Samacheer Kalvi Class 12 Physics Solutions Chapter 5 Electromagnetic Waves

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 5 Electromagnetic Waves

12th Physics Guide Electromagnetic Waves Text Book Back Questions and Answers

Part I:

Text Book Evaluation:

I. Multiple Choice Questions:

 

Question 1. The dimension of \( \frac { 1 }{ \mu_{0} \varepsilon_{0} } \)
(a) [LT-1]
(b) [L2 T-2]
(c) [L-1T]
(d) [L-2 T2]
Answer: (b) [L2 T-2]
Solution:
Dimension of \( \mu_{0} = \text{MLT}^{-2}\text{A}^{-2} \)
Dimension of \( \varepsilon_{0} = \text{M}^{-1}\text{L}^{-3} \text{T}^{4}\text{A}^{2} \)
\( \implies \) Dimension of \( \frac { 1 }{ \mu_{0} \varepsilon_{0} } = \frac { 1 }{ \text{MLT}^{-2}\text{A}^{-2} \text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2} } \)
\( \implies \frac { 1 }{ \mu_{0} \varepsilon_{0} } = \frac { 1 }{ \text{L}^{-2}\text{T}^{2} } = \text{L}^{2}\text{T}^{-2} \)
In simple words: The expression \( \frac { 1 }{ \mu_{0} \varepsilon_{0} } \) represents the square of the speed of light. Since speed has dimensions of length divided by time ([LT⁻¹]), the square of speed will have dimensions of [L²T⁻²].

🎯 Exam Tip: Remember that \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \), where c is the speed of light. This directly helps to determine the dimension of the given expression.

 

Question 2. If the amplitude of the magnetic field is \( 3 \times 10^{-6} \) T, then amplitude of the electric field for a electromagnetic waves is
(a) 100 Vm-1
(b) 300 Vm-1
(c) 600 Vm-1
(d) 900 Vm-1
Answer: (d) 900 Vm-1
Solution:
Velocity of electromagnetic wave \( \text{C} = \frac { \text{E}_{0} }{ \text{B}_{0} } \)
\( \implies \text{E}_{0} = \text{C} \times \text{B}_{0} \)
\( \implies \text{E}_{0} = 3 \times 10^{8} \times 3 \times 10^{-6} \)
\( \implies \text{E}_{0} = 900 \text{ Vm}^{-1} \)
In simple words: To find the strength of the electric field, you multiply the speed of light by the given strength of the magnetic field. This formula is a key relationship in electromagnetic waves.

🎯 Exam Tip: Always use the speed of light in vacuum (c = \( 3 \times 10^8 \text{ m/s} \)) for calculations involving electromagnetic waves unless specified otherwise. Make sure to use consistent units.

 

Question 3. Which electromagnetic radiation is used for viewing objects through fog?
(a) Micro wave
(b) Gamma rays
(c) X-rays
(d) Infrared
Answer: (d) Infrared
Solution:
Infrared rays can pass through mist, fog, cloud, etc., because they have longer wavelengths compared to visible light, which helps them penetrate through scattering particles more effectively.
In simple words: Infrared light can see through fog because its waves are long enough to pass around tiny water droplets. Other types of light waves are too short and get scattered away.

🎯 Exam Tip: When considering wave penetration through obstacles, remember that waves with longer wavelengths tend to diffract more and scatter less, allowing them to pass through certain media more easily.

 

Question 4. Which of the following is false for electromagnetic waves?
(a) transverse
(b) mechanical waves
(c) longitudinal
(d) produced by accelerating charges
Answer: (c) longitudinal
Solution:
Electromagnetic waves do not need any medium for their propagation. They are transverse waves and are produced by accelerating charges. Therefore, they are non-mechanical and always travel perpendicular to the oscillations of the electric and magnetic fields.
In simple words: Electromagnetic waves are always transverse, meaning they wiggle side-to-side, not back-and-forth like longitudinal waves. They also don't need anything to travel through.

🎯 Exam Tip: Understand the key characteristics of electromagnetic waves: they are transverse, propagate in a vacuum at the speed of light, and consist of oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation.

 

Question 5. Consider an oscillator which has a charged particle oscillating about its mean position with a frequency of 300 MHz. The wavelength of electromagnetic waves produced by this oscillator is
(a) 1 m
(b) 10 m
(c) 100 m
Answer: (a) 1 m
Solution:
Given frequency \( \text{f} = 300 \text{ MHz} = 3 \times 10^8 \text{ Hz} \).
Velocity of electromagnetic wave \( \text{C} = 3 \times 10^8 \text{ m/s} \).
The formula relating speed, frequency, and wavelength is \( \text{C} = \text{f} \lambda \).
\( \implies \lambda = \frac { \text{C} }{ \text{f} } \)
\( \implies \lambda = \frac { 3 \times 10^8 }{ 3 \times 10^8 } \)
\( \implies \lambda = 1 \text{ m} \)
In simple words: To find the wavelength, you divide the speed of light by the frequency of the oscillator. This gives you the length of one complete wave.

🎯 Exam Tip: Remember the fundamental wave equation \( c = f\lambda \). This formula is essential for calculating any of the three quantities if the other two are known for electromagnetic waves.

 

Question 6. The electric and the magnetic fields, associated with an electromagnetic wave propagating along negative x axis can be represented by
(a) \( \overrightarrow{\mathrm{E}} = \text{E}_{0} \hat{i} \) and \( \vec{B} = \text{B}_{0} \hat{k} \)
(b) \( \overrightarrow{\mathrm{E}} = \text{E}_{0} \hat{k} \) and \( \vec{B} = \text{B}_{0} \hat{j} \)
(c) \( \overrightarrow{\mathrm{E}} = \text{E}_{0} \hat{i} \) and \( \vec{B} = \text{B}_{0} \hat{i} \)
(d) \( \overrightarrow{\mathrm{E}} = \text{E}_{0} \hat{j} \) and \( \vec{B} = \text{B}_{0} \hat{i} \)
Answer: (d) \( \overrightarrow{\mathrm{E}} = \text{E}_{0} \hat{j} \) and \( \vec{B} = \text{B}_{0} \hat{i} \)
Solution:
For an electromagnetic wave, the electric field \( \overrightarrow{\mathrm{E}} \), magnetic field \( \vec{B} \), and direction of propagation \( \vec{v} \) are mutually perpendicular. The direction of propagation is given by the cross product \( \vec{v} \propto \overrightarrow{\mathrm{E}} \times \vec{B} \).
Given that the wave propagates along the negative x-axis ( \( -\hat{i} \) direction).
Let's check option (d): \( \overrightarrow{\mathrm{E}} = \text{E}_{0} \hat{j} \) and \( \vec{B} = \text{B}_{0} \hat{i} \).
Then \( \overrightarrow{\mathrm{E}} \times \vec{B} = (\text{E}_{0} \hat{j}) \times (\text{B}_{0} \hat{i}) = \text{E}_{0}\text{B}_{0} (\hat{j} \times \hat{i}) = \text{E}_{0}\text{B}_{0} (-\hat{k}) \).
This indicates propagation along the negative z-axis, which is inconsistent with the question asking for negative x-axis propagation. However, as per instructions, we follow the provided answer.
In simple words: For an electromagnetic wave, the electric and magnetic fields are always at right angles to each other, and both are at right angles to the direction the wave travels. This makes them perpendicular to each other.

🎯 Exam Tip: Remember the right-hand rule for the direction of wave propagation: \( \vec{k} \propto \vec{E} \times \vec{B} \), where \( \vec{k} \) is the propagation vector. This ensures consistency between field orientations and propagation direction.

 

Question 7. In an electromagnetic wave in free space the rms value of the electric field is 3 Vm-1. The peak value of the magnetic field is
(b) \( 1.0 \times 10^{-8} \) T
(c) \( 2.828 \times 10^{-8} \) T
(d) \( 2.0 \times 10^{-8} \) T
Answer: (a) \( 1.414 \times 10^{-8} \) T
Solution:
Given RMS value of electric field \( \text{E}_{\text{rms}} = 3 \text{ Vm}^{-1} \).
The peak value of the electric field \( \text{E}_{0} = \text{E}_{\text{rms}} \times \sqrt{2} = 3\sqrt{2} \text{ Vm}^{-1} \).
The relation between peak electric field, peak magnetic field, and speed of light is \( \text{E}_{0} = \text{C} \times \text{B}_{0} \).
\( \implies \text{B}_{0} = \frac { \text{E}_{0} }{ \text{C} } \)
\( \implies \text{B}_{0} = \frac { 3\sqrt{2} }{ 3 \times 10^8 } \)
\( \implies \text{B}_{0} = \sqrt{2} \times 10^{-8} \)
\( \implies \text{B}_{0} \approx 1.414 \times 10^{-8} \text{ T} \)
In simple words: First, convert the RMS electric field to the peak electric field by multiplying by the square root of 2. Then, divide this peak electric field by the speed of light to find the peak magnetic field.

🎯 Exam Tip: Always distinguish between RMS and peak values for fields in electromagnetic waves. Remember that \( \text{E}_0 = \text{E}_{\text{rms}}\sqrt{2} \) and \( \text{B}_0 = \text{B}_{\text{rms}}\sqrt{2} \).

 

Question 8. An electromagnetic wave is propagating in a medium with a velocity \( \overrightarrow{\mathrm{v}} = \text{v}\hat{i} \). The instantaneous oscillating electric field of this e.m. wave is along +y-axis, then the direction of oscillating magnetic field of the e.m. wave will be along.
(a) - y direction
(b) - x direction
(c) + z direction
(d) - z direction
Answer: (c) + z direction
In simple words: If the wave moves along the x-axis and the electric field wiggles along the y-axis, then the magnetic field must wiggle along the z-axis because all three are at right angles to each other.

🎯 Exam Tip: Use the vector cross product relationship \( \vec{v} \propto \vec{E} \times \vec{B} \). If \( \vec{v} \) is in \( \hat{i} \) direction and \( \vec{E} \) is in \( \hat{j} \) direction, then \( \hat{i} \propto \hat{j} \times \vec{B} \). This implies \( \vec{B} \) must be in the \( \hat{k} \) (z-axis) direction, as \( \hat{j} \times \hat{k} = \hat{i} \).

 

Question 9. If the magnetic monopole exists, then which of the Maxwell's equation to be modified?
(a) \( \oint_{\text{S}} \vec{E} . d\vec{A} = \frac { \text{Q}_{\text{enclosed}} }{ \varepsilon_{0} } \)
(b) \( \oint_{\text{S}} \vec{B} . d\vec{A} = 0 \)
(d) \( \oint_{\text{L}} \vec{E} . d\vec{l} = - \frac { d }{ d t } \Phi_{B} \)
Answer: (b) \( \oint_{\text{S}} \vec{B} . d\vec{A} = 0 \)
Solution:
Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. This implies that magnetic monopoles (isolated north or south poles) do not exist. If magnetic monopoles were to exist, this equation would need to be modified to include a magnetic charge term, similar to how electric charge appears in Gauss's law for electricity. This ensures consistency with the existence of isolated magnetic charges.
In simple words: If magnetic monopoles, which are like single north or south poles, were found, then the rule that says magnetic lines always form closed loops would have to change. This specific rule, Gauss's law for magnetism, would need to be updated.

🎯 Exam Tip: Maxwell's equations describe the behavior of electric and magnetic fields. Understand what each equation implies and how the existence of magnetic monopoles would specifically affect Gauss's law for magnetism.

 

Question 10. Fraunhofer lines are an example of ............ spectrum.
(a) line emission
(b) line absorption
(c) band emission
(d) band absorption
Answer: (b) line absorption
In simple words: Fraunhofer lines are dark lines seen in the sun's spectrum. They are formed when cooler gases in the sun's outer atmosphere absorb specific colors (wavelengths) of light that are coming from the hotter inner part of the sun.

🎯 Exam Tip: Remember that Fraunhofer lines are a key piece of evidence for the composition of the Sun's atmosphere and are a classic example of absorption spectra.

 

Question 11. Which of the following is an electromagnetic wave?
(a) \( \alpha \) – rays
(b) \( \beta \) – rays
(c) \( \gamma \) – rays
(d) all of them
Answer: (c) \( \gamma \) – rays
Solution:
\( \alpha \) – rays are Helium Nuclei (\( \text{}_{2}\text{He}^{4} \)).
\( \beta \) – rays are Electrons (\( \text{e}^{-1} \)).
\( \gamma \) – rays are Quantum of electromagnetic energy. This means gamma rays are a form of electromagnetic radiation, unlike alpha and beta rays which are particles.
In simple words: Gamma rays are a type of electromagnetic wave, just like radio waves or visible light, but with much higher energy. Alpha and beta rays, however, are tiny particles, not waves.

🎯 Exam Tip: Differentiate between particle radiations (like alpha and beta rays) and electromagnetic radiations (like gamma rays, X-rays, UV, visible light, IR, microwave, radio waves). Electromagnetic waves are packets of energy (photons).

 

Question 12. Which one of them is used to produce a propagating electromagnetic wave?
(a) an accelerating charge
(b) a charge moving with constant velocity
(c) a stationary charge
(d) an uncharged particle
Answer: (a) an accelerating charge
Solution:
A stationary charge produces only a static electric field. A charge moving with constant velocity produces both a static electric field and a constant magnetic field, but no electromagnetic wave. Only an accelerating charge creates changing electric and magnetic fields that propagate as an electromagnetic wave. The acceleration causes continuous changes in the fields, leading to their propagation.
In simple words: An electromagnetic wave is made when an electric charge speeds up or slows down. If the charge just sits still or moves at a steady speed, it won't make these waves.

🎯 Exam Tip: A fundamental principle of electromagnetism is that accelerating charges are the source of electromagnetic radiation. This is a common conceptual question.

 

Question 13. Let \( \text{E} = \text{E}_{0} \sin [10^6\text{x} - \omega\text{t}] \) be the electric field of plane electromagnetic wave, the value of \( \omega \) is
(a) \( 0.3 \times 10^{-4} \text{ rad s}^{-1} \)
(b) \( 3 \times 10^{-14} \text{ rad s}^{-1} \)
(c) \( 0.3 \times 10^{14} \text{ rad s}^{-1} \)
(d) \( 3 \times 10^{14} \text{ rad s}^{-1} \)
Answer: (d) \( 3 \times 10^{14} \text{ rad s}^{-1} \)
Solution:
The given electric field equation is \( \text{E} = \text{E}_{0} \sin [10^6\text{x} - \omega\text{t}] \).
The standard form of an electromagnetic wave equation is \( \text{E} = \text{E}_{0} \sin (\text{kx} - \omega\text{t}) \).
Comparing the given equation with the standard form, we get the wave number \( \text{k} = 10^6 \text{ rad m}^{-1} \).
We know that the speed of light \( \text{c} = \frac { \omega }{ \text{k} } \).
\( \implies \omega = \text{ck} \)
\( \implies \omega = (3 \times 10^8 \text{ m/s}) \times (10^6 \text{ rad m}^{-1}) \)
\( \implies \omega = 3 \times 10^{14} \text{ rad s}^{-1} \)
Therefore, the angular frequency \( \omega \) is \( 3 \times 10^{14} \text{ rad s}^{-1} \). This value represents how fast the wave oscillates in time.
In simple words: The number next to 'x' in the wave equation is the wave number (k). We know that speed of light (c) is equal to omega (angular frequency) divided by k. So, we multiply the speed of light by the wave number to find omega.

🎯 Exam Tip: For wave equations in the form \( E = E_0 \sin(kx - \omega t) \), remember that \( k \) is the wave number and \( \omega \) is the angular frequency. The speed of the wave \( c = \omega / k \).

 

Question 14. Which of the following is NOT true for electromagnetic waves?
(a) it transports energy
(b) it transports momentum
(c) it transports angular momentum
(d) in a vacuum, it travels at different speeds which depend on its frequency
Answer: (d) in a vacuum, it travels at different speeds which depend on its frequency
Solution:
Reason:
All electromagnetic waves with different frequencies travel with the same velocity, \( 3 \times 10^8 \text{ ms}^{-1} \), in a vacuum. This is a fundamental property of light. However, in a medium, their speed varies depending on their wavelength (and thus frequency), leading to phenomena like dispersion. The constant speed in vacuum is why option (d) is false.
In simple words: All electromagnetic waves, no matter their color or type, always move at the same super-fast speed in empty space. Only when they pass through materials does their speed change depending on their frequency.

🎯 Exam Tip: A crucial concept for electromagnetic waves is that their speed in a vacuum is a universal constant (c). Any statement suggesting different speeds for different frequencies in a vacuum is incorrect.

 

Question 15. The electric and magnetic fields of an electromagnetic wave are
(c) in phase and perpendicular to each other
(d) out of phase and not perpendicular to each other
Answer: (a) in phase and perpendicular to each other
Solution:
In an electromagnetic wave, such as \( \text{E}_{y} = \text{E}_{0} \sin (\text{ky} - \omega\text{t}) \) and \( \text{B}_{z} = \text{B}_{0} \sin (\text{kz} - \omega\text{t}) \), the electric and magnetic components oscillate in phase. This means they reach their maximum and minimum values at the same time and in the same place. Additionally, they are always perpendicular to each other and to the direction of propagation, with a phase difference \( \Delta\Phi = 0 \).
In simple words: The electric field and the magnetic field in an electromagnetic wave always swing up and down together, at the same time, and they are always at right angles to each other.

🎯 Exam Tip: Visualize the electric and magnetic fields in an electromagnetic wave as oscillating perpendicularly to each other and to the wave's direction of travel. Crucially, they oscillate *in phase*, meaning their peaks and troughs align.

 

II. Short Questions and Answers:

 

Question 1. What is displacement current?
Answer: The displacement current is a type of current that appears in regions where the electric field and electric flux are changing over time. It is not a flow of actual charges but rather a conceptual current introduced by Maxwell to ensure Ampere's law holds true in all situations, especially in circuits with capacitors. This current helps to explain the propagation of electromagnetic waves.
In simple words: Displacement current is like a pretend current that flows when the electric field is changing. It helps explain how electric and magnetic fields work together.

🎯 Exam Tip: Define displacement current as a current that arises due to a changing electric flux, and explain its importance in Maxwell's modification of Ampere's circuital law.

 

Question 2. What are electromagnetic waves?
Answer:
1. Electromagnetic waves are special types of waves that do not need any material (like air or water) to travel through; they can move through empty space. They are also known as non-mechanical waves. They travel at the speed of light in a vacuum.
2. They are transverse waves, meaning the oscillations of the electric and magnetic fields are perpendicular to the direction the wave is moving. For example, if the wave moves forward, the fields wiggle up-and-down and side-to-side.
In simple words: Electromagnetic waves are waves that carry energy through space without needing a medium. They are transverse, meaning they wiggle sideways, and they travel at the speed of light.

🎯 Exam Tip: When defining electromagnetic waves, highlight their key features: they are non-mechanical, transverse, and travel at the speed of light in a vacuum.

 

Question 3. Write down the integral form of modified Ampere's circuital law.
Answer: The integral form of Maxwell's modified Ampere's circuital law is:
\( \oint_{\text{L}} \vec{B} . d\vec{l} = \mu_{0}(\text{I}_{\text{enclosed}} + \text{I}_{\text{d}}) \)
\( \implies \oint_{\text{L}} \vec{B} . d\vec{l} = \mu_{0}\text{I}_{\text{c}} + \mu_{0}\varepsilon_{0} \frac { d\Phi_{E} }{ d t } \)
This law relates the magnetic field around any closed path to the sum of the conduction current (\( \text{I}_{\text{c}} \)) and the displacement current (\( \text{I}_{\text{d}} \)) passing through the surface bounded by that path. The displacement current term, \( \mu_{0}\varepsilon_{0} \frac { d\Phi_{E} }{ d t } \), was added by Maxwell to account for changing electric fields.
In simple words: This law says that the magnetic field around a loop is created not just by electric currents flowing through wires, but also by electric fields that are changing.

🎯 Exam Tip: Clearly state both terms in the modified Ampere's law: the conduction current and the displacement current, and mention \( \mu_0 \) and \( \varepsilon_0 \) as permeability and permittivity of free space.

 

Question 4. Write notes an Gauss' law in magnetism.
Answer: Gauss's law for magnetism states that the surface integral of the magnetic field over any closed surface is always zero. Mathematically, it is written as:
\( \oint_{\text{S}} \vec{B} . d\vec{A} = 0 \)
Where \( \vec{B} \) is the magnetic field and \( d\vec{A} \) is an infinitesimal area vector. This law implies that magnetic field lines are always continuous and form closed loops, never starting or ending at a point. This means that isolated magnetic poles (magnetic monopoles) do not exist, as any magnetic pole would always be part of a dipole. The total magnetic flux coming out of a closed surface must be equal to the total magnetic flux going into it.
In simple words: This law means you can never find a single magnetic north pole or south pole on its own. Magnetic lines always loop back, so if you draw a box around a magnet, the amount of magnetic field going in will always equal the amount coming out.

🎯 Exam Tip: State the mathematical form of Gauss's law for magnetism and clearly explain its physical implication: the non-existence of magnetic monopoles.

 

Question 5. Give two uses for each of the following.
(i) IR radiation
(ii) Microwaves and
(iii) UV radiation.
Answer:
(i) IR radiation:
1. Infrared radiation is used in remote controls for televisions and other electronic devices to send signals.
2. It is also used in night vision cameras because warm objects emit infrared radiation, allowing viewing in low light.
(ii) Microwaves:
1. Microwaves are used in radar systems for aircraft navigation and speed detection. They are very precise.
2. They are used in microwave ovens to heat food quickly by making water molecules vibrate.
(iii) UV radiation:
1. Ultraviolet radiation is used to destroy bacteria and sterilize medical equipment because it damages DNA.
2. It is also used in water purification systems to kill harmful microorganisms.
In simple words: Infrared is for TV remotes and seeing at night. Microwaves are for radar and cooking food. UV light is used to kill germs and clean water.

🎯 Exam Tip: Provide at least two distinct and specific applications for each type of electromagnetic radiation, demonstrating a clear understanding of their properties.

 

Question 6. What are Fraunhofer lines? How are they useful in the identification elements present in the sun?
Answer: When the spectrum of light obtained from the Sun is carefully examined, it shows a large number of dark lines within its continuous range of colors. These dark lines are known as Fraunhofer lines. They represent a line absorption spectrum. These lines are formed when the cooler gases in the Sun's outer atmosphere absorb specific wavelengths of light emitted from the hotter inner core. By comparing the wavelengths of these dark lines with the known absorption spectra of various elements on Earth, scientists can identify the elements present in the Sun's atmosphere. Each element has a unique set of absorption lines, acting like a fingerprint.
In simple words: Fraunhofer lines are dark lines in the sun's light spectrum. They tell us what elements are in the sun's outer layer, because each element absorbs certain colors of light.

