Electromagnetic Waves

Electromagnetic waves are unique in that they consist of oscillating electric and magnetic fields and do not require a medium to travel, which allows them to propagate through a vacuum.

These waves differ from mechanical waves as they are not longitudinal. Instead, electromagnetic waves are transverse, meaning that the oscillations of the fields are perpendicular to the direction of wave propagation.

Electromagnetic waves carry both energy and momentum. The energy carried by a wave is given by the equation:

$ E = h \nu $

where $ h $ is Planck's constant and $ \nu $ is the frequency of the wave.

The momentum carried by electromagnetic waves can be described using the equation:

$ p = \frac{h}{\lambda} $

where $ p $ is the momentum and $ \lambda $ is the wavelength of the wave. This relationship illustrates how electromagnetic waves convey momentum as they move through space.

To find the wavelength of a wave, you can use the formula:

$\lambda = \frac{c}{f}$

where:

$\lambda$ is the wavelength,

$c$ is the speed of light (in vacuum, which is $3 \times 10^8 \, \text{m/s}$),

$f$ is the frequency.

Given that:

$f = 3 \, \text{GHz} = 3 \times 10^9 \, \text{Hz}$

Plug the values into the formula:

$\lambda = \frac{3 \times 10^8 \, \text{m/s}}{3 \times 10^9 \, \text{Hz}} = 0.1 \, \text{m}$

Thus, the wavelength of the wave is 0.1 m, which corresponds to Option A.

UV - Water purification

X-rays - Diagnostic tool in medicine

Microwave - Communication, Radar

Infrared wave - Improving visibility in foggy days.

Pressure $ = {l \over c}$

$ \Rightarrow {F \over A} = {l \over c}$

$ \Rightarrow l = {{7.2 \times {{10}^{ - 9}} \times 3 \times {{10}^8}} \over {36 \times {{10}^{ - 4}}}}$ W/m²

$ = 600$ W/m²

$ \Rightarrow l = 0.06$ W/cm²

In an EM wave :

1. $\overrightarrow E \,\, \bot \,\,\overrightarrow B $

2. $\overrightarrow V \equiv \overrightarrow E \times \overrightarrow B $

3. Energy is equally divided

4. $\left| {\overrightarrow V } \right| = \left| {\overrightarrow E } \right|/\left| {\overrightarrow B } \right|$

Ratio $ = {{|q\overrightarrow E |} \over {|q\overrightarrow v \times \overrightarrow B |}}$

$ = {E \over {vB}} = {{{v_{wave}}} \over v}$

$\Rightarrow$ Ratio $ = {{3 \times {{10}^8}} \over {3 \times {{10}^7}}} = 10$

$v = {E \over B}$ and $K = {1 \over 2}m{v^2}$

$ \Rightarrow \sqrt {{{2K} \over m}} \times B = E$

$ \Rightarrow E = \sqrt {{{2 \times 728 \times 1.6 \times {{10}^{ - 19}}} \over {9.1 \times {{10}^{ - 31}}}}} \times 12 \times {10^{ - 3}}$

$ = 192000$ V/m

Speed of light $c = {\omega \over k} = {{1.5 \times {{10}^{11}}} \over {0.5 \times {{10}^3}}} = 3 \times {10^8}$ m/sec

So, ${E_0} = {B_0}c$

$ = 2 \times {10^{ - 8}} \times 3 \times {10^8}$

$ = 6$ V/m

Direction will be along z-axis.

$c = 3 \times {10^8}$ m/sec

$B = {E \over c} = {{540} \over {3 \times {{10}^8}}} = 18 \times {10^{ - 7}}\,T$

${Z_C} = {V \over I}$

$ \Rightarrow {1 \over {\omega C}} = {{230} \over {6.9}}\,M\,\Omega $

$ \Rightarrow C = {{6.9} \over {230\,\omega }}\,\mu F$

$ = {{6.9} \over {230 \times 600}}\,\mu F$

$C = 50\,pF$

Given the electric field expression:

$E_z = 300 \sin(5\pi \times 10^3 x - 3\pi \times 10^{11} t) \, \text{Vm}^{-1}$

The amplitude of the electric field ($E_0$) is $300 \, \text{V/m}$.

