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Electromagnetic Waves NCERT Solutions, One-Shot Study Material, and Important Questions
This chapter bridges electricity and magnetism, unlocking the secrets of light, radio, and X-rays. It's a highly scoring, conceptual chapter crucial for CBSE Board Exams 2026 and competitive tests like JEE/NEET. By mastering these updated NCERT solutions, you'll secure easy marks!
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Chapter NameElectromagnetic Waves (Chapter 8)
SubjectPhysics
Class12
BoardCBSE (2026-27 Syllabus)
DifficultyEasy to Moderate
Exam Weightage~3-4 Marks (Combined with Optics for 18 Marks)
Learning Objectives
After completing this chapter, students will be able to:
- Understand the concept of Displacement Current and Maxwell's correction to Ampere's law.
- Explain the transverse nature and characteristics of Electromagnetic Waves.
- Calculate the speed, frequency, and wavelength of EM waves using mathematical relations.
- Identify and categorize different regions of the Electromagnetic Spectrum (Radio waves to Gamma rays).
- State the practical applications of different types of EM waves in daily life.
Key Concepts, Definitions & Formulas
- Displacement Current ($I_d$): The current that comes into existence due to a time-varying electric field.
$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$$ - Maxwell's Equations: Four equations that form the foundation of classical electromagnetism, uniting electricity and magnetism.
- Ampere-Maxwell Law: Shows that magnetic fields are produced both by conduction currents and changing electric fields.
$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$
- Ampere-Maxwell Law: Shows that magnetic fields are produced both by conduction currents and changing electric fields.
- Speed of EM Waves ($c$): In a vacuum, EM waves travel at the speed of light.
$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^8 \text{ m/s}$$ - Relation between $E$ and $B$: The amplitudes of the electric and magnetic fields are related by the speed of light.
$$c = \frac{E_0}{B_0}$$ - Electromagnetic Spectrum: The orderly distribution of EM waves according to their wavelength or frequency (Gamma, X-Rays, UV, Visible, IR, Micro, Radio).
Trick to remember: "Rich Men In Vegas Use X-ray Glasses".
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Full NCERT Solutions
Question 8.1: Figure shows a capacitor made of two circular plates each of radius $12 \text{ cm}$, and separated by $5.0 \text{ cm}$. The capacitor is being charged by an external source. The charging current is constant and equal to $0.15 \text{ A}$.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain.
Answer:
Step 1: Identify given values.
Radius, $R = 12 \text{ cm} = 0.12 \text{ m}$
Separation, $d = 5.0 \text{ cm} = 0.05 \text{ m}$
Current, $I = 0.15 \text{ A}$
Step 2: Calculate Capacitance (a).
$$C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 (\pi R^2)}{d}$$
$$C = \frac{8.85 \times 10^{-12} \times 3.14 \times (0.12)^2}{0.05} = 8.0 \times 10^{-12} \text{ F} = 8.0 \text{ pF}$$
Step 3: Calculate the rate of change of potential difference (a).
We know $Q = CV \implies I = \frac{dQ}{dt} = C \frac{dV}{dt}$
$$\frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8.0 \times 10^{-12}} = 1.87 \times 10^{10} \text{ V/s}$$
Step 4: Obtain displacement current (b).
The displacement current ($I_d$) is exactly equal to the conduction current ($I_c$) charging the capacitor.
$$I_d = I_c = 0.15 \text{ A}$$
Step 5: Verify Kirchhoff's Rule (c).
Yes, Kirchhoff's first rule is perfectly valid at each plate if we consider the total current as the sum of conduction current and displacement current ($I = I_c + I_d$). The current entering the plate ($I_c$) equals the current leaving it ($I_d$).
Step 1: Identify given values.