🎯 Exam Tip: Explain that Fraunhofer lines are absorption lines, and their specific wavelengths act as "fingerprints" to identify elements present in the Sun's atmosphere.

 

Question 7. Write notes on Ampere – Maxwell law.
Answer: Maxwell modified Ampere's circuital law to account for changing electric fields, which produce magnetic fields, similar to how changing magnetic fields produce electric fields (Faraday's law). The original Ampere's law was incomplete when applied to charging capacitors. Maxwell's modified Ampere's law states:
\( \oint \vec{B} . d\vec{l} = \mu_{0}\text{I}_{\text{c}} + \mu_{0}\varepsilon_{0} \frac { d\Phi_{E} }{ d t } \)
Here, \( \text{I}_{\text{c}} \) is the conduction current (actual flow of charge) and \( \mu_{0}\varepsilon_{0} \frac { d\Phi_{E} }{ d t } \) is the displacement current (\( \text{I}_{\text{d}} \)), which arises from a changing electric flux (\( \Phi_{E} \)). This modification ensured that the law is consistent with charge conservation and explained the existence of electromagnetic waves. The total current enclosed by the surface becomes the sum of conduction current and displacement current.
In simple words: Ampere-Maxwell's law explains that magnetic fields are made by two things: actual electric currents and changing electric fields. Maxwell added the changing electric field part to make the law complete.

🎯 Exam Tip: Clearly state Maxwell's modification to Ampere's law, including both conduction and displacement current terms. Emphasize that this modification was crucial for the theoretical prediction of electromagnetic waves.

 

Question 8. Why are e.m waves non-mechanical?
Answer: Electromagnetic (e.m.) waves are considered non-mechanical because they do not require a material medium (like air, water, or solids) for their propagation. Unlike mechanical waves (such as sound waves or water waves), which need particles to vibrate and transmit energy, e.m. waves can travel through a vacuum, such as outer space. This ability to propagate without a medium is due to the self-sustaining oscillations of electric and magnetic fields that constitute the wave. These fields generate each other as they propagate.
In simple words: Electromagnetic waves are called "non-mechanical" because they don't need anything to travel through. They can go through empty space, unlike sound waves that need air or water to move.

🎯 Exam Tip: The defining characteristic of non-mechanical waves is their ability to travel through a vacuum. Mentioning this and contrasting it with mechanical waves will earn full marks.

 

III. Long Questions and Answers:

 

Question 1. Write down Maxwell equations in integral form.
Answer: Electrodynamics, the study of electric and magnetic phenomena, is concisely summarized by four fundamental equations, collectively known as Maxwell's Equations. These equations describe how electric and magnetic fields are generated and interact.
1. **Gauss's Law for Electricity:** This equation relates the net electric flux through any closed surface to the net electric charge enclosed within that surface. It states that electric fields originate from electric charges.
\( \oint_{\text{S}} \vec{E} . d\vec{A} = \frac { Q_{\text{enclosed}} }{ \varepsilon_{0} } \)
2. **Gauss's Law for Magnetism:** This law states that the net magnetic flux through any closed surface is always zero. This implies that magnetic field lines are continuous and form closed loops, meaning isolated magnetic poles (monopoles) do not exist.
\( \oint_{\text{S}} \vec{B} . d\vec{A} = 0 \)
3. **Faraday's Law of Electromagnetic Induction:** This law describes how a changing magnetic flux through a surface induces an electromotive force (and thus an electric field) around a closed loop bounding that surface. This is the principle behind generators.
\( \oint_{\text{L}} \vec{E} . d\vec{l} = - \frac { d }{ d t } \Phi_{B} \)
4. **Ampere-Maxwell Law:** This is the modified form of Ampere's circuital law, which relates the magnetic field around a closed loop to both the conduction current passing through the surface and the rate of change of electric flux through that surface. The inclusion of the displacement current term, \( \mu_{0}\varepsilon_{0} \frac { d\Phi_{E} }{ d t } \), was a crucial addition by Maxwell, ensuring consistency and predicting electromagnetic waves.
\( \oint_{\text{L}} \vec{B} . d\vec{l} = \mu_{0}\text{I}_{\text{enclosed}} + \mu_{0}\varepsilon_{0} \frac { d }{ d t } \int_{\text{S}} \vec{E} . d\vec{A} \)
These four integral equations together form the foundation of classical electromagnetism, explaining everything from static fields to electromagnetic radiation.
In simple words: Maxwell's equations are four main rules for electricity and magnetism. They tell us how electric charges make electric fields, why magnetic poles always come in pairs, how changing magnetism makes electricity, and how both electric current and changing electric fields make magnetism.

🎯 Exam Tip: List each of Maxwell's four equations by name and provide their integral form accurately. Briefly explain the physical meaning of each equation to show a comprehensive understanding.

 

Question 2. Explain the types of electromagnetic waves.
Answer: Electromagnetic waves form a spectrum of various types, all traveling at the speed of light in a vacuum but differing in wavelength and frequency. Here are some key types:
**Microwaves:**
Microwaves are produced by special electronic oscillators in electric circuits. Their wavelengths range from about 1 millimeter (\( 1 \times 10^{-3} \text{ m} \)) to 30 centimeters (\( 3 \times 10^{-1} \text{ m} \)). They have frequencies from \( 3 \times 10^{11} \text{ Hz} \) down to \( 1 \times 10^{9} \text{ Hz} \). Microwaves exhibit reflection and polarization. They are widely used in radar systems for aircraft navigation and speed detection, in microwave ovens for cooking, and for long-distance wireless communication via satellites. They help us warm our food and navigate safely.
**X-rays:**
X-rays are created when high-speed electrons suddenly decelerate as they hit a target material, often one with a high atomic number. They are also produced during electronic transitions of inner-shell electrons within atoms. X-rays have very short wavelengths, typically ranging from \( 10^{-13} \text{ m} \) to \( 10^{-8} \text{ m} \), and high frequencies from \( 3 \times 10^{21} \text{ Hz} \) to \( 1 \times 10^{16} \text{ Hz} \). They possess high penetrating power and are used extensively to study crystal structures, detect fractures in bones, and identify flaws in metal products. They are essential tools in medicine and material science.
**Radio Waves:**
Radio waves are produced by oscillating charges in electric circuits. They have the longest wavelengths in the electromagnetic spectrum, ranging from about 10 centimeters (\( 1 \times 10^{-1} \text{ m} \)) to tens of thousands of meters (\( 1 \times 10^4 \text{ m} \)). Their frequencies span from \( 3 \times 10^{9} \text{ Hz} \) down to \( 3 \times 10^{4} \text{ Hz} \). Radio waves obey laws of reflection and diffraction. They are used for radio and television communication, and in cellular phones for transmitting voice communication across different frequency bands. They enable global wireless communication.
**Visible Light:**
Visible light is emitted by incandescent bodies (like hot filaments) and by excited atoms in gases. Its wavelength range is very narrow, from about 400 nanometers (\( 4 \times 10^{-7} \text{ m} \)) to 700 nanometers (\( 7 \times 10^{-7} \text{ m} \)). The corresponding frequency range is \( 7 \times 10^{14} \text{ Hz} \) to \( 4 \times 10^{14} \text{ Hz} \). Visible light interacts with matter through reflection, refraction, interference, diffraction, polarization, and the photoelectric effect, and it enables our sense of sight. It allows us to perceive the world around us.
In simple words: The electromagnetic spectrum includes different waves like microwaves (for cooking and radar), X-rays (for seeing bones), radio waves (for communication), and visible light (what we see). They all travel fast but have different lengths and energies.

🎯 Exam Tip: For each type of wave, provide its source, typical wavelength/frequency range, and at least two distinct practical applications. A good answer includes a brief explanation of how each wave interacts with matter.

 

Question 3. Discuss the Hertz experiment.
Answer: Heinrich Hertz's experiment, conducted in 1888, was the first experimental confirmation of the existence of electromagnetic waves, which Maxwell had previously predicted theoretically. This pivotal experiment validated Maxwell's theory and demonstrated that light is indeed an electromagnetic wave. The experiment involved a spark-gap oscillator as a transmitter and a simple loop antenna as a receiver.
Construction:
1. The Hertz apparatus consisted of two large metal spheres (A and B) connected to two smaller metal electrodes (S1 and S2). These electrodes were positioned very close together, forming a spark gap. The larger spheres increased the capacitance, allowing more energy storage.
2. The ends of the spheres (A and B) were connected to an induction coil (H.T. Coil), which could generate a very high electromotive force (emf) when discharged.
3. The receiver was a simple metallic loop with a small gap (x, y) at one point, forming a "ring electrode". This loop was designed to detect any electromagnetic oscillations.


Working:
1. The induction coil produced a very high electromotive force (emf), which was applied across the electrodes S1 and S2. This created an intense electric field in the small gap between them.
2. When the potential difference across S1 and S2 became very high, the air in the gap ionized, leading to an electric discharge and the production of a spark. This spark created rapidly oscillating charges.
3. These oscillating charges generated electromagnetic waves, which traveled outwards from the transmitter.
4. When these waves reached the receiver loop, they induced oscillating currents in it. If the receiver was properly oriented and tuned, a tiny spark was observed across the gap (x, y) of the ring electrode.
5. Hertz demonstrated that these waves could be reflected, refracted, and diffracted, just like light. He also showed that they traveled at the speed of light. This was a crucial step in proving the electromagnetic nature of light.
In simple words: Hertz's experiment used a spark to create electromagnetic waves, which then made another spark in a receiver loop. This showed that light is an electromagnetic wave and proved Maxwell's theory.

🎯 Exam Tip: When describing the Hertz experiment, ensure you clearly explain the function of both the transmitter and the receiver, how electromagnetic waves were generated, and how their existence was detected. Mention the key properties of waves (reflection, refraction) that Hertz confirmed.

 

Question 4. Explain the Maxwell's modification Ampere's circuital law.
Answer: Maxwell modified Ampere's circuital law by adding a term called displacement current. The original law only considered conduction current. The modified law accounts for both conduction current and the changing electric flux in a region, which together produce a magnetic field. This change made the law more complete, especially for cases like a charging capacitor where no actual conduction current flows between the plates.
In simple words: Maxwell made Ampere's law better by adding a "displacement current" part. This new part includes how changing electric fields also create magnetic fields, not just moving charges.

🎯 Exam Tip: Remember to clearly state the original Ampere's law and then explain how Maxwell's addition of displacement current makes it more general, especially for changing electric fields.

 

Question 5. Explain the importance of Maxwell's correction.
Answer: Maxwell's correction to Ampere's law was crucial for several reasons.
1. Ampere's law alone suggested that only electric current could create a magnetic field. If this were strictly true, electromagnetic radiation would not exist.
2. Maxwell's addition, which involved a term for changing electric flux, allowed for the existence of time-varying electric fields that could produce magnetic fields. This completed the set of equations needed to describe electromagnetic waves.
3. This correction explained the existence of electromagnetic waves, which could travel through empty space at the speed of light. It proved that light itself is an electromagnetic wave.
4. It ensured that the laws of electromagnetism were consistent, especially in situations where charge is not conserved, like inside a capacitor during charging.
In simple words: Maxwell's change to Ampere's law was very important because it showed that changing electric fields can also make magnetic fields. This helped explain how light and other electromagnetic waves can exist and travel.

🎯 Exam Tip: When explaining Maxwell's correction, highlight its role in unifying electricity and magnetism and predicting electromagnetic waves, emphasizing its consistency with charge conservation.

 

Question 6. Write down the properties of electromagnetic waves.
Answer: Electromagnetic waves have several key properties:
1. They are produced by accelerated charges.
2. They do not need any medium to travel, meaning they can propagate through a vacuum. This is why they are called non-mechanical waves.
3. The electric field vector, the magnetic field vector, and the direction of wave propagation are all mutually perpendicular to each other. This shows that they are transverse in nature.
4. They travel in free space or a vacuum with a constant velocity, \( C = 3 \times 10^8 \text{ ms}^{-1} \). This speed is given by the expression \( C = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \).
5. In any medium other than a vacuum, electromagnetic waves travel at a velocity less than the speed of light in free space.
6. They are not affected or bent by electric and magnetic fields.
7. They can undergo interference, diffraction, and polarization.
8. The energy density of an electromagnetic wave is given by \( U = \varepsilon_{0}E^2 = \frac{1}{\mu_{0}}B^2 \).
9. Electromagnetic waves carry both energy and linear momentum. The linear momentum \( p \) is equal to \( \frac{\text{Energy}}{\text{Speed}} \).
10. If an electromagnetic wave of energy \( U \) is fully absorbed by a surface, the momentum transferred to the surface is \( P = \frac{U}{C} \). If it is completely reflected, the momentum transferred is \( \Delta p = \frac{2U}{C} \).
11. The rate at which energy flows across a unit area is called the pointing vector for electromagnetic waves, \( \vec{S} = \frac{1}{\mu_{0}} (\vec{E} \times \vec{B}) \).
12. Besides energy and linear momentum, electromagnetic waves also carry angular momentum.
In simple words: Electromagnetic waves are made by fast-moving charges and can travel without air or water. Their electric and magnetic parts wiggle at right angles to each other and to the direction they are moving. They always travel at the speed of light in empty space and are not bent by electric or magnetic forces. They carry energy and push things.

🎯 Exam Tip: When listing properties, clearly distinguish between their origin (accelerated charges), nature (transverse, non-mechanical), speed (speed of light in vacuum), and interactions (not deflected by fields, carry energy and momentum).

 

Question 7. The source of electromagnetic wave is accelerating electric charges.
Answer: The statement is correct. Electromagnetic waves are indeed produced by accelerating electric charges.
**Explanation:**
1. Stationary charges only create a static electric field around them.
2. Charges moving at a constant velocity produce both electric and magnetic fields, but these fields are not time-varying and do not propagate as waves.
3. When a charge accelerates (meaning its velocity changes, either in speed or direction), it creates a changing electric field.
4. This changing electric field then creates a changing magnetic field that is perpendicular to it.
5. This changing magnetic field, in turn, generates a changing electric field, and so on. This continuous interplay between oscillating electric and magnetic fields creates a self-propagating electromagnetic wave.
6. For example, if a charge accelerates along the x-axis, it produces a changing electric field along the y-axis, and a linked magnetic field along the z-axis. Both fields oscillate and propagate as a wave.
7. In free space or a vacuum, the ratio between the amplitudes of the electric field (\( E_0 \)) and the magnetic field (\( B_0 \)) is equal to the speed of light (\( C \)).
8. The energy of electromagnetic waves comes from the energy of the oscillating (accelerating) charge that produced them.
In simple words: Electromagnetic waves start when an electric charge moves faster or slower, or changes direction. This fast-moving charge creates moving electric and magnetic fields that push each other along, making a wave that travels.

🎯 Exam Tip: Focus on the "accelerating" aspect. Remember that stationary charges produce only electric fields, and charges moving at constant velocity produce both static electric and magnetic fields, but neither produces propagating electromagnetic waves.

 

Question 8. Explain the types of emission spectrum.
Answer: An emission spectrum is produced when a self-luminous source emits light, and this light is then separated into its different wavelengths. Each substance has its own unique emission spectrum.
There are three main types of emission spectra:
1. **Continuous Emission Spectra:** This type of spectrum contains all visible colors, blending smoothly from one to the next, similar to a rainbow. It is produced by hot, dense sources like incandescent solids (e.g., a glowing light bulb filament), liquids, or gases under high pressure.
2. **Line Emission Spectra:** This spectrum consists of sharp, distinct lines of specific wavelengths or frequencies, separated by dark spaces. It is produced when light from a hot gas (at low pressure) passes through a prism. Each line corresponds to light emitted by excited atoms transitioning between energy levels. Examples include the spectra of atomic hydrogen and helium.
3. **Band Emission Spectra:** This spectrum features several closely spaced spectral lines that overlap to form specific bands, separated by dark spaces. It arises when molecules are excited and is characteristic of molecules rather than individual atoms. It is useful for studying the structure of molecules. Examples include spectra from hydrogen gas or ammonia gas in a discharge tube.
In simple words: Emission spectra show the light given off by a hot object. There are three types: "continuous" (like a rainbow, all colors), "line" (only specific bright lines of color), and "band" (groups of lines close together). Each type tells us about the object making the light.

🎯 Exam Tip: Clearly define what an emission spectrum is, then describe each type (continuous, line, band) with simple examples and how they are formed. Focus on the source (solid/liquid, hot gas, molecules) for each type.

 

Question 9. Explain the types of the absorption spectrum.
Answer: An absorption spectrum is observed when light passes through a substance that absorbs certain wavelengths, and the remaining light is then dispersed by a prism. The spectrum obtained shows dark lines or bands against a continuous background, corresponding to the absorbed wavelengths. This spectrum is characteristic of the absorbing substance.
There are three main types of absorption spectra:
1. **Continuous Absorption Spectrum:** This occurs when a medium absorbs all wavelengths except for a specific range. For example, when white light passes through a blue glass plate, the plate absorbs all colors except blue, resulting in a continuous blue spectrum.
2. **Line Absorption Spectrum:** This spectrum appears as dark lines superimposed on a continuous background. It is produced when white light from an incandescent source passes through a cold, dilute gas. The atoms in the cold gas absorb specific wavelengths of light that they would normally emit if they were hot. A famous example is the Fraunhofer lines in the solar spectrum, which are dark lines caused by absorption by elements in the Sun's outer atmosphere.
3. **Band Absorption Spectrum:** This spectrum consists of dark bands within a continuous spectrum. It is observed when light passes through solutions of organic and inorganic compounds, or through vapours like iodine. The molecules in these substances absorb light over specific wavelength ranges, creating these dark bands.
In simple words: An absorption spectrum shows dark lines or bands where light has been soaked up by a substance. It can be "continuous" (most light absorbed, only one color gets through), "line" (dark lines where specific colors were absorbed by atoms), or "band" (dark bands where molecules absorbed light).

🎯 Exam Tip: When discussing absorption spectra, differentiate them from emission spectra by emphasizing the dark lines/bands against a bright background. Explain the role of the absorbing medium and the type of source light (usually white light).

 

IV. Numerical Problems:

 

Question 1. Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.
Answer: For a charging capacitor, the displacement current \( I_d \) between the plates is equal to the conduction current \( I_c \) flowing in the connecting wires.
Given conduction current \( I_c = 5 \, \text{A} \).
Therefore, the displacement current \( I_d = I_c = 5 \, \text{A} \).
This happens because the changing electric flux between the plates acts like a current.
In simple words: In a charging capacitor, the special "displacement current" inside the capacitor is exactly the same as the regular current flowing in the wires connected to it. So if the wire current is 5 A, the displacement current is also 5 A.

🎯 Exam Tip: Remember that for a parallel plate capacitor in a circuit, the displacement current between the plates is always equal to the conduction current in the wires connected to it, ensuring continuity of current.

 

Question 2. A transmitter consists of an LC circuit with an inductance of 1 µH and capacitance of 1 µF. What is the wavelength of the electromagnetic waves it emits?
Answer: Given values are:
Inductance \( L = 1 \, \mu\text{H} = 1 \times 10^{-6} \, \text{H} \)
Capacitance \( C = 1 \, \mu\text{F} = 1 \times 10^{-6} \, \text{F} \)
The speed of electromagnetic waves \( c = 3 \times 10^8 \, \text{m/s} \)

First, calculate the frequency \( \nu \) of oscillation for the LC circuit using the formula:
\( \nu = \frac{1}{2\pi\sqrt{LC}} \)
\( \implies \nu = \frac{1}{2\pi\sqrt{(1 \times 10^{-6} \, \text{H})(1 \times 10^{-6} \, \text{F})}} \)
\( \implies \nu = \frac{1}{2\pi\sqrt{1 \times 10^{-12}}} \)
\( \implies \nu = \frac{1}{2\pi \times 10^{-6}} \, \text{Hz} \)

Now, calculate the wavelength \( \lambda \) using the wave speed formula:
\( c = \nu \lambda \)
\( \implies \lambda = \frac{c}{\nu} \)
\( \implies \lambda = \frac{3 \times 10^8 \, \text{m/s}}{1 / (2\pi \times 10^{-6} \, \text{Hz})} \)
\( \implies \lambda = 3 \times 10^8 \times 2\pi \times 10^{-6} \, \text{m} \)
\( \implies \lambda = 3 \times 10^8 \times 2 \times 3.14159 \times 10^{-6} \, \text{m} \)
\( \implies \lambda = 18.84954 \times 10^2 \, \text{m} \)
\( \implies \lambda = 1884.954 \, \text{m} \)
The calculated wavelength is approximately \( 18.85 \times 10^2 \, \text{m} \), or \( 1885 \, \text{m} \).

The provided solution directly calculates \( \lambda = C \times 2\pi\sqrt{LC} \).
\( \lambda = 3 \times 10^8 \times 2\pi \times 10^{-6} = 1884.954 \, \text{m} \)
This matches the given result \( 18.86 \times 10^2 \, \text{m} \).
In simple words: We find the natural frequency of the LC circuit first. Then, using the speed of light, we can calculate the wavelength of the waves this circuit produces. The wavelength comes out to be about 1885 meters.

🎯 Exam Tip: Remember the formulas for resonance frequency (\( \nu = \frac{1}{2\pi\sqrt{LC}} \)) and wave speed (\( c = \nu \lambda \)). Ensure correct unit conversions, especially for micro-H and micro-F.

 

Question 3. A pulse of light of duration \( 10^{-6} \) s is absorbed completely by a small object initially at rest. If the power of the pulse is \( 60 \times 10^{-3} \) W, calculate the final momentum of the object.
Answer: Given values:
Duration of light pulse \( \Delta t = 10^{-6} \, \text{s} \)
Power of the pulse \( P = 60 \times 10^{-3} \, \text{W} \)
Speed of light \( C = 3 \times 10^8 \, \text{m/s} \)

First, calculate the total energy \( U \) of the light pulse:
\( U = P \times \Delta t \)
\( \implies U = (60 \times 10^{-3} \, \text{W}) \times (10^{-6} \, \text{s}) \)
\( \implies U = 60 \times 10^{-9} \, \text{J} \)

When light is completely absorbed by an object, the momentum \( p \) imparted to the object is related to the energy \( U \) by the formula:
\( p = \frac{U}{C} \)
\( \implies p = \frac{60 \times 10^{-9} \, \text{J}}{3 \times 10^8 \, \text{m/s}} \)
\( \implies p = 20 \times 10^{-17} \, \text{kg m/s} \)
So, the final momentum of the object is \( 20 \times 10^{-17} \, \text{kg m/s} \). This small momentum is often negligible but can be significant in space.
In simple words: We first find the total energy in the light pulse by multiplying its power by its duration. Then, we divide this energy by the speed of light to find the momentum it transferred to the object. The final momentum of the object will be this calculated value.

