Chapter 33: Electromagnetic Waves

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Electromagnetic waves, such as radio waves used in spacecraft communication, are a direct consequence of Maxwell’s equations. Building on the concepts from Chapters 31–32, this chapter explores the nature and applications of electromagnetic waves. For JEE Main, JEE Advanced, and NEET students, mastering electromagnetic waves is essential, as they frequently appear in problems involving wave properties, energy transport, and the electromagnetic spectrum. This chapter, Electromagnetic Waves, covers generation of electromagnetic waves, wave propagation and properties, electromagnetic spectrum, and energy transport and Poynting vector, providing detailed explanations, derivations, solved examples, and practical applications to ensure conceptual clarity and problem-solving proficiency.

33.1 Generation of Electromagnetic Waves

Electromagnetic waves are generated by accelerating charges, a key concept for JEE/NEET.

Maxwell’s Prediction

Maxwell’s equations predict that a changing electric field produces a changing magnetic field, and vice versa, leading to self-sustaining electromagnetic waves. A time-varying electric field $\overrightarrow{E}$ induces a magnetic field $\overrightarrow{B}$ (Faraday’s law), and a time-varying $\overrightarrow{B}$ induces an $\overrightarrow{E}$ (Ampere-Maxwell law).

Accelerating Charges

An accelerating charge produces a changing electric field, which in turn generates a changing magnetic field, forming an electromagnetic wave that propagates outward at the speed of light $c$.

Wave Equation

Maxwell’s equations in free space lead to the wave equation for $\overrightarrow{E}$ and $\overrightarrow{B}$:

$${\mathrm{\nabla }}^2\overrightarrow{E}={\mu }_0{ϵ}_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2},{\mathrm{\nabla }}^2\overrightarrow{B}={\mu }_0{ϵ}_0\frac{{\partial }^2\overrightarrow{B}}{\partial t^2}$$

• The wave speed $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}\approx 3\times {10}^8\text{m/s}$.

Derivation: Wave Equation for Electromagnetic Waves
Start with Maxwell’s equations in free space (no charges or currents):

• Faraday’s law: $\mathrm{\nabla }\times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$
• Ampere-Maxwell law: $\mathrm{\nabla }\times \overrightarrow{B}={\mu }_0{ϵ}_0\frac{\partial \overrightarrow{E}}{\partial t}$
• Gauss’s law for electricity: $\mathrm{\nabla }\cdot \overrightarrow{E}=0$
• Gauss’s law for magnetism: $\mathrm{\nabla }\cdot \overrightarrow{B}=0$
  Take the curl of Faraday’s law:

$$\mathrm{\nabla }\times (\mathrm{\nabla }\times \overrightarrow{E})=\mathrm{\nabla }\times (-\frac{\partial \overrightarrow{B}}{\partial t})⟹\mathrm{\nabla }(\mathrm{\nabla }\cdot \overrightarrow{E})-{\mathrm{\nabla }}^2\overrightarrow{E}=-\frac{\partial }{\partial t}(\mathrm{\nabla }\times \overrightarrow{B})$$

Since $\mathrm{\nabla }\cdot \overrightarrow{E}=0$, this simplifies to:

$$-{\mathrm{\nabla }}^2\overrightarrow{E}=-\frac{\partial }{\partial t}(\mathrm{\nabla }\times \overrightarrow{B})$$

Substitute $\mathrm{\nabla }\times \overrightarrow{B}={\mu }_0{ϵ}_0\frac{\partial \overrightarrow{E}}{\partial t}$ from the Ampere-Maxwell law:

$${\mathrm{\nabla }}^2\overrightarrow{E}={\mu }_0{ϵ}_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}$$

This is the wave equation for $\overrightarrow{E}$, with speed $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}$. A similar derivation for $\overrightarrow{B}$ yields the same result.

Derivation: Wave Speed in Rocket Communication
For a spacecraft using electromagnetic waves, $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}=\frac{1}{\sqrt{(4\pi \times {10}^{-7})\times (8.85\times {10}^{-12})}}\approx 3\times {10}^8\text{m/s}$, ensuring rapid signal transmission (your interest, April 19, 2025).

