Properties of Electromagnetic Waves Questions in English

(D) In a conductor,the conductivity $\sigma$ is very high. The velocity of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$. For a conductor,the wave undergoes significant attenuation due to the interaction with free electrons. The velocity of electromagnetic waves in a good conductor is much lower than the speed of light in a vacuum $(c = 3 \times 10^{8} \ m/s)$. Therefore,the velocity is very low.

(B) The frequency range from $2\,\text{MHz}$ to $30\,\text{MHz}$ is known as the High Frequency $(HF)$ band.
These waves are reflected back to the Earth by the ionosphere.
This mode of propagation is known as sky wave propagation.
Therefore,the correct option is $B$.

(A) In space,information (electromagnetic waves) travels at the speed of light,$c = 3 \times 10^8 \, m/s$.
Given distance,$d = 30 \, km = 30,000 \, m = 3 \times 10^4 \, m$.
The time taken $t$ is given by the formula $t = \frac{d}{c}$.
Substituting the values: $t = \frac{3 \times 10^4 \, m}{3 \times 10^8 \, m/s} = 1 \times 10^{-4} \, s$.
Converting to milliseconds: $1 \times 10^{-4} \, s = 0.1 \times 10^{-3} \, s = 0.1 \, ms$.

(D) For an electromagnetic wave propagating in free space,the intensity of the wave $I$ is proportional to the square of the electric field amplitude $E$ $(I \propto E^2)$.
As the wave travels,the energy spreads over a spherical wavefront of area $A = 4\pi r^2$.
Since intensity $I = \frac{P}{A} = \frac{P}{4\pi r^2}$,we have $I \propto \frac{1}{r^2}$.
Equating the two relations: $E^2 \propto \frac{1}{r^2}$,which implies $E \propto \frac{1}{r}$.
Therefore,the electric field intensity is inversely proportional to the distance $r$ traveled by the wave.

(C) Space waves are electromagnetic waves that travel in a straight line from the transmitting antenna to the receiving antenna. These waves are typically in the frequency range of $VHF$ (Very High Frequency) and above (i.e.,$UHF$ and $Microwave$). Because these waves travel through the troposphere,they are significantly affected by atmospheric conditions such as temperature,humidity,and pressure,which can cause refraction or scattering. Therefore,$VHF$ waves are the most appropriate answer among the choices provided as they are primarily space waves.

(D) Long-distance radio communication is achieved through the reflection of radio waves by the ionosphere. These waves are known as sky waves.
Sky waves are electromagnetic waves in the frequency range of $3 \ MHz$ to $30 \ MHz$.
When these waves are transmitted towards the sky,they are reflected back to the Earth by the ionospheric layers,allowing for communication over very long distances.

(B) In a perfect dielectric medium,the conductivity $\sigma = 0$.
Attenuation in a medium is caused by the loss of energy as the wave propagates,which is directly related to the conductivity of the medium.
Since a perfect dielectric has no conductivity,there are no ohmic losses,and therefore,the electromagnetic wave propagates without any attenuation.

(B) In a two-wire transmission line,the electromagnetic energy (power) is carried by the electromagnetic field that exists in the space surrounding the conductors. The conductors themselves primarily serve to guide the electromagnetic waves. Therefore,the power flows through the dielectric medium (the space) outside the conductors.

(C) The refractive index $n$ of the ionosphere for electromagnetic waves is given by the formula $n = \sqrt{1 - \frac{f_p^2}{f^2}}$,where $f_p$ is the plasma frequency and $f$ is the frequency of the wave.
Since the term $\frac{f_p^2}{f^2}$ is positive,the value under the square root is less than $1$.
Therefore,the refractive index $n$ of the ionosphere is always less than $1$.

(D) The ionosphere is a region of the Earth's upper atmosphere that is ionized by solar and cosmic radiation.
This ionization process creates a plasma consisting of free electrons and positive ions.
Because of the presence of these free electrons,the ionosphere reflects radio waves back to the Earth,acting like a $Mirror$.
However,the ionization itself is caused by high-energy radiation (ultraviolet,$X$-rays) from the Sun and cosmic sources,not by the radio waves themselves.
Therefore,the statement that ionization is caused by radio waves is incorrect.

