Electromagnetic Waves

To match the given List I (EM-Wave) with List II (Wavelength Range), we need to understand the typical wavelength ranges associated with each type of electromagnetic (EM) wave mentioned in List I.

Here's the matching based on known wavelength ranges:

• Infra-red: Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. The wavelength range for infrared is typically from about 700 nm (nanometers) to 1 mm (millimeter), which means it matches with option (III) - 1 mm to 700 nm.


• Ultraviolet: Ultraviolet (UV) light sits between visible light and X-rays along the electromagnetic spectrum, with wavelengths shorter than visible light but longer than X-rays. The typical wavelength range for UV light is from 400 nm down to 10 nm, so it matches with option (II) - 400 nm to 1 nm (even though the strict lower bound should be 10 nm, this is the closest option provided).


• X-rays: X-rays have a wavelength range shorter than UV rays and longer than gamma rays, typically ranging from 10 nm down to 0.01 nm (or $10^{-2}$ nm). Thus, X-rays match with option (IV) - 1 nm to $10^{-3}$ nm, as this is the closest range provided that fits the typical scope of X-ray wavelengths.


• Gamma rays: Gamma rays have the shortest wavelength on the electromagnetic spectrum, with wavelength less than 10 picometers (pm), which equates to less than $10^{-3}$ nm. This corresponds to option (I) - $<10^{-3}$ nm.

So, based on the above analysis, the correct option for the matching would be:

• (A)-(III)


• (B)-(II)


• (C)-(IV)


• (D)-(I)

Therefore, the correct answer is Option B: (A)-(III), (B)-(II), (C)-(IV), (D)-(I).

Wavelengths are as

Gamma rays $<1 \mathrm{~nm}$

x-ray < $(1-10)$ nm

Infrared $<(700-10^5) \mathrm{~nm}$

Microwave $<(10^5-10^8) \mathrm{~nm}$

To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field $\overrightarrow{\mathrm{E}}$ and the magnetic field $\overrightarrow{\mathrm{B}}$ in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following relation holds:

$\overrightarrow{\mathrm{B}} = \frac{\overrightarrow{\mathrm{E}} \times \hat{\mathrm{k}}}{\mathrm{c}}$

where:

• $\overrightarrow{\mathrm{E}}$ is the electric field.
• $\hat{\mathrm{k}}$ is the unit vector in the direction of propagation of the wave.
• $\mathrm{c}$ is the speed of light in a vacuum.

Given the electric field:

$\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$

The wave is propagating in the $z$-direction, so $\hat{\mathrm{k}} = \hat{z}$. The unit vector $\hat{\mathrm{i}}$ represents the $x$-direction.

The magnetic field induction is given by:

$\overrightarrow{\mathrm{B}} = \frac{(\hat{\mathrm{i}} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})) \times \hat{\mathrm{z}}}{\mathrm{c}}$

The cross product $\hat{\mathrm{i}} \times \hat{\mathrm{z}}$ yields $\hat{\mathrm{j}}$ (the unit vector in the $y$-direction):

$\overrightarrow{\mathrm{B}} = \hat{\mathrm{j}} \frac{40 \cos \omega(\mathrm{t}-z / \mathrm{c})}{\mathrm{c}}$

Therefore, the magnetic field induction is:

$\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z/\mathrm{c})$

The correct answer is:

Option A:

$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

The speed of electromagnetic waves in air (and in a vacuum) is approximately the speed of light, which we denote as $ c $. The speed of light $ c $ is $ 3 \times 10^8 $ meters per second. The relationship between the speed of light $ c $, the frequency $ f $, and the wavelength $ \lambda $ of an electromagnetic wave is given by the equation:

$ c = f \lambda $

Here, we are given that the frequency $ f $ of the electromagnetic wave is 60 MHz (megahertz), which can be converted to hertz (Hz) by multiplying by $ 10^6 $ (because 1 MHz = $ 10^6 $ Hz).

$ f = 60 \times 10^6 \text{ Hz} $

To find the wavelength $ \lambda $ in meters, we can rearrange the wave equation to solve for $ \lambda $:

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

Substitute the given values of $ c $ and $ f $ into the equation to find the wavelength:

$ \lambda = \frac{3 \times 10^8 \text{ m/s}}{60 \times 10^6 \text{ Hz}} $

Now, simplify the equation by dividing the numbers:

$ \lambda = \frac{3}{60} \times 10^{8-6} \text{ m} $

$ \lambda = \frac{1}{20} \times 10^2 \text{ m} $

$ \lambda = 5 \text{ m} $

Therefore, the wavelength of the wave is 5 meters, corresponding to option B. Additionally, in an electromagnetic wave, the electric and magnetic field vectors are indeed mutually perpendicular to each other and to the direction of propagation, which in this case is the $ z $ direction.