Solved Example: A JEE Main problem asks for the speed of electromagnetic waves in vacuum.

• Solution:
  $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}=\frac{1}{\sqrt{(4\pi \times {10}^{-7})\times (8.85\times {10}^{-12})}}\approx 3.00\times {10}^8\text{m/s}$.
  • JEE Tip: $c$ is the speed of light; use ${\mu }_0$ and ${ϵ}_0$ in SI units. Common error: Forgetting the square root.

Solved Example: A NEET problem involves an oscillating charge with frequency $f={10}^6\text{Hz}$. Find the wavelength of the emitted wave.

• Solution:
  Wavelength $\lambda =\frac{c}{f}=\frac{3\times {10}^8}{{10}^6}=300\text{m}$.
  • NEET Tip: Use $c=3\times {10}^8\text{m/s}$ for vacuum; $\lambda$ in meters. Common error: Using incorrect $c$.

Solved Example: A JEE Advanced problem asks for the frequency of an electromagnetic wave with $\lambda =500\text{nm}$.

• Solution:
  $f=\frac{c}{\lambda }=\frac{3\times {10}^8}{500\times {10}^{-9}}=6\times {10}^{14}\text{Hz}$.
  • JEE Tip: Convert $\lambda$ to meters; $f$ in Hz. Common error: Incorrect unit conversion.

Solved Example: A JEE Main problem involves a wave with $f=2\times {10}^9\text{Hz}$. Find $\lambda$.

• Solution:
  $\lambda =\frac{c}{f}=\frac{3\times {10}^8}{2\times {10}^9}=0.15\text{m}$.
  • JEE Tip: This is a microwave frequency; ensure consistent units. Common error: Forgetting to convert frequency.

Application: Electromagnetic waves are generated in antennas, lasers, and rocketry (e.g., spacecraft communication, aligning with your interest, April 19, 2025).

33.2 Wave Propagation and Properties

Electromagnetic waves have specific properties and propagation characteristics, critical for JEE/NEET.

Transverse Nature

Electromagnetic waves are transverse: $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular to each other and to the direction of propagation $\overrightarrow{k}$. For a wave traveling in the $z$-direction:

$$\overrightarrow{E}=E_0\sin (kz-\omega t)\hat{x},\overrightarrow{B}=B_0\sin (kz-\omega t)\hat{y}$$

• $E_0=c{B}_0$, where $c$ is the speed of light.

Speed and Polarization

• Speed in vacuum: $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}\approx 3\times {10}^8\text{m/s}$.
• Polarization: Direction of $\overrightarrow{E}$ defines the polarization (e.g., linear, circular).

Wave Number and Frequency

• Wave number $k=\frac{2\pi }{\lambda }$.
• Angular frequency $\omega =2\pi f$.
• Relation: $\omega =kc$.

Derivation: Relationship Between $\overrightarrow{E}$ and $\overrightarrow{B}$ in an Electromagnetic Wave
Consider a plane electromagnetic wave propagating in the $z$-direction: $\overrightarrow{E}=E_x(z,t)\hat{x}$, $\overrightarrow{B}=B_y(z,t)\hat{y}$. From Faraday’s law, $\mathrm{\nabla }\times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$:

$$(\mathrm{\nabla }\times \overrightarrow{E}{)}_y=-\frac{\partial E_x}{\partial z}=-\frac{\partial B_y}{\partial t}$$

From the Ampere-Maxwell law, $\mathrm{\nabla }\times \overrightarrow{B}={\mu }_0{ϵ}_0\frac{\partial \overrightarrow{E}}{\partial t}$:

$$(\mathrm{\nabla }\times \overrightarrow{B}{)}_x=\frac{\partial B_y}{\partial z}={\mu }_0{ϵ}_0\frac{\partial E_x}{\partial t}$$

Assume $E_x=E_0\sin (kz-\omega t)$, $B_y=B_0\sin (kz-\omega t)$. Then:

• $\frac{\partial E_x}{\partial z}=k{E}_0\cos (kz-\omega t)$, $\frac{\partial B_y}{\partial t}=-\omega B_0\cos (kz-\omega t)$, so $k{E}_0=\omega B_0$.
• $\frac{\partial B_y}{\partial z}=k{B}_0\cos (kz-\omega t)$, $\frac{\partial E_x}{\partial t}=-\omega E_0\cos (kz-\omega t)$, so $k{B}_0=-{\mu }_0{ϵ}_0(-\omega E_0)$, or $k{B}_0={\mu }_0{ϵ}_0\omega E_0$.
  From the first equation, $B_0=\frac{k}{\omega }{E}_0=\frac{E_0}{c}$ (since $\omega =kc$), so $E_0=c{B}_0$.

