Proof of Maxwell's First Equation: Modified Ampère's Law & Displacement Current (with MCQs)

Why does a magnetic field circle the gap of a charging capacitor, where no charge flows? Answering that puzzle led James Clerk Maxwell to add the displacement current to Ampère’s law — the step that predicted electromagnetic waves and, ultimately, radio, mobile phones and Wi-Fi. The derivation of ∇ × H = J + ∂D/∂t is a classic question in electromagnetic-field theory exams (university finals, BCS EEE/Telecom written and viva). This post explains it step by step with an animation and solved examples.

⚡ Key takeaways

• Maxwell’s first equation (as numbered in many textbooks) is the modified Ampère’s circuital law: ∇ × H = J + ∂D/∂t.
• Integral form: ∮ H·dl = ∫ (J + ∂D/∂t)·dS.
• Displacement current density J_d = ∂D/∂t = ε ∂E/∂t (A/m²). In a capacitor, displacement current = conduction current in the wires.
• Without J_d, Ampère’s law contradicts the continuity equation ∇·J = −∂ρ_v/∂t.
• A changing E-field produces H, a changing H produces E (Faraday) → electromagnetic waves travelling at c = 1/√(μ₀ε₀).

📑 Contents

1. Maxwell’s four equations at a glance
2. Ampère’s circuital law and its problem
3. Proof (derivation) of the first equation
4. Integral form using Stokes’ theorem
5. Displacement current in a capacitor (animation)
6. Physical significance
7. Conduction vs displacement current
8. Solved problems
9. Practice MCQs
10. বাংলায় দ্রুত রিভিশন

1. Maxwell’s Four Equations at a Glance

Law	Differential (point) form	Integral form	Meaning
Modified Ampère’s law	∇ × H = J + ∂D/∂t	∮ H·dl = ∫(J + ∂D/∂t)·dS	Currents and changing E-fields create H
Faraday’s law	∇ × E = −∂B/∂t	∮ E·dl = −d/dt ∫ B·dS	Changing B induces E (EMF)
Gauss’s law (electric)	∇ · D = ρv	∮ D·dS = Qenc	Charges are sources of D
Gauss’s law (magnetic)	∇ · B = 0	∮ B·dS = 0	No magnetic monopoles

Constitutive relations: D = εE, B = μH, J = σE. Note: textbooks number the equations differently; some call Faraday’s law the first. For a full list in different media see Maxwell’s Equations for Different Media.

2. Ampère’s Circuital Law and Its Problem

Ampère’s circuital law states that the line integral of H around any closed path equals the current enclosed: ∮ H·dl = I_enc. Using Stokes’ theorem, its point form is ∇ × H = J.

This works perfectly for steady (DC) currents. But take the divergence of both sides. The divergence of a curl is always zero, so

∇ · (∇ × H) = 0 = ∇ · J

The equation of continuity (conservation of charge), however, says

∇ · J = − ∂ρ_v / ∂t

which is not zero whenever charge density changes with time — for example on the plates of a charging capacitor. So ∇ × H = J cannot be complete for time-varying fields. Something is missing.

💡 The capacitor paradoxDraw a loop around the wire feeding a capacitor. Choose a flat surface bounded by the loop that the wire pierces — enclosed current = I. Now stretch the surface like a bag so it passes between the plates — no conduction current crosses it, enclosed current = 0. Same loop, two different answers! Maxwell resolved this by recognising that the changing electric field between the plates acts as a current.

3. Proof (Derivation) of Maxwell’s First Equation

1. Assume the complete law has an extra term: ∇ × H = J + J_d.
2. Take divergence: 0 = ∇·J + ∇·J_d ⇒ ∇·J_d = −∇·J.
3. From continuity: −∇·J = ∂ρ_v/∂t, so ∇·J_d = ∂ρ_v/∂t.
4. From Gauss’s law ρ_v = ∇·D, so ∂ρ_v/∂t = ∂(∇·D)/∂t = ∇·(∂D/∂t).
5. Therefore ∇·J_d = ∇·(∂D/∂t), and the simplest choice is J_d = ∂D/∂t.

