Electromagnetic Waves

The standard wave equation for the electric field component of an electromagnetic wave is given as $\mathrm{E}= \mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$, where $\mathrm{E}_0$ is the amplitude of electric field, $\omega$ is the angular frequency and k is the wave number.

The give equation for the electric field of electromagnetic wave is $=\sqrt{377} \sin \left(6.27 \times 10^3 t-2.09 \times\right. \left.10^{-5} \mathrm{x}\right) \mathrm{N} / \mathrm{C}$

So, electric field amplitude is $\mathrm{E}_0=\sqrt{377} \mathrm{~N} / \mathrm{C}$.

The total energy density (energy per unit volume) of an electromagnetic wave is the sum of the electric field energy density ( $\mathrm{u}_{\mathrm{E}}$ ) and magnetic field energy density ( $\mathrm{u}_{\mathrm{B}}$ ).

$ \mathrm{u}=\mathrm{u}_{\mathrm{E}}+\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2+\frac{1}{2} \mu_0 \mathrm{~B}^2 $

In an electromagnetic wave traveling in a vacuum, the instantaneous electric and magnetic fields are related by the speed of light (c):

$ \mathrm{E}=\mathrm{cB} \Rightarrow \mathrm{~B}=\frac{\mathrm{E}}{\mathrm{c}} $

So, the energy density for magnetic field is,

$ u_B=\frac{1}{2} \mu_0\left(\frac{E}{c}\right)^2=\frac{E^2}{2 \mu_0 c^2} $

Since $c=\frac{1}{\sqrt{\mu_0} \epsilon_0} \Rightarrow c^2=\frac{1}{\mu_0 \epsilon_0} \Rightarrow \mu_0 c^2=\frac{1}{\epsilon_0}$.

$ \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2 $

Thus, the energy is equally shared: $\mathrm{u}_{\mathrm{E}}=\mathrm{u}_{\mathrm{B}}$.

The total instantaneous energy density is:

$ \mathrm{u}=\mathrm{u}_{\mathrm{E}}+\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2+\frac{1}{2} \epsilon_0 \mathrm{E}^2=\epsilon_0 \mathrm{E}^2 $

Since the electric field oscillates sinusoidally $\left(E=E_0 \sin \omega t\right)$, the average value of $E^2$ over one cycle is $\frac{E_0^2}{2}$.

$ \overline{\mathrm{u}}=\epsilon_0\left(\frac{\mathrm{E}_0^2}{2}\right)=\frac{1}{2} \epsilon_0 \mathrm{E}_0^2 $

Intensity is defined as the energy crossing a unit area per unit time. It is related to the average energy density by the speed of the wave $c$ :

$ \mathrm{I}=\overline{\mathrm{u}} \times \mathrm{c} $

$\Rightarrow $ $ I=\left(\frac{1}{2} \epsilon_0 E_0^2\right) c $

The characteristic impedance of free space, $\mathrm{Z}_0=\sqrt{\frac{\mu_0}{\epsilon_0}}$ (approx $377 \Omega$ ).

$I=\frac{1}{2} \epsilon_0 E_0^2\left(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\right)$

$\Rightarrow $ $I=\frac{E_0^2}{2}\left(\frac{\epsilon_0}{\sqrt{\mu_0 \epsilon_0}}\right)=\frac{E_0^2}{2} \sqrt{\frac{\epsilon_0}{\mu_0}}=\frac{(\sqrt{377})^2}{2} \times \frac{1}{377}=\frac{1}{2}$

$\Rightarrow $ $ \frac{1}{2}=\frac{1}{\alpha} \Rightarrow \alpha=2 $

Therefore, the value of $\alpha$ is 2.

The general standard mathematical form for a plane wave traveling in the positive x -direction is:

$ \mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}) $

Where :

• A is the amplitude (maximum displacement).

• k is the angular wave number, related to the wavelength $(\lambda)$ by $\mathrm{k}=\frac{2 \pi}{\lambda}$.

• $\omega$ is the angular frequency, related to the time period (T) by $\omega=\frac{2 \pi}{T}$.