The velocity ($v$) of the wave in the medium is given by the ratio of the coefficients of time and displacement in the wave equation, which can be calculated as:

$v = \frac{\text{Coefficient of } t}{\text{Coefficient of } x} = \frac{3\pi \times 10^{11}}{5\pi \times 10^3} = \frac{3}{5} \times 10^8 \, \text{m/s}$

The relationship between the electric field amplitude and the magnetic field amplitude in an electromagnetic wave is given by:

$B_0 = \frac{E_0}{v}$

Substituting the values for $E_0$ and $v$ into this formula:

$B_0 = \frac{300}{\frac{3}{5} \times 10^8}$

$B_0 = \frac{300 \times 5}{3 \times 10^8}$

$B_0 = \frac{1500}{3 \times 10^8}$

$B_0 = 5 \times 10^{-6} \, \text{T}$

Therefore, the amplitude of the magnetic field ($B_0$) is $5 \times 10^{-6}$ T.

In first 3 options speed of light is 3 $\times$ 10⁸ m/sec and in the fourth option it is 4 $\times$ 10⁸ m/sec.

Using

E = CB

We can check the option is B.

E₀ = 2.25 V/m

B₀ = 1.5 $\times$ 10^$-$8 T

$ \Rightarrow {{{E_0}} \over {{B_0}}} = 1.5 \times {10^8}$ m/s

$\Rightarrow$ Refractive index = 2

Distance to be travelled = 6 km

Time taken $ = {{6 \times {{10}^3}} \over {1.5 \times {{10}^8}}} = 4 \times {10^{ - 5}}$ s

In a material medium speed of light is given by $v = {1 \over {\sqrt {{\varepsilon _0}{\varepsilon _r}{\mu _0}{\mu _r}} }}$. So statement 2 is false.

UV rays are used to sterilize surgical material.

Microwaves are used in radar system.

Infrared are used for green house effect and

X-rays are used to study crystal structure.

Wavelength of microwave is maximum then visible light then X-rays and then gamma rays so the correct order will be

$\lambda$_gamma-ray < $\lambda$_Xray < $\lambda$_visible < $\lambda$_microwave

$\overrightarrow E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}$

${E_{0x}} = 301.6$

${E_{0y}} = + 452.4$

${E_0} = \sqrt {E_{0x}^2 + E_{0y}^2} $

Now, ${{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C} = {{\sqrt {E_{0x}^2 + E_{0y}^2} } \over C}$

Also, $\widehat B = \widehat C \times \widehat E \Rightarrow \widehat k \times {{({E_{0x}}\widehat i + {E_{0y}}\widehat j)} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}$

$\widehat B = {{ - {E_{0x}}\widehat j - {E_{0y}}\widehat i} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}$

$\overrightarrow B = - {{{E_{0x}}} \over C}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over C}\sin (kz - \omega t){\widehat a_x}$

$\overrightarrow H = {{\overrightarrow B } \over {{\mu _0}}}$

$\overrightarrow H = - {{{E_{0x}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_x}$

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

$ = {1 \over 2} 8.5 \times {10^{ - 12}} \times {(56.5)^2} \times 3 \times {10^8}$

$ = 4.24$ W/m²

H = 4.5 $\times$ 10^$-$2

So B = $\mu$₀$\mu$H

Thus $E = {c \over n}B$ (where n $\Rightarrow$ refractive index)

So $E = {{3 \times {{10}^8} \times 4\pi \times {{10}^{ - 7}} \times 1.61 \times 4.5 \times {{10}^{ - 2}}} \over {\sqrt {1.61 \times 6.44} }}$

$E = 8.48$

Given, frequency of electric field of EM wave,

$ f=30 \mathrm{MHz}=3 \times 10^7 \mathrm{~Hz} $

Amplitude, $E_0=150 \mathrm{~V} / \mathrm{m}$

∴ Amplitude of magnetic field vector,

$ B_0=\frac{E_0}{c}=\frac{150}{3 \times 10^8}=5 \times 10^7 \mathrm{~T} $

Angular frequency,

$ \begin{aligned} \omega & =2 \pi f=22 \pi \times 3 \times 10^7 \\ & =6 \pi \times 10^7 \mathrm{rad} / \mathrm{s} \end{aligned} $

