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

Section 4 — Maxwell's Equations & EM Waves

GE-2 / PHYS7021 unit 4 · 5 lecture-hours · 9 solved questions

4.1 Theory you need

∇·E = ρ/ε0 · ∇·B = 0 · ∇×E = −∂B/∂t · ∇×B = μ0J + μ0ε0∂E/∂t

The last term μ0ε0∂E/∂t is Maxwell's displacement current (Jd = ε0∂E/∂t) — added to save continuity of current. In free space the equations give the wave equation ∇2E = μ0ε0∂2E/∂t2 with speed c = 1/√(μ0ε0) ≈ 3×108 m/s. In a dielectric medium v = 1/√(με) = c/n, so refractive index n = c/v and λ′ = λ/n.

v (+y) E (vertical) B (horizontal), E ⊥ B ⊥ v

4.2 Solved past questions — exam-style answers

1. What do you mean by conduction current and displacement current?
2022
  1. Conduction current: actual flow of free charges through a conductor; Jc = σE; it produces Joule heating.
  2. Displacement current: the quantity Jd = ε0∂E/∂t (Id = ε0dΦE/dt) introduced by Maxwell; it is not a flow of charge but a changing electric field (e.g., between capacitor plates).
  3. Like a real current, it produces a magnetic field and restores the continuity equation ∇·J + ∂ρ/∂t = 0.

∴ J_c = σE (charge flow) ; J_d = ε₀ ∂E/t (changing field)

2. E = E0cos ωt of frequency 1015 Hz applied to a conductor (σ = 107 mho/m). Ratio of conduction to displacement current?
2019

Given: f = 1015 Hz, σ = 107 mho/m. Formula: Jc/Jd = σE/(ε0ωE) = σ/(2πfε0).

  1. Denominator: 2π × 1015 × 8.85×10−12 = 5.56×104.
  2. Ratio = 107 / 5.56×104 ≈ 180.
  3. Hence in a good conductor the conduction current dominates (≈180 times).

∴ J_c/J_d ≈ 180

3. Can we apply Ampère's law for non-steady currents? What is Maxwell's correction?
2021
  1. No. For non-steady currents ∮B·dl = μ0I gives different values for different surfaces bounded by the same loop (e.g., charging capacitor: no conduction current crosses the gap), violating charge continuity.
  2. Maxwell's correction: add the displacement current Id = ε0dΦE/dt between the plates.
  3. Corrected law: ∮B·dl = μ0(Ic + Id), i.e. ∇×B = μ0J + μ0ε0∂E/∂t.

∴ Ampère's law needs +μ₀ε₀∂E/t term for non-steady currents

4. Write down Maxwell's equations (free space / with symbols / inside material / with the modification of Ampère's law).
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  1. (i) ∇·E = ρ/ε0 — Gauss's law: electric charges are the sources of E.
  2. (ii) ∇·B = 0 — no magnetic monopoles.
  3. (iii) ∇×E = −∂B/∂t — Faraday: a changing B creates E.
  4. (iv) ∇×B = μ0J + μ0ε0∂E/∂t — Ampère with Maxwell's modification: both conduction current and displacement current create B.
  5. Inside material: use D, H: ∇·D = ρfree; ∇·B = 0; ∇×E = −∂B/∂t; ∇×H = Jfree + ∂D/∂t.
  6. In free space (ρ = 0, J = 0) these four equations predict electromagnetic waves travelling at c.

∴ The four equations above (with Maxwell's correction in iv)

5. From Maxwell's equations derive the EM wave equation in free space; express the speed in terms of ε0 and μ0.
×220182022
  1. In free space take curl of (iii): ∇×(∇×E) = −∂(∇×B)/∂t.
  2. LHS = ∇(∇·E) − ∇2E = −∇2E (since ∇·E = 0).
  3. RHS = −μ0ε0∂2E/∂t2 (using (iv) with J = 0).
  4. Hence ∇2E = μ0ε0∂2E/∂t2 — the wave equation with speed v = 1/√(μ0ε0) = 2.998×108 m/s = c.

∴ v = 1/√(μ₀ε₀) = c

6. Define Poynting vector (and its unit).
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  1. Definition: S = (1/μ0) E×B = E×H — the energy flux density of the EM wave.
  2. It gives the power flowing per unit area, directed along the propagation (E×B direction).
  3. Unit: watt per square metre (W/m2).

∴ S = (1/μ₀)E×B , unit W/m²

7. Velocity of light in air 3×108 m/s: find velocity and wavelength in water (n = 1.33) / glass (n = 1.658) for λ = 5893 Å; relation of refractive index to velocities in two media.
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Formula: n = c/v, and between two media n21 = v1/v2 = λ1/λ2 (frequency unchanged).

  1. Water: v = c/n = 3×108/1.33 = 2.26×108 m/s.
  2. Glass: v = 3×108/1.658 = 1.81×108 m/s.
  3. λ′ in glass = λ/n = 5893/1.658 ≈ 3554 Å.

∴ v_water = 2.26×10⁸ m/s ; v_glass = 1.81×10⁸ m/s ; λ′ ≈ 3554 Å

8. Write down the main properties of electromagnetic waves.
2023
  1. Transverse: E and B are perpendicular to each other and to the direction of propagation.
  2. Need no material medium; travel in vacuum with c = 3×108 m/s.
  3. Ratio of amplitudes E0/B0 = c.
  4. Carry energy and momentum (Poynting vector); exert radiation pressure.
  5. Not deflected by electric or magnetic fields; obey superposition; produced by accelerated charges.

∴ Transverse, c in vacuum, E/B = c, carry energy & momentum

9. Show that E = E0cos(ky − ωt)k and B = B0cos(ky − ωt)î represent an electromagnetic field.
2023

Given: E along ẑ, B along x̂, both varying as cos(ky − ωt) ⇒ travel along +y.

  1. Divergences: ∇·E = ∂Ez/∂z = 0 ✔ ; ∇·B = ∂Bx/∂x = 0 ✔.
  2. Faraday: (∇×E)x = −∂Ez/∂y = kE0 sin(ky−ωt); −∂B/∂t = −ωB0 sin(ky−ωt) x̂.
  3. These match when kE0 = ωB0, i.e. E0/B0 = ω/k = c ✔.
  4. The Ampère–Maxwell equation is likewise satisfied with c = 1/√(μ0ε0); E ⊥ B ⊥ . Hence a valid EM wave travelling along +y.

∴ All Maxwell equations satisfied when E₀/B₀ = c ⇒ valid EM wave

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