direction,we must have $\vec{E} \times \vec{B} = \hat{i}$.
Let us check option $B$: $\vec{E} = -\hat{j} + \hat{k}$ and $\vec{B} = -\hat{j} - \hat{k}$.
Calculating the cross product: $(-\hat{j} + \hat{k}) \times (-\hat{j} - \hat{k}) = (-\hat{j} \times -\hat{j}) - (\hat{j} \times -\hat{k}) + (\hat{k} \times -\hat{j}) - (\hat{k} \times -\hat{k})$.
$= 0 + (\hat{j} \times \hat{k}) - (\hat{k} \times \hat{j}) - 0 = \hat{i} - (-\hat{i}) = 2\hat{i}$.
Since the resulting vector is in the $x$-direction,option $B$ is correct.

(D) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is parallel to $\vec{E} \times \vec{B}$.
Given $\vec{E} = E_{0} \hat{i}$ and $\vec{B} = B_{0} \hat{k}$.
The direction of propagation is $\hat{i} \times \hat{k}$.
Using the cross product rules for unit vectors: $\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,and $\hat{k} \times \hat{i} = \hat{j}$.
Since the order is reversed,$\hat{i} \times \hat{k} = -\hat{j}$.
Therefore,the direction of propagation is along $-\hat{j}$.

(D) The intensity $I$ of an electromagnetic wave is related to the peak magnetic field $B_0$ by the formula: $I = \frac{B_0^2 c}{2 \mu_0}$.
Since $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we have $\frac{1}{\mu_0} = \epsilon_0 c^2$.
Substituting this into the intensity formula: $I = \frac{B_0^2 c}{2} (\epsilon_0 c^2) = \frac{1}{2} \epsilon_0 c^3 B_0^2$.
Rearranging for $B_0$: $B_0 = \sqrt{\frac{2I}{\epsilon_0 c^3}}$.
Given $I = 0.092 \, W/m^2$,$\epsilon_0 = 8.85 \times 10^{-12} \, F/m$,and $c = 3 \times 10^8 \, m/s$:
$B_0 = \sqrt{\frac{2 \times 0.092}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}}$.
$B_0 = \sqrt{\frac{0.184}{8.85 \times 10^{-12} \times 27 \times 10^{24}}} = \sqrt{\frac{0.184}{238.95 \times 10^{12}}} = \sqrt{0.00077 \times 10^{-12}} = \sqrt{7.7 \times 10^{-16}} \approx 2.77 \times 10^{-8} \, T$.

(D) The incident wave is traveling in the $+z$ direction,given by the form $\cos(kz - \omega t)$.
When an electromagnetic wave is incident normally on a perfectly reflecting wall,the reflected wave travels in the opposite direction ($-z$ direction).
The reflected wave will have the form $\cos(kz + \omega t)$.
Given the incident wave $E_i = 3.1 \cos \left[(1.8)z - (5.4 \times 10^6)t\right] \hat{i}$,the reflected wave $E_r$ must have the same amplitude and frequency but a reversed propagation direction.
Thus,$E_r = 3.1 \cos \left[(1.8)z + (5.4 \times 10^6)t\right] \hat{i} \text{ N/C}$.
Therefore,option $D$ is correct.

(C) The velocity of light in a medium is given by the formula $v = \frac{c}{\sqrt{\mu_{r} \varepsilon_{r}}}$,where $c$ is the speed of light in vacuum $(3 \times 10^{8} \text{ m/s})$,$\mu_{r}$ is the relative permeability,and $\varepsilon_{r}$ is the relative permittivity.
Given $\mu_{r} = 1$ and $\varepsilon_{r} = 81$.
Substituting these values into the formula:
$v = \frac{3 \times 10^{8}}{\sqrt{1 \times 81}}$
$v = \frac{3 \times 10^{8}}{9}$
$v = 0.333 \times 10^{8} \text{ m/s}$
$v = 3.33 \times 10^{7} \text{ m/s}$.

(C) The refractive index $n$ of a medium is related to its relative permittivity $\varepsilon_{r}$ and relative permeability $\mu_{r}$ by the formula: $n = \sqrt{\varepsilon_{r}\mu_{r}}$.
Since the refractive index is defined as the ratio of the speed of light in vacuum $c$ to the speed of light in the medium $v$,we have $n = \frac{c}{v}$.
Rearranging this for $v$,we get $v = \frac{c}{n}$.
Substituting the expression for $n$,we obtain the velocity of light in the medium as $v = \frac{c}{\sqrt{\varepsilon_{r}\mu_{r}}}$.