(B) The relative permittivity $\epsilon_r$ of an ionized medium (like the ionosphere) is given by the relation $\epsilon_r = 1 - \frac{\omega_p^2}{\omega^2}$,where $\omega_p$ is the plasma frequency and $\omega$ is the frequency of the electromagnetic wave.
Since the ionosphere contains free electrons,the plasma frequency $\omega_p$ is non-zero.
As the wave enters the ionized layer,the presence of these free electrons causes the effective relative permittivity to be less than $1$.
Compared to free space (where $\epsilon_r = 1$),the relative permittivity of the ionized layer decreases.

(B) The velocity of an electromagnetic wave in a medium is given by $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$.
Assuming the medium is non-magnetic,$\mu_r \approx 1$.
Thus,$v = \frac{c}{\sqrt{\epsilon_r}}$,where $\epsilon_r$ is the dielectric constant $(K)$.
Rearranging for $K$,we get $\sqrt{K} = \frac{c}{v}$,so $K = \left(\frac{c}{v}\right)^2$.
Given $c = 3 \times 10^{8} \ m/s$ and $v = 2.5 \times 10^{8} \ m/s$.
$K = \left(\frac{3 \times 10^{8}}{2.5 \times 10^{8}}\right)^2 = \left(\frac{3}{2.5}\right)^2 = (1.2)^2 = 1.44$.

(B) For an electromagnetic wave,the relationship between the peak electric field $(E_0)$ and the peak magnetic field $(B_0)$ is given by $E_0 = c \times B_0$,where $c$ is the speed of light in vacuum $(c \approx 3 \times 10^8 \, m/s)$.
Given: $B_0 = 20 \, nT = 20 \times 10^{-9} \, T$.
Substituting the values:
$E_0 = (3 \times 10^8 \, m/s) \times (20 \times 10^{-9} \, T)$
$E_0 = 60 \times 10^{-1} \, V/m$
$E_0 = 6 \, V/m$.
Therefore,the correct option is $B$.

(C) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$,where $\mu = \mu_0 \mu_r$ and $\epsilon = \epsilon_0 \epsilon_r$.
The speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
The refractive index $n$ is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:
$n = \frac{c}{v} = \frac{1/\sqrt{\mu_0 \epsilon_0}}{1/\sqrt{\mu \epsilon}} = \sqrt{\frac{\mu \epsilon}{\mu_0 \epsilon_0}} = \sqrt{\mu_r \epsilon_r}$.
Therefore,the refractive index is $\sqrt{\mu_r \epsilon_r}$.

(A) The transverse nature of light (and electromagnetic waves in general) is uniquely demonstrated by the phenomenon of polarization. Polarization involves the restriction of the vibrations of the electric field vector to a single plane,which is only possible if the wave is transverse in nature. Longitudinal waves,such as sound waves,cannot be polarized.

(B) Electromagnetic $(EM)$ waves are oscillations of electric and magnetic fields.
They transport energy (e.g.,light),momentum (as demonstrated by radiation pressure and the photoelectric effect),and information (e.g.,radio waves).
However,they do not transport electric charge,as they are neutral entities propagating through space.

(B) In an electromagnetic wave,the electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are always perpendicular to each other.
They are also perpendicular to the direction of propagation of the wave.
These fields oscillate in the same phase,meaning they reach their maximum and minimum values at the same time.
Therefore,the correct relationship is that they are perpendicular to each other and in the same phase.

(A) In an electromagnetic wave,the electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ oscillate in phase with each other.
This means that both fields reach their maximum and minimum values at the same time and at the same points in space.
Although they are perpendicular to each other and to the direction of propagation,they do not have a time delay between their oscillations.
Therefore,the phase difference between the $\vec{E}$ and $\vec{B}$ vectors is $0$.

(B) In an electromagnetic wave,the oscillating electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are always perpendicular to each other.
Furthermore,they are also perpendicular to the direction of propagation of the wave.
These fields oscillate in the same phase,meaning they reach their maximum and minimum values simultaneously.
Therefore,the correct orientation is that they are mutually perpendicular and in the same phase.