Radius, $R = 12 \text{ cm} = 0.12 \text{ m}$
Separation, $d = 5.0 \text{ cm} = 0.05 \text{ m}$
Current, $I = 0.15 \text{ A}$
Step 2: Calculate Capacitance (a).
$$C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 (\pi R^2)}{d}$$
$$C = \frac{8.85 \times 10^{-12} \times 3.14 \times (0.12)^2}{0.05} = 8.0 \times 10^{-12} \text{ F} = 8.0 \text{ pF}$$
Step 3: Calculate the rate of change of potential difference (a).
We know $Q = CV \implies I = \frac{dQ}{dt} = C \frac{dV}{dt}$
$$\frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8.0 \times 10^{-12}} = 1.87 \times 10^{10} \text{ V/s}$$
Step 4: Obtain displacement current (b).
The displacement current ($I_d$) is exactly equal to the conduction current ($I_c$) charging the capacitor.
$$I_d = I_c = 0.15 \text{ A}$$
Step 5: Verify Kirchhoff's Rule (c).
Yes, Kirchhoff's first rule is perfectly valid at each plate if we consider the total current as the sum of conduction current and displacement current ($I = I_c + I_d$). The current entering the plate ($I_c$) equals the current leaving it ($I_d$).
Question 8.2: A parallel plate capacitor made of circular plates each of radius $R = 6.0 \text{ cm}$ has a capacitance $C = 100 \text{ pF}$. The capacitor is connected to a $230 \text{ V}$ ac supply with an angular frequency of $300 \text{ rad/s}$.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point $3.0 \text{ cm}$ from the axis between the plates.
Answer:
Step 1: Identify given values.
$R = 0.06 \text{ m}$, $C = 100 \times 10^{-12} \text{ F}$, $V_{rms} = 230 \text{ V}$, $\omega = 300 \text{ rad/s}$
Step 2: Calculate RMS conduction current (a).
$$I_{rms} = \frac{V_{rms}}{X_C} = V_{rms} (\omega C)$$
$$I_{rms} = 230 \times 300 \times 100 \times 10^{-12} = 6.9 \times 10^{-6} \text{ A} = 6.9 \text{ \mu A}$$
Step 3: Analyze displacement current (b).
Yes, the conduction current in the connecting wires is always equal to the displacement current in the region between the plates.
Step 4: Determine the amplitude of B (c).
Distance from axis, $r = 3.0 \text{ cm} = 0.03 \text{ m}$.
First, find the peak current:
$$I_0 = \sqrt{2} I_{rms} = 1.414 \times 6.9 \text{ \mu A} = 9.76 \text{ \mu A}$$
Using Ampere-Maxwell law for a point inside the capacitor ($r < R$):
$$B_0 = \frac{\mu_0 I_0 r}{2\pi R^2}$$
$$B_0 = \frac{4\pi \times 10^{-7} \times 9.76 \times 10^{-6} \times 0.03}{2\pi \times (0.06)^2} = 1.63 \times 10^{-11} \text{ T}$$
Step 1: Identify given values.
$R = 0.06 \text{ m}$, $C = 100 \times 10^{-12} \text{ F}$, $V_{rms} = 230 \text{ V}$, $\omega = 300 \text{ rad/s}$
Step 2: Calculate RMS conduction current (a).
$$I_{rms} = \frac{V_{rms}}{X_C} = V_{rms} (\omega C)$$
$$I_{rms} = 230 \times 300 \times 100 \times 10^{-12} = 6.9 \times 10^{-6} \text{ A} = 6.9 \text{ \mu A}$$
Step 3: Analyze displacement current (b).
Yes, the conduction current in the connecting wires is always equal to the displacement current in the region between the plates.
Step 4: Determine the amplitude of B (c).
Distance from axis, $r = 3.0 \text{ cm} = 0.03 \text{ m}$.