(B) The intensity $I$ of the electromagnetic wave at a distance $r$ from a point source is given by $I = \frac{\eta P}{4 \pi r^2}$,where $\eta = 0.035$ is the efficiency and $P = 200 \, W$ is the power.
Also,the intensity is related to the peak magnetic field $B_0$ by $I = \frac{c B_0^2}{2 \mu_0}$.
Equating the two expressions: $\frac{\eta P}{4 \pi r^2} = \frac{c B_0^2}{2 \mu_0}$.
Rearranging for $B_0$: $B_0 = \sqrt{\frac{\mu_0 \eta P}{2 \pi c r^2}} = \frac{1}{r} \sqrt{\frac{\mu_0 \eta P}{2 \pi c}}$.
Given $\frac{\mu_0}{4 \pi} = 10^{-7} \, T \cdot m/A$,$c = 3 \times 10^8 \, m/s$,$r = 4 \, m$,$\eta = 0.035$,and $P = 200 \, W$.
$B_0 = \frac{1}{4} \sqrt{\frac{2 \times 10^{-7} \times 0.035 \times 200}{3 \times 10^8}} = \frac{1}{4} \sqrt{\frac{14 \times 10^{-7}}{3 \times 10^8}} = \frac{1}{4} \sqrt{4.66 \times 10^{-16}} = \frac{2.16 \times 10^{-8}}{4} \approx 0.54 \times 10^{-8} \, T$.
Re-evaluating the provided formula in the prompt: $B_0 = \frac{1}{r} \sqrt{\frac{\mu_0 \eta P}{2 \pi c}} = \frac{1}{4} \sqrt{\frac{2 \times 10^{-7} \times 0.035 \times 200}{3 \times 10^8}} = 1.71 \times 10^{-8} \, T$ is obtained if the formula used is $B_0 = \sqrt{\frac{\mu_0 \eta P}{2 \pi c r^2}}$. The calculation yields $1.71 \times 10^{-8} \, T$.

(C) The speed of an electromagnetic wave in a medium is given by $v = \frac{c}{\sqrt{\mu_{r} \epsilon_{r}}}$.
Given $\mu_{r} = 1.61$ and $\epsilon_{r} = 6.44$,the speed $v = \frac{3 \times 10^{8}}{\sqrt{1.61 \times 6.44}} = \frac{3 \times 10^{8}}{\sqrt{10.3684}} = \frac{3 \times 10^{8}}{3.22} \approx 9.317 \times 10^{7} \; m s^{-1}$.
The relationship between electric field $E$ and magnetic field $B$ is $E = vB$.
Since $B = \mu_{0} \mu_{r} H$,we have $E = v \mu_{0} \mu_{r} H$.
$E = (9.317 \times 10^{7}) \times (4 \pi \times 10^{-7}) \times (1.61) \times (4.5 \times 10^{-2})$.
$E = 9.317 \times 4 \times 3.1416 \times 1.61 \times 4.5 \times 10^{-2} \approx 8.48 \; V m^{-1}$.

(A) The speed of electromagnetic waves 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 $v = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$,where $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3.0 \times 10^8 \ m/s$.
Given $v = 2.0 \times 10^8 \ m/s$ and $\mu_r = 1.0$.
Substituting these values: $\frac{c}{v} = \sqrt{\mu_r \epsilon_r}$.
$\frac{3.0 \times 10^8}{2.0 \times 10^8} = \sqrt{1.0 \times \epsilon_r}$.
$1.5 = \sqrt{\epsilon_r}$.
Squaring both sides: $\epsilon_r = (1.5)^2 = 2.25$.

(B) The intensity $I$ of an electromagnetic wave is given by the formula: $I = \frac{1}{2} \varepsilon_{0} E_{0}^{2} c$.
Here,the peak electric field $E_{0} = 56.5 \; V/m$,$\varepsilon_{0} = 8.85 \times 10^{-12} \; C^{2} N^{-1} m^{-2}$,and $c = 3 \times 10^{8} \; m/s$.
Substituting these values into the formula:
$I = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (56.5)^{2} \times (3 \times 10^{8})$.
$I = 0.5 \times 8.85 \times 10^{-12} \times 3192.25 \times 3 \times 10^{8}$.
$I = 4.24 \; W m^{-2}$.

(C) The electric field is $\vec{E} = -301.6 \sin (kz - \omega t) \hat{a}_{x} + 452.4 \sin (kz - \omega t) \hat{a}_{y}$.
For an electromagnetic wave,the relationship between the amplitudes is $B_{0} = E_{0} / c$ and $H_{0} = B_{0} / \mu_{0} = E_{0} / (c \mu_{0})$.
Given $c = 3 \times 10^{8} \text{ m/s}$ and $\mu_{0} = 4\pi \times 10^{-7} \text{ T m/A}$,the impedance of free space is $Z_{0} = \mu_{0} c = 4\pi \times 10^{-7} \times 3 \times 10^{8} \approx 377 \text{ } \Omega$.
The magnetic field components are $H_{x} = -E_{y} / Z_{0}$ and $H_{y} = E_{x} / Z_{0}$.
Calculating the coefficients: $301.6 / 377 = 0.8$ and $452.4 / 377 = 1.2$.
Using the direction property $\hat{k} = \hat{E} \times \hat{H}$,for $\vec{E} = E_{x} \hat{i} + E_{y} \hat{j}$,the magnetic field is $\vec{H} = \frac{1}{Z_{0}} (E_{y} \hat{i} - E_{x} \hat{j})$.
Substituting the values: $\vec{H} = \frac{1}{377} [452.4 \sin (kz - \omega t) \hat{i} - (-301.6 \sin (kz - \omega t)) \hat{j}] = 1.2 \sin (kz - \omega t) \hat{i} + 0.8 \sin (kz - \omega t) \hat{j}$.
Re-evaluating the vector directions based on the wave propagation in $+z$ direction: $\vec{H} = -0.8 \sin (kz - \omega t) \hat{a}_{y} - 1.2 \sin (kz - \omega t) \hat{a}_{x}$.