Electromagnetic Waves

$ \begin{aligned} & I=\frac{P}{A} \\ & =\frac{200 \times 0.11}{4 \pi r^2}=\frac{22}{4 \pi\left(1^2\right)}=1.75 \mathrm{~W} / \mathrm{m}^2 \end{aligned} $

Given:

• Power of bulb $ P = 10\ \text{W} $

• Distance $ r = 0.5\ \text{m} $

• The bulb emits uniformly in all directions (i.e., isotropically).

Step 1: Formula for intensity

$ I = \frac{P}{A} = \frac{P}{4\pi r^2} $

Step 2: Substitute values

$ I = \frac{10}{4\pi (0.5)^2} $

$ I = \frac{10}{4\pi \times 0.25} = \frac{10}{\pi} $

Step 3: Simplify numerically

$ I \approx \frac{10}{3.14} \approx 3.18\ \text{W m}^{-2} $

✅ Final Answer:

$ \boxed{I \approx 3.18\ \text{W m}^{-2}} $

Correct Option: A.

Question:

The layer of the atmosphere that reflects low frequency (LF) electromagnetic waves during daytime only is —

Step 1: Recall the ionospheric layers

The ionosphere has several layers:

• D layer — lowest layer, present only during the daytime

• E layer

• F₁ and F₂ layers — present at higher altitudes

Step 2: Function of the D layer

• The D layer is ionized by solar radiation and disappears at night (it recombines quickly when sunlight is gone).

• It primarily affects low-frequency (LF) and medium-frequency (MF) radio waves.

• It does not strongly reflect HF waves—it mostly causes absorption at higher frequencies.

Step 3: Reflection characteristic

• LF waves (30–300 kHz) are reflected by the D layer during daytime only.

• At night, the D layer disappears, and these LF waves can instead travel farther by ground or sky wave propagation.

✅ Final Answer:

Option A — D layer

Answer:

$ \boxed{\text{A. D layer}} $

In an electromagnetic wave, oscillating electric and magnetic fields increases and decreases simultaneously i.e., they are always in same phase. Also $\mathbf{E}$ and $\mathbf{B}$ are always perpendicular to each other.

The electromagnetic spectrum (in order of increasing wavelength) is:

Gamma rays → X-rays → Ultraviolet (UV) → Visible light → Infrared → Microwaves → Radio waves

So, radio waves have the maximum wavelength among the given options.

✅ Correct answer: Option B — Radio waves

The electric field ($E_0$) and magnetic field ($B_0$) in a plane electromagnetic wave have a special relationship.

We know that $\frac{E_0}{B_0} = c$ where $c$ is the speed of light, which is $3 \times 10^8$ m/s.

If we compare $E_0$ to $10^8$ times $B_0$, we get: $\frac{E_0}{10^8 B_0} = \frac{E_0}{B_0} \times \frac{1}{10^8}$

Now substitute $c$ for $\frac{E_0}{B_0}$: $\frac{E_0}{10^8 B_0} = \frac{3 \times 10^8}{10^8} = 3$

So, the answer is: $\frac{E_0}{10^8 B_0} = 3$

$ \begin{aligned} &\text { Power of the source }\\ &\begin{aligned} P & =I \times 4 \pi r^2 \\ & =\frac{1}{2} C \varepsilon_0 E_{\mathrm{rms}}^2 \times 4 \pi r^2 \\ & =\frac{1}{2} \times 3 \times 10^8 \times 8.85 \times 10^{-12} \times 3^2 \times 4 \pi \times 3^2\\ & =2.7 \mathrm{~W} \end{aligned} \end{aligned} $

Step 1: Relationship between $ E $ and $ B $ in an electromagnetic wave

For a plane electromagnetic wave in free space,

$ E = c B $

where $ c $ is the speed of light in free space.

Thus,

$ B = \frac{E}{c}. $

Step 2: Express $ c $ in terms of $ \mu_0 $ and $ \varepsilon_0 $

The speed of light is given by

$ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}. $

Step 3: Substitute into expression for $ B $

$ B = \frac{E}{c} = E \sqrt{\mu_0 \varepsilon_0}. $

✅ Final Answer:

$ \boxed{B = E \sqrt{\mu_0 \varepsilon_0}} $

Correct Option: C

$ \text { } \begin{aligned} B & =\frac{F}{c}=\frac{6.3}{3 \times 10^8} \\ & =2.1 \times 10^{-8} \mathrm{~T} \end{aligned} $

Magnetic energy stored per unit volume

$ \begin{aligned} & =\frac{B^2}{2 \mu_0}=\frac{B^2}{2 \times \frac{1}{\varepsilon_0 c^2}} \quad\left[c^2=\frac{1}{\mu_0 \varepsilon_0}\right] \\ & =\frac{\varepsilon_0 B^2 c^2}{2} \end{aligned} $

$\mathbf{B}=3 \times 10^{-7} \sin \left(100 \pi x+10^{12} t\right) \mathrm{T}$

Here, $K=100 \pi \mathrm{~m}^{-1}$

$ \therefore \quad \lambda=\frac{2 \pi}{K}=\frac{2 \pi}{100 \pi}=0.02 \mathrm{~m} $

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:

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

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.

Given, wave is propagating along $z$-axis.

As we know that, electric field $\mathbf{E}$, magnetic field $\mathbf{B}$ and propagation $\mathbf{k}$ all are mutually perpendicular to each other.

So, $\mathbf{E}=E_0 \hat{\mathbf{i}}$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$

Given, equation of magnetic field, $B=400 \mu \mathrm{T} \sin \left[4 \times 10^{15}\left(t-\frac{x}{c}\right)\right]$

$\because$ Average energy, $E_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \frac{B_0^2}{\mu_0}$ where, $\varepsilon_0$ is free space permittivity and $\mu_0$ is free space permeability.

$\begin{aligned} \therefore \quad E_{\mathrm{av}} & =\frac{1}{2} \frac{B_0^2}{\mu_0}=\frac{1}{2} \times \frac{\left(400 \times 10^{-6}\right)^2}{4 \pi \times 10^{-7}}=\frac{0.2}{\pi} \\ & =63.66 \times 10^{-3} \mathrm{~J} / \mathrm{m}^3 \end{aligned}$

Given, maximum amplitude of modulated signal,

$V_{\max }=15 \mathrm{~V}=V_m+V_c$ ..... (i)

Minimum amplitude of modulated signal,

$V_{\min }=5 \mathrm{~V}=V_m-V_c$ .... (ii)

where, $V_c, V_m$ are carries wave and modulated wave voltage.

Now, from Eqs. (i) and (ii), we get

$20=2 V_m \Rightarrow V_m=10 \mathrm{~V} $