Reasoning Chain: How Maxwell Predicted Light is an EM Wave

Step 1 — Observe the inconsistency Ampere's original law: ∮B·dl = μ_{0}I_c Problem: Draw two surfaces bounded by the same Amperian loop around a wire connected to a charging capacitor.

• Surface 1 (cuts the wire): enclosed current = I_c → gives B ≠ 0
• Surface 2 (passes through capacitor gap): enclosed current = 0 → gives B = 0 Same loop, two surfaces, two different answers — a fatal inconsistency.

Step 2 — Identify what's different about Surface 2 The capacitor gap has no charges crossing, but there IS a changing electric field. As the capacitor charges, E between the plates increases. This means dΦ_E/dt ≠ 0.

Step 3 — Introduce displacement current Define: Id = ε_{0}(dΦ_E/dt). Modify Ampere's law: ∮B·dl = μ_{0}(I_c + Id) = μ_{0}(I_c + ε_{0}dΦ_E/dt) Now both surfaces give the same answer (since Id through Surface 2 = I_c through Surface 1).

Step 4 — Write the wave equations With the four equations complete, Maxwell combined Faraday's law and the modified Ampere law:

• From Faraday: ∂E/∂x = –∂B/∂t
• From Ampere-Maxwell (in free space, I_c = 0): ∂B/∂x = –μ_{0}ε_{0}(∂E/∂t) Differentiating the first and substituting the second yields: ∂^{2}E/∂$x^{2}$ = μ_{0}ε_{0} (∂^{2}E/∂$t^{2}$)

Step 5 — Recognise the wave equation This is the standard wave equation ∂^{2}y/∂$x^{2}$ = (1/$v^{2}$)(∂^{2}y/∂$t^{2}$), with: $v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = \frac{1}{\sqrt{(4\pi \times 10^{-7})(8.85 \times 10^{-12})}} \approx 3 \times 10^8 \text{ m/s}$

Step 6 — The revelation The known speed of light was ~$3 \times 10^{8}$ m/s (measured by Foucault and Fizeau). Maxwell's calculated wave speed matches exactly. Conclusion: LIGHT IS AN ELECTROMAGNETIC WAVE. This was one of the greatest intellectual achievements in physics — uniting electricity, magnetism, and optics into one framework.