Class 12 Physics NCERT REPRINT 2026-27

Chapter 8: Electromagnetic Waves

Complete concise study notes, displacement current, Ampere-Maxwell law, Maxwell's 4 equations, nature of EM waves, complete electromagnetic spectrum, solved examples 8.1–8.2, and exercises 8.1–8.10.

01 / Exam-Focused Notes

Chapter 8: Electromagnetic Waves

Complete official NCERT textbook coverage with high-precision vector graphs: Inconsistency in Ampere's Law & Maxwell's Displacement Current ($i_d = \varepsilon_0 d\Phi_E/dt$), The 4 Maxwell's Equations, Sources & Transverse Nature of EM Waves, Wave Speed ($c = 1/\sqrt{\mu_0 \varepsilon_0} = E_0/B_0$), Energy Equipartition ($u_E = u_B$), and Complete Electromagnetic Spectrum Classification (Radio, Micro, IR, Visible, UV, X-rays, Gamma rays).

100% SYLLABUS
Chapter Navigation • NCERT Sections 8.1 – 8.4
8.1-8.2Displacement Current & Maxwell Eq. 8.3Nature & Speed of EM Waves 8.4Electromagnetic Spectrum
8.1 & 8.2
NCERT Sections

Introduction & Maxwell's Displacement Current

Inconsistency in Ampere's Law & Resolution

8.2.1 Need for Displacement Current ($i_d$) & Ampere-Maxwell Law

  • Inconsistency in Ampere's Circuital Law:
    Applying $\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 i(t)$ to a loop enclosing the wire outside a charging capacitor gives non-zero $B(2\pi r) = \mu_0 i(t)$. But applying it to a pot-like or tiffin-box surface with bottom between capacitor plates (where no conduction current passes) gives zero magnetic field.
    Contradiction: Calculated one way, magnetic field is non-zero; calculated another way, it is zero.
  • Maxwell's Resolution & Displacement Current:
    Between plates of area $A$ with charge $Q(t)$, uniform electric field is $E = \frac{Q}{\varepsilon_0 A} \implies$ electric flux $\Phi_E = E A = \frac{Q}{\varepsilon_0}$.
    As capacitor charges, rate of change of flux is: $$\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0}\frac{dQ}{dt} = \frac{i(t)}{\varepsilon_0} \implies i_d \equiv \varepsilon_0 \frac{d\Phi_E}{dt} = i(t)$$
$$i_{\text{total}} = i_c + i_d = i_c + \varepsilon_0 \frac{d\Phi_E}{dt}, \qquad \oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \quad \text{[Ampere-Maxwell Law]}$$
  • Continuity of Current: Outside plates: $i_c = i, i_d = 0$. Inside dielectric/gap: $i_c = 0, i_d = i$. Total current is continuous across all circuit boundaries.
  • Physical Significance: Time-varying electric fields produce magnetic fields, providing symmetry to Faraday's law (where time-varying magnetic fields produce electric fields).
Fig 8.1: Conduction Current (ic) and Displacement Current (id) in Charging Capacitor
i_c (t) +Q(t) dE/dt > 0 i_d = ε₀ (dΦ_E / dt) -Q(t) i_c (t) Total Current is Continuous Everywhere: i = i_c (outside) = i_d (inside)
Fundamental Unification

8.2.2 The Four Maxwell's Equations in Vacuum

Law / Equation Mathematical Statement Physical Meaning
1. Gauss's Law (Electrostatics) ∮ E · dA = Q / ε₀ Electric charges act as sources and sinks of electric field.
2. Gauss's Law (Magnetism) ∮ B · dA = 0 Isolated magnetic monopoles do not exist; field lines form closed loops.
3. Faraday's Law of Induction ∮ E · dl = −dΦB / dt Time-varying magnetic flux produces an induced electric field.
4. Ampere-Maxwell Law ∮ B · dl = μ₀ i_c + μ₀ε₀ (dΦE / dt) Conduction current and time-varying electric flux both produce magnetic fields.
8.3
NCERT Section

