Physics 2021 Paper I 50 marks Calculate

Q7

(a) A region 1, z < 0, has a dielectric material with εᵣ = 3·2 and a region 2, z > 0 has a dielectric material with εᵣ = 2·0. Let the displacement vector in the region 1 be, D⃗₁ = – 30 aₓ + 50 aᵧ + 70 aᵤ nCm⁻². Assume the interface charge density is zero. Find in the region 2, the D⃗₂ and P⃗₂, where P⃗₂ is the electric polarization vector in the region 2. (20 marks) (b) Calculate the skin depth of electromagnetic waves of 1 MHz incident on a good conductor having σ = 5·8 × 10⁷ Sm⁻¹. Assume that inside the conductor μ = μ₀ = 4π × 10⁻⁷ Hm⁻¹. (10 marks) (c) The spectral composition of solar radiation is similar to that of a black body radiator whose maximum emission corresponds to the wavelength 0·48 μm. Find the mass lost by the Sun every second due to radiation. Evaluate the time interval during which the mass of the Sun reduces by 1 per cent. Given : Stefan Boltzmann constant = 5·669 × 10⁻⁸ W m⁻² K⁻⁴, radius of the Sun = 6·957 × 10⁸ m, surface temperature of the Sun = 5772 K and mass of the Sun is 1·9885 × 10³⁰ kg. (20 marks)

हिंदी में प्रश्न पढ़ें

(a) क्षेत्र 1, z < 0 परावैद्युत पदार्थ εᵣ = 3·2 का बना है और क्षेत्र 2, z > 0 परावैद्युत पदार्थ εᵣ = 2·0 का है । मान लीजिए कि विस्थापन सदिश क्षेत्र 1 में D⃗₁ = – 30 aₓ + 50 aᵧ + 70 aᵤ nCm⁻² है । मान लीजिए कि अंतरापृष्ठ आवेश घनत्व शून्य है । क्षेत्र 2 में D⃗₂ और P⃗₂ ज्ञात कीजिए, जहाँ P⃗₂ क्षेत्र 2 में वैद्युत ध्रुवण सदिश है । (20 अंक) (b) एक सुचालक, जिसका σ = 5·8 × 10⁷ Sm⁻¹ है, पर 1 MHz की विद्युत-चुंबकीय तरंगें आपतित होती हैं । इस सुचालक के लिए त्वचा गहराई की गणना कीजिए । मान लीजिए कि सुचालक के अंदर μ = μ₀ = 4π × 10⁻⁷ Hm⁻¹ है । (10 अंक) (c) सौर विकिरण का वर्णक्रम (स्पेक्ट्रल) संयोजन एक कृष्णिका विकिरक के समान है जिसके अधिकतम उत्सर्जन का तरंगदैर्ध्य 0·48 μm है । विकिरण के कारण सूर्य की द्रव्यमान क्षति प्रति सेकंड ज्ञात कीजिए । उस समय अंतराल की गणना कीजिए जिसमें सूर्य का द्रव्यमान 1% घट जाता है । दिया गया है : स्टीफन बोल्ट्ज़मान नियतांक = 5·669 × 10⁻⁸ W m⁻² K⁻⁴, सूर्य की त्रिज्या = 6·957 × 10⁸ m, सूर्य के पृष्ठ का ताप = 5772 K और सूर्य का द्रव्यमान 1·9885 × 10³⁰ kg है । (20 अंक)

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How this answer will be evaluated

Approach

Calculate the required quantities systematically across all three parts, allocating approximately 40% of effort to part (a) given its 20 marks, 20% to part (b), and 40% to part (c). Begin each part by stating the relevant governing equations, show step-by-step derivations with proper unit handling, and conclude with physically meaningful interpretations of the numerical results.

