Difficulty: Intermediate to Advanced | Prerequisites: Faraday's Law, Ampere's Law with Maxwell's displacement current (Part 3), AC circuits (Part 4).
Electromagnetic (EM) waves are the payoff of the entire course. Maxwell showed that changing electric fields create magnetic fields and vice versa, and that these self-sustaining oscillations propagate through space at the speed of light. This final topic brings together everything: the relationship between E and B fields, how energy is transported (Poynting vector), and how light's oscillation direction can be filtered (polarisation). Every form of light, radio, Wi-Fi, and X-ray is an electromagnetic wave, so this material connects directly to optics, communications, and modern physics.
EM waves are transverse waves with perpendicular E and B fields, travelling at c = 3 × 10⁸ m/s. The Poynting vector gives the power per unit area carried by the wave. Unpolarised light passing through an ideal polariser loses half its intensity; after that, Malus's Law (I = I₀ cos² θ) governs subsequent polarisers.
Electromagnetic wave
A self-propagating transverse wave consisting of oscillating electric and magnetic fields perpendicular to each other and to the direction of travel. In simple terms, light is a wave of electric and magnetic fields that does not need a medium to travel through.
Speed of light (c)
c = 1/√(μ₀ε₀) ≈ 3.00 × 10⁸ m/s. The speed at which all electromagnetic waves travel in vacuum.
Poynting vector (S)
S = (1/μ₀)(E × B). It gives the rate and direction of energy transport per unit area (units: W/m²). In simple terms, the Poynting vector tells you how much power the wave carries and which way it is going.
Intensity (I)
The time-averaged power per unit area of the wave: I = S_avg = E_max² / (2μ₀c). Also written as I = ½ ε₀cE_max² or I = E_max B_max / (2μ₀). Measured in W/m².
Polarisation
The direction in which the electric field oscillates. Unpolarised light has E oscillating in all directions perpendicular to the propagation direction. A polariser transmits only the component along its transmission axis.
Malus's Law
When polarised light of intensity I₀ passes through a polariser whose axis is at angle θ to the polarisation direction: I = I₀ cos² θ.
In an electromagnetic wave, E and B oscillate in phase, perpendicular to each other and to the propagation direction.
Their magnitudes are related by: E = cB, or equivalently B = E/c.
If E = E_max sin(kz − ωt), then B = B_max sin(kz − ωt), with B_max = E_max / c.
The wave propagates in the direction of E × B.
Given: E = 300 sin(kz − ωt) V/m.
Peak magnetic field: B_max = E_max / c = 300 / (3 × 10⁸) = 1.0 × 10⁻⁶ T (1 μT).
Intensity: I = E_max² / (2μ₀c) = 300² / (2 × 4π × 10⁻⁷ × 3 × 10⁸).
Numerator: 90,000.
Denominator: 2 × 4π × 10⁻⁷ × 3 × 10⁸ = 2 × (1.2566 × 10⁻⁶) × (3 × 10⁸) ≈ 753.98.
I = 90,000 / 753.98 ≈ 119.4 W/m².
S = (1/μ₀)(E × B) points in the direction of wave propagation.
The magnitude gives the instantaneous power per unit area.
The time-averaged value (intensity) is I = S_avg = E_max B_max / (2μ₀) = E_max² / (2μ₀c).
The Poynting vector is not the direction of E or B, and it is not the polarisation angle. It represents the energy flow.
Unpolarised light contains E-field oscillations in all directions perpendicular to propagation, with random orientations.
When unpolarised light passes through an ideal polariser, exactly half the intensity is transmitted: I = I₀/2. This holds regardless of the polariser's orientation, because on average half the light is aligned with any chosen axis.
After passing through the first polariser, the light is polarised. If it then passes through a second polariser at angle θ to the first, the transmitted intensity follows Malus's Law: I = I₀ cos² θ.
At θ = 0°, all light passes. At θ = 90°, no light passes (crossed polarisers).