First, find the peak current:
$$I_0 = \sqrt{2} I_{rms} = 1.414 \times 6.9 \text{ \mu A} = 9.76 \text{ \mu A}$$
Using Ampere-Maxwell law for a point inside the capacitor ($r < R$):
$$B_0 = \frac{\mu_0 I_0 r}{2\pi R^2}$$
$$B_0 = \frac{4\pi \times 10^{-7} \times 9.76 \times 10^{-6} \times 0.03}{2\pi \times (0.06)^2} = 1.63 \times 10^{-11} \text{ T}$$
Question 8.3: What physical quantity is the same for X-rays of wavelength $10^{-10} \text{ m}$, red light of wavelength $6800 \text{ \AA}$, and radiowaves of wavelength $500 \text{ m}$?
Answer:
Step 1: Identify the underlying principle.
The speed of light in a vacuum ($c \approx 3 \times 10^8 \text{ m/s}$) is the exact same for X-rays, red light, and radiowaves.
Final Answer: All electromagnetic waves travel at this universal speed in a vacuum, regardless of their wavelength or frequency.
Step 1: Identify the underlying principle.
The speed of light in a vacuum ($c \approx 3 \times 10^8 \text{ m/s}$) is the exact same for X-rays, red light, and radiowaves.
Final Answer: All electromagnetic waves travel at this universal speed in a vacuum, regardless of their wavelength or frequency.
Question 8.4: A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is $30 \text{ MHz}$, what is its wavelength?
Answer:
Step 1: Determine directions.
In an EM wave, the electric field vector ($\mathbf{E}$) and magnetic field vector ($\mathbf{B}$) are mutually perpendicular to each other, and also perpendicular to the direction of wave propagation. Since the wave travels along the z-direction, $\mathbf{E}$ and $\mathbf{B}$ must lie in the x-y plane.
Step 2: Calculate the wavelength.
Given $f = 30 \text{ MHz} = 30 \times 10^6 \text{ Hz}$.
$$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{30 \times 10^6} = 10 \text{ m}$$
Step 1: Determine directions.
In an EM wave, the electric field vector ($\mathbf{E}$) and magnetic field vector ($\mathbf{B}$) are mutually perpendicular to each other, and also perpendicular to the direction of wave propagation. Since the wave travels along the z-direction, $\mathbf{E}$ and $\mathbf{B}$ must lie in the x-y plane.
Step 2: Calculate the wavelength.
Given $f = 30 \text{ MHz} = 30 \times 10^6 \text{ Hz}$.
$$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{30 \times 10^6} = 10 \text{ m}$$
Question 8.5: A radio can tune in to any station in the $7.5 \text{ MHz}$ to $12 \text{ MHz}$ band. What is the corresponding wavelength band?
Answer:
Step 1: Identify given frequencies.
$f_1 = 7.5 \times 10^6 \text{ Hz}$, $f_2 = 12 \times 10^6 \text{ Hz}$
Step 2: Calculate wavelength for lower frequency limit.
$$\lambda_1 = \frac{c}{f_1} = \frac{3 \times 10^8}{7.5 \times 10^6} = 40 \text{ m}$$
Step 3: Calculate wavelength for upper frequency limit.
$$\lambda_2 = \frac{c}{f_2} = \frac{3 \times 10^8}{12 \times 10^6} = 25 \text{ m}$$
Final Answer: The corresponding wavelength band is $25 \text{ m}$ to $40 \text{ m}$.
Step 1: Identify given frequencies.
$f_1 = 7.5 \times 10^6 \text{ Hz}$, $f_2 = 12 \times 10^6 \text{ Hz}$
Step 2: Calculate wavelength for lower frequency limit.
$$\lambda_1 = \frac{c}{f_1} = \frac{3 \times 10^8}{7.5 \times 10^6} = 40 \text{ m}$$
Step 3: Calculate wavelength for upper frequency limit.
$$\lambda_2 = \frac{c}{f_2} = \frac{3 \times 10^8}{12 \times 10^6} = 25 \text{ m}$$
Final Answer: The corresponding wavelength band is $25 \text{ m}$ to $40 \text{ m}$.
