EMR eqs

Question 1

Maxwell’s eqs (+ free space)

Answer 1

∮ₛ E.dS = Qₑₙ꜀/∈₀
∮ₛ B.dS = 0
∮꜀ E.dL = -d/dt ∫ₛB.dS
∮꜀ B.dL = µ₀Iₑₙ꜀ = µ₀∫ₛ(j+∈₀ ∂E/∂t).dA
———————————-
∇.E = ρ/ϵ₀
∇.B = 0
∇ × E = −∂B/∂t
∇ × B = µ₀(J+ϵ₀ ∂E/∂t)
———–Free Space————————–
∇.E = 0
∇.B = 0
∇ × E = −∂B/∂t
∇ × B = µ₀ϵ₀ ∂E/∂t

Gauss, no magnetic monopoles, Faraday, Ampere

Question 2

Maxwell in a medium

Answer 2

∇ · D = ρf
∇ · B = 0
∇ × E = –∂B/∂t
∇ × H = jf + ∂D/∂t
{}{}{}{}{}{}{}{}{}{}{}
–∇ · P = ρb
P · n̂ = σb
∇ × M = Jb
M · n̂ = K
{}{}{}{}{}{}{}{}{}{}{}
P = ε₀χₑ E
D = ε₀εᵣ E
M = χₘH
H = B/μ₀μᵣ
{}{}{}{}{}{}{}{}{}{}{}
μᵣ = 1 + χₘ
εᵣ = 1 + χₑ

Question 3

Electric and magnetic fields in relation to each other

Answer 3

D = ε₀E + P

H = 1/μ₀ B – M

Question 4

Lorentz force

Answer 4

F = q(E + v × B)

Question 5

Static potentials

Answer 5

Static:
E = −∇φ
⇒ ∇²φ = −ρ/ε₀

B = ∇ × A
⇒ ∇²A = −μ₀J

Question 6

Dynamic potentials

Answer 6

Dynamic:
E = −∇Φ − ∂A/∂t
B = ∇ × A
^ uses Coulomb gauge ∇ · A = 0
Lorentz gauge:
∇ · A = –μ₀ε₀ ∂Φ/∂t

Question 7

Inhomogeneous wave equations relating potentials, charges and currents.

Answer 7

∇²A − μ₀ε₀ ∂²A/∂t² = −μ₀J
∇²Φ − μ₀ε₀ ∂²Φ/∂t² = −ρ/ε₀

Question 8

Plane wave equations

Answer 8

∇²E = 1/c² ∂²E/∂t²

∇²B = 1/c² ∂²B/∂t²

where c = 1/√µ₀ε₀’

{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}

Solutions are
E = E₀exp(i(ωt±kz))
B = B₀exp(i(ωt±kz))

Question 9

Relations in plane waves

Answer 9

c = ω/k

cB₀ = E₀

B = 1/ω k × E

Question 10

Refractive index

Answer 10

n = √εᵣμᵣ’

μᵣ is normally 1

Question 11

Energy in an EM field

Answer 11

Energy per unit volume

UE = ε₀/2 E²
(= ε₀/2 E₀² cos²(ωt−kz))

UB = 1/2µ₀ B²
(= 1/2µ₀ B₀² cos²(ωt−kz))

Total energy
U = ∫ UE + UB dV

In plane EM wave, equal energy density in E and B fields, UE = UB

Question 12

Poynting vector

Answer 12

S = 1/µ₀ (E × B)

Question 13

Power from energy

Answer 13

Time average of energy, ⟨U⟩

Question 14

Size of Poynting vector for plane wave

Answer 14

If |E| = c|B|

|S| = |E|²/μ₀c

Question 15

Delayed time and potentials

Answer 15

τ = t − |r−r′|/c

Φ(r, t) = 1/4πε₀ ∫ᵥ ρ(r′,τ)/|r−r′|dV′

A(r, t) = μ₀/4π ∫ᵥ J(r′,τ)/|r−r′|dV′

Question 16

E, B and S for accelerated charges

Answer 16

E⊥ = qa⊥/4πε₀rc²
B = r×E/c
|S(r̂,t)| = q²a²(τ)sin²θ/16π²ε₀r²c³

Question 17

Larmor formula

Answer 17

P(t) = q²a²(τ)/6πε₀c³

Question 18

Oscillating current Larmor

Answer 18

qa(t′) = μa(t′)l
= μ dv/dt′ l
= d(μv)/dt′ l

μ = current per unit length

μv can be written in the form
I(t′) = I₀ cos(ωt)

