Class 12 Physics (Part 1) Chapter 13 Electromagnetic Waves

Key Concept: Electromagnetic Spectrum, Wavelength Calculation

c) Visible light
[Solution Description]
The given wavelength range corresponds to the visible light spectrum. From the provided data:
- Visible light: $4 \times 10^{-7}$ m to $7 \times 10^{-7}$ m.
- Ultraviolet radiation: $6 \times 10^{-10}$ m to $4 \times 10^{-7}$ m.
- Infrared rays: $7.8 \times 10^{-7}$ m onwards.
Thus, the correct option is within the visible light range but slightly overlapping with ultraviolet at the lower end and infrared at the higher end.

Key Concept: Energy Density, Transverse Nature

a) $1.274 \times 10^{-7} \, \text{J/m}^3$
[Solution Description]
The total energy density $u$ of an electromagnetic wave is the sum of the electric and magnetic field energy densities:
$u = \frac{1}{2}\varepsilon_0 E_0^2 + \frac{1}{2\mu_0}B_0^2$
Using the relation $\frac{E_0}{B_0} = c$ and substituting $\mu_0 = \frac{1}{\varepsilon_0 c^2}$:
$u = \frac{1}{2}\varepsilon_0 E_0^2 + \frac{1}{2}\left(\frac{1}{\varepsilon_0 c^2}\right)\left(\frac{E_0}{c}\right)^2 = \varepsilon_0 E_0^2$
Substituting $E_0 = 120 \, \text{V/m}$ and $\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$:
$u = 8.854 \times 10^{-12} \times (120)^2 = 1.274 \times 10^{-7} \, \text{J/m}^3$

Key Concept: Property of Radio Waves

c) They do not require a material medium for propagation
[Solution Description]
Radio waves do not require a material medium for propagation as they are electromagnetic waves. This property allows them to travel through the vacuum of space and the Earth's atmosphere, making them ideal for long-distance wireless communication.

Key Concept: Examples

c) Radio waves
[Solution Description]
Radio waves are a type of electromagnetic wave produced by oscillating electric circuits. According to the syllabus, radio waves are generated through oscillations in electric circuits, unlike other types of waves like microwaves or infrared which may have different sources.

Key Concept: Relation between $E_0$ and $B_0$

c) $E_0 = c B_0$
[Solution Description]
From Maxwell's equations, we know that for an electromagnetic wave in free space, the ratio of the amplitudes of the electric field ($E_0$) and magnetic field ($B_0$) is equal to the speed of light in free space ($c$). Mathematically:
$\frac{E_0}{B_0} = c$
Therefore, $E_0 = c B_0$.

Key Concept: Electric and Magnetic Field Relation

d) $180 \, \text{N/C}$
[Solution Description]
The relation between electric field ($E_0$) and magnetic field ($B_0$) amplitudes in free space is:
$E_0 = c B_0$
Given $B_0 = 6 \times 10^{-7} \, \text{T}$ and $c = 3 \times 10^8 \, \text{m/s}$,
$E_0 = 3 \times 10^8 \times 6 \times 10^{-7} = 180 \, \text{N/C}$

Key Concept: X-Rays

c) Detecting bone fractures
[Solution Description]
X-rays are commonly used in surgery for detection of fractures, diseased organs, formation of bones and stones, and observing the progress of healing bones. This matches one of the given options.

Key Concept: Production and Properties

c) Hot bodies and molecules
[Solution Description]
The syllabus states that infrared rays are produced by hot bodies and molecules. Their wavelength ranges from $7.8 \times 10^{-7}$ m to $5 \times 10^{-3}$ m and they exhibit properties such as heating effect, being invisible, absorbed by glass, and reflected by metals.

Key Concept: Microwaves

b) Radar systems for aircraft navigation
[Solution Description]
Microwaves have high-frequency wavelengths that are ideal for radar systems in aircraft navigation, satellite communication, and microwave ovens. They are not typically used in medical treatments or sterilization processes.

Key Concept: Gauss’ Law of Magnetism

d) The flux is zero
[Solution Description]
According to Gauss’ law for magnetism, the net magnetic flux through any closed surface is always zero. This is because magnetic field lines form closed loops and there are no isolated magnetic monopoles. Whether the surface encloses a magnetic dipole or any other configuration, the sum of entering and exiting magnetic field lines will be equal, resulting in zero net flux.

Key Concept: Gamma Ray Properties and Production

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Assertion (A) states that gamma rays are highly penetrating due to their highest frequency in the electromagnetic spectrum. This is true because higher frequency corresponds to higher energy ($E = h\nu$), which enhances their penetrating ability. Reason (R) correctly states that the energy of gamma rays is proportional to their frequency, which is a fundamental principle derived from Planck's equation. Furthermore, Reason (R) logically explains Assertion (A), as the high frequency (and thus high energy) is indeed the cause of their penetrating nature. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Faraday’s Law of Electromagnetic Induction

c) The induced emf is equal to the negative rate of change of magnetic flux.
[Solution Description]
Faraday’s Law states that the induced electromotive force (emf) in a closed loop is equal to the negative rate of change of magnetic flux through the loop. Mathematically, it is expressed as:
$\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$
Here, $\Phi_B$ is the magnetic flux through the surface bounded by the loop.

Key Concept: Wavelength Range, Energy of UV photon

b) $6.63 \times 10^{-19}$ J
[Solution Description]
To find the energy of one photon, we use the formula:
$E = h \nu$.
Given $\nu = 1 \times 10^{15}$ Hz and $h = 6.63 \times 10^{-34}$ Js.
Substituting the values:
$E = 6.63 \times 10^{-34} \times 1 \times 10^{15} = 6.63 \times 10^{-19}$ J.

Key Concept: Electromagnetic Spectrum - X-Rays

c) Detection of fractures and diseased organs
[Solution Description]
X-rays are used in surgery for detecting fractures, diseased organs, formation of bones and stones, and observing the progress of healing bones. They are also used in radiotherapy to cure diseases like cancer.

Key Concept: Magnetic Energy Density

b) $4.97 \times 10^{-8} \, \text{J/m}^3$
[Solution Description]
The average magnetic energy density is given by:
$u_B = \frac{1}{4\mu_0} B_0^2$
Substituting the given values:
$u_B = \frac{1}{4 \times 4\pi \times 10^{-7}} \times (5 \times 10^{-7})^2$
$u_B = \frac{1}{16\pi \times 10^{-7}} \times 25 \times 10^{-14}$
$u_B = \frac{25 \times 10^{-7}}{16\pi} \approx 4.97 \times 10^{-8} \, \text{J/m}^3$

Key Concept: Magnetic Field in Electromagnetic Waves

c) It is perpendicular to both the electric field and the direction of propagation
[Solution Description]
In an electromagnetic wave, the magnetic field $\overrightarrow{B}$ is always perpendicular to the electric field $\overrightarrow{E}$ and also to the direction of wave propagation. This is a fundamental property of transverse electromagnetic waves.

Key Concept: Electromagnetic Spectrum Applications

c) Microwaves
[Solution Description]
According to the syllabus, microwaves are commonly used in radar systems due to their ability to penetrate through clouds, smoke, and light rain. They have wavelengths ranging from $1 \times 10^{-3}$ m to 1 m and frequencies between $3 \times 10^8$ Hz and $3 \times 10^{11}$ Hz. Microwaves are produced by special vacuum tubes like klystrons and magnetrons.

Key Concept: Radiation Pressure

d) $p = \frac{u}{c}$
[Solution Description]
Radiation pressure ($p$) is given by $p = \frac{u}{c}$, where $u$ is the energy density and $c$ is the speed of light. This follows from the momentum transfer per unit area per unit time when the wave is absorbed.

Key Concept: Definition of Displacement Current

a) $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$
[Solution Description]
The displacement current $I_d$ is given by the rate of change of electric flux $\Phi_E$ through a surface, multiplied by the permittivity of free space $\epsilon_0$. The correct formula is:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$

Key Concept: Energy density, RMS values

c) $2.83 \times 10^{-7}$ T
[Solution Description]
First, we calculate the average energy density using the electric field:
$u = \frac{1}{2}\varepsilon_0 E_0^2 = \frac{1}{2}(8.85 \times 10^{-12})(120)^2 = 6.37 \times 10^{-8}$ J/m^3. Since both electric and magnetic energy densities are equal, total energy density is twice this value. The rms magnetic field can be found using the magnetic energy density formula: $B_{rms} = \sqrt{\mu_0 u}$, where $\mu_0 = 4\pi \times 10^{-7}$ Tm/A. Calculating gives $B_{rms} = \sqrt{(4\pi \times 10^{-7})(6.37 \times 10^{-8})} = 2.83 \times 10^{-7}$ T.

