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Electromagnetic Waves

🎓 Class 12📖 Physics Part-I📖 9 notes🧠 15 Q&A⏱️ ~14 min
Alternating CurrentChapter 8 of 8

Electromagnetic WavesStudy Notes

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8.1 INTRODUCTION

Explanation

8.1 INTRODUCTION

In earlier chapters, we learned that an electric current produces a magnetic field and that two current-carrying wires exert forces on each other due to their magnetic fields. Additionally, a changing magnetic field produces an electric field, as seen in Faraday's law of electromagnetic induction. The question arises: does a changing electric field also produce a magnetic field? James Clerk Maxwell answered this affirmatively, proposing that not only electric currents but also time-varying electric fields generate magnetic fields. While applying Ampere's circuital law to a charging capacitor, Maxwell noticed an inconsistency. The law, as originally formulated, did not account for the magnetic field in the region between the capacitor plates where no conduction current flows. To resolve this, Maxwell introduced the concept of displacement current, an additional current term arising from the changing electric field. This addition made Ampere's law consistent and symmetric with Faraday's law. Maxwell formulated a set of four equations, now known as Maxwell's equations, that unify electricity, magnetism, and optics. These equations describe how electric and magnetic fields are generated by charges, currents, and changes in each other. One of the most profound predictions from Maxwell's equations is the existence of electromagnetic waves—coupled oscillations of electric and magnetic fields that propagate through space at a speed very close to the speed of light (3 × 10^8 m/s). This led to the remarkable conclusion that light itself is an electromagnetic wave, unifying the previously separate domains of electricity, magnetism, and optics. Heinrich Hertz experimentally demonstrated electromagnetic waves in 1885, confirming Maxwell's prediction. Subsequently, the technological exploitation of these waves revolutionized communication. This chapter begins by discussing the need for displacement current and its implications, followed by a qualitative description of electromagnetic waves and their properties. Finally, the chapter covers the electromagnetic spectrum, spanning from gamma rays with extremely short wavelengths to long radio waves with very large wavelengths.

  • Electric currents produce magnetic fields; changing magnetic fields produce electric fields.
  • Maxwell introduced displacement current to resolve inconsistencies in Ampere's law.
  • Maxwell's equations unify electricity, magnetism, and optics.
  • Electromagnetic waves are coupled oscillations of electric and magnetic fields propagating through space.
  • Speed of electromagnetic waves equals the speed of light, leading to the conclusion that light is an electromagnetic wave.
  • Hertz experimentally confirmed electromagnetic waves in 1885.
  • 📌 Displacement current: A current term introduced by Maxwell due to time-varying electric fields, acting as a source of magnetic field.
  • 📌 Maxwell's equations: A set of four fundamental equations describing electromagnetism.
  • 📌 Electromagnetic waves: Self-sustaining oscillations of electric and magnetic fields propagating through space.

8.2 DISPLACEMENT CURRENT

Explanation

8.2 DISPLACEMENT CURRENT

Ampere's circuital law states that the line integral of the magnetic field B around a closed loop equals μ₀ times the conduction current passing through any surface bounded by the loop. However, when applied to a charging capacitor, this law leads to a contradiction. Consider a parallel plate capacitor connected to a time-varying current i(t). Outside the capacitor, the conduction current flows in the wires, producing a magnetic field B at a point P. Using Ampere's law with a loop encircling the wire, the magnetic field is given by B(2πr) = μ₀ i(t), where r is the radius of the circular loop. Now, consider a different surface bounded by the same loop but passing between the capacitor plates where no conduction current flows. Ampere's law then predicts zero magnetic field, contradicting the previous result. The resolution lies in recognizing that the changing electric field between the capacitor plates acts like a current, called the displacement current. The electric field E between the plates is related to the charge Q on the plates as E = Q/(ε₀ A), where A is the plate area and ε₀ is the permittivity of free space. The electric flux through the surface between the plates is Φ_E = E × A = Q/ε₀. Differentiating with respect to time, dΦ_E/dt = (1/ε₀) dQ/dt = i/ε₀. Multiplying both sides by ε₀, we get ε₀ dΦ_E/dt = i, which identifies the displacement current i_d = ε₀ dΦ_E/dt. Thus, the total current passing through any surface bounded by the loop is the sum of the conduction current i_c and the displacement current i_d. Ampere's law is generalized to the Ampere-Maxwell law: ∮ B · dl = μ₀ i_c + μ₀ ε₀ dΦ_E/dt This generalization restores consistency and symmetry in Maxwell's equations. The displacement current produces magnetic fields just like conduction current. Inside the capacitor, conduction current is zero but displacement current is non-zero; outside the capacitor, conduction current exists but displacement current is zero. This concept has profound implications, including the prediction of electromagnetic waves. The symmetry between the laws of induction (changing magnetic fields produce electric fields) and the Ampere-Maxwell law (changing electric fields produce magnetic fields) implies that time-varying electric and magnetic fields sustain each other and propagate through space as waves.