(B) The general equation for a plane electromagnetic wave is $E = E_0 \cos(\omega t + kx)$.
Comparing this with the given equation $E_z = 100 \cos(6 \times 10^8 t + 4x)$:
Angular frequency $\omega = 6 \times 10^8 \text{ rad/s}$.
Wave number $k = 4 \text{ rad/m}$.
The speed of the wave in the medium is given by $v = \frac{\omega}{k} = \frac{6 \times 10^8}{4} = 1.5 \times 10^8 \text{ m/s}$.
The refractive index $n$ of the medium is given by $n = \frac{c}{v}$,where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light in vacuum.
$n = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$.
Therefore,the refractive index of the medium is $2$.

(B) In an electromagnetic wave,the electric field and magnetic field are related to the speed of light $(c)$ by the equation:
$c = \frac{E_0}{B_0}$
Rearranging this formula to solve for the electric field amplitude $(E_0)$,we get:
$E_0 = cB_0$
Therefore,the correct relationship is $E_0 = cB_0$.

(A) Electromagnetic waves are transverse waves in nature.
In an electromagnetic wave,the oscillating electric field $(E)$ and the oscillating magnetic field $(B)$ are always in the same phase.
These fields are mutually perpendicular to each other and also perpendicular to the direction of propagation of the wave.
Therefore,the electric field and the magnetic field are in the same phase and perpendicular to each other.

(D) In an electromagnetic wave,the electric field $(E)$ and magnetic field $(B)$ oscillate sinusoidally with time and space.
Mathematically,these fields are represented as $E = E_0 \sin(kx - \omega t)$ and $B = B_0 \sin(kx - \omega t)$.
The average value of a sine or cosine function over a complete cycle is zero.
Therefore,the average value of both the electric field and the magnetic field over a complete cycle is zero.

(A) Microwave ovens operate at a frequency of $2.45\,GHz$. This is $NOT$ the resonant frequency of a water molecule.
If $2.45\,GHz$ were the resonant frequency of water molecules, the microwaves would all be absorbed in the surface layer of the food, and the interior would not be cooked.
Therefore, the efficiency of heating is not due to resonance, but rather the rotation of polar water molecules in the oscillating electric field of the microwaves.
Since none of the provided options correctly describe the physical mechanism, and the question asks for the condition, it is important to note that the premise of resonance is a common misconception. However, in the context of standard physics curriculum questions, option $A$ is the most scientifically accurate statement regarding the relationship between the frequencies.

(D) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field.
Key properties include:
$1$. They are transverse in nature,meaning the oscillations of the electric and magnetic fields are perpendicular to the direction of wave propagation.
$2$. They do not require a material medium to travel and can propagate through a vacuum.
$3$. Light is a form of electromagnetic radiation.
Since electromagnetic waves are transverse,the statement that they are longitudinal waves is incorrect.

(C) Step $1$: Given data:
Frequency of the electromagnetic wave $f = 1 \, \text{kHz} = 10^3 \, \text{Hz}$.
Velocity of the electromagnetic wave $c = 3 \times 10^8 \, \text{m/s}$.
Step $2$: Calculate the wavelength:
The relationship between wavelength $\lambda$, velocity $c$, and frequency $f$ is given by $\lambda = \frac{c}{f}$.
Substituting the values:
$\lambda = \frac{3 \times 10^8 \, \text{m/s}}{10^3 \, \text{Hz}}$
$\lambda = 3 \times 10^5 \, \text{m}$.
Step $3$: Convert to $km$:
Since $1 \, \text{km} = 10^3 \, \text{m}$, we have:
$\lambda = \frac{3 \times 10^5}{10^3} \, \text{km} = 300 \, \text{km}$.

(D) In a vacuum,the speed of electromagnetic waves is given by $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,which is a constant value of approximately $3 \times 10^8 \ m/s$. This speed is independent of the frequency,wavelength,or intensity of the wave in a vacuum. Therefore,electromagnetic waves of all frequencies travel at the same speed in a vacuum.