Key Concept: Faraday’s Law of Electromagnetic Induction

b) $0.0157 \, \text{V}$
[Solution Description]
Faraday's Law states:
$e = -\frac{d\Phi_B}{dt}$
The magnetic flux $\Phi_B$ through the loop is:
$\Phi_B = B \cdot A = B \cdot \pi r^2$
Given $r = 0.1 \, \text{m}$, $A = \pi (0.1)^2 = 0.0314 \, \text{m}^2$.
Since $\frac{dB}{dt} = 0.5 \, \text{T/s}$, the rate of change of flux is:
$\frac{d\Phi_B}{dt} = A \cdot \frac{dB}{dt} = 0.0314 \times 0.5 = 0.0157 \, \text{Wb/s}$
Therefore, the induced emf is:
$|e| = 0.0157 \, \text{V}$

Key Concept: Infrared Production

c) From hot bodies and molecular transitions
[Solution Description]
Infrared radiation is produced by hot bodies and through rotational and vibrational transitions in molecules, as mentioned in the syllabus.

Key Concept: Gauss’ Law Interpretation

b) The electric flux is zero if no charge is present inside the surface.
[Solution Description]
Gauss’ law states that the total electric flux through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the net charge enclosed by the surface, mathematically represented as:
$\oint \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}$
This holds true regardless of the shape or size of the closed surface and the distribution of charges inside it.

Key Concept: Transverse nature of EM waves

c) The directions of variations of electric and magnetic fields are perpendicular to each other and also perpendicular to the direction of propagation.
[Solution Description]
Electromagnetic waves are transverse in nature, meaning the electric field ($\vec{E}$), magnetic field ($\vec{B}$), and direction of propagation ($\vec{k}$) are all mutually perpendicular to each other. Additionally, the variations in both fields occur simultaneously, reaching their maximum values at the same place and time.

Key Concept: Energy Density in Electromagnetic Waves

b) 1
[Solution Description]
The average electric energy density ($\overline{u_e}$) and average magnetic energy density ($\overline{u_m}$) are given by:
$\overline{u_e} = \frac{1}{4} \varepsilon_0 E_0^2, \quad \overline{u_m} = \frac{1}{4 \mu_0} B_0^2.$
Using the relation $\frac{E_0}{B_0} = c$ and $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$, we can write:
$\overline{u_e} = \frac{1}{4} \varepsilon_0 E_0^2 = \frac{1}{4} \varepsilon_0 c^2 B_0^2 = \frac{1}{4 \mu_0} B_0^2 = \overline{u_m}.$
Hence, the ratio $\frac{\overline{u_e}}{\overline{u_m}}$ is 1.

Key Concept: Speed of Electromagnetic Waves in Medium

b) $1.5 \times 10^8 \, \text{m/s}$
[Solution Description]
The speed of the electromagnetic wave in the medium is given by $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$. Substituting $c = 3 \times 10^8 \, \text{m/s}$, $\mu_r = 1$, and $\epsilon_r = 4$, we get $v = \frac{3 \times 10^8}{\sqrt{1 \times 4}} = 1.5 \times 10^8 \, \text{m/s}$.

Key Concept: Transverse Nature of Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in an electromagnetic wave propagating along the X-axis, the electric field \$\vec{E}\$ has no component along the direction of propagation. This is true because, as derived from Gauss's law for electricity, the net electric flux through a closed surface in a charge-free region must be zero. For a plane wave, this leads to the conclusion that \$E_x = 0\$, meaning the electric field is perpendicular to the direction of propagation. The reason correctly explains why the assertion is true, as it provides the fundamental principle (Gauss's law) that leads to the transverse nature of the electric field in electromagnetic waves.

Key Concept: Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that electromagnetic waves are transverse in nature, which is true as their electric and magnetic fields oscillate perpendicular to the direction of propagation. The reason correctly explains this by stating that both $\overrightarrow{E}$ and $\overrightarrow{B}$ fields are perpendicular to each other and to the propagation direction. Hence, the reason is the correct explanation for the assertion.

Key Concept: Ampere-Maxwell Law: Total Current

b) $I_c + I_d$
[Solution Description]
The Ampere-Maxwell law states that the line integral of the magnetic field around a closed loop is proportional to the sum of the conduction current and displacement current passing through the loop:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d)$
Therefore, the total current enclosed is $I_c + I_d$.

Key Concept: Communication

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Solution description placeholder.

Key Concept: Propagation direction

b) The wave is transverse
[Solution Description]
If $\overrightarrow{E}$ has no component along the X-axis, it means the electric field is transverse to the direction of propagation (X-axis). Similarly, $\overrightarrow{B}$ will also have no component along the X-axis, confirming the transverse nature of electromagnetic waves.

Key Concept: Gamma Rays

d) Providing information about atomic nuclei
[Solution Description]
Gamma rays are primarily used to provide information about the structure of atomic nuclei.

Key Concept: Ampere-Maxwell Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion states the Ampere-Maxwell Law, which is correct. The reason explains that the total current consists of conduction and displacement currents, which is also correct. Moreover, the reason correctly explains why the assertion is true since the law explicitly includes both types of currents in its formulation.

Key Concept: Characteristics of Electromagnetic Waves

b) $3.0 \times 10^8 \, \text{m s}^{-1}$
[Solution Description]
The speed of electromagnetic waves in free space is given by $c = 3.0 \times 10^8 \, \text{m s}^{-1}$, as per the relation $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$. This is a fundamental property of electromagnetic waves.

Key Concept: Electromagnetic Spectrum - Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that X-rays are used in medical imaging to detect fractures and diseases, which is true as they are commonly employed in radiology. The Reason explains that X-rays have high penetrating power and can pass through soft tissues but are absorbed by bones, which is also true and correctly explains why they are useful for medical imaging. Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Electromagnetic Spectrum, Frequency and Wavelength Relationship

c) X-rays
[Solution Description]
First, calculate the wavelength using $\lambda = \frac{c}{\nu}$ where $\nu = 2 \times 10^{18}$ Hz.
$\lambda = \frac{3 \times 10^8}{2 \times 10^{18}} = 1.5 \times 10^{-10} \text{ m}$
From the syllabus, this falls under the X-ray wavelength range ($10^{-13}$ m to $3 \times 10^{-8}$ m).

Key Concept: Transverse Nature of Electromagnetic Waves

c) Perpendicular to each other and also perpendicular to the direction of propagation
[Solution Description]
Electromagnetic waves are transverse in nature. The electric field $\vec{E}$ and magnetic field $\vec{B}$ are perpendicular to each other and also perpendicular to the direction of propagation.

Key Concept: Electric Field and Displacement Current

d) It remains the same
[Solution Description]
The displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$. The electric flux $\Phi_E = E \cdot A = \frac{q}{\epsilon_0}$ does not depend on the area $A$, since $E = \frac{q}{\epsilon_0 A}$ and hence $\Phi_E = \frac{q}{\epsilon_0}$ remains unchanged. Therefore, even if the area is doubled, $\Phi_E$ and its time derivative remain the same, so the displacement current $I_d$ also remains unchanged.

Key Concept: Uses of X-rays and Gamma Rays

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that X-rays are used in radiotherapy to treat cancer because they can destroy cancerous cells, which is true as per the syllabus. The Reason claims that X-rays have higher energy than gamma rays, making them more effective for deep-seated tumors. However, this is false because gamma rays have higher energy than X-rays. Thus, Assertion is true, but Reason is false.

Key Concept: Logical Reasoning

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that if all prime numbers greater than 2 are odd (which is true), then every even integer greater than 2 can be expressed as the sum of two primes. This conclusion is known as the Goldbach conjecture, which is mentioned in the reason. While the reason correctly states the Goldbach conjecture, it does not logically explain why the assertion's premise (primes > 2 being odd) would lead to its conclusion. The Goldbach conjecture remains unproven despite primes > 2 being odd. Thus, both the assertion and reason are true, but the reason is not the correct explanation for the assertion.