$\begin{aligned} & \frac{1}{2} \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{B}^2}{2 \mu_0} \\ & \because \mathrm{E}=\mathrm{CB} \text { and } \mathrm{C}=\frac{1}{\mu_0 \varepsilon_0} \end{aligned}$

The average energy density of an electromagnetic wave is given by the formula:

$U = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$

For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density. Therefore, we can focus on just one part to find the total energy density. The formula for the energy density due to the electric field is:

$U_E = \frac{1}{2} \varepsilon_0 E^2$

Given that the electric field amplitude ($E$) is $50 \, \mathrm{Vm}^{-1}$ and the permittivity of free space ($\varepsilon_0$) is $8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2$, we can calculate the energy density due to the electric field as follows:

$U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 = 1.10625 \times 10^{-8} \, \mathrm{J/m^3}$

Since the energy density contributions from the electric and magnetic fields are equal, the total average energy density of the electromagnetic wave is just this value.

$\begin{aligned} & \text { Given } \vec{E}=E_0 \cos (\omega t-k z) \hat{i} \\ & \vec{B}=\frac{E_0}{C} \cos (\omega t-k z) \hat{j} \\ & \hat{C}=\hat{E} \times \hat{B} \end{aligned}$

$\begin{aligned} \frac{E}{B} & =C \\ \frac{E}{B} & =3 \times 10^8 \\ B & =\frac{E}{3 \times 10^8}=\frac{9.6}{3 \times 10^8} \\ B & =3.2 \times 10^{-8} T \\ \hat{B} & =\hat{v} \times \hat{E} \\ & =\hat{i} \times \hat{j}=\hat{k} \end{aligned}$

So,

$\overrightarrow{\mathrm{B}}=3.2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}$

$\begin{aligned} & \text { Radiation pressure }=\frac{I}{\mathrm{~V}} \\ & =\frac{\mathrm{I} \cdot \mu}{\mathrm{c}} \\ & =\frac{6 \times 10^8 \times 3}{3 \times 10^8} \\ & =6 \mathrm{~N} / \mathrm{m}^2 \end{aligned}$

$\begin{aligned} & \mathrm{I}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \times \mathrm{c} \\ & \mathrm{I}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8 \\ & \mathrm{I}=53.1 \mathrm{~W} / \mathrm{m}^2 \end{aligned}$

Given,

$ B_{0} $ (Peak value of magnetic field)

$ =30 \times 10^{-9} \mathrm{~T} $

Using the relation of $ B_{0} $ and $ E_{0} $ (Peak value of electric field)

$ E_{0}=c \times B_{0} $

where $ c $ is speed of light

$ \begin{array}{l} E_{0}=30 \times 10^{-9} \times 3 \times 10^{8} \\\\ E_{0}=9 \mathrm{~V} / \mathrm{m} \end{array} $

We know that the momentum transferred by EM wave $=\frac{U}{c}$ Also we know that, $\frac{E_0}{B_0}=c$

$ \therefore \text { Momentum transferred }=\frac{U}{E_0} \times B_0 $

Given that,

$ B_0=270 \mathrm{nT}=270 \times 10^{-9} \mathrm{~T} $

Speed of light, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$

Here, $E_0=B_0 \times c$

$ \begin{aligned} & E_0=270 \times 10^{-9} \times 3 \times 10^8=810 \times 10^{-1} \\ & E_0=81 \mathrm{NC}^{-1} \end{aligned} $

The amplitude of the electric fields part of the wave is $81 \mathrm{NC}^{-1}$.

For a wave moving in the +z‐direction we need

$\mathbf E\perp\mathbf B\,,\quad\mathbf E\times\mathbf B\;\text{along }+z,$

so in components

$\mathbf E=(E_x,E_y,0),\;\mathbf B=(B_x,B_y,0)\implies \mathbf E\times\mathbf B=(0,0\;E_xB_y - E_yB_x)$

and we require

$E_xB_y - E_yB_x>0\,. $

Let’s check each option (we write only the key step):

• A: $\mathbf E=(-2,-3,0)$, $\mathbf B=(3,-2,0)$

$E_xB_y - E_yB_x =(-2)(-2)-(-3)(3)=4+9=13>0.$

⇒ +z

• B: $(3,4)\times(4,-3)\to3·(-3) - 4·4 = -25<0$

• C: $(1,-2)\times(-2,-1)\to1·(-1) - (-2)·(-2) = -5<0$

• D: $(-2,3)\times(1,2)\to(-2)·2 - 3·1 = -7<0$

Only Option A gives a positive z‐component, so the correct choice is A.