∇ × H = J + ∂D/∂t (Maxwell’s first equation, point form)

J_d = ∂D/∂t is called the displacement current density (A/m²). For time-harmonic fields (e^jωt), ∂/∂t → jω, giving ∇ × H = (σ + jωε)E.

4. Integral Form Using Stokes’ Theorem

Integrate the point form over an open surface S bounded by a closed path L:

∫_S(∇ × H)·dS = ∫_S(J + ∂D/∂t)·dS

By Stokes’ theorem, the left side equals ∮_L H·dl. Hence

∮_L H·dl = ∫_S J·dS + ∫_S (∂D/∂t)·dS = I_c + I_d

In words: the magnetomotive force around a closed path equals the total (conduction + displacement) current through any surface bounded by it.

5. Displacement Current in a Capacitor (Animation)

An AC source drives current through the wires (orange arrows). Charges pile up on the plates, so the electric field between them (green arrows) grows and shrinks. No charge crosses the gap, but the changing E-field constitutes a displacement current exactly equal to the wire current — and it produces the same circular magnetic field (purple loops) around the gap.

Animation 1 – Conduction current in the wire, displacement current in the gap, and the same H-field around both.

Proof for a parallel-plate capacitor (area A, gap d): D = εE = εV/d, so I_d = A ∂D/∂t = (εA/d) dV/dt = C dV/dt = I_c. ✔

6. Physical Significance

• A time-varying electric field produces a magnetic field, just as a time-varying magnetic field produces an electric field (Faraday). Together they let fields sustain each other in empty space — electromagnetic waves.
• Combining the curl equations in free space gives the wave equation ∇²E = μ₀ε₀ ∂²E/∂t², with speed c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s — Maxwell’s prediction that light is an electromagnetic wave (confirmed by Hertz in 1888).
• It makes Ampère’s law consistent with conservation of charge.
• It explains current through capacitors, radiation from antennas and propagation in waveguides and free space.

7. Conduction vs Displacement Current

Basis	Conduction current	Displacement current
Cause	Actual flow of free charges	Time-varying electric field (flux)
Density	Jc = σE	Jd = ε ∂E/∂t (= jωεE for sinusoids)
Exists in	Conductors	Dielectrics, free space, capacitors (also conductors, but negligible)
Obeys Ohm’s law?	Yes	No
Under DC steady state	Exists	Zero

The ratio J_c/J_d = σ/(ωε) (the loss tangent) classifies a medium: σ/ωε ≫ 1 → good conductor; σ/ωε ≪ 1 → good dielectric. See also What Is a Medium in Electromagnetic Fields?

8. Solved Problems

Problem 1. A 1 µF capacitor has v = 100 sin(1000t) V across it. Find the displacement current.

i_d = C dv/dt = 10⁻⁶ × 100 × 1000 cos(1000t) = 0.1 cos(1000t) A — equal to the conduction current in the leads.

Problem 2. In free space E = 10 sin(ωt − βz) a_x V/m with f = 100 MHz. Find the amplitude of J_d.

J_d = ε₀ ∂E/∂t = ε₀ × 10ω cos(ωt − βz) a_x; amplitude = 8.854 × 10⁻¹² × 10 × 2π × 10⁸ ≈ 0.0556 A/m².

Problem 3. Compare conduction and displacement currents at 1 MHz in (a) copper (σ = 5.8 × 10⁷ S/m, ε_r = 1) and (b) sea water (σ = 4 S/m, ε_r = 81).

1. (a) σ/(ωε) = 5.8 × 10⁷ / (2π × 10⁶ × 8.854 × 10⁻¹²) ≈ 1.0 × 10¹² → displacement current negligible.
2. (b) σ/(ωε) = 4 / (2π × 10⁶ × 81 × 8.854 × 10⁻¹²) ≈ 890 → sea water behaves as a good conductor at 1 MHz (why submarines need very low frequencies).