Key Concept: Average Energy Density of Electric and Magnetic Fields, Equality of Energy Densities

b) $2.21 \times 10^{-8} \, \text{J/m}^3$
[Solution Description]
Since $u_E = u_B$ for an electromagnetic wave, we can use the formula for electric energy density to find $u_B$:
$u_E = \frac{1}{4} \varepsilon_0 E_0^2$.
Substituting values:
$u_E = \frac{1}{4}(8.85 \times 10^{-12})(100)^2 = 2.21 \times 10^{-8} \, \text{J/m}^3$.
Hence, $u_B = 2.21 \times 10^{-8} \, \text{J/m}^3$.

Key Concept: Phase and propagation of EM waves

c) $\left(\frac{1}{3} \times 10^{-6}\right) \sin(6\pi \times 10^8 t - 2\pi x)$ T
[Solution Description]
The magnetic field can be found using the relation $\frac{\vec{E}}{\vec{B}} = c$ and noting that $\vec{E}$ and $\vec{B}$ are in phase. Here, $E_y = E_0 \sin(\omega t - kx)$, where $E_0 = 100$ V/m, $\omega = 6\pi \times 10^8$ rad/s, and $k = 2\pi$ m$^{-1}$. The corresponding magnetic field amplitude is $B_0 = \frac{E_0}{c} = \frac{100}{3 \times 10^8} = \frac{1}{3} \times 10^{-6}$ T. Thus, $B_z(x,t) = B_0 \sin(\omega t - kx) = \left(\frac{1}{3} \times 10^{-6}\right) \sin(6\pi \times 10^8 t - 2\pi x)$ T.

Key Concept: Faraday's Law, Gauss' Law of Magnetism

a) $\frac{B_0 \pi r^2}{\tau} e^{-t/\tau}$
[Solution Description]
The magnetic flux through the loop is:
$\Phi_B = \vec{B} \cdot \vec{A} = B_0 e^{-t/\tau} \cdot \pi r^2.$
By Faraday’s law, the induced emf is:
$e = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}\left(B_0 \pi r^2 e^{-t/\tau}\right) = \frac{B_0 \pi r^2}{\tau} e^{-t/\tau}.$

Key Concept: Wavelength and Frequency Ranges, Temperature Range of Electromagnetic Spectrum

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the temperature associated with gamma rays is higher than that of radio waves. From the given data, gamma rays have a wavelength range of $10^{-14}$ m to $10^{-10}$ m, leading to a temperature range of $2.9 \times 10^{10}$ K to $2.9 \times 10^{7}$ K. Radio waves, on the other hand, have a wavelength range of $1 \times 10^{-1}$ m to $10^{4}$ m, resulting in a temperature range of $2.9 \times 10^{-3}$ K to $2.9 \times 10^{-8}$ K. Clearly, gamma rays are at a much higher temperature than radio waves.
The Reason provides the formula for the temperature-wavelength relationship: $T = \frac{0.29 \times 10^{-2}}{\lambda_m}$ K. This formula shows that temperature is indeed inversely proportional to wavelength, which explains why shorter wavelengths (like gamma rays) correspond to higher temperatures and longer wavelengths (like radio waves) correspond to lower temperatures.
Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Production of Infrared Radiation

c) From hot bodies and molecular transitions
[Solution Description]
The syllabus states: "Production : From hot bodies, and by rotational and vibrational transitions in molecules." So, infrared radiation is mainly produced by hot bodies or molecular transitions.

Key Concept: Transverse Nature of Electromagnetic Waves

c) The electric field has no component along the X-axis.
[Solution Description]
The transverse nature of electromagnetic waves implies that $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular to the direction of propagation. From Gauss's law, we derived $E_x = 0$ for a plane wave, meaning the electric field has no component along the propagation direction (X-axis). Therefore, $\overrightarrow{E}$ must lie in the YZ-plane.

Key Concept: Wavelength Range

b) $4 \times 10^{-7} \, \text{m}$ to $7 \times 10^{-7} \, \text{m}$
[Solution Description]
The visible light spectrum ranges from $4 \times 10^{-7} \, \text{m}$ to $7 \times 10^{-7} \, \text{m}$.

Key Concept: Speed in Medium

b) $1 \times 10^8 \, \text{m s}^{-1}$
[Solution Description]
The speed of an electromagnetic wave in a medium is given by:
$v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$
Substituting $\mu_r = 1$, $\epsilon_r = 9$, and $c = 3 \times 10^8 \, \text{m s}^{-1}$:
$v = \frac{3 \times 10^8}{\sqrt{1 \times 9}} = \frac{3 \times 10^8}{3} = 1 \times 10^8 \, \text{m s}^{-1}$

Key Concept: Electromagnetic Spectrum Frequency

b) $7 \times 10^{14}$ Hz to $4 \times 10^{14}$ Hz
[Solution Description]
The frequency range for visible light is given as $7 \times 10^{14}$ Hz to $4 \times 10^{14}$ Hz. This corresponds to wavelengths between $4 \times 10^{-7}$ m (violet) and $7 \times 10^{-7}$ m (red).

Key Concept: Infrared Radiation

b) Providing electrical energy to satellites
[Solution Description]
Infrared radiation has various uses, including providing electrical energy to satellites via solar cells, producing dehydrated fruits, treating muscular strain, and in solar water heaters. It is also used in weather forecasting through infrared photography and in warfare to see through haze or fog. However, it is not used for studying molecular structures, which is an application of ultraviolet radiation.

Key Concept: Transverse Nature, Wave Characteristics

d) The electric and magnetic fields are mutually perpendicular and perpendicular to the direction of propagation.
[Solution Description]
The transverse nature of electromagnetic waves implies that both the electric field $\overrightarrow{E}$ and the magnetic field $\overrightarrow{B}$ are perpendicular to the direction of wave propagation. Additionally, they are perpendicular to each other and reach their maximum values simultaneously. The ratio of their magnitudes in free space is $E/B = c$, where $c$ is the speed of light.

Key Concept: Microwaves

c) They can be transmitted as a beam in a particular direction
[Solution Description]
Microwaves are used in radar systems because they can be transmitted as a beam in a particular direction without significant diffraction by obstacles of normal dimensions. This property allows precise detection and tracking of objects like aircraft.

Key Concept: Industrial Applications - Infrared Radiation

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that infrared radiation is used in solar water heaters to convert solar energy into thermal energy. This is true because infrared radiation is a part of the electromagnetic spectrum that carries heat energy, and solar water heaters utilize this property to heat water.
The reason claims that infrared radiation has higher penetration power compared to visible light, making it suitable for heating applications. However, this is false because infrared radiation does not necessarily have higher penetration power than visible light; instead, it is primarily absorbed by objects, causing them to heat up.
Therefore, Assertion is true, but Reason is false.

Key Concept: Properties of radio waves

d) Reflection and diffraction
[Solution Description]
The syllabus states that radio waves exhibit properties such as reflection and diffraction.

Key Concept: Transverse Nature of EM Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that electromagnetic waves are transverse in nature, which is true as per the syllabus. The Reason explains why they are transverse by stating that the electric and magnetic fields are perpendicular to each other and to the direction of propagation, which is also true and correctly explains the Assertion.

Key Concept: Electromagnetic Spectrum

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The electromagnetic spectrum orders waves by wavelength or frequency. Gamma rays have wavelengths in the range $1 \times 10^{-14} \, \text{m}$ to $1 \times 10^{-10} \, \text{m}$, while ultraviolet radiation has wavelengths in the range $6 \times 10^{-10} \, \text{m}$ to $4 \times 10^{-7} \, \text{m}$. Therefore, gamma rays indeed have shorter wavelengths. The reason correctly identifies the production mechanisms for both gamma rays and ultraviolet radiation, but the difference in their production methods does not explain why gamma rays have shorter wavelengths. Thus, both statements are true, but the Reason does not correctly explain the Assertion.

Key Concept: Ampere-Maxwell’s Circuital Law

c) A changing electric field contributes to the magnetic field via displacement current.
[Solution Description]
The Ampere-Maxwell law includes both conduction current ($I$) and displacement current ($I_d$), where $I_d = \epsilon_0 \frac{dE}{dt} A$. Thus, a changing electric field also produces a magnetic field.

Key Concept: Wavelength-frequency relation

b) $6 \times 10^{-8}$ m
[Solution Description]
The wavelength $\lambda$ can be calculated using the formula $\lambda = \frac{c}{\nu}$, where $c$ is the speed of light ($3 \times 10^8$ m/s) and $\nu$ is the frequency. Substituting the given frequency into the equation:
$\lambda = \frac{3 \times 10^8}{5 \times 10^{15}} = 6 \times 10^{-8} \, \text{m}$
The correct wavelength is $6 \times 10^{-8}$ m.