🎯 Exam Tip: Remember that momentum transferred by light is \( p = U/C \) for complete absorption and \( p = 2U/C \) for complete reflection. Make sure to use the correct energy (Power x time) for \( U \).

 

Question 4. Let an electromagnetic wave propagation along the x-direction, the magnetic field oscillates at a frequency of \( 10^{10} \) Hz and has an amplitude of \( 10^{-5} \) T, acting along the y-direction. Then, compute the wavelength of the wave. Also, write down the expression for the electric field in this case.
Answer: Given data:
Frequency \( f = 10^{10} \, \text{Hz} \)
Amplitude of magnetic field \( B_0 = 10^{-5} \, \text{T} \)
Speed of light \( C = 3 \times 10^8 \, \text{m/s} \)
Wave propagates along the x-direction, magnetic field oscillates along the y-direction.

**1. Wavelength \( \lambda \):**
We use the relation \( C = f\lambda \).
\( \implies \lambda = \frac{C}{f} \)
\( \implies \lambda = \frac{3 \times 10^8 \, \text{m/s}}{10^{10} \, \text{Hz}} \)
\( \implies \lambda = 3 \times 10^{-2} \, \text{m} \)
The wavelength of the electromagnetic wave is \( 3 \times 10^{-2} \, \text{m} \).

**2. Expression for the electric field (\( E_z \)):**
For an electromagnetic wave, the electric field \( E \) and magnetic field \( B \) are perpendicular to each other and to the direction of propagation. If propagation is along x-axis and \( \vec{B} \) is along y-axis, then \( \vec{E} \) must be along z-axis.

First, find the amplitude of the electric field \( E_0 \):
\( E_0 = C \times B_0 \)
\( \implies E_0 = (3 \times 10^8 \, \text{m/s}) \times (10^{-5} \, \text{T}) \)
\( \implies E_0 = 3 \times 10^3 \, \text{N/C} \)

Next, calculate the angular frequency \( \omega \) and wave number \( k \):
\( \omega = 2\pi f = 2\pi (10^{10}) \, \text{rad/s} \)
\( k = \frac{2\pi}{\lambda} = \frac{2\pi}{3 \times 10^{-2}} = \frac{200\pi}{3} \, \text{m}^{-1} \)

The general expression for an electric field oscillating along the z-direction and propagating along the x-direction is:
\( E_z = E_0 \sin(kx - \omega t) \)
\( \implies E_z = 3 \times 10^3 \sin\left(\frac{200\pi}{3}x - 2\pi (10^{10})t\right) \, \text{N/C} \)
This expression describes the electric field's oscillation over space and time.
In simple words: First, we find the wavelength by dividing the speed of light by the given frequency. Then, because the wave travels in x-direction and the magnetic field is in y-direction, the electric field must be in z-direction. We calculate its strength using the speed of light and the magnetic field strength, then write the wave equation using this strength, the wavelength, and the frequency.

🎯 Exam Tip: Remember the relationship \( C = f\lambda \) for wavelength and \( E_0 = C B_0 \) for electric field amplitude. Ensure the electric field direction is perpendicular to both magnetic field and propagation direction, using the right-hand rule (\( \vec{E} \times \vec{B} \) gives propagation direction).

 

Question 5. If the relative permeability and relative permittivity of a medium is 1.0 and 2.25, respectively. Find the speed of the electromagnetic wave in this medium.
Answer: Given data:
Relative permeability \( \mu_r = 1.0 \)
Relative permittivity \( \varepsilon_r = 2.25 \)
Speed of light in vacuum \( C = 3 \times 10^8 \, \text{m/s} \)

The speed of an electromagnetic wave \( V \) in a medium is given by the formula:
\( V = \frac{C}{\sqrt{\mu_r \varepsilon_r}} \)
\( \implies V = \frac{3 \times 10^8 \, \text{m/s}}{\sqrt{1.0 \times 2.25}} \)
\( \implies V = \frac{3 \times 10^8 \, \text{m/s}}{\sqrt{2.25}} \)
\( \implies V = \frac{3 \times 10^8 \, \text{m/s}}{1.5} \)
\( \implies V = 2 \times 10^8 \, \text{m/s} \)
The speed of the electromagnetic wave in this medium is \( 2 \times 10^8 \, \text{m/s} \). This speed is lower than the speed of light in a vacuum because of the medium's properties.
In simple words: To find how fast light travels in a material, we divide the speed of light in empty space by the square root of how much the material affects magnetic fields (relative permeability) and electric fields (relative permittivity). For this material, the light travels at two-thirds the speed of light in empty space.

🎯 Exam Tip: Remember the formula for the speed of light in a medium \( V = \frac{C}{\sqrt{\mu_r \varepsilon_r}} \). Make sure to correctly calculate the square root of the product of relative permeability and permittivity.

 

Part II:

 

I. Matching Type Questions:

 

Question a.

(1) A \( \rightarrow \) b, B \( \rightarrow \) c, C \( \rightarrow \) d, D \( \rightarrow \) a
(2) A \( \rightarrow \) d, B \( \rightarrow \) a, C \( \rightarrow \) b, D \( \rightarrow \) c
(3) A \( \rightarrow \) c, B \( \rightarrow \) d, C \( \rightarrow \) b, D \( \rightarrow \) a
(4) A \( \rightarrow \) c, B \( \rightarrow \) a, C \( \rightarrow \) d, D \( \rightarrow \) b
Answer: (3) A \( \rightarrow \) c, B \( \rightarrow \) d, C \( \rightarrow \) b, D \( \rightarrow \) a
In simple words: Match J.C. Maxwell with predicting waves, Ampere's Law with its integral form, changing electric fields with displacement current, and Hertz with proving waves exist.

🎯 Exam Tip: For matching questions, connect each scientist or concept to their primary contribution or relevant law. J.C. Maxwell predicted EM waves, Hertz experimentally confirmed them, Ampere's Law involves current integral, and changing electric fields lead to displacement current.

 

Question b.

(1) A \( \rightarrow \) d, B \( \rightarrow \) c, C \( \rightarrow \) a, D \( \rightarrow \) b
(2) A \( \rightarrow \) c, B \( \rightarrow \) a, C \( \rightarrow \) d, D \( \rightarrow \) b
(3) A \( \rightarrow \) b, B \( \rightarrow \) C, C \( \rightarrow \) d, D \( \rightarrow \) a
(4) A \( \rightarrow \) d, B \( \rightarrow \) a, C \( \rightarrow \) b, D \( \rightarrow \) C
Answer: (1) A \( \rightarrow \) d, B \( \rightarrow \) c, C \( \rightarrow \) a, D \( \rightarrow \) b
In simple words: Connect Gauss law with electric flux and charge, Faraday's law with changing magnetic flux and electric fields, Ampere's circuital law with current and magnetic fields, and Maxwell's modified Ampere's law with both conduction and displacement currents.

🎯 Exam Tip: It's important to remember the integral forms of Maxwell's equations and how they relate to the fundamental laws. Pay attention to the signs and symbols for electric and magnetic fields, flux, and current.

 

Question c.

(1) A \( \rightarrow \) c, B \( \rightarrow \) d, C \( \rightarrow \) b, D \( \rightarrow \) a
(2) A \( \rightarrow \) b, B \( \rightarrow \) c, C \( \rightarrow \) d, D \( \rightarrow \) a
(3) A \( \rightarrow \) d, B \( \rightarrow \) c, C \( \rightarrow \) b, D \( \rightarrow \) a
(4) A \( \rightarrow \) d, B \( \rightarrow \) c, C \( \rightarrow \) b, D \( \rightarrow \) c
Answer: (3) A \( \rightarrow \) d, B \( \rightarrow \) c, C \( \rightarrow \) b, D \( \rightarrow \) a
In simple words: Match microwaves with aircraft navigation, ultraviolet with molecular studies, X-rays with crystal structures, and infrared with night vision.

🎯 Exam Tip: Know the practical applications of different parts of the electromagnetic spectrum. Microwaves are used in radar (like aircraft navigation), X-rays for internal structures (crystals), UV for molecular studies and sterilization, and infrared for heat detection (night vision).

 

Question d.

(1) A \( \rightarrow \) d, B \( \rightarrow \) a, C \( \rightarrow \) b, D \( \rightarrow \) c
(2) A \( \rightarrow \) c, B \( \rightarrow \) a, C \( \rightarrow \) d, D \( \rightarrow \) b
(3) A \( \rightarrow \) d, B \( \rightarrow \) c, C \( \rightarrow \) a, D \( \rightarrow \) b
(4) A \( \rightarrow \) b, B \( \rightarrow \) d, C \( \rightarrow \) a, D \( \rightarrow \) c
Answer: (4) A \( \rightarrow \) b, B \( \rightarrow \) d, C \( \rightarrow \) a, D \( \rightarrow \) c
In simple words: Match the formula \( \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \) with the velocity of light in vacuum. Match \( \sqrt{\varepsilon_r \mu_r} \) with the refractive index of a medium. Match \( \frac{1}{\mu_0} B^2 \) with the energy density of the electromagnetic wave. Match \( \frac{1}{2} \frac{1}{\mu_0} B^2 \) with the average energy density of the electromagnetic wave.

🎯 Exam Tip: Pay close attention to constants (\( \mu_0, \varepsilon_0 \)) versus relative values (\( \mu_r, \varepsilon_r \)) and whether the term is for instantaneous or average energy density. The factor of \( 1/2 \) often indicates an average value.

 

Question e.

(1) A \( \rightarrow \) d, B \( \rightarrow \) a, C \( \rightarrow \) b, D \( \rightarrow \) c
Answer: (1) A \( \rightarrow \) d, B \( \rightarrow \) a, C \( \rightarrow \) b, D \( \rightarrow \) c
In simple words: Match a carbon arc with continuous emission, atomic helium with line emission, molecular structure with band emission, and the sun's spectrum with line absorption (Fraunhofer lines).

🎯 Exam Tip: Understand the origin and type of spectra. Hot, dense sources (like a carbon arc) produce continuous spectra. Excited atoms (like helium) produce line emission. Molecules produce band spectra. Light from a hot source passing through a cooler gas (like the sun's atmosphere) results in line absorption.

 

II. Fill in the blanks:

 

Question 1. Electromagnetic waves are ________ in nature.
Answer: Transverse
In simple words: Electromagnetic waves, like light, wiggle up and down or side to side as they move forward, which means they are transverse.

🎯 Exam Tip: Remember that in transverse waves, the oscillations (of electric and magnetic fields) are perpendicular to the direction of wave propagation.

 

Question 2. The phase difference between electric and magnetic field vectors in an electromagnetic wave ________.
Answer: Zero
In simple words: The electric and magnetic parts of an electromagnetic wave rise and fall together at the same time and in the same place; they are "in phase."

🎯 Exam Tip: A key characteristic of electromagnetic waves is that the electric and magnetic field components oscillate in phase, meaning their peaks and troughs occur at the same points in space and time.

 

Question 3. The angle between electric and magnetic component in an electromagnetic wave is ________.
Answer: \( \pi/2 \) (or) 90°
In simple words: In an electromagnetic wave, the electric and magnetic parts always wiggle at a perfect right angle (90 degrees) to each other.

🎯 Exam Tip: Always recall that the electric and magnetic field vectors in an electromagnetic wave are mutually perpendicular, forming a 90-degree angle with each other.

 

Question 4. Linear momentum of electromagnetic wave is ________.
Answer: \( U/C \)
In simple words: The push that an electromagnetic wave carries is equal to its total energy divided by the speed of light.

🎯 Exam Tip: The linear momentum \( p \) carried by an electromagnetic wave is directly proportional to its energy \( U \) and inversely proportional to the speed of light \( C \), so \( p = U/C \).

 

Question 5. The rate of flow of energy crossing in unit area is known as ___________.
Answer: Pointing vector
In simple words: The speed at which energy moves across a certain space is called the pointing vector. It shows the direction and amount of energy flow.

🎯 Exam Tip: Remember that the pointing vector describes the direction and magnitude of electromagnetic energy flux.

 

Question 6. The expression for intensity of electromagnetic wave is equal to ___________.
Answer: Power / Surface Area
In simple words: The intensity of an electromagnetic wave tells you how much power is spread over a certain area. It's like how bright a light is over a surface.

🎯 Exam Tip: Intensity is a key concept in wave physics, linking power output to how it spreads out in space.

 

Question 7. The electromagnetic wave from the sun which is absorbed by atmospheric ozone is ___________.
Answer: Ultraviolet rays
In simple words: The ozone layer in the sky protects us by soaking up harmful ultraviolet rays from the sun. This is why it's so important for life on Earth.

🎯 Exam Tip: Always associate the ozone layer with its protective role against UV radiation.

 

Question 8. The electromagnetic wave produced by Hertz is ___________.
Answer: Radio waves
In simple words: Heinrich Hertz was the first person to create and detect radio waves. His experiments proved that electromagnetic waves really exist.

🎯 Exam Tip: Hertz's experiments were crucial for proving Maxwell's theory and launching the age of radio communication.

 

Question 9. When white light passing through chlorophyll it gives ___________.
Answer: Band absorption spectrum
In simple words: Chlorophyll, the green stuff in plants, soaks up certain colors of light. When white light shines through it, you'll see a dark band where those colors are missing from the spectrum.

🎯 Exam Tip: Remember that absorption spectra are unique to substances and show which wavelengths of light they absorb.

 

Question 10. The range of visible light is ___________.
Answer: \( 4 \times 10^{-7} \text{ m to } 7 \times 10^{-7} \text{ m} \)
In simple words: The colors we can see with our eyes, from red to violet, fall within a very specific range of wavelengths, from about 400 nanometers to 700 nanometers.

🎯 Exam Tip: Visible light is a small part of the electromagnetic spectrum, covering wavelengths that our eyes can detect.

III. Choose the Odd One Out:

 

Question 1. If n is a natural number, then \( 5^{2n} - 1 \) is always divisible by
(a) Velocity of light in vacuum
(b) \( 3 \times 10^8 \text{ ms}^{-1} \)
(c) \( \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \)
(d) \( \sqrt{\mu_r \varepsilon_r} \)
Answer: (d) \( \sqrt{\mu_r \varepsilon_r} \)
In simple words: The first three options all describe the speed of light in empty space. The last option, \( \sqrt{\mu_r \varepsilon_r} \), represents the refractive index of a material, which is a different concept.

🎯 Exam Tip: Understand the different ways to express the speed of light in vacuum and how the refractive index relates to material properties.

 

Question 2. If n is a natural number, then \( 5^{2n} - 1 \) is always divisible by
(a) Fraunhofer
(b) J.C. Maxwell
(c) Hertz
(d) Michael
Answer: (a) Fraunhofer
In simple words: Maxwell and Hertz are famous scientists who greatly contributed to the theory and experimental proof of electromagnetic waves, along with Michael Faraday. Fraunhofer, on the other hand, is associated with the dark lines observed in the solar spectrum, a different physical phenomenon.

🎯 Exam Tip: Distinguish between key scientists and the phenomena or laws they are associated with in physics.

 

Question 3. If n is a natural number, then \( 5^{2n} - 1 \) is always divisible by
(a) X-rays
(b) Gamma rays
(c) Microwave
(d) Ultraviolet ray
Answer: (c) Microwave
In simple words: X-rays, Gamma rays, and Ultraviolet rays are all forms of high-energy electromagnetic radiation often used in medical imaging or sterilization. Microwaves, however, are a lower-energy form of radiation typically used for cooking and communication.

🎯 Exam Tip: Be familiar with the different regions of the electromagnetic spectrum and their typical applications.

 

Question 4. If n is a natural number, then \( 5^{2n} - 1 \) is always divisible by
(a) Ampere - Maxwell law
(b) \( \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dt}}=\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}} \)
(c) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{ds}}=\mu_{0}\left(\mathrm{I}_{\mathrm{c}}+\mathrm{I}_{\mathrm{d}}\right) \)
(d) \( \oint \overrightarrow{B .} \overrightarrow{dl} = \mu_{0} I_{0}+\mu_{0} \varepsilon_{0} \frac{d}{d t} \int_{S} \vec{E} \cdot d \vec{A} \)
Answer: (b) \( \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dt}}=\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}} \)
In simple words: Options (a), (c), and (d) all represent correct forms or concepts related to Ampere-Maxwell's law or its components, dealing with magnetic fields and currents. Option (b) shows an integral of the electric field over time, which is not the correct form of Faraday's law or any standard Maxwell's equation.

🎯 Exam Tip: Pay close attention to the vector quantities and differential elements in integral forms of Maxwell's equations (e.g., \( \overrightarrow{\mathrm{d}l} \) for line integrals and \( \overrightarrow{\mathrm{dA}} \) for surface integrals).

 

Question 5. If n is a natural number, then \( 5^{2n} - 1 \) is always divisible by
(a) \( \varepsilon_{0}=\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}} \)
(b) Displacement current
(c) \( \mu_0 I_0 \)
(d) \( \varepsilon_{0} \frac{\mathrm{d}}{\mathrm{dt}} \int_{\mathrm{S}} \overrightarrow{\mathrm{E} \cdot \mathrm{d}} \overrightarrow{\mathrm{A}} \)
Answer: (c) \( \mu_0 I_0 \)
In simple words: The first, second, and fourth options all relate to the concept or calculation of displacement current, which involves a changing electric flux. Option (c), \( \mu_0 I_0 \), represents a term for conduction current in Ampere's law, making it the odd one out.

🎯 Exam Tip: Remember that displacement current is proportional to the rate of change of electric flux, not directly to conduction current.

IV. Choose the Incorrect Pair:

 

Question 1. Choose the Incorrect Pair:
(a) Electromagnetic wave - Transverse
(b) components is an electromagnetic wave - Speed of electromagnetic wave
(c) Hertz - Ultraviolet radiation
(d) Momentum imparted by electromagnetic \( p = \frac{\mathrm{U}}{\mathrm{C}} \)
Answer: (c) Hertz - Ultraviolet radiation
In simple words: Heinrich Hertz's famous experiments involved creating and detecting radio waves, which are a low-frequency type of electromagnetic radiation. He did not work with ultraviolet radiation.

🎯 Exam Tip: Connect key experiments or scientists to the specific types of electromagnetic waves or phenomena they studied.

 

Question 2. Choose the Incorrect Pair:
(a) Atomic spectrum - pure line spectrum
(b) Solar spectrum - line spectrum
(c) Molecules - Band Spectrum
(d) Carbon arc - Continuous emission spectrum
Answer: (b) Solar spectrum - line spectrum
In simple words: The solar spectrum shows dark lines, which are called Fraunhofer lines. These are actually a type of *absorption* spectrum, not a pure line *emission* spectrum. The sun's atmosphere absorbs specific wavelengths of light.

🎯 Exam Tip: Differentiate between emission spectra (bright lines from excited atoms) and absorption spectra (dark lines where light has been absorbed).

 

Question 3. Choose the Incorrect Pair:
(a) Refractive index of a medium - \( \mu = \sqrt{\varepsilon_{\mathrm{r}} \mu_{\mathrm{r}}} \)
(b) Pointing vector - \( \overrightarrow{S}=C^2 \varepsilon_{0}(\vec{B} \times \vec{E}) \)
(c) To detect faults, cracks, flaws, and holes - X-rays
(d) Fraunhofer lines - Sun's atmosphere
Answer: (b) Pointing vector - \( \overrightarrow{S}=C^2 \varepsilon_{0}(\vec{B} \times \vec{E}) \)
In simple words: The standard formula for the Poynting vector, which shows the direction and rate of energy flow in an electromagnetic field, is \( \overrightarrow{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) \). The formula provided in option (b) is incorrect in its constant and the order of the cross product.

🎯 Exam Tip: Memorize the correct expression for the Poynting vector, including the constant and the order of the electric and magnetic field vectors.

 

Question 4. Choose the Incorrect Pair:
(a) Net electric flux and net charge - Gauss law
(b) Electric field and magnetic flux - Faraday's law
(c) Magnetic field around any closed path to the conduction current - Maxwell's law
(d) Magnetic field around any closed path to the conduction current and displacement current through that path - Ampere - Maxwell's law
Answer: (c) Magnetic field around any closed path to the conduction current - Maxwell's law
In simple words: Ampere's law, in its original form, connects a magnetic field to the conduction current only. Maxwell's law is the modified version that includes both conduction current and displacement current. So, linking a magnetic field to *only* conduction current as "Maxwell's law" is incomplete and therefore incorrect.

🎯 Exam Tip: Remember that Maxwell's full set of equations unified electricity and magnetism, notably by adding the displacement current to Ampere's Law.

V. Choose the correct pair:

 

Question 1. Choose the correct pair:
(a) Wavelength range of radio wave \( 3 \times 10^9 \text{ m to } 3 \times 10^4 \text{ m} \)
(b) Wavelength range of ultraviolet radiation \( - 5 \times 10^{17} \text{ m to } 7 \times 10^{14} \text{ m} \)
(c) Wavelength range of Gamma radiation \( - 10^{-14} \text{ m to } 1 \times 10^{-10} \text{ m} \)
(d) Wavelength range of Infrared radiation \( - 4 \times 10^{14} \text{ to } 6 \times 10^{10} \text{ m} \)
Answer: (c) Wavelength range of Gamma radiation \( - 10^{-14} \text{ m to } 1 \times 10^{-10} \text{ m} \)
In simple words: Gamma rays have the shortest wavelengths in the electromagnetic spectrum, typically ranging from \( 10^{-14} \) meters to \( 10^{-10} \) meters. This means they carry a lot of energy.

🎯 Exam Tip: Familiarize yourself with the approximate wavelength and frequency ranges for all major types of electromagnetic radiation.

 

Question 2. Choose the correct pair:
(a) Incandescent solids - Band absorption spectra
(b) Spectrum from carbon arc - Line spectrum
(c) Spectra of atomic hydrogen - Line absorption spectrum
(d) the characteristics of substance - Absorption spectrum
Answer: (d) the characteristics of substance - Absorption spectrum
In simple words: Every unique chemical substance has its own specific pattern of light it absorbs. This absorption spectrum acts like a fingerprint, allowing scientists to identify the substance.

🎯 Exam Tip: Understand that absorption spectra are used for chemical analysis and identification due to their unique patterns for each element or compound.

 

Question 3. Choose the correct pair:
(a) Maxwell's prediction on the electromagnetic wave was verified by - Ampere
(b) Rate of change in the magnetic field produces electric field - Gauss
(c) Total electric flux is equal to \( 1 / \varepsilon_0 \) times net charge enclosed by the surface - Gauss law
(d) An example of mechanical wave - Electromagnetic waves
Answer: (c) Total electric flux is equal to \( 1 / \varepsilon_0 \) times net charge enclosed by the surface - Gauss law
In simple words: Gauss's law for electricity states that the total electric field passing through a closed surface is directly related to the total electric charge contained inside that surface. This is a fundamental law in electromagnetism.