Electromagnetic Waves

Propagation & Transverse Nature

8.3.1 Sources, Transverse Character & Mathematical Equations

  • Sources of EM Waves:
    • Stationary charges produce only static electric fields ($E$).
    • Uniformly moving charges (steady currents) produce static magnetic fields ($B$).
    • Accelerating / oscillating charges radiate electromagnetic waves with wave frequency equal to oscillation frequency $\nu$.
  • Transverse Nature of EM Waves:
    $\mathbf{E}$ and $\mathbf{B}$ vectors oscillate perpendicularly to each other and perpendicular to the direction of propagation vector $\mathbf{k}$ ($\mathbf{E} \times \mathbf{B} \parallel \mathbf{k}$).
    For wave propagating in $+z$-direction with $\mathbf{E}$ along $x$-axis and $\mathbf{B}$ along $y$-axis: $$E_x(z, t) = E_0 \sin(k z - \omega t), \qquad B_y(z, t) = B_0 \sin(k z - \omega t)$$ where $k = \frac{2\pi}{\lambda}$ is wave number and $\omega = 2\pi\nu$ is angular frequency.
$$c = \frac{\omega}{k} = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^8\text{ m s}^{-1}, \qquad \frac{E_0}{B_0} = c \implies E_0 = c B_0$$
Fig 8.4: Transverse Electromagnetic Wave: Mutually Perpendicular E and B Fields
Propagation (k || +z) → Electric Field E (x) E_x B_y E ⊥ B ⊥ k • E₀ / B₀ = c = 1 / √(μ₀ ε₀)
  • Speed in Material Medium: $$v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}} = \frac{c}{n}$$ where $n = \sqrt{\mu_r \varepsilon_r}$ is refractive index.
  • Energy Density & Equipartition:
    Electric energy density $u_E = \frac{1}{2}\varepsilon_0 E^2$; Magnetic energy density $u_B = \frac{B^2}{2\mu_0}$.
    Since $E = c B$ and $c = 1/\sqrt{\mu_0 \varepsilon_0}$: $$u_B = \frac{(E/c)^2}{2\mu_0} = \frac{E^2 \mu_0 \varepsilon_0}{2\mu_0} = \frac{1}{2}\varepsilon_0 E^2 = u_E$$ Total average energy density is shared equally between electric and magnetic fields: $$u_{\text{avg}} = \langle u_E \rangle + \langle u_B \rangle = \frac{1}{2}\varepsilon_0 E_0^2 = \frac{B_0^2}{2\mu_0}$$
Example 8.1

Finding Magnetic Field Vector from Electric Field

A plane EM wave of frequency $25\text{ MHz}$ travels in free space along the $+x$-direction. At a particular point, $\mathbf{E} = 6.3\,\hat{\mathbf{j}}\text{ V/m}$. What is $\mathbf{B}$ at this point?

Magnitude of Magnetic Field ($B$): $$B = \frac{E}{c} = \frac{6.3\text{ V/m}}{3.0 \times 10^8\text{ m s}^{-1}} = 2.1 \times 10^{-8}\text{ T}$$
Direction of Vector $\mathbf{B}$: Wave travels in $+x$-direction ($\hat{\mathbf{i}}$) and $\mathbf{E}$ is along $+y$-direction ($\hat{\mathbf{j}}$). $$\hat{\mathbf{E}} \times \hat{\mathbf{B}} = \hat{\mathbf{k}}_{\text{prop}} \implies \hat{\mathbf{j}} \times \hat{\mathbf{B}} = \hat{\mathbf{i}} \implies \hat{\mathbf{B}} = +\hat{\mathbf{k}} \quad (\text{along }+z\text{-axis})$$
Answer: $\mathbf{B} = 2.1 \times 10^{-8}\,\hat{\mathbf{k}}\text{ T}$.
Example 8.2

Harmonic EM Wave Parameters and Field Expressions

The magnetic field in a plane EM wave is $B_y = (2 \times 10^{-7}\text{ T})\sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)$.
(a) What is wavelength $\lambda$ and frequency $\nu$?
(b) Write an expression for electric field $E_z$.

(a) Wavelength & Frequency: Comparing with $B_y = B_0 \sin(k x + \omega t)$: $$k = 0.5 \times 10^3\text{ m}^{-1} \implies \lambda = \frac{2\pi}{k} = \frac{2 \times 3.1416}{500} = 1.26 \times 10^{-2}\text{ m} = 1.26\text{ cm}$$ $$\omega = 1.5 \times 10^{11}\text{ rad s}^{-1} \implies \nu = \frac{\omega}{2\pi} = \frac{1.5 \times 10^{11}}{6.283} = 2.39 \times 10^{10}\text{ Hz} = 23.9\text{ GHz}$$
(b) Electric Field Expression: $$E_0 = c B_0 = (3.0 \times 10^8\text{ m s}^{-1}) \times (2 \times 10^{-7}\text{ T}) = 60\text{ V m}^{-1}$$ Wave propagates along $-x$-axis; $B$ is along $+y$-axis $\implies \mathbf{E}$ is along $+z$-axis: $$E_z = 60 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)\text{ V m}^{-1}$$
Answers: (a) $\lambda = 1.26\text{ cm}, \nu = 23.9\text{ GHz}$; (b) $E_z = 60 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)\text{ V/m}$.
8.4
NCERT Section

