Electromagnetic Waves

Wave equations from Maxwell's equations, properties of EM waves

Electromagnetic Waves

Wave Equation from Maxwell

In vacuum, taking curl of Faraday's law and using Ampere-Maxwell:

∇2E⃗=μ0ϵ0∂2E⃗∂t2\nabla^2\vec{E} = \mu_0\epsilon_0\frac{\partial^2\vec{E}}{\partial t^2}

∇2B⃗=μ0ϵ0∂2B⃗∂t2\nabla^2\vec{B} = \mu_0\epsilon_0\frac{\partial^2\vec{B}}{\partial t^2}

These are wave equations with speed:

c=1μ0ϵ0=3.00×108 m/sc = \frac{1}{\sqrt{\mu_0\epsilon_0}} = 3.00 \times 10^8 \text{ m/s}

(Speed of light!)

Plane Wave Solution

Electric field: E⃗=E0sin⁡(kx−ωt)j^\vec{E} = E_0\sin(kx - \omega t)\hat{j}

Magnetic field: B⃗=B0sin⁡(kx−ωt)k^\vec{B} = B_0\sin(kx - \omega t)\hat{k}

where:

  • k=2π/λk = 2\pi/\lambda (wave number)
  • ω=2πf\omega = 2\pi f (angular frequency)
  • ω=ck\omega = ck (dispersion relation)

Properties of EM Waves

  1. Transverse: E⃗\vec{E} and B⃗\vec{B} perpendicular to propagation direction

  2. Perpendicular to each other: E⃗⊥B⃗\vec{E} \perp \vec{B}

  3. In phase: E⃗\vec{E} and B⃗\vec{B} oscillate together

  4. Right-hand rule: E⃗×B⃗\vec{E} \times \vec{B} points in propagation direction

  5. Field magnitudes related: EB=c\frac{E}{B} = c

Energy in EM Wave

Energy density: u=ϵ0E2=B2μ0u = \epsilon_0 E^2 = \frac{B^2}{\mu_0}

(Equal contributions from E⃗\vec{E} and B⃗\vec{B})

Intensity (average power per area): I=⟨S⟩=12ϵ0cE02=E022μ0cI = \langle S \rangle = \frac{1}{2}\epsilon_0 cE_0^2 = \frac{E_0^2}{2\mu_0 c}

or in terms of B0B_0: I=cB022μ0I = \frac{cB_0^2}{2\mu_0}

Momentum and Radiation Pressure

EM wave carries momentum:

p=Ucp = \frac{U}{c}

Momentum density: g=uc=Sc2g = \frac{u}{c} = \frac{S}{c^2}

Radiation pressure:

Complete absorption: P=IcP = \frac{I}{c}

Complete reflection: P=2IcP = \frac{2I}{c}

Polarization

Linear polarization: E⃗\vec{E} oscillates in fixed plane

Circular polarization: E⃗\vec{E} rotates, constant magnitude

Unpolarized: Random polarization directions

Malus's law: Intensity through polarizer: I=I0cos⁡2θI = I_0\cos^2\theta

where θ\theta is angle from polarization axis.

Electromagnetic Spectrum

All EM waves travel at cc in vacuum, differ only in frequency/wavelength:

  • Radio waves: f<109f < 10^9 Hz
  • Microwaves: 10910^9 - 101210^{12} Hz
  • Infrared: 101210^{12} - 101410^{14} Hz
  • Visible: 4×10144 \times 10^{14} - 7×10147 \times 10^{14} Hz
  • Ultraviolet: 101510^{15} - 101710^{17} Hz
  • X-rays: 101710^{17} - 101910^{19} Hz
  • Gamma rays: f>1019f > 10^{19} Hz

Standing EM Waves

Boundary conditions (e.g., in cavity) create standing waves:

Ey=2E0sin⁡(kx)cos⁡(ωt)E_y = 2E_0\sin(kx)\cos(\omega t)

Nodes: E=0E = 0 at x=0,λ/2,λ,...x = 0, \lambda/2, \lambda, ...