🎯 Exam Tip: Clearly remember which law corresponds to which fundamental relationship in electromagnetism.

 

Question 4. Choose the correct pair:
(a) Velocity of X-ray in vacuum is - \( 3 \times 10^8 \text{ ms}^{-1} \)
(b) Velocity of X-ray in medium - \( 3 \times 10^8 \text{ ms}^{-1} \)
(c) Electromagnetic waves - Deflected by both electric and magnetic field
(d) The electric component of electromagnetic wave - \( E_z = E_0 \sin(kz - \omega t) \text{ ms}^{-1} \)
Answer: (a) Velocity of X-ray in vacuum is - \( 3 \times 10^8 \text{ ms}^{-1} \)
In simple words: All types of electromagnetic waves, including X-rays, travel at the same speed in a vacuum. This speed is universally known as the speed of light, which is approximately \( 3 \times 10^8 \text{ meters per second} \).

🎯 Exam Tip: Remember that the speed of all electromagnetic waves in vacuum is constant, `c`, regardless of their frequency or wavelength.

VI. Assertion and Reason:

 

Question 1. Assertion (A): The velocity of electromagnetic waves in a medium is always less than the velocity of electromagnetic wave in free space (or) vacuum. Reason (R): The wavelength of the electromagnetic wave decreases when traveling into a denser medium but frequency does not change.
Answer: (c) A and R are correct
In simple words: Both the statement (A) and the reason (R) are true. Electromagnetic waves slow down when they enter a denser material, and because their frequency stays the same, their wavelength must become shorter.

🎯 Exam Tip: Understand the relationship \( v = f\lambda \) and how \( v \) and \( \lambda \) change when light enters a medium, while \( f \) remains constant.

 

Question 2. Assertion (A): Refractive index of air is \( \mu=\sqrt{\mu_{0}} \). Reason (R): Since the dielectric constant of air equal to one (\( \varepsilon_r = 1 \) for air). The Refractive index of air \( \mu=\sqrt{\mu_{r}} \).
Answer: (b) A is wrong R is correct
In simple words: The refractive index of a material depends on both its relative permeability (\( \mu_r \)) and relative permittivity (\( \varepsilon_r \)). For air, the relative permittivity is very close to 1. If \( \varepsilon_r \) is 1, then the refractive index \( \mu \) would be approximately \( \sqrt{\mu_r} \), which matches the statement in Reason (R). However, Assertion (A) is incorrect because it misses the permittivity term (\( \varepsilon_0 \)) in the formula.

🎯 Exam Tip: Always use the complete formula \( n = \sqrt{\mu_r \varepsilon_r} \) for refractive index and recall that for air, \( \mu_r \approx 1 \) and \( \varepsilon_r \approx 1 \), so \( n \approx 1 \).

 

Question 3. Assertion (A): \( \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} \mathrm{A}}=\frac{\mathrm{Q}_{\text {enclosed }}}{\epsilon_{0}} \). Reason (R): \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \mathrm{A}} \) since isolated charges cannot exist since isolated magnetic monopole exist.
Answer: (a) A is correct R is wrong
In simple words: Assertion (A) correctly states Gauss's law for electric fields, which relates the total electric flux through a surface to the charge inside. Reason (R) is incorrectly phrased and logically inconsistent; it attempts to explain Gauss's law for magnetism (where magnetic flux is zero because magnetic monopoles don't exist) but contains a contradiction in its wording.

🎯 Exam Tip: Understand the clear distinction between Gauss's law for electric fields (related to electric charge) and Gauss's law for magnetic fields (related to the non-existence of magnetic monopoles).

 

Question 4. Assertion (A): All medium other than air has relative permittivity less than one. Reason (R): Electromagnetic wave travels faster in a medium than in air.
Answer: (d) A and R are wrong
In simple words: Both the assertion and the reason are false. Most materials have a relative permittivity greater than one. Also, electromagnetic waves always travel slower in any medium compared to their speed in air or a vacuum.

🎯 Exam Tip: Remember that relative permittivity is a measure of how a material stores electrical energy and is typically greater than one, causing EM waves to slow down.

VII. Choose the correct statement:

 

Question 1. Choose the correct statement:
(a) Maxwell's law of induction states that Time-varying electric field produces a magnetic field
(b) Faraday's law of induction states that Time-varying electric field produces a magnetic field
(c) Hertz experimentally proved accelerated charges produce an only magnetic field
(d) Ampere's law states that total flux over any surface enclosing a charge q is always zero.
Answer: (a) Maxwell's law of induction states that Time-varying electric field produces a magnetic field
In simple words: Maxwell's key contribution was showing that a changing electric field, like a changing electric current, can create a magnetic field. This idea is crucial for understanding how electromagnetic waves work.

🎯 Exam Tip: Connect Maxwell's contribution to the concept of displacement current, where changing electric fields generate magnetic fields, completing the symmetry with Faraday's law.

 

Question 2. Choose the correct statement:
(a) Electromagnetic waves need a medium for its propagation
(b) waves are mostly transverse, it can also longitudinal wave
(c) Electromagnetic waves are produced by any accelerated charges
(d) The ratio between electric energy and magnetic energy in an electromagnetic wave is 2:1
Answer: (c) Electromagnetic waves are produced by any accelerated charges
In simple words: When an electric charge moves and its speed or direction changes (it accelerates), it creates disturbances in electric and magnetic fields that travel as electromagnetic waves. This is the fundamental way these waves are generated.

🎯 Exam Tip: Remember the primary source of electromagnetic waves: accelerating charges. This concept is fundamental to understanding their generation.

 

Question 3. Choose the correct statement:
(a) Electromagnetic waves have only linear momentum no angular momentum.
(b) The energy density of the electromagnetic wave is \( u = \frac{\beta^{2}}{2 \mu_{0}} \)
(c) The average energy density of electromagnetic wave is \( u = \frac{\beta^{2}}{\mu_{0}} \)
(d) If the electromagnetic wave incident on a surface, the momentum imparted as the surface is \( p = \frac{\mathrm{u}}{\mathrm{c}} \)
Answer: (d) If the electromagnetic wave incident on a surface, the momentum imparted as the surface is \( p = \frac{\mathrm{u}}{\mathrm{c}} \)
In simple words: When an electromagnetic wave hits a surface and is completely absorbed, it transfers energy \( u \) and also momentum \( p \). The amount of momentum transferred is equal to the energy divided by the speed of light \( c \).

🎯 Exam Tip: Recall that electromagnetic waves carry both energy and momentum, and know the formula for momentum transfer upon absorption or reflection.

 

Question 4. Choose the correct statement:
(a) Cellular phones used infrared radiation.
(b) Molecular spectrum is sharp at one end and fades out at the other end.
(c) When light from the incandescent lamp passed through cold gas, the spectrum obtained is the Band absorption spectrum.
(d) Fraunhofer lines help to identify the elements in the earth's atmosphere.
Answer: (b) Molecular spectrum is sharp at one end and fades out at the other end.
In simple words: A characteristic feature of a molecular band spectrum is that it has a very clear, sharp edge on one side and then gradually becomes fainter on the other side. This unique appearance helps identify molecules.

🎯 Exam Tip: Understand the distinctive features of band spectra, which are produced by molecules, in contrast to line spectra produced by atoms.

VIII. Choose the Incorrect Statement:

 

Question 1. Choose the Incorrect Statement:
(a) According to Faraday change in the magnetic field at a point with respect to time produces an electric field at that point
(b) Amperes law is the line integral of the magnetic field \( \vec{B} \) around any closed loop is equal to \( \mu_0 \) times the net current I threading through the area enclosed by the loop.
(c) Total current enclosed by a surface is the sum of conduction current and displacement current \( I = I_c + I_d \)
(d) When a constant current is applied between the plates of capacitor \( I_c = 0 \) and hence \( I = I_d \)
Answer: (d) When a constant current is applied between the plates of capacitor \( I_c = 0 \) and hence \( I = I_d \)
In simple words: If a constant current is flowing in the circuit connected to a capacitor, there is conduction current in the wires, and a changing electric field between the plates causes displacement current. The statement that conduction current \( I_c = 0 \) when current is "applied" to the capacitor is incorrect, as current flows in the external wires to charge the capacitor.

🎯 Exam Tip: Distinguish between conduction current (flow of charges in wires) and displacement current (changing electric flux in the capacitor gap) and their relationship during capacitor charging.

 

Question 2. Choose the Incorrect Statement:
(a) Velocity of an electromagnetic wave in a medium of refractive index 1.5 is \( 2.25 \times 10^8 \text{ ms}^{-1} \)
(b) If the velocity of the electromagnetic wave is \( 2 \times 10^8 \text{ ms}^{-1} \) the refractive index of the medium is 2
(c) If the velocity of light is \( 2.25 \times 10^8 \text{ ms}^{-1} \) the medium must have a refractive index of 1.33
(d) If the refractive index of a medium increases the velocity of electromagnetic wave decreases.
Answer: (c) If the velocity of light is \( 2.25 \times 10^8 \text{ ms}^{-1} \) the medium must have a refractive index of 1.33
In simple words: This statement is considered incorrect in the context of the problem, possibly due to a subtle interpretation of "velocity of light" or an intended miscalculation. However, mathematically, if an electromagnetic wave travels at \( 2.25 \times 10^8 \text{ m/s} \) in a medium, its refractive index would indeed be approximately 1.33 (since \( n = c/v \), so \( 3 \times 10^8 / 2.25 \times 10^8 = 1.33 \)).

🎯 Exam Tip: Always double-check calculations for refractive index \( n = c/v \), where \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light in vacuum.

 

Question 3. Choose the Incorrect Statement:
(a) Electromagnetic wave \( I = \frac{\text{Energy (U)}}{\text{Time (T)}} \)
(b) Intensity of electromagnetic wave \( I = \frac{\text{Power (P)}}{\text{Surface Area (A)}} \)
(c) Energy density in an electric field is \( \frac{1}{2} \varepsilon_0 E^2 \)
(d) Energy density in a magnetic field is \( \frac{1}{2} \mu_0 B^2 \)
Answer: (d) Energy density in a magnetic field is \( \frac{1}{2} \mu_0 B^2 \)
In simple words: The energy density stored in a magnetic field is correctly given by the formula \( \frac{B^2}{2\mu_0} \). The option (d) incorrectly places \( \mu_0 \) in the numerator instead of the denominator.

🎯 Exam Tip: Carefully remember the constants and their positions (numerator/denominator) in the formulas for electric and magnetic energy densities.

 

Question 4. Choose the Incorrect Statement:
(a) Electromagnetic waves carry not only energy and momentum but also angular momentum in
(b) The dark line in the solar spectrum are known as Fraunhofer lines
(c) Solar spectrum is the best example for line absorption spectrum
(d) The instantaneous magnitude of the electric and magnetic field vectors in an electromagnetic wave are related by \( B = Ec \)
Answer: (d) The instantaneous magnitude of the electric and magnetic field vectors in an electromagnetic wave are related by \( B = Ec \)
In simple words: In an electromagnetic wave, the instantaneous magnitudes of the electric field (E) and magnetic field (B) are related by \( E = cB \) or \( B = E/c \), where \( c \) is the speed of light. The given relation \( B = Ec \) is incorrect.

🎯 Exam Tip: Clearly recall the fundamental relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave.

IX. Choose the best answer:

 

Question 1. The speed of electromagnetic waves in vacuum is given by
(a) \( \mu_0 \varepsilon_0 \)
(b) \( \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \)
(c) \( \sqrt { \frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } \)
(d) \( \frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } \)
Answer: (d) \( \frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } \)
In simple words: The speed of light in a vacuum, often denoted as \( c \), is a fundamental constant. It is defined by the inverse square root of the product of the permeability of free space (\( \mu_0 \)) and the permittivity of free space (\( \varepsilon_0 \)).

🎯 Exam Tip: This formula is a crucial definition in electromagnetism, linking fundamental constants to the speed of light.

 

Question 2. Electromagnetic waves are
(a) Neither longitudinal nor transverse
(b) Longitudinal
(c) transverse
(d) both longitudinal and transverse
Answer: (c) transverse
In simple words: In an electromagnetic wave, the electric and magnetic fields oscillate perpendicular to each other and also perpendicular to the direction the wave is traveling. This defines it as a transverse wave.

🎯 Exam Tip: Remember that electromagnetic waves are always transverse, meaning their oscillations are perpendicular to their direction of propagation.

 

Question 3. If E and B be the electric and magnetic field vectors of an electromagnetic wave, then the propagation of the wave is along the direction of
(a) E
(b) B
(c) E x B
(d) Bx E
Answer: (c) E x B
In simple words: The direction in which an electromagnetic wave travels is given by the cross product of the electric field vector (\( \vec{E} \)) and the magnetic field vector (\( \vec{B} \)). This is also the direction of the Poynting vector.

🎯 Exam Tip: The direction of wave propagation is consistently given by the right-hand rule for \( \vec{E} \times \vec{B} \).

 

Question 4. To produce a displacement of 8.854 mA between the parallel plate capacitor in 0.2 \( \mu \)s, the change in electric flux must be
(a) 200 Wb
(b) 20 Wb
(c) 2 Wb
(d) 0.2 Wb
Answer: (a) 200 Wb
In simple words: Displacement current is caused by a changing electric flux. To find out how much the electric flux changes, we multiply the displacement current by the time period and divide by the permittivity of free space.
Solution:
The displacement current \( I_d \) is given by:
\( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \)
We need to find the change in electric flux, \( d\Phi_E \). We can rearrange the formula:
\( d\Phi_E = \frac{I_d \times dt}{\varepsilon_0} \)
Given:
\( I_d = 8.854 \text{ mA} = 8.854 \times 10^{-3} \text{ A} \)
\( dt = 0.2 \mu \text{s} = 0.2 \times 10^{-6} \text{ s} \)
Permittivity of free space \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ F/m} \)
Substitute the values:
\( d\Phi_E = \frac{8.854 \times 10^{-3} \text{ A} \times 0.2 \times 10^{-6} \text{ s}}{8.854 \times 10^{-12} \text{ F/m}} \)
\( d\Phi_E = 0.2 \times 10^{-3} \times 10^6 = 0.2 \times 10^3 = 200 \text{ Wb} \)
Therefore, the change in electric flux must be 200 Wb.
In simple words: When a current of 8.854 mA flows for 0.2 microseconds through a capacitor, the electric flux between its plates must change by 200 Webers. This shows how displacement current is linked to changing electric fields.

🎯 Exam Tip: Remember the relationship between displacement current, permittivity, and the rate of change of electric flux, which is crucial for solving problems involving capacitors in circuits.

 

Question 5. Maxwell's modified Ampere's Law is
(a) \( \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} s}=\mu_{0}\left(\mathrm{I}_{\mathrm{c}}+\mathrm{I}_{\mathrm{d}}\right) \)
(b) \( \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} s}=\mu_{0} \mathrm{I}_{\mathrm{c}} \)
(c) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} s}=\mu_{0} \mathrm{I}_{\mathrm{c}} \)
(d) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} s}=\mu_{0}\left(\mathrm{I}_{\mathrm{c}}+\mathrm{I}_{\mathrm{d}}\right) \)
Answer: (d) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} s}=\mu_{0}\left(\mathrm{I}_{\mathrm{c}}+\mathrm{I}_{\mathrm{d}}\right) \)
In simple words: This equation, also known as Ampere-Maxwell's law, tells us that a magnetic field around a closed loop is created by both the actual flow of charge (conduction current, \( I_c \)) and a changing electric field (displacement current, \( I_d \)).

🎯 Exam Tip: Ensure you remember the correct form of Ampere-Maxwell's law, including both conduction and displacement current terms, and that it involves a line integral of the magnetic field.

 

Question 6. Hertz produced an electromagnetic wave of frequency \( 5 \times 10^7 \text{ Hz} \) means, the wavelength of the wave is
(a) 150 m
(b) 15 m
(c) 6 m
(d) 60 m
Answer: (c) 6 m
In simple words: To find the length of the wave, we divide the speed of light by the wave's frequency. This calculation shows that a wave with a frequency of \( 5 \times 10^7 \text{ Hz} \) has a wavelength of 6 meters.
Solution:
The relationship between wavelength \( \lambda \), speed of light \( c \), and frequency \( f \) is given by:
\( \lambda = \frac{c}{f} \)
Given:
Speed of light \( c = 3 \times 10^8 \text{ m/s} \)
Frequency \( f = 5 \times 10^7 \text{ Hz} \)
Substitute the values into the formula:
\( \lambda = \frac{3 \times 10^8 \text{ m/s}}{5 \times 10^7 \text{ Hz}} \)
\( \lambda = \frac{30}{5} = 6 \text{ m} \)
Therefore, the wavelength of the electromagnetic wave is 6 m.
In simple words: If an electromagnetic wave vibrates 50 million times a second, its wavelength in free space will be 6 meters long. This is how frequency and wavelength are related for all light waves.

🎯 Exam Tip: Always use the formula \( \lambda = c/f \) for electromagnetic waves, where `c` is the speed of light in vacuum.

 

Question 7. In Hertz experiment, the energy of the electromagnetic wave is
(a) Kinetic energy of the oscillating charge
(b) Potential energy of the charge
(c) Electric energy between plates
(d) Magnetic energy of the connecting wires
Answer: (a) Kinetic energy of the oscillating charge
In simple words: In Hertz's experiment, the energy carried by the electromagnetic waves originated from the movement of the charged particles in the oscillator. These particles had kinetic energy as they rapidly moved back and forth.

🎯 Exam Tip: Remember that electromagnetic waves are produced by accelerating charges, and their energy is derived from the energy of these oscillating charges.

 

Question 8. A medium has a refractive index of 1.5 with relative permittivity of 2 has a relative magnetic permeability
(a) 11.25
(b) 112.5
(c) 1.125
(d) 2.125
Answer: (c) 1.125
In simple words: The refractive index of a material is determined by both its relative permittivity and relative magnetic permeability. Knowing two of these values lets us calculate the third.
Solution:
The refractive index \( n \) of a medium is related to its relative magnetic permeability \( \mu_r \) and relative permittivity \( \varepsilon_r \) by the formula:
\( n = \sqrt{\mu_r \varepsilon_r} \)
Given:
Refractive index \( n = 1.5 \)
Relative permittivity \( \varepsilon_r = 2 \)
We need to find the relative magnetic permeability \( \mu_r \).
Substitute the given values into the formula:
\( 1.5 = \sqrt{\mu_r \times 2} \)
To solve for \( \mu_r \), first square both sides of the equation:
\( (1.5)^2 = \mu_r \times 2 \)
\( 2.25 = 2 \mu_r \)
Now, divide by 2:
\( \mu_r = \frac{2.25}{2} \)
\( \mu_r = 1.125 \)
Therefore, the relative magnetic permeability of the medium is 1.125.
In simple words: For a material with a refractive index of 1.5 and a relative permittivity of 2, its relative magnetic permeability must be 1.125. This helps us understand how the material affects electromagnetic waves.

🎯 Exam Tip: Always remember the formula \( n = \sqrt{\mu_r \varepsilon_r} \) and be careful with squaring and square roots during calculations.

 

Question 9. The electromagnetic radiation most prevalent in the atmosphere is
(a) Visible light
(b) Infrared
(c) UV
(d) Radio waves
Answer: (b) Infrared
In simple words: Infrared radiation is the most common electromagnetic radiation found in the Earth's atmosphere. It is often felt as heat.

🎯 Exam Tip: Remember the different types of electromagnetic radiation and their properties, especially their prevalence in the atmosphere.

 

Question 10. An electromagnetic wave of energy U is completely transferred to a surface, the momentum imparted on the surface is
(a) U
(b) Uc
(c) U/c
(d) c/U
Answer: (c) U/c
In simple words: When an electromagnetic wave completely transfers its energy to a surface, the momentum it gives to that surface is found by dividing its energy by the speed of light. This is a fundamental concept in radiation pressure.

🎯 Exam Tip: For problems involving total energy transfer, the momentum imparted is always the energy divided by the speed of light.

 

Question 11. The average energy density of an electromagnetic wave of magnetic field \( 4\pi \times 10^{-7} \) T is
(a) \( 2\pi \times 10^{-7} \text{ Jm}^{-3} \)
(b) \( 4\pi \times 10^{-7} \text{ Jm}^{-3} \)
(c) \( 19.878 \times 10^{-6} \text{ Jm}^{-3} \)
(d) \( 6.626 \times 10^{-7} \text{ Jm}^{-3} \)
Answer: (d) \( 6.626 \times 10^{-7} \text{ Jm}^{-3} \)
Solution:
The formula for average magnetic energy density is \( = \frac{1}{2} \frac{B^2}{\mu_0} \).
Given magnetic field \( B = 4\pi \times 10^{-7} \text{ T} \).
The permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \).
Now, we put these values into the formula:
\( = \frac{1}{2} \times \frac{(4\pi \times 10^{-7})^2}{4\pi \times 10^{-7}} \)
\( = \frac{1}{2} \times \frac{16\pi^2 \times 10^{-14}}{4\pi \times 10^{-7}} \)
\( = \frac{1}{2} \times 4\pi \times 10^{-7} \)
\( = 2\pi \times 10^{-7} \text{ Jm}^{-3} \)
Wait, there's a discrepancy in the source answer (d) and the calculation using the magnetic field value. Let's re-examine the question's provided options or if the question meant a different type of energy density.
The question asks for the energy density for a magnetic field of \( 4\pi \times 10^{-7} \text{ T} \).
The calculation for magnetic energy density \( U_B = \frac{B^2}{2\mu_0} \) gives:
\( U_B = \frac{(4\pi \times 10^{-7})^2}{2 \times 4\pi \times 10^{-7}} \)
\( = \frac{16\pi^2 \times 10^{-14}}{8\pi \times 10^{-7}} \)
\( = 2\pi \times 10^{-7} \text{ Jm}^{-3} \)
\( \approx 2 \times 3.14159 \times 10^{-7} \)
\( \approx 6.283 \times 10^{-7} \text{ Jm}^{-3} \)
This matches option (a). However, the source provided (d) as the answer which is \( 6.626 \times 10^{-7} \text{ Jm}^{-3} \). The provided solution shows \( = \frac{1}{2} \times \frac{1}{4\pi \times 10^{-7}} (4\pi \times 10^{-7})^2 = 2\pi \times 10^{-7} \text{ Jm}^{-3} \). This result is clearly (a). Let's follow the calculation shown in the provided solution itself which leads to (a). It seems there might be an error in the given answer option (d) in the source. I will output the answer based on the correct calculation shown in the solution, which is (a). The source has a typo in its stated answer option.
Answer: (a) \( 2\pi \times 10^{-7} \text{ Jm}^{-3} \)
Solution:
The average energy density \( \) for an electromagnetic wave due to its magnetic field is given by the formula:
\( = \frac{1}{2} \frac{B^2}{\mu_0} \)
Here, \( B \) is the magnetic field strength and \( \mu_0 \) is the permeability of free space.
Given: Magnetic field \( B = 4\pi \times 10^{-7} \text{ T} \)
Value of \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \)
Substitute these values into the formula:
\( = \frac{1}{2} \times \frac{(4\pi \times 10^{-7})^2}{4\pi \times 10^{-7}} \)
\( \implies = \frac{1}{2} \times \frac{16\pi^2 \times 10^{-14}}{4\pi \times 10^{-7}} \)
\( \implies = \frac{1}{2} \times 4\pi \times 10^{-7} \)
\( \implies = 2\pi \times 10^{-7} \text{ Jm}^{-3} \)
This value matches option (a). Magnetic energy density quantifies the amount of energy stored per unit volume within a magnetic field.
In simple words: The energy stored in the magnetic field of an electromagnetic wave, per unit volume, can be calculated using a simple formula with the magnetic field strength and a constant called permeability of free space. When you do the math, it comes out to \( 2\pi \times 10^{-7} \) joules per cubic meter.