The Electromagnetic Spectrum

Complete Spectrum Classification

8.4.1 Classification, Wavelength Ranges, Production & Applications (Table 8.1)

Type Wavelength Range Production Detection Key Applications
Radio waves > 0.1 m Rapid acceleration of electrons in LC circuits & aerials Receiver aerials AM/FM radio, television, mobile phone communication (UHF).
Microwaves 0.1 m to 1 mm Klystrons, magnetrons, Gunn diodes Point contact diodes Aircraft radar navigation, speed guns, microwave ovens (water resonance).
Infrared (IR) 1 mm to 700 nm Thermal vibrations of molecules & hot bodies Thermopiles, bolometers, IR film "Heat waves", greenhouse effect, physical therapy, TV remotes, night vision.
Visible Light 700 nm to 400 nm Atomic electron transitions to lower energy states Human eye, photocells, film Human vision, optical instruments, photography ($4 \times 10^{14}$ to $7 \times 10^{14}\text{ Hz}$).
Ultraviolet (UV) 400 nm to 1 nm Inner shell electron transitions, hot bodies, welding arcs Photocells, photographic film LASIK eye surgery, water purifiers (kills germs), ozone layer absorption.
X-rays 1 nm to 10⁻³ nm Bombardment of heavy metal target by high-energy electrons Photographic film, Geiger tubes Medical radiography / diagnostic imaging, cancer radiotherapy, crystal analysis.
Gamma rays (γ) < 10⁻³ nm Nuclear reactions, radioactive decay of unstable nuclei Geiger tubes, scintillation counters Destroying cancer cells in oncology, nuclear physics research.
02 / Interactive Assessment

Concept Check & Board Mastery Quiz

15 targeted MCQs covering displacement current, Maxwell's equations, transverse wave properties, velocity of light, and the electromagnetic spectrum with in-depth explanations.

15 MCQS
0 / 15 Answered
Score: 0 (0%)
03 / Textbook Solutions

NCERT Exercises 8.1 – 8.10

Complete stepwise solutions for every textbook exercise question in NCERT Class 12 Physics Chapter 8 Electromagnetic Waves Reprint 2026-27.

10 QUESTIONS
Ex 8.1 Displacement Current in Charging Capacitor

8.1 A capacitor is made of two circular plates of radius $R = 12\text{ cm}$ separated by $d = 5.0\text{ cm}$. 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.

(a) $C = 8.01\text{ pF}, \frac{dV}{dt} = 1.87 \times 10^9\text{ V s}^{-1}$ • (b) $i_d = 0.15\text{ A}$ • (c) Yes, valid when generalized to include displacement current.
(a) Capacitance ($C$) & $dV/dt$:
Area $A = \pi R^2 = \pi \times (0.12\text{ m})^2 = 0.04524\text{ m}^2$.
$$C = \frac{\varepsilon_0 A}{d} = \frac{(8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}) \times 0.04524\text{ m}^2}{0.05\text{ m}} = 8.01 \times 10^{-12}\text{ F} = 8.01\text{ pF}$$ Since $q = C V \implies i = C \frac{dV}{dt}$: $$\frac{dV}{dt} = \frac{i}{C} = \frac{0.15\text{ A}}{8.01 \times 10^{-12}\text{ F}} = 1.87 \times 10^9\text{ V s}^{-1}$$
(b) Displacement Current ($i_d$):
$$i_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \frac{d(E A)}{dt} = \varepsilon_0 A \frac{d}{dt}\left(\frac{V}{d}\right) = C \frac{dV}{dt} = 0.15\text{ A}$$ (Displacement current inside the dielectric gap equals conduction current in the lead wires).
(c) Kirchhoff's First Rule:
Yes, Kirchhoff's junction rule is strictly valid at each plate provided the total current ($i = i_c + i_d$) is taken into account: conduction current flowing into the plate equals displacement current leaving the plate into the gap.
Ex 8.2 AC Capacitor Magnetic Field between Plates

8.2 A parallel plate capacitor made of circular plates of radius $R = 6.0\text{ cm}$ has capacitance $C = 100\text{ pF}$, connected to a $230\text{ V}$ AC supply with $\omega = 300\text{ rad s}^{-1}$.
(a) What is the rms value of the conduction current?
(b) Is conduction current equal to displacement current?
(c) Determine the amplitude of $B$ at a point $r = 3.0\text{ cm}$ from the axis between the plates.