Rewrite qa(t′)
qa(t′) = −I₀lω sin(ωt′)
= −I₀lω sin(ω(t−r/c))
= −I₀lω sin(ωt−kr))

P(t) = I₀²l²ω²/6πε₀c³ sin²(ωt−kr))

Question 19

Hertzian Larmor

Answer 19

Oscillating current Larmor:
P(t) = I₀²l²ω²/6πε₀c³ sin²(ωt−kr))

For Hertzian dipole, I₀lω = P₀ω²
P(t) = P₀²ω⁴/6πε₀c³ sin²(ωt−kr))

Question 20

Radiation resistance

Answer 20

Rrad = l²ω²/6πε₀c³

Can be written pther ways by remembering
ω = 2πc/λ and
c² = 1/(ε₀μ₀)

Question 21

Free-space impedance

Answer 21

Z₀ = μ₀c = 1/ε₀c ≃ 377ohm. (2.40)

Z₀ is often given approximate form:
2πZ₀/3 ≃ 80π²

Question 22

Half wave antenna fields

Answer 22

E_θ(r,t) = –I₀/2πε₀c cos[(πcosθ)/2]/rsinθ

B_φ = Eθ/c

Question 23

Dispersion relation in a dielectric

Answer 23

k² = εᵣε₀μᵣμ₀ω² – iμᵣμ₀σω

Question 24

Derive n from oscillatory behaviour

Answer 24

Oscillator under Lorentz force
ẍ + βẋ + ω₀x = q/m [E + ẋ × B]

Solved by
x = (q/m)E / (ω₀²–ω²)+iβω

p = qx
P = ∑ᵢ N fᵢpᵢ

P = χₑε₀E
χₑ = εᵣ – 1
n = √εᵣ’
n = nᵣ – inᵢ

Question 25

Derive the complex dispersion relation

Answer 25

Plane wave travelling in Z direction
E = (Eₓ, 0, 0)
B = (0, Bᵧ, 0)

Eₓ = E₀exp(i(ωt−kz))
Bᵧ = B₀exp(i(ωt−kz))

Faraday’s law
−ikE₀exp(i(ωt−kz)) = −iωB₀exp(i(ωt−kz))
B₀/E₀ = k/ω

Ampere’s and Ohm’s laws
−ikB₀exp(i(ωt−kz)) = −(iωεᵣε₀μᵣμ₀ + μᵣμ₀σ) E₀exp(i(ωt−kz))

Combine using B₀ = E₀/c
k² = ω²εᵣε₀μᵣμ₀ − iωμᵣμ₀σ

Question 26

Summarise the boundary conditions at the surface of a conductor

Answer 26

n̂ · E = ρₛ/ε₀ → normal discontinuous
n̂ × E = 0 → transverse continuous
n̂ · B = 0 → normal continuous
n̂ × B = μ₀J_z → transverse discontinuous

Question 27

Plasma frequency

Answer 27

ωₚ = √Ne²/ε₀m’

Question 28

DC conductivity in plasmas

Answer 28

σ₀ = Ne²/mγ꜀ = ε₀ωₚ²/γ꜀
where γ꜀ = 1/τ꜀ is collision rate (frequency)

Question 29

Frequency-dependent conductivity σ(ω) in terms of σ₀

Answer 29

σ(ω) = σ₀/(1+ iω/γ꜀)

Question 30

Skin depth and wavelength for poor conductors

Answer 30

Condition:
σ/ωεᵣε₀ ≪ 1

Dieletric dispersion relation
k꜀² = εᵣμᵣ ω²/c² (1 − iσ/ωεᵣε₀)

Taylor expand √1 + x’ ≃ 1 + x/2

Skin depth
δ = 1/|kᵢ| = 2/σ √εᵣε₀/μᵣμ₀’

Wavelength
λ = 2πc/ω 1/√εᵣμᵣ’