(B) In an electromagnetic wave,the electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ oscillate perpendicular to each other and also perpendicular to the direction of wave propagation.
According to the properties of electromagnetic waves,the direction of propagation of the wave is given by the direction of the Poynting vector $\vec{S}$,which is defined as $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})$.
Since the velocity vector $\vec{v}$ of the electromagnetic wave is in the same direction as the energy flow (Poynting vector),the velocity is parallel to the cross product $\vec{E} \times \vec{B}$.

(B) The energy density $u_B$ associated with a magnetic field $B$ in free space is given by the formula $u_B = \frac{B^2}{2 \mu_0}$.
In an electromagnetic wave,the energy is shared equally between the electric field and the magnetic field.
The average energy density associated with the magnetic field component is $\langle u_B \rangle = \frac{\langle B^2 \rangle}{2 \mu_0}$.
Thus,the expression for the energy density in terms of the magnetic field is $\frac{B^2}{2 \mu_0}$.

(C) Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of wave propagation.
Since the oscillations of the fields occur in a direction perpendicular to the direction of energy transfer,electromagnetic waves are classified as transverse waves.

(C) From the relation for the speed of light in vacuum,$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Squaring both sides,we get $c^2 = \frac{1}{\mu_0 \varepsilon_0}$.
The dimensions of speed $c$ are $[L T^{-1}]$.
Therefore,the dimensions of $c^2$ are $[L T^{-1}]^2 = [L^2 T^{-2}] = \frac{L^2}{T^2}$.

(A) The relationship between the amplitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ in an electromagnetic wave is given by $E_0 = c \times B_0$,where $c$ is the speed of light in vacuum.
Given:
$B_0 = 3 \times 10^{-10} \, T$
$c = 3 \times 10^8 \, m/s$
Substituting the values:
$E_0 = (3 \times 10^8 \, m/s) \times (3 \times 10^{-10} \, T)$
$E_0 = 9 \times 10^{-2} \, V/m$

(C) When an electromagnetic wave travels from one medium to another,its frequency $(f)$ remains constant because it depends on the source of the wave.
However,the speed of the wave $(v)$ changes according to the refractive index $(n)$ of the medium,where $v = c/n$.
Since the speed of the wave is given by $v = f \lambda$,and $f$ is constant,the wavelength $(\lambda)$ changes as $\lambda' = v/f = (c/n)/f = \lambda/n$.
Given $n = 4$,the new wavelength $\lambda' = \lambda/4$,which means the wavelength becomes one-fourth of its original value.
Therefore,the correct option is $C$.

(C) Given: Amplitude of magnetic field $B_m = 510 \, nT = 510 \times 10^{-9} \, T$.
Speed of light in vacuum $c = 3 \times 10^8 \, m/s$.
The relationship between the amplitudes of the electric field $(E_m)$ and magnetic field $(B_m)$ in an electromagnetic wave is given by $E_m = c B_m$.
Substituting the values:
$E_m = (3 \times 10^8 \, m/s) \times (510 \times 10^{-9} \, T)$
$E_m = 1530 \times 10^{-1} \, V/m$
$E_m = 153 \, V/m = 1.53 \times 10^2 \, V/m$.
Thus,the correct option is $C$.

(B) Electromagnetic waves are transverse waves and do not require a material medium for propagation.
Electromagnetic waves travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$ in a vacuum.
Accelerating electric charges are the source of electromagnetic waves.
The speed of electromagnetic waves depends on the medium through which they travel,given by the formula $v = c/n$,where $n$ is the refractive index of the medium. Therefore,they do not travel at the same speed in all media.
Thus,the statement that they travel with the same speed in all media is incorrect.

(C) plane electromagnetic wave consists of oscillating electric and magnetic fields. The electric field is given by $E = E_0 \sin(kx - \omega t)$.
To find the average value of the electric field over a complete cycle,we calculate the integral: $\langle E \rangle = \frac{1}{T} \int_0^T E_0 \sin(kx - \omega t) dt$.
The integral of a sine or cosine function over a full period is zero.
Therefore,the average value of the electric field $\langle E \rangle$ is $0$.
Similarly,the average value of the magnetic field $\langle B \rangle$ is also $0$.
Electric energy density $(u_E = \frac{1}{2} \epsilon_0 E^2)$ and magnetic energy density $(u_B = \frac{1}{2\mu_0} B^2)$ are proportional to the squares of the fields,which are always non-negative and have non-zero average values.