The speed of propagation of traveling wave is given as,

$ v=\frac{\omega}{k} $

From Maxwell's equations, the speed of an electromagnetic wave in a vacuum (c) depends on the fundamental constants of free space:

$ c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} $

Where $\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space.

When light travels through a material medium, its speed (v) changes depending on the magnetic and electrical properties of that specific medium :

$ v=\frac{1}{\sqrt{\mu \epsilon}} $

Where $\mu$ is the absolute permeability and $\epsilon$ is the absolute permittivity of the medium.

Relative to a vacuum :

• $\mu=\mu_0 \mu_{\mathrm{r}}$ (where $\mu_{\mathrm{r}}$ is relative permeability)

• $\epsilon=\epsilon_0 \epsilon_{\mathrm{r}}$ (where $\epsilon_{\mathrm{r}}$ is relative permittivity, also called the dielectric constant, often denoted as K )

$\mathrm{v}=\frac{1}{\sqrt{\left(\mu_0 \mu_{\mathrm{r}}\right)\left(\epsilon_0 \epsilon_{\mathrm{r}}\right)}}$

$\Rightarrow $ $v=\frac{1}{\sqrt{\mu_0 \epsilon_0}} \cdot \frac{1}{\sqrt{\mu_r \epsilon_r}}$

$\Rightarrow $ $ v=\frac{c}{\sqrt{\mu_r \epsilon_r}} $

As the medium is non-magnetic so its relative permeability is exactly 1 ( $\mu_{\mathrm{r}}=1$ ).

$ \mathrm{v}=\frac{\mathrm{c}}{\sqrt{1 \times \epsilon_{\mathrm{r}}}} $

$\Rightarrow $ $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\epsilon_{\mathrm{r}}}}$

$\Rightarrow $ $\sqrt{\epsilon_{\mathrm{r}}}=\frac{\mathrm{c}}{\mathrm{v}}$

$\Rightarrow $ $ \epsilon_{\mathrm{r}}=\left(\frac{\mathrm{c}}{\mathrm{v}}\right)^2 $

From the given electric field equation :

$ E_y=20 \sin \left(3 \times 10^6 x-4.5 \times 10^{14} t\right) $

By comparing this to the standard form $\mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \sin (\mathrm{kx}-\omega \mathrm{t})$ :

• $\mathrm{k}=3 \times 10^6 \mathrm{rad} / \mathrm{m}$

• $\omega=4.5 \times 10^{14} \mathrm{rad} / \mathrm{s}$

The speed of the wave in the medium (v) :

$ \mathrm{v}=\frac{\omega}{\mathrm{k}} $

$\Rightarrow $ $\mathrm{v}=\frac{4.5 \times 10^{14}}{3 \times 10^6}$

$\Rightarrow $ $ \mathrm{v}=1.5 \times 10^8 \mathrm{~m} / \mathrm{s} $

We are given $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$.

$ \epsilon_{\mathrm{r}}=\left(\frac{\mathrm{c}}{\mathrm{v}}\right)^2 $

$\Rightarrow $ $\epsilon_{\mathrm{r}}=\left(\frac{3 \times 10^8}{1.5 \times 10^8}\right)^2$

$\Rightarrow $ $ \epsilon_{\mathrm{r}}=(2)^2=4 $

Therefore, the dielectric constant of the medium is 4 .

Hence, the correct answer is 4 .

According to Ohm's Law in point form, the conduction current density is proportional to the electric field (E) :

$ \mathrm{J}_{\mathrm{c}}=\sigma \mathrm{E} $

The maximum value occurs when the electric field is at its peak $\left(\mathrm{E}_0\right)$ :

$ \left(\mathrm{J}_{\mathrm{c}}\right)_{\max }=\sigma \mathrm{E}_0 $

Displacement current density arises from a time-varying electric field. For an electromagnetic wave, the electric field can be represented as $E=E_0 \sin (\omega t)$.