Question 8.6: A charged particle oscillates about its mean equilibrium position with a frequency of $10^9 \text{ Hz}$. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
Step 1: Relate particle oscillation to wave frequency.
The frequency of the electromagnetic wave produced is always equal to the frequency of the oscillating charged particle.
Final Answer: Therefore, the frequency of the EM wave is $10^9 \text{ Hz}$.
Step 1: Relate particle oscillation to wave frequency.
The frequency of the electromagnetic wave produced is always equal to the frequency of the oscillating charged particle.
Final Answer: Therefore, the frequency of the EM wave is $10^9 \text{ Hz}$.
Question 8.7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0 = 510 \text{ nT}$. What is the amplitude of the electric field part of the wave?
Answer:
Step 1: Identify the given values.
$B_0 = 510 \text{ nT} = 510 \times 10^{-9} \text{ T}$
Step 2: Apply the relationship formula.
Using the relation between $E_0$ and $B_0$:
$$E_0 = c \times B_0 = (3 \times 10^8) \times (510 \times 10^{-9}) = 153 \text{ N/C} \text{ (or V/m)}$$
Step 1: Identify the given values.
$B_0 = 510 \text{ nT} = 510 \times 10^{-9} \text{ T}$
Step 2: Apply the relationship formula.
Using the relation between $E_0$ and $B_0$:
$$E_0 = c \times B_0 = (3 \times 10^8) \times (510 \times 10^{-9}) = 153 \text{ N/C} \text{ (or V/m)}$$
Question 8.8: Suppose that the electric field amplitude of an electromagnetic wave is $E_0 = 120 \text{ N/C}$ and that its frequency is $\nu = 50.0 \text{ MHz}$.
(a) Determine $B_0$, $\omega$, $k$, and $\lambda$.
(b) Find expressions for $E$ and $B$.
Answer:
Step 1: Identify given values.
$E_0 = 120 \text{ N/C}$, $f = 50 \times 10^6 \text{ Hz}$
Step 2: Calculate wave parameters (a).
* $B_0$ (Magnetic field amplitude): $$B_0 = \frac{E_0}{c} = \frac{120}{3 \times 10^8} = 400 \times 10^{-9} \text{ T} = 400 \text{ nT}$$ * $\omega$ (Angular frequency): $$\omega = 2\pi f = 2 \times 3.14 \times 50 \times 10^6 = 3.14 \times 10^8 \text{ rad/s}$$ * $\lambda$ (Wavelength): $$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{50 \times 10^6} = 6.0 \text{ m}$$ * $k$ (Propagation constant): $$k = \frac{2\pi}{\lambda} = \frac{2 \times 3.14}{6.0} = 1.05 \text{ rad/m}$$
Step 3: Write expressions for E and B (b).
Assuming the wave propagates in the positive x-direction, the Electric field ($\mathbf{E}$) can be along the y-axis and Magnetic field ($\mathbf{B}$) along the z-axis.
$$\mathbf{E} = E_0 \sin(kx - \omega t) \mathbf{\hat{j}} = 120 \sin(1.05x - 3.14 \times 10^8 t) \mathbf{\hat{j}} \text{ V/m}$$ $$\mathbf{B} = B_0 \sin(kx - \omega t) \mathbf{\hat{k}} = (400 \times 10^{-9}) \sin(1.05x - 3.14 \times 10^8 t) \mathbf{\hat{k}} \text{ T}$$
Step 1: Identify given values.
$E_0 = 120 \text{ N/C}$, $f = 50 \times 10^6 \text{ Hz}$
Step 2: Calculate wave parameters (a).