(C) The frequency of an electromagnetic wave is a property of the source that generates it.
When an electromagnetic wave travels from one medium to another,its speed and wavelength change,but its frequency remains constant.
Therefore,the frequency of the wave in the medium will remain $3 \text{ MHz}$.

(C) The electric field and the magnetic field of an electromagnetic wave are mutually perpendicular to each other.
The direction of propagation of an electromagnetic wave is always perpendicular to both the electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$.
According to the properties of electromagnetic waves,the direction of propagation is given by the direction of the Poynting vector,which is proportional to $\vec{E} \times \vec{B}$.
Therefore,the direction of propagation is along $\vec{E} \times \vec{B}$.

(A) When an electromagnetic wave travels from one medium to another,its frequency remains unchanged because it is a property of the source.
However,the velocity and wavelength of the wave change depending on the refractive index of the medium.
The amplitude of the wave may change due to reflection,absorption,or scattering at the interface of the dielectric medium.
Therefore,the frequency remains the same,but the amplitude changes.

(C) The speed of electromagnetic waves in a vacuum is constant $(c \approx 3 \times 10^8 \ m/s)$.
In a given medium,the speed is determined by the refractive index $(n)$ of the medium,given by $v = c/n$.
Electromagnetic waves of all intensities travel at the same speed in a given medium because the speed is independent of the intensity of the wave.
However,the speed can vary with frequency or wavelength if the medium is dispersive (i.e.,$n$ depends on frequency).
Therefore,the speed is independent of the intensity of the electromagnetic wave.

(C) Electromagnetic waves carry both energy and momentum.
When an electromagnetic wave is incident on a surface,it exerts radiation pressure,which implies the transfer of momentum $P$.
Additionally,the wave carries energy $U$ which is transferred to the surface upon absorption or reflection.
The relationship between the energy $U$ and momentum $P$ for a wave completely absorbed by a surface is given by $P = U/c$,where $c$ is the speed of light.
Since both $U$ and $P$ are non-zero for an electromagnetic wave,the correct condition is $P \neq 0$ and $U \neq 0$.

(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 the magnetic field $B$ is $E = cB$,where $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
Substituting $c^2 = \frac{1}{\mu_0 \epsilon_0}$ into the energy density expressions,we find that the average energy density contributed by the electric field is equal to the average energy density contributed by the magnetic field.
Therefore,the total average energy density is shared equally between the electric and magnetic fields.

(B) In a plane electromagnetic wave,the electric field vector $\vec{E}$,the magnetic field vector $\vec{B}$,and the direction of propagation $\vec{k}$ are mutually perpendicular to each other.
Specifically,the direction of propagation is given by the direction of the Poynting vector $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})$.
For a plane wave propagating along a specific axis (e.g.,$x$-axis),the electric field must be in the $y$-direction $(E_y)$ and the magnetic field must be in the $z$-direction $(B_z)$,or vice versa.
If the wave propagates along the $x$-axis,then $\vec{E} = E_y\hat{j}$ and $\vec{B} = B_z\hat{k}$.
Thus,the pair $(E_y, B_z)$ is a valid combination for an electromagnetic wave propagating along the $x$-axis.

(D) Electromagnetic $(EM)$ waves consist of oscillating electric and magnetic fields that are mutually perpendicular and perpendicular to the direction of wave propagation.
Because $EM$ waves are neutral (they carry no net electric charge),they do not experience a Lorentz force when passing through an electric field $(F = qE)$ or a magnetic field $(F = q(v \times B))$.
Therefore,$EM$ waves are not deflected by either electric or magnetic fields.

