(a) In a partially conducting medium, εᵣ = 18.5, μᵣ = 800 and σ = 1 S m⁻¹. Find α, β, η and the velocity u, for a frequency of…

(a) Interpreting the given phase term as β z, the field is E(z,t) = 50 e^(−α z) cos(ωt − β z) a_y V/m.

Using μ₀ = 4π×10⁻⁷ H/m and ε₀ = 8.854×10⁻¹² F/m:

ε = ε₀ εᵣ = 8.854×10⁻¹² × 18.5 = 1.638×10⁻¹⁰ F/m μ = μ₀ μᵣ = 4π×10⁻⁷ × 800 = 1.005×10⁻³ H/m ω = 2π×10⁹ = 6.283×10⁹ rad/s

Loss tangent: p = σ/(ωε) = 1/(6.283×10⁹ × 1.638×10⁻¹⁰) = 0.9716

For a lossy dielectric, using γ = √(jωμ(σ + jωε)) = α + jβ:

α = ω√(με/2) [√(1 + p²) − 1]^(1/2) β = ω√(με/2) [√(1 + p²) + 1]^(1/2)

Now √(με) = 4.058×10⁻⁷ s/m, so ω√(με) = 6.283×10⁹ × 4.058×10⁻⁷ = 2549.7 rad/m √(1 + p²) = 1.3943

Thus α = 2549.7 × 0.4440 = 1132 Np/m β = 2549.7 × 1.09414 = 2789.7 rad/m

Phase velocity: u = ω/β = 6.283×10⁹/2789.7 = 2.252×10⁶ m/s

Intrinsic impedance: η = √(μ/ε)/(1 − jp)^(1/2)

√(μ/ε) = 2477 Ω |η| = 2477/(1 + p²)^(1/4) = 2098 Ω θ = (1/2) tan⁻¹(p) = 0.3855 rad = 22.09°

So η = 2098 ∠22.09° Ω = 1944 + j789 Ω

For E along y and propagation along +z, Maxwell’s equations give H = −(1/η) E a_x

Therefore H(z,t) = −(50/2098) e^(−1132 z) cos(6.283×10⁹ t − 2789.7 z − 0.3855) a_x A/m

i.e. H(z,t) = −0.02383 e^(−1132 z) cos(2π×10⁹ t − 2789.7 z − 0.3855) a_x A/m.

(b) Negative temperature arises from the thermodynamic definition 1/T = (∂S/∂U)_(V,N). If entropy S decreases as internal energy U increases, then ∂S/∂U < 0 and hence T < 0. It is not a temperature below absolute zero; a negative-temperature system is hotter than any positive-temperature system. Heat flows from a negative-temperature body to a positive-temperature body.

For negative temperature to be meaningful, the following restrictions are essential:

• The system must have a bounded energy spectrum. If energy can increase without limit, the partition function diverges for T < 0 and no equilibrium negative temperature exists.
• The system must be in internal thermal equilibrium among a restricted set of degrees of freedom, usually spin or other bounded degrees of freedom.
• The ordinary unbounded translational or kinetic degrees must be effectively decoupled or absent; otherwise they absorb energy and destroy negative temperature.
• The population distribution must be inverted: higher-energy states are more populated than lower-energy states. For T < 0, the Boltzmann factor exp(−E/kT) increases with E.
• The system must be sufficiently isolated from positive-temperature reservoirs, or prepared by a fast non-adiabatic process such as laser pumping or rapid spin inversion.
• The density of states must fall sufficiently or the number of states must be finite so that the partition function converges.

Thus negative temperature is a stable thermodynamic concept only for bounded, internally equilibrated systems with population inversion.

(c) Laplace’s equation in cylindrical polar coordinates is (1/r) ∂/∂r(r ∂V/∂r) + (1/r²) ∂²V/∂φ² + ∂²V/∂z² = 0.

Let V(r,φ,z) = R(r) Φ(φ) Z(z).

Substituting and dividing by RΦZ: (1/(rR)) d/dr(r dR/dr) + (1/(r²Φ)) d²Φ/dφ² + (1/Z) d²Z/dz² = 0.

Separate the z-part: (1/Z) d²Z/dz² = k_z².

Hence d²Z/dz² − k_z² Z = 0.

The remaining equation is (1/(rR)) d/dr(r dR/dr) + (1/(r²Φ)) d²Φ/dφ² + k_z² = 0.

Multiply by r²: (r/R) d/dr(r dR/dr) + (1/Φ) d²Φ/dφ² + k_z² r² = 0.

Separate the φ-part: (1/Φ) d²Φ/dφ² = −n².

Thus d²Φ/dφ² + n² Φ = 0.

Single-valuedness of V requires Φ(φ + 2π) = Φ(φ), so n = 0, ±1, ±2, ...

The radial equation becomes r² d²R/dr² + r dR/dr + (k_z² r² − n²) R = 0.

Therefore the separated differential equations are:

d²Z/dz² − k_z² Z = 0 d²Φ/dφ² + n² Φ = 0 r² d²R/dr² + r dR/dr + (k_z² r² − n²) R = 0

The radial equation is Bessel’s equation of order n. If the separation constant is chosen with the opposite sign, the radial equation becomes the modified Bessel equation.