∴ Propagation constant,

$ k=\frac{2 \pi}{\lambda}=\frac{2 \pi}{c / f}=\frac{2 \pi f}{c}=\frac{\omega}{c}=\frac{6 \pi \times 10^7}{3 \times 10^8}=\frac{\pi}{5} $

Since, $\mathbf{E}, \mathbf{B}$ and direction of propagation are mutually perpendicular to each other, hence, for the given situation $\mathbf{B}$ will be directed along $Z$-axis ∴ Expression of magnetic field,

$ \begin{aligned} B & =B_0 \sin [k x-\omega t] \hat{\mathbf{z}} \\ \Rightarrow \quad B & =5 \times 10^{-7} \sin \left[\frac{\pi}{5} x-6 \pi \times 10^7 t\right] \hat{\mathbf{z}} \\ & =5 \times 10^{-7} \sin \left[\pi\left(\frac{x}{5}-6 \times 10^7 t\right)\right] \hat{\mathbf{z}} \end{aligned} $

Average power delivered by sun to the top of earth's atmosphere,

$ \begin{gathered} P_{\text {avg }}=\frac{6}{\pi} \times 10^3 \mathrm{~W} / \mathrm{m}^2 \\ \frac{1}{2} \varepsilon_0 E_0^2 c=\frac{6}{\pi} \times 10^3 \\ E_0=\sqrt{\frac{6}{\pi} \times 10^3 \times \frac{2}{\varepsilon_0 \cdot c}} \\ =\sqrt{\frac{6}{\pi} \times 10^3 \times \frac{2}{8.85 \times 10^{-12} \times\left(3 \times 10^8\right)}} \\ \end{gathered} $

$ \begin{aligned} & =12 \times 10^2 \mathrm{~N} / \mathrm{C} \\ \therefore \quad B_0 & =\frac{E_0}{c} \\ & =\frac{12 \times 10^2}{3 \times 10^8}=4 \times 10^{-6} \mathrm{~T} \end{aligned} $

Intensity of laser beam,

$ I=2.1 \times 10^{15} \mathrm{~W} / \mathrm{m}^2 $

We know that, intensity of a EM wave

$ \begin{aligned} I & =\varepsilon_0 E_{\mathrm{rms}}^2 c \\ & =\varepsilon_0\left(\frac{E_0}{\sqrt{2}}\right)^2 c=\varepsilon_0 \frac{E_0^2}{2} c \\ & =\varepsilon_0 \frac{\left(B_0 c\right)^2}{2} c \quad\left[\because E_0=B_0 c\right] \end{aligned} $

$ \begin{aligned} & I=\varepsilon_0 \frac{B_0^2 c^3}{2} \\ & \Rightarrow B_0=\sqrt{\frac{2 I}{\varepsilon_0 c^3}}=\sqrt{\frac{2 I}{\varepsilon_0 \cdot \frac{1}{\mu_0 \varepsilon_0} \cdot c}} \\ & {\left[\because c^3=c^2 \cdot c=\frac{1}{\mu_0 \varepsilon_0} \cdot c\right] } \end{aligned} $

         $ \begin{aligned} & =\sqrt{\frac{2 \mu_0 I}{c}} \\ & =\sqrt{\frac{2 \times 4 \pi \times 10^{-7} \times 2.1 \times 10^{15}}{3 \times 10^8}} \\ & =\sqrt{17.6}=4.195 \\ & =4.2 \mathrm{~T} \end{aligned} $

Given, power of bulb, $P=100 \mathrm{~W}$

$ r=5 \mathrm{~m} $

Power of visible radiation,

$ \begin{aligned} P^{\prime} & =20 \% \text { of } P \\ & =\frac{20}{100} \times 100=20 \mathrm{~W} \end{aligned} $