Maxwell defined displacement current density as :

$ \mathrm{J}_{\mathrm{d}}=\epsilon \frac{\partial \mathrm{E}}{\partial \mathrm{t}} $

$\Rightarrow $ $ \mathrm{J}_{\mathrm{d}}=\epsilon \frac{\partial}{\partial \mathrm{t}}\left(\mathrm{E}_0 \sin (\omega \mathrm{t})\right)=\epsilon \omega \mathrm{E}_0 \cos (\omega \mathrm{t}) $

The maximum value occurs when the cosine term is 1 :

$ \left(\mathrm{J}_{\mathrm{d}}\right)_{\max }=\epsilon \omega \mathrm{E}_0 $

The ratio of the maximum values is:

Ratio $=\frac{\left(\mathrm{J}_{\mathrm{c}}\right)_{\text {max }}}{\left(\mathrm{J}_{\mathrm{d}}\right)_{\text {max }}}=\frac{\sigma \mathrm{E}_0}{\epsilon \omega \mathrm{E}_0}=\frac{\sigma}{\epsilon \omega}$

The conductivity $(\sigma)$ of the medium is, $\sigma=10 \mathrm{mho} / \mathrm{m}$

The angular frequency $(\omega)$ is $2 \pi \mathrm{f}=2 \pi \times\left(100 \times 10^6\right)=2 \pi \times 10^8 \mathrm{rad} / \mathrm{s}$

Since no specific medium permittivity is given, we use the permittivity of free space ( $\epsilon_0$ ).

Given $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9$, then $\epsilon_0=\frac{1}{36 \pi \times 10^9}$.

Substituting these values into the ratio:

Ratio $=\frac{10}{\frac{1}{36 \pi \times 10^9} \times 2 \pi \times 10^8}=\frac{10 \times 36 \pi \times 10^9}{2 \pi \times 10^8}$

Ratio $=180 \times 10^1=1800$

Therefore, the ratio of maximum conduction current density to maximum displacement current density is 1800 .

Hence, the correct answer is $\mathbf{1 8 0 0}$.

$ \frac{dV}{dt} = \frac{I}{C} $

Given that the displacement current is $ I = 0.25 \times 10^{-3} \, \text{A} $ and the capacitance is $ C = 2.5 \times 10^{-6} \, \text{F}, $ the rate of change of the potential difference is calculated as follows:

$ \frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times \frac{10^{-3}}{10^{-6}} = 0.1 \times 10^3 = 100 \, \text{V/s}. $

Thus, the magnitude of the rate of change of the potential difference is $ 100 \, \text{V/s} $.

To determine the displacement current through a hypothetical plane surface within a parallel plate capacitor, follow these steps:

Current Density Calculation:

The current density ($ J_d $) is the current ($ I $) divided by the area ($ A $) of the capacitor plates.

Given: $ I = 6 \, \text{A} $ and $ A = 16 \, \text{cm}^2 $.

$ J_d = \frac{I}{A} = \frac{6 \, \text{A}}{16 \, \text{cm}^2} $

Displacement Current Through the Hypothetical Surface:

The hypothetical surface has an area $ A_0 = 3.2 \, \text{cm}^2 $.

The displacement current through this smaller area ($ I_{small} $) is the current density multiplied by the area of the hypothetical surface.

$ I_{small} = J_d \times A_0 = \frac{6 \, \text{A}}{16 \, \text{cm}^2} \times 3.2 \, \text{cm}^2 $

Calculation:

Simplify the expression to find the displacement current:

$ I_{small} = \left(\frac{6}{16}\right) \times 3.2 = 1.2 \, \text{A} = 1200 \, \text{mA} $

Thus, the displacement current through the hypothetical plane surface is 1200 mA.

As the cube has unit volume so energy density, $u=\frac{E}{1}=E$

where, $E=$ total energy

Energy of a photon, $E_1=h \nu$
So for n photon, $E=n h \nu$

Here, the question mentions "average energy contained in the electromagnetic waves within the same volume" which means we have to consider standing wave in a cavity (since the photons are in a cube).