(a) $I_{\text{rms}} = 6.9\,\mu\text{A}$ • (b) Yes, $i_c = i_d$ at all instants • (c) $B_0 = 1.63 \times 10^{-11}\text{ T}$.
(a) RMS Conduction Current:
$X_C = \frac{1}{\omega C} = \frac{1}{300 \times 100 \times 10^{-12}} = \frac{10^8}{3}\,\Omega = 3.33 \times 10^7\,\Omega$.
$$I_{\text{rms}} = \frac{V}{X_C} = \frac{230\text{ V}}{(10^8/3)\,\Omega} = 6.9 \times 10^{-6}\text{ A} = 6.9\,\mu\text{A}$$
(b) Equality: Yes, conduction current in lead wires is equal to displacement current inside capacitor at every instant ($i_c = i_d$). Peak value $i_0 = \sqrt{2} I_{\text{rms}} = 1.414 \times 6.9 \times 10^{-6}\text{ A} = 9.76 \times 10^{-6}\text{ A}$.
(c) Magnetic Field Amplitude at $r = 3\text{ cm}$ ($r < R$):
Applying Ampere-Maxwell law for radius $r$: $$\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 i_d(r) \implies B_0 (2\pi r) = \mu_0 \left(\frac{\pi r^2}{\pi R^2} i_0\right) \implies B_0 = \frac{\mu_0 r i_0}{2\pi R^2}$$ $$B_0 = \frac{(4\pi \times 10^{-7}) \times (0.03\text{ m}) \times (9.76 \times 10^{-6}\text{ A})}{2\pi \times (0.06\text{ m})^2} = \frac{2 \times 10^{-7} \times 0.03 \times 9.76 \times 10^{-6}}{3.6 \times 10^{-3}} = 1.63 \times 10^{-11}\text{ T}$$
Ex 8.3 Universal Speed of EM Waves

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 $500\text{ m}$?

The speed of propagation in vacuum ($c = 3.0 \times 10^8\text{ m s}^{-1}$) is identical for all electromagnetic waves.
All electromagnetic waves, regardless of their wavelength or frequency, travel with the exact same fundamental speed in free space/vacuum: $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3.0 \times 10^8\text{ m s}^{-1}$$
Ex 8.4 EM Wave Direction & Wavelength

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?

$\mathbf{E}$ and $\mathbf{B}$ lie in the $xy$-plane and are mutually perpendicular • Wavelength $\lambda = 10.0\text{ m}$.
Field Vector Directions: Because EM waves are transverse, $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to the propagation direction ($+z$). Hence, $\mathbf{E}$ and $\mathbf{B}$ oscillate in the $xy$-plane such that $\mathbf{E} \perp \mathbf{B}$ and $\mathbf{E} \times \mathbf{B} \parallel +\hat{\mathbf{k}}$ (e.g. $\mathbf{E}$ along $x$-axis and $\mathbf{B}$ along $y$-axis).
Wavelength Calculation: $$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8\text{ m s}^{-1}}{30 \times 10^6\text{ Hz}} = 10.0\text{ m}$$
Ex 8.5 Radio Band Wavelength Range

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?

Wavelength Band: $25\text{ m}$ to $40\text{ m}$.
Maximum Wavelength ($\lambda_1$): $$\lambda_1 = \frac{c}{\nu_1} = \frac{3.0 \times 10^8\text{ m s}^{-1}}{7.5 \times 10^6\text{ Hz}} = 40.0\text{ m}$$
Minimum Wavelength ($\lambda_2$): $$\lambda_2 = \frac{c}{\nu_2} = \frac{3.0 \times 10^8\text{ m s}^{-1}}{12 \times 10^6\text{ Hz}} = 25.0\text{ m}$$
Ex 8.6 Frequency of Radiated EM Wave

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?

