(a) (i) In free space, an electric field (E⃗) is given by the following expression : E⃗ = 10 cos (ω t - 100 x)ĵ V/m Find the…

(a)(i)

Given E = 10 cos(ωt − 100x) ĵ V/m. Comparing with E = E₀ cos(ωt − kx) ĵ, we get k = 100 rad/m. For a free-space plane wave, c = ω/k = 1/√(μ₀ε₀) = 3.00×10⁸ m/s. Hence ω = c k = (3.00×10⁸)(100) = 3.00×10¹⁰ rad/s.

Maxwell–Ampère law in free space gives J_d = ∂D/∂t = ε₀ ∂E/∂t. ∂E/∂t = −10ω sin(ωt − 100x) ĵ. Thus J_d = −ε₀(10)(3.00×10¹⁰) sin(ωt − 100x) ĵ = −2.656 sin(3.00×10¹⁰ t − 100x) ĵ A/m². This is the displacement current density. Through area A, the total displacement current is I_d = J_d A. Validity: free space, no conduction current.

(a)(ii)

For a plane electromagnetic wave in free space, |E| = c|B|. The Poynting vector is S = (1/μ₀) E × B, so |S| = |E||B|/μ₀ = cB²/μ₀. Given B = 55×10⁻⁸ T,

|S| = (3.00×10⁸)(55×10⁻⁸)²/(4π×10⁻⁷) = 1815/(8π) W/m² ≈ 72.2 W/m². If 55×10⁻⁸ T is the peak amplitude of a sinusoidal wave, the time-averaged magnitude is half, ≈36.1 W/m².

(b)

At equilibrium, the Gibbs free energy G of a closed system at fixed T and P is minimum. Consider transfer of δn_i of component i from phase α to phase β. Then δG = μ_i^β δn_i − μ_i^α δn_i = (μ_i^β − μ_i^α)δn_i. At equilibrium δG = 0 for arbitrary small δn_i, hence μ_i^α = μ_i^β. If they were unequal, spontaneous transfer would lower G until equality. Thus every component that can cross a phase boundary has the same chemical potential in all coexisting phases; T and P are also equal at equilibrium.

For a binary liquid–vapour system, let components be 1 and 2, liquid phase L and vapour phase V. The equilibrium conditions are μ₁^L(T,P,x₁) = μ₁^V(T,P,y₁), μ₂^L(T,P,x₂) = μ₂^V(T,P,y₂), with x₁ + x₂ = 1 and y₁ + y₂ = 1, where x_i and y_i are liquid and vapour mole fractions. Writing μ_i^L = μ_i^L,0(T,P) + RT ln a_i^L, μ_i^V = μ_i^V,0(T,P) + RT ln a_i^V, equating gives a_i^V/a_i^L = exp[(μ_i^L,0 − μ_i^V,0)/RT] ≡ K_i. For an ideal solution, μ_i^L = μ_i*(T) + RT ln x_i, where μ_i* is the chemical potential of pure liquid i. Pure liquid i is itself in equilibrium with its vapour at the saturation pressure p_i^sat, so μ_i*(T) = μ_i°(T) + RT ln(p_i^sat/P°). For an ideal vapour, μ_i^V = μ_i°(T) + RT ln(y_i P/P°). Equating μ_i^L and μ_i^V (neglecting the small pressure dependence of the liquid) gives y_i P = x_i p_i^sat, i = 1,2. Combining, P = x₁p₁^sat + x₂p₂^sat. This is Raoult’s law.

(c)

Consider blackbody radiation in a cavity of volume V at temperature T. Periodic boundary conditions give allowed wave vectors with spacing 2π/L. The number of wave-vector states in the shell k to k + dk is dN_k = V/(2π)³ · 4πk² dk. Photons have two transverse polarizations, so dN = 2 · V/(2π)³ · 4πk² dk = Vk²/π² dk. Using k = 2πν/c, dk = 2π/c dν, dN = 8πVν²/c³ dν. Thus the mode density per unit volume per unit frequency is g(ν) = 8πν²/c³.

Photons are bosons with chemical potential μ = 0 because photon number is not conserved. The Bose–Einstein occupation number for energy hν is n̄(ν) = 1/[exp(hν/k_B T) − 1]. Therefore the energy density in frequency interval dν is u(ν,T)dν = g(ν) hν n̄(ν)dν, giving u(ν,T) = (8πhν³/c³) · 1/[exp(hν/k_B T) − 1]. This is Planck’s radiation law. The spectral radiance is B(ν,T) = (2hν³/c²) · 1/[exp(hν/k_B T) − 1]. Integrating over all frequencies gives u(T) = ∫₀^∞ u(ν,T)dν = (8π⁵k_B⁴/15h³c³)T⁴ = aT⁴.

The classical Rayleigh–Jeans law treats mode energies as continuous and assigns equipartition energy k_B T per mode: u_RJ(ν,T) = 8πν²k_B T/c³. It agrees with Planck only for hν ≪ k_B T, since then exp(hν/k_B T) − 1 ≈ hν/k_B T. At high frequencies, Rayleigh–Jeans diverges as ν², the ultraviolet catastrophe. Planck’s law is exponentially suppressed as exp(−hν/k_B T), so total energy is finite. Thus quantum statistics introduces energy quantization hν, removes the ultraviolet catastrophe, gives the correct Wien displacement and Stefan–Boltzmann laws, and reduces to the classical limit only at low frequencies.