(a) In an inertial reference frame S there is only a uniform electric field E⃗ = 8 kVm⁻¹. Find the magnitude of E⃗' and B⃗' in…

(a) Fields E′ and B′ in the moving frame S′

In S, E = 8×10³ V/m and B = 0. The speed is v = 0.6c, so β = 0.6 and γ = 1/√(1 − 0.36) = 1.25. Resolve E along and perpendicular to v, with α = 45°: E∥ = E cos α and E⊥ = E sin α.

The Lorentz transformation of the fields, with B = 0, gives

E′∥ = E∥, E′⊥ = γE⊥, B′∥ = 0, B′⊥ = −γ(v × E)/c².

Hence

|E′| = E√(cos²α + γ² sin²α) = 8×10³ √(0.5 + 0.78125) = 8×10³ × 1.132 = 9.06×10³ V/m (= 1000√82 V/m).

|B′| = γvE sin α/c² = γβE sin α/c = 1.25 × 0.6 × 8×10³ × 0.7071/(3×10⁸) = 1.41×10⁻⁵ T.

So E′ ≈ 9.06 kV/m and B′ ≈ 14.1 μT. B′ is perpendicular to both v and E.

(b) Heat generated in the coil after the switch is opened

Before opening, the steady current in the coil is I₀ = E/R = 3.0/1.0 = 3.0 A (the inductor offers no opposition to a steady current, and R₀ is across the source with the same voltage).

After S is opened the source is disconnected and the coil (L, R) and R₀ form a closed loop of resistance R + R₀ = 3.0 Ω. The coil current decays as i = I₀e^(−(R + R₀)t/L), and the energy stored in the inductor is dissipated in the loop:

Q_total = ½LI₀² = ½ × 2.0×10⁻⁶ × 9 = 9.0 μJ.

This is shared in proportion to resistance, since the same current flows through both:

Q_coil = [R/(R + R₀)] Q_total = (1/3) × 9.0 μJ = 3.0 μJ,

and 6.0 μJ appears in R₀.

(c) Thermodynamic processes of a perfect gas

(i) Isothermal: T constant, so PV = constant (Boyle's law) and ΔU = 0, so Q = W. For a reversible process, W = ∫P dV = nRT ln(V₂/V₁) = nRT ln(P₁/P₂).

(ii) Adiabatic: Q = 0, so ΔU = −W. For a reversible process PV^γ = constant, TV^(γ−1) = constant, with γ = C_p/C_v. Then W = (P₁V₁ − P₂V₂)/(γ − 1) = nR(T₁ − T₂)/(γ − 1) = nC_v(T₁ − T₂). The gas cools in expansion, and the adiabatic curve is steeper than the isotherm.

(iii) Isobaric: P constant, so V/T = constant (Charles's law). W = P(V₂ − V₁) = nR(T₂ − T₁), Q = nC_p ΔT and ΔU = nC_v ΔT, which gives C_p − C_v = R.

(iv) Isochoric: V constant, so P/T = constant (Gay-Lussac's law). Since dV = 0, W = 0 and Q = ΔU = nC_v ΔT.