1. Maxwell’s Equations Faraday Induction (Serway)
Experimental Observations
Faraday Induction in point form
Maxwell’s Displacement Current (Serway)
Ampere’s Law with Displacement Current
Maxwell’s Equations (complete)

2. Faraday Induction ϕ 𝑚 = 𝑩∙𝑑𝑺 𝑒𝑚𝑓=− 𝑑 ϕ 𝑚 𝑑𝑡
Consider a single turn of wire, through which an
externally-applied magnetic flux is present. The flux
varies with time.
ϕ 𝑚 = 𝑩∙𝑑𝑺
A current, Iind , is generated in the wire loop as
a result of the changing magnetic flux.
𝑒𝑚𝑓=− 𝑑 ϕ 𝑚 𝑑𝑡

3. Faraday Induction I (Serway)
“Physics for Scientists and Engineers”, Serway/Jewett, 8th Edition, page 895

4. Replacing Current with Electric Field
Current in the wire is caused by electric field, E, that pushes the
charge around the wire. In this illustration, the wire is removed,
and the electric field remains. An integration contour, C, is also
shown, that coincides with E.
𝐸∙𝑑𝐿 =𝑒𝑚𝑓=− 𝑑 ϕ 𝑚 𝑑𝑡

5. Faraday Induction II (Serway)
Experimentally-verified postulate
(note ds here is a path)

6. Faraday Induction Basic Law: Combining: Stoke’s Theorem on left
𝑬∙𝑑𝑳=𝑒𝑚𝑓=− 𝑑ϕ 𝑑𝑡 ϕ= 𝑆 𝑩∙𝑑𝑺
Combining:
𝑬∙𝑑𝑳=− 𝑑 𝑑𝑡 𝑆 𝑩∙𝑑𝑺
Stoke’s Theorem on left
𝑆 𝛻×𝑬∙𝑑𝑺 =− 𝑑 𝑑𝑡 𝑆 𝜕𝑩 𝜕𝑡 ∙𝑑𝑺
Equating Integrands
𝛻×𝐸=− 𝜕𝑩 𝜕𝑡
Note: when 𝜕𝐵 𝜕𝑡 =0 becomes previous 3rd Maxwell Equation

7. Displacement Current (Serway)

8. Displacement Current – The Problem
Start with Ampere’s Law
𝛻×𝑯=𝑱
Take Divergence of both sides
𝛻∙𝛻×𝑯≡𝟎=𝛻∙𝑱
Requires steady-state current at all times.
𝛻∙𝑱=𝟎
Not possible!

9. Displacement Current – The Solution
Replace J with J + G:
𝛁×𝑯=𝑱+𝑮
Taking Divergence of both sides:
𝛻∙𝛻×𝑯≡0=𝛻∙𝑱+𝛁∙𝑮
From Continuity Equation:
𝛁∙𝑮=−𝛻∙𝑱= 𝜕𝜌 𝜕𝑡
Using Gauss’s Law:
𝛁∙𝑮= 𝜕 𝜵∙𝑫 𝜕𝑡 =𝛻∙ 𝜕𝑫 𝜕𝑡 → 𝐺= 𝜕𝑫 𝜕𝑡
Ampere’s Law (complete)
𝛁×𝑯=𝑱+ 𝜕𝑫 𝜕𝑡

10. Maxwell’s Equations in Point Form (Complete)
Ampere’s Circuital Law
Faraday’s Law of Induction
Gauss’ Law for the electric field
Gauss’s Law for the magnetic field
Wave setup and propagation
Speed of light, EM spectrum

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