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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 05 (Electromagnetic Waves)

Chapter 08: Electromagnetic Waves

Master Revision Dossier • Displacement Current, Maxwell's Equations, Wave Propagation, Energy Dynamics & Complete Spectrum

1. Definitive Formula & Physical Quantities Matrix

Concept Mathematical Formulation SI Unit / Dim Boundary Conditions & Notes
Displacement Current $$I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 A \frac{dE}{dt}$$ $\text{Ampere (A)}$
$[\text{A}]$		 Exists inside capacitor gap during charging/discharging. Equal in magnitude to conduction current ($I_c \equiv I_d$). Zero when fully charged DC.
Ampere-Maxwell Law $$\oint \vec{B}\cdot d\vec{l} = \mu_0 (I_c + I_d) = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$ $\text{Tesla}\cdot\text{m}$ Resolves inconsistency of Ampere's circuital law. Establishes that time-varying electric flux generates a circulating magnetic field.
Wave Speed in Vacuum $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = \frac{E_0}{B_0} = \frac{\omega}{k}$$ $\text{m/s}$
$[\text{L}\text{T}^{-1}]$		 $c \approx 3.0 \times 10^8\text{ m/s}$. In material medium: $v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}} = \frac{c}{n}$. Independent of frequency.
Field Amplitudes Ratio $$\frac{E_0}{B_0} = \frac{E_{\text{rms}}}{B_{\text{rms}}} = c$$ $\text{m/s}$ $E_0 = c B_0$. Electric field amplitude is numerical factor of $3 \times 10^8$ larger than magnetic field amplitude in SI units.
Average Energy Density $$u_{\text{avg}} = \frac{1}{2}\varepsilon_0 E_0^2 = \varepsilon_0 E_{\text{rms}}^2 = \frac{B_0^2}{2\mu_0} = \frac{B_{\text{rms}}^2}{\mu_0}$$ $\text{J/m}^3$
$[\text{M}^1\text{L}^{-1}\text{T}^{-2}]$		 Exact 50%-50% equal energy distribution between electric and magnetic vectors: $\langle u_E \rangle \equiv \langle u_B \rangle = \frac{1}{4}\varepsilon_0 E_0^2$.
Poynting Vector & Intensity $$\vec{S} = \frac{1}{\mu_0}(\vec{E}\times\vec{B}), \quad I = \langle S \rangle = u_{\text{avg}} c = \frac{1}{2}\varepsilon_0 c E_0^2$$ $\text{W/m}^2$
$[\text{M}^1\text{T}^{-3}]$		 Direction of $\vec{S}$ gives direction of energy propagation ($\hat{c} = \hat{E}\times\hat{B}$). Time average of $|\vec{S}|$ represents intensity $I$.
Radiation Momentum & Force $$p = \frac{U}{c} \text{ (Abs)}, \quad p = \frac{2U}{c} \text{ (Ref)}$$ $\text{kg}\cdot\text{m/s}$ Radiation force $F = \frac{IA}{c}$ (Complete absorption); $F = \frac{2IA}{c}$ (Complete reflection / perfect mirror).
Radiation Pressure $$P_{\text{rad}} = \frac{I}{c} \text{ (Abs)}, \quad P_{\text{rad}} = \frac{2I}{c} \text{ (Ref)}$$ $\text{N/m}^2 \text{ (Pa)}$ Mechanical pressure exerted by photons colliding on exposed surface area. Twice as large on reflecting surfaces.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 8.1: Deduction of Displacement Current & Consistency of Ampere's Law

1. Consider a parallel plate capacitor of plate area $A$ charging with conduction current $I_c = \frac{dq}{dt}$.

2. For a flat surface $S_1$ spanning an Amperian loop $C$ outside the plates: $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_c$. For a pot-shaped surface $S_2$ passing through the capacitor gap: $\oint \vec{B}\cdot d\vec{l} = 0$ (no physical wire crosses $S_2$). This is Ampere's inconsistency.