* $B_0$ (Magnetic field amplitude): $$B_0 = \frac{E_0}{c} = \frac{120}{3 \times 10^8} = 400 \times 10^{-9} \text{ T} = 400 \text{ nT}$$ * $\omega$ (Angular frequency): $$\omega = 2\pi f = 2 \times 3.14 \times 50 \times 10^6 = 3.14 \times 10^8 \text{ rad/s}$$ * $\lambda$ (Wavelength): $$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{50 \times 10^6} = 6.0 \text{ m}$$ * $k$ (Propagation constant): $$k = \frac{2\pi}{\lambda} = \frac{2 \times 3.14}{6.0} = 1.05 \text{ rad/m}$$
Step 3: Write expressions for E and B (b).
Assuming the wave propagates in the positive x-direction, the Electric field ($\mathbf{E}$) can be along the y-axis and Magnetic field ($\mathbf{B}$) along the z-axis.
$$\mathbf{E} = E_0 \sin(kx - \omega t) \mathbf{\hat{j}} = 120 \sin(1.05x - 3.14 \times 10^8 t) \mathbf{\hat{j}} \text{ V/m}$$ $$\mathbf{B} = B_0 \sin(kx - \omega t) \mathbf{\hat{k}} = (400 \times 10^{-9}) \sin(1.05x - 3.14 \times 10^8 t) \mathbf{\hat{k}} \text{ T}$$
Question 8.9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = h\nu$ to obtain the photon energy in units of eV for different parts of the electromagnetic spectrum.
Answer:
Step 1: Set up the formula.
Using $E = h\nu = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34} \text{ Js}$. To convert Joules to eV, we divide by $1.6 \times 10^{-19}$.
The simplified formula becomes: $$E (\text{in eV}) \approx \frac{1240}{\lambda \text{ (in nm)}}$$
Step 2: Calculate for different spectrum parts.
* Radio waves: ($\lambda \sim 1 \text{ m}$): Very low energy ($\sim 10^{-6} \text{ eV}$) * Visible light: ($\lambda \sim 500 \text{ nm}$): Moderate energy ($\sim 2 \text{ to } 3 \text{ eV}$) * Gamma rays: ($\lambda \sim 10^{-12} \text{ m}$): Extremely high energy ($\sim 10^6 \text{ eV}$)
Step 1: Set up the formula.
Using $E = h\nu = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34} \text{ Js}$. To convert Joules to eV, we divide by $1.6 \times 10^{-19}$.
The simplified formula becomes: $$E (\text{in eV}) \approx \frac{1240}{\lambda \text{ (in nm)}}$$
Step 2: Calculate for different spectrum parts.
* Radio waves: ($\lambda \sim 1 \text{ m}$): Very low energy ($\sim 10^{-6} \text{ eV}$) * Visible light: ($\lambda \sim 500 \text{ nm}$): Moderate energy ($\sim 2 \text{ to } 3 \text{ eV}$) * Gamma rays: ($\lambda \sim 10^{-12} \text{ m}$): Extremely high energy ($\sim 10^6 \text{ eV}$)
Question 8.10: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \text{ Hz}$ and amplitude $48 \text{ V/m}$.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field.
Answer:
Step 1: Identify given values.
$f = 2.0 \times 10^{10} \text{ Hz}$, $E_0 = 48 \text{ V/m}$.
Step 2: Solve part (a) Wavelength.
$$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{2 \times 10^{10}} = 0.015 \text{ m}$$
Step 3: Solve part (b) Magnetic field amplitude.
$$B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7} \text{ T}$$
Step 4: Prove energy density equality (c).
Average energy density of electric field: $$U_E = \frac{1}{4} \varepsilon_0 E_0^2$$ Average energy density of magnetic field: $$U_B = \frac{B_0^2}{4\mu_0}$$ We know $E_0 = c B_0$ and $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies c^2 = \frac{1}{\mu_0 \varepsilon_0}$.
Substitute $E_0$ in $U_E$:
$$U_E = \frac{1}{4} \varepsilon_0 (c B_0)^2 = \frac{1}{4} \varepsilon_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right) B_0^2 = \frac{B_0^2}{4\mu_0} = U_B$$
Hence, proved $U_E = U_B$.
