Axiom Physics | Class 12 - Characteristics & Propagation of EM Waves

⚡ 01. The Spark (Analogy)

Socho do acrobats ko jo hawa mein ek dusre ko bari-bari se toss kar rahe hain bina zameen ko chuye: pehla acrobat doosre ko hawa mein uchhalta hai, aur doosra aage badhkar pehle ko aage fenkta hai. Electromagnetic Wave theek yahi self-sustaining leapfrog process hai! Ek oscillating electric field ek magnetic field generate karta hai (Ampere-Maxwell Law), aur woh badalta hua magnetic field turant agle point par ek naya electric field paida kar deta hai (Faraday's Law). Dono fields ek dusre ko generate karte hue vacuum ke sannate mein $3 \times 10^8\text{ m/s}$ ki cosmic speed se aage badhte hain—inhe travel karne ke liye kisi physical material medium ya hawa ki zarurat nahi hoti!

📐 02. The Core Law & Formulae

1. Harmonic Wave Equations & Velocity

For wave propagating along the $+x$ axis:

$$E_y(x,t) = E_0 \sin(kx - \omega t), \quad B_z(x,t) = B_0 \sin(kx - \omega t)$$ $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = \frac{E_0}{B_0} = \frac{\omega}{k} \approx 3.0 \times 10^8\text{ m/s}$$

In material medium: $v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{n}$ where $n = \sqrt{\mu_r \varepsilon_r}$ is refractive index.

2. Energy Density, Momentum & Pressure

Equal energy partition between electric and magnetic vectors:

$$u_{\text{avg}} = \frac{1}{2}\varepsilon_0 E_{\text{rms}}^2 + \frac{B_{\text{rms}}^2}{2\mu_0} = \varepsilon_0 E_{\text{rms}}^2 = \frac{1}{2}\varepsilon_0 E_0^2$$ $$I = u_{\text{avg}} c = \frac{1}{2}\varepsilon_0 c E_0^2, \quad p = \frac{U}{c}$$

Radiation Pressure on surface: $P_{\text{rad}} = \frac{I}{c}$ (Complete absorption); $P_{\text{rad}} = \frac{2I}{c}$ (Complete reflection).

03. Interactive Visualizer: 3D Orthogonal Field Wave Simulator ($\vec{E} \perp \vec{B} \perp \vec{c}$)

Peak Electric Field ($E_0$) 300 V/m

Propagation Speed / Frequency 1.0x Wave Velocity

Peak Magnetic Field ($B_0 = E_0/c$) 1.00 μT

B₀ = E₀ / (3.0 × 10⁸ m/s)

Poynting Flux Intensity ($I$):

119.4 W/m²

Total Energy Density ($u_{\text{avg}}$):

3.98 × 10⁻⁷ J/m³

Radiation Pressure (Absorbing):

3.98 × 10⁻⁷ N/m²

Electric vs Magnetic Ratio:

u_E = u_B (50% / 50%)

📝 04. Comprehensive Stepwise Derivations Blueprint

Primary Board Derivation

Mathematical Relation Between Amplitudes ($E_0 = c B_0$) & Speed of Light ($c = 1/\sqrt{\mu_0\varepsilon_0}$)

Fig 8.2A: Mutual spatial orthogonality of Electric field $\vec{E}$ ($y$-axis), Magnetic field $\vec{B}$ ($z$-axis), and propagation direction $\vec{c}$ ($x$-axis).

1. In free space (vacuum), charge density $\rho = 0$ and conduction current $I_c = 0$. Maxwell's third and fourth equations in differential/wave format relate the rate of change of space and time:

$$\frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{\partial t} \quad \text{--- [Eq. 1 from Faraday's Law]}$$ $$-\frac{\partial B_z}{\partial x} = \mu_0 \varepsilon_0 \frac{\partial E_y}{\partial t} \quad \text{--- [Eq. 2 from Ampere-Maxwell Law]}$$

2. Differentiating Eq. 1 with respect to $x$ and substituting into Eq. 2 yields the classical three-dimensional wave equation:

$$\frac{\partial^2 E_y}{\partial x^2} = \mu_0 \varepsilon_0 \frac{\partial^2 E_y}{\partial t^2}$$

3. Comparing with the standard mechanical wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}$, the velocity of propagation in vacuum is:

$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = \frac{1}{\sqrt{(4\pi \times 10^{-7}) \times (8.854 \times 10^{-12})}} \approx 2.998 \times 10^8\text{ m/s}$$

4. Substituting trial harmonic solutions $E_y = E_0\sin(kx - \omega t)$ and $B_z = B_0\sin(kx - \omega t)$ into Eq. 1:

$$k E_0 \cos(kx - \omega t) = \omega B_0 \cos(kx - \omega t) \implies \frac{E_0}{B_0} = \frac{\omega}{k} = c$$

$$E_0 = c B_0 \quad \text{or} \quad E_{\text{rms}} = c B_{\text{rms}}$$

Derivation B

Proof of Equal Energy Distribution between Electric and Magnetic Fields

1. The instantaneous electrostatic energy density is $u_E = \frac{1}{2}\varepsilon_0 E^2$ and magnetic energy density is $u_B = \frac{B^2}{2\mu_0}$.

