EMR eqs Flashcards
Maxwell’s eqs (+ free space)
∮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
Maxwell in a medium
∇ · D = ρf
∇ · B = 0
∇ × E = –∂B/∂t
∇ × H = jf + ∂D/∂t
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–∇ · P = ρb
P · n̂ = σb
∇ × M = Jb
M · n̂ = K
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P = ε₀χₑ E
D = ε₀εᵣ E
M = χₘH
H = B/μ₀μᵣ
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μᵣ = 1 + χₘ
εᵣ = 1 + χₑ
Electric and magnetic fields in relation to each other
D = ε₀E + P
H = 1/μ₀ B – M
Lorentz force
F = q(E + v × B)
Static potentials
Static:
E = −∇φ
⇒ ∇²φ = −ρ/ε₀
B = ∇ × A
⇒ ∇²A = −μ₀J
Dynamic potentials
Dynamic:
E = −∇Φ − ∂A/∂t
B = ∇ × A
^ uses Coulomb gauge ∇ · A = 0
Lorentz gauge:
∇ · A = –μ₀ε₀ ∂Φ/∂t
Inhomogeneous wave equations relating potentials, charges and currents.
∇²A − μ₀ε₀ ∂²A/∂t² = −μ₀J
∇²Φ − μ₀ε₀ ∂²Φ/∂t² = −ρ/ε₀
Plane wave equations
∇²E = 1/c² ∂²E/∂t²
∇²B = 1/c² ∂²B/∂t²
where c = 1/√µ₀ε₀’
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Solutions are
E = E₀exp(i(ωt±kz))
B = B₀exp(i(ωt±kz))
Relations in plane waves
c = ω/k
cB₀ = E₀
B = 1/ω k × E
Refractive index
n = √εᵣμᵣ’
μᵣ is normally 1
Energy in an EM field
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
Poynting vector
S = 1/µ₀ (E × B)
Power from energy
Time average of energy, ⟨U⟩
Size of Poynting vector for plane wave
If |E| = c|B|
|S| = |E|²/μ₀c
Delayed time and potentials
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′
E, B and S for accelerated charges
E⊥ = qa⊥/4πε₀rc²
B = r×E/c
|S(r̂,t)| = q²a²(τ)sin²θ/16π²ε₀r²c³
Larmor formula
P(t) = q²a²(τ)/6πε₀c³
Oscillating current Larmor
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))
Hertzian Larmor
Oscillating current Larmor:
P(t) = I₀²l²ω²/6πε₀c³ sin²(ωt−kr))
For Hertzian dipole, I₀lω = P₀ω²
P(t) = P₀²ω⁴/6πε₀c³ sin²(ωt−kr))
Radiation resistance
Rrad = l²ω²/6πε₀c³
Can be written pther ways by remembering
ω = 2πc/λ and
c² = 1/(ε₀μ₀)
Free-space impedance
Z₀ = μ₀c = 1/ε₀c ≃ 377ohm. (2.40)
Z₀ is often given approximate form:
2πZ₀/3 ≃ 80π²
Half wave antenna fields
E_θ(r,t) = –I₀/2πε₀c cos[(πcosθ)/2]/rsinθ
B_φ = Eθ/c
Dispersion relation in a dielectric
k² = εᵣε₀μᵣμ₀ω² – iμᵣμ₀σω
Derive n from oscillatory behaviour
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