Step 1: Identify given values.
$f = 2.0 \times 10^{10} \text{ Hz}$, $E_0 = 48 \text{ V/m}$.
Step 2: Solve part (a) Wavelength.
$$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{2 \times 10^{10}} = 0.015 \text{ m}$$
Step 3: Solve part (b) Magnetic field amplitude.
$$B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7} \text{ T}$$
Step 4: Prove energy density equality (c).
Average energy density of electric field: $$U_E = \frac{1}{4} \varepsilon_0 E_0^2$$ Average energy density of magnetic field: $$U_B = \frac{B_0^2}{4\mu_0}$$ We know $E_0 = c B_0$ and $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies c^2 = \frac{1}{\mu_0 \varepsilon_0}$.
Substitute $E_0$ in $U_E$:
$$U_E = \frac{1}{4} \varepsilon_0 (c B_0)^2 = \frac{1}{4} \varepsilon_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right) B_0^2 = \frac{B_0^2}{4\mu_0} = U_B$$
Hence, proved $U_E = U_B$.
Extra Important Questions (Board Style)
Multiple Choice Questions (MCQs)
1. Which of the following electromagnetic waves has the highest frequency?
a) Microwaves
b) X-rays
c) Gamma rays
d) Ultraviolet rays
Answer: (c) Gamma rays
2. Displacement current is continuously produced by:
a) A constant electric field
b) A constant magnetic field
c) A time-varying electric field
d) A time-varying magnetic field
Answer: (c) A time-varying electric field
3. In an EM wave, the phase difference between the electric and magnetic field vectors is:
a) $\pi/2$
b) $\pi/4$
c) $\pi$
d) Zero
Answer: (d) Zero (They oscillate in the same phase)
Short Answer Questions
4. Name the electromagnetic waves used in LASIK eye surgery and mention their wavelength range.
Answer: Ultraviolet (UV) rays are used in LASIK eye surgery. Their wavelength range is $400 \text{ nm}$ to $1 \text{ nm}$.
5. Why are microwaves considered suitable for radar systems used in aircraft navigation?
Answer: Microwaves have shorter wavelengths, which allows them to travel as a focused beam without diffracting (bending) much, making them ideal for precise radar location.
6. Identify the part of the EM spectrum that produces a heating effect.
Answer: Infrared (IR) waves. They are readily absorbed by water molecules in objects, increasing thermal motion and producing a heating effect.
7. Write the mathematical expression for the speed of an EM wave in a material medium.
Answer: $$v = \frac{1}{\sqrt{\mu \varepsilon}}$$ where $\mu$ is the permeability and $\varepsilon$ is the permittivity of the medium.
Long Answer Questions
8. State the four properties of electromagnetic waves. Write Maxwell's four equations that form the theoretical basis of EM waves.
Answer:
Properties:
1. They do not require a material medium to propagate.
2. They are transverse in nature.
3. They travel with the speed of light ($c$) in a vacuum.
4. They carry energy and momentum, which are shared equally between the electric and magnetic fields.
Maxwell's Equations:
1. Gauss's Law for Electricity: $$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\varepsilon_0}$$ 2. Gauss's Law for Magnetism: $$\oint \mathbf{B} \cdot d\mathbf{A} = 0$$ 3. Faraday's Law: $$\oint \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt}$$ 4. Ampere-Maxwell Law: $$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$
Properties:
1. They do not require a material medium to propagate.
2. They are transverse in nature.
3. They travel with the speed of light ($c$) in a vacuum.
4. They carry energy and momentum, which are shared equally between the electric and magnetic fields.
Maxwell's Equations:
1. Gauss's Law for Electricity: $$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\varepsilon_0}$$ 2. Gauss's Law for Magnetism: $$\oint \mathbf{B} \cdot d\mathbf{A} = 0$$ 3. Faraday's Law: $$\oint \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt}$$ 4. Ampere-Maxwell Law: $$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$
9. What is displacement current? Derive the expression for it using Ampere-Maxwell's modified law during the charging of a capacitor.