3. Electric field between capacitor plates at any instant: $E = \frac{\sigma}{\varepsilon_0} = \frac{q}{\varepsilon_0 A}$. Electric flux through surface $S_2$:

$$\Phi_E = E A = \left(\frac{q}{\varepsilon_0 A}\right) A = \frac{q}{\varepsilon_0} \implies q = \varepsilon_0 \Phi_E$$

4. Differentiating with respect to time $t$:

$$I_c = \frac{dq}{dt} = \varepsilon_0 \frac{d\Phi_E}{dt} = I_d \implies \oint \vec{B}\cdot d\vec{l} = \mu_0 (I_c + I_d)$$

Blueprint 8.2: Electromagnetic Wave Speed and Amplitude Ratio ($E_0 = c B_0$)

1. In vacuum ($\rho = 0, I_c = 0$), Maxwell's equations relating spatial and temporal variations of plane wave ($E_y, B_z$ propagating along $+x$):

$$\frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{\partial t} \quad \text{and} \quad -\frac{\partial B_z}{\partial x} = \mu_0 \varepsilon_0 \frac{\partial E_y}{\partial t}$$

2. Differentiating and combining gives the wave equation: $\frac{\partial^2 E_y}{\partial x^2} = \mu_0 \varepsilon_0 \frac{\partial^2 E_y}{\partial t^2}$. Comparing with standard wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2 y}{\partial t^2}$:

$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3.0 \times 10^8\text{ m/s}$$

3. Substituting $E_y = E_0\sin(kx - \omega t)$ and $B_z = B_0\sin(kx - \omega t)$: $k E_0 \cos(kx - \omega t) = \omega B_0 \cos(kx - \omega t)$:

$$\frac{E_0}{B_0} = \frac{\omega}{k} = c \implies E_0 = c B_0 \quad \text{and} \quad E_{\text{rms}} = c B_{\text{rms}}$$

Blueprint 8.3: Proof of Equal Energy Distribution ($\langle u_E \rangle \equiv \langle u_B \rangle$)

1. Time-averaged electric energy density: $\langle u_E \rangle = \frac{1}{2}\varepsilon_0 \langle E^2 \rangle = \frac{1}{2}\varepsilon_0 \left(\frac{E_0^2}{2}\right) = \frac{1}{4}\varepsilon_0 E_0^2$.

2. Time-averaged magnetic energy density: $\langle u_B \rangle = \frac{\langle B^2 \rangle}{2\mu_0} = \frac{B_0^2}{4\mu_0}$.

3. Substitute $B_0 = E_0/c$ and $c^2 = \frac{1}{\mu_0\varepsilon_0}$ into the magnetic energy density:

$$\langle u_B \rangle = \frac{(E_0/c)^2}{4\mu_0} = \frac{E_0^2}{4\mu_0 c^2} = \frac{E_0^2}{4\mu_0 \left(\frac{1}{\mu_0\varepsilon_0}\right)} = \frac{1}{4}\varepsilon_0 E_0^2 \equiv \langle u_E \rangle \implies u_{\text{total}} = \frac{1}{2}\varepsilon_0 E_0^2 = \varepsilon_0 E_{\text{rms}}^2$$

