Are you an RBSE Class 12 student aiming to master Physics? Chapter 8, Electromagnetic Waves, is a crucial unit for your board exams, covering fundamental concepts and requiring practice with numerical problems. This comprehensive guide provides you with all the essential concepts, formulas, and step-by-step NCERT solutions to secure a top rank!
🌟 Chapter 8 Overview: Electromagnetic Waves
This chapter, based on the NCERT curriculum adopted by the Rajasthan Board (RBSE), introduces the revolutionary concept of the displacement current and the unified theory of electromagnetism given by James Clerk Maxwell.
Key Concepts to Master:
- Maxwell’s Equations: The four fundamental equations (Gauss’s Law for Electrostatics, Gauss’s Law for Magnetism, Faraday’s Law of Induction, and Ampere-Maxwell Law) that form the basis of all electromagnetic phenomena.
- Displacement Current (Id): Understanding the need for the displacement current (Id=ϵ0dtdΦE) to generalize Ampere’s Circuital Law and ensure the consistency of electromagnetic theory.
- Electromagnetic Waves (EM Waves):
- Production: How accelerated charges produce time-varying electric and magnetic fields, which propagate as EM waves.
- Properties: Transverse nature, mutual perpendicularity of E and B fields, and their direction of propagation.
- Speed in Vacuum: The speed of light c=μ0ϵ0
1≈3×108 m/s.
- Relation between E and B: The ratio of the amplitude of the electric field (E0) to the magnetic field (B0) is equal to the speed of light: c=B0E0.
- Electromagnetic Spectrum: A systematic classification of EM waves according to their frequency or wavelength (Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays).
- Know the range of wavelength/frequency for each part.
- Memorize the sources and at least two practical applications for each type.
💡 Important Formulas (For Quick Revision)
| Concept | Formula | Notes |
| Speed of EM Wave in Vacuum | c=μ0ϵ0 | c≈3×108 m/s |
| Speed in Medium | v=μϵ | μ=μrμ0, ϵ=ϵrϵ0 |
| Relation between E and B | E0=cB0 or Erms=cBrms | Ratio is c |
| Wave Equation | c=νλ | ν is frequency, λ is wavelength |
| Displacement Current | Id=ϵ0dtdΦE | ΦE is electric flux |
| Ampere-Maxwell Law | ∮B⋅dl=μ0(Ic+Id) | Ic: Conduction current, Id: Displacement current |
| Energy Density | u=ue+um=21ϵ0E2+21μ0B2 | ue=um (Electric and Magnetic densities are equal) |
📚 NCERT Exercise Solutions: Chapter 8 (Step-by-Step)
Here are detailed solutions for the most commonly asked and crucial numerical problems from NCERT Class 12 Physics Chapter 8 (Electromagnetic Waves), as followed by the RBSE.
Problem 1: Capacitor and Displacement Current
Question: A capacitor made of two circular plates (radius r=12 cm, separation d=5.0 cm) is being charged by a constant current I=0.15 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 valid at each plate of the capacitor?
Solution:
(a) Capacitance (C): $$C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 (\pi r^2)}{d}$$Given: r=0.12 m, d=0.05 m, ϵ0=8.85×10−12 F/m.
C=0.058.85×10−12×π×(0.12)2≈8.0×10−11 F
Rate of change of potential difference (dV/dt): We know q=CV. Differentiating with respect to time (t): $$\frac{dq}{dt} = C \frac{dV}{dt}$$Since the charging current $I = \frac{dq}{dt}$:$$\frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8.0 \times 10^{-11}} \approx \mathbf{1.875 \times 10^9 \text{ V/s}}$$
(b) Displacement Current (Id): In the region between the capacitor plates, the conduction current (Ic) is zero, but the displacement current (Id) exists. During charging, the displacement current is equal to the conduction current:
Id=Ic=0.15 A
(c) Kirchhoff’s First Rule (Junction Rule): Yes, Kirchhoff’s first rule is valid at each plate only if we consider the total current as the sum of conduction current (Ic) and displacement current (Id).
- Incoming Conduction Current (Ic) at one plate is balanced by Outgoing Displacement Current (Id) leaving that plate. The total current Ic+Id is conserved throughout the circuit.
Problem 2: E and B Field Amplitudes
Question: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0=510 nT. What is the amplitude of the electric field part of the wave?
Solution:
The amplitude of the electric field (E0) and the magnetic field (B0) in a vacuum are related by the speed of light (c):
E0=cB0
Given: B0=510 nT=510×10−9 T. Speed of light c=3×108 m/s.
E0=(3×108 m/s)×(510×10−9 T)
E0=1530×10−1=153 N/C
Problem 3: Wavelength and Frequency
Question: A plane electromagnetic wave travels in a vacuum along the z-direction. The frequency of the wave is 30 MHz. What is its wavelength?
Solution:
The speed of the wave (c), frequency (ν), and wavelength (λ) are related by the equation:
c=νλ⟹λ=νc
Given: ν=30 MHz=30×106 Hz. Speed of light c=3×108 m/s.
λ=30×106 Hz3×108 m/s
λ=303×102=101×100=10 m
🚀 Pro-Tips for RBSE Board Exam Preparation
- Focus on Theory: The Electromagnetic Spectrum (properties, uses, order of λ and ν) is a highly scoring topic. Ensure you can write 2-3 points for each type of wave (Gamma, X-ray, UV, Visible, IR, Microwave, Radio).
- Derivation Practice: Practice the derivation of Displacement Current and the subsequent generalization of Ampere’s Circuital Law (Ampere-Maxwell Law).
- Numerical Types: The most common numerical problems involve:
- Calculating E0 from B0 (or vice-versa).
- Calculating C, Id, and dtdV for a charging capacitor.
- Finding the wavelength or frequency using c=νλ.
- Conceptual Questions: Be prepared for questions on why E and B are perpendicular, the source of EM waves, and the direction of propagation (Poynting vector).
| Chapter 8 | (Open) |
By systematically covering these concepts and practicing the NCERT solutions, you will ensure a perfect score in RBSE Class 12 Physics Chapter 8.