Electromagnetic Waves

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Exercise 8.1Electromagnetic Spectrum and Properties of EM Waves

Figure 8.6 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. The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge 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.

Solution

(a) Radius r = 12 cm = 0.12 m, d = 5.0 cm = 0.05 m Area A = πr² = π × (0.12)² = π × 0.0144 = 4.524 × 10⁻² m² Capacitance: C = ε₀A/d = 8.85 × 10⁻¹² × 4.524 × 10⁻² / 0.05 = 8.85 × 4.524 × 10⁻¹⁴ / 0.05 = 40.04 × 10⁻¹⁴ / 0.05 = 8.01 × 10⁻¹² F ≈ 8 pF Rate of change of V: dV/dt = I/C = 0.15 / (8 × 10⁻¹²) = 1.875 × 10¹⁰ V/s (b) The displacement current equals the conduction current: I_d = I = 0.15 A (c) Yes, Kirchhoff's junction rule is valid at each plate. At each plate: the conduction current arriving = displacement current leaving (or vice versa). Conduction current I flows into the plate, and displacement current I_d = I flows between the plates. This ensures current continuity and satisfies the junction rule.

A parallel plate capacitor (Figure 8.7) made of circular plates each of radius R = 6 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with angular frequency ω = 300 rad/s. (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 cm from the axis between the plates.

Solution

(a) Impedance of capacitor: X_C = 1/(ωC) = 1/(300 × 100 × 10⁻¹²) = 1/(3 × 10⁻⁸) = 3.33 × 10⁷ Ω I_rms = V_rms/X_C = 230 / (3.33 × 10⁷) = 6.9 × 10⁻⁶ A ≈ 6.9 μA (b) Yes, displacement current = conduction current at all times. Maxwell's modification of Ampere's law ensures that the total current (conduction + displacement) is continuous. I_displacement = I_conduction = 6.9 μA. (c) At point r = 3 cm = 0.03 m from axis (inside the capacitor, R = 6 cm): Using modified Ampere's law for displacement current: B × 2πr = μ₀ × I_d × (πr²/πR²) [only fraction of displacement current enclosed] B = μ₀I_d × r / (2πR²) I₀ = I_rms × √2 = 6.9 × 10⁻⁶ × √2 ≈ 9.76 × 10⁻⁶ A B₀ = μ₀I₀r / (2πR²) = (4π × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03) / (2π × (0.06)²) = (4π × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03) / (2π × 0.0036) = (2 × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03) / 0.0036 = (5.856 × 10⁻¹³) / 0.0036 = 1.63 × 10⁻¹⁰ T

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?

Solution

The physical quantity that is the same for all electromagnetic waves in vacuum is the speed of propagation: c = 3 × 10⁸ m/s. All electromagnetic waves — X-rays, visible light, and radio waves — travel at the same speed c in vacuum, regardless of their wavelength or frequency. This is the universal speed of electromagnetic radiation in vacuum.

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?

Solution

The wave travels along z-direction. In an electromagnetic wave: • Electric field E, magnetic field B, and the direction of propagation k are mutually perpendicular • Therefore E and B must lie in the x-y plane (perpendicular to z) • E and B are also perpendicular to each other So E could be along x-axis and B along y-axis (or vice versa). Wavelength at f = 30 MHz = 30 × 10⁶ Hz: λ = c/f = 3 × 10⁸ / (30 × 10⁶) = 10 m

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

Solution

c = 3 × 10⁸ m/s For f₁ = 7.5 MHz = 7.5 × 10⁶ Hz: λ₁ = c/f₁ = 3 × 10⁸ / (7.5 × 10⁶) = 40 m For f₂ = 12 MHz = 12 × 10⁶ Hz: λ₂ = c/f₂ = 3 × 10⁸ / (12 × 10⁶) = 25 m The corresponding wavelength band is 25 m to 40 m.

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

Solution

An oscillating charged particle produces electromagnetic waves with the same frequency as its oscillation. Frequency of EM waves = frequency of oscillation = 10⁹ Hz The wavelength of these waves: λ = c/f = 3 × 10⁸ / 10⁹ = 0.3 m These are radio waves (microwave boundary).

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

B₀ = 510 nT = 510 × 10⁻⁹ T = 5.1 × 10⁻⁷ T Relationship between E and B in an EM wave: E₀ = cB₀ = 3 × 10⁸ × 5.1 × 10⁻⁷ = 15.3 × 10¹ = 153 N/C The amplitude of the electric field E₀ = 153 N/C.

Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is f = 50 MHz. (a) Determine B₀, ω, k, and λ. (b) Find expressions for E and B.

Solution

E₀ = 120 N/C, f = 50 × 10⁶ Hz (a) B₀ = E₀/c = 120 / (3 × 10⁸) = 4 × 10⁻⁷ T = 400 nT ω = 2πf = 2π × 50 × 10⁶ = π × 10⁸ = 3.14 × 10⁸ rad/s λ = c/f = 3 × 10⁸ / (50 × 10⁶) = 6 m k = 2π/λ = 2π/6 = π/3 ≈ 1.05 rad/m (b) If wave travels along z-direction, E along x, B along y: E = E₀ cos(kz – ωt) x̂ = 120 cos(1.05z – 3.14 × 10⁸t) x̂ N/C B = B₀ cos(kz – ωt) ŷ = 4 × 10⁻⁷ cos(1.05z – 3.14 × 10⁸t) ŷ T

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Electromagnetic Waves

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