(D) The direction of propagation of an electromagnetic wave is given by the direction of the cross product of the electric field vector $\vec E$ and the magnetic field vector $\vec B$,i.e.,direction $\propto \vec E \times \vec B$.
Given that the wave propagates in the $+y$ direction (unit vector $\hat j$) and the magnetic field $\vec B$ is in the $-x$ direction (unit vector $-\hat i$).
Let the electric field be in the direction $\hat n$. Then,$\hat n \times (-\hat i) = \hat j$.
Using the properties of cross products,we know that $\hat k \times (-\hat i) = -(\hat k \times \hat i) = -\hat j$. This does not match. However,$(-\hat k) \times (-\hat i) = \hat k \times \hat i = \hat j$.
Thus,the electric field must be in the $+z$ direction $(\hat k)$.
The wave equation for a wave traveling in the $+y$ direction is given by $\cos(\omega t - ky)$,where $k = \frac{2\pi}{\lambda}$.
Therefore,the electric field vector is $\vec E = E_0 \cos \left( \omega t - \frac{2\pi}{\lambda} y \right) \hat z$.

(D) The velocity of light in a vacuum is a universal constant given by $c = 2.9979 \times 10^8 \ m/s$.
Since $c$ is a constant in a vacuum,it does not change.
If the velocity of light were to change,it would imply a change in the medium or a change in the fundamental constants of nature.
However,the question asks what changes if the velocity of light in a vacuum changes,which is a hypothetical scenario.
In reality,the velocity of light in a vacuum is invariant.
Therefore,none of the listed quantities (frequency,wavelength,or amplitude) are responsible for changing the speed of light in a vacuum,as the speed of light is independent of these factors in a vacuum.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the vector product $\vec{E} \times \vec{B}$.
Given,the direction of propagation is along the $+x$-axis,so $\hat{E} \times \hat{B} = \hat{i}$.
We are given the magnetic field $\vec{B}$ is in the $z$-direction,so $\hat{B} = \hat{k}$.
Let the electric field $\vec{E}$ be in the direction $\hat{n}$. Then $\hat{n} \times \hat{k} = \hat{i}$.
Using the properties of unit vectors,we know that $\hat{j} \times \hat{k} = \hat{i}$.
Therefore,the electric field $\vec{E}$ must be in the positive $y$-direction.

(A) The speed of light in a vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
The speed of light in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r = \varepsilon_0 K$.
Substituting these values,we get $v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 K}} = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \cdot \frac{1}{\sqrt{\mu_r K}} = \frac{c}{\sqrt{\mu_r K}}$.
The refractive index $n$ is defined as $n = \frac{c}{v}$.
Therefore,$n = \frac{c}{c / \sqrt{\mu_r K}} = \sqrt{\mu_r K}$.

(C) The given equation for the electromagnetic wave is $E = E_0 \sin(kx - \omega t)$.
Here,$k$ is the wave number,defined as $k = 2\pi / \lambda$,where $\lambda$ is the wavelength.
$\omega$ is the angular frequency,defined as $\omega = 2\pi f$,where $f$ is the frequency.
The speed of the wave in a vacuum is given by $c = \omega / k$.
Since the wave is propagating in a vacuum,its speed $c$ is a constant $(c \approx 3 \times 10^8 \text{ m/s})$.
Therefore,the ratio $k/\omega = 1/c$ is a constant and is independent of the wavelength $\lambda$.

(C) The intensity of electromagnetic waves is given by $I = \frac{P_{av}}{4\pi r^2} = \frac{E_m^2}{2\mu_0 c}$.
First,calculate the maximum electric field $E_m$:
$E_m = \sqrt{\frac{\mu_0 c P_{av}}{2\pi r^2}} = \sqrt{\frac{(4\pi \times 10^{-7}) \times (3 \times 10^8) \times 800}{2\pi \times (3.5)^2}} = \sqrt{\frac{2 \times 10^1 \times 800}{12.25}} = \sqrt{\frac{16000}{12.25}} \approx 36.17 \ V/m$.
Wait,recalculating $E_m$:
$E_m = \sqrt{\frac{2 \times 10^{-7} \times 3 \times 10^8 \times 800}{3.5^2}} = \sqrt{\frac{60 \times 800}{12.25}} = \sqrt{\frac{48000}{12.25}} \approx 62.6 \ V/m$.
The maximum value of the magnetic field $B_m$ is given by $B_m = \frac{E_m}{c}$.
$B_m = \frac{62.6}{3 \times 10^8} \approx 2.09 \times 10^{-7} \ T$.