Average energy density for a standing wave : $\langle u\rangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 E_0{ }^2+\frac{1}{2} \frac{B_0{ }^2}{\mu_0}\right)$

The factor $\frac{1}{2}$ comes from averaging $\sin ^2(k x)$

over the volume. $\left\langle\sin ^2(k x)\right\rangle=\frac{1}{2}$

For an $E M$ wave $E=C B$

$ \text { so } E_0=C B_0 $

Hence, $\langle u\rangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0\left(c B_0\right)^2+\frac{B_0^2}{2 \mu_0}\right)$

$ =\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 c^2 B_0^2+\frac{B_0^2}{2 \mu_0}\right) $

$ \begin{aligned} \Rightarrow\langle u\rangle=\frac{1}{2} & {\left[\frac{1}{2}\left(\frac{1}{\mu_0 c^2}\right) c^2 B_0^2+\frac{B_0^2}{2 \mu_0}\right]\left(\begin{array}{l} As, c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \\ \text { so } \varepsilon_0=\frac{1}{\mu_0 c^2} \end{array}\right.} \\ \Rightarrow\langle u\rangle & =\frac{1}{2}\left[\frac{B_0^2}{2 \mu_0}+\frac{B_0^2}{2 \mu_0}\right] \\ & \Rightarrow\langle u\rangle=\frac{B_0^2}{2 \mu_0} \end{aligned} $

Given that this is equal to the photon energy density. so $\frac{B_0^2}{2 \mu_0}=n h \nu$

Total energy of photons.

$ \begin{aligned} E & =n h \nu \\ & =35 \times 10^7 \times 6 \times 10^{-34} \times 1.15 \\ \Rightarrow E & =210 \times 10^{-12} \mathrm{~J} \end{aligned} $

For $E M$ waves : $E=\frac{B_0{ }^2}{2 \mu_0}$

So $\frac{B_0^2}{2 \mu_0}=210 \times 10^{-12}$

$ \begin{aligned} & \Rightarrow B_0^2=2 \times 4 \times \frac{22}{7} \times 10^{-7} \times 210 \times 10^{-12} \\ & \Rightarrow B_0^2=5280 \times 10^{-19} \\ & \Rightarrow B_0^2=528 \times 10^{-18} \\ & \Rightarrow B_0=\sqrt{528} \times 10^{-9} \\ & \Rightarrow B_0=22.978 \times 10^{-9} \mathrm{~T} \\ & \\ & \text { = } \alpha \times 10^{-9} \mathrm{~T} \text { (given) } \\ &So, \alpha=22.98 \text { Ans. } \end{aligned} $

We know that $v=\frac{c}{n}=\frac{c}{\sqrt{\varepsilon_{r}}}$

Putting the values:

$0.2 c=\frac{c}{\sqrt{\varepsilon_{r}}}$

$\Rightarrow \sqrt{\varepsilon_{r}}=5$

$\Rightarrow$ Required ratio $=\frac{\varepsilon_{r}}{n}=\frac{\varepsilon_{r}}{\sqrt{\varepsilon_{r}}}=\sqrt{\varepsilon_{r}}=5$

$\Rightarrow x=5$

$ \begin{aligned} & \text { Force }=\int P d A \cos \theta \\\\ & =\frac{2 \mathrm{I}}{\mathrm{C}} \int \mathrm{dA} \cos \theta=\frac{2 \mathrm{I}}{\mathrm{C}} \pi \mathrm{R}^2=2 \frac{\mathrm{p}_0}{4 \pi \mathrm{R}^2} \cdot \frac{\pi \mathrm{R}^2}{\mathrm{C}} \\\\ & =\frac{\mathrm{p}_0}{2 \mathrm{C}}=\frac{24}{2 \times 3 \times 10^8}=4 \times 10^{-8} \mathrm{~N} \end{aligned} $

$\mathbf{P}^{\prime}=10 \%$ of $110 \mathbf{W}$

$=\frac{10}{100} \times 110 \mathrm{~W}$

$=11 \mathrm{~W}$

$\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_1^2}-\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_2^2}$

$=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25}\right]$

$=\frac{11}{4 \pi} \times \frac{24}{25}$

$=\frac{264}{\pi} \times 10^{-2}=84 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2$

$4.425\,\mu A = {{{E_0}A} \over d} \times {{dV} \over {dt}}$

$ \Rightarrow d = {{8.85 \times {{10}^{ - 12}} \times 40 \times {{10}^{ - 4}}} \over {4.425 \times {{10}^{ - 6}}}} \times {10^6}$

$ \Rightarrow d = 8 \times {10^{ - 3}}$ m

$ \Rightarrow x = 8$

$I = {1 \over 2}{\varepsilon _0}E_0^2\,.\,c = {1 \over 2}{\varepsilon _0}{(c{B_0})^2}c$