  • Ampere's circuital law without displacement current leads to contradictions in charging capacitors.
  • Displacement current arises from time-varying electric flux between capacitor plates.
  • Displacement current i_d = ε₀ dΦ_E/dt acts as a source of magnetic field like conduction current.
  • Generalized Ampere-Maxwell law includes both conduction and displacement currents.
  • Displacement current restores symmetry between electric and magnetic fields in Maxwell's equations.
  • Displacement current enables the existence of electromagnetic waves.
  • 📌 Conduction current (i_c): Current due to flow of charges in conductors.
  • 📌 Displacement current (i_d): Current equivalent due to changing electric field, i_d = ε₀ dΦ_E/dt.
  • 📌 Ampere-Maxwell law: Generalized Ampere's law including displacement current.

MAXWELL'S EQUATIONS IN VACUUM

Formula

MAXWELL'S EQUATIONS IN VACUUM

Maxwell's equations in vacuum summarize the fundamental laws of electromagnetism, relating electric and magnetic fields to their sources—charges and currents—and describing how these fields change in time and space. These four equations are: 1. Gaus

Practice QuestionsElectromagnetic Waves

Includes NCERT exercise questions with answers

Q1.8.1 Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 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 (junction rule) valid at each plate of the capacitor? Explain.

Answer:

Given: Radius of plates, r = 12 cm = 0.12 m Separation between plates, d = 5.0 cm = 0.05 m Charging current, I = 0.15 A (a) Calculate capacitance C: Capacitance of parallel plate capacitor, C = ε₀ A / d Area, A = π r² = π × (0.12)² = π × 0.0144 = 0.04524 m² ε₀ = 8.854 × 10⁻¹² F/m So, C = (8.854 × 10⁻¹²) × 0.04524 / 0.05 = 8.01 × 10⁻¹² F = 8.01 pF Rate of change of potential difference (dV/dt): Current I = C (dV/dt) ⇒ dV/dt = I / C = 0.15 / (8.01 × 10⁻¹²) = 1.87 × 10¹⁰ V/s (b) Displacement current I_d: Displacement current I_d = ε₀ (dΦ_E/dt), where Φ_E = Electric flux = E × A But I_d = I (since charging current is constant), so displacement current equals conduction current = 0.15 A (c) Kirchhoff's first rule (junction rule) states that the total current entering a junction equals the total current leaving it. At the capacitor plates, conduction current flows in the wires but no conduction current flows across the gap; instead, displacement current flows. Since displacement current equals conduction current, Kirchhoff's first rule is valid at each plate of the capacitor.

Explanation:

Step 1: Calculate area of plates. Step 2: Use formula for capacitance. Step 3: Use relation I = C dV/dt to find rate of change of voltage. Step 4: Displacement current equals conduction current for charging capacitor. Step 5: Kirchhoff's first rule holds because displacement current accounts for current continuity.

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Q2.8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s⁻¹. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Answer:

Given: Radius R = 6.0 cm = 0.06 m Capacitance C = 100 pF = 100 × 10⁻¹² F Voltage amplitude V₀ = 230 V (rms given, so V_rms = 230 V) Angular frequency ω = 300 rad/s (a) RMS conduction current I_rms: For capacitor, current leads voltage by 90° I = C dV/dt = C ω V₀ cos(ω t) RMS current I_rms = ω C V_rms I_rms = 300 × 100 × 10⁻¹² × 230 = 6.9 × 10⁻⁶ A = 6.9 μA (b) The conduction current in the wires is equal in magnitude to the displacement current between the plates because the displacement current accounts for the changing electric field in the capacitor gap. So, yes, conduction current equals displacement current. (c) Magnetic field amplitude B at distance r = 3.0 cm = 0.03 m inside the capacitor: Displacement current I_d = I_rms = 6.9 μA Magnetic field inside capacitor at radius r from axis: B = (μ₀ I_d r) / (2 π R²) μ₀ = 4π × 10⁻⁷ T·m/A B = (4π × 10⁻⁷) × (6.9 × 10⁻⁶) × 0.03 / (2 π × (0.06)²) Simplify π: B = (4 × 10⁻⁷) × 6.9 × 10⁻⁶ × 0.03 / (2 × 0.0036) = (8.28 × 10⁻¹²) / 0.0072 = 1.15 × 10⁻⁹ T = 1.15 nT