Key points expected

  • Part (a): Apply boundary conditions for dielectric interface—tangential E continuous (E₁ₜ = E₂ₜ) and normal D continuous (D₁ₙ = D₂ₙ) with σ = 0; decompose D⃗₁ into normal (z) and tangential (x-y) components
  • Part (a): Calculate D⃗₂ using ε₂/ε₁ ratio for tangential components and equality for normal component; then find P⃗₂ = D⃗₂ − ε₀E⃗₂ = D⃗₂(1 − 1/εᵣ₂)
  • Part (b): Use skin depth formula δ = √(2/ωμσ) = 1/√(πfμσ) for good conductor; substitute f = 10⁶ Hz, μ = 4π×10⁻⁷ H/m, σ = 5.8×10⁷ S/m
  • Part (c): Apply Wien's displacement law λₘT = b to verify T ≈ 5772 K, then use Stefan-Boltzmann law P = σAT⁴ for total radiated power
  • Part (c): Calculate mass loss rate using E = mc² equivalence (ṁ = P/c²) and time for 1% mass loss as Δt = 0.01M☉/ṁ
  • Explicit unit conversions: μm to m, MHz to Hz, nC to C, and proper handling of SI prefixes throughout

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies all governing principles: for (a) distinguishes normal vs tangential boundary conditions and recognizes Dₙ continuity; for (b) identifies good conductor approximation (σ >> ωε); for (c) correctly applies Wien's law verification and Stefan-Boltzmann radiation law with mass-energy equivalenceIdentifies most principles correctly but confuses boundary conditions in (a) or uses approximate skin depth formula without justification; in (c) may skip Wien's law verification or misapply Stefan-Boltzmann to volume instead of surfaceFundamental errors: applies D continuity to tangential components, uses wrong skin depth formula (δ = 1/α without derivation), or treats Sun as point source; confuses Poynting vector with energy density
Derivation rigour20%10Shows complete derivations: for (a) explicitly decomposes vectors and derives E₂ from D₁/ε₁; for (b) derives from Maxwell's equations or at least states δ = √(2/ωμσ); for (c) derives ṁ = 4πR²σT⁴/c² with all intermediate steps; justifies all approximationsShows key steps but skips some algebraic manipulation or assumes known formulas without brief derivation; in (b) may quote skin depth without showing origin; in (c) may jump directly to final formulaNo derivations shown—only final formulas quoted; or contains major algebraic errors such as incorrect vector decomposition, wrong exponent in Stefan-Boltzmann, or dimensional inconsistency in final expressions
Diagram / FBD10%5Clear diagram for (a) showing z = 0 interface, labeling regions 1 and 2 with εᵣ values, indicating D⃗₁ and decomposed components (D₁ₙ, D₁ₜ), and showing coordinate system; optional sketch for (c) showing Sun's radiative emission geometryBasic diagram for (a) showing interface and regions but lacks vector decomposition or proper labeling; or describes geometry verbally without figureNo diagram provided where clearly needed; or incorrect diagram showing wrong interface orientation, confused normal/tangential directions, or irrelevant free-body diagrams inappropriate for electromagnetism
Numerical accuracy35%17.5All calculations correct with proper significant figures: (a) D⃗₂ = (−18.75aₓ + 31.25aᵧ + 70aᵤ) nC/m², P⃗₂ = (−9.375aₓ + 15.625aᵧ + 35aᵤ) nC/m²; (b) δ ≈ 6.6×10⁻⁵ m or 66 μm; (c) ṁ ≈ 4.3×10⁹ kg/s, t₁% ≈ 1.4×10¹¹ years (≈10¹⁰ years acceptable with rounding); shows unit cancellation explicitlyCorrect method but arithmetic errors: wrong component values in (a) due to ε ratio error; order-of-magnitude error in (b) due to Hz vs rad/s confusion; in (c) correct power calculation but error in c² or percentage calculation yielding wrong timescaleMajor numerical errors: incorrect boundary condition application giving D⃗₂ = D⃗₁; skin depth off by orders of magnitude; mass loss rate or timescale completely unrealistic (e.g., seconds or exceeding age of universe by factor >10³); no units or inconsistent units
Physical interpretation15%7.5Interprets all results physically: explains why D changes direction across interface due to different εᵣ; notes skin depth implies EM wave penetration ~66 μm in copper at 1 MHz, relevant for RF shielding; comments that Sun's mass loss rate is small compared to total mass, explaining long stellar lifetime (~10¹⁰ years comparable to universe age)Brief comment on one or two results without connecting to broader physics; may state numbers without physical significance or make generic statements about 'electromagnetic phenomena'No interpretation provided; or nonsensical physical claims (e.g., negative mass loss, D⃗₂ magnitude exceeding D⃗₁ violating energy conservation, skin depth larger than conductor thickness without comment)

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