Key Concept: Radiotherapy

c) Gamma rays
[Solution Description]
Gamma rays are used in radiotherapy for treating cancer because they have high penetration power and are ionizing, which means they can destroy malignant cells effectively.

Key Concept: Speed of Electromagnetic Waves in a Medium

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The speed of an electromagnetic wave in a medium is given by $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$, where $c$ is the speed of light in free space. For the speed in the medium to be less than that in free space, the denominator $\sqrt{\mu_r \epsilon_r}$ must be greater than 1. This implies $\mu_r \epsilon_r > 1$. However, while this condition holds for many media, it is not universally true for all media. For example, some meta-materials may have $\mu_r \epsilon_r \leq 1$. Therefore, the assertion is generally true for most conventional media, but the reason is not universally correct.

Key Concept: Relation between Electric and Magnetic Fields

a) $1.67 \times 10^{-7} \, \text{T}$
[Solution Description]
The relationship between electric field $E$ and magnetic field $B$ in an electromagnetic wave is:
$E = cB$
Rearranging to find $B$:
$B = \frac{E}{c}$
Substituting the given values:
$B = \frac{50 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}}$
Simplifying:
$B \approx 1.67 \times 10^{-7} \, \text{T}$

Key Concept: X-Rays in Radiology

d) CT scan
[Solution Description]
CT (Computed Tomography) scans use X-rays to produce detailed cross-sectional images of internal structures, unlike standard radiography which only provides 2D images.

Key Concept: Displacement Current Definition

b) The current equivalent of the rate of change of electric flux
[Solution Description]
Displacement current ($I_d$) is the current equivalent of the rate of change of electric flux between the plates of a capacitor. It is given by:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.
This current arises due to the changing electric field and completes the circuit when conduction current stops inside the capacitor.

Key Concept: Propagation of Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that electromagnetic waves can propagate through a vacuum, which is true as they do not require a material medium for propagation. The reason explains that their propagation relies on oscillating electric and magnetic fields perpendicular to each other, eliminating the need for a medium. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Diagnostic Imaging

d) Magnetic Resonance Imaging (MRI)
[Solution Description]
Magnetic Resonance Imaging (MRI) uses radio waves along with strong magnetic fields to generate detailed images of the internal structures of the body. This makes MRI a crucial tool in modern diagnostic medicine.

Key Concept: Electromagnetic Spectrum

a) Gamma rays
[Solution Description]
The electromagnetic spectrum in order of increasing wavelength is: Gamma rays < X-rays < Ultraviolet light < Visible light < Infrared rays < Microwaves < Radio waves. Thus, gamma rays have the shortest wavelength.

Key Concept: Electromagnetic Spectrum Wavelength

d) Gamma rays
[Solution Description]
The electromagnetic spectrum is arranged in order of increasing wavelength or decreasing frequency. Gamma rays have the shortest wavelength, ranging from $1 \times 10^{-14}$ m to $1 \times 10^{-10}$ m.

Key Concept: Maxwell's Equations, Displacement Current and Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion (A) states that the displacement current is crucial for electromagnetic wave production, which is correct because it completes Ampere’s circuital law by accounting for the changing electric flux, enabling the self-propagation of electromagnetic waves. The Reason (R) correctly explains that displacement current ensures current continuity in circuits with capacitors and arises from a time-varying electric field, illustrating its role in completing the explanation for (A). Therefore, both Assertion and Reason are true, and Reason explains Assertion correctly.

Key Concept: Energy Density in Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the average energy density of an electromagnetic wave is equally divided between its electric and magnetic field components. This is true because, for a sinusoidal electromagnetic wave, the average electric energy density $\overline{u_e} = \frac{1}{4} \varepsilon_0 E_0^2$ and the average magnetic energy density $\overline{u_m} = \frac{1}{4} \frac{B_0^2}{\mu_0}$ are equal when we use the relation $\frac{E_0}{B_0} = c$ and $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$. The reason correctly explains why the two energy densities are equal by invoking the fundamental relationship between $E_0$ and $B_0$ in an electromagnetic wave.

Key Concept: Energy Density in Electromagnetic Waves

c) $u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$
[Solution Description]
The average energy density in an electromagnetic wave is the sum of the energy densities of the electric and magnetic fields. It is given by the formula:
$u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$

Key Concept: Electroagnetic Wave Speed

c) $3 \times 10^8 \, \text{m/s}$
[Solution Description]
The relationship between the electric field amplitude $E_0$ and the magnetic field amplitude $B_0$ in an electromagnetic wave is given by $E_0 = c B_0$, where $c$ is the speed of light in vacuum. Rearranging, we get $c = \frac{E_0}{B_0}$. Substituting the given values:
$c = \frac{150}{5 \times 10^{-7}} = 3 \times 10^8 \, \text{m/s}.$
This confirms the universal speed of electromagnetic waves in free space.

Key Concept: Speed of Electromagnetic Waves

c) $3.0 \times 10^8 \, \text{m s}^{-1}$
[Solution Description]
The speed of electromagnetic waves in free space is given by $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, which equals $3.0 \times 10^8 \, \text{m s}^{-1}$. The correct answer is the option matching this value.

Key Concept: Transverse nature, Gauss's law for electricity

a) The X-component of $\vec{E}$ must be zero
[Solution Description]
Using Gauss's law for electricity, we consider a small volume in the path of the wave. Since there is no enclosed charge, the net flux through the volume must be zero. The electric field does not vary with $y$ or $z$, so the contributions from the faces perpendicular to Y and Z-axes cancel out. For the faces perpendicular to the X-axis, if $E_x \neq 0$, the flux would not cancel unless $E_x$ is constant, which contradicts the wave nature. Therefore, $E_x = 0$.

Key Concept: Displacement Current

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the magnetic field exists between the plates of a charging capacitor due to the displacement current. This is correct because Ampere's law is modified by Maxwell to include displacement current, which accounts for the changing electric field. The reason explains the formula for displacement current, which is accurate. Thus, both are true, and the reason correctly explains the assertion.

Key Concept: Gauss’ Law, Electric Flux

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Understand Assertion (A). It states that if no charge is enclosed within a closed surface, the electric flux through it is zero. This aligns with Gauss’ Law, which states $\int \vec{E} \cdot d\vec{S} = \frac{q}{\varepsilon_0}$. If $q = 0$ (no enclosed charge), then the flux $\int \vec{E} \cdot d\vec{S} = 0$. Thus, Assertion (A) is true.
Step 2: Understand Reason (R). It correctly states Gauss’ Law, which relates the net electric flux through a closed surface to the enclosed charge. Therefore, Reason (R) is also true.
Step 3: Check if Reason explains Assertion. Since Gauss’ Law directly implies that zero enclosed charge results in zero flux, Reason (R) provides the correct explanation for Assertion (A).
Conclusion: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Properties, Uses

d) Penetration capability through atmospheric obstructions
[Solution Description]
Infrared rays can penetrate fog and mist effectively, unlike visible light, making them suitable for surveillance and imaging in adverse weather conditions. This property is crucial in military applications.

Key Concept: Frequency range, Production, Uses

c) Weather forecasting through infrared photography
[Solution Description]
Infrared rays are produced by rotational and vibrational transitions in molecules, which is utilized in weather forecasting through infrared photography. This property helps in detecting molecular movements and temperature variations.

Key Concept: Transverse Nature, Gauss’ Law for Electricity

c) The wave is purely transverse.
[Solution Description]
According to Gauss' law for electricity, the electric flux through a closed surface enclosing no charge is zero. For a plane electromagnetic wave propagating along the X-axis, the electric field $\overrightarrow{E}$ cannot have a component along the X-axis because a static field cannot propagate as a wave. Therefore, $\overrightarrow{E}$ must be perpendicular to the X-axis, confirming that the wave is transverse in nature.

Key Concept: Transverse nature of electromagnetic waves

d) Neither \$
vec\{E}\$ nor \$
vec\{B}\$ has a component along the X-axis
[Solution Description]
Since electromagnetic waves are transverse, both \$
vec\{E}\$ and \$
vec\{B}\$ must be perpendicular to the direction of propagation (X-axis). Therefore, neither field has a component along the X-axis.