Average intensity of radiation,

$ \begin{aligned} I & =\frac{P^{\prime}}{4 \pi r^2} \\ & =\frac{20}{4 \pi \times 5^2}=\frac{5}{25 \pi} \mathrm{~W} / \mathrm{m}^2 \\ & =\frac{\alpha}{25 \pi} \mathrm{~W} / \mathrm{m}^2\,\,\,\text { (given) } \\ \therefore \alpha & =5 \end{aligned} $

Equations of EM waves are given as,

$ \begin{aligned} & \mathbf{E}(t)=\mathbf{E}_m \sin (k x-\omega t) \\ & \mathbf{B}(t)=\mathbf{B}_m \sin (k x-\omega t) \end{aligned} $

If $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ are unit vectors along $\mathbf{E}(t)$ and $\mathbf{B}(t)$ respectively, then

$ \hat{\mathbf{x}}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{1^2+1^2}}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}} $

and $\hat{\mathbf{y}}=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{1^2+1^2}}=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}$

∴ Unit vector in the direction of propagation of EM wave is $\hat{\mathbf{z}}$, then

$ \begin{aligned} \hat{\mathbf{z}} & =\hat{\mathbf{x}} \times \hat{\mathbf{y}} \\ & =\left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}\right) \times\left(\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}\right) \\ & =\frac{1}{2}[\hat{\mathbf{i}} \times \hat{\mathbf{i}}-\hat{\mathbf{i}} \times \hat{\mathbf{j}}+\hat{\mathbf{j}} \times \hat{\mathbf{i}}-\hat{\mathbf{j}} \times \hat{\mathbf{j}}] \\ & =\frac{1}{2}[0-\hat{\mathbf{k}}-\hat{\mathbf{k}}-0]=\frac{1}{2} \times(-2 \hat{\mathbf{k}}) \\ & =-\hat{\mathbf{k}} \end{aligned} $

Hence, unit vector along the direction of propagation of EM wave will be $-\hat{\mathbf{k}}$.

Given, power of incident beam,

$ P=10 \mathrm{~W} $

Energy of EM wave, $E=\frac{h c}{\lambda}$

or Power or EM wave,

$ \begin{aligned} P & =\frac{E}{t}=\frac{h c}{\lambda t} \\ \Rightarrow \quad P & =\frac{h}{\lambda} \cdot\left(\frac{c}{t}\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

But we know that, de-Broglie wavelngth

$ \lambda=\frac{h}{p} $

where, $p=$ momentum

$ \Rightarrow \quad p=\frac{h}{\lambda} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

$ \begin{aligned} P & =p \frac{c}{t}=\frac{p}{t} \cdot c=F c \\ \Rightarrow \quad F & =\frac{P}{c} \end{aligned} $

$ \begin{aligned} & \therefore \text { Total effective force } \\ & \quad=F_{\text {absorbed }}+F_{\text {reflected }} \\ & \quad=70 \% \text { of } \frac{P}{c}+2 \times 30 \% \text { of } \frac{P}{c} \\ & \quad=0.7 \frac{P}{c}+0.6 \frac{P}{c} \\ & \quad=1.3 \frac{P}{c}=\frac{1.3 \times 10}{3 \times 10^8}=4.33 \times 10^{-8} \mathrm{~N} \end{aligned} $

Expression for the magnetic field of EM wave is given as

$ \mathbf{B}=\left(2 \times 10^{-8}\right) \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathbf{k}} \mathrm{T} $

Magnitude of magnetic field vector,

$ B_0=2 \times 10^{-8} \mathrm{~T} $

Magnitude of electric field vector of EM wave,

$ E_0=B_0 c $

where, $c$ is speed of light

$ \begin{aligned} & =2 \times 10^{-8} \times 3 \times 10^8 \\ & =6 \mathrm{~V} / \mathrm{m} \quad\left(\because c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right) \end{aligned} $

If $\hat{\mathbf{n}}$ be the unit vector along the direction of electric field, then we know that unit vector along electric field vector $\mathbf{x}$

Unit vector along magnetic field vector $=$ Unit vector along the direction of propagation of EM wave.