Derivation: Wave Propagation in Rocket Signal
A spacecraft signal ($E_0=100\text{V/m}$) has $B_0=\frac{E_0}{c}=\frac{100}{3\times {10}^8}\approx 3.33\times {10}^{-7}\text{T}$, ensuring proper wave propagation (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves a wave with $E_0=50\text{V/m}$. Find $B_0$.

• Solution:
  $B_0=\frac{E_0}{c}=\frac{50}{3\times {10}^8}\approx 1.67\times {10}^{-7}\text{T}$.
  • JEE Tip: $E_0$ and $B_0$ are related by $c$; units are V/m and T. Common error: Forgetting $c$.

Solved Example: A NEET problem involves a wave with $\lambda =600\text{nm}$. Find $k$.

• Solution:
  $k=\frac{2\pi }{\lambda }=\frac{2\pi }{600\times {10}^{-9}}\approx 1.05\times {10}^7{\text{m}}^{-1}$.
  • NEET Tip: $k$ is the wave number; $\lambda$ in meters. Common error: Using incorrect units.

Solved Example: A JEE Advanced problem involves a wave with $f=5\times {10}^{14}\text{Hz}$. Find $\omega$.

• Solution:
  $\omega =2\pi f=2\pi \times (5\times {10}^{14})\approx 3.14\times {10}^{15}\text{rad/s}$.
  • JEE Tip: $\omega$ in rad/s; $f$ in Hz. Common error: Forgetting $2\pi$.

Solved Example: A JEE Main problem involves a wave with $E_0=120\text{V/m}$. Find $B_0$.

• Solution:
  $B_0=\frac{120}{3\times {10}^8}=4\times {10}^{-7}\text{T}$.
  • JEE Tip: Ensure $E_0$ in V/m, $B_0$ in T; $c$ connects them. Common error: Incorrect $c$.

Application: Wave properties apply to optics, communication, and rocketry (e.g., spacecraft signal propagation, aligning with your interest, April 19, 2025).

33.3 Electromagnetic Spectrum

The electromagnetic spectrum categorizes waves by frequency, a key concept for JEE/NEET.

Spectrum Overview

Electromagnetic waves span a range of frequencies:

• Radio Waves: ${10}^4$–${10}^9\text{Hz}$ (communication).
• Microwaves: ${10}^9$–${10}^{12}\text{Hz}$ (radar, heating).
• Infrared: ${10}^{12}$–${10}^{14}\text{Hz}$ (thermal imaging).
• Visible Light: $4\times {10}^{14}$–$7.5\times {10}^{14}\text{Hz}$ (red to violet).
• Ultraviolet (UV): ${10}^{15}$–${10}^{16}\text{Hz}$ (sterilization).
• X-Rays: ${10}^{16}$–${10}^{19}\text{Hz}$ (medical imaging).
• Gamma Rays: $>{10}^{19}\text{Hz}$ (nuclear processes).

Wavelength and Frequency

• $\lambda f=c$, where $c=3\times {10}^8\text{m/s}$.
• Higher frequency means shorter wavelength.

Derivation: Wavelength Range for Visible Light
Visible light ranges from $f=4\times {10}^{14}\text{Hz}$ (red) to $7.5\times {10}^{14}\text{Hz}$ (violet). Using $\lambda =\frac{c}{f}$:

• Red: $\lambda =\frac{3\times {10}^8}{4\times {10}^{14}}=750\times {10}^{-9}\text{m}=750\text{nm}$.
• Violet: $\lambda =\frac{3\times {10}^8}{7.5\times {10}^{14}}=400\times {10}^{-9}\text{m}=400\text{nm}$.
  Thus, visible light spans 400–750 nm.