First find the wavelength:

$\lambda=\frac{c}{f}=\frac{3\times10^8\rm\,m/s}{3\times10^6\rm\,Hz}=100\rm\,m$

A typical monopole (quarter‐wave) antenna is

$\frac{\lambda}{4}=\frac{100\rm\,m}{4}=25\rm\,m$

So the correct choice is Option D: 25 m.

When energy $ E $ in the form of radiation falls normally on a perfectly reflecting surface, the momentum transferred to the surface can be calculated as follows:

Energy of the Radiation:

The energy of radiation is given by the equation:

$ E = \frac{h c}{\lambda} $

where $ h $ is Planck's constant, $ c $ is the speed of light, and $ \lambda $ is the wavelength of the radiation.

Momentum of the Incident Radiation:

The momentum $ p_i $ of the incident radiation is given by:

$ p_i = \frac{E}{c} $

which simplifies to:

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

Momentum of the Reflected Radiation:

Since the surface is perfectly reflecting, the momentum of the reflected radiation $ p_r $ is:

$ p_r = -p_i $

Momentum Transferred to the Surface:

The momentum transferred ($\Delta p$) to the surface is the difference between the incident and reflected momentum:

$ \Delta p = p_i - p_r = \frac{E}{c} - \left(-\frac{E}{c}\right) $

Simplifying further:

$ \Delta p = \frac{E}{c} + \frac{E}{c} = \frac{2E}{c} $

Thus, the momentum transferred to the surface is $\frac{2E}{c}$.

The correct answer is Option C: X-rays.

Here's why:

When X-rays are directed at a solid (especially a crystalline material), they are diffracted by the planes of atoms in the crystal.

This diffraction pattern, known as X-ray diffraction (XRD), is used to determine the arrangement of atoms within the material.

Techniques like X-ray crystallography rely on X-rays because their wavelengths are similar in scale to the interatomic distances, allowing for precise structural investigations.

Thus, using X-rays, scientists can deduce the internal structure of solids.

Given that the root mean square (rms) value of the electric field of an electromagnetic wave is $ E_{\text{rms}} = 660 \, \text{N/C} $, the average energy density $ E_{\text{avg}} $ of the electromagnetic wave can be calculated using the formula:

$ E_{\text{avg}} = \varepsilon_0 E_{\text{rms}}^2 $

Where:

$\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$ (the permittivity of free space)

Plugging in the values, we have:

$ E_{\text{avg}} = 8.85 \times 10^{-12} \times (660)^2 $

Calculating this:

$ E_{\text{avg}} = 3.85 \times 10^{-6} \, \text{J/m}^3 $

Thus, the average energy density of the electromagnetic wave is $ 3.85 \times 10^{-6} \, \text{J/m}^3 $.

Given:

Wavelength, $\lambda = 500 \, \text{m}$

Power, $P = 10 \, \text{kW} = 10^4 \, \text{W}$

To find the number of photons emitted per second by the transmitter, we use the formula for power related to energy and photon emission:

$ P = \frac{E}{t} \quad \text{and} \quad E = \frac{n \cdot h \cdot c}{\lambda} $

This leads to:

$ P = \frac{n \cdot h \cdot c}{\lambda \cdot t} \quad \Rightarrow \quad \frac{n}{t} = \frac{P \times \lambda}{h \times c} $

Substitute the known values:

$ \frac{n}{t} = \frac{10^4 \times 500}{6.64 \times 10^{-34} \times 3 \times 10^8} $

Simplify the expression:

$ \frac{n}{t} = \frac{500}{6.64 \times 3} \times 10^{30} $

Simplifying further gives:

$ \frac{n}{t} = \frac{50}{6.64 \times 3} \times 10^{31} \approx 2.5 \times 10^{31} $

Thus, the order of the number of photons emitted per second is:

$ \text{Order} = 10^{31} $

Accelerating charges can produce electromagnetic waves because a change in the velocity of charges causes a variation in the electromagnetic waves.

Stationary charges do not change electric field with time. Charges in uniform motion produces steady electric and magnetic fields, but no changing fields. Stationary magnet produces a static magnetic field with no time variation. So, due to these reasons they can not produce EM waves.