Explanation:

Step 1: Calculate rms current using I_rms = ω C V_rms. Step 2: Understand displacement current equals conduction current in capacitor. Step 3: Use formula for magnetic field inside capacitor due to displacement current. Step 4: Substitute values and simplify to find B.

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Q3.8.3 What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?

Answer:

The physical quantity that is the same for all electromagnetic waves, regardless of their wavelength, is the speed of the wave in vacuum. All electromagnetic waves travel at the speed of light, c = 3 × 10⁸ m/s.

Explanation:

Electromagnetic waves of all wavelengths travel at the same speed in vacuum, which is the speed of light.

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Q4.8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Answer:

In a plane electromagnetic wave traveling along the z-direction, the electric field vector (E) and magnetic field vector (B) are perpendicular to each other and both are perpendicular to the direction of propagation (z-axis). Thus, E and B lie in the x-y plane. Frequency, ν = 30 MHz = 30 × 10⁶ Hz Wavelength, λ = c / ν = (3 × 10⁸ m/s) / (30 × 10⁶ Hz) = 10 m

Explanation:

Electromagnetic waves are transverse waves with E and B perpendicular to propagation direction and to each other. Wavelength is calculated using λ = c/ν.

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Q5.8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Answer:

Frequency range: 7.5 MHz to 12 MHz Convert to Hz: 7.5 MHz = 7.5 × 10⁶ Hz 12 MHz = 12 × 10⁶ Hz Wavelength λ = c / ν For 7.5 MHz: λ = 3 × 10⁸ / 7.5 × 10⁶ = 40 m For 12 MHz: λ = 3 × 10⁸ / 12 × 10⁶ = 25 m Therefore, the wavelength band is from 25 m to 40 m.

Explanation:

Use the relation λ = c/ν to convert frequency band to wavelength band.

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Q6.8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Answer:

The frequency of the electromagnetic waves produced by the oscillator is the same as the frequency of oscillation of the charged particle. Therefore, frequency of EM waves = 10⁹ Hz.

Explanation:

Oscillating charges produce electromagnetic waves at the same frequency as their oscillation.

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Q7.8.7 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?

Answer:

Given: B₀ = 510 nT = 510 × 10⁻⁹ T Speed of light, c = 3 × 10⁸ m/s Relation between E₀ and B₀ in EM wave: E₀ = c B₀ Calculate: E₀ = 3 × 10⁸ × 510 × 10⁻⁹ = 153 V/m

Explanation:

Use the relation E₀ = c B₀ to find electric field amplitude from magnetic field amplitude.

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Q8.8.8 Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine B₀, ω, k, and λ. (b) Find expressions for E and B.

Answer:

(a) Given: E₀ = 120 N/C ν = 50.0 MHz = 50 × 10⁶ Hz Calculate B₀: B₀ = E₀ / c = 120 / (3 × 10⁸) = 4.0 × 10⁻⁷ T Angular frequency ω: ω = 2π ν = 2π × 50 × 10⁶ = 3.14 × 10⁸ rad/s Wavenumber k: k = ω / c = (3.14 × 10⁸) / (3 × 10⁸) = 1.05 m⁻¹ Wavelength λ: λ = 2π / k = 2π / 1.05 ≈ 6.0 m (b) Expressions for E and B fields (assuming wave propagates along z and E along x): E = E₀ sin(k z - ω t) î B = B₀ sin(k z - ω t) ĵ

Explanation:

Step 1: Calculate B₀ using E₀ = c B₀. Step 2: Calculate angular frequency ω = 2πν. Step 3: Calculate wavenumber k = ω/c. Step 4: Calculate wavelength λ = 2π/k. Step 5: Write expressions for E and B assuming standard wave form.

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