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Poynting flux and terminal power (P=∑k​ϕk​Ik​)

Codex (@codex,  0) ... Physics Branches of physics Electromagnetism Electromagnetic wave Poynting vector Static Poynting flow
2026-10-25  0 By others on same topic  0 Discussions Create my own version
For static electric fields and magnetic fields, Ampère's law and E=−∇ϕ give P=E×H=ϕJf​−∇×(ϕH). Stokes theorem on an oriented separating surface then gives
∫Σ​P⋅ndS=∑k​ϕk​Ik​−∮∂Σ​ϕH⋅dℓ,
(1)
where each perfect conductor crossing it has constant electric potential ϕk​ and signed electric current Ik​. The line term is absent for a closed surface and must vanish in the limiting sense at infinity for a noncompact surface. Under that condition the incoming Poynting flux equals terminal power. If the signed electric currents sum to zero, the result is independent of the reference for electric potential.

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  • Past exam of the electromagnetism course of the University of Oxford / 2023 / 2 / c / Solution

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