(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 the inertial reference frame S' moving with a constant velocity v⃗ relative to the frame S at an angle α = 45^° to the vect

Question

(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 the inertial reference frame S' moving with a constant velocity v⃗ relative to the frame S at an angle α = 45^° to the vector E⃗. The velocity of the frame S' is 0.6 times the velocity of light c. (20 marks)

(b) In the given circuit, L = 2·0 μH, R = 1·0 Ω, R₀ = 2·0 Ω and E = 3·0 V. Find the amount of heat generated in the coil after the switch S is disconnected. The internal resistance of the source is negligible. (10 marks)

(c) Explain the characteristics of the following thermodynamic processes for a perfect gas :
(i) Isothermal process
(ii) Adiabatic process
(iii) Isobaric process
(iv) Isochoric process

Obtain the expression for the work done by the gas during the above processes. (20 marks)

UPSC Answer Check · Accepted Answer

(a) Let E = E₀ x̂, with E₀ = 8 kV m⁻¹ = 8×10³ V m⁻¹. The velocity is v = 0.6c, so β = 0.6 and γ = 1/√(1−β²) = 1/√(1−0.36) = 1/0.8 = 5/4. Since α = 45° is the angle between v and E, the components of E parallel and perpendicular to v are E_∥ = E₀ cos 45° and E_⊥ = E₀ sin 45°.

Using the Lorentz field transformation, with B = 0 in S:
E′_∥ = E_∥, E′_⊥ = γ E_⊥,
B′_∥ = 0, B′_⊥ = −γ(v×E)/c².

Therefore,
|E′| = √(E₀² cos²45° + γ² E₀² sin²45°)
= E₀ √(1/2 + γ²/2)
= 8×10³ √(1/2 + (25/16)/2)
= 8×10³ √(41/32)
= 1000√82 V m⁻¹ ≈ 9.06×10³ V m⁻¹.

Also,
|B′| = γ v E₀ sin45° / c² = γβ E₀ sin45° / c
= (5/4)(3/5)(8×10³)(√2/2)/(3×10⁸)
= √2×10⁻⁵ T ≈ 1.41×10⁻⁵ T.

Final: |E′| = 1000√82 V m⁻¹ ≈ 9.06 kV m⁻¹; |B′| = √2×10⁻⁵ T ≈ 14.1 μT.

(b) When switch S is closed in steady state, the inductor acts as a short circuit. Hence the voltage across the branch L, R is E, so the initial current through the coil is
I₀ = E/R = 3.0/1.0 = 3.0 A.

After S is disconnected, the source is removed. The L, R branch and the R₀ branch form a closed discharge loop with total resistance R + R₀ = 1.0 + 2.0 = 3.0 Ω. Let i(t) be the loop current. By Kirchhoff’s voltage law,
L di/dt + (R+R₀)i = 0
⇒ i(t) = I₀ e^[−(R+R₀)t/L].

Total heat dissipated in the loop is
Q_total = ∫₀^∞ i²(R+R₀) dt = 1/2 L I₀²
= 1/2 × 2.0×10⁻⁶ × (3.0)²
= 9.0×10⁻⁶ J.

Heat generated in the coil resistance R is
Q_coil = ∫₀^∞ i²R dt = R/(R+R₀) Q_total
= (1.0/3.0)(9.0×10⁻⁶)
= 3.0×10⁻⁶ J.

Final: Heat generated in the coil = 3.0 μJ.

(c)(i) Isothermal process: T is constant. For a perfect gas, PV = nRT = constant, which is Boyle’s law. Since U depends only on T, ΔU = 0, so Q = W. The work done by the gas is
W = ∫ P dV = ∫ nRT dV/V
= nRT ln(V_f/V_i) = nRT ln(P_i/P_f).
Valid for a quasi-static reversible isothermal process. The P-V curve is a rectangular hyperbola.

(c)(ii) Adiabatic process: No heat exchange occurs: Q = 0. For a perfect gas, PV^γ = constant, TV^(γ−1) = constant, and T^γ P^(1−γ) = constant, where γ = C_P/C_V. By the first law, ΔU = −W. For a reversible adiabatic expansion,
W = ∫ P dV = (P_i V_i − P_f V_f)/(γ−1)
= nR(T_i − T_f)/(γ−1)
= nC_V(T_i − T_f).
During expansion the gas cools; during compression it heats. The adiabatic curve is steeper than an isotherm.

(c)(iii) Isobaric process: P is constant. For a perfect gas, V/T = constant, which is Charles’s law. The heat absorbed is Q = nC_P ΔT, and ΔU = nC_V ΔT. The work done by the gas is
W = P(V_f − V_i) = nR(T_f − T_i).
Using Q = ΔU + W gives C_P − C_V = R.

(c)(iv) Isochoric process: V is constant. For a perfect gas, P/T = constant, which is Gay-Lussac’s law. Since dV = 0, no work is done by the gas:
W = ∫ P dV = 0.
Thus Q = ΔU = nC_V ΔT.

(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…

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