$ \begin{aligned}i.e.\,\, \hat{\mathbf{n}} \times \hat{\mathbf{k}} & =-\hat{\mathbf{j}} \\clearly,\,\, \hat{\mathbf{n}} & =\hat{\mathbf{i}} \end{aligned} $

Hence, expression of electric field vector of EM wave,

$ \mathbf{E}=6 \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathbf{i}} \mathrm{V} / \mathrm{m} $

Given, Magnetic field in a plane EM wave,

$B=\left(3 \times 10^{-7} \mathrm{~T}\right) \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \mathrm{T}$

Here, $B_0=3 \times 10^{-7} \mathrm{~T}$

Amplitude of electric field,

$\begin{aligned} E_0 & =B_0 c \\ & =3 \times 10^{-7} \times 3 \times 10^8 \\ \Rightarrow \quad E_0 & =90 \mathrm{Vm}^{-1} \end{aligned}$

From the given equation of $B$, it is clear that EM wave is travelling along negative direction of $X$-axis i.e.

$\therefore$ If $\hat{\mathbf{n}}$ be the unit vector along the direction of electric field, then

$\hat{\mathbf{n}} \times \hat{j}=-\hat{i}$

Clearly, $\hat{\mathbf{n}}=\hat{k}$

Hence.

$\begin{aligned} E & =E_0 \sin (k x+\omega t) \hat{k} \\ & =90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1} \end{aligned}$

The correct answer is Option A: Space Waves. Here's why:

UHF (Ultra High Frequency) waves, which range from 300 MHz to 3 GHz, typically propagate using space waves. Let's break down the different types of wave propagation:

Space Waves:

• Travel in a straight line directly from the transmitting antenna to the receiving antenna.
• These waves are not affected by the Earth's surface or atmosphere significantly.
• They are commonly used for satellite communication, terrestrial microwave links, and short-distance line-of-sight communication.

Surface Waves:

• These waves travel along the surface of the Earth.
• They are primarily used for low-frequency (LF) and very low frequency (VLF) radio waves, which have wavelengths longer than UHF.

Ground Waves:

• Similar to surface waves, they travel along the ground.
• They are primarily used for medium-frequency (MF) radio waves and are affected by the Earth's conductivity.

Sky Waves:

• These waves are reflected by the ionosphere, a layer of charged particles in the upper atmosphere.
• They are primarily used for long-distance communication with high-frequency (HF) radio waves.

In summary: UHF frequencies, being much higher than those used for ground or sky waves, primarily propagate via space waves, which travel directly from the transmitter to the receiver without significant interaction with the Earth's surface or the atmosphere.

Given:

• Intensity of light (I) = 12 W/m²


• Surface area (A) = 4 cm² = 4 × 10⁻⁴ m²

For a perfectly black surface, the radiation pressure (p) is calculated using the formula:

$ p = \frac{I}{c} $

where $ c $ is the speed of light in vacuum, approximately $ 3 \times 10^8 $ m/s.

Substituting the given values:

$ p = \frac{12}{3 \times 10^8} $

$ p = 4 \times 10^{-8} \, \text{Pa} $

Hence, the radiation pressure on the surface is $ 4 \times 10^{-8} \, \text{Pa} $.

Electric field $(E)$ and magnetic field $(B)$ of an electromagnetic wave are given as

$E=E_0 \sin (k x-\omega t)$

and $B=B_0 \sin (k x-\omega t)$

We know that,

$\begin{aligned} E_0 & =C B_0 \\ K & =\frac{2 \pi}{\lambda} \text { and } \omega=2 f \pi \\ \therefore \quad \frac{E_0}{B_0} & =C=f \lambda \\ \Rightarrow \quad \frac{E_0}{B_0} & =f \lambda \\ & =\frac{\omega}{2 \pi} \cdot \frac{2 \pi}{k} \quad\left(\because f=\frac{\omega}{2 \pi} \text { and } \lambda=\frac{2 \pi}{k}\right) \\ \Rightarrow \quad \frac{E_0}{B_0} & =\frac{\omega}{k} \\ \Rightarrow \quad E_0 k & =B_0 \omega \end{aligned}$

We know that, Modulation index,

$\mu=\frac{A_m}{A_c} \quad \text{.... (i)}$

Where, $A_m=$ amplitude of message signal $A_c=$ amplitude of carrier signal When $A_m$ is increased by $1 \%$ then, new,

$A_m^{\prime}=A_m+\frac{A_m}{100}=\frac{101 A_m}{100}$

Similarly, $A_c$ is increased by $3 \%$.