$ \Rightarrow I = {1 \over 2}{\varepsilon _0}{c^3}B_0^2$

$ \Rightarrow 0.22 = {1 \over 2}\left( {8.85 \times {{10}^{ - 12}}} \right){\left( {3 \times {{10}^8}} \right)^3}B_0^2$

$ \Rightarrow {B_0} \simeq 43 \times {10^{ - 9}}$ T

$E = 50\sin \left( {\omega t - {\omega \over c}.\,x} \right)$

Energy density = ${1 \over 2}{ \in _0}E_0^2$

Energy of volume $V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$

${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$

$V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}$

= .0005 $\times$ 10⁶ (c.m)³

= 500 (c.m)³

$|B|\, = {{|E|} \over C} = {6 \over {3 \times {{10}^8}}}$

= 2 $\times$ 10^$-$8 T

$\therefore$ x = 2

E₀ = 200

$I = {1 \over 2}{\varepsilon _0}E_0^2.C$

Radiation pressure

$P = {{2I} \over C}$

$ = \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$

$ = {\varepsilon _0}E_0^2$

$ = 8.85 \times {10^{ - 12}} \times {200^2}$

$ = 8.85 \times {10^{ - 8}} \times 4$

$ = {{354} \over {{{10}^9}}}$

$I = {1 \over 2}C{ \in _0}{E^2}$

${E^2} \propto I$

$I = {{Power} \over {Area}}$

${E^2} \propto {P \over A}$

$E \propto \sqrt P $

${{E'} \over E} = \sqrt {{{60} \over {100}}} $

$E' = \sqrt {{3 \over 5}} E$

So the value of x = 3

Given f = 9 $\times$ 10² Hz

$ \in $ = $ \in $₀$ \in $_r

$ \in $ = 80 $ \in $₀

So $ \in $_r = 80

$\rho $ = 0.25 $\Omega$m

V(t) = V₀ sin (2$\pi$ft)

${I_d} = {{dq} \over {dt}} = {{cdv} \over {dt}}$

${I_d} = {{{ \in _0}{ \in _r}A} \over d}{d \over {dt}}({v_0}\sin (2\pi ft))$

${I_d} = {{{ \in _0}{ \in _r}A} \over d}{V_0}(2\pi f)\cos (2\pi ft)$ .......... (1)

& ${I_c} = {V \over R}$

${I_c} = {{{V_0}\sin (2\pi ft)} \over {\rho {d \over A}}} = {{A{v_0}\sin (2\pi ft)} \over {\rho d}}$ ....... (2)

divide equation (1) and (2)

${{{I_d}} \over {{I_c}}} = { \in _0}{ \in _r}2\pi f(\rho )\cot (2\pi ft)$

${{{I_d}} \over {{I_c}}} = {1 \over {4\pi \times 9 \times {{10}^9}}} \times 80 \times 2\pi \times 9 \times {10^2} \times (0.25) \times \cot (2\pi \times 9 \times {10^2} \times {1 \over {800}})$

$ = {{{{10}^3}} \over {{{10}^9}}}\left( {\cot \left( {{{9\pi } \over 4}} \right)} \right)$

$ = {{{{10}^3}} \over {{{10}^9}}}$

${{{I_d}} \over {{I_c}}} = {1 \over {{{10}^6}}}$

${I_c} = {10^6}{I_d}$

So x = 6

Pressure = ${{Intensity} \over C}$ (for absorbing surface)

I = P $\times$ C

I = ${{2.5 \times {{10}^{ - 6}}} \over {30c{m^2}}}$ N $\times$ 3 $\times$ 10⁸ m/s

I = 25 W/cm²

Intensity of electro magnetic wave is,

$I = {1 \over 2}c{\varepsilon _0}E_0^2 = {P \over {4\pi {r^2}}}$

${1 \over 2}4\pi {\varepsilon _0} \times c \times E_0^2 = {P \over {{r^2}}}$

${1 \over 2} \times {{3 \times {{10}^5} \times E_0^2} \over {9 \times {{10}^9}}} = {{1000 \times 1.25} \over {{{(2)}^2}}} \times {1 \over {100}}$