Key Concept: Electromagnetic Spectrum, Properties and Uses of Gamma Rays

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that gamma rays are highly penetrating due to their highest energy in the electromagnetic spectrum, which is true as gamma rays have the smallest wavelengths ($1 \times 10^{-14}$ to $1 \times 10^{-10}$ m) and highest frequencies ($3 \times 10^{22}$ to $3 \times 10^{18}$ Hz). The reason provides the correct explanation by stating the relationship between energy and frequency ($E = h\nu$) and since gamma rays have the highest frequency, they indeed possess the highest energy. Hence, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Speed of Electromagnetic Wave in Free Space

d) $v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
[Solution Description]
The speed of an electromagnetic wave in free space is given by:
$v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
where $\mu_0$ is the permeability of free space and $\varepsilon_0$ is the permittivity of free space. This speed is equal to the speed of light $c$.

Key Concept: Wavelength range, Properties, Uses

d) Generate significant heating effect when absorbed
[Solution Description]
Infrared radiation has a wavelength range of $8 \times 10^{-7}$ to $5 \times 10^{-3}$ m and exhibits strong heating effects, making it ideal for applications like solar water heaters where heat absorption is required. Its penetration ability through fog also aids in efficient energy transfer.

Key Concept: Electromagnetic Spectrum - Energy of Photon

c) \$3.31 \times 10^{-18} J\$
[Solution Description]
The energy (\$E\$) of a photon can be calculated using Planck's equation:
$E = h\nu$
where \$h = 6.62 \times 10^{-34} Js\$ (Planck's constant) and \$\nu = 5 \times 10^{15} Hz\$.
Substituting the values:
$E = 6.62 \times 10^{-34} \times 5 \times 10^{15} = 3.31 \times 10^{-18} \, \text{J}.$
Thus, the energy of the photon is \$3.31 \times 10^{-18} J\$.

Key Concept: Transverse Nature, Energy Density

b) $4.43 \times 10^{-10} \, \text{J/m}^3$
[Solution Description]
The average energy density for an electromagnetic wave can be calculated using its electric field component. Given $E_0 = 10 \, \text{V/m}$, the root mean square (rms) value of the electric field is:
$E_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{V/m}$
Now, plug in $\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$ into the energy density formula:
$u_E = \frac{1}{2} \epsilon_0 E_{\text{rms}}^2 = \frac{1}{2} \times 8.85 \times 10^{-12} \times \left( \frac{10}{\sqrt{2}} \right)^2$
Simplify the calculation step-by-step:
$u_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times \frac{100}{2} = 2.2125 \times 10^{-10} \, \text{J/m}^3$
Since the energy is equally divided between electric and magnetic fields, the total energy density is twice this value:
$u = 2 \times 2.2125 \times 10^{-10} = 4.425 \times 10^{-10} \, \text{J/m}^3$

Key Concept: Faraday's law, Transverse nature of EM waves

b) The electric field must be purely in the Y-direction, orthogonal to both $\overrightarrow{B}$ and the propagation direction.
[Solution Description]
Faraday's law states $\nabla \times \overrightarrow{E} = -\frac{\partial \overrightarrow{B}}{\partial t}$. For $\overrightarrow{B} = B_z \hat{k}$, varying with $x$ and $t$, the curl of $\overrightarrow{E}$ must have a non-zero Y-component: $\frac{\partial E_z}{\partial x} - \frac{\partial E_x}{\partial z} = -\frac{\partial B_z}{\partial t}$. Since $E_x = 0$ (from transverse nature) and $\overrightarrow{E}$ is independent of $z$ (plane wave), this simplifies to $\frac{\partial E_z}{\partial x} = -\frac{\partial B_z}{\partial t}$. Thus, $\overrightarrow{E}$ must have a time-varying Z-component perpendicular to $\overrightarrow{B}$ and the propagation direction.

Key Concept: Microwaves properties and uses in radar

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that microwaves are used in radar systems for aircraft navigation, which is true as per the syllabus. The reason provides that microwaves have wavelengths of the order of $10^{-2}$ m, which is also correct as mentioned in the syllabus. Moreover, the reason correctly explains why microwaves are used in radar systems since their wavelength allows them to not be diffracted by obstacles of normal dimensions, making them ideal for directional transmission in radar. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Uses of X-Rays

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion states that X-rays are used to detect fractures in bones, which is true as they are commonly used in medical imaging for this purpose. The reason states that X-rays can penetrate soft tissues but are absorbed by dense materials like bones, which is also true and correctly explains why X-rays are suitable for detecting bone fractures (the contrast between absorbed and transmitted rays creates the image). Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Displacement Current

d) $\epsilon_0 \frac{d\Phi_E}{dt}$
[Solution Description]
Displacement current arises due to a time-varying electric field. The formula for displacement current is given by:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$
where $\epsilon_0$ is the permittivity of free space and $\Phi_E$ is the electric flux.

Key Concept: Energy density in electromagnetic waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The average energy density of an electromagnetic wave is given by $u = \frac{1}{4} \varepsilon_0 E_0^2 + \frac{1}{4 \mu_0} B_0^2$. It is known that $\frac{1}{4} \varepsilon_0 E_0^2$ represents the average electric energy density ($u_e$) and $\frac{1}{4 \mu_0} B_0^2$ represents the average magnetic energy density ($u_m$). For electromagnetic waves in vacuum, the relation $E_0 = c B_0$ holds, leading to $u_e = u_m$. Thus, the total average energy density is equally divided between electric and magnetic fields.

Key Concept: Radio waves, displacement current

b) 0.2 A
[Solution Description]
According to Maxwell's equations, the displacement current $I_d$ between capacitor plates equals the charging current $I_c$. Therefore, $I_d = I_c = 0.2$ A regardless of the plate dimensions.

Key Concept: Properties and Uses of Ultraviolet Rays

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that ultraviolet (UV) rays are used for sterilization due to their ability to kill bacteria and viruses, which is true as per the syllabus. The reason explains that UV rays ionize gases and disrupt the DNA of microorganisms. This is also true because UV radiation damages the genetic material of microbes, preventing them from replicating. Moreover, ionization contributes to this effect, making the reason a correct explanation of the assertion.

Key Concept: Radio waves wavelength and frequency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, radio waves indeed have a wavelength range of $1 \times 10^{-3}$ to $3 \times 10^{-1}$ meters and a corresponding frequency range of $3 \times 10^{11}$ to $1 \times 10^9$ Hz. These ranges are inversely related by the equation $\lambda = \frac{c}{\nu}$, where $\lambda$ is wavelength, $c$ is the speed of light, and $\nu$ is frequency. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Gauss’ Law Application

b) Infinite plane of charge
[Solution Description]
Gauss’ Law simplifies the calculation of electric fields for highly symmetric charge distributions such as point charges, infinite planes, and spherical shells due to the symmetry allowing simplification of the integral $\int \vec{E} \cdot d\vec{S}$.

Key Concept: Transverse Wave Characteristics

c) $\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$, and both are perpendicular to the direction of propagation
[Solution Description]
Electromagnetic waves are transverse, meaning both $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular to the direction of propagation (X-axis in this case). Additionally, $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular to each other. Their magnitudes are related by $|\overrightarrow{E}| = c |\overrightarrow{B}|$, where $c$ is the speed of light. This relationship ensures they satisfy Maxwell's equations.

Key Concept: Gauss' Law of Electricity, Faraday's Law

b) $\epsilon_0 E_0 a^3$
[Solution Description]
To find the net charge enclosed within the cube, we use Gauss' law:
$\oint \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}.$
The cube has six faces. Only the faces perpendicular to the $x$-axis will have non-zero flux since $\vec{E}$ is along $\hat{i}$. For the face at $x = 0$, $\vec{E} = \vec{0}$. For the face at $x = a$, $\vec{E} = E_0 a \hat{i}$. The area vector for this face is $d\vec{S} = a^2 \hat{i}$. Thus, the flux through this face is:
$\Phi_E = \vec{E} \cdot d\vec{S} = E_0 a \cdot a^2 = E_0 a^3.$
The total flux is only through this face. Using Gauss' law:
$E_0 a^3 = \frac{q}{\epsilon_0} \implies q = \epsilon_0 E_0 a^3.$

Key Concept: Basic Concept

d) They can travel through vacuum without any medium
[Solution Description]
Electromagnetic waves are transverse waves that can travel through vacuum because they do not require any material medium for their propagation, as mentioned in the syllabus. The correct answer is the option stating this property.

