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Lecture 11: Stokes Theorem

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- 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

- Curl is incompressible
- Consider two surfaces (S1 + S2) 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

by Stokes Theorem

CONSERVATIVE = IRROTATIONAL

- These are examples (see Lecture 12) of second-order differential operators in vector calculus

- So this is a conservative field, so we should be able to find a potential

- All can be made consistent if

z

(0,0,1)

C2

quarter circle

C1

C3

y

(0,1,0)

x

- Check via direct integration

- Magnetic Field due to a steady current I
- “Ampere’s Law”
- Since
- obeys a continuity equation with no sources or sinks (see Lecture 13)

applying Stoke’s Theorem

units: A m-2

Differential form of Ampere’s Law

- 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

|B|

a r

- 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

R2

d

R1

- 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).