3. Complete Electromagnetic Spectrum (Wavelength, Frequency, Production, Detection & 3 Uses)

Wave Type Wavelength Range Frequency Range Production Mechanism Detection Method 3 High-Yield Uses
1. Radio Waves > $0.1\text{ m}$
($10^3\text{m} - 0.1\text{m}$)		 $10^4 - 10^9\text{ Hz}$
($500\text{kHz} - 1000\text{MHz}$)		 Accelerated charges in conducting wires; LC tuned oscillators. Receiver aerials, tuned LC antenna circuits. 1. AM/FM radio broadcasting.
2. Television transmission (VHF/UHF).
3. Cellular mobile phone communication.
2. Microwaves $0.1\text{ m} - 1.0\text{ mm}$
($10^{-1} - 10^{-3}\text{ m}$)		 $1\text{ GHz} - 300\text{ GHz}$
($10^9 - 3\times 10^{11}\text{ Hz}$)		 Special vacuum tubes: Klystron, Magnetron, Gunn diodes. Point-contact crystal diodes, semiconductor detectors. 1. Aircraft & defense RADAR navigation.
2. Microwave ovens (resonant water heating).
3. Satellite telecommunication & Wi-Fi.
3. Infrared (Heat) $1\text{ mm} - 700\text{ nm}$
($10^{-3} - 7\times 10^{-7}\text{ m}$)		 $3\times 10^{11} - 4\times 10^{14}\text{ Hz}$ Thermal vibration of molecules in hot bodies and atoms. Thermopiles, bolometers, infrared photographic film. 1. Remote controls for TV and appliances.
2. Night-vision military surveillance & haze photography.
3. Muscular pain physiotherapy lamps.
4. Visible Light $700\text{ nm} - 400\text{ nm}$
($7\times 10^{-7} - 4\times 10^{-7}\text{ m}$)		 $4\times 10^{14} - 7.5\times 10^{14}\text{ Hz}$ Transitions of valence electrons in excited incandescent atoms. Human retinas, photographic emulsion, photocells. 1. Human physiological ocular vision (VIBGYOR).
2. High-speed optical fiber communications.
3. Biological photosynthesis in green plants.
5. Ultraviolet (UV) $400\text{ nm} - 1\text{ nm}$
($4\times 10^{-7} - 10^{-9}\text{ m}$)		 $7.5\times 10^{14} - 10^{16}\text{ Hz}$ Very hot bodies (Sun), electric arcs, mercury vapor discharge lamps. Photocells, fluorescent screens, photographic paper. 1. Water purification (bactericidal action).
2. Detection of forged currency & signatures.
3. LASIK precision corneal eye surgery.
6. X-Rays $1\text{ nm} - 10^{-3}\text{ nm}$
($10^{-9} - 10^{-12}\text{ m}$)		 $10^{16} - 3\times 10^{19}\text{ Hz}$ Bombardment of heavy metal target (Tungsten) by fast electrons. Photographic film, GM counters, ionization chambers. 1. Diagnostic radiography (bone fractures & CT scans).
2. Crystallography & lattice structure analysis.
3. Airport luggage security scanners.
7. Gamma Rays ($\gamma$) < $10^{-3}\text{ nm}$
($< 10^{-12}\text{ m}$)		 > $3\times 10^{19}\text{ Hz}$
($10^{19} - 10^{23}\text{ Hz}$)		 Radioactive decay of atomic nuclei; nuclear de-excitation. Ionization chambers, scintillation counters. 1. Oncology radiotherapy (killing cancer cells).
2. Preservation of packaged food (sterilization).
3. Probing fundamental nuclear energy levels.

4. High-Yield "Red Alert" Exam Trap Matrix

⚠️ Trap 1: Displacement Current in Steady DC

In a fully charged capacitor connected to a DC battery, the electric field is constant ($d\Phi_E/dt = 0$).
Therefore, displacement current is strictly ZERO ($I_d = 0$). $I_d$ exists ONLY during charging or discharging when electric flux varies with time.

⚠️ Trap 2: X-Rays vs Gamma Rays Overlap Duality

An X-ray and a $\gamma$-ray can have identical wavelength ($\sim 10^{-11}\text{ m}$) and identical speed ($c$).
They are distinguished ONLY by their ORIGIN:
• X-rays originate from atomic electron transitions or sudden electron decelerations.
• Gamma rays originate exclusively from nuclear de-excitation inside radioactive nuclei.