Quantity | Expression |
|---|---|
Speed of light | c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s |
E-B relationship | E_max = cB_max |
Poynting vector | S = (1/μ₀)(E × B) |
Intensity (time-averaged) | I = E_max² / (2μ₀c) |
Unpolarised light through polariser | I = I₀ / 2 |
Malus's Law | I = I₀ cos² θ |
Wave equation | E = E_max sin(kz − ωt), B = B_max sin(kz − ωt) |
Wave number | k = 2π/λ |
Angular frequency | ω = 2πf = kc |
Polarised sunglasses use a polarising filter oriented to block horizontally polarised glare reflected from roads and water. Microwave ovens exploit the fact that water molecules absorb EM radiation at microwave frequencies, converting field energy into thermal energy. Radio antennas are designed so that the Poynting vector of the transmitted wave is directed towards the receiver.
Students often think the Poynting vector gives the direction of the electric field. It does not: S points in the direction of energy propagation, which is perpendicular to both E and B.
A common error is applying Malus's Law (I₀ cos² θ) to unpolarised light hitting the first polariser. Malus's Law only applies to already-polarised light. For unpolarised light through the first polariser, the rule is I = I₀/2.
Students sometimes assume B_max has the same order of magnitude as E_max. It does not: B_max = E_max/c, so B is many orders of magnitude smaller in SI units (though the energy is shared equally between the fields).
Confusing intensity with amplitude: intensity is proportional to E_max², not E_max.
⚠️ The Poynting vector question ("what does S represent?") is a standard 3-mark multiple-choice item. The answer is: the rate and direction of energy transport per unit area.
⚠️ The unpolarised-light-through-a-polariser question (I = I₀/2) is tested frequently. Do not use cos² θ for the first polariser.
⚠️ The E-to-B and intensity calculation is a common 5-mark short-answer. Know B_max = E_max/c and I = E_max²/(2μ₀c).
⚠️ Make sure you can convert between E_max and B_max quickly. The factor is always c = 3 × 10⁸ m/s.
Fill in the blank: The peak magnetic field of an EM wave is related to the peak electric field by B_max = _______.
True or false: The Poynting vector points in the direction of the electric field.
Fill in the blank: When unpolarised light of intensity I₀ passes through an ideal polariser, the transmitted intensity is _______.
True or false: The intensity of an EM wave is proportional to the square of the peak electric field.
Fill in the blank: After passing through a first polariser, polarised light hits a second polariser at angle θ. The transmitted intensity is I = _______.
Answers: 1. E_max / c. 2. False (it points in the direction of wave propagation). 3. I₀/2. 4. True. 5. I₀ cos² θ (Malus's Law, where I₀ here is the intensity after the first polariser).
Q: What does the Poynting vector represent in an electromagnetic wave?
A: The rate and direction of energy transport per unit area (power per unit area, in W/m²). It is defined as S = (1/μ₀)(E × B).
Q: An unpolarised light beam of intensity I₀ passes through an ideal polariser. What is the transmitted intensity?
A: I = I₀/2. Unpolarised light has equal components in all transverse directions, so on average half the intensity is transmitted.
Q: A light wave has E = 300 sin(kz − ωt) V/m. Calculate B_max.
A: B_max = E_max / c = 300 / (3 × 10⁸) = 1.0 × 10⁻⁶ T.
Q: For the same wave, calculate the intensity.
A: I = E_max² / (2μ₀c) = 90,000 / (2 × 4π × 10⁻⁷ × 3 × 10⁸) ≈ 119.4 W/m².
EM waves are the culmination of Maxwell's equations, which tie together Gauss's Law (Part 1), Faraday's Law (Part 3), and Ampere's Law with the displacement current. The wave speed c = 1/√(μ₀ε₀) echoes the RLC resonant frequency ω₀ = 1/√(LC), highlighting the deep analogy between circuits and fields. Polarisation connects to optics topics like Brewster's angle and birefringence, which appear in more advanced courses.
electromagnetic wave, EM wave, Maxwell's equations, speed of light, c, Poynting vector, energy flux, intensity, power per unit area, polarisation, polarization, unpolarised light, polariser, polarizer, Malus's Law, transmission axis, E-field, B-field, wave equation, transverse wave, amplitude, frequency, wavelength, wave number, angular frequency, PHYS 212, University Physics electricity and magnetism