Transcript notes16 2317 - University of Houston

ECE 2317
Applied Electricity and Magnetism
Prof. D. Wilton
ECE Dept.
Notes 16
Notes prepared by the EM group,
University of Houston.
Curl of a Vector
V  x, y, z   arbitrary vector function
curl V  vector function
z
S
Cz
S
y
S
1
S 0 S
1
y  curl V  lim
S 0 S
1
z  curl V  lim
S 0 S
x  curl V  lim

Cx

Cy

Cz
V  dr
V  dr
V  dr
 V  dr  ofcirculation
V on C
Cx, y , z
x, y, z
x  curl V  circulation per
unit area about
x, etc.
Cy
x
Cx
Note: Paths are defined
according to the “right-hand rule”
Curl of a Vector (cont.)
“curl meter”
 xˆ , yˆ , zˆ
Assume that V represents
the velocity of a fluid.
 curl V  velocityof rotation (in the sense indicated)
Curl Calculation
z
Path Cx :
4
1
z 2
3

Cx
V  dr 
Cx
y
y
 y 
V
dx

V
dy

V
dz

V
, 0  z
y
z
z  0,
 Cx x
 2

y 

 Vz  0, 
, 0  z
2


z 

 Vy  0, 0,   y
2 

z 

 Vy  0, 0,
 y
2 

(side 1)
(side 2)
(side 3)
(side 4)
Curl Calculation (cont.)
  y 
y  

V
0,
,
0

V
0,

, 0
 z
 z 2
2





V

dr


y

z


 Cx
y




 
z 
z  

Vy  0, 0, 2   Vy  0, 0,  2  



  z y   
z




Vy
Vz
 S
 S
y
z
 Vz Vy 
C V  dr  S  y  z 
x
Though above calculation is for
a path about the origin, just add
(x,y,z) to all arguments above to
obtain the same result for a path
about any point (x,y,z) .
Curl Calculation (cont.)
 Vz Vy 
C V  dr  S  y  z 
x
From the curl definition:
Hence
1
s  0 S
x  curl V  lim
Vz Vy
x  curl V 

y
z

Cx
V  dr
Curl Calculation (cont.)
Similarly,
 Vx Vz 
V

dr


S



C

z

x


y
 Vx Vz 
y  curl V  



z

x


 Vy Vx 
C V  dr  S  x  y 
z
 Vy Vx 
z  curl V  



x

y


Hence,
 Vz Vy 
 Vx Vz   Vy Vx 
curl V  x 



 y

 z
z 
x   x
y 
 z
 y
x
Note the cyclic nature of the three terms:
y
z
Del Operator
 


x  y z 
y
z 
 x

 

 
  V   x  y  z   xVx  yVy  zVz
y
z 
 x

x
y
z

x
Vx

y
Vy

z
Vz

 Vz Vy 
 Vz Vx   Vy Vx 
 x



 y

 z

y

z

x

z

x

y

 



Del Operator (cont.)
Hence,
curl V  V
Example
V  x  3xy z   y  2 x  z   z  2 xz 
2
 V 
2
3
x
y
z
x
y
z

x
Vx

y
Vy

z
Vz

x
3xy 2 z

y
2x2 z3

z
2 xz
  V  x  0  3z
2

  y  2 z  3xy   z  4 x  6 xyz 
2
Example
V  x y
 Vz Vy 
 Vz Vx
 V  x 


y



z 
z
 x
 y
  Vy Vx 


 z
y 
  x
  V  z  1
y
x
Example (cont.)
  V  z  1
   V   z  1  0
y
x
Summary of Curl Formulas
 Vz Vy 
 Vz Vx   Vy Vx 
 V  x 



 y

 z

y

z

x

z

x

y

 



 1 Vz V 
 V Vz 
1    V  V
V   



  
  z 
z 
 
  

  
 z
1   V sin   V
V 


r sin  








1  1 Vr   rV  
1    rV  Vr 


r  
  

r  sin  
r 
r  r
 



Stokes’s Theorem
S (open)
n
C
n : chosen from “right-hand rule” applied to the surface
    V   n dS   V  dr
S
C
“The surface integral of circulation per unit area equals the total circulation.”
Proof
Divide S into rectangular patches that are normal to x, y, or z axes.
ni
r
i
n
Ci
S
S
 n  x, y, or z 
i
C
Independently consider the left and right
hand sides (LHS and RHS) of Stokes’s theorem:
LHS :
   V   n dS    V 
S
i
ri
 ni S
Proof (cont.)
S
C
LHS :
   V   n dS   V 
1
ni     V   lim
s  0 S
ni    V  ri
 n i S
i
S
1

S
ri

Ci

Ci
V  dr
V  dr
  V  r  ni S   V  dr
i
Ci
e.g, ni  x, y, z
Proof (cont.)
S
C
Hence,
   V 
i
ri
 n i S      V   n ds
S
   V  dr   V  dr
i
Ci
C
     V   n ds   V  dr
S
C
C
(Interior edge integrals cancel)
Example
y
CC
Verify Stokes’s theorem
for V  x y
= a,
z= const
 V  dr   V
CB
C
C
CA
dx  V y dy
C

x
x
 x dy
C
 I C A  I CB  I CC
V  x y
ICA  0
( dy = 0 )
I CC  0
(x=0)
(dz = 0)
Example (cont.)
IB 
 x dy
a
y a y
a 2 1  y  
IB  
 sin   
2
2
 a  

y 0
2
CB
y
=a
B
CB
x
a 2 1
 sin 1
2
a2   
  
2 2
A
a
I B   a 2  y 2 dy
0
I
 a2
4
2
Example (cont.)
Alternative evaluation
(use cylindrical coordinates):
B
I B   V  dr
A

B
  V   d   ˆ a d  z dz

A


2
 V a d
0
Now use:
 
V  V    y x    x y  
 x cos ,
x  a cos 
or
V   a cos  cos
 a cos 
2
Example (cont.)