Used in:

  • Lasers
  • Microwave ovens
  • Radio antennas

Doppler Effect

Source moving with velocity vv:

Moving toward observer: f′=fcc−vf' = f\frac{c}{c - v}

Moving away: f′=fcc+vf' = f\frac{c}{c + v}

For v≪cv \ll c: Δff≈vc\frac{\Delta f}{f} \approx \frac{v}{c}

📚 Practice Problems

1Problem 1medium

❓ Question:

A plane electromagnetic wave in vacuum has electric field amplitude E₀ = 600 V/m and frequency f = 5.0 × 10¹⁴ Hz. Find: (a) the magnetic field amplitude B₀, (b) the wavelength λ, and (c) the intensity I of the wave.

💡 Show Solution

Given:

  • E₀ = 600 V/m
  • f = 5.0 × 10¹⁴ Hz
  • c = 3.0 × 10⁸ m/s
  • μ₀ = 4π × 10⁻⁷ T·m/A
  • ε₀ = 8.85 × 10⁻¹² F/m

(a) Magnetic field amplitude:

In EM waves: E0=cB0E_0 = cB_0

B0=E0c=6003.0×108B_0 = \frac{E_0}{c} = \frac{600}{3.0 \times 10^8}

B0=2.0×10−6 T=2.0 μTB_0 = \boxed{2.0 \times 10^{-6} \text{ T} = 2.0 \text{ μT}}

(b) Wavelength:

λ=cf=3.0×1085.0×1014\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}}

λ=6.0×10−7 m=600 nm\lambda = \boxed{6.0 \times 10^{-7} \text{ m} = 600 \text{ nm}}

This is orange/yellow visible light!

(c) Intensity:

I=12ε0cE02I = \frac{1}{2}\varepsilon_0 c E_0^2

I=12(8.85×10−12)(3.0×108)(600)2I = \frac{1}{2}(8.85 \times 10^{-12})(3.0 \times 10^8)(600)^2

I=12(8.85×10−12)(3.0×108)(3.6×105)I = \frac{1}{2}(8.85 \times 10^{-12})(3.0 \times 10^8)(3.6 \times 10^5)

I=477 W/m2I = \boxed{477 \text{ W/m}^2}

Alternatively: I=E0B02μ0=(600)(2.0×10−6)2(4π×10−7)=477I = \frac{E_0 B_0}{2\mu_0} = \frac{(600)(2.0 \times 10^{-6})}{2(4\pi \times 10^{-7})} = 477 W/m² ✓

2Problem 2medium

❓ Question:

A plane electromagnetic wave in vacuum has electric field amplitude E₀ = 600 V/m and frequency f = 5.0 × 10¹⁴ Hz. Find: (a) the magnetic field amplitude B₀, (b) the wavelength λ, and (c) the intensity I of the wave.

💡 Show Solution

Given:

  • E₀ = 600 V/m
  • f = 5.0 × 10¹⁴ Hz
  • c = 3.0 × 10⁸ m/s
  • μ₀ = 4π × 10⁻⁷ T·m/A
  • ε₀ = 8.85 × 10⁻¹² F/m

(a) Magnetic field amplitude:

In EM waves: E0=cB0E_0 = cB_0

B0=E0c=6003.0×108B_0 = \frac{E_0}{c} = \frac{600}{3.0 \times 10^8}

B0=2.0×10−6 T=2.0 μTB_0 = \boxed{2.0 \times 10^{-6} \text{ T} = 2.0 \text{ μT}}

(b) Wavelength:

λ=cf=3.0×1085.0×1014\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}}

λ=6.0×10−7 m=600 nm\lambda = \boxed{6.0 \times 10^{-7} \text{ m} = 600 \text{ nm}}

This is orange/yellow visible light!

(c) Intensity:

I=12ε0cE02I = \frac{1}{2}\varepsilon_0 c E_0^2

I=12(8.85×10−12)(3.0×108)(600)2I = \frac{1}{2}(8.85 \times 10^{-12})(3.0 \times 10^8)(600)^2

I=12(8.85×10−12)(3.0×108)(3.6×105)I = \frac{1}{2}(8.85 \times 10^{-12})(3.0 \times 10^8)(3.6 \times 10^5)

I=477 W/m2I = \boxed{477 \text{ W/m}^2}

Alternatively: I=E0B02μ0=(600)(2.0×10−6)2(4π×10−7)=477I = \frac{E_0 B_0}{2\mu_0} = \frac{(600)(2.0 \times 10^{-6})}{2(4\pi \times 10^{-7})} = 477 W/m² ✓

3Problem 3hard

❓ Question:

A laser beam with intensity I = 1.0 × 10⁴ W/m² is incident normally on a perfectly reflecting mirror of area A = 2.0 cm². Find: (a) the radiation pressure on the mirror, (b) the force on the mirror, and (c) compare this to the force if the mirror were perfectly absorbing.

