(a) (i) Explain the T-s diagram for the reversible Carnot cycle and hence obtain the expression for the efficiency of the Carnot…

(a)(i) On a T-s diagram, the reversible Carnot cycle appears as two horizontal isotherms and two vertical isentropes.

• 1→2: isothermal reversible expansion at T_H; entropy increases from s₁ to s₂; heat absorbed Q_H = T_H(s₂ - s₁).
• 2→3: adiabatic reversible expansion; entropy constant, so s₃ = s₂, temperature falls T_H to T_C.
• 3→4: isothermal reversible compression at T_C; entropy decreases from s₂ to s₁; heat rejected Q_C = T_C(s₂ - s₁).
• 4→1: adiabatic reversible compression; entropy constant, temperature rises T_C to T_H.

Net work: W = Q_H - Q_C = (T_H - T_C)(s₂ - s₁). Efficiency: η = W/Q_H = [(T_H - T_C)(s₂ - s₁)]/[T_H(s₂ - s₁)] = 1 - T_C/T_H. Thus η = 1 - T_C/T_H, valid for a reversible Carnot engine with Kelvin temperatures.

(a)(ii) For m gram of solid, dQ = m Cᵥ dT = m A T³ dT. Q = ∫₃₀₀⁵⁰⁰ m A T³ dT = m A [T⁴/4]₃₀₀⁵⁰⁰ = (m A/4)(500⁴ - 300⁴) = (m A/4)(6.25×10¹⁰ - 8.1×10⁹) = (m A/4)(5.44×10¹⁰) = 1.36×10¹⁰ m A. So Q = 1.36×10¹⁰ m A, in the same energy unit as Cᵥ, if A is per gram per K⁴.

(b) Let n̂ be the unit normal directed from medium 1 to medium 2, σ the free surface charge density, and K the free surface current density. Use Maxwell’s integral laws. For a pillbox of vanishing thickness, Gauss’s law for D gives ∮ D·dA = Q_f ⟹ n̂·(D₂ - D₁) = σ. Similarly, ∮ B·dA = 0 gives n̂·(B₂ - B₁) = 0. For a small rectangular loop crossing the boundary, Faraday’s law gives, as area → 0, n̂ × (E₂ - E₁) = 0. Ampere–Maxwell’s law gives, with displacement-current flux vanishing as area → 0, n̂ × (H₂ - H₁) = K. Therefore the general boundary conditions are:

• n̂·(D₂ - D₁) = σ, so D₂n - D₁n = σ.
• n̂·(B₂ - B₁) = 0, so B₂n = B₁n.
• n̂ × (E₂ - E₁) = 0, so E₂t = E₁t.
• n̂ × (H₂ - H₁) = K, equivalently H₂t - H₁t = K × n̂. Thus normal D jumps by σ, normal B is continuous, tangential E is continuous, and tangential H jumps by K. If σ = 0 and K = 0, D_n and H_t are also continuous.

(c)(i) Given εᵣ = 4, μᵣ = 1, σ = 0. v = 1/√(με) = c/√(εᵣ μᵣ) = c/2 = 1.5×10⁸ m/s. ω = 2πf = 2π×10⁸ rad/s. β = ω/v = (2π×10⁸)/(1.5×10⁸) = 4π/3 rad/m. Take Eₓ = E₀ cos(ωt - βz + φ) aₓ. At t = 0, z = 1/8 m, Eₓ = +10⁻⁴ V/m = maximum. So -β/8 + φ = 2πn. Since β/8 = π/6, choose φ = π/6. Thus E = 10⁻⁴ cos(2π×10⁸ t - (4π/3)z + π/6) aₓ V/m.

(c)(ii) Intrinsic impedance: η = √(μ/ε) = √(μ₀/(4ε₀)) = η₀/2 = 60π Ω. For propagation along a_z, H = (1/η) a_z × E = (Eₓ/η) a_y. Hence H = (10⁻⁴/(60π)) cos(2π×10⁸ t - (4π/3)z + π/6) a_y A/m. The magnetic-field amplitude is 10⁻⁴/(60π) A/m ≈ 5.31×10⁻⁷ A/m.

(c)(iii) At t = 10⁻⁸ s, ωt = (2π×10⁸)(10⁻⁸) = 2π rad. For positive maximum, cos(phase) = 1: 2π - (4π/3)z + π/6 = 2πn. This gives z = 13/8 - (3/2)n m, n integer. For positive z, the locations are z = 1/8 + (3/2)k m, k = 0, 1, 2, … i.e. z = 1/8, 13/8, 25/8, … m. The spacing is the wavelength λ = 2π/β = 3/2 m. Since the period is T = 1/f = 10⁻⁸ s, the pattern at t = 10⁻⁸ s is the same as at t = 0; in an unbounded medium the maxima lie at z = 1/8 + 1.5k m for any integer k.