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Faraday's law -- How is the RHS required for all surfaces?
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[QUOTE="Delta2, post: 6501509, member: 189563"] do you mean why the right side is independent of the surface s as long as we keep the curve c , the boundary of the surface the same? This is a consequence of Stoke's Theorem (or curl theorem). Since ##\nabla\cdot \vec{B}=0## (Gauss's law for magnetism) we can set ##\vec{B}=\nabla\times\vec{A}##. Stokes theorem tell us that $$\iint_S \vec{B}\cdot d\vec{S}=\iint _S(\nabla\times \vec{A})\cdot d\vec{S}=\oint_C\vec{A}\cdot d\vec{l}$$ so the surface integral of B over S will equal the line integral of A over the curve C, thus it remains constant for all surfaces S with the same boundary C. [/QUOTE]
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Faraday's law -- How is the RHS required for all surfaces?
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