Question 31

Skin depth for good conductors

Answer 31

Condition:
σ/ωεᵣε₀ ≫ 1

Skin depth
δ = 1/|kᵢ| = √2/μᵣμ₀ωσ’

Question 32

Phase change at a boundary

Answer 32

n₂ > n₁
(air to glass)
π phase shift

n₂ < n₁
(glass to air)
no phase shift

Question 33

Fields with normal incidence on a dielectric

Answer 33

B = E/v = nE/c = E√εᵣε₀μᵣμ₀’

Question 34

Reflected and transmitted field amplitudes, normal reflection in a dielectric

Answer 34

• Hₜ₁ = Hₜ₂ (no current at the boundary), so
  Bᵢ₀/μ₁ + Bᵣ₀/μ₁ = Bₜ₀/μ₂
• Know B = E/v = E√ε₀εᵣμ₀μᵣ’ so
  B/μᵣ = E√ε₀εᵣμ₀μᵣ’ 1/μᵣ = 1/c √εᵣ/μᵣ’ E
• Sub in
  √ε₁/μ₁’ Eᵢ₀ + √ε₁/μ₁’ Eᵣ₀ = √ε₂/μ₂’ Et₀
• No magnetic properties in dielectrics,
  μ₁ = μ₂ = 1
• Take √ε₁’ = n₁ and √ε₂’ = n₂, use boundary condition Eᵢ₀ − Eᵣ₀ = Eₜ₀
  n₁(Eᵢ₀ + Eᵣ₀) = n₂Eₜ₀
  = n₂(Eᵢ₀ − Eᵣ₀)
• Reflected electric wave amplitude
  Eᵣ₀/Eᵢ₀ = n₂−n₁/n₂+n₁
• For magnetic field, use boundary conditions, Bᵢ₀ + Bᵣ₀ = Bₜ₀, Eᵢ₀ + Eᵣ₀ = Eₜ₀, as well as general dielectric relation B = 1/c √ε’ E
• ** Reflected magnetic field amplitude**
  Bᵣ₀/Bᵢ₀ = n₂−n₁/n₂+n₁
• Transmitted fraction of amplitudes
  Bᵣ₀/Bᵢ₀ + Bₜ₀/Bᵢ₀ = 1 and
  Eᵣ₀/Eᵢ₀ + Eₜ₀/Eᵢ₀ = 1
  get
  Bₜ₀/Bᵢ₀ = Eₜ₀/Eᵢ₀ = 2n₁/n₂ + n₁

Do NOT confuse reflected and transmitted field amplitudes with intensity.₁

Question 35

Intensity of transmitted and reflected waves, normal to a dielectric

Answer 35

〈Sᵢ〉= 1/μ₀〈Eᵢ × Bᵢ〉
where Eᵢ = Eᵢ₀ cos(ωt−kz)ẑ
and Bᵢ = Bᵢ₀ cos(ωt−kz)ẑ

B = 1/c √εᵣ’ E

〈Sᵢ〉= ε₀c/2 Eᵢ₀² n₁
where 1/2 is average of cos² and n₁ = √ε₁’

Can write Poynting vector using coefficients calculated for the field amplitudes,
〈Sᵣ〉= ε₀c/2 Eᵢ₀² (n₂−n₁)²/(n₂+n₁)² n₁
〈Sₜ〉= ε₀c/2 Eᵢ₀² 4n₁²/(n₂+n₁)² n₂

Reflected and transmitted portions of intensity are:
* T =〈Sₜ〉/〈Sᵢ〉
= (4n₁²/(n₂+n₁)² n₂) / n₁
= 4n₁n₂/(n₂+n₁)²
* R =〈Sᵣ〉/〈Sᵢ〉
= ((n₂−n₁)²/(n₂+n₁)² n₁) / n₁
= (n₂−n₁)²/(n₂+n₁)²

• R + T = 1

Question 36

Refractive index of a conductor

Answer 36

n₂ = √εᵣ’
= √−iσ/ωε₀’
≃ √σ/2ωε₀’ (1−i)

Question 37

Reflection and transmission coeffiscients for low frequency plasmas

Answer 37

Refractive index
n₂ ≃ √−iσ/ωε₀’
= √σ/2ωε₀’ (1−i)
= α(1−i)
with α = √σ/2ωε₀’ ≫ 1 for good conductors.