Key Concept: Speed of EM Waves in Media, Relative Permittivity and Permeability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given:
$\mu_r = 1$, $\epsilon_r = 4$, $c = 3 \times 10^8 \, \text{m/s}$
The speed of the wave in the medium ($v$) is given by:
$v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$
Substituting the values:
$v = \frac{3 \times 10^8}{\sqrt{1 \times 4}} = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \, \text{m/s}$
The Assertion (A) is true as the calculated speed matches the given value.
The Reason (R) is also true because the formula $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$ shows that the speed is indeed inversely proportional to the square root of the product of $\mu_r$ and $\epsilon_r$.
Moreover, the Reason correctly explains the Assertion since it provides the mathematical basis for the given speed in the medium.

Key Concept: Speed of Electromagnetic Waves in a Medium

c) 2
[Solution Description]
Using the formula for the speed of an electromagnetic wave in a medium:
$v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$
Rearranging to solve for $\epsilon_r$:
$\epsilon_r = \frac{c^2}{v^2 \mu_r}$
Substituting the given values:
$\epsilon_r = \frac{(3 \times 10^8)^2}{(1.5 \times 10^8)^2 \times 2} = \frac{9 \times 10^{16}}{2.25 \times 10^{16} \times 2} = \frac{9}{4.5} = 2$
Thus, the relative permittivity is 2.

Key Concept: Average Energy Density Formula

b) $\epsilon_0 E_0^2$
[Solution Description]
The average energy density ($u$) of an electromagnetic wave in terms of the electric field amplitude $E_0$ is given by:
$u = \epsilon_0 E_0^2$
This is derived from the fact that the total average energy density is the sum of the average electric and magnetic energy densities, which are equal in free space.

Key Concept: Relation between Electric and Magnetic Fields

a) $1 \times 10^{-6} \, \text{T}$
[Solution Description]
The relation between $E_0$ and $B_0$ is:
$E_0 = c B_0$
Rearranging to solve for $B_0$:
$B_0 = \frac{E_0}{c} = \frac{300}{3 \times 10^8} = 1 \times 10^{-6} \, \text{T}$

Key Concept: Uses of X-Rays

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that X-rays are used in the detection of fractures and diseased organs, which is true as they are commonly employed in medical imaging for these purposes. The reason provided is also accurate; X-rays can penetrate soft tissues easily but are absorbed by denser materials such as bones and tumors. Moreover, the reason correctly explains why X-rays are suitable for this application. Therefore, both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Gauss's Law for Electricity

c) The net electric flux is always zero.
[Solution Description]
Gauss's law states $\int \overrightarrow{E} \cdot d\overrightarrow{A} = 0$ for a charge-free region. Since $\overrightarrow{E}$ is parallel to the Y-axis and varies only with $x$, the flux contributions through faces normal to the Y-axis (BCGF and ADHE) cancel out, as do those normal to the Z-axis (ABFE and DCGH). The net flux through any closed surface is zero.

Key Concept: Maxwell's Equations, Gauss's Law of Electricity, Faraday's Law

b) $\epsilon_0 E_0 \pi r^2$
[Solution Description]
The displacement current $I_d$ is given by $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux. The electric flux through the circular surface is:
$\Phi_E = \int \vec{E} \cdot d\vec{A} = E_0 t \cdot \pi r^2$
Taking the derivative with respect to time:
$\frac{d\Phi_E}{dt} = E_0 \pi r^2$
Thus, the displacement current is:
$I_d = \epsilon_0 E_0 \pi r^2$

Key Concept: Verbal Communication

b) Face-to-face conversation
[Solution Description]
Verbal communication involves spoken words. The most common form is face-to-face conversation as it allows direct interaction between individuals.

Key Concept: Visible Light Spectrum

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states the correct wavelength range of visible light ($4 \times 10^{-7}$ m to $7 \times 10^{-7}$ m). The reason also correctly states the corresponding frequency range calculated using $c = \lambda \nu$, where $c$ is the speed of light ($3 \times 10^8$ m/s). For $\lambda = 4 \times 10^{-7}$ m, $\nu = \frac{3 \times 10^8}{4 \times 10^{-7}} = 7.5 \times 10^{14}$ Hz. For $\lambda = 7 \times 10^{-7}$ m, $\nu = \frac{3 \times 10^8}{7 \times 10^{-7}} \approx 4.29 \times 10^{14}$ Hz. However, the reason does not explain why this specific range makes it "visible," so it is NOT the correct explanation of the assertion.

Key Concept: Average Energy Density

b) $4.425 \times 10^{-8} \, \text{J/m}^3$
[Solution Description]
The average energy density $\overline{u}$ in an electromagnetic wave is given by:
$\overline{u} = \frac{1}{2} \varepsilon_0 E_0^2$
Substituting the given values:
$\overline{u} = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times (100 \, \text{V/m})^2$
Simplifying:
$\overline{u} = \frac{1}{2} \times 8.85 \times 10^{-12} \times 10^4 \, \text{J/m}^3$
$\overline{u} = 4.425 \times 10^{-8} \, \text{J/m}^3$

Key Concept: Inconsistency of Ampere’s Circuital Law

d) Because the law ignores displacement current
[Solution Description]
Ampere's circuital law predicts zero magnetic field between the plates since it only considers conduction current, which is zero here. However, experimentally, a magnetic field exists. This discrepancy arises because the law does not account for the time-varying electric field (displacement current) between the plates.

Key Concept: Electromagnetic wave characteristics

d) $\frac{E}{B} = c$
[Solution Description]
According to the syllabus, for an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related as $E/B = c$, where $c$ is the speed of light. This means the ratio of the electric field magnitude to the magnetic field magnitude is equal to the speed of light.

Key Concept: Electromagnetic Spectrum, Applications

d) Ultraviolet radiation
[Solution Description]
Ultraviolet radiation has enough energy to cause photoelectric emission, which is useful in burglar alarms. It is also harmful to the human body upon prolonged exposure, as it can damage skin cells. The wavelength range of ultraviolet radiation is $6 \times 10^{-10}$ m to $4 \times 10^{-7}$ m.

Key Concept: Average Energy Density, Electric and Magnetic Field Relationship

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the total average energy density ($u$) of an electromagnetic wave is equally divided between the electric ($u_e$) and magnetic ($u_m$) fields. Mathematically, this means $u_e = u_m$. From the given formula for average energy densities:
$u_e = \frac{1}{4} \varepsilon_0 E_0^2$
$u_m = \frac{1}{4 \mu_0} B_0^2$
Substituting the relation $E_0 = cB_0$ and $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ into $u_e$, we get:
$u_e = \frac{1}{4} \varepsilon_0 (cB_0)^2 = \frac{1}{4} \varepsilon_0 \left(\frac{B_0^2}{\mu_0 \varepsilon_0}\right) = \frac{1}{4\mu_0} B_0^2 = u_m$
This confirms the equality $u_e = u_m$, validating the assertion. The reason correctly states the relationship $E_0 = cB_0$ and explains why the energies are equal. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Transverse Nature of Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in an electromagnetic wave, the electric field $\overrightarrow{E}$ has no component along the direction of propagation. This is true because, as shown in the derivation using a parallelepiped and Gauss' law, the only possible solution for a propagating wave is that $E_x = 0$. The reason given is also true because Gauss' law indeed states that the total electric flux through a closed surface enclosing no charge must be zero. Furthermore, the reason correctly explains why the assertion is true because it is through applying Gauss' law that we deduce the absence of a longitudinal component of $\overrightarrow{E}$.

Key Concept: Electromagnetic Spectrum - Gamma Rays

a) Highly-penetrating and chargeless
[Solution Description]
Gamma rays are highly-penetrating, chargeless, and harmful to the human body. They have properties like chemical reaction on photographic plates, fluorescence, ionization, and diffraction.

Key Concept: Transverse Nature of Electromagnetic Waves, Gauss's Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For an electromagnetic wave propagating along the X-axis, we analyze the electric field $\vec{E}$. Using Gauss’s law, the total electric flux through a closed surface is zero when no charge is enclosed. If $\vec{E}$ had an X-component, the flux terms from the faces perpendicular to the X-axis would not cancel out, violating Gauss’s law. Hence, $\vec{E}$ must be transverse to the direction of propagation. The reason correctly explains why the assertion is true.

Key Concept: Radio waves, wavelength-frequency relationship

b) 3.125 m
[Solution Description]
The relationship between wavelength ($\lambda$) and frequency ($\nu$) is given by: $\lambda = \frac{c}{\nu}$ where $c = 3 \times 10^8$ m/s. For $\nu = 96 \times 10^6$ Hz, $\lambda = \frac{3 \times 10^8}{96 \times 10^6} = 3.125$ m.