💡 Show Solution

Given:

  • I = 1.0 × 10⁴ W/m²
  • A = 2.0 cm² = 2.0 × 10⁻⁴ m²
  • c = 3.0 × 10⁸ m/s
  • Perfectly reflecting

(a) Radiation pressure (reflecting):

For perfect reflection: P=2IcP = \frac{2I}{c}

P=2(1.0×104)3.0×108P = \frac{2(1.0 \times 10^4)}{3.0 \times 10^8}

P=6.67×10−5 PaP = \boxed{6.67 \times 10^{-5} \text{ Pa}}

(b) Force on mirror:

F=PA=(6.67×10−5)(2.0×10−4)F = PA = (6.67 \times 10^{-5})(2.0 \times 10^{-4})

F=1.33×10−8 N=13.3 nNF = \boxed{1.33 \times 10^{-8} \text{ N} = 13.3 \text{ nN}}

Very small! But significant for:

  • Solar sails in space
  • Optical tweezers (manipulating particles)
  • Radiation pressure from Sun on comet tails

(c) Perfectly absorbing:

For perfect absorption: Pabs=IcP_{abs} = \frac{I}{c}

Pabs=1.0×1043.0×108=3.33×10−5 PaP_{abs} = \frac{1.0 \times 10^4}{3.0 \times 10^8} = 3.33 \times 10^{-5} \text{ Pa}

Fabs=(3.33×10−5)(2.0×10−4)F_{abs} = (3.33 \times 10^{-5})(2.0 \times 10^{-4})

Fabs=6.67×10−9 N=6.67 nNF_{abs} = \boxed{6.67 \times 10^{-9} \text{ N} = 6.67 \text{ nN}}

Reflection produces twice the force as absorption!

Why? Momentum change:

  • Absorption: Δp = p (from p to 0)
  • Reflection: Δp = 2p (from +p to -p)

Physics: EM waves carry momentum p=U/cp = U/c where U is energy

4Problem 4hard

❓ Question:

A laser beam with intensity I = 1.0 × 10⁴ W/m² is incident normally on a perfectly reflecting mirror of area A = 2.0 cm². Find: (a) the radiation pressure on the mirror, (b) the force on the mirror, and (c) compare this to the force if the mirror were perfectly absorbing.

💡 Show Solution

Given:

  • I = 1.0 × 10⁴ W/m²
  • A = 2.0 cm² = 2.0 × 10⁻⁴ m²
  • c = 3.0 × 10⁸ m/s
  • Perfectly reflecting

(a) Radiation pressure (reflecting):

For perfect reflection: P=2IcP = \frac{2I}{c}

P=2(1.0×104)3.0×108P = \frac{2(1.0 \times 10^4)}{3.0 \times 10^8}

P=6.67×10−5 PaP = \boxed{6.67 \times 10^{-5} \text{ Pa}}

(b) Force on mirror:

F=PA=(6.67×10−5)(2.0×10−4)F = PA = (6.67 \times 10^{-5})(2.0 \times 10^{-4})

F=1.33×10−8 N=13.3 nNF = \boxed{1.33 \times 10^{-8} \text{ N} = 13.3 \text{ nN}}

Very small! But significant for:

  • Solar sails in space
  • Optical tweezers (manipulating particles)
  • Radiation pressure from Sun on comet tails

(c) Perfectly absorbing:

For perfect absorption: Pabs=IcP_{abs} = \frac{I}{c}

Pabs=1.0×1043.0×108=3.33×10−5 PaP_{abs} = \frac{1.0 \times 10^4}{3.0 \times 10^8} = 3.33 \times 10^{-5} \text{ Pa}

Fabs=(3.33×10−5)(2.0×10−4)F_{abs} = (3.33 \times 10^{-5})(2.0 \times 10^{-4})

Fabs=6.67×10−9 N=6.67 nNF_{abs} = \boxed{6.67 \times 10^{-9} \text{ N} = 6.67 \text{ nN}}

Reflection produces twice the force as absorption!