R = |n₂−n₁|² / |n₂+n₁|²
=|( (α−n₁)−iα / (α+n₁)−iα )|²
using |a−ib| = √a²+b²’
= (1− n₁/α)²+1 / (1+ n₁/α)²+1
= ( 2 − 2n₁/α + n₁²/α² ) / ( 2 + 2n₁/α + n₁²/α² )

Use α ≫ 1

R = (1 − n₁/α) (1 + n₁/α)⁻¹

Taylor expand 1−x/1+x ≃ 1−2x+…
≃ 1 − 2n₁/α

In a conductor, ωε₀/σ ≪ 1, and
* R ≃ 1 − 2n₁√2ωε₀/σ’
* T = 1 − R ≃ 2n₁√2ωε₀/σ’

Question 38

Reflection and transmission for high frequency waves

Answer 38

Dispersion relation in a good conductor
k = ± 1/c √ω²−ωₚ²’

• ω ≪ ωₚ
  Evanescent wave, does not penetrate. εᵣ < 0, k imaginary, T = 0 and R = 1
• ω > ωₚ
  Metal transparent. εᵣ > 0, k is real, T→1, R→0 at high frequencies.

Question 39

S- and P- polarisation

Answer 39

Orthogonal components
* S-polarisation has E is normal to the plane of incidence
* P-polarisation has E is in the plane of incidence

Question 40

Derive the Fresnel equation for P-polarised light

Answer 40

Take μ₁ = μ₂ = 1.
* B is perpendicular to plane of incidence, boundary condition is
Bᵢ₀ + Bᵣ₀ = Bₜ₀
Recall B = n E/c
n₁Eᵢ₀ + n₁Eᵣ₀ = n₂Eₜ₀
* Parallel E component is continuous at boundary,
Eᵢ₀ cos θᵢ − Eᵣ₀ cosθᵢ = Eₜ₀ cosθₜ

Eₜ₀ = n₁/n₂ (Eᵢ₀ + Eᵣ₀)

Using cosθₜ = √1−sin²θₜ’
and Snell’s Law
get reflection component Fresnel equation
ρ‖ = Eᵣ₀/Eᵢ₀ |‖
= (n₂cosθᵢ − n₁cosθₜ) / (n₁cosθₜ + n₂cosθᵢ)
= (n²cosθᵢ − √n²−sin²θᵢ’) / (n²cosθᵢ + √n²−sin²θᵢ’)

Only valid for a non-magnetic medium,
μ₁ = μ₂ = 1

Question 41

Derive the Fresnel equation for S-polarised light

Answer 41

S-polarised component of E is perpendicular to plane of incidence, so no cosθ terms in continuity equations.
* First boundary condition using E‖ is continuous:
Eᵢ₀ − Eᵣ₀ = Eₜ₀
* Second boundary condition using H‖ is continuous and H‖ = B‖/μᵣ:
Bᵢ₀/μ₁ cosθᵢ − Bᵣ₀/μ₁ cosθᵢ = Bₜ₀/μ₂ cosθt
where μ₁ and μ₂ are relative permeabilities, set to μ₁ = 1 and μ₂ = 1.

Use H = nE/c to find
n₁/c Eᵢ₀ cosθᵢ − n₁/c Eᵣ₀ cosθᵢ = n₂/c Eₜ₀ cos θₜ

Eᵣ/Eᵢ = − (n₁ cosθᵢ − n₂ cosθₜ)/(n₁ cosθᵢ + n₂ cosθₜ)

Using cos θₜ = √1−sin²θₜ’
and Snell’s law sinθₜ = (n₁/n₂) sinθᵢ
get reflected component Fresnel equation
ρ⊥ ≡ Eᵣ₀/Eᵢ₀ |⊥
= (n₁cosθᵢ − n₂cosθₜ) / (n₁cosθᵢ + n₂cosθₜ)
= (cosθᵢ − √n²−sin²θᵢ’) / (cosθᵢ + √n²−sin²θᵢ’)

where n = n₂/n₁

The Fresnel equation for the transmitted component can be deduced by τ⊥ = 1 − ρ⊥.