Key Concept: Faraday's Law of Induction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that the induced emf in a closed loop is proportional to the rate of change of magnetic flux through the loop, which is correctly described by Faraday's Law of Induction: $e = -\frac{d\Phi_B}{dt}$. The reason (R) explains that according to Faraday's law, a changing magnetic flux induces an electric field, which is also correct as it relates to Maxwell's third equation $\int \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$. Furthermore, the reason provides the correct explanation for the assertion since the induced emf arises due to the electric field generated by the changing magnetic flux.

Key Concept: Electromagnetic Spectrum, Temperature and Wavelength

c) 290 K
[Solution Description]
Convert $\lambda_m$ to cm: $1 \times 10^{-5}$ m = $1 \times 10^{-3}$ cm.
Using the formula $T = \frac{0.29}{\lambda_m}$,
$T = \frac{0.29}{1 \times 10^{-3}} = 290 \text{ K}$
This temperature corresponds to the infrared region, as per the syllabus examples.

Key Concept: Faraday's Law, Induced Electric Field

b) $0.01 e^{-0.5}$ V/m
[Solution Description]
Magnetic flux through the loop: $\Phi_B = B(t) \cdot l^2 = 0.4e^{-0.5t} \times (0.2)^2 = 0.016 e^{-0.5t}$ Wb.
Rate of change of flux: $\frac{d\Phi_B}{dt} = 0.016 \times (-0.5)e^{-0.5t} = -0.008 e^{-0.5t}$ Wb/s.
At $t = 1$ s: $\frac{d\Phi_B}{dt} = -0.008 e^{-0.5(1)} = -0.008 e^{-0.5}$ Wb/s.
From Faraday’s Law, $E \cdot (4l) = -\frac{d\Phi_B}{dt} = 0.008 e^{-0.5}$ V.
Thus, $E = \frac{0.008 e^{-0.5}}{4 \times 0.2} = 0.01 e^{-0.5}$ V/m.

Key Concept: Wave Propagation, Spectrum Identification

c) Visible light
[Solution Description]
The wavelength $\lambda = 5 \times 10^{-7} \, \text{m}$ corresponds to visible light (approximately $400-700 \, \text{nm}$). The frequency can be calculated as $f = \frac{c}{\lambda} = \frac{3 \times 10^8}{5 \times 10^{-7}} = 6 \times 10^{14} \, \text{Hz}$. The given magnetic field equation also matches the frequency calculation since $\omega = 1.2 \times 10^{15} \, \text{rad/s}$ implies $f = \frac{\omega}{2\pi} \approx 1.9 \times 10^{14} \, \text{Hz}$ (Note: There might be an inconsistency here, but the wavelength clearly indicates visible light).

Key Concept: Wave properties and Maxwell's equations

c) $3 \times 10^8$ m/s
[Solution Description]
The speed of the wave is given by $v = \nu\lambda = \frac{\omega}{2\pi} \cdot \lambda = \frac{4\pi \times 10^{14}}{2\pi} \cdot 500 \times 10^{-9} = 2 \times 10^{14} \times 500 \times 10^{-9} = 10^8$ m/s. However, in free space, the speed of an electromagnetic wave $v$ is equal to the speed of light $c = 3 \times 10^8$ m/s. Therefore, the ratio $\frac{E}{B} = v = c = 3 \times 10^8$ m/s.

Key Concept: Electric and Magnetic Field Relationship

c) $E = cB$
[Solution Description]
In free space, the magnitudes of the electric and magnetic fields in electromagnetic waves are related by $\frac{E}{B} = c$, where $c$ is the speed of light. The correct answer is the option stating this relationship.

Key Concept: Production, Properties

b) Oscillating currents in vacuum tubes; polarization
[Solution Description]
Step 1: From the syllabus, microwaves are produced by oscillating currents in special vacuum tubes or electromagnetic oscillators in electric circuits.
Step 2: The wavelength $3 \times 10^{-2}$ m falls within the microwave range ($1 \times 10^{-3}$ m to $3 \times 10^{-1}$ m), confirming their classification.
Step 3: The key property utilized here is reflection, as microwaves can be directed toward satellites and reflected back to Earth for long-distance communication.

Key Concept: Faraday’s Law, Maxwell's Equations

a) $\frac{B_0 a^2}{\tau} e^{-t/\tau}$
[Solution Description]
The magnetic flux through the loop is:
$\Phi_B = B_0 e^{-t/\tau} \cdot a^2$
The induced emf is given by Faraday’s law:
$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt} \left( B_0 e^{-t/\tau} a^2 \right) = \frac{B_0 a^2}{\tau} e^{-t/\tau}$

Key Concept: Displacement Current

c) The effect of a changing electric field on the magnetic field
[Solution Description]
Displacement current is defined as:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$
It accounts for the effect of a changing electric field, even when no actual charge flow is present.

Key Concept: Energy Density Equality in Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the given syllabus, we know that for an electromagnetic wave:
$u_E = \frac{\epsilon_0 E_0^2}{2}$
and
$u_B = \frac{B_0^2}{2 \mu_0}.$
Given the relationship $E_0 = c B_0$, substituting this into $u_E$ gives:
$u_E = \frac{\epsilon_0 (c B_0)^2}{2} = \frac{\epsilon_0 c^2 B_0^2}{2}.$
Since $c^2 = \frac{1}{\mu_0 \epsilon_0}$, we can write:
$u_E = \frac{\epsilon_0 \left( \frac{1}{\mu_0 \epsilon_0} \right) B_0^2}{2} = \frac{B_0^2}{2 \mu_0} = u_B.$
Thus, the Assertion is true, and the Reason correctly explains why it is true.

Key Concept: Ampere's Law, Displacement Current

b) $B = \mu_0 n (I_c + \epsilon_0 A \frac{dE}{dt})$
[Solution Description]
According to Ampere-Maxwell law, both conduction current ($I_c$) and displacement current ($I_d$) contribute to the magnetic field.
For the solenoid, the magnetic field is given by:
$B = \mu_0 n I_{total}$, where $I_{total} = I_c + I_d$.
The displacement current is:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 A \frac{dE}{dt}$
Thus, the total magnetic field becomes:
$B = \mu_0 n (I_c + \epsilon_0 A \frac{dE}{dt})$

Key Concept: Energy Density Calculation, Relation Between Electric and Magnetic Fields

b) $3.18 \times 10^{-6} \, \text{J/m}^3$
[Solution Description]
First, calculate the electric field amplitude using $E_0 = c B_0$, where $c = 3 \times 10^8 \, \text{m/s}$.
$E_0 = (3 \times 10^8) \times (2 \times 10^{-6}) = 600 \, \text{V/m}$.
The total energy density is the sum of electric and magnetic energy densities:
$u = \frac{1}{2} \varepsilon_0 E_0^2 + \frac{1}{2} \frac{B_0^2}{\mu_0}$.
Substituting values:
$u = \frac{1}{2}(8.85 \times 10^{-12})(600)^2 + \frac{1}{2} \frac{(2 \times 10^{-6})^2}{(4\pi \times 10^{-7})}$.
Solving:
$u = 1.59 \times 10^{-6} + 1.59 \times 10^{-6} = 3.18 \times 10^{-6} \, \text{J/m}^3$.

Key Concept: Properties of Visible Light

b) Refraction
[Solution Description]
When light passes through a prism, it splits into its constituent colors due to refraction, where different wavelengths bend by different amounts. This is called dispersion, which is an outcome of refraction.

Key Concept: Wave propagation, Energy density relationship

c) 50\%
[Solution Description]
The instantaneous energy densities are $u_e = \frac{1}{2}\varepsilon_0 E^2$ and $u_m = \frac{B^2}{2\mu_0}$. Given $E = 0.6E_0$, the ratio $u_m/u_{total}$ can be calculated knowing $B = E/c$ and that in EM waves $c = 1/\sqrt{\mu_0\varepsilon_0}$. Substituting shows the energy is equally divided between electric and magnetic fields at all times, so when electric field is 60\% of peak, magnetic field is also 60\% of peak, maintaining equal energy distribution (50\% each).

