This document discusses a course on electromagnetic theory taught by Arpan Deyasi. It covers topics related to vector differentiation, including the vector differential operator in Cartesian, cylindrical and spherical coordinates. It also covers differentiation of scalar functions, including calculating gradients, directional derivatives and finding normals to surfaces. Finally, it discusses differentiation of vector functions, specifically divergence, which represents the volume density of the net outward flux from a vector field.

1. Course: Electromagnetic Theory
paper code: EI 503
Course Coordinator: Arpan Deyasi
Department of Electronics and Communication Engineering
RCC Institute of Information Technology
Kolkata, India
Topics: Scalar and Vector Differentiation
Arpan Deyasi
Electromagnetic
Theory

2. Vector Differential Operator
Del is called the vector differential operator 
It is the first order differentiation
In Cartesian coordinate ˆ
ˆ ˆ
i j k
x y z
  
 = + +
  
In Cylindrical coordinate
1
ˆ
ˆ ẑ
z
 
  
  
 = + +
  
In Spherical coordinate
1 1
ˆ ˆ
ˆ
sin
r
r r r
 
  
  
 = + +
  
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Electromagnetic
Theory

3. Differentiation of Scalar Function
Φ (x, y, z) is defined and differentiable scalar function at each
point (x, y, z) in a certain region of space
ˆ
ˆ ˆ
i j k
x y z
   
  
 = + +
  
Gradient of Φ Vector function
Significance: rate of change of scalar function in a region of space
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Electromagnetic
Theory

4. Differentiation of Scalar Function
n̂
Let is the arbitrary vector in a region of space where gradient
of any scalar function Φ exists
ˆ ˆ
ˆ ˆ ˆ ˆ
ˆ
. . . .
n i i j j k k
x y z
  

  
 = + +
  
ˆ
.n
x y z
  

  
 = + +
  
Directional Derivative of Φ in the direction of n
Significance: rate of change of scalar function Φ in the direction of n
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Electromagnetic
Theory

5. Differentiation of Scalar Function
Prob 1:
2 3 2
( , , ) 3
x y z x y y z
 = − find
( )
( )
2 2 2 3
( 1,2,1) ( 1,2,1)
ˆ
ˆ ˆ
6 3 3 2
xyi x y z j y zk

− −
 = + − −
Soln:
( )
2 3 2
ˆ
ˆ ˆ 3
i j k x y y z
x y z

 
  
 = + + −
 
  
 
( )
2 2 2 3 ˆ
ˆ ˆ
6 3 3 2
xyi x y z j y zk

 = + − −
( 1,2,1)

−

( 1,2,1)
ˆ
ˆ ˆ
12 9 16
i j k

−
 = − − −
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Electromagnetic
Theory

6. Differentiation of Scalar Function
Prob 2: ( , , ) ln
x y z r
 = find 

Soln:
ˆ
ˆ ˆ
r xi yj zk
= + +
( )
2 2 2
1
ln ln
2
r x y z
 = = + +
( )
2 2 2
1
ln ln
2
r x y z
  
 = =  + +
 
( )
2 2 2
1 ˆ ln ....
2
i x y z
x


 
 
 = + + +
 
 

 
Arpan Deyasi
Electromagnetic
Theory

7. Differentiation of Scalar Function
( )
2 2 2
1 1
ˆ.2 . ....
2
i x
x y z

 
 
 = +
+ +
 
 
( )
2 2 2
ˆ
ˆ ˆ
ix jy kz
x y z

 
+ +
 
 =
+ +
 
 
2
r
r

 =
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8. Differentiation of Scalar Function
Prob 3: ( , , ) n
x y z r
 = find 

Soln:
ˆ
ˆ ˆ
r xi yj zk
= + +
( )
/2
2 2 2 n
n
r x y z
 = = + +
( )
/2
2 2 2 n
n
r x y z

 =  =  + +
( )
/2
2 2 2 n
n
r x y z

 =  =  + +
( )
/2
2 2 2
ˆ
ˆ ˆ
n
i j k x y z
x y z

 
  
 = + + + +
 
  
 
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Electromagnetic
Theory

9. Differentiation of Scalar Function
( )
/2
2 2 2
ˆ .........
n
i x y z
x


 
 = + + +
 

 
( )
( /2 1)
2 2 2
ˆ .2 .........
2
n
n
i x y z x

−
 
 = + + +
 
 
( )
( )
( /2 1)
2 2 2
ˆ .........
n
inx x y z

−
 = + + +
2 ( /2 1)