Answer:
Step 1: Definition.
Displacement current is the equivalent current created by a changing electric field across space.
Step 2: Derivation during capacitor charging.
When charging a capacitor, conduction current stops at the plates. However, the varying electric field between plates creates an electric flux:
$$\Phi_E = E \cdot A = \left(\frac{Q}{A\varepsilon_0}\right) \cdot A = \frac{Q}{\varepsilon_0}$$
Differentiating with respect to time $t$, we get:
$$\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dQ}{dt}$$
Since $I = \frac{dQ}{dt}$, we find:
$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$$
This bridges the gap in Ampere's original circuit law.
Step 1: Definition.
Displacement current is the equivalent current created by a changing electric field across space.
Step 2: Derivation during capacitor charging.
When charging a capacitor, conduction current stops at the plates. However, the varying electric field between plates creates an electric flux:
$$\Phi_E = E \cdot A = \left(\frac{Q}{A\varepsilon_0}\right) \cdot A = \frac{Q}{\varepsilon_0}$$
Differentiating with respect to time $t$, we get:
$$\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dQ}{dt}$$
Since $I = \frac{dQ}{dt}$, we find:
$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$$
This bridges the gap in Ampere's original circuit law.
10. Draw a neat, labeled diagram of a plane electromagnetic wave propagating in the X-direction. Show the directions of the electric and magnetic fields.
Answer:
(Visual description for exam drawing)
Step 1: Draw an x, y, and z axis. The wave progresses along the positive x-axis.
Step 2: Draw a sine wave oscillating strictly in the xy-plane — label this $\mathbf{E}$ (Electric field).
Step 3: Draw a second sine wave oscillating strictly in the xz-plane, perfectly in phase with the first wave — label this $\mathbf{B}$ (Magnetic field).
(Visual description for exam drawing)
Step 1: Draw an x, y, and z axis. The wave progresses along the positive x-axis.
Step 2: Draw a sine wave oscillating strictly in the xy-plane — label this $\mathbf{E}$ (Electric field).
Step 3: Draw a second sine wave oscillating strictly in the xz-plane, perfectly in phase with the first wave — label this $\mathbf{B}$ (Magnetic field).
Case-Based Questions
11-13. Read the following paragraph and answer the questions.
The electromagnetic spectrum encompasses a continuous range of frequencies and wavelengths of electromagnetic radiation. From low-frequency radio waves used in communication to high-frequency gamma rays emitted by radioactive nuclei, each band has specific properties and applications. Earth's atmosphere heavily filters the spectrum, allowing visible light and some radio waves to pass while blocking harmful UV and X-rays.
11. Which EM wave is predominantly used for mobile phone communication?
Answer: Microwaves and Radio waves (specifically UHF band).
12. Why are infrared waves commonly known as "heat waves"?
Answer: Because they induce resonant vibrations in molecules (especially water), releasing energy in the form of heat.
13. Name the wave used for studying the internal crystal structure of solids.
Answer: X-rays (due to their short wavelength, which is comparable to atomic spacing, making them perfect for crystallography).
Answer: Microwaves and Radio waves (specifically UHF band).
12. Why are infrared waves commonly known as "heat waves"?
Answer: Because they induce resonant vibrations in molecules (especially water), releasing energy in the form of heat.
13. Name the wave used for studying the internal crystal structure of solids.
Answer: X-rays (due to their short wavelength, which is comparable to atomic spacing, making them perfect for crystallography).
Assertion-Reason Questions
(Options: A = Both True & R is correct explanation, B = Both True & R is not correct, C = A is True & R is False, D = A is False & R is True)
14. Assertion (A): Electromagnetic waves are transverse in nature.
Reason (R): The electric and magnetic fields are perpendicular to each other and perpendicular to the direction of propagation.
