University of Burdwan · CBCS · PYQs 2018–2024

PHYS7021 · Electricity, Magnetism & Wave Optics

Sem-VII Physics Minor — syllabus-wise theory ➜ solved past questions ➜ model answers · Indian-textbook style · mobile friendly

⚠ Borderline Questions & Answers

These appeared in GE-2/GE-4 exams but are NOT listed verbatim in the PHYS7021 syllabus text. Keep them as backup (examiners sometimes repeat them); confirm with your lecturer.

GE-2 side

B1. Dipole (3 cm, 5 μC) in E = 2×104 N/C → maximum & minimum torque.
2018

Given: 2a = 3 cm = 0.03 m, q = 5 μC, E = 2×104 N/C. Formula: τ = pE sinθ, p = q·2a.

  1. p = 5×10−6 × 0.03 = 1.5×10−7 C·m.
  2. τmax (θ = 90°) = pE = 1.5×10−7 × 2×104 = 3×10−3 N·m.
  3. τmin (θ = 0°) = 0.

∴ τ_max = 3×10⁻³ N·m ; τ_min = 0

B2. Explain what happens when a dipole is placed in a uniform electric field.
2023
  1. Forces +qE and −qE are equal and opposite ⇒ no net force (dipole does not translate).
  2. They form a couple ⇒ torque τ = pE sinθ, which rotates the dipole.
  3. It turns until p aligns with E (stable equilibrium, U = −pE).

∴ Torque pE sinθ aligns the dipole; no translation

B3. Dipole (±1 μC, 2 cm) in 2.5×104 N/C → torque to rotate it by 30°.
2023

Given: q = 1 μC, 2a = 0.02 m ⇒ p = 2×10−8 C·m; E = 2.5×104 N/C; θ = 30°.

  1. τ = pE sin30° = 2×10−8 × 2.5×104 × 0.5.
  2. = 2.5×10−4 N·m.

∴ τ = 2.5×10⁻⁴ N·m

B4. Lorentz force: q = 1.6×10−19 C, v = 3î+2ĵ m/s, E = 6î+6ĵ+3k̂ V/m, B = ĵ+2k̂ T.
2019

Formula: F = q(E + v×B).

  1. v×B = (3î+2ĵ)×(ĵ+2k̂) = 3(î×ĵ) + 6(î×k̂) + 2(ĵ×ĵ) + 4(ĵ×k̂) = 3k̂ − 6ĵ + 0 + 4î = 4î − 6ĵ + 3k̂.
  2. E + v×B = (6+4)î + (6−6)ĵ + (3+3)k̂ = 10î + 6k̂.
  3. F = 1.6×10−19 × √(102+62) = 1.6×10−19 × √136 ≈ 1.87×10−18 N, along (10î+6k̂)/√136.

∴ F ≈ 1.87×10⁻¹⁸ N along (10î+6k̂)/√136

B5. Work done by a magnetic field on a moving charge?
2022
  1. Magnetic force F = qv×B is always perpendicular to velocity.
  2. Power = F·v = 0 at every instant.
  3. Hence work = 0; the field changes the direction of velocity, never the speed.

∴ Work = 0

B6. Force between two parallel wires 10 cm long, 2 cm apart, carrying 20 A and 30 A.
2022

Given: L = 0.1 m, d = 0.02 m, I1 = 20 A, I2 = 30 A. Formula: F = (μ0/2π)(I1I2/d)L.

  1. F = (2×10−7) × (20×30/0.02) × 0.1 = (2×10−7) × (3×104) × 0.1.
  2. = 6×10−4 N (attractive if currents are in the same direction).

∴ F = 6×10⁻⁴ N

B7. Parallel-plate capacitor A = 0.25 m2, d = 1 cm, 10 V → force of attraction between plates.
2023

Given: A = 0.25 m2, d = 0.01 m, V = 10 V.

  1. E = V/d = 10/0.01 = 1000 V/m.
  2. F = ½ε0E2A = 0.5 × 8.85×10−12 × 106 × 0.25.
  3. ≈ 1.1×10−6 N.

∴ F ≈ 1.1×10⁻⁶ N

GE-4 side

B8. Visibility of interference fringes / numerical for amplitudes 2 cm & 1 cm.
×220192023

Formula: V = (Imax − Imin)/(Imax + Imin); intensity ∝ (amplitude)2.

  1. Imax ∝ (2+1)2 = 9; Imin ∝ (2−1)2 = 1.
  2. V = (9−1)/(9+1) = 8/10 = 0.8.

∴ V = 0.8

B9. Resolving power of a grating (show R = nN) / of an optical instrument (definition).
×220222024
  1. Definition: resolving power = λ/dλ — the ability to separate two close wavelengths.
  2. Grating: by Rayleigh's criterion the principal maximum of one wavelength must fall on the first minimum of the other ⇒ R = nN (order × total illuminated rulings).
  3. Optical instrument: e.g., telescope R = D/1.22λ (D = aperture); microscope R = 2μ sinθ/1.22λ.

∴ R = λ/dλ ; grating R = nN

B10. What is optical activity?
2023
  1. The property of certain substances (sugar solution, quartz, camphor, turpentine) to rotate the plane of polarization of light passing through them.
  2. Measured with a polarimeter; specific rotation θ = α/(l·c) (α = observed rotation, l = path length, c = concentration).

∴ Rotation of polarization plane; e.g. sugar solutions

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