9. Practice MCQs (click to see answer)

1. Maxwell’s modification of Ampère’s law added: (a) Conduction current (b) Displacement current (c) Convection current (d) Eddy current

✔ (b) Displacement current ∂D/∂t.

2. Displacement current density is: (a) σE (b) ∂D/∂t (c) ∂B/∂t (d) ρv

✔ (b) ∂D/∂t.

3. The point form of the modified Ampère’s law is: (a) ∇ × E = −∂B/∂t (b) ∇·D = ρ (c) ∇ × H = J + ∂D/∂t (d) ∇·B = 0

✔ (c).

4. Ampère’s law without displacement current violates: (a) Coulomb’s law (b) Equation of continuity (c) Ohm’s law (d) Lenz’s law

✔ (b) Equation of continuity.

5. Displacement current flows through: (a) Resistor (b) Capacitor (c) Inductor (d) Battery only

✔ (b) Capacitor (dielectric gap).

6. ∇·B = 0 indicates: (a) Magnetic monopoles exist (b) Magnetic monopoles do not exist (c) Charge is conserved (d) E is conservative

✔ (b) No isolated magnetic poles.

7. For a good conductor, σ/ωε is: (a) ≪ 1 (b) = 1 (c) ≫ 1 (d) 0

✔ (c) ≫ 1.

8. The theorem used to convert the point form into integral form of the first equation is: (a) Gauss’s divergence theorem (b) Stokes’ theorem (c) Green’s theorem (d) Thevenin’s theorem

✔ (b) Stokes’ theorem.

9. Under DC steady-state conditions, displacement current is: (a) Maximum (b) Equal to conduction current (c) Zero (d) Infinite

✔ (c) Zero (∂D/∂t = 0).

10. Velocity of EM waves in free space is: (a) √(μ₀ε₀) (b) 1/√(μ₀ε₀) (c) μ₀/ε₀ (d) 1/(μ₀ε₀)

✔ (b) 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s.

10. বাংলায় দ্রুত রিভিশন

📝 পরীক্ষার আগে এক নজরে

• ম্যাক্সওয়েলের প্রথম সমীকরণ (সংশোধিত অ্যাম্পিয়ারের সূত্র): ∇ × H = J + ∂D/∂t; ইন্টিগ্রাল রূপ: ∮H·dl = ∫(J + ∂D/∂t)·dS।
• শুধু ∇ × H = J লিখলে ∇·J = 0 হয়, যা চার্জ সংরক্ষণের সমীকরণ (∇·J = −∂ρ/∂t)-এর সাথে সাংঘর্ষিক। তাই ম্যাক্সওয়েল ডিসপ্লেসমেন্ট কারেন্ট J_d = ∂D/∂t যোগ করেন।
• ক্যাপাসিটরের প্লেটের মাঝে কোনো চার্জ না চললেও পরিবর্তনশীল তড়িৎক্ষেত্র ডিসপ্লেসমেন্ট কারেন্ট তৈরি করে, যা তারের কারেন্টের সমান (C dV/dt)।
• তাৎপর্য: পরিবর্তনশীল E থেকে H এবং পরিবর্তনশীল H থেকে E — ফলে তড়িৎচুম্বকীয় তরঙ্গ তৈরি হয়, বেগ c = 1/√(μ₀ε₀)।
• σ/ωε ≫ 1 হলে ভালো পরিবাহী, ≪ 1 হলে ভালো ডাইইলেকট্রিক।

Related Posts

• Maxwell’s Equations for Different Media
• What Is a Medium in Electromagnetic Fields?
• Dielectric Materials: Polarization & Properties
• Equivalent Circuit of a Transmission Line

References: Matthew N. O. Sadiku, Elements of Electromagnetics; W. H. Hayt & J. A. Buck, Engineering Electromagnetics; David J. Griffiths, Introduction to Electrodynamics. Images and animations: original work by Totalecer.