🎯 Exam Tip: Remember the formula for magnetic energy density \( U_B = \frac{B^2}{2\mu_0} \) and electric energy density \( U_E = \frac{1}{2}\varepsilon_0 E^2 \). Ensure you use the correct physical constants and units.

 

Question 12. The greenhouse effect is caused by
(a) UV rays
(b) X-rays
(c) Gamma rays
(d) IR rays
Answer: (d) IR rays.
In simple words: The greenhouse effect, which warms the Earth, is mainly caused by infrared (IR) radiation. This type of radiation gets trapped by gases in our atmosphere.

🎯 Exam Tip: Understand which types of electromagnetic radiation are involved in different atmospheric phenomena. Infrared radiation is key to the greenhouse effect.

 

Question 13. The velocity of electromagnetic wave in free space or vacuum is
(a) \( 3 \times 10^{-8} \text{ ms}^{-1} \)
(b) \( \sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}} \text{ ms}^{-1} \)
(c) \( \frac{1}{\mu_{r} \varepsilon_{r}} \text{ ms}^{-1} \)
(d) \( \left(\varepsilon_{0} \mu_{0}\right)^{-1 / 2} \text{ ms}^{-1} \)
Answer: (d) \( \left(\varepsilon_{0} \mu_{0}\right)^{-1 / 2} \text{ ms}^{-1} \)
In simple words: The speed of light in empty space, or vacuum, is determined by a special formula involving two fundamental constants: the permittivity and permeability of free space. This speed is a constant value.

🎯 Exam Tip: Know the fundamental relationship \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \) for the speed of light in vacuum. It's often expressed as \( (\mu_0 \varepsilon_0)^{-1/2} \).

 

Question 14. The velocity of an electromagnetic wave through a medium which a permeability in free space \( 4\pi \times 10^{-7} \text{ Hm}^{-1} \) and permittivity in free space of \( 8.854 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2} \) is
(a) \( 2 \times 10^8 \text{ ms}^{-1} \)
(b) \( 3 \times 10^8 \text{ ms}^{-1} \)
(c) \( 2.25 \times 10^8 \text{ ms}^{-1} \)
(d) \( 1.5 \times 10^8 \text{ ms}^{-1} \)
Answer: (b) \( 3 \times 10^8 \text{ ms}^{-1} \)
Solution:
The velocity of an electromagnetic wave in free space \( C \) is given by:
\( C = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \)
Given:
Permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1} \)
Permittivity of free space \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2} \)
Substitute the values:
\( C = \frac{1}{\sqrt{4\pi \times 10^{-7} \times 8.854 \times 10^{-12}}} \)
\( C = \frac{1}{\sqrt{1.112 \times 10^{-16} \times \pi}} \)
\( C \approx \frac{1}{\sqrt{3.5 \times 10^{-16}}} \)
\( C \approx \frac{1}{1.87 \times 10^{-8}} \)
\( C \approx 0.534 \times 10^8 \text{ ms}^{-1} \)
Let's re-calculate: \( 4\pi \times 10^{-7} \times 8.854 \times 10^{-12} \approx 12.566 \times 8.854 \times 10^{-19} \approx 111.26 \times 10^{-19} \approx 1.1126 \times 10^{-17} \).
So, \( C = \frac{1}{\sqrt{1.1126 \times 10^{-17}}} = \frac{1}{\sqrt{0.11126 \times 10^{-16}}} \approx \frac{1}{0.3335 \times 10^{-8}} \approx 3 \times 10^8 \text{ ms}^{-1} \).
This calculation confirms option (b). The velocity of an electromagnetic wave in vacuum is a universal constant, often denoted as \( c \).
In simple words: The speed of an electromagnetic wave in empty space is a fixed value, which can be found using specific electrical and magnetic constants. When you put these numbers into the formula, the speed comes out to be about \( 3 \times 10^8 \) meters per second, which is the speed of light.

🎯 Exam Tip: Always remember that the speed of light in a vacuum \( c \) is a fundamental constant, approximately \( 3 \times 10^8 \text{ m/s} \), and can be calculated using \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \).

 

Question 15. The peak value of an electric field in an electromagnetic wave is \( 3\sqrt{2} \text{ Vm}^{-1} \). The peak value of the magnetic field is
(a) \( 1.414 \times 10^{-8} \text{ T} \)
(b) \( 1.0 \times 10^{-8} \text{ T} \)
(c) \( 2.828 \times 10^{-8} \text{ T} \)
(d) \( 2.0 \times 10^{-8} \text{ T} \)
Answer: (a) \( 1.414 \times 10^{-8} \text{ T} \)
Solution:
The relationship between the peak electric field \( E_0 \), peak magnetic field \( B_0 \), and the speed of light \( C \) in an electromagnetic wave is given by:
\( E_0 = C \times B_0 \)
So, to find the peak magnetic field \( B_0 \), we rearrange the formula:
\( B_0 = \frac{E_0}{C} \)
Given: Peak electric field \( E_0 = 3\sqrt{2} \text{ Vm}^{-1} \)
Speed of light \( C = 3 \times 10^8 \text{ ms}^{-1} \)
Substitute the values:
\( B_0 = \frac{3\sqrt{2}}{3 \times 10^8} \)
\( \implies B_0 = \frac{\sqrt{2}}{10^8} \)
\( \implies B_0 = 1.414 \times 10^{-8} \text{ T} \)
This value corresponds to option (a). This relationship is fundamental to understanding electromagnetic wave propagation. The electric and magnetic fields oscillate perpendicularly to each other and to the direction of wave travel.
In simple words: In an electromagnetic wave, the strength of the electric field and the magnetic field are linked by the speed of light. If you know the peak electric field, you can find the peak magnetic field by dividing the electric field by the speed of light. For the given values, this calculates to \( 1.414 \times 10^{-8} \) Tesla.

🎯 Exam Tip: Remember the relation \( E_0 = C B_0 \) for the peak values of electric and magnetic fields in an electromagnetic wave. \( \sqrt{2} \) is approximately 1.414.

 

Question 16. Which of the following is not an electromagnetic wave
(a) x-ray
(b) \( \gamma \)-ray
(c) \( \beta \)-ray
(d) y-ray
Answer: (c) \( \beta \)-ray
In simple words: X-rays, gamma-rays, and y-rays are all types of electromagnetic waves, which means they are made of vibrating electric and magnetic fields. Beta-rays, however, are streams of electrons or positrons, not electromagnetic waves.

🎯 Exam Tip: Be able to distinguish between different types of radiation. Electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Beta-rays are particles (electrons/positrons).

 

Question 17. The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to
(a) the speed of light in vacuum
(b) reciprocal of the speed of light in vacuum
(c) the ratio of magnetic permeability to electric susceptibility in vacuum
(d) unity
Answer: (b) reciprocal of the speed of light in vacuum.
In simple words: The ratio of the magnetic field's strength to the electric field's strength in an electromagnetic wave in empty space is equal to one divided by the speed of light. This means the magnetic field is weaker than the electric field by a factor of the speed of light.

🎯 Exam Tip: Understand the relationship \( E_0 = C B_0 \). Therefore, the ratio \( B_0/E_0 = 1/C \), which is the reciprocal of the speed of light.

 

Question 18. The current in the electric circuit which arises due to the flow of electrons in the connecting wires of the circuit in a defined closed path is called
(a) alternating current
(b) direct current
(c) conduction current
(d) displacement current
Answer: (c) conduction current
In simple words: The current created by the actual movement of electrons through a wire in a complete circuit is known as conduction current. This is the normal type of electric current we usually talk about.

🎯 Exam Tip: Differentiate between conduction current (flow of charges in a conductor) and displacement current (current due to changing electric flux).

 

Question 19. The conduction current is the same as the displacement current when the source is
(a) ac only
(b) dc only
(c) either ac or dc
(d) neither dc nor ac
Answer: (c) either ac or dc
In simple words: In a circuit like a capacitor, the conduction current flowing into the plates is always equal to the displacement current between the plates, whether the source is alternating current (AC) or direct current (DC). This keeps the total current continuous.

🎯 Exam Tip: For a capacitor being charged or discharged, the total current (conduction + displacement) is always continuous. In the gap of a capacitor, the displacement current equals the conduction current in the wires.

 

Question 20. If a variable frequency ac source connected to a capacitor then with a decrease in frequency, the displacement current will
(a) increase
(b) decrease
(c) remains constant
(d) first decrease then increase
Answer: (b) decrease
Solution:
For a capacitor connected to an AC source, the current \( I \) flowing through it is given by:
\( I = \frac{E}{X_C} \)
Where \( E \) is the voltage and \( X_C \) is the capacitive reactance.
The capacitive reactance \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi \nu C} \)
So, the current \( I = E \times (2\pi \nu C) = 2\pi \nu CE \)
This shows that \( I \propto \nu \), where \( \nu \) is the frequency.
As the conduction current \( I_c \) is equal to the displacement current \( I_d \) in the capacitor gap (for a closed loop), \( I_d \) will also be proportional to the frequency.
Therefore, if the frequency \( \nu \) of the AC source decreases, the displacement current \( I_d \) will also decrease. This is because a lower frequency means the electric field changes more slowly, leading to a smaller displacement current.
In simple words: When a capacitor is connected to an alternating current (AC) source, the displacement current through the capacitor is directly linked to the frequency of the source. So, if the frequency goes down, the displacement current will also go down because the electric field is changing more slowly.

🎯 Exam Tip: Remember that for a capacitor, both conduction current and displacement current are directly proportional to the frequency of the AC source. A lower frequency means a slower change in the electric field, leading to a smaller displacement current.

 

Question 21. An electromagnetic wave has a wavelength of 10 cm. It is in the
(a) Visible region
(b) Radio region
(c) UV region
(d) X-ray region
Answer: (b) Radio region.
In simple words: An electromagnetic wave with a wavelength of 10 centimeters falls into the radio wave part of the spectrum. Radio waves have longer wavelengths than visible light, UV, or X-rays.

🎯 Exam Tip: Familiarize yourself with the electromagnetic spectrum, knowing the typical wavelength and frequency ranges for each type of wave (radio, microwave, infrared, visible, UV, X-ray, gamma ray).

 

Question 22. The displacement current was first postulated by
(a) Maxwell
(b) Marconi
(c) Ampere
(d) Hertz
Answer: (a) Maxwell
In simple words: James Clerk Maxwell was the first to propose the idea of displacement current. This concept was crucial for making Ampere's law complete and for developing the theory of electromagnetic waves.

🎯 Exam Tip: Maxwell's contributions to electromagnetism, including the concept of displacement current, are foundational. Remember his role in unifying electricity, magnetism, and light.

 

Question 23. Ampere's circuital law holds good for
(a) conduction current
(b) displacement current
(c) both (a) and (b)
(d) None of the options
Answer: (c) both (a) and (b)
Solution:
Ampere's original circuital law only considered conduction current:
\( \oint \vec{B} \cdot \vec{dl} = \mu_0 I_{\text{conduction}} \)
However, Maxwell modified Ampere's law to include displacement current, making it applicable to time-varying electric fields as well:
\( \oint \vec{B} \cdot \vec{dl} = \mu_0 (I_{\text{conduction}} + I_{\text{displacement}}) \)
The displacement current is given by \( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \).
Therefore, the modified Ampere's circuital law (also known as Ampere-Maxwell law) holds good for both conduction current and displacement current. This modification was crucial for the prediction of electromagnetic waves. In steady-state conditions, \( I_d = 0 \) and the law reduces to Ampere's original law.
In simple words: The updated version of Ampere's circuital law, called the Ampere-Maxwell law, works for both regular current (conduction current) that flows through wires and the current that appears when electric fields change (displacement current). This makes the law applicable in all situations.

🎯 Exam Tip: Distinguish between Ampere's original law and Maxwell's modification. The modified law, often called Ampere-Maxwell's law, includes both conduction and displacement currents, making it universally applicable.

 

Question 24. Which of the following has maximum frequency?
(a) X-rays
(b) IR rays
(c) UV rays
(d) \( \gamma \)-rays
Answer: (d) \( \gamma \)-rays
In simple words: Among the options given, gamma-rays have the highest frequency in the electromagnetic spectrum. Higher frequency means more energy.

🎯 Exam Tip: Remember the order of electromagnetic waves in terms of increasing frequency: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma ray.

 

Question 25. Displacement current is due to
(a) continuous when an electric field is changing in the circuit
(b) continuous when the magnetic field is changing in the circuit
(c) continuous in both types of fields
(d) continuous through wires and resistance only
Answer: (a) continuous when an electric field is changing in the circuit
Solution:
Displacement current arises from a changing electric flux through a surface. According to Maxwell, \( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). This means whenever the electric field is changing over time in a region, a displacement current is produced. It is not about a continuous electric field but a continuously *changing* electric field.
In simple words: Displacement current happens when the electric field in a circuit is constantly changing. It's like a special kind of current that flows even when there are no actual charges moving, as long as the electric field is varying.

🎯 Exam Tip: Understand that displacement current is proportional to the *rate of change* of electric flux, not just the presence of an electric field.

 

Question 26. The displacement current is
(a) \( \varepsilon_0 \frac{d\Phi_E}{dt} \)
(b) \( \frac{\varepsilon_{0}}{R} \frac{d\Phi_{E}}{dt} \)
(c) \( \varepsilon E / R \)
(d) \( EqC / R \)
Answer: (a) \( \varepsilon_0 \frac{d\Phi_E}{dt} \)
In simple words: Displacement current is defined as the permittivity of free space multiplied by the rate at which the electric flux changes over time. This formula tells us how a changing electric field generates a "current" effect.

🎯 Exam Tip: This is the fundamental definition of displacement current. Memorize this formula as it is central to Maxwell's equations.

 

Question 27. The frequency of a wave is \( 6 \times 10^{15} \) Hz. The wave is
(a) Visible region
(b) Microwave
(c) X-ray
(d) UV rays
Answer: (d) UV rays
In simple words: A wave that vibrates \( 6 \times 10^{15} \) times per second falls into the ultraviolet (UV) part of the electromagnetic spectrum. UV light has higher frequencies than visible light.

🎯 Exam Tip: Recall the frequency ranges for different electromagnetic waves. Ultraviolet light typically has frequencies in the range of \( 10^{15} \) to \( 10^{17} \) Hz.

 

Question 28. If \( \overrightarrow{\mathbf{E}} \) and \( \overrightarrow{\mathbf{B}} \) represent the electric and magnetic field vectors of an electromagnetic wave, then the direction of propagation of the electromagnetic wave, is along
(a) \( \overrightarrow{\mathbf{E}} \)
(b) \( \overrightarrow{\mathbf{B}} \)
(c) \( \overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{E}} \)
(d) \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \)
Answer: (d) \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \)
Solution:
In an electromagnetic wave, the electric field vector \( \overrightarrow{\mathbf{E}} \), the magnetic field vector \( \overrightarrow{\mathbf{B}} \), and the direction of propagation are mutually perpendicular. The direction of propagation is given by the Poynting vector, which is parallel to \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \). This means the wave travels in the direction of the cross product of the electric and magnetic field vectors. If \( \overrightarrow{\mathbf{E}} \) is along the y-axis and \( \overrightarrow{\mathbf{B}} \) is along the z-axis, then \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \) will be along the x-axis, which is the direction of wave propagation.
In simple words: The direction an electromagnetic wave travels is determined by using the right-hand rule with the electric and magnetic fields. You point your fingers in the direction of the electric field and curl them towards the magnetic field, and your thumb will show the wave's direction. This is represented mathematically by the cross product \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \).

🎯 Exam Tip: Always remember that the direction of propagation of an electromagnetic wave is given by the cross product \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \). The three vectors \( \overrightarrow{\mathbf{E}} \), \( \overrightarrow{\mathbf{B}} \), and \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \) form a mutually perpendicular system.

 

Question 29. According to Maxwell's hypothesis, a changing electric field gives rise to
(a) an e.m.f
(b) electric displacement field
(c) magnetic field
(d) pressure gradient
Answer: (c) magnetic field
Solution:
Maxwell proposed that a changing electric field creates a magnetic field. This is the essence of displacement current and is symmetrical to Faraday's law, which states that a changing magnetic field creates an electric field. This fundamental insight was critical for the existence of electromagnetic waves.
In simple words: Maxwell suggested that when an electric field changes over time, it creates a magnetic field. This idea was very important because it showed how electricity and magnetism are linked and how waves like light can travel.

🎯 Exam Tip: Maxwell's extension of Ampere's law, introducing displacement current, established that a time-varying electric field is a source of a magnetic field, just as a time-varying magnetic field is a source of an electric field (Faraday's law).

 

Question 30. An electromagnetic wave propagating along the north has its electric field vector point towards
(a) north
(b) east
(c) west
(d) downwards
Answer: (b) east
In simple words: If an electromagnetic wave is moving north, its electric field and magnetic field must be at right angles to that direction. If the electric field points east, the magnetic field would point downwards (or west if electric field was up), keeping all three directions perpendicular to each other.

🎯 Exam Tip: Remember that in an electromagnetic wave, the electric field, magnetic field, and direction of propagation are mutually perpendicular. If propagation is North, E and B must be in the East-West and Up-Down planes.

 

Question 31. Vertical straight conduct carries a current vertically upwards. A point P lies to the east of it at a small distance and another point Q lies to the west at the same distance. The magnetic field at P is
(a) Greater than that at Q
(b) Same as at Q
(c) Less than at Q
(d) Greater or less than at Q depending upon the strength of the current.
Answer: (b) Same as at Q
In simple words: For a straight wire carrying current, the strength of the magnetic field at any point depends only on how far away it is from the wire. So, if points P and Q are the same distance away, the magnetic field strength will be the same at both places, even if their directions are opposite.

🎯 Exam Tip: According to Ampere's law and the Biot-Savart law, the magnitude of the magnetic field around an infinitely long straight conductor depends only on the current and the perpendicular distance from the wire.

 

Question 32. Gamma rays are used in the treatment of
(a) Cancer
(b) Polio
(c) AIDS
(d) Tuberculosis
Answer: (a) Cancer
In simple words: Gamma rays are used in medicine to treat cancer because their high energy can kill cancer cells. This treatment is known as radiotherapy.

🎯 Exam Tip: Recall the applications of different parts of the electromagnetic spectrum. Gamma rays are often used in medical treatments due to their high penetrating power and ability to damage cells.

 

Question 33. Which of the following types of radiations are radiated by an oscillating electric charge?
(a) Electric
(b) Magnetic
(c) Thermoelectric
(d) Electromagnetic
Answer: (d) Electromagnetic
In simple words: An electric charge that vibrates back and forth produces electromagnetic radiation. These waves are made of both electric and magnetic fields that travel together.

🎯 Exam Tip: Remember that accelerating charges (like oscillating charges) are the source of electromagnetic waves. Stationary charges produce only electric fields, and charges moving at constant velocity produce static magnetic fields but not radiating waves.

 

Question 34. If \( \overrightarrow{\mathbf{E}} \) and \( \overrightarrow{\mathbf{B}} \) are the electric and magnetic field vectors of e.m. waves then the direction of propagation of e.m. wave is along the direction of
(a) \( \overrightarrow{\mathbf{E}} \)
(b) \( \overrightarrow{\mathbf{B}} \)
(c) \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \)
(d) None of these
Answer: (c) \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \)
Solution:
The direction of propagation of an electromagnetic wave is perpendicular to both the electric field vector \( \overrightarrow{\mathbf{E}} \) and the magnetic field vector \( \overrightarrow{\mathbf{B}} \). This direction is given by the vector cross product \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \). This means that if \( \overrightarrow{\mathbf{E}} \) is along one axis and \( \overrightarrow{\mathbf{B}} \) is along another, the wave propagates along the third perpendicular axis. This is often visualized with the right-hand rule. The Poynting vector, which represents the direction and magnitude of energy flow, is also in this direction.
In simple words: The way an electromagnetic wave moves through space is shown by the cross product of its electric and magnetic field vectors. Imagine \( \overrightarrow{\mathbf{E}} \) and \( \overrightarrow{\mathbf{B}} \) forming a plane; the wave travels perpendicular to this plane.

🎯 Exam Tip: The cross product \( \overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}} \) defines the direction of energy flow (Poynting vector) and thus the direction of propagation for an electromagnetic wave.

 

Question 35. According to Maxwell's equation, the velocity of light in any medium is expressed as
(a) \( \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \)
(b) \( \frac{1}{\sqrt{\mu \varepsilon}} \)
(c) \( \sqrt{\frac{\mu}{\varepsilon}} \)
(d) \( \sqrt{\frac{\mu_{0}}{\varepsilon}} \)
Answer: (b) \( \frac{1}{\sqrt{\mu \varepsilon}} \)
Solution:
The velocity of light \( v \) in any medium is given by the formula \( v = \frac{1}{\sqrt{\mu \varepsilon}} \), where \( \mu \) is the permeability of the medium and \( \varepsilon \) is the permittivity of the medium. For free space (vacuum), \( \mu \) becomes \( \mu_0 \) and \( \varepsilon \) becomes \( \varepsilon_0 \), so the velocity becomes \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \). This formula demonstrates how the properties of the medium affect the speed of light within it. The ratio of the speed of light in vacuum to the speed in a medium gives the refractive index of that medium.
In simple words: The speed of light when it travels through any material, not just empty space, depends on two properties of that material: its permeability and its permittivity. The formula shows that light travels slower in materials than in a vacuum because these properties are different.

🎯 Exam Tip: Differentiate between the speed of light in vacuum \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \) and the speed of light in a medium \( v = \frac{1}{\sqrt{\mu \varepsilon}} \), where \( \mu = \mu_r \mu_0 \) and \( \varepsilon = \varepsilon_r \varepsilon_0 \).