Answer: (A) Both A and R are true, and R is the correct explanation of A.
15. Assertion (A): Sound waves traveling through air are an example of electromagnetic waves.
Reason (R): Sound waves require a material medium for their propagation.
Answer: (D) Assertion is false (Sound is a mechanical wave, not an EM wave). Reason is true.
Common Mistakes Students Make
- Mixing up E and B relation: Students often write $B_0 = c \cdot E_0$. The correct formula is $E_0 = c \cdot B_0$. Just remember: the Electric field amplitude ($E_0$) is always massively larger than the Magnetic field amplitude ($B_0$)!
- Forgetting Unit Conversions: A wavelength given in angstroms ($1 \text{ \AA} = 10^{-10} \text{ m}$) or nanometers ($1 \text{ nm} = 10^{-9} \text{ m}$) must be converted into standard SI units before applying formulas.
- Confusion over Phase: In EM waves, Electric and Magnetic fields oscillate in phase. Many students wrongly assume there's a phase difference of $\pi/2$ just because they are geometrically perpendicular.
- Spectrum Order: Mixing up the order of the EM spectrum based on increasing frequency vs. increasing wavelength.
Exam Preparation Tips
- Last-Minute Revision: Write down the 7 EM waves in order of increasing frequency, along with one common use and one production method for each. This guarantees you 2 solid marks!
- Equation Mastery: Practice writing the sinusoidal equations of $\mathbf{E}$ and $\mathbf{B}$. Remember that the direction of propagation is given by the cross product $\mathbf{E} \times \mathbf{B}$.
- Time Management: For numericals involving $10^8$ or $10^{-12}$ powers, group the powers of 10 separately to avoid calculation errors and save time during the exam.
FAQ Section
Is Chapter 8 Electromagnetic Waves important for boards?
Yes, while it's a short chapter, it practically guarantees ~3-4 marks through straightforward conceptual questions and simple numericals. It's a highly scoring section.
Where can I download the updated NCERT PDF for Class 12 Physics?
You can find the updated NCERT Class 12 Physics textbook PDFs directly on the official NCERT website (ncert.nic.in) under the "Publications" tab.
Which questions are most important in Chapter 8?
NCERT exercise questions 8.1 (Displacement current), 8.8 (Writing E and B equations), and 8.10 are extremely important for board exams.
What is the weightage of this chapter in CBSE 2026?
Electromagnetic Waves (Chapter 8) is combined with Ray and Wave Optics for an 18-mark unit block. Individually, expect it to carry a weightage of around 3 to 4 marks.
How can I memorize the Electromagnetic Spectrum easily?
Use a simple mnemonic based on increasing frequency: "Radio Microwaves Infrared Visible Ultraviolet X-rays Gamma" → "Raging Martians Invaded Venus Using X-ray Guns".
Conclusion: Understanding Electromagnetic Waves isn't just about passing the CBSE Physics exam—it's about understanding how your Wi-Fi, radio, and the sun's light actually reach you! Revise the spectrum order regularly, practice the displacement current derivation, and solve Past Year Questions (PYQs) to solidify your confidence. Download these notes, stay focused, and you will undoubtedly crush your 2026 Board Exams!
More Class 12 Physics Chapters
Ch 1: Electric Charges and Fields
Ch 2: Electrostatic Potential and Capacitance
Ch 3: Current Electricity
Ch 4: Moving Charges and Magnetism
Ch 5: Magnetism and Matter
Ch 6: Electromagnetic Induction
Ch 7: Alternating Current
Ch 8: Electromagnetic Waves
Ch 9: Ray Optics
Ch 10: Wave Optics
Ch 11: Dual Nature of Radiation and Matter
Ch 12: Atoms
Ch 13: Nuclei
Class 12 Mock QUIZs
Ch 14: Semiconductor Electronics