⚠️ Trap 3: Vacuum Speed Invariance (1:1 Ratio)

*Board Trap:* "Compare the speed of a $100\text{ kHz}$ radio wave with a $10^{22}\text{ Hz}$ gamma ray in vacuum."
*Correct Answer:* RATIO IS EXACTLY 1:1! All EM waves travel with identical velocity $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3.0 \times 10^8\text{ m/s}$ in vacuum, independent of frequency, wavelength, or intensity.

⚠️ Trap 4: Equal Energy Partition (1:1 Ratio)

Although $E_0 = c B_0$ (meaning $E_0$ is $3 \times 10^8$ times larger numerically than $B_0$), energy is shared equally: $$\langle u_E \rangle = \frac{1}{4}\varepsilon_0 E_0^2 \equiv \langle u_B \rangle = \frac{B_0^2}{4\mu_0}$$ Exactly 50% of total wave energy is transported by the electric field, and 50% by the magnetic field.

5. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Capacitor Charging & Induced Magnetic Field Exemplar 01

A parallel plate capacitor with circular plates of radius $R = 12\text{ cm}$ is charged by a current of $0.15\text{ A}$. Calculate (i) displacement current, (ii) rate of change of electric flux, and (iii) magnetic field induced at $r = 6.0\text{ cm}$ from the central axis.

Solution:

1. $I_d = I_c = 0.15\text{ A}$.

2. $\frac{d\Phi_E}{dt} = \frac{I_d}{\varepsilon_0} = \frac{0.15}{8.854 \times 10^{-12}} \approx 1.69 \times 10^{10}\text{ V}\cdot\text{m/s}$.

3. At $r = 6.0\text{ cm} = 0.06\text{ m}$ ($r \le R = 0.12\text{ m}$):

$$B = \frac{\mu_0 I_d r}{2\pi R^2} = \frac{(4\pi \times 10^{-7}) \times 0.15 \times 0.06}{2\pi \times (0.12)^2} = \frac{2 \times 10^{-7} \times 0.009}{0.0144} = 1.25 \times 10^{-7}\text{ Tesla} = 0.125\text{ }\mu\text{T}$$
Problem 2: Wave Field Amplitudes, Intensity & Radiation Force Exemplar 02

A plane electromagnetic wave has electric field amplitude $E_0 = 120\text{ V/m}$ and frequency $50\text{ MHz}$. (i) Find $B_0, \omega$, and $k$. (ii) Calculate wave intensity $I$. (iii) Determine radiation force on a completely absorbing surface of area $1.5\text{ m}^2$.

Solution:

$$B_0 = \frac{E_0}{c} = \frac{120}{3 \times 10^8} = 4.0 \times 10^{-7}\text{ T} = 0.40\text{ }\mu\text{T}$$ $$\omega = 2\pi f = 2\pi(50 \times 10^6) = 100\pi \times 10^6 \approx 3.14 \times 10^8\text{ rad/s}$$ $$k = \frac{\omega}{c} = \frac{3.14 \times 10^8}{3 \times 10^8} \approx 1.05\text{ rad/m}$$

Intensity and radiation force:

$$I = \frac{1}{2}\varepsilon_0 c E_0^2 = \frac{1}{2}(8.854 \times 10^{-12})(3 \times 10^8)(120)^2 \approx 19.12\text{ W/m}^2$$ $$F = \frac{I A}{c} = \frac{19.12 \times 1.5}{3 \times 10^8} \approx 9.56 \times 10^{-8}\text{ Newtons}$$

6. Printable Pocket Flashcard Grid

Displacement Current

$I_d = \varepsilon_0\frac{d\Phi_E}{dt}$

Wave Speed

$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = \frac{E_0}{B_0}$

Energy Partition

$u_E = u_B = \frac{1}{4}\varepsilon_0 E_0^2$

Radiation Pressure

$P = \frac{I}{c}$ (Abs), $\frac{2I}{c}$ (Ref)