 

Question 36. The electromagnetic waves do not transport
(a) energy
(b) charge
(c) momentum
(d) information
Answer: (b) charge
In simple words: Electromagnetic waves carry energy, momentum, and can transmit information, but they do not carry any electric charge themselves. They are vibrations of fields, not particles with charge.

🎯 Exam Tip: Electromagnetic waves are energy carriers, but they are massless and chargeless. They transmit energy and momentum without transporting matter.

 

Question 37. The amplitudes of electric and magnetic fields related to each other are
(a) \( E_0 = B_0 \)
(b) \( E_0 = cB_0 \)
(c) \( E_0 = \frac{B_0}{c} \)
(d) \( E_0 = \frac{c}{B_0} \)
Answer: (b) \( E_0 = cB_0 \)
In simple words: The maximum strength of the electric field in an electromagnetic wave is equal to the speed of light multiplied by the maximum strength of the magnetic field. This means the electric field is always much stronger than the magnetic field.

🎯 Exam Tip: This is a key relationship in electromagnetic waves, connecting the peak amplitudes of the electric and magnetic fields with the speed of light. It's often used in calculations.

 

Question 38. In an electromagnetic wave, the direction of the magnetic induction \( \overrightarrow{\mathbf{B}} \) is
(a) parallel to the electric field \( \overrightarrow{\mathrm{E}} \)
(b) perpendicular to the electric field \( \overrightarrow{\mathrm{E}} \)
(c) antiparallel to the pointing vector \( \overrightarrow{\mathrm{S}} \)
(d) random
Answer: (b) perpendicular to the electric field \( \overrightarrow{\mathrm{E}} \)
In simple words: In an electromagnetic wave, the magnetic field always vibrates at a right angle (90 degrees) to the electric field. Both of them are also at right angles to the direction the wave is traveling.

🎯 Exam Tip: Remember the orthogonality: \( \overrightarrow{\mathbf{E}} \), \( \overrightarrow{\mathbf{B}} \), and the direction of propagation are all mutually perpendicular in an electromagnetic wave.

 

Question 39. The speed of the electromagnetic wave is the same for
(a) odd frequencies
(b) even frequencies
(c) all frequencies
(d) all intensities
Answer: (c) all frequencies
Solution:
In a vacuum, all electromagnetic waves, regardless of their frequency or wavelength, travel at the same speed, which is the speed of light \( c \). This constant speed is a defining characteristic of electromagnetic waves in free space. The speed of light is only affected by the medium it travels through, not by its frequency or intensity. However, different frequencies will experience different speeds in a *dispersive medium*. The question refers to "the electromagnetic wave" in general terms, implying vacuum unless specified, where speed is constant for all frequencies.
In simple words: In empty space, all types of electromagnetic waves, whether they are radio waves or X-rays, travel at the exact same speed. This speed is constant, no matter how high or low their frequency is.

🎯 Exam Tip: A crucial property of electromagnetic waves in a vacuum is that their speed is independent of their frequency or wavelength. This is a common point of confusion; be sure to recall this fundamental concept.

 

Question 40. A plane electromagnetic wave is incident on a material surface. If the wave delivers momentum p and energy E, then
(a) p = 0, E = 0
(b) p \( \ne \) 0, E \( \ne \) 0
(c) p \( \ne \) 0, E = 0
(d) p = 0, E \( \ne \) 0
Answer: (b) p \( \ne \) 0, E \( \ne \) 0
Solution:
Electromagnetic waves carry both energy and momentum. When an electromagnetic wave is incident on a material surface, it transfers both its energy and momentum to that surface. Therefore, neither the momentum (p) nor the energy (E) transferred will be zero. This transfer of momentum results in radiation pressure exerted by the wave on the surface. This effect is used in technologies like solar sails.
In simple words: When light or any electromagnetic wave hits a surface, it pushes on that surface and also gives it some energy. So, both momentum and energy are transferred, meaning neither of them is zero.

🎯 Exam Tip: Remember that electromagnetic waves possess both energy and momentum. The transfer of this momentum to a surface is what causes radiation pressure.

 

Question 41. We consider the radiation emitted by the human body. Which one of the following statements is true?
a) The radiation emitted is in the infrared region
b) The radiation is emitted only during the day
c) The radiation is emitted during the summers and absorbed during winters
d) The radiation is emitted lies in the ultraviolet region and hence is not visible
Answer: a) The radiation emitted is in the infrared region
Solution:
All objects with a temperature above absolute zero (0 Kelvin) emit thermal radiation. The human body, being at a temperature of around 37°C (310 K), primarily emits radiation in the infrared region of the electromagnetic spectrum. This happens continuously, day and night, regardless of the season. Infrared radiation is not visible to the human eye but can be detected by special cameras (thermal cameras).
In simple words: The human body, because it is warm, continuously gives off a type of radiation called infrared. This is why thermal cameras can see people in the dark, as infrared light is a form of heat energy.

🎯 Exam Tip: Understand the concept of black-body radiation and Wien's displacement law, which explains that the peak wavelength of emitted radiation shifts with temperature. Human body temperature results in peak emission in the infrared range.

 

Question 42. The decreasing order of the wavelength of the infrared, microwave, ultraviolet, and gamma rays are
a) microwave, infrared, ultraviolet, gamma rays
b) infrared, microwave, ultraviolet, gamma rays
c) gamma rays, ultraviolet, infrared, microwaves
d) microwaves, gamma rays, infrared, ultraviolet
Answer: a) microwave, infrared, ultraviolet, gamma rays
Solution:
To arrange them in decreasing order of wavelength, we need to know their relative positions in the electromagnetic spectrum.
The electromagnetic spectrum in order of decreasing wavelength (or increasing frequency/energy) is:
Radio waves > Microwaves > Infrared > Visible light > Ultraviolet > X-rays > Gamma rays.
From the given options: microwave, infrared, ultraviolet, gamma rays.
Arranging these in decreasing order of wavelength:
Microwaves (longest wavelength among the options)
Infrared
Ultraviolet
Gamma rays (shortest wavelength among the options)
Thus, the order is microwave, infrared, ultraviolet, gamma rays. Understanding the electromagnetic spectrum helps to remember the order. Each type of wave has distinct properties and uses.
In simple words: To list these types of waves from the longest wavelength to the shortest, we follow the order of the electromagnetic spectrum. Microwaves have the longest wavelength, followed by infrared, then ultraviolet, and finally, gamma rays have the shortest wavelength.

🎯 Exam Tip: Memorize the complete electromagnetic spectrum in order of increasing/decreasing wavelength (or frequency). This is a very common type of question. Use mnemonics if helpful.

 

Question 43. Name the Em waves used for studying the molecular structure and also in forensic investigations.
a) UV rays
b) Gamma rays
c) \( \lambda \)-rays
d) X-rays
Answer: a) UV rays
In simple words: Ultraviolet (UV) rays are used to study how molecules are put together and are also helpful in police investigations, like finding hidden evidence or examining documents.

🎯 Exam Tip: Connect the applications of different electromagnetic waves with their properties. UV light, for instance, interacts with molecules in ways that reveal their structure and presence.

 

Question 44. Radio waves diffract around the building, although light waves do not, The reason is that radio waves
a) travel with a speed larger than c
b) have a much larger wavelength than light
c) are not electromagnetic waves
d) None of the options
Answer: b) have a much larger wavelength than light
Solution:
Diffraction is the bending of waves around obstacles or through openings. The extent of diffraction is significant when the wavelength of the wave is comparable to or larger than the size of the obstacle or opening. Radio waves have wavelengths ranging from meters to kilometers, which are much larger than the typical size of buildings. Visible light, on the other hand, has wavelengths in the order of nanometers, which are much smaller than buildings. Therefore, radio waves diffract significantly around buildings, allowing us to receive radio signals even behind obstacles, while light waves do not appear to diffract around everyday objects. This is a practical demonstration of wave phenomena.
In simple words: Radio waves can bend around buildings because their wavelengths are very long, similar to the size of a building. Light waves have much shorter wavelengths, so they don't bend noticeably around buildings, which is why we can't see around corners.

🎯 Exam Tip: Remember that the degree of diffraction depends on the ratio of the wavelength to the size of the obstacle. Significant diffraction occurs when the wavelength is comparable to or larger than the obstacle size.

 

Question 45. Microwaves are detected by
a) bolometer
b) point contact diodes
c) thermopiles
d) the eye
Answer: b) point contact diodes
In simple words: Microwaves are detected using special electronic devices called point contact diodes. These diodes are very sensitive to the high frequencies of microwaves.

🎯 Exam Tip: Associate different detection methods with specific regions of the electromagnetic spectrum. Point contact diodes are common detectors for microwaves.

 

Question 46. Which of the following electromagnetic waves has the longest wavelength?
a) X-rays
b) visible light
c) radio waves
d) microwaves
Answer: c) radio waves
In simple words: Out of X-rays, visible light, radio waves, and microwaves, radio waves have the longest wavelength. They are at the lowest energy end of the electromagnetic spectrum.

🎯 Exam Tip: Always remember the order of the electromagnetic spectrum in terms of wavelength. Radio waves are at the extreme long-wavelength end.

 

Question 47. Which of the following is of the shortest wavelength?
(a) X-rays
(b) y-rays
(c) microwaves
(d) radio waves
Answer: (b) y-rays
In simple words: Gamma rays (y-rays) have the highest energy and shortest wavelength among all electromagnetic waves. This makes them very powerful.

🎯 Exam Tip: Remember the electromagnetic spectrum order. Wavelength decreases from radio waves to gamma rays, while frequency and energy increase in that direction.

 

Question 48. The range of wavelength of visible light is
(a) 10 \( \dot{A} \) to 100 \( \dot{A} \)
(b) 4000 \( \dot{A} \) to 8000 \( \dot{A} \)
(c) 8000 \( \dot{A} \) to 10,000 \( \dot{A} \)
(d) 10,000 \( \dot{A} \) to 15,000 \( \dot{A} \)
Answer: (b) 4000 \( \dot{A} \) to 8000 \( \dot{A} \)
In simple words: Visible light, which is the light we can see, has wavelengths that fall within a specific range, usually from 4000 Angstroms (violet) to 8000 Angstroms (red). This range allows us to perceive different colors.

🎯 Exam Tip: Knowing the approximate wavelength ranges for different parts of the electromagnetic spectrum, especially visible light, is crucial for answering such questions.

 

Question 49. Which of the following rays has minimum frequency?
(a) U.V. rays
(b) X-rays
(c) y-rays
(d) infrared rays
Answer: (d) infrared rays
In simple words: Infrared rays have the lowest frequency among the given options. Low frequency means they have longer wavelengths and less energy compared to UV, X-rays, or gamma rays.

🎯 Exam Tip: Frequency and wavelength are inversely related. Minimum frequency implies maximum wavelength. Recall the order of the electromagnetic spectrum to identify the correct answer.

 

Question 50. An accelerated electron would produce
(a) y-rays
(b) β-rays
(c) x-rays
(d) e.m. waves
Answer: (d) e.m. waves
In simple words: When an electron speeds up or changes direction (accelerates), it creates disturbances in the electric and magnetic fields around it. These disturbances travel outwards as electromagnetic waves, which include radio waves, light, X-rays, and more.

🎯 Exam Tip: A fundamental principle in electromagnetism is that accelerating charges produce electromagnetic waves. This is how many forms of radiation are generated.

 

Question 51. Which of the following is the infrared wavelength?
(a) \( 10^{-4} \) cm
(b) \( 10^{-5} \) cm
(c) \( 10^{-6} \) cm
(d) \( 10^{-7} \) cm
Answer: (a) \( 10^{-4} \) cm
In simple words: The infrared range of light has wavelengths generally between \( 8 \times 10^{-5} \) cm and \( 3 \times 10^{-3} \) cm. From the given options, \( 10^{-4} \) cm fits best into this range, making it a typical infrared wavelength. Infrared radiation is often associated with heat.

🎯 Exam Tip: Familiarize yourself with the approximate wavelength and frequency ranges for each major part of the electromagnetic spectrum, as these are common recall questions.

 

Question 52. The wavelength of X-rays is of the order of
(a) 1 meter
(b) 1 cm
(c) 1 micron
(d) 1 angstrom
Answer: (d) 1 angstrom
In simple words: X-rays have very tiny wavelengths, usually around 1 Angstrom. This extremely small wavelength allows them to penetrate many materials, which is why they are used for medical imaging.

🎯 Exam Tip: Wavelengths decrease significantly as you move from radio waves to gamma rays. X-rays are near the shorter end of the spectrum, indicating a very small wavelength.

 

Question 53. Maxwell's modified form of Ampere's circuital law is
(a) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} s}=0 \)
(b) \( \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I \)
(c) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}}=\mu_{0} \mathrm{I}+\frac{1}{\varepsilon_{0}} \frac{\mathrm{d} q}{\mathrm{dt}} \)
(d) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\mu_{0} \varepsilon_{0} \frac{\mathrm{d} \varphi}{\mathrm{d} t} \)
Answer: (d) \( \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\mu_{0} \varepsilon_{0} \frac{\mathrm{d} \varphi}{\mathrm{d} t} \)
In simple words: Maxwell changed Ampere's law by adding a "displacement current" term. This new term accounts for magnetic fields created by changing electric fields, not just by moving charges, making the law complete.

🎯 Exam Tip: Understanding Maxwell's correction to Ampere's law is fundamental. It introduces the concept of displacement current, which is crucial for electromagnetic wave theory.

 

Question 54. Radio waves do not penetrate the band of
(a) ionosphere
(b) mesosphere
(c) troposphere
(d) stratosphere
Answer: (a) ionosphere
In simple words: Radio waves are reflected by the ionosphere, a layer of Earth's atmosphere that contains many ions and electrons. This reflection allows radio signals to travel long distances around the curve of the Earth. If they did not reflect, they would escape into space.

🎯 Exam Tip: The ionosphere's ability to reflect radio waves is a key concept in radio communication. It's important to remember which atmospheric layer interacts with which type of electromagnetic radiation.

 

Question 55. Which of the following is responsible for the "Greenhouse effect"?
(a) infrared rays
(b) ultraviolet rays
(c) X-rays
(d) radio waves
Answer: (a) infrared rays
In simple words: The greenhouse effect happens because certain gases in the Earth's atmosphere trap infrared rays. These trapped rays cause the Earth's temperature to rise, similar to how a greenhouse works to keep plants warm.

🎯 Exam Tip: The greenhouse effect is directly linked to the absorption of infrared radiation by greenhouse gases. This is a critical environmental concept often tested in physics and environmental science.

 

II. Short Questions and Answers:

 

Question 1. Name the scientist who first predicted the existence of electromagnetic waves.
Answer: James Clerk Maxwell was the first scientist to predict the existence of electromagnetic (EM) waves mathematically. He developed a set of equations that described how electric and magnetic fields interact and propagate. This was a major breakthrough in physics.

🎯 Exam Tip: Always remember Maxwell's name in connection with the prediction and theory of electromagnetic waves, as his equations form the bedrock of classical electromagnetism.

 

Question 2. Give the uses of IR?
Answer:
1. Infrared radiation helps satellites get electrical energy using solar cells.
2. It is used to produce dehydrated fruits by removing moisture.
3. It keeps plants warm in greenhouses.
4. It provides heat therapy for muscular pain and sprains.
5. It acts as a signal carrier in TV remotes.
6. It is used in night vision photography to see in the dark.
In simple words: Infrared rays are useful for many things like powering satellites, drying fruits, keeping greenhouses warm, treating muscle pain, and helping us see in the dark with special cameras.

🎯 Exam Tip: When listing uses for any type of electromagnetic radiation, try to provide a diverse set of examples from different fields (e.g., technology, medicine, daily life) to show broad understanding.

 

Question 3. What oscillates in electromagnetic waves?
Answer: In electromagnetic waves, the electric and magnetic fields oscillate. These oscillations happen at right angles to each other and also at right angles to the direction the wave is moving. This constant change and interaction keep the wave propagating. The wave does not need any material to travel, it can move through empty space.

🎯 Exam Tip: Emphasize that both electric and magnetic fields oscillate, and they are mutually perpendicular to each other and to the direction of propagation. This defines the transverse nature of EM waves.

 

Question 4. Explain the concept of the intensity of electromagnetic waves.
Answer: The intensity of an electromagnetic wave tells us how much energy crosses a certain area in a certain amount of time. It is measured as the power (energy per unit time) flowing through a unit area, where the area is placed perpendicular to the direction the wave is traveling.
\( \mathrm{I}=\frac{\text { Power }(\mathrm{P})}{\text { Surface area }(\mathrm{A})} \)
In simple words: Intensity is how much energy an electromagnetic wave carries per second through a specific area. It's like how strong the light feels on your skin; a brighter light has higher intensity because it carries more energy.

🎯 Exam Tip: Define intensity clearly with its relationship to power and area. Knowing the formula and its units (\( \text{W/m}^2 \)) can help you score full marks.

 

Question 5. What is the electromagnetic spectrum?
Answer: The electromagnetic spectrum is a complete arrangement of all types of electromagnetic radiation. These radiations are organized based on their wavelengths or frequencies, creating different groups like radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Each group has its own unique properties, but they all travel at the speed of light.

🎯 Exam Tip: When defining the electromagnetic spectrum, it's crucial to mention that it's an ordered arrangement based on wavelength/frequency and includes all forms of electromagnetic radiation.

 

Question 6. What is the role of ozone in the atmosphere?
Answer: The ozone layer, located high up in the Earth's atmosphere (stratosphere), acts like a protective shield. Its main job is to absorb harmful ultraviolet (UV) radiation coming from the sun. By absorbing these rays, the ozone layer stops them from reaching the Earth's surface, which protects living things from damage. UV radiation can cause skin cancer, eye damage, and harm to plants.

🎯 Exam Tip: Focus on the ozone layer's role in absorbing UV radiation and its importance for protecting life on Earth. This is a common environmental and physics concept.

 

Question 7. Write down the uses of radio waves?
Answer: Radio waves are used in many communication systems. They are essential for radio and television broadcasting, allowing us to listen to music and watch programs. They are also used in cellular phones to transmit voice and data signals, especially in the ultra-high frequency band. Radio waves can travel long distances, making them ideal for these applications.

🎯 Exam Tip: Provide specific examples for radio wave uses, such as radio/TV communication and cellular phones, to illustrate practical applications.

 

XI. Three Mark Questions:

 

Question 1. Write short notes on (a) Infrared radiation (b) Ultraviolet radiation (c) Gamma radiation).
Answer:
(a) Infrared radiation:
Infrared radiation (IR) is produced by hot objects and by molecules undergoing rotation and vibration. Its wavelength ranges from \( 8 \times 10^{-7} \) m to \( 5 \times 10^{3} \) m, with frequencies from \( 4 \times 10^{14} \) Hz to \( 6 \times 10^{10} \) Hz. IR is used to power satellites with solar cells, dehydrate fruits, keep greenhouses warm, for heat therapy, in TV remotes as a signal carrier, and for night vision photography. These rays are often called heat waves.

(b) Ultraviolet radiation:
Ultraviolet radiation (UV) is produced by the Sun, electric arcs, and ionized gases, as well as by electron transitions in atoms. Its wavelength range is \( 6 \times 10^{-10} \) m to \( 4 \times 10^{-7} \) m, and its frequency ranges from \( 5 \times 10^{17} \) Hz to \( 7 \times 10^{14} \) Hz. UV radiation has less penetrating power and can be absorbed by atmospheric ozone. It is harmful to the human body and is used to destroy bacteria, sterilize surgical instruments, in burglar alarms, to detect invisible writing and fingerprints, and to study molecular structures. Too much exposure to UV can cause skin damage and eye problems.

(c) Gamma rays:
Gamma rays are produced by transitions within atomic nuclei and by the decay of certain elementary particles. Their wavelength range is \( 1 \times 10^{-14} \) m to \( 1 \times 10^{-10} \) m, with frequencies from \( 3 \times 10^{22} \) Hz to \( 3 \times 10^{18} \) Hz. Gamma rays have very high penetrating power, even more than X-rays and UV radiation, and carry no electric charge. They are harmful to the human body. Gamma rays provide information about the structure of atomic nuclei and are used in radiotherapy for treating cancer and tumors, and in sterilizing medical equipment to kill harmful microorganisms.
In simple words: Infrared is heat radiation, used for warmth and remotes. Ultraviolet comes from the sun, can be harmful, and is used for sterilization. Gamma rays are from atomic nuclei, very powerful and penetrating, used in cancer treatment but also dangerous.

🎯 Exam Tip: For each type of radiation, remember its source, typical wavelength/frequency range, two or three key uses, and at least one significant property or effect.

 

Question 2. Give four basic properties of electromagnetic waves.
Answer: The basic properties of electromagnetic waves are:
(i) Electromagnetic waves are generated by accelerating electric charges and do not need any physical medium to travel. They can move through empty space.
(ii) The electric field (\( \vec{E} \)) and magnetic field (\( \vec{B} \)) oscillations are perpendicular to each other. They are also perpendicular to the direction the wave travels. This shows that electromagnetic waves are transverse waves.
(iii) All electromagnetic waves travel at the same speed in free space, which is the speed of light, \( C = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} = 3 \times 10^{8} \) m/s. This speed is constant for all electromagnetic waves in a vacuum.
(iv) The electric field (\( \vec{E} \)) and magnetic field (\( \vec{B} \)) components oscillate in the same phase, meaning they reach their maximum and minimum values at the same time and place. The ratio of their amplitudes is equal to the speed of light, \( \frac{E_{0}}{B_{0}}=C \).
In simple words: EM waves are made by fast-moving charges, don't need a medium, have electric and magnetic fields that wiggle at right angles to each other and to the travel direction, and all travel at the speed of light in empty space.

🎯 Exam Tip: When describing properties, use clear and concise language. Ensure you cover their origin, transverse nature, constant speed in vacuum, and the phase relationship between electric and magnetic fields.

 

Question 3. Radio waves and gamma rays both are transverse in nature and electromagnetic in character and have the same speed in vacuum. In what respects are they different?
Answer: Radio waves and gamma rays differ in several key ways, despite sharing some fundamental properties:

In simple words: Even though both are electromagnetic waves, radio waves come from vibrating atoms and have long wavelengths and low energy, so they don't penetrate much. Gamma rays come from changes inside atomic nuclei, have very short wavelengths and high energy, and can pass through almost anything.

🎯 Exam Tip: For comparison questions, a table format is excellent for clearly showing differences. Focus on key distinguishing factors like origin, wavelength/frequency, and penetrating power.

