Maxwell's Equations & EM Waves: Master Notes

Chapter Overview

Maxwell unified electricity, magnetism, and optics into a single foundational theory. In IIT-JEE (Advanced), top problems focus on displacement current in charging capacitors ($I_d={\epsilon }_0{\dot{\mathrm{\Phi }}}_E$), Poynting energy flux ($\overrightarrow{S}=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$), EM wave propagation parameters ($E_0/B_0=c$), and radiation pressure on matter.

---

1. The Four Maxwell Equations

Law	Differential Form	Physical Meaning
Gauss's Law (Electrostatics)	$\mathrm{\nabla }\cdot \overrightarrow{E}=\frac{\rho }{{\epsilon }_0}$	Electric charges are sources/sinks of $\overrightarrow{E}$
Gauss's Law (Magnetism)	$\mathrm{\nabla }\cdot \overrightarrow{B}=0$	Magnetic monopoles do not exist
Faraday-Lenz Law	$\mathrm{\nabla }\times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$	Time-varying $\overrightarrow{B}$ creates circular $\overrightarrow{E}$
Ampère-Maxwell Law	$\mathrm{\nabla }\times \overrightarrow{B}={\mu }_0(\overrightarrow{J}+{\epsilon }_0\frac{\partial \overrightarrow{E}}{\partial t})$	Conduction current + Displacement current create $\overrightarrow{B}$

---

2. Displacement Current ($I_d$) Inside a Charging Capacitor

During the charging of a circular parallel plate capacitor of radius $R$:

Multi-Mode DiagramMaxwell Equations & Displacement Current

Option 1: Publication-Grade Scientific Vector SVG

Displacement current density J_d = epsilon_0 * (dE/dt) inside capacitor gap matching conduction current I_d = I_c.

Ampere-Maxwell Law: $\oint \vec{B}\cdot d\vec{l} = \mu_0\left(I_c + \varepsilon_0 \frac{d\Phi_E}{dt}\right)$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate induced magnetic field ring from charging displacement current.

Charging rate dV/dt: 5 V/μs

Displacement Current ($I_d$):4.40 mA

Option 3: Interactive 3D Spatial Viewport

3D electric and magnetic field loops in capacitor.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Displacement Current:$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$

Option 5: Reactive Parametric Physics with Live Calculus

Magnetic field calculator inside capacitor gap.

Radial distance r: 5 cm

: 45

Field B(r) ($B(r)$):$2.00 μT$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

• Total Displacement Current across plates:$$I_d={\epsilon }_0\frac{d{\mathrm{\Phi }}_E}{dt}={\epsilon }_0\frac{d(E\cdot \pi R^2)}{dt}=I_{\text{conduction}}$$
• Magnetic Field inside the plates at distance $r\le R$ from axis:$$\oint \overrightarrow{B}\cdot d\overrightarrow{\ell }={\mu }_0{I}_{d,\text{enc}}⟹B(2\pi r)={\mu }_0\left(I\frac{\pi r^2}{\pi R^2}\right)$$$$\boxed{{\displaystyle B(r)=\frac{{\mu }_{0}Ir}{2\pi R^2}(r\le R)}}$$

---

3. Electromagnetic Wave Properties & Energetics

$$\overrightarrow{E}(x,t)=E_0\sin (kx-\omega t)\hat{j},\overrightarrow{B}(x,t)=B_0\sin (kx-\omega t)\hat{k}$$

• Propagation Velocity: $c=\frac{\omega }{k}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\approx 3\times {10}^8\text{ m/s}$
• Field Amplitude Ratio: $\boxed{{\displaystyle \frac{E_0}{B_0}=\frac{E_{\text{rms}}}{B_{\text{rms}}}=c}}$
• Poynting Vector (Energy Flux Density):$$\overrightarrow{S}=\frac{1}{{\mu }_0}(\overrightarrow{E}\times \overrightarrow{B})⟹\text{Intensity }I=⟨S⟩=\frac{1}{2}c{\epsilon }_0{E}_0^2=c{u}_{\text{total}}$$
• Energy Equipartition: The total energy is shared equally between electric and magnetic fields:$$u_E=\frac{1}{2}{\epsilon }_0{E}_{\text{rms}}^2=u_B=\frac{B_{\text{rms}}^2}{2{\mu }_0}=\frac{1}{2}{u}_{\text{total}}$$