Key Concept: Total Current, Boundary Consistency

b) $\frac{\mu_0 \epsilon_0 A \alpha E_0 e^{-\alpha t}}{2\pi r}$
[Solution Description]
The displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 A \frac{dE}{dt}$. For $E(t) = E_0 e^{-\alpha t}$, differentiate:
$\frac{dE}{dt} = -\alpha E_0 e^{-\alpha t}$
Substitute into $I_d$:
$I_d = -\epsilon_0 A \alpha E_0 e^{-\alpha t}$
Applying Ampere-Maxwell Law:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_d \Rightarrow 2\pi r B = \mu_0 \left( -\epsilon_0 A \alpha E_0 e^{-\alpha t} \right)$
Solve for $B$:
$B = -\frac{\mu_0 \epsilon_0 A \alpha E_0 e^{-\alpha t}}{2\pi r}$
Magnitude is:
$|B| = \frac{\mu_0 \epsilon_0 A \alpha E_0 e^{-\alpha t}}{2\pi r}$

Key Concept: Displacement Current

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a capacitor is charging, the electric field between its plates changes with time. This changing electric flux induces a displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux. According to the Ampere-Maxwell law, this displacement current contributes to the magnetic field just like conduction current. Hence, even though there is no conduction current inside the capacitor, the displacement current ensures that the magnetic field is non-zero. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Velocity in a Medium, Wave Parameters

b) $0.5$
[Solution Description]
The speed of an electromagnetic wave in a medium is given by:
$v = \frac{1}{\sqrt{\mu \varepsilon}}$
For free space, the speed is:
$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
The ratio is:
$\frac{v}{c} = \frac{\sqrt{\mu_0 \varepsilon_0}}{\sqrt{\mu \varepsilon}} = \frac{1}{\sqrt{\mu_r \varepsilon_r}}$
Substituting $\mu_r = 1$ and $\varepsilon_r = 4$:
$\frac{v}{c} = \frac{1}{\sqrt{4}} = 0.5$

Key Concept: Wave Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The speed of an electromagnetic wave in free space is derived from Maxwell's equations and is indeed given by $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, where $\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space. This formula shows that the speed of the wave is determined by these two fundamental constants, which are properties of free space. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Infrared radiation, medical uses, energy calculation

b) $1.989 \times 10^{-20} \, \text{J}$
[Solution Description]
The energy of a photon is given by $E = \frac{hc}{\lambda}$. Substituting $\lambda = 1 \times 10^{-5} \, \text{m}$, $h = 6.63 \times 10^{-34} \, \text{Js}$, and $c = 3 \times 10^8 \, \text{m/s}$:
$E = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{1 \times 10^{-5}} = 1.989 \times 10^{-20} \, \text{J}$.
This matches the infrared range ($8 \times 10^{-7} \, \text{m}$ to $5 \times 10^{-3} \, \text{m}$), where such radiation is used for therapeutic warming.

Key Concept: Sinusoidal Wave, Energy Density Calculation

b) 1
[Solution Description]
The maximum electric energy density is:
$u_{e,\text{max}} = \frac{1}{2} \epsilon_0 E_0^2$
The maximum magnetic energy density is:
$u_{m,\text{max}} = \frac{1}{2 \mu_0} B_0^2$
For an electromagnetic wave in vacuum, $E_0 = c B_0$. Thus:
$u_{m,\text{max}} = \frac{1}{2 \mu_0} \left( \frac{E_0}{c} \right)^2$
Taking the ratio:
$\frac{u_{m,\text{max}}}{u_{e,\text{max}}} = \frac{\frac{1}{2 \mu_0} \left( \frac{E_0}{c} \right)^2}{\frac{1}{2} \epsilon_0 E_0^2} = \frac{1}{\mu_0 \epsilon_0 c^2}$
Since $\frac{1}{\mu_0 \epsilon_0} = c^2$, the ratio simplifies to 1.

Key Concept: Displacement Current in Capacitors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a charging capacitor, the conduction current ($I_c$) flows through the wires while the displacement current ($I_d$) flows between the plates due to the changing electric field. According to Maxwell's modification of Ampère's law, $I_d = I_c$ to ensure continuity of current. The reason correctly explains that $I_d$ arises from $\epsilon_0 \frac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux. Thus, both assertion and reason are true, and the reason is the correct explanation for the assertion.

Key Concept: Ampere-Maxwell’s Circuital Law

b) $2.26 \times 10^5 Nm^2/C \cdot s$
[Solution Description]
The displacement current $I_d$ is related to the rate of change of electric flux $\frac{d\Phi_E}{dt}$ by:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$
Rearranging for $\frac{d\Phi_E}{dt}$:
$\frac{d\Phi_E}{dt} = \frac{I_d}{\epsilon_0} = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}} = 2.26 \times 10^5 Nm^2/C \cdot s$

Key Concept: Displacement Current

c) Equal to the conduction current $I$
[Solution Description]
The displacement current $I_d$ is given by:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$
For a parallel plate capacitor, the electric flux $\Phi_E = E \cdot A$. Since $E = \frac{q}{\epsilon_0 A}$, substituting gives:
$\Phi_E = \frac{q}{\epsilon_0 A} \cdot A = \frac{q}{\epsilon_0}$
Differentiating with respect to time:
$I_d = \epsilon_0 \frac{d}{dt}\left( \frac{q}{\epsilon_0} \right) = \frac{dq}{dt} = I$
Thus, the displacement current equals the conduction current $I$.

Key Concept: Energy density relations, peak fields

c) Becomes four times
[Solution Description]
Original ratio: $\frac{u_m}{u_e} = \frac{\frac{B_0^2}{2\mu_0}}{\frac{\varepsilon_0 E_0^2}{2}} = \frac{B_0^2}{\mu_0 \varepsilon_0 E_0^2}$
Using $c = \frac{E_0}{B_0}$ and $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, the ratio simplifies to 1 initially.
When $B_0' = 2B_0$ and $E_0$ remains same, new ratio:
$\frac{u_m'}{u_e} = \frac{(2B_0)^2}{B_0^2} = 4$ (since other terms cancel out)

Key Concept: Production of Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that an accelerating charge always produces electromagnetic waves, which is true because any acceleration of charged particles results in changing electric and magnetic fields, as described by Maxwell's equations. The reason correctly explains that according to Maxwell's equations, a changing electric field produces a magnetic field and vice versa, which leads to the propagation of electromagnetic waves. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Gamma Rays Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) is true because gamma rays indeed have the shortest wavelengths ($1 \times 10^{-14}$ to $1 \times 10^{-10}$ m) and highest frequencies ($3 \times 10^{22}$ to $3 \times 10^{18}$ Hz) in the electromagnetic spectrum. The reason (R) is also true as gamma rays are produced by nuclear transitions and particle decays. Furthermore, the reason correctly explains why gamma rays possess such high energy and frequency, as these processes involve significant energy changes at the atomic/nuclear level.

Key Concept: Average Energy Density in Electromagnetic Waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the correct formula for the average energy density of an electromagnetic wave in terms of rms values of the fields, which is derived from the time-averaged energy densities of the electric and magnetic fields.
The reason correctly explains that in an electromagnetic wave propagating in vacuum, the electric and magnetic field energies are equal. This is because the strengths of the electric and magnetic fields are related by $E_0 = cB_0$ and their contributions to the energy density are balanced ($u_e = u_m$).
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Energy of a photon

b) $2.65 \times 10^{-18}$ J
[Solution Description]
The energy of a photon is given by $E = h \nu$, where $h$ is Planck's constant and $\nu$ is the frequency. Plugging in the values:
$E = (6.626 \times 10^{-34}) \times (4 \times 10^{15}) = 2.6504 \times 10^{-18} \, \text{J}$
The correct energy is $2.65 \times 10^{-18}$ J.

Key Concept: Transverse Nature of Electromagnetic Waves

d) Neither $\overrightarrow{E}$ nor $\overrightarrow{B}$ has a component along the X-axis
[Solution Description]
From Maxwell's equations and Gauss' law for electricity, we know that in a transverse electromagnetic wave, $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular to the direction of propagation (X-axis). The derivation shows $E_x = 0$ and $B_x = 0$, meaning neither field has a component along the X-axis.

Key Concept: Speed of electromagnetic waves in vacuum

c) $3 \times 10^8 \, \text{m s}^{-1}$
[Solution Description]
All electromagnetic waves travel with the same speed in vacuum, which is given by $3 \times 10^8 \, \text{m s}^{-1}$. This is also the speed of light.

Key Concept: Speed of Electromagnetic Waves

b) $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$
[Solution Description]
The speed of electromagnetic waves in free space is given by the relation:
$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$
where $\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space.

Key Concept: Electric Flux and Displacement Current