$E_0^2 = {{60 \times 1000 \times 1.25} \over {4 \times 100}} = {{125 \times 3} \over 2}$

$E_0^2 = {{375} \over 2} = 187.5$

${E_0} = 13.69$

${E_0} \approx 137 \times {10^{ - 1}}$ v/m

Firstly, we know that the intensity (I) of a wave is defined as the power (P) per unit area (A). For a spherical wave emanating from a point source, the area of the sphere is $4\pi r^2$ where r is the distance from the source. So,

$I = \frac{P}{4\pi r^2} \tag{1}$ ........(1)

This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :

$I = \frac{1}{2} c \varepsilon_0 E^2 \tag{2}$ .........(2)

where $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=$ speed of light in vacuum and $\varepsilon_0$ is the permittivity of free space.

From equations (1) and (2), we can solve for E and square root it to find the peak value of the electric field $E_{\text{peak}}$ :

$E_{\text{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$

Since $ \varepsilon_0=\frac{1}{\mu_0 c^2} $

Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 \times 8\, \text{W} = 0.8\, \text{W}$.

So, substituting $P = 0.8\, \text{W}$, $r = 10\, \text{m}$, $c = 3 \times 10^8\, \text{m/s}$, and $\mu_0 = 4\pi \times 10^{-7}\, \text{T m/A}$, we have :

$E_{\text{peak}} = \sqrt{\frac{2 \times 0.8 \times 4\pi \times 10^{-7} \times 3 \times 10^8}{4\pi \times 100}} = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}}$

Comparing with the expression in the question, we find that $x = 2$.

Therefore, the answer is $x = 2$.

Given wavelength of an x-ray beam = 10$\mathop A\limits^o $

$ \because $ E = ${{hc} \over \lambda } = m{c^2}$

$ \Rightarrow $ m = ${h \over {c\lambda }}$

The mass of a fictitious particle having the same energy as that of the x-ray photons = ${x \over 3}$h kg

$ \therefore $ ${x \over 3}h = {h \over {c\lambda }}$

$x = {3 \over {c\lambda }}$

$ = {3 \over {3 \times {{10}^8} \times 10 \times {{10}^{ - 10}}}}$

x = 10

Given, frequency of wave, f = 3 GHz = 3 $\times$ 10⁹ Hz

Relative permittivity, $\varepsilon $_r = 2.25

Since, f = C/$\lambda$

$ \Rightarrow \lambda = {c \over f} = {{3 \times {{10}^8}} \over {3 \times {{10}^9}}} = 0.1$ m

$\because$ $\lambda$_m (wavelength of wave in a medium) = $\lambda$/$\mu$ and we know that, $\mu = \sqrt {{\mu _r}{\varepsilon _r}} $

As, dielectric is non-magnetic, $\mu$_r = 1

$ \Rightarrow \mu = \sqrt {2.25} = 1.5$

$ \Rightarrow {\lambda _m} = {{0.1} \over {1.5}} = {1 \over {15}} = 0.0667$ m

= 6.67 cm = 667 $\times$ 10^$-$2 cm

Given, $\mu$_r = $\varepsilon $_r = 2

where, $\mu$_r is relative permeability, $\varepsilon $_r is relative permittivity.

Speed of electromagnetic wave v is given by

$v = {c \over n}$

where, n = refractive index = $\sqrt {{\mu _r}{\varepsilon _r}} = \sqrt 4 = 2$

$ \Rightarrow v = {{3 \times {{10}^8}} \over 2}$ = 15 $\times$ 10⁷ m/s

$\because$ x $\times$ 10⁷ = 15 $\times$ 10⁷

$\Rightarrow$ x = 15

I = ${1 \over 2}$$\varepsilon $₀$E_0^2$c

$ \Rightarrow $ E₀ = $\sqrt {{{2I} \over {{\varepsilon _0}c}}} $

$ \therefore $ E_rms = ${{{E_0}} \over {\sqrt 2 }}$ = $\sqrt {{I \over {{\varepsilon _0}c}}} $

= $\sqrt {{{{{315} \over \pi }} \over {8.86 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} $

= 194