---

3.1 Radiation Pressure ($P_{\text{rad}}$)

• Perfect Absorbing Surface: $P_{\text{rad}}=\frac{I}{c},F_{\text{rad}}=\frac{P_{\text{power}}}{c}$
• Perfect Reflecting Surface: $P_{\text{rad}}=\frac{2I}{c},F_{\text{rad}}=\frac{2P_{\text{power}}}{c}$
• Oblique Incidence at Angle $\theta$ to Normal:$$P_{\text{rad, absorbing}}=\frac{I{\cos }^2\theta }{c},P_{\text{rad, reflecting}}=\frac{2I{\cos }^2\theta }{c}$$

---

4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 (Paper 1) — Magnetic Field inside Charging Capacitor

Question:
A parallel plate circular capacitor of radius $R=10\text{ cm}$ is charged by a constant current $I=2.0\text{ A}$. Find the magnetic field $B$ between the plates at a radial distance $r=5\text{ cm}$ from the central axis.

Step-by-Step Solution:

Using the Ampère-Maxwell law for a loop of radius $r<R$:

$$B(r)=\frac{{\mu }_{0}Ir}{2\pi R^2}=\left(\frac{{\mu }_0}{2\pi }\right)\frac{Ir}{R^2}$$

Given $\frac{{\mu }_0}{4\pi }={10}^{-7}⟹\frac{{\mu }_0}{2\pi }=2\times {10}^{-7}\text{ T}\cdot \text{m/A}$:

$$B=(2\times {10}^{-7})\times \frac{2.0\times 0.05}{(0.10{)}^2}=(2\times {10}^{-7})\times \frac{0.10}{0.01}=2\times {10}^{-7}\times 10=\boxed{{\displaystyle 2.0\times {10}^{-6}\text{ T}=2.0\mu \text{T}}}$$

---

PYQ 2: JEE Advanced 2020 — Solar Radiation Pressure on Dust Grain

Question:
A perfectly absorbing spherical dust grain of radius $r$ and density $\rho =3000{\text{ kg/m}}^3$ is at distance $R$ from the Sun. The Sun's total radiant power is $P_{\text{sun}}=3.8\times {10}^{26}\text{ W}$ and mass is $M_{\text{sun}}=2.0\times {10}^{30}\text{ kg}$. Find the critical radius $r_{\text{crit}}$ of the grain such that radiation repulsion exactly balances solar gravitational attraction. ($G=6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$).

Step-by-Step Solution:

1. Gravitational Attraction:$$F_g=\frac{G{M}_{\text{sun}}{m}_{\text{grain}}}{R^2}=\frac{G{M}_{\text{sun}}\left(\frac{4}{3}\pi r^3\rho \right)}{R^2}$$
2. Radiation Repulsion Force:$$F_{\text{rad}}=\frac{\text{Power received}}{c}=\frac{\left(\frac{P_{\text{sun}}}{4\pi R^2}\right)(\pi r^2)}{c}=\frac{P_{\text{sun}}{r}^2}{4c{R}^2}$$
3. Balancing $F_{\text{rad}}=F_g$:$$\frac{P_{\text{sun}}{r}^2}{4c{R}^2}=\frac{G{M}_{\text{sun}}\left(\frac{4}{3}\pi r^3\rho \right)}{R^2}$$Canceling $r^2$ and $R^2$:$$\frac{P_{\text{sun}}}{4c}=\frac{4\pi }{3}G{M}_{\text{sun}}\rho r⟹\boxed{{\displaystyle r_{\text{crit}}=\frac{3P_{\text{sun}}}{16\pi cG{M}_{\text{sun}}\rho }}}$$(Any grain with $r<r_{\text{crit}}$ is blown out of the solar system by light pressure!)

---

5. High-Yield Formula Sheet

Entity	Formula	Notes
Displacement Current	$I_d={\epsilon }_0\frac{d{\mathrm{\Phi }}_E}{dt}$	Between plates
Wave Speed	$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=3\times {10}^8\text{ m/s}$	Vacuum speed
Field Ratio	$E_0=c{B}_0$	Transverse fields
Poynting Intensity	$I=\frac{1}{2}c{\epsilon }_0{E}_0^2$	Energy flux
Reflecting Pressure	$P_{\text{rad}}=\frac{2I}{c}$	Normal incidence