Derivation: Spectrum in Rocket Navigation
A spacecraft uses radio waves ($f={10}^8\text{Hz}$), so $\lambda =\frac{3\times {10}^8}{{10}^8}=3\text{m}$, ideal for long-range communication (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves a wave with $f={10}^{10}\text{Hz}$. Identify the type.

• Solution:
  $f={10}^{10}\text{Hz}$ falls in the microwave range (${10}^9$–${10}^{12}\text{Hz}$).
  • JEE Tip: Compare frequency to spectrum ranges; microwaves are ${10}^9$–${10}^{12}\text{Hz}$. Common error: Confusing with radio waves.

Solved Example: A NEET problem involves a wave with $\lambda =300\text{nm}$. Identify the type.

• Solution:
  $f=\frac{c}{\lambda }=\frac{3\times {10}^8}{300\times {10}^{-9}}={10}^{15}\text{Hz}$, which is in the UV range (${10}^{15}$–${10}^{16}\text{Hz}$).
  • NEET Tip: UV is just beyond visible light; calculate $f$ to confirm. Common error: Assuming visible light.

Solved Example: A JEE Advanced problem involves a gamma ray with $f={10}^{20}\text{Hz}$. Find $\lambda$.

• Solution:
  $\lambda =\frac{3\times {10}^8}{{10}^{20}}=3\times {10}^{-12}\text{m}$.
  • JEE Tip: Gamma rays have high frequency, short wavelength; units in m. Common error: Incorrect exponent.

Solved Example: A JEE Main problem involves a wave with $f=5\times {10}^{11}\text{Hz}$. Identify the type.

• Solution:
  $f=5\times {10}^{11}\text{Hz}$ is in the microwave range (${10}^9$–${10}^{12}\text{Hz}$).
  • JEE Tip: Microwaves are between radio and infrared; check frequency range. Common error: Misidentifying as infrared.

Application: The spectrum applies to communication, medical imaging, and rocketry (e.g., spacecraft radio communication, aligning with your interest, April 19, 2025).

33.4 Energy Transport and Poynting Vector

Electromagnetic waves carry energy, described by the Poynting vector, a key concept for JEE/NEET.

Energy Density

The energy density of an electromagnetic wave has electric and magnetic contributions:

$$u_E=\frac{1}{2}{ϵ}_0{E}^2,u_B=\frac{1}{2}\frac{B^2}{{\mu }_0},u=u_E+u_B$$

• For an electromagnetic wave, $u_E=u_B$, so $u={ϵ}_0{E}^2$.

Poynting Vector

The Poynting vector $\overrightarrow{S}$ represents the energy flux (power per unit area):

$$\overrightarrow{S}=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$$

• Units: W/m².
• Average intensity: $S_{\text{avg}}=\frac{1}{2}{ϵ}_{0}c{E}_0^2$.

Intensity

The intensity $I$ of a wave is the average power per unit area:

$$I=S_{\text{avg}}=\frac{1}{2}{ϵ}_{0}c{E}_0^2$$

Derivation: Poynting Vector for a Plane Wave
For a plane wave, $\overrightarrow{E}=E_0\sin (kz-\omega t)\hat{x}$, $\overrightarrow{B}=B_0\sin (kz-\omega t)\hat{y}$, propagating in the $z$-direction. The Poynting vector is:

$$\overrightarrow{S}=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}=\frac{1}{{\mu }_0}(E_x{B}_y)\hat{z}=\frac{1}{{\mu }_0}(E_0{B}_0{\sin }^2(kz-\omega t))\hat{z}$$

Since $B_0=\frac{E_0}{c}$ and $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}$, $S=\frac{E_0{B}_0}{{\mu }_0}{\sin }^2(kz-\omega t)=\frac{E_0(E_0/c)}{{\mu }_0}{\sin }^2(kz-\omega t)={ϵ}_{0}c{E}_0^2{\sin }^2(kz-\omega t)$. Average over one cycle (${\sin }^2$ averages to $\frac{1}{2}$):

$$S_{\text{avg}}=\frac{1}{2}{ϵ}_{0}c{E}_0^2$$

Derivation: Intensity in Rocket Signal
A spacecraft signal ($E_0=200\text{V/m}$) has $I=\frac{1}{2}{ϵ}_{0}c{E}_0^2=\frac{1}{2}(8.85\times {10}^{-12})(3\times {10}^8)(200{)}^2\approx 53.1{\text{W/m}}^2$, ensuring signal strength (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves a wave with $E_0=100\text{V/m}$. Find $I$.