Why? Momentum change:

  • Absorption: Δp = p (from p to 0)
  • Reflection: Δp = 2p (from +p to -p)

Physics: EM waves carry momentum p=U/cp = U/c where U is energy

5Problem 5hard

❓ Question:

A plane EM wave traveling in the +x direction has electric field E⃗=E0sin⁡(kx−ωt)j^\vec{E} = E_0 \sin(kx - \omega t)\hat{j} where E₀ = 300 V/m, k = 1.0 × 10⁷ m⁻¹, and ω = 3.0 × 10¹⁵ rad/s. Find: (a) the magnetic field vector, (b) verify these satisfy v = ω/k = c, and (c) the Poynting vector and its time-averaged value.

💡 Show Solution

Given:

  • E⃗=E0sin⁡(kx−ωt)j^\vec{E} = E_0 \sin(kx - \omega t)\hat{j}
  • E₀ = 300 V/m
  • k = 1.0 × 10⁷ m⁻¹
  • ω = 3.0 × 10¹⁵ rad/s
  • Direction: +x
  • c = 3.0 × 10⁸ m/s
  • μ₀ = 4π × 10⁻⁷ T·m/A

(a) Magnetic field:

For EM wave: E⃗⊥B⃗⊥\vec{E} \perp \vec{B} \perp direction

E is in +y direction, wave travels in +x, so B must be in ±z direction.

Using right-hand rule (E⃗×B⃗\vec{E} \times \vec{B} points in propagation direction): j^×k^=i^\hat{j} \times \hat{k} = \hat{i} ✓

Amplitude: B0=E0/c=300/(3.0×108)=1.0×10−6B_0 = E_0/c = 300/(3.0 \times 10^8) = 1.0 \times 10^{-6} T

B⃗=B0sin⁡(kx−ωt)k^=(1.0×10−6)sin⁡(kx−ωt)k^ T\boxed{\vec{B} = B_0 \sin(kx - \omega t)\hat{k} = (1.0 \times 10^{-6})\sin(kx - \omega t)\hat{k} \text{ T}}

(b) Verify wave speed:

v=ωk=3.0×10151.0×107v = \frac{\omega}{k} = \frac{3.0 \times 10^{15}}{1.0 \times 10^7}

v=3.0×108 m/s=cv = 3.0 \times 10^8 \text{ m/s} = c ✓

Also check: c=1μ0ε0=3.0×108c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3.0 \times 10^8 m/s ✓

(c) Poynting vector:

S⃗=1μ0E⃗×B⃗\vec{S} = \frac{1}{\mu_0}\vec{E} \times \vec{B}

S⃗=1μ0E0B0sin⁡2(kx−ωt)(j^×k^)\vec{S} = \frac{1}{\mu_0}E_0 B_0 \sin^2(kx - \omega t)(\hat{j} \times \hat{k})

S⃗=E0B0μ0sin⁡2(kx−ωt)i^\vec{S} = \frac{E_0 B_0}{\mu_0}\sin^2(kx - \omega t)\hat{i}

S⃗=(300)(1.0×10−6)4π×10−7sin⁡2(kx−ωt)i^\vec{S} = \frac{(300)(1.0 \times 10^{-6})}{4\pi \times 10^{-7}}\sin^2(kx - \omega t)\hat{i}

S⃗=(239)sin⁡2(kx−ωt)i^ W/m2\boxed{\vec{S} = (239)\sin^2(kx - \omega t)\hat{i} \text{ W/m}^2}

Time-averaged value:

⟨sin⁡2(kx−ωt)⟩=12\langle \sin^2(kx - \omega t) \rangle = \frac{1}{2}

⟨S⃗⟩=12E0B0μ0i^=12ε0cE02i^\langle \vec{S} \rangle = \frac{1}{2}\frac{E_0 B_0}{\mu_0}\hat{i} = \frac{1}{2}\varepsilon_0 c E_0^2 \hat{i}

⟨S⃗⟩=120i^ W/m2\boxed{\langle \vec{S} \rangle = 120\hat{i} \text{ W/m}^2}

This is the intensity I!