2. The average electric energy density over a full harmonic cycle is:

$$\langle u_E \rangle = \frac{1}{2}\varepsilon_0 \langle E_0^2 \sin^2(kx - \omega t) \rangle = \frac{1}{2}\varepsilon_0 E_0^2 \left(\frac{1}{2}\right) = \frac{1}{4}\varepsilon_0 E_0^2$$

3. The average magnetic energy density over a full cycle is:

$$\langle u_B \rangle = \frac{B_0^2}{4\mu_0}$$

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

$$\langle u_B \rangle = \frac{(E_0/c)^2}{4\mu_0} = \frac{E_0^2}{4\mu_0 c^2} = \frac{E_0^2}{4\mu_0 \left(\frac{1}{\mu_0\varepsilon_0}\right)} = \frac{1}{4}\varepsilon_0 E_0^2$$

$$\langle u_E \rangle \equiv \langle u_B \rangle = \frac{1}{4}\varepsilon_0 E_0^2 = \frac{1}{4}\frac{B_0^2}{\mu_0}$$

5. Total average electromagnetic energy density $u_{\text{avg}}$:

$$u_{\text{avg}} = \langle u_E \rangle + \langle u_B \rangle = \frac{1}{2}\varepsilon_0 E_0^2 = \varepsilon_0 E_{\text{rms}}^2 = \frac{B_0^2}{2\mu_0} = \frac{B_{\text{rms}}^2}{\mu_0}$$

Core Fact: Although the numerical value of $E_0$ is huge compared to $B_0$ ($E_0 = 3\times 10^8 B_0$), exactly 50% of total wave energy is transported by the electric field, and the remaining 50% by the magnetic field.

Derivation C

Poynting Vector $\vec{S}$, Intensity & Momentum Transfer

1. The energy flowing per unit area per second normal to propagation direction is defined by the Poynting Vector ($\vec{S}$):

$$\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) \quad \left(\text{Unit: }\frac{\text{Watt}}{\text{m}^2}\right)$$

2. Time-averaged value of $|\vec{S}|$ represents Wave Intensity ($I$):

$$I = \langle S \rangle = u_{\text{avg}} c = \left(\frac{1}{2}\varepsilon_0 E_0^2\right) c = \frac{E_0 B_0}{2\mu_0}$$

3. If electromagnetic radiation carrying total energy $U$ is incident on a surface, it delivers linear momentum $p$:

$$p = \frac{U}{c} \quad (\text{For complete absorption}), \qquad p = \frac{2U}{c} \quad (\text{For complete reflection})$$

4. Consequently, electromagnetic waves exert mechanical Radiation Pressure ($P_{\text{rad}}$) on surfaces:

$$P_{\text{rad}} = \frac{I}{c} \quad (\text{Perfect Absorber}), \qquad P_{\text{rad}} = \frac{2I}{c} \quad (\text{Perfect Reflector / Mirror})$$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: The Source of Electromagnetic Waves

• Stationary Charge ($v = 0$): Produces electrostatic field only ($\vec{E}$).
• Uniformly Moving Charge ($v = \text{const}$): Produces both static $\vec{E}$ and steady $\vec{B}$, but radiates NO electromagnetic waves!
• Accelerated / Oscillating Charge ($a \ne 0$): RADIATES electromagnetic waves continuously! An LC circuit produces EM waves because electrons undergo continuous oscillatory acceleration.

Trap 2: "B-Field is Weak, So Electric Field Carries All Energy" Myth

*Exam Trap Question:* Since $B_0 = E_0/c \approx 10^{-8} E_0$, does the electric field carry $99.99\%$ of wave energy?
*Correct Answer:* NO! Both carry exactly 50% energy! Energy density formula contains $\mu_0$ in denominator ($\mu_0 \approx 10^{-6}$): $$u_E = \frac{1}{4}\varepsilon_0 E_0^2 \equiv u_B = \frac{B_0^2}{4\mu_0}$$ Energy sharing is strictly 1:1!

Trap 3: Cross-Product Direction of Wave Propagation

The unit vector along the propagation of an electromagnetic wave is given strictly by: $$\hat{c} = \hat{E} \times \hat{B} \quad (\text{NOT } \hat{B} \times \hat{E}!)$$ *If $\vec{E}$ points along $+\hat{j}$ and $\vec{B}$ points along $+\hat{k}$*: $\vec{c}$ travels along $\hat{j}\times\hat{k} = \mathbf{+\hat{i}}$ (along $+x$ axis).