Hence
IB 
2

a 2 cos 2 d
0

 a2
 1  cos2 
0  2  d
2

2
sin
2






 a2  

4 0
2
 
a  
4
2
a 2
I
4
Example (cont.)
Now Use Stokes’s Theorem:
I   V  dr      V   z ds
C
V xy
(nˆ  z)
S
 Vz Vy 
 Vz Vx   Vy Vx 
 V  x 



 y

 z

y

z

x

z

x

y

 



  V  z 1
1
I   z 1  z dS   dS A   a 2 
4
S
S
I
 a2
4
Rotation Property of Curl
n
1
V  dr
 V   n  lim

S 0 S
C
(constant)
S (planar)
C
The component of curl in any direction
measures the rotation (circulation) about
that direction
Rotation Property of Curl (cont.)
n
Proof:
Stokes’s Th.:
   V   n ds   V  dr
S
But
C
   V   n ds    V   n S
S
Hence
(constant)
   V   n S   V  dr
C
Taking the limit:
1
V  dr
 V   n  lim

S 0 S
C
S (planar)
C
Vector Identity
 V   0
Proof:
 Vz Vy 
 Vz Vx   Vy Vx 
 V  x 



 y

 z

y

z

x

z

x

y

 



Ax Ay Az
   V  


x
y
z
A
  2Vz  2Vy    2Vz  2Vx    2Vy  2Vx 






 x y x z   y x y z   z x z y 
 

 

0
Vector Identity
 V   0
Visualization:
V
Edge integrals cancel
when summed over
closed box!
nˆ i Si
Ci
Flux of V out of V
1
   V  
V

face i
  V  nˆ i Si
 1
V dr

Si C
i
0
Example
Find curl of E:
3
2
1
q
s0
Infinite sheet of charge
(side view)
l0
Infinite line charge
Point charge
Example (cont.)
1
x
 s0 
E  xˆ 

2

 0
s0
 Ez E y 
 Ez Ex   E y Ex 
 E  x 



 y

 z
z 
z   x
y 
 x
 y
  E  x  0  0  y 0  0  z 0  0
0
Example (cont.)
2
 0 
E  

2



0


l0
 1 Ez E 
 E Ez 
1     E  E
 E   



  
  z 
z 
 
  

  
 z
0




Example (cont.)
3
q
1
 E 
r sin 
 q 
E  r
2 
 4  0 r 
   E sin   E 
1  1 Er   rE  
1    rE  Er 





r  
  





r
sin




r
r

r








0
By superposition, the result  E  0 ,
must be true for any general charge distribution
Faraday’s Law (Differential Form)
Stokes’s Th.:
    E   n dS   E  dr  0
S
(in statics)
C
n
Let S  S
Hence
S
small planar surface
    E   n dS  0
S
Let S  0:
n     E  S  0
n    E   0
Faraday’s Law (cont.)
n    E   0
Let n  xˆ , yˆ , zˆ :
x    E   0
y    E   0
z    E   0
Hence
 E  0
n
S
Faraday’s Law (Summary)

E dr  0
Integral form of Faraday’s law
C
Stokes’s
theorem
curl
definition
 E  0
Differential (point) form of Faraday’s law
Path Independence
V  0
Assume
A
B
C1
I1   V  d r
C2
I2   V  d r
C1
C2
I1  I 2
Path Independence (cont.)
Proof
B
A
C
C = C2 - C1

C  C2 C1
V  d r      V   n dS  0
S
I 2  I1  0
S is any surface that is
attached to C.
(proof complete)
Path Independence (cont.)
V  0
Stokes’s theorem
Definition of curl
path independence
 V dr  0
C
Summary of Electrostatics
  D  v
 E  0
D  0 E
Faraday’s Law: Dynamics
In statics,
 E  0
Experimental Law
(dynamics):
B
 E  
t
Faraday’s Law: Dynamics (cont.)
B
 E  
t
Bz
zˆ     E   
0
t
(assume that Bz increases with time)
y
magnetic field Bz (increasing with time)
x
electric field E
Faraday’s Law: Integral Form
B
 E  
t
Apply Stokes’s theorem:
    E   nˆ dS   E  d r
S
C
 B 
 
  nˆ dS
t 
S
Faraday’s Law (Summary)
 B 
C E  d r  S  t   nˆ dS
Integral form of Faraday’s law
Stokes’s Theorem
B
 E  
t
Differential (point) form of Faraday’s law
Faraday’s Law (Experimental Setup)
+
y
V >0
-
x
Note: the voltage drop
along the wire is zero
magnetic field B (increasing with time)
Faraday’s Law (Experimental Setup)
+
A
y
 B 
C E  d r  S  t   nˆ dS
Bz

dS  0
t
S
V >0
-
B
C
x
S
B
V   E dr 
A
Hence
 E dr
C
V 0
(nˆ   z)
Note: the
voltage drop
along the wire
is zero
Differential Form of
Maxwell’s Equations
  D  v
B
 E  
t
B  0
D
 H  J 
t
electric Gauss law
Faraday’s law
magnetic Gauss law
Ampere’s law
Integral Form of
Maxwell’s Equations
 D  nˆ dS    dV
v
S
electric Gauss law
V
d
C E  d r   dt S B  nˆ dS
 B  nˆ dS  0
Faraday’s law
magnetic Gauss law
S
d
C H  d r  S J  nˆ dS  dt S D  nˆ dS
Ampere’s law