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)
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
Faraday Induction I (Serway) “Physics for Scientists and Engineers”, Serway/Jewett, 8th Edition, page 895
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
Faraday Induction II (Serway) Experimentally-verified postulate (note ds here is a path)
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
Displacement Current (Serway)
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!
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
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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