The average value of electric energy density in an electromagnetic wave can be determined as follows:

The formula for average electric energy density is given by:

$ u_c = \frac{1}{2} \varepsilon_0 E_{\mathrm{rms}}^2 $

Where $ E_{\mathrm{rms}} $ is the root mean square value of the electric field, which is expressed as:

$ E_{\mathrm{rms}} = \frac{E_0}{\sqrt{2}} $

Substituting the expression for $ E_{\mathrm{rms}} $ into the formula for $ u_c $, we have:

$ u_c = \frac{1}{2} \varepsilon_0 \left(\frac{E_0}{\sqrt{2}}\right)^2 $

Simplifying this, we get:

$ u_c = \frac{1}{4} \varepsilon_0 E_0^2 $

Thus, the average electric energy density is $\frac{1}{4} \varepsilon_0 E_0^2$.

The speed of an electromagnetic wave in a medium can be described by the equation:

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

Given that the relative permeability ($\mu_r$) of the medium is 1, the equation simplifies to:

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

where:

$ v $ is the speed of the wave in the medium,

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

$ \varepsilon_r $ is the relative permittivity.

We need to find the relative permittivity ($\varepsilon_r$) of the medium when the wave speed ($v$) is $2 \times 10^8 \, \text{m/s}$:

$ \varepsilon_r = \frac{c^2}{v^2} = \frac{(3 \times 10^8)^2}{(2 \times 10^8)^2} $

Calculating this gives:

$ \varepsilon_r = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25 \approx 2.28 $

Therefore, the relative permittivity is approximately 2.28.

To solve for the upper sideband frequency and the bandwidth of an amplitude-modulated signal, given:

Message signal frequency ($ f_m $): 3 kHz

Carrier frequency ($ f_c $): 1 MHz (1000 kHz)

Calculating the Upper Sideband Frequency

The upper sideband frequency is calculated as follows:

$ f_{\text{USB}} = f_c + f_m = 1000\, \text{kHz} + 3\, \text{kHz} = 1003\, \text{kHz} = 1.003\, \text{MHz} $

Calculating the Bandwidth

The bandwidth of an amplitude-modulated signal is the difference between the upper sideband and the lower sideband frequencies:

$ \text{Bandwidth} = (f_c + f_m) - (f_c - f_m) = 2f_m = 2 \times 3\, \text{kHz} = 6\, \text{kHz} $

Thus, the upper sideband frequency is 1.003 MHz, and the bandwidth is 6 kHz.

Solar radiation is a form of electromagnetic radiation. Here’s why Option C is the correct choice:

Nature of Solar Radiation:

Solar radiation includes visible light, ultraviolet light, infrared, and other types of electromagnetic waves.

Transverse vs. Longitudinal Waves:

Transverse Waves: In these waves, the oscillations (the electric and magnetic fields in the case of EM waves) are perpendicular to the direction of propagation.

Longitudinal Waves: In these, the oscillations occur in the same direction as the wave travels. Sound waves are a common example, but electromagnetic waves are not longitudinal.

Mechanical vs. Electromagnetic Waves:

Mechanical Waves: These require a medium (like air or water) to propagate (e.g., sound waves).

Electromagnetic Waves: These do not need a medium and can travel through a vacuum, like the solar radiation from the sun.

Since solar radiation consists of electromagnetic waves that are transverse in nature, the correct option is:

$\textbf{Option C: Transverse EM wave}.$

To find the energy contained in a volume of an electromagnetic wave, we need to calculate the energy density and then multiply it by the volume.

For an electromagnetic wave, the energy density $u$ is given by:

$u = \frac{1}{2} \varepsilon_0 E^2$

where $\varepsilon_0$ is the vacuum permittivity and $E$ is the electric field amplitude.

First, let's find the amplitude of the electric field. In the given equation:

$\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}$

The amplitude $E$ is $20 \mathrm{~N/C}$.

Now, we can find the energy density:

$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} \cdot (20 \mathrm{~N/C})^2$

$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot 400$

$u = 1.77 \times 10^{-9} \mathrm{J/m^3}$

The energy contained in a volume of $5 \times 10^{-4} \mathrm{m^3}$ will be:

$E_\text{total} = u \cdot V$

$E_\text{total} = 1.77 \times 10^{-9} \mathrm{J/m^3} \cdot 5 \times 10^{-4} \mathrm{m^3}$

$E_\text{total} = 8.85 \times 10^{-13} \mathrm{J}$

The energy of an electromagnetic wave contained in a small volume oscillates with double the frequency of the wave.

