Lecture 04. Curl and Divergence

Transcription

1 Lecture 04 Curl and Divergence

2 UCF Curl of Vector Field (1) F c d l F C Curl (or rotor) of a vector field a n curlf F d l lim c s s 0 F s a n C a n : normal direction of s follow right-hand rule

3 UCF Curl of Vector Field () We can take a to be a, a, a and get (curl F) if a = a (eample) n z n z z,, F d l F z d + F + z d c + + = (,, ) (,, ) + F (, +, z) d + F(,, z) d + + +Δ/ -Δ/ Δ a n (,,z) Δ l s = -Δ/ +Δ/ = [ F (,, z) F (, +, z)] d + [ F ( +,, z) F (,, z)] d + +

4 UCF Curl of Vector Field (3) Finite difference -Δ/ +Δ/ c Note that Define a a a U( + ) U( ) U F l d ( ) d d F d l + F c az curlf lim = = s 0 s Likewise a z curl F =, z curl a z F= z z z A F= A = a + a + a A Az z curl F F F z Del operator a a a z curlf = z F F F z F= F

5 UCF Curl of Vector Field (4) In general F = ha h a ha hhh u u u hf hf hf Matlab: curl (onl numerical) Self-defined: rot (smbolic) Cartesian clindrical spherical u 1 u u 3 h 1 h h 3 z ρ Φ z 1 ρ 1 r θ Φ 1 r rsinθ

6 UCF Surface Integration ψ = A d s flu In electromagnetics: ψ = e D d s electric flu ψ = d s m B electric flu densit displace vector ds = dsa n magnetic flu magnetic flu densit I = J d s current volume current densit

7 UCF Stokes s Theorem ( F) ds= F dl c S

8 UCF Eample 1.9 e109.m

9 UCF Eample 1.10 Given a vector field A = a a, verif Stokes s Theorem over one-quarter of a circle whose radius is 3. e110.m

10 UCF Eample 1.10 Solutions: A dl = A d + A d A= a a check stokes's theorem A d l = A d l , arc 0 3 = d + d 3 0 A = 0 + = A = 0 Matlab B= A= [ B, B, B ] z 3 9 z 0 0 B ds= B dd = ( B d) d a n = a z z e110.m

11 UCF Divergence of Vector Field (1) Closed surface integration: F d s F No source inside d s = 0 Source is inside F d s 0 Divergence of a vector field: divf = lim v 0 F v d s a n divf = F= + + z Proof

12 UCF Divergence of Vector Field () Proof: outward flu through face 1 is z ψ1 = F ( az ) dd F(,, z ) for face z ψ = Fa zdd F(,, z + ) z z ψ1+ ψ = [ Fz (,, + ) Fz (,, )] ( z) = z z z Likewise ψ3 + ψ4 = z -Δ/ ψ5 + ψ6 = z ψ = ψ1+ ψ + ψ3 + ψ4 + ψ5 + ψ6 = ( + + ) z z z Δz +Δ/ Δ Face 3 -Δ/ d ( + + ) z z lim F s = lim = + + = F v 0 v v 0 z z Δ Face Face 6 Face 1 Face 5 Face 4 +Δ/ z+δz/ z-δz/

13 UCF Divergence of Vector Field (3) In general 1 F = + + hhh u u u ( Fh h ) ( Fhh ) ( Fhh ) Matlab: divergence (onl numerical) Self-defined: div (smbolic)

14 UCF Gauss s Divergence Theorem V Fdv = F ds S

15 UCF Eample 1.11 Assume that vector field: A0 A= a r r eists in a region surrounding the origin of a spherical coordinate sstem. Find the value of the closed-surface integral over the unit sphere.

16 UCF Eample 1.11 Solution: ds r sinθdθdφa = r The closed-surface integral is given b A = r sin d d = 4πA φ= 0 θ= 0 r φ= π θ= π 0 A ds a ( θ θ φ ) r a r rdθ In this integral, we have used the differential surface area in spherical coordinates that has a unit vector a r. If the vector A had an additional components directed in the a θ or a φ directions, their contribution to this surface integral would be zero, since the scalar product of these terms will be equal to zero. 0 e111.m rsinθdφ

17 UCF Eample 1.1 A= a + a + za 5 4 Assume that the vector field z is defined in a region 1 3, 4 and 1 z Find the value of the closed-surface integral A ds over the surface of this region. e11.m

18 UCF Eample 1.1 Solution: The region has si surfaces. On the front surface, =3 and is the onl component perpendicular to that surface. Therefore, A ds front = A = 3 ddz = 5 3ddz = 15ddz 4 4 ( ) A ds front = 15ddz = 15 dz = On the back surface, =1 and is the onl component perpendicular to that surface. Therefore, A ds back = A = 1 ( ddz) = 5 1ddz = 5ddz 4 4 ( ) A ds back = 5ddz = 5 dz = z z A= a + a + a 1 3, 4 and 1 z

19 UCF Eample 1.1 On the right surface, and is the onl component perpendicular to that surface. Therefore, A ds = A ddz right = 4 = ddz = ddz A ds right = 16ddz = 16 dz = On the left surface, and is the onl component perpendicular to that surface. Therefore, A ds = A ( ddz) left = ( ) ( ddz) 4ddz = = A ds left = 4ddz = 4 dz = z z A= a + a + a 1 3, 4 and 1 z

20 UCF Eample 1.1 On the top surface, and is the onl component perpendicular to that surface. Therefore, A ds = A dd top z z= = 4 dd = 8dd 4 3 A ds = 8dd = 96 top 1 On the bottom surface, and is the onl component perpendicular to that surface. Therefore, A ds = A ( dd) top z z= 1 = 4 ( 1)( dd) = 4dd 4 3 A ds = 4dd = 48 top 1 The closed surface integral can be obtained b the summation of the above si surface integrals as: A d s = = 468 e11.m intsurf.m