$\because$ New value of amplitude of carrier wave,

$\begin{aligned} A_c^{\prime} & =A_c+3 \% \text { of } A_c \\ & =A_c+\frac{3 A_c}{100} \\ & =\frac{103 A_c}{100} \end{aligned}$

New modulation index,

$\begin{aligned} \mu^{\prime} & =\frac{A_m^{\prime}}{A_c^{\prime}}=\frac{101 A_m / 100}{103 A_c / 100} \\ & =\frac{101}{103} \cdot \frac{A_m}{A_c} \\ & =\frac{101}{103} \mu \quad \text{[From Eq. (i)]} \end{aligned}$

$\begin{aligned} \% \text { change in modulation index } & =\frac{\mu^{\prime}-\mu}{\mu} \times 100 \\ & =\frac{\frac{101}{103} \mu-\mu}{\mu} \times 100 \\ & =-2 \% \end{aligned}$

Given, Electric field along $X$-axis

$E=8.7 \mathrm{~Vm}^{-1}$

$\therefore$ Magnetic field along $Y$ - axis,

$\begin{aligned} B & =\frac{E}{c} \quad[\because E=c B] \\ & =\frac{8.7}{3 \times 10^8}=29 \times 10^{-8} \mathrm{~T} \end{aligned}$

Given, Average power per unit area,

$I=9240 \mathrm{Wm}^{-2}$

We know that, average power per unit area delivered by an electromagnetic wave is known as intensity of EM wave which is given as

$\begin{aligned} I & =\frac{C B^2}{2 \mu_0} \\ \Rightarrow \quad B & =\sqrt{\frac{2 \mu_0 I}{c}}=\sqrt{\frac{2 \times 4 \pi \times 10^{-7} \times 9240}{3 \times 10^8}} \\ & =87.98 \times 10^{-7} \mathrm{~T}=8.798 \times 10^{-6} \mathrm{~T} \\ & =8.798 \mu \mathrm{T} \approx 8.8 \mu \mathrm{T} \end{aligned}$

Given, Intensity of light beam,

$I=10^{-3} \mathrm{~N} / \mathrm{m}^2$

Area,

$\begin{aligned} & A=20 \mathrm{~cm}^2=2 \times 10^{-3} \mathrm{~m}^2 \\ & \alpha=45^{\circ} \end{aligned}$

For a fully reflecting surface, force exerted by the beam on the surface, is given as

$\begin{aligned} F= & \frac{2 I A \cos \alpha}{c} \\ & =\frac{2 \times 10^{-3} \times 2 \times 10^{-3} \cos 45^{\circ}}{3 \times 10^8} \\ & =\frac{4 \times 10^{-6}}{3 \times 10^8} \times \frac{1}{\sqrt{2}}=9.42 \times 10^{-15} \mathrm{~N} \\ & \simeq 9.4 \times 10^{-15} \mathrm{~N} \end{aligned}$

The maximum number of TV signals, that can transmitted along a co-axial cable is 125.

The speed of light in a medium is given by the equation:

$ v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} $

where:

• $c$ is the speed of light in vacuum (approximately $3 \times 10^8$ m/s),
• $\varepsilon_r$ is the relative permittivity of the medium (in this case, distilled water, and is given as 81),
• $\mu_r$ is the relative permeability of the medium (given as 1 for distilled water).

Substituting the given values into the equation, we get:

$ v = \frac{3 \times 10^8 \, \text{m/s}}{\sqrt{81 \times 1}} $

$ v = \frac{3 \times 10^8 \, \text{m/s}}{9} $

$ v = 3.33 \times 10^7 \, \text{m/s} $

So, the speed of light in distilled water is $3.33 \times 10^7$ m/s.