6Problem 6hard

❓ Question:

A plane EM wave traveling in the +x direction has electric field E⃗=E0sin⁡(kx−ωt)j^\vec{E} = E_0 \sin(kx - \omega t)\hat{j} where E₀ = 300 V/m, k = 1.0 × 10⁷ m⁻¹, and ω = 3.0 × 10¹⁵ rad/s. Find: (a) the magnetic field vector, (b) verify these satisfy v = ω/k = c, and (c) the Poynting vector and its time-averaged value.

💡 Show Solution

Given:

  • E⃗=E0sin⁡(kx−ωt)j^\vec{E} = E_0 \sin(kx - \omega t)\hat{j}
  • E₀ = 300 V/m
  • k = 1.0 × 10⁷ m⁻¹
  • ω = 3.0 × 10¹⁵ rad/s
  • Direction: +x
  • c = 3.0 × 10⁸ m/s
  • μ₀ = 4π × 10⁻⁷ T·m/A

(a) Magnetic field:

For EM wave: E⃗⊥B⃗⊥\vec{E} \perp \vec{B} \perp direction

E is in +y direction, wave travels in +x, so B must be in ±z direction.

Using right-hand rule (E⃗×B⃗\vec{E} \times \vec{B} points in propagation direction): j^×k^=i^\hat{j} \times \hat{k} = \hat{i} ✓

Amplitude: B0=E0/c=300/(3.0×108)=1.0×10−6B_0 = E_0/c = 300/(3.0 \times 10^8) = 1.0 \times 10^{-6} T

B⃗=B0sin⁡(kx−ωt)k^=(1.0×10−6)sin⁡(kx−ωt)k^ T\boxed{\vec{B} = B_0 \sin(kx - \omega t)\hat{k} = (1.0 \times 10^{-6})\sin(kx - \omega t)\hat{k} \text{ T}}

(b) Verify wave speed:

v=ωk=3.0×10151.0×107v = \frac{\omega}{k} = \frac{3.0 \times 10^{15}}{1.0 \times 10^7}

v=3.0×108 m/s=cv = 3.0 \times 10^8 \text{ m/s} = c ✓

Also check: c=1μ0ε0=3.0×108c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3.0 \times 10^8 m/s ✓

(c) Poynting vector:

S⃗=1μ0E⃗×B⃗\vec{S} = \frac{1}{\mu_0}\vec{E} \times \vec{B}

S⃗=1μ0E0B0sin⁡2(kx−ωt)(j^×k^)\vec{S} = \frac{1}{\mu_0}E_0 B_0 \sin^2(kx - \omega t)(\hat{j} \times \hat{k})

S⃗=E0B0μ0sin⁡2(kx−ωt)i^\vec{S} = \frac{E_0 B_0}{\mu_0}\sin^2(kx - \omega t)\hat{i}

S⃗=(300)(1.0×10−6)4π×10−7sin⁡2(kx−ωt)i^\vec{S} = \frac{(300)(1.0 \times 10^{-6})}{4\pi \times 10^{-7}}\sin^2(kx - \omega t)\hat{i}

S⃗=(239)sin⁡2(kx−ωt)i^ W/m2\boxed{\vec{S} = (239)\sin^2(kx - \omega t)\hat{i} \text{ W/m}^2}

Time-averaged value:

⟨sin⁡2(kx−ωt)⟩=12\langle \sin^2(kx - \omega t) \rangle = \frac{1}{2}

⟨S⃗⟩=12E0B0μ0i^=12ε0cE02i^\langle \vec{S} \rangle = \frac{1}{2}\frac{E_0 B_0}{\mu_0}\hat{i} = \frac{1}{2}\varepsilon_0 c E_0^2 \hat{i}

⟨S⃗⟩=120i^ W/m2\boxed{\langle \vec{S} \rangle = 120\hat{i} \text{ W/m}^2}

This is the intensity I!

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