Here's why: The electromagnetic wave consists of oscillating electric and magnetic fields. The energy density of the wave is proportional to the square of the amplitude of these fields. Since the square of a sinusoidal function oscillates with twice the frequency of the original function, the energy density of the electromagnetic wave also oscillates with twice the frequency of the wave.

The energy density associated with the electric field $\vec{E}$ and magnetic field $\vec{B}$ of an electromagnetic wave in free space is given by:

For the electric field: $U_{E} = \frac{1}{2} \epsilon_{0} E^2$

For the magnetic field: $U_{B} = \frac{1}{2} \frac{B^2}{\mu_{0}}$

These expressions match Option A:

$U_{E} = \frac{\epsilon{0} E^{2}}{2}, U_{B} = \frac{B^{2}}{2 \mu{0}}$

The average energy density of an electromagnetic wave is equally shared between the electric field and the magnetic field. This means that the average electric energy density is equal to the average magnetic energy density.

The average electric energy density ($u_E$) is given by:

$ u_E = \frac{1}{2} \varepsilon_0 E^2 $

and the average magnetic energy density ($u_B$) is given by:

$ u_B = \frac{1}{2\mu_0} B^2 $

where $\varepsilon_0$ is the permittivity of free space, $\mu_0$ is the permeability of free space, $E$ is the electric field strength, and $B$ is the magnetic field strength.

In an electromagnetic wave, $E$ and $B$ are related by the equation:

$ E = cB $

where $c$ is the speed of light in vacuum.

Substituting this into the energy density equations gives:

$ u_E = \frac{1}{2} \varepsilon_0 (cB)^2 $

and

$ u_B = \frac{1}{2\mu_0} B^2 $

Since $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$, we can rewrite $u_E$ as:

$ u_E = \frac{1}{2} \frac{1}{\mu_0} B^2 = u_B $

Therefore, the ratio of the average electric energy density to the average magnetic energy density is 1.

A point source of $100 \mathrm{~W}$ emits light with $5 \%$ efficiency. At a distance of $5 \mathrm{~m}$ from the source, the intensity produced by the electric field component is evaluated as follows:

The intensity at a distance of $5$ meters can be calculated as:

Intensity at 5 m $= \frac{5}{4\pi \times 5^2}\left(\frac{W}{m^2}\right)$

$= \frac{1}{20\pi}\left(\frac{W}{m^2}\right)$

Since the intensity due to the electric field component is half of the total intensity:

Intensity due to electric field $= \frac{1}{40\pi}\left(\frac{W}{m^2}\right)$

Thus, we arrive at our result:

$= \frac{1}{40\pi}\left(\frac{W}{m^2}\right)$

Statement I is correct as photon do not carry any charge, hence cannot feel force from either fields.

Statement II is wrong as $E_0=cB_0$

${E_0} = {{{B_0}} \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$

A-R

X-rays are used to study the crystal structure by Bragg's law method.

B-Q

UV rays are used to study the finger print in forensic lab.

C-S

Radio waves are used in TV communication for sending the signals' to large distance.

D-P

IR rays are used in TV remotes to send signal for short distance.

Hence, A-R, B-Q, C-S, D-P is correct match.

Given, Intensity of EM radiation $=0.6 \mathrm{~W} / \mathrm{m}^2$ The radiation pressure is given by,

$ \begin{aligned} P & =\frac{\text { Intensity }}{\text { Speed of light }}=\frac{0.6}{3 \times 10^8} \\ & =2 \times 10^{-9} \mathrm{~N} / \mathrm{m}^2 \end{aligned} $

Given, speed of electromagnetic wave in medium

$ \begin{aligned} & v=1.5 \times 10^8 \mathrm{~ms}^{-1} \\ & \varepsilon_r=2 \end{aligned} $

We know that,

$ \begin{aligned} & & v & =\frac{c}{\sqrt{\mu_r \varepsilon_r}} \\ & \Rightarrow & \sqrt{\mu_r \varepsilon_r} & =\frac{c}{v}=\frac{3 \times 10^8}{1.5 \times 10^8}=2 \\ & \Rightarrow & \mu_r \varepsilon_r & =4 \\ & \Rightarrow & \mu_r & =\frac{4}{\varepsilon_r}=\frac{4}{2}=2 \end{aligned} $

We know that, magnetic susceptibility

$ \chi=\mu_r-1=2-1=1 $