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Lecture 11: Stokes Theorem Consider a surface S, embedded in a vector field Assume it is bounded by a rim (not necessarily planar) For each small loop For whole loop (given that all interior boundaries cancel in the normal way) OUTER RIM SURFACE INTEGRAL OVER ANY SURFACE WHICH SPANS RIM

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Consequences Curl is incompressible Consider two surfaces (S 1 + S 2 ) with same rim must be the the same for both surfaces (by Stokes Theorem) For closed surface formed by Also for a conservative field using the divergence theorem CONSERVATIVE = IRROTATIONAL These are examples (see Lecture 12) of second-order differential operators in vector calculus by Stokes Theorem

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Example So this is a conservative field, so we should be able to find a potential All can be made consistent if

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Another Example quarter circle (0,1,0) y x z (0,0,1) C1C1 C3C3 C2C2 Check via direct integration

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Magnetic Field due to a steady current I “Ampere’s Law” Since obeys a continuity equation with no sources or sinks (see Lecture 13) Physical Example applying Stoke’s Theorem units: A m -2 Differential form of Ampere’s Law

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Now make the wire ‘fat’ (of radius a), carrying a total current I Inside the wire Outside the wire Inside, has a similar field to in solid-body rotation, where z axis along the wire, r is radial distance from the z axis a r |B|

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Outside wire: z component from loops in the xy plane z component is zero, because and x and y components are also zero in the yz and xz planes Therefore R1R1 R2R2 dd Note that region where field is conservative ( ) is not simply connected: loop cannot be shrunk to zero without going inside wire (where field is not conservative).

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