Frequency of EM Wave $\nu = 10^9\text{ Hz} = 1\text{ GHz}$.
According to Maxwell's electromagnetic theory, the frequency of the radiated electromagnetic wave is strictly equal to the oscillation frequency of the accelerating charge source ($\nu = 10^9\text{ Hz}$).
Ex 8.7 Electric Field Amplitude from Magnetic Field

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?

Electric Field Amplitude $E_0 = 153\text{ N/C} = 153\text{ V/m}$.
Formula: $$E_0 = c B_0$$
Calculation: $$E_0 = (3.0 \times 10^8\text{ m s}^{-1}) \times (510 \times 10^{-9}\text{ T}) = 153\text{ V m}^{-1}$$
Ex 8.8 Complete EM Wave Parameters & Equations

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 $\mathbf{E}$ and $\mathbf{B}$.

(a) $B_0 = 400\text{ nT}, \omega = 3.14 \times 10^8\text{ rad/s}, k = 1.05\text{ rad/m}, \lambda = 6.0\text{ m}$
(b) $\mathbf{E} = 120\sin(1.05 x - 3.14 \times 10^8 t)\hat{\mathbf{j}}\text{ V/m}, \mathbf{B} = 4 \times 10^{-7}\sin(1.05 x - 3.14 \times 10^8 t)\hat{\mathbf{k}}\text{ T}$.
(a) Parameter Calculations:
$$B_0 = \frac{E_0}{c} = \frac{120\text{ N/C}}{3.0 \times 10^8\text{ m s}^{-1}} = 4.0 \times 10^{-7}\text{ T} = 400\text{ nT}$$ $$\omega = 2\pi\nu = 2 \times 3.1416 \times (50 \times 10^6\text{ s}^{-1}) = 3.14 \times 10^8\text{ rad s}^{-1}$$ $$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8}{50 \times 10^6} = 6.0\text{ m}$$ $$k = \frac{2\pi}{\lambda} = \frac{2 \times 3.1416}{6.0\text{ m}} = 1.05\text{ rad m}^{-1}$$
(b) Field Vector Expressions: Assuming propagation along $+x$-axis with $\mathbf{E}$ along $y$-axis and $\mathbf{B}$ along $z$-axis: $$\mathbf{E}(x, t) = (120\text{ N/C})\sin(1.05 x - 3.14 \times 10^8 t)\,\hat{\mathbf{j}}$$ $$\mathbf{B}(x, t) = (4.0 \times 10^{-7}\text{ T})\sin(1.05 x - 3.14 \times 10^8 t)\,\hat{\mathbf{k}}$$
Ex 8.9 Photon Energy Scale Across EM Spectrum

8.9 Use $E = h\nu$ to obtain photon energy in units of $\text{eV}$ for different parts of the EM spectrum. How are the different energy scales related to the sources of radiation?

Energy Scale Formula: $E = h\nu = \frac{h c}{\lambda}$ where $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}, h = 6.63 \times 10^{-34}\text{ J s}$.
  • Radio waves ($\lambda = 3\text{ m}, \nu = 10^8\text{ Hz}$): $E \approx 4.1 \times 10^{-7}\text{ eV} \implies$ low-energy free electron acceleration in circuits.
  • Microwaves ($\lambda = 3\text{ cm}, \nu = 10^{10}\text{ Hz}$): $E \approx 4.1 \times 10^{-5}\text{ eV} \implies$ molecular rotational transitions.
  • Infrared ($\lambda = 3\,\mu\text{m}, \nu = 10^{14}\text{ Hz}$): $E \approx 0.41\text{ eV} \implies$ molecular vibrational states.
  • Visible Light ($\lambda = 600\text{ nm}, \nu = 5 \times 10^{14}\text{ Hz}$): $E \approx 2.07\text{ eV} \implies$ outer-shell atomic electronic transitions.
  • Ultraviolet ($\lambda = 30\text{ nm}, \nu = 10^{16}\text{ Hz}$): $E \approx 41.4\text{ eV} \implies$ deep atomic valence ionization.
  • X-rays ($\lambda = 0.1\text{ nm}, \nu = 3 \times 10^{18}\text{ Hz}$): $E \approx 12.4\text{ keV} \implies$ inner-shell electron transitions in heavy atoms.
  • Gamma rays ($\lambda = 10^{-12}\text{ m}, \nu = 3 \times 10^{20}\text{ Hz}$): $E \approx 1.24\text{ MeV} \implies$ nuclear energy level transitions.
Ex 8.10 Equipartition of EM Energy Density

8.10 In a plane EM wave, electric field oscillates at frequency $\nu = 2.0 \times 10^{10}\text{ Hz}$ with amplitude $E_0 = 48\text{ V m}^{-1}$.
(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 $\mathbf{E}$ field equals the average energy density of the $\mathbf{B}$ field.