 

Question 4. A plane electromagnetic wave travels, in a vacuum, along the y-direction. Write (i) the ratio of the magnitudes, and (ii) the directions of its electric and magnetic field vectors.
Answer:
(i) The ratio of the magnitudes of the electric field (\( E \)) to the magnetic field (\( B \)) in a vacuum is equal to the speed of light (\( c \)): \( \frac{E}{B} = c \).
(ii) For an electromagnetic wave traveling along the y-direction, its electric field vector oscillates along the z-axis, and its magnetic field vector oscillates along the x-axis. The direction of wave propagation is given by the cross product of \( \vec{E} \times \vec{B} \), which in this case would be \( \hat{k} \times \hat{i} = \hat{j} \), matching the y-direction of propagation. The electric and magnetic fields are always perpendicular to each other and to the direction of propagation.
In simple words: For a wave going along the y-axis, the electric field wiggles up-down (z-axis) and the magnetic field wiggles side-to-side (x-axis). The strength of the electric field divided by the magnetic field is always the speed of light.

🎯 Exam Tip: Remember the right-hand rule for the directions: if the wave propagates along \( \vec{v} \), then \( \vec{E} \), \( \vec{B} \), and \( \vec{v} \) are mutually perpendicular, and \( \vec{v} \) is in the direction of \( \vec{E} \times \vec{B} \).

 

Question 5. How does a charge q oscillating at a certain frequency produce electromagnetic waves?
Answer:
1. When a charged particle (q) oscillates at a specific frequency (\( \nu \)), it constantly accelerates.
2. An accelerating charge produces an oscillating electric field around it.
3. This oscillating electric field, in turn, generates an oscillating magnetic field perpendicular to itself.
4. The oscillating magnetic field then creates an oscillating electric field, and this process continues. This self-sustaining cycle leads to the propagation of an electromagnetic wave.
5. This wave travels outwards from the oscillating charge. The electric and magnetic fields of this wave are always perpendicular to each other and to the direction of the wave's travel, making it a transverse wave.
In simple words: When a charged particle jiggles back and forth, it creates wobbly electric and magnetic fields. These wobbly fields keep creating each other and spread out as an electromagnetic wave. This is like how throwing a pebble in water creates ripples.

🎯 Exam Tip: The core idea is that an *accelerating* charge produces EM waves. Emphasize the mutual generation of oscillating electric and magnetic fields.

 

Question 6. Two students A and B prepare the following table about the electromagnetic waves. Rewrite this table in its corrected form:

Student	Direction of Propagation	Electric field	Magnetic field	Peak value of Electric field	Peak value of Magnetic field
A	Along y-axis	Along x-axis	Along z-axis	E	B = E/c
B	Along x-axis	Along y-axis	Along z-axis	E = cB	B

Answer:
Correction for Student 'A':

Direction of Propagation	Electric field	Magnetic field	Peak value of Electric field	Peak value of Magnetic field
Along y-axis	Along x-axis	Along z-axis	E	\( B = E/c \)

No correction for student 'B'
In simple words: The table describes how electric and magnetic fields are oriented when an electromagnetic wave travels. For wave A, if it moves along the y-axis, the electric field (E) is along the x-axis, and the magnetic field (B) is along the z-axis. The strength of B is E divided by the speed of light. Student A's table had an error in the peak value of B. Student B's table is already correct, showing that if the wave moves along the x-axis, E is along y, and B is along z, with E = cB.

🎯 Exam Tip: Remember that for an electromagnetic wave, the electric field, magnetic field, and direction of propagation are mutually perpendicular. Also, \( E = cB \) where \( c \) is the speed of light.

 

XII. Conceptual Questions:

 

Question 1. Why can light travel in a vacuum, whereas sound cannot do so?
Answer:
1. Light waves are electromagnetic waves. This means they are made of vibrating electric and magnetic fields. They do not require any physical medium, like air or water, for their propagation. Light can travel through empty space, such as the vacuum of outer space.
2. Sound waves, on the other hand, are mechanical waves. Mechanical waves need a material medium (like solids, liquids, or gases) to travel because they propagate by making particles in the medium vibrate. In a vacuum, there are no particles to vibrate, so sound cannot travel.
In simple words: Light can travel through empty space because it's an electromagnetic wave, which means it carries its own electric and magnetic fields. Sound cannot travel through empty space because it's a mechanical wave that needs something to vibrate, like air or water.

🎯 Exam Tip: The key distinction is between electromagnetic waves (like light) that don't need a medium and mechanical waves (like sound) that do. Clearly state this fundamental difference.

 

Question 2. State the reason why microwaves are best suited for long-distance transmission of signals.
Answer:
1. Microwaves have very short wavelengths, typically in the range of millimeters. This means they can be focused into narrow beams.
2. Because of their short wavelengths, microwaves are not easily diffracted (bent) around obstacles that are larger than their wavelength, unlike longer radio waves. This prevents the signal from spreading out too much.
3. This characteristic allows them to be used to transmit signals in a very specific direction, as needed for long-distance communication without significant loss of signal strength. Their ability to carry a lot of information also makes them suitable for satellite communication and radar.
In simple words: Microwaves have short wavelengths, so they can be sent in narrow, focused beams. This helps them travel long distances without spreading out or getting easily blocked, making them good for things like satellite TV and radar.

🎯 Exam Tip: Highlight the importance of short wavelength for microwaves, which enables them to be highly directional (less diffraction) and ideal for point-to-point communication over long distances.

 

Question 3. Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiation. Name the radiations and write the range of their frequency.
Answer: Welders need protection from large amounts of harmful ultraviolet (UV) radiation produced by the intense light of the welding arc. The wavelength of UV radiation ranges from 1 nanometer (nm) to 400 nanometers. To convert this to frequency, we use the formula \( f = c/\lambda \), where \( c = 3 \times 10^8 \) m/s.
For \( \lambda = 1 \) nm (\( 1 \times 10^{-9} \) m), \( f = (3 \times 10^8) / (1 \times 10^{-9}) = 3 \times 10^{17} \) Hz.
For \( \lambda = 400 \) nm (\( 400 \times 10^{-9} \) m), \( f = (3 \times 10^8) / (400 \times 10^{-9}) = 7.5 \times 10^{14} \) Hz.
So, the frequency range of UV radiation is approximately from \( 7.5 \times 10^{14} \) Hz to \( 3 \times 10^{17} \) Hz.
In simple words: Welders wear special goggles to block ultraviolet (UV) radiation from the bright welding light. This UV radiation has frequencies from about \( 7.5 \times 10^{14} \) Hz to \( 3 \times 10^{17} \) Hz, which can harm eyes.

🎯 Exam Tip: Identify UV radiation as the primary danger. Provide the wavelength range and, if asked, calculate or state the corresponding frequency range using \( f = c/\lambda \).

 

Question 4. How do you convince yourself that electromagnetic waves carry energy and momentum?
Answer:
1. **Energy:** When an electromagnetic wave, like light, interacts with matter, its oscillating electric and magnetic fields cause the charged particles (electrons and protons) in the matter to oscillate. These oscillating charges gain kinetic energy from the wave. This process demonstrates that the electromagnetic wave transfers energy to the matter, for example, when sunlight warms your skin or heats water in a solar heater.
2. **Momentum:** Electromagnetic waves also carry momentum. When an electromagnetic wave hits a surface, it exerts a tiny but measurable pressure, called radiation pressure, on that surface. This pressure is evidence that the wave is transferring momentum to the surface, similar to how a moving object imparts momentum when it collides with something else. For instance, solar sails for spacecraft propulsion are designed to utilize this radiation pressure.
In simple words: We know EM waves carry energy because they can heat things up, like the sun warming our hands. We know they carry momentum because they can push on objects, like how light from the sun gently pushes on spacecraft in space.

🎯 Exam Tip: Give distinct, simple examples for both energy and momentum. For energy, think of heating effects. For momentum, think of radiation pressure or force exerted on a surface.

 

Question 5. State the condition under which a microwave oven heats up a food item containing water molecules most efficiently.
Answer:
1. For a microwave oven to heat food most efficiently, the frequency of the microwaves generated by the oven must precisely match the resonant frequency of the water molecules present in the food item.
2. When this frequency match (resonance) occurs, the water molecules absorb energy from the microwaves very effectively. This absorbed energy is then converted into the kinetic energy of the molecules, causing them to move faster.
3. The increased kinetic energy directly leads to a significant rise in the temperature of the food item, cooking it quickly and thoroughly.
In simple words: A microwave oven heats food best when its waves vibrate at the same speed (frequency) that water molecules naturally like to vibrate. This makes the water molecules absorb the most energy and get hot quickly.

🎯 Exam Tip: The key concept here is "resonance." Emphasize that the microwave frequency must match the natural (resonant) frequency of water molecules for efficient energy absorption and heating.

 

Question 6. Long-distance radio broadcasts use short-wave bands. Why?
Answer: Long-distance radio broadcasts use short-wave bands because these radio waves are easily reflected back to Earth by a layer in the atmosphere called the ionosphere. The ionosphere acts like a natural mirror for short waves. This reflection allows the radio signals to bounce off the ionosphere and travel over the horizon, covering vast distances around the curvature of the Earth that would otherwise be out of reach. Longer waves (like medium wave) also reflect, but short waves offer more reliable long-distance communication.

🎯 Exam Tip: The critical point is the reflection of short-wave radio signals by the ionosphere, which enables them to travel beyond the line of sight.

 

Question 7. It is necessary to use satellites for long-distance TV transmission. Why?
Answer:
1. TV signals use high frequencies, which are not reflected by the Earth's ionosphere. Instead, they pass right through it into space.
2. Additionally, ground-wave transmission (signals traveling along the Earth's surface) for these high-frequency TV signals is only possible over very limited distances.
3. Therefore, to transmit TV signals over long distances, especially across continents or oceans, it is necessary to use satellites. Satellites in geostationary orbit receive signals from Earth, amplify them, and then re-transmit them back to a much wider area on Earth, overcoming the curvature of the Earth and the limitations of atmospheric reflection.
In simple words: TV signals use very high frequencies that don't bounce off the atmosphere like radio waves do. So, to send them far, we use satellites in space. These satellites catch the signals and send them back down to cover huge areas on Earth.

🎯 Exam Tip: The key reasons are that high-frequency TV signals penetrate the ionosphere (not reflected) and ground-wave propagation is limited, necessitating satellites for global coverage.

 

Question 8. Optical and radio telescopes built on the ground but x-ray astronomy is possible only from satellites orbiting the earth. Why?
Answer:
1. The Earth's atmosphere is transparent to visible light and radio waves. This means that these waves can pass through the atmosphere and reach ground-based optical and radio telescopes without being absorbed or blocked significantly.
2. However, the Earth's atmosphere strongly absorbs X-rays. This protective layer prevents harmful X-rays from reaching the ground, which is good for life on Earth, but it makes X-ray astronomy impossible from the surface.
3. Therefore, X-ray astronomy can only be done using satellites that orbit high above the Earth's atmosphere, typically at heights like 36,000 km, where the atmosphere is very thin or non-existent, allowing X-rays from space to be detected directly.
In simple words: We can use normal telescopes on Earth for visible light and radio waves because they pass through our atmosphere. But X-rays from space cannot get through the atmosphere, so we must send X-ray telescopes into orbit on satellites to observe them.

🎯 Exam Tip: The transparency/opacity of Earth's atmosphere to different parts of the electromagnetic spectrum is the central concept. Emphasize that X-rays are absorbed by the atmosphere.

 

Question 9. The small ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer: The small ozone layer in the stratosphere is vital for human survival because it absorbs the majority of harmful ultraviolet (UV) radiation from the sun. UV radiation can cause severe damage to living organisms, including skin cancer, cataracts, and damage to DNA in humans and animals, as well as harming plant life. Without the ozone layer, much more UV radiation would reach the Earth's surface, making life as we know it extremely difficult, if not impossible. Its protective action is essential for maintaining a habitable planet.

🎯 Exam Tip: Reiterate the ozone layer's role in absorbing UV radiation and explicitly mention the negative health and environmental impacts that would occur without it.

 

Question 10. If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
Answer: If the Earth did not have an atmosphere, its average surface temperature would be much lower than it is now. This is because the atmosphere, through a process called the Greenhouse effect, traps infrared radiation that the Earth radiates after being warmed by the sun. Gases like carbon dioxide and water vapor in the lower atmosphere act as a blanket, preventing this heat from escaping quickly into space. Without an atmosphere, there would be no Greenhouse effect, and the Earth's heat would radiate away much faster, leading to a significantly colder planet.

🎯 Exam Tip: Connect the absence of an atmosphere directly to the absence of the greenhouse effect and the resulting inability to retain heat, leading to a lower temperature.

 

Question 11. Some scientists have predicted that a global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a 'nuclear winter'. What might be the basis of this prediction?
Answer: The prediction of a 'nuclear winter' after a global nuclear war is based on the idea that the massive amounts of dust, soot, and smoke generated by widespread fires and explosions would rise high into the atmosphere. These particles would then spread globally, forming thick clouds that would block a significant portion of sunlight from reaching the Earth's surface. This reduction in solar radiation would cause a drastic and prolonged drop in global temperatures, leading to conditions similar to a severe winter worldwide, hence the term 'nuclear winter.' Such conditions would severely impact agriculture, leading to famine and widespread ecosystem collapse.

🎯 Exam Tip: The core idea of 'nuclear winter' is the blocking of solar light by atmospheric debris (soot, dust) after a nuclear war, leading to a severe global temperature drop.

 

XIII. Additional Problems (Two Marks):

 

Question 1. A parallel plate capacitor has circular plates, each of radius 5.0 cm. It is being charged so that the electric field in the gap between its plates rises steadily at \( 10^{12} \) V \( \text{m}^{-1} \text{s}^{-1} \). What is the displacement current?
Answer:
Given data:
Radius of plates, \( r = 5.0 \) cm \( = 5 \times 10^{-2} \) m
Rate of change of electric field, \( \frac{dE}{dt} = 10^{12} \text{ V m}^{-1} \text{s}^{-1} \)
Permittivity of free space, \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{m}^{-2} \)

The displacement current \( I_d \) is given by:
\( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \)
Where \( \Phi_E = E \cdot A \) is the electric flux and \( A \) is the area of the circular plates.
So, \( \frac{d\Phi_E}{dt} = \frac{d}{dt}(E \cdot A) = A \frac{dE}{dt} \)
Area of the circular plate, \( A = \pi r^2 = \pi (5 \times 10^{-2})^2 = \pi (25 \times 10^{-4}) \text{ m}^2 \)

Now, substitute the values into the displacement current formula:
\( I_d = \varepsilon_0 \cdot \pi r^2 \cdot \frac{dE}{dt} \)
\( I_d = (8.854 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{m}^{-2}) \times (3.14 \times (5 \times 10^{-2})^2 \text{ m}^2) \times (10^{12} \text{ V m}^{-1} \text{s}^{-1}) \)
\( I_d = (8.854 \times 10^{-12}) \times (3.14 \times 25 \times 10^{-4}) \times (10^{12}) \)
\( I_d = 8.854 \times 3.14 \times 25 \times 10^{-4} \)
\( I_d = 0.069581 \text{ A} \)
\( I_d \approx 0.07 \text{ A} \)
\( I_d = 70 \text{ mA} \)
In simple words: The displacement current is calculated using the rate at which the electric field changes and the area of the capacitor plates. We use a formula that includes the permittivity of free space, the area, and the given rate of change of the electric field to find that the current is about 70 mA.

🎯 Exam Tip: Remember the formula for displacement current \( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \) and how to calculate electric flux \( \Phi_E \) for a given geometry. Pay attention to units and power of ten calculations.

 

Question 2. The oscillating magnetic field in a plane electromagnetic wave is given by \( B_y = (8 \times 10^{-6}) \sin [2 \times 10^{11}t + 300\pi x] \text{ T} \)
(i) calculate the wavelength of the electromagnetic wave.
(ii) Write down the expression for the oscillating electric field.
Answer:
Given the magnetic field expression: \( B_y = (8 \times 10^{-6}) \sin [2 \times 10^{11}t + 300\pi x] \text{ T} \)
The standard equation for a plane electromagnetic wave is \( B_y = B_0 \sin [kx - \omega t] \) or \( B_y = B_0 \sin [kx + \omega t] \) for a wave propagating in the negative x-direction.

Comparing the given equation \( B_y = (8 \times 10^{-6}) \sin [300\pi x + 2 \times 10^{11}t] \) with the standard form, we have:
Wave number \( k = 300\pi \text{ rad/m} \)
Angular frequency \( \omega = 2 \times 10^{11} \text{ rad/s} \)
Peak magnetic field \( B_0 = 8 \times 10^{-6} \text{ T} \)

(i) Calculate the wavelength (\( \lambda \)):
We know that \( k = \frac{2\pi}{\lambda} \).
So, \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{300\pi} = \frac{1}{150} \text{ m} \)
\( \lambda = 0.00666... \text{ m} \)
\( \lambda \approx 0.0067 \text{ m} \) or \( 0.67 \text{ cm} \)

(ii) Write down the expression for the oscillating electric field:
For an electromagnetic wave, the electric field \( E \) and magnetic field \( B \) are perpendicular to each other and to the direction of propagation. The speed of the wave is \( c = \frac{\omega}{k} \).
Also, the amplitude of the electric field \( E_0 \) is related to the amplitude of the magnetic field \( B_0 \) by \( E_0 = c B_0 \).
First, let's find the speed of light from \( \omega \) and \( k \):
\( c = \frac{2 \times 10^{11} \text{ rad/s}}{300\pi \text{ rad/m}} \approx 2.12 \times 10^8 \text{ m/s} \) (This is not exactly \( 3 \times 10^8 \text{ m/s} \), indicating it might be in a medium or there's a slight approximation in the numbers, but we'll use the fundamental speed of light \( c = 3 \times 10^8 \text{ m/s} \) for \( E_0 = cB_0 \) unless specified it's not in vacuum). Assuming it's in free space (as is typical unless otherwise stated for these equations), we use \( c = 3 \times 10^8 \text{ m/s} \).

\( E_0 = c B_0 = (3 \times 10^8 \text{ m/s}) \times (8 \times 10^{-6} \text{ T}) = 2400 \text{ V/m} \)

Since the wave propagates along the negative x-direction (indicated by \( +kx \) and \( +\omega t \) if we write it as \( \sin(kx+\omega t) \)), and the magnetic field is along the y-direction, the electric field must be perpendicular to both x and y. So, the electric field will be along the z-direction.
The expression for the oscillating electric field is:
\( E_z = E_0 \sin [kx + \omega t] \)
\( E_z = 2400 \sin [300\pi x + 2 \times 10^{11}t] \text{ V/m} \)
In simple words: First, we find the wavelength by using the wave number from the given equation, which comes out to be about 0.67 cm. Next, we calculate the maximum strength of the electric field using the speed of light and the maximum strength of the magnetic field. Since the magnetic field wiggles along the y-axis and the wave travels along the x-axis, the electric field must wiggle along the z-axis. We then write the electric field equation using these values.

🎯 Exam Tip: Be able to identify \( B_0 \), \( k \), and \( \omega \) from the wave equation. Remember the relationships \( \lambda = \frac{2\pi}{k} \) and \( E_0 = cB_0 \). Use the right-hand rule (\( \vec{E} \times \vec{B} \) gives propagation direction) to determine the direction of the electric field vector.

 

Question 3. The oscillating electric field of an electromagnetic wave is given by:
\( E_y = 30 \sin [2 \times 10^{11}t + 300\pi x] \text{ Vm}^{-1} \)
(a) Obtain the value of the wavelength of the electromagnetic wave.
(b) Write down the expression for the oscillating magnetic field.
Answer:
Given data:
Electric field expression: \( E_y = 30 \sin [2 \times 10^{11}t + 300\pi x] \text{ Vm}^{-1} \)
The standard equation for a plane electromagnetic wave is \( E_y = E_0 \sin [kx \pm \omega t] \).

Comparing the given equation \( E_y = 30 \sin [300\pi x + 2 \times 10^{11}t] \) with the standard form, we have:
Peak electric field \( E_0 = 30 \text{ V/m} \)
Wave number \( k = 300\pi \text{ rad/m} \)
Angular frequency \( \omega = 2 \times 10^{11} \text{ rad/s} \)

(a) Obtain the value of the wavelength (\( \lambda \)) of the electromagnetic wave.
We know that \( k = \frac{2\pi}{\lambda} \).
So, \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{300\pi} = \frac{1}{150} \text{ m} \)
\( \lambda = 0.00666... \text{ m} \)
\( \lambda \approx 0.0067 \text{ m} \) or \( 0.67 \text{ cm} \)

(b) Write down the expression for the oscillating magnetic field.
For an electromagnetic wave in free space, the amplitude of the magnetic field \( B_0 \) is related to the amplitude of the electric field \( E_0 \) by \( B_0 = \frac{E_0}{c} \), where \( c = 3 \times 10^8 \text{ m/s} \).
\( B_0 = \frac{30 \text{ V/m}}{3 \times 10^8 \text{ m/s}} = 10 \times 10^{-8} \text{ T} = 10^{-7} \text{ T} \)

The wave propagates along the negative x-direction (due to the \( +kx \) and \( +\omega t \) terms). The electric field is along the y-direction. According to the right-hand rule (\( \vec{E} \times \vec{B} \) gives propagation direction), if \( \vec{E} \) is along y and propagation is along -x, then \( \vec{B} \) must be along z-direction (since \( \hat{j} \times \hat{k} = \hat{i} \), and we need \( -\hat{i} \)). So, the magnetic field is along the z-direction.
The expression for the oscillating magnetic field is:
\( B_z = B_0 \sin [kx + \omega t] \)
\( B_z = 10^{-7} \sin [300\pi x + 2 \times 10^{11}t] \text{ T} \)
In simple words: First, we find the wavelength using the wave number from the electric field equation, which is about 0.67 cm. Then, we find the maximum strength of the magnetic field by dividing the maximum electric field by the speed of light. Since the electric field is along the y-axis and the wave travels along the negative x-axis, the magnetic field will be along the z-axis. We then write its equation using these values.

🎯 Exam Tip: This question is similar to the previous one, but with E-field given. Ensure you correctly identify \( k \) and \( \omega \), use \( \lambda = \frac{2\pi}{k} \), and determine \( B_0 = \frac{E_0}{c} \). The direction of \( \vec{B} \) is perpendicular to both \( \vec{E} \) and the propagation direction.

 

Question 3. The oscillating electric field of an electromagnetic wave is given by:
(a) Obtain the value of the wavelength of the electromagnetic wave.
(b) Write down the expression for the oscillating magnetic field.
Answer:
Given data:
\( E_y = 30 \sin [2 \times 10^{11}t + 300\pi x] Vm^{-1} \)
The standard equation is \( E_y = E_0 \sin [2\pi(\frac{x}{\lambda}+\frac{t}{T})] \).
Comparing these, we get \( \frac{2 \pi}{\lambda} = 300\pi \) and \( E_0 = 30 Vm^{-1} \).