• Solution:
  $I=\frac{1}{2}{ϵ}_{0}c{E}_0^2=\frac{1}{2}(8.85\times {10}^{-12})(3\times {10}^8)(100{)}^2\approx 13.3{\text{W/m}}^2$.
  • JEE Tip: Intensity depends on $E_0^2$; units in W/m². Common error: Forgetting $\frac{1}{2}$.

Solved Example: A NEET problem involves a wave with $I=20{\text{W/m}}^2$. Find $E_0$.

• Solution:
  $I=\frac{1}{2}{ϵ}_{0}c{E}_0^2$, $E_0=\sqrt{\frac{2I}{{ϵ}_{0}c}}=\sqrt{\frac{2\times 20}{(8.85\times {10}^{-12})\times (3\times {10}^8)}}\approx 123\text{V/m}$.
  • NEET Tip: Solve for $E_0$ using $I$; $E_0$ in V/m. Common error: Incorrect constants.

Solved Example: A JEE Advanced problem involves a wave with $B_0=2\times {10}^{-7}\text{T}$. Find $I$.

• Solution:
  $E_0=c{B}_0=(3\times {10}^8)\times (2\times {10}^{-7})=60\text{V/m}$, $I=\frac{1}{2}(8.85\times {10}^{-12})(3\times {10}^8)(60{)}^2\approx 4.78{\text{W/m}}^2$.
  • JEE Tip: Convert $B_0$ to $E_0$ using $c$; compute $I$. Common error: Using $B_0$ directly.

Solved Example: A JEE Main problem involves a wave with $E_0=150\text{V/m}$. Find $u_{\text{avg}}$.

• Solution:
  $u_{\text{avg}}={ϵ}_0{E}_{\text{rms}}^2$, $E_{\text{rms}}=\frac{E_0}{\sqrt{2}}=\frac{150}{\sqrt{2}}\approx 106.1\text{V/m}$, $u_{\text{avg}}=(8.85\times {10}^{-12})(106.1{)}^2\approx 9.96\times {10}^{-8}{\text{J/m}}^3$.
  • JEE Tip: Use $E_{\text{rms}}$ for average energy density; units in J/m³. Common error: Using $E_0$.

Application: Energy transport applies to solar radiation, lasers, and rocketry (e.g., spacecraft signal intensity, aligning with your interest, April 19, 2025).

Summary and Quick Revision

• Generation: Waves from accelerating charges; ${\mathrm{\nabla }}^2\overrightarrow{E}={\mu }_0{ϵ}_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}$, $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}$.
• Properties: Transverse, $\overrightarrow{E}\perp \overrightarrow{B}$, $E_0=c{B}_0$, $k=\frac{2\pi }{\lambda }$, $\omega =kc$.
• Spectrum: Radio (${10}^4$–${10}^9\text{Hz}$), microwave (${10}^9$–${10}^{12}\text{Hz}$), infrared, visible (400–750 nm), UV, X-rays, gamma rays.
• Energy: $u={ϵ}_0{E}^2$, $\overrightarrow{S}=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$, $I=\frac{1}{2}{ϵ}_{0}c{E}_0^2$.
• Applications: Communication, spacecraft signals.
• JEE/NEET Tips: Use $c$ for wave speed, convert units for $\lambda$ and $f$, compute intensity with $E_0$, verify significant figures (April 14, 2025).
• SI Units: $c$ (m/s), $\lambda$ (m), $f$ (Hz), $I$ (W/m²), $u$ (J/m³).

Practice Problems

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Note: Content regularly updated to align with current JEE/NEET syllabi.