Trap 4: Why Optical Effects are Dominated by Electric Vector

Lorentz force on an electron in an atom: $F_E = e E$ and $F_B = e v B = e v (E/c) = (v/c) F_E$. Because typical orbital electron speeds in materials are $v \ll c$, magnetic force is negligible compared to electric force ($F_B \approx 10^{-4} F_E$). Therefore, the Electric Vector $\vec{E}$ is called the Light Vector responsible for all optical vision, photographic, and polarization effects.

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: 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 $\vec{E}$ field equals the average energy density of the $\vec{B}$ field.

Solution:

1. Frequency $f = 2.0 \times 10^{10}\text{ Hz}$, $c = 3.0 \times 10^8\text{ m/s}$. Wavelength:

$$\lambda = \frac{c}{f} = \frac{3.0 \times 10^8\text{ m/s}}{2.0 \times 10^{10}\text{ Hz}} = 1.5 \times 10^{-2}\text{ m} = 1.5\text{ cm} \quad (\text{Microwaves})$$

2. Magnetic field amplitude ($B_0$):

$$B_0 = \frac{E_0}{c} = \frac{48\text{ V/m}}{3.0 \times 10^8\text{ m/s}} = 1.6 \times 10^{-7}\text{ Tesla} = 0.16\text{ }\mu\text{T}$$

3. Energy densities:

$$\langle u_E \rangle = \frac{1}{4}\varepsilon_0 E_0^2 = \frac{1}{4} \times (8.854 \times 10^{-12}) \times (48)^2 \approx 5.10 \times 10^{-9}\text{ J/m}^3$$ $$\langle u_B \rangle = \frac{B_0^2}{4\mu_0} = \frac{(1.6 \times 10^{-7})^2}{4 \times (4\pi \times 10^{-7})} = \frac{2.56 \times 10^{-14}}{5.026 \times 10^{-6}} \approx 5.10 \times 10^{-9}\text{ J/m}^3$$

$$\langle u_E \rangle \equiv \langle u_B \rangle \approx 5.10 \times 10^{-9}\text{ J/m}^3 \quad (\text{Equal partition proved})$$

JEE/NEET Competitive Trap

Problem: Light with an energy flux of $18\text{ W/cm}^2$ falls on a non-reflecting completely absorbing surface at normal incidence. If the surface has an area of $20\text{ cm}^2$, find the total momentum delivered to the surface in $30\text{ minutes}$ and determine the average radiation force exerted on it. What would be the force if the surface were a perfect mirror?

Solution:

1. Convert intensity and area into SI units:

$$I = 18\text{ W/cm}^2 = 18 \times 10^4\text{ W/m}^2 = 1.8 \times 10^5\text{ W/m}^2$$ $$A = 20\text{ cm}^2 = 20 \times 10^{-4}\text{ m}^2 = 2.0 \times 10^{-3}\text{ m}^2$$

Total energy incident in $t = 30\text{ min} = 1800\text{ s}$:

$$U = I \cdot A \cdot t = (1.8 \times 10^5) \times (2.0 \times 10^{-3}) \times 1800 = 360 \times 1800 = 6.48 \times 10^5\text{ Joules}$$

2. Momentum delivered to absorbing surface ($p = U/c$):

$$p = \frac{U}{c} = \frac{6.48 \times 10^5\text{ J}}{3.0 \times 10^8\text{ m/s}} = 2.16 \times 10^{-3}\text{ kg}\cdot\text{m/s}$$

3. Average radiation force exerted on absorbing surface:

$$F_{\text{abs}} = \frac{p}{t} = \frac{2.16 \times 10^{-3}\text{ N}\cdot\text{s}}{1800\text{ s}} = 1.20 \times 10^{-6}\text{ N} \quad \left(\text{or } F = \frac{I A}{c} = \frac{360}{3\times 10^8} = 1.2\times 10^{-6}\text{ N}\right)$$

4. If the surface were a perfectly reflecting mirror, momentum change doubles ($\Delta p = 2U/c$):

$$F_{\text{ref}} = 2 \times F_{\text{abs}} = 2 \times (1.20 \times 10^{-6}\text{ N}) = 2.40 \times 10^{-6}\text{ N}$$

🗂️ 07. Quick Revision Flashcard Core Formulae

Wave Speed

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

Energy Density

$u = \frac{1}{2}\varepsilon_0 E_0^2$ ($50\% E, 50\% B$)

Poynting Vector

$\vec{S} = \frac{1}{\mu_0}(\vec{E}\times\vec{B})$

Radiation Pressure

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

Characteristics & Propagation of Electromagnetic Waves