(a) To find the wavelength \( \lambda \):
\( \frac{2 \pi}{\lambda} = 300\pi \)
\( \implies \lambda = \frac{2\pi}{300\pi} \)
\( \implies \lambda = \frac{1}{150} \text{ m} \)
\( \implies \lambda = 0.67 \text{ cm} \)

(b) To write the expression for the oscillating magnetic field:
We know the speed of light \( c = 3 \times 10^8 \text{ m/s} \). We can find the magnetic field amplitude \( B_0 \) using \( c = \frac{E_0}{B_0} \).
\( B_0 = \frac{E_0}{c} \)
\( B_0 = \frac{30 \text{ Vm}^{-1}}{3 \times 10^8 \text{ m/s}} \)
\( B_0 = 10^{-7} \text{ T} \)
The magnetic field is perpendicular to the direction of propagation (x-axis) and also perpendicular to the electric field (y-axis). So, it acts along the z-axis.
Thus, the expression for the oscillating magnetic field is:
\( B_z = 10^{-7} \sin [2 \times 10^{11}t + 300\pi x] \text{ T} \)
In simple words: We first found the repeating length (wavelength) of the wave using the given equation. Then, we used the speed of light to calculate how strong the magnetic part of the wave is. Since electric and magnetic fields are always at right angles to each other and to the wave's path, we could write the full magnetic field equation.

🎯 Exam Tip: Remember to correctly identify the wave number (k) and angular frequency (ω) from the given equation to find wavelength and amplitude. Ensure the units are consistent throughout the calculation.

 

Question 4. Radiation of energy E falls normally on a perfectly reflecting surface. Find the momentum transferred to the surface.
Answer:
When radiation with energy \( E \) falls on a surface, it carries momentum. The initial momentum (\( P \)) of this radiation is given by the formula:
\( P = \frac{E}{C} \)
Here, \( C \) is the speed of light. Electromagnetic waves always carry momentum when they travel.
For a perfectly reflecting surface, the radiation does not get absorbed; instead, it bounces back completely. When it reflects, its momentum changes direction. The change in momentum is twice the initial momentum because the momentum reverses direction.
Therefore, the total momentum transferred to the surface is the change in momentum:
\( \Delta p = P_{initial} - P_{final} = \frac{E}{C} - (-\frac{E}{C}) = \frac{2E}{C} \)
So, the momentum transferred to the surface is \( \frac{2E}{C} \).
In simple words: Light has energy and also pushes things, like a small force. If light hits a mirror and bounces back completely, it gives the mirror a push that is twice as strong as if it just stopped there. This push is called momentum transfer.

🎯 Exam Tip: Always remember that for a perfectly reflecting surface, the change in momentum is twice the incident momentum, as the momentum reverses direction.

 

Question 5. A charged particle oscillates about its mean equilibrium position with a frequency of \( 10^9 \) Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
According to Maxwell's theory, an oscillating or accelerating charged particle produces electromagnetic waves. The frequency of these electromagnetic waves is exactly the same as the frequency of the oscillating charged particle that created them. This is a fundamental principle of how electromagnetic radiation is generated.
Since the charged particle oscillates with a frequency of \( 10^9 \) Hz, the electromagnetic waves produced will also have the same frequency.
Therefore, the frequency of the electromagnetic waves produced by the oscillator is \( 10^9 \) Hz.
In simple words: When a charged particle wiggles back and forth, it creates light waves. The speed at which the particle wiggles (its frequency) is the same as the speed at which the light waves wiggle.

🎯 Exam Tip: The frequency of electromagnetic waves produced by an oscillating charge is always equal to the oscillation frequency of the charge itself.

 

Question 6. The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is \( B_0 = 510 \) nT. What is the amplitude of the electric field part of the wave?
Answer:
We are given the amplitude of the magnetic field \( B_0 = 510 \) nT. We need to convert this to Tesla (T) for calculations:
\( B_0 = 510 \times 10^{-9} \text{ T} \)
In a vacuum, the amplitude of the electric field (\( E_0 \)) and the amplitude of the magnetic field (\( B_0 \)) of an electromagnetic wave are related by the speed of light (\( c \)). The speed of light in a vacuum is approximately \( c = 3 \times 10^8 \text{ m/s} \).
The relationship is given by:
\( E_0 = c \times B_0 \)
Now, we can substitute the values:
\( E_0 = (3 \times 10^8 \text{ m/s}) \times (510 \times 10^{-9} \text{ T}) \)
\( E_0 = 1530 \times 10^{-1} \text{ Vm}^{-1} \)
\( E_0 = 153 \text{ Vm}^{-1} \)
So, the amplitude of the electric field part of the wave is \( 153 \text{ Vm}^{-1} \). This relationship shows how strong the electric and magnetic parts of the wave are compared to each other.
In simple words: We know how strong the magnetic part of the light wave is. We use the speed of light to figure out how strong its electric part is, because they are always linked in a fixed way.

🎯 Exam Tip: Always remember the fundamental relation \( c = E_0/B_0 \) for electromagnetic waves in vacuum and ensure unit consistency (nanoTesla to Tesla conversion) for accurate calculations.

 

Question 7. Suppose that the electric field amplitude of an electromagnetic wave is \( E_0 = 120 \) NC\(^{-1}\) and that its frequency is \( \nu = 50.0 \) MHz.
(a) Determine, \( B_0 \), \( \omega \), \( k \) and \( \lambda \).
(b) Find expressions for \( \overrightarrow{E} \) and \( \overrightarrow{B} \).
Answer:
Given:
Electric field amplitude \( E_0 = 120 \text{ NC}^{-1} \)
Frequency \( \nu = 50.0 \text{ MHz} = 50 \times 10^6 \text{ Hz} \)
Speed of light \( c = 3 \times 10^8 \text{ m/s} \)

(a) Determine \( B_0 \), \( \omega \), \( k \), and \( \lambda \):
1. Magnetic field amplitude \( B_0 \):
\( B_0 = \frac{E_0}{c} \)
\( B_0 = \frac{120 \text{ NC}^{-1}}{3 \times 10^8 \text{ m/s}} \)
\( B_0 = 4 \times 10^{-7} \text{ T} \)

2. Angular frequency \( \omega \):
\( \omega = 2\pi\nu \)
\( \omega = 2 \times 3.14 \times (50 \times 10^6 \text{ Hz}) \)
\( \omega = 314 \times 10^6 \text{ rad/s} \)
\( \omega \approx 3.14 \times 10^8 \text{ rad/s} \)

3. Wavelength \( \lambda \):
\( \lambda = \frac{c}{\nu} \)
\( \lambda = \frac{3 \times 10^8 \text{ m/s}}{50 \times 10^6 \text{ Hz}} \)
\( \lambda = 6.00 \text{ m} \)

4. Wave number \( k \):
\( k = \frac{2\pi}{\lambda} \)
\( k = \frac{2 \times 3.14}{6.00 \text{ m}} \)
\( k \approx 1.05 \text{ rad/m} \)

(b) Find expressions for \( \overrightarrow{E} \) and \( \overrightarrow{B} \):
Let's assume the wave is propagating along the x-axis. For an electromagnetic wave propagating in the x-direction, the electric field \( \overrightarrow{E} \) will be along the y-axis, and the magnetic field \( \overrightarrow{B} \) will be along the z-axis (or vice versa, but this is a common convention).
The general forms are:
\( \overrightarrow{E} = E_0 \sin(kx - \omega t) \hat{j} \)
\( \overrightarrow{B} = B_0 \sin(kx - \omega t) \hat{k} \)
Substituting the calculated values:
\( \overrightarrow{E} = 120 \sin(1.05x - 3.14 \times 10^8 t) \hat{j} \text{ NC}^{-1} \)
\( \overrightarrow{B} = 4 \times 10^{-7} \sin(1.05x - 3.14 \times 10^8 t) \hat{k} \text{ T} \)
In simple words: We are given how strong the electric part of a light wave is and how often it wiggles. From this, we first find out how strong the magnetic part is, how fast it wiggles in circles, how wavy it is, and the length of one wave. Then, we write down the full equations for both the electric and magnetic parts, showing their direction and how they change over space and time.

🎯 Exam Tip: Remember that \( k = 2\pi/\lambda \) and \( \omega = 2\pi\nu \). When writing the full expressions for \( \overrightarrow{E} \) and \( \overrightarrow{B} \), ensure the directions are mutually perpendicular to each other and to the direction of wave propagation.

 

Question 8. About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation:
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m
Answer:
A light bulb acts like a point source, sending out light in all directions evenly. Only 5% of its power is converted to visible light.
Total power of the bulb = 100 W
Power converted to visible radiation \( P_{visible} = 5\% \text{ of } 100 \text{ W} = 0.05 \times 100 \text{ W} = 5 \text{ W} \)
The average intensity (\( I \)) of visible radiation at a distance \( r \) from the bulb is given by the formula:
\( I = \frac{\text{Power}}{\text{Surface Area}} = \frac{P_{visible}}{4\pi r^2} \)
Here, \( 4\pi r^2 \) is the surface area of a sphere with radius \( r \).

(a) At a distance of 1 m from the bulb (\( r = 1 \text{ m} \)):
\( I_{(1\text{ m})} = \frac{5 \text{ W}}{4\pi (1 \text{ m})^2} \)
\( I_{(1\text{ m})} = \frac{5}{4\pi} \approx \frac{5}{12.56} \)
\( I_{(1\text{ m})} \approx 0.398 \text{ Wm}^{-2} \)
\( I_{(1\text{ m})} \approx 0.4 \text{ Wm}^{-2} \)

(b) At a distance of 10 m from the bulb (\( r = 10 \text{ m} \)):
\( I_{(10\text{ m})} = \frac{5 \text{ W}}{4\pi (10 \text{ m})^2} \)
\( I_{(10\text{ m})} = \frac{5}{4\pi \times 100} = \frac{5}{400\pi} \approx \frac{5}{1256} \)
\( I_{(10\text{ m})} \approx 0.00398 \text{ Wm}^{-2} \)
\( I_{(10\text{ m})} \approx 0.004 \text{ Wm}^{-2} \)
In simple words: A light bulb sends out light in all directions, but only a small part of its power is light we can see. We calculate how much of this visible light energy spreads out over a bigger and bigger area as you move away from the bulb. The farther you are, the less bright it seems because the light is spread over a much larger space.

🎯 Exam Tip: Remember that intensity decreases with the square of the distance (\( I \propto 1/r^2 \)) because the light spreads over a spherical area. Always calculate the actual visible power first before applying the intensity formula.

 

Question 1. Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
Answer:
Maxwell extended Ampere's circuital law to make it consistent with the conservation of charge and to account for changing electric fields. Here's how:

1. **Ampere's Circuital Law:** According to Ampere's law, the line integral of the magnetic field (\( \overrightarrow{B} \)) around a closed loop (\( C \)) is proportional to the total current (\( I \)) passing through the area bounded by that loop.
\( \oint_{C} \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 I \) (Equation 1)

2. **Problem with Charging Capacitor:** Imagine a parallel plate capacitor being charged. If we draw a loop \( C_1 \) outside the capacitor plates, a conduction current \( I \) flows through the wire. Ampere's law gives:
\( \oint_{C_1} \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 I \) (Equation 2)
Now, consider a different loop \( C_2 \) between the capacitor plates, enclosing the same wire but located in the gap where there is no conduction current. According to Ampere's law, the current \( I \) inside this loop is zero, so:
\( \oint_{C_2} \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 (0) = 0 \) (Equation 3)
This creates a problem: if \( C_1 \) and \( C_2 \) are infinitesimally close to each other, the magnetic field should be almost the same for both loops. This means \( \oint_{C_1} \overrightarrow{B} \cdot \overrightarrow{dl} \) should be equal to \( \oint_{C_2} \overrightarrow{B} \cdot \overrightarrow{dl} \). However, Equation 2 gives a non-zero value, while Equation 3 gives zero, which is inconsistent.

3. **Maxwell's Correction (Displacement Current):** Maxwell solved this inconsistency by proposing that a changing electric field also acts as a source of a magnetic field. He introduced the concept of "displacement current" (\( I_d \)). Between the capacitor plates, as the capacitor charges, the electric field changes with time. This changing electric field is equivalent to a current, even though no actual charges are moving.
The displacement current (\( I_d \)) is defined as:
\( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \)
where \( \Phi_E \) is the electric flux through the surface.

4. **Generalized Ampere's Law:** Maxwell modified Ampere's law to include this displacement current, making it consistent everywhere. The total current (\( I_{total} \)) is now the sum of the conduction current (\( I_c \)) and the displacement current (\( I_d \)):
\( I_{total} = I_c + I_d \)
So, the modified Ampere's circuital law is:
\( \oint_{C} \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 (I_c + I_d) \)
Substituting the expression for \( I_d \):
\( \oint_{C} \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \)
This generalized law correctly predicts the magnetic field in regions with changing electric fields, like inside a charging capacitor, thus resolving the inconsistency.
In simple words: Ampere's law helps find magnetic fields from electric currents. But when a capacitor charges, the electric current seems to disappear between the plates, making the law incomplete. Maxwell added a "displacement current" term for changing electric fields. This new term acts like a current and makes the law work everywhere, even where no real charges are flowing.

🎯 Exam Tip: Clearly state the inconsistency of original Ampere's law for a charging capacitor, define displacement current, and then present the full generalized Ampere-Maxwell law with both current terms.

 

Question 2. Show that during the charging of a parallel plate capacitor, the rate of change of charge on each plate equals \( \varepsilon_0 \) times the rate of change of electric flux (\( \Phi_E \)) linked with it. what is the name given to the term \( \varepsilon_0 \frac{d\Phi_E}{dt} \)?
Answer:
Consider a parallel plate capacitor with plate area \( A \) and charge \( q \) on its plates at any instant \( t \) during charging. Let \( \sigma \) be the surface charge density.
The electric field (\( E \)) between the plates of the capacitor is given by:
\( E = \frac{\sigma}{\varepsilon_0} \)
Since \( \sigma = \frac{q}{A} \), we can write the electric field as:
\( E = \frac{q}{\varepsilon_0 A} \)
The electric flux (\( \Phi_E \)) through the area between the capacitor plates is given by:
\( \Phi_E = E \cdot A \)
Substitute the expression for \( E \):
\( \Phi_E = \left(\frac{q}{\varepsilon_0 A}\right) A \)
\( \Phi_E = \frac{q}{\varepsilon_0} \)
Now, we need to find the rate of change of electric flux with respect to time:
\( \frac{d\Phi_E}{dt} = \frac{d}{dt}\left(\frac{q}{\varepsilon_0}\right) \)
Since \( \varepsilon_0 \) is a constant, we can take it out of the differentiation:
\( \frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dq}{dt} \)
Rearranging this equation, we get:
\( \frac{dq}{dt} = \varepsilon_0 \frac{d\Phi_E}{dt} \)
This shows that the rate of change of charge on each plate (\( \frac{dq}{dt} \)) is equal to \( \varepsilon_0 \) times the rate of change of electric flux (\( \frac{d\Phi_E}{dt} \)) linked with the capacitor.

The term \( \varepsilon_0 \frac{d\Phi_E}{dt} \) is called the **displacement current (\( I_d \))**.
In simple words: When a capacitor charges, the amount of charge on its plates changes over time. We found that how quickly this charge changes is directly related to how quickly the electric push (flux) between the plates changes, with a special constant called epsilon-nought. This changing electric push is what we call the "displacement current."

🎯 Exam Tip: Start with the definition of the electric field between capacitor plates and electric flux. The derivation is straightforward if these initial relations are correct. Clearly state the name of the term at the end.

 

Question 3. Consider a plane em. a wave traveling with speed c in the positive z-direction.
(i) Use Faraday's law to show the \( E = cB \)
(ii) Use modified Ampere's circuital law to show that \( c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \)
Answer:
Let the electromagnetic wave travel in the positive z-direction. We can assume the electric field (\( \overrightarrow{E} \)) is in the x-direction and the magnetic field (\( \overrightarrow{B} \)) is in the y-direction.
So, \( \overrightarrow{E} = E(z,t) \hat{i} \) and \( \overrightarrow{B} = B(z,t) \hat{j} \).

(i) **Using Faraday's Law to show \( E = cB \):**
Faraday's Law of electromagnetic induction states that the line integral of the electric field around a closed loop is equal to the negative rate of change of magnetic flux through the area bounded by the loop:
\( \oint \overrightarrow{E} \cdot \overrightarrow{dl} = -\frac{d\Phi_B}{dt} \)
Consider a rectangular loop in the x-z plane with length \( dl \) along the x-axis and \( dz \) along the z-axis. The sides are parallel to E. The electric field contributes only along the x-direction.
The line integral \( \oint \overrightarrow{E} \cdot \overrightarrow{dl} \) becomes \( E \cdot dl \).
The magnetic flux \( \Phi_B = \int \overrightarrow{B} \cdot \overrightarrow{dA} \). For our loop, \( \Phi_B = B \cdot (dl \cdot dz) \).
So, \( E \cdot dl = -\frac{d}{dt} (B \cdot dl \cdot dz) \)
\( E = -\frac{dB}{dt} dz \)
For a wave traveling in the z-direction, we know \( E = E_0 \sin(kz - \omega t) \) and \( B = B_0 \sin(kz - \omega t) \).
\( E_x = cB_y \)
More rigorously, for a plane wave \( E_x = E_0 \sin(kz - \omega t) \) and \( B_y = B_0 \sin(kz - \omega t) \), Faraday's law in differential form for these components is \( \frac{\partial E_x}{\partial z} = -\frac{\partial B_y}{\partial t} \).
Differentiating gives: \( k E_0 \cos(kz - \omega t) = \omega B_0 \cos(kz - \omega t) \)
\( k E_0 = \omega B_0 \)
\( E_0 = \frac{\omega}{k} B_0 \)
Since \( \frac{\omega}{k} \) is the wave speed, \( c \), we have \( E_0 = c B_0 \). This means \( E = cB \).

(ii) **Using Modified Ampere's Circuital Law to show \( c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \):**
The modified Ampere's law is:
\( \oint \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \)
In free space (vacuum), there is no conduction current (\( I_c = 0 \)). So the law simplifies to:
\( \oint \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \)
Consider a rectangular loop in the y-z plane with length \( dl \) along the y-axis and \( dz \) along the z-axis. The magnetic field contributes only along the y-direction.
The line integral \( \oint \overrightarrow{B} \cdot \overrightarrow{dl} \) becomes \( B \cdot dl \).
The electric flux \( \Phi_E = \int \overrightarrow{E} \cdot \overrightarrow{dA} \). For our loop, \( \Phi_E = E \cdot (dl \cdot dz) \).
So, \( B \cdot dl = \mu_0 \varepsilon_0 \frac{d}{dt} (E \cdot dl \cdot dz) \)
\( B = \mu_0 \varepsilon_0 \frac{dE}{dt} dz \)
Again, using the differential form for a plane wave: \( \frac{\partial B_y}{\partial z} = -\mu_0 \varepsilon_0 \frac{\partial E_x}{\partial t} \)
Differentiating gives: \( k B_0 \cos(kz - \omega t) = \mu_0 \varepsilon_0 \omega E_0 \cos(kz - \omega t) \)
\( k B_0 = \mu_0 \varepsilon_0 \omega E_0 \)
\( B_0 = \mu_0 \varepsilon_0 \frac{\omega}{k} E_0 \)
\( B_0 = \mu_0 \varepsilon_0 c E_0 \)
We already found \( E_0 = cB_0 \). Substitute \( E_0 \) in this equation:
\( B_0 = \mu_0 \varepsilon_0 c (cB_0) \)
\( B_0 = \mu_0 \varepsilon_0 c^2 B_0 \)
\( 1 = \mu_0 \varepsilon_0 c^2 \)
\( c^2 = \frac{1}{\mu_0 \varepsilon_0} \)
\( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \)
This important equation shows that the speed of an electromagnetic wave in vacuum is determined by the electric and magnetic properties of free space.
In simple words: For a light wave moving forward, we can use two main rules of electricity and magnetism. First, Faraday's law tells us that a changing magnetic field creates an electric field, and this helps us see that the strength of the electric part of the wave is linked to the magnetic part by the wave's speed. Second, Ampere's modified law says that a changing electric field also creates a magnetic field. By putting these two ideas together, we find a formula for the speed of light that uses constants for electric and magnetic fields in empty space.

🎯 Exam Tip: When using integral forms of Maxwell's equations, clearly define the orientation of the loops and surfaces. For plane waves, differential forms are often more direct. Remember \( \omega/k = c \).

 

Question 4. A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.
Answer:
Consider a parallel plate capacitor with plate area \( A \) and separation \( d \). It is being charged by an external AC source. Let \( q \) be the charge on the positive plate at any instant \( t \).

1. **Electric Field in the Capacitor:**
The electric field (\( E \)) between the plates of a parallel plate capacitor is given by:
\( E = \frac{\sigma}{\varepsilon_0} \)
where \( \sigma \) is the surface charge density, \( \sigma = \frac{q}{A} \).
So, \( E = \frac{q}{\varepsilon_0 A} \)

2. **Electric Flux Through the Capacitor:**
The electric flux (\( \Phi_E \)) through the region between the plates is the product of the electric field and the area:
\( \Phi_E = E \cdot A \)
Substitute the expression for \( E \):
\( \Phi_E = \left(\frac{q}{\varepsilon_0 A}\right) A \)
\( \Phi_E = \frac{q}{\varepsilon_0} \)

3. **Displacement Current (\( I_d \)):**
The displacement current is defined as:
\( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \)
Substitute the expression for \( \Phi_E \):
\( I_d = \varepsilon_0 \frac{d}{dt}\left(\frac{q}{\varepsilon_0}\right) \)
Since \( \varepsilon_0 \) is a constant, it cancels out:
\( I_d = \frac{dq}{dt} \)

4. **Conduction Current (\( I_c \)):**
The current charging the capacitor through the connecting wires is the conduction current (\( I_c \)). This current is defined as the rate at which charge flows to the positive plate:
\( I_c = \frac{dq}{dt} \)

5. **Conclusion:**
Comparing the expressions for displacement current and conduction current, we find:
\( I_d = \frac{dq}{dt} \) and \( I_c = \frac{dq}{dt} \)
Therefore, \( I_d = I_c \).
This shows that the displacement current inside the capacitor is exactly equal to the conduction current flowing through the external wires that are charging the capacitor. This is essential for the consistency of Ampere-Maxwell's law.
In simple words: When you charge a capacitor, electricity flows in the wires. Inside the capacitor, no charges actually move, but the electric field changes. We show that the "displacement current" caused by this changing electric field is exactly the same as the real current flowing in the wires. This makes sure that the total current is conserved and Maxwell's equations work correctly everywhere.

🎯 Exam Tip: Clearly define the electric field and flux first. The key step is the differentiation of flux to find displacement current, which then directly matches the definition of conduction current.