(a) Two inductors having inductances L₁ and L₂ are connected in parallel. The inductors have a mutual inductance M. Derive the…

(a) Effective inductance of two parallel inductors with mutual inductance

Let the supply voltage be V, the branch currents I₁ and I₂ with I = I₁ + I₂, and resistances negligible. For the coupling M, take the aiding case (M positive, with the usual dot convention). By Kirchhoff's voltage law applied to each branch, with the same V across both: L₁ dI₁/dt + M dI₂/dt = V M dI₁/dt + L₂ dI₂/dt = V The determinant is Δ = L₁L₂ − M². By Cramer's rule, dI₁/dt = (L₂ − M)V/Δ and dI₂/dt = (L₁ − M)V/Δ. Adding: dI/dt = (L₁ + L₂ − 2M)V/Δ, so V = L_eff dI/dt with L_eff = (L₁L₂ − M²)/(L₁ + L₂ − 2M) (aiding). If the windings oppose, M → −M: L_eff = (L₁L₂ − M²)/(L₁ + L₂ + 2M). Check: for M = 0, L_eff = L₁L₂/(L₁ + L₂), the usual parallel formula; the expression has the dimension of inductance. Because M² ≤ L₁L₂, L_eff is non-negative.

(b)(i) Joule-Kelvin coefficient

When a gas expands slowly through a porous plug or throttle, from pressure P₁ to P₂ without heat exchange, the enthalpy stays constant. The Joule-Kelvin coefficient is the rate of change of temperature with pressure at constant enthalpy: μ_JK = (∂T/∂P)_H, in K/Pa (or K per atm). Since dP < 0 on expansion, μ_JK > 0 means cooling and μ_JK < 0 means heating.

(b)(ii) Van der Waals gas and the inversion temperature

From dH = T dS + V dP = Cp dT + [V − T(∂V/∂T)_P] dP, at constant H: μ_JK = [T(∂V/∂T)_P − V]/Cp. For one mole of a van der Waals gas, P = RT/(V − b) − a/V². At constant P, 0 = R dT/(V − b) + [−RT/(V − b)² + 2a/V³] dV, so (∂V/∂T)_P = [R/(V − b)] / [RT/(V − b)² − 2a/V³]. Substituting, μ_JK = V[2a(V − b)² − RTbV²] / {Cp[RTV³ − 2a(V − b)²]}. For a dilute gas (V ≫ b), this reduces to μ_JK ≈ (1/Cp)(2a/RT − b). Inversion: μ_JK = 0 gives 2a(V − b)² = RTbV², so T_i = (2a/Rb)(1 − b/V)², and at low pressure T_i ≈ 2a/(Rb) = 27T_c/4. Conditions: if T < T_i, μ_JK > 0, and expansion cools the gas (attractive forces dominate, the term 2a/RT exceeds b); if T > T_i, μ_JK < 0, and expansion heats it (repulsion, the b term, dominates); at T = T_i there is no change. An ideal gas (a = b = 0) has μ_JK = 0. Hydrogen (T_i ≈ 200 K) and helium (≈ 40 K) heat on throttling at room temperature unless pre-cooled, while nitrogen (≈ 620 K) cools.

(c) Fresnel's equations and Brewster's law for E parallel to the plane of incidence

Let the interface be z = 0 and the plane of incidence the xz-plane; media of index n₁ and n₂ are non-magnetic dielectrics; θ₁ and θ₂ are the angles of incidence and refraction (n₁ sin θ₁ = n₂ sin θ₂). The incident, reflected and transmitted amplitudes are Eᵢ, Eᵣ, Eₜ, in the plane of incidence, with H along y. Boundary conditions at z = 0: the tangential components of E and H are continuous (the interface carries no free charge or current). Tangential E: (Eᵢ − Eᵣ) cos θ₁ = Eₜ cos θ₂. Tangential H, with H = nE/(cμ₀) in a non-magnetic medium: n₁(Eᵢ + Eᵣ) = n₂Eₜ. Eliminate Eₜ: n₂ cos θ₁ (Eᵢ − Eᵣ) = n₁ cos θ₂ (Eᵢ + Eᵣ), so r∥ = Eᵣ/Eᵢ = (n₂ cos θ₁ − n₁ cos θ₂)/(n₂ cos θ₁ + n₁ cos θ₂) = tan(θ₁ − θ₂)/tan(θ₁ + θ₂), t∥ = Eₜ/Eᵢ = 2n₁ cos θ₁/(n₂ cos θ₁ + n₁ cos θ₂). These are Fresnel's equations for parallel polarisation. Brewster's law: r∥ = 0 requires n₂ cos θ₁ = n₁ cos θ₂. With Snell's law, sin θ₁ cos θ₁ = sin θ₂ cos θ₂, i.e. sin 2θ₁ = sin 2θ₂, so θ₁ + θ₂ = 90° (for n₁ ≠ n₂). Then sin θ₂ = cos θ₁ and tan θ_B = n₂/n₁. At the Brewster angle the reflected and refracted rays are at right angles; the dipoles in medium 2, oscillating along the refracted E-direction, cannot radiate along the reflected direction. So reflected light from unpolarised incident light is completely polarised with E perpendicular to the plane of incidence. For air to glass (n = 1.5), θ_B ≈ 56.3°.