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Content Ampere-Maxwell Law
Ampere-Maxwell Law![]() ∮ 𝐵⋅𝑑 𝑙=0 But when the 𝐵-field is moving, 𝜇0 ∮ 𝐵⋅𝑑 𝑙≠0 Electric flux through this surface: 1𝜀0 𝑄𝐴𝐴= 𝑄𝜀0 The time derivative of electric flux is: 𝑑𝑑𝑡 𝑑𝑑𝑡 𝑄𝜀0≡ 1𝜀0𝐼 So the changing flux acts like a current inside the capacitor. And therefore the line integral of magnetic field is. ∮ 𝐵⋅𝑑 𝑙=𝜇0 ∑𝐼enclosed+𝜀0 𝑑𝑑𝑡 where the first right term contributes outside the capacitor and the second right term contributes inside the capacitor. Maxwell's Equations
Everything there is to know about electricity and magnetism is contained in these four laws plus the force law: 𝐹=𝑞 𝐸+𝑞 𝑣× 𝐵 Differential Form of Gauss's LawGauss's Law:∮ 𝐸⋅ 𝑛𝑑𝐴= 1𝜀0 ∑𝑄enclosed ![]()
𝐸⋅ 𝑛𝑑𝐴∆𝑉
1𝜀0 ∑𝑄enclosed∆𝑉≡ 1𝜀0𝜌 ⇒
𝐸⋅ 𝑛𝑑𝐴∆𝑉
(𝐸2−𝐸1)∆𝑦∆𝑧∆𝑥∆𝑦∆𝑧=
(𝐸2−𝐸1)∆𝑥≡ ∂𝐸𝑥∂𝑥= 1𝜀0𝜌 for a general case where 𝐸 can point in any direction: ∂𝐸𝑥∂𝑥+ ∂𝐸𝑦∂𝑦+ ∂𝐸𝑧∂𝑧≡ ∇⋅ 𝐸= 1𝜀0𝜌 The parallel derivative of Gauss's Law differential form where ∇≡ ∂∂𝑥, ∂∂𝑦, ∂∂𝑧 Differential Form of Ampere's LawAmpere-Maxwell Law:∮ 𝐵⋅𝑑 𝑙=𝜇0 ∑𝐼enclosed+𝜀0 𝑑𝑑𝑡 ![]() 𝐽⋅ 𝑛∆𝐴.
𝐵⋅𝑑 𝑙∆𝐴
𝐽⋅ 𝑛∆𝐴∆𝐴
𝑑𝑑𝑡 ∫𝜇0𝜀0=
𝜇0𝐽𝑧∆𝐴∆𝐴+
𝑑𝑑𝑡 𝐸𝑧∆𝐴∆𝐴𝜇0𝜀0 ⇒
𝐵⋅𝑑 𝑙∆𝐴 𝐽𝑧+ 𝑑𝐸𝑧𝑑𝑡𝜇0𝜀0 ⇒
𝐵⋅𝑑 𝑙∆𝐴
(𝐵1,𝑥−𝐵3,𝑥)∆𝑥+(𝐵2,𝑦−𝐵4,𝑦)∆𝑦∆𝑥∆𝑦=𝜇0 𝐽𝑧+ 𝑑𝐸𝑧𝑑𝑡𝜇0𝜀0 ⇒
𝐵⋅𝑑 𝑙∆𝐴
(𝐵1,𝑥−𝐵3,𝑥)∆𝑦
(𝐵2,𝑦−𝐵4,𝑦)∆𝑥=− ∂𝐵𝑥∂𝑦+ ∂𝐵𝑦∂𝑥=𝜇0 𝐽𝑧+ 𝑑𝐸𝑧𝑑𝑡𝜇0𝜀0 crossed derivatives: − ∂𝐵𝑥∂𝑦+ ∂𝐵𝑦∂𝑥 For a loop in any direction, this can be re-expressed as: ∂𝐵𝑧∂𝑦− ∂𝐵𝑦∂𝑧 ∂𝐵𝑥∂𝑧− ∂𝐵𝑧∂𝑥 ∂𝐵𝑦∂𝑥− ∂𝐵𝑥∂𝑦 ∇× 𝐵=𝜇0 𝐽+𝜀0 ∂ Differential Form of Maxwell's EquationsDivergence for enclosed flux:
Source and Referencehttps://www.youtube.com/watch?v=fkfnDopQBYQ&list=PLZ6kagz8q0bvxaUKCe2RRvU_h7wtNNxxi&index=26©sideway ID: 200200502 Last Updated: 2/5/2020 Revision: 0 Latest Updated Links
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