(a) $\lambda = 1.5\text{ cm}$ • (b) $B_0 = 1.6 \times 10^{-7}\text{ T} = 160\text{ nT}$ • (c) Proved: $\langle u_E \rangle = \langle u_B \rangle = \frac{1}{4}\varepsilon_0 E_0^2$.
(a) Wavelength: $$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8\text{ m s}^{-1}}{2.0 \times 10^{10}\text{ Hz}} = 1.5 \times 10^{-2}\text{ m} = 1.5\text{ cm}$$
(b) Magnetic Field Amplitude: $$B_0 = \frac{E_0}{c} = \frac{48\text{ V m}^{-1}}{3.0 \times 10^8\text{ m s}^{-1}} = 1.6 \times 10^{-7}\text{ T} = 160\text{ nT}$$
(c) Equipartition Proof:
Average electric energy density: $\langle u_E \rangle = \frac{1}{4}\varepsilon_0 E_0^2$.
Average magnetic energy density: $\langle u_B \rangle = \frac{B_0^2}{4\mu_0}$.
Substitute $B_0 = E_0/c$ and $c = 1/\sqrt{\mu_0\varepsilon_0} \implies c^2 = 1/(\mu_0\varepsilon_0)$: $$\langle u_B \rangle = \frac{E_0^2}{4\mu_0 c^2} = \frac{E_0^2 (\mu_0\varepsilon_0)}{4\mu_0} = \frac{1}{4}\varepsilon_0 E_0^2 = \langle u_E \rangle$$ Hence proved.
04 / Rapid Reference

Chapter Summary & Points to Ponder

6 core summary principles, Maxwell's Equations table, complete EM spectrum reference, and official NCERT Points to Ponder for Chapter 8.

100% SYLLABUS

Displacement Current

i_d = ε₀ (dΦE / dt)

Arises from time-varying electric flux; acts as a source of magnetic field identical to conduction current.

Ampere-Maxwell Law

∮ B · dl = μ₀ i_c + μ₀ε₀ (dΦE/dt)

Resolves inconsistency in Ampere's law, establishing complete current continuity across capacitors.

Transverse Character

E ⊥ B ⊥ k • E₀ / B₀ = c

Self-sustaining sinusoidal oscillations of electric and magnetic fields perpendicular to wave propagation.

Speed of Light in Vacuum & Medium

c = 1/√(μ₀ε₀) • v = 1/√(με)

Fundamental universal constant $c \approx 3 \times 10^8\text{ m s}^{-1}$; medium speed depends on $\mu_r$ and $\varepsilon_r$.

Equipartition of Energy

⟨u_E⟩ = ⟨u_B⟩ = ¼ ε₀ E₀²

Total average energy density $u = \frac{1}{2}\varepsilon_0 E_0^2 = \frac{B_0^2}{2\mu_0}$ is equally split between $E$ and $B$ fields.

EM Spectrum Order (Increasing ν)

Radio < Micro < IR < Visible < UV < X-ray < γ

Wavelength ranges from $>0.1\text{ m}$ (Radio) down to $<10^{-12}\text{ m}$ (Gamma rays).

NCERT Official

Points to Ponder

  1. Same Speed in Vacuum: All electromagnetic waves travel through vacuum with the exact same speed $c = 3 \times 10^8\text{ m s}^{-1}$. The differences in their interaction with matter arise purely from differences in frequency $\nu$ and wavelength $\lambda$.
  2. Source Scale and Wavelength: The radiating source size often correlates with the radiated wavelength: gamma rays ($\sim 10^{-14}\text{ m}$) originate from atomic nuclei, X-rays from inner electron shells of heavy atoms, and radio waves from macroscopic electrical circuits and antennas.
  3. Why Infrared Waves are "Heat Waves": Infrared frequencies resonate with vibrational modes of whole atoms and molecules (such as $\text{H}_2\text{O}$ and $\text{CO}_2$), increasing internal thermal energy and temperature upon absorption.
  4. Human Vision Evolution: The peak sensitivity of the human eye ($\sim 550\text{ nm}$) precisely matches the peak wavelength distribution emitted by the Sun, reflecting evolutionary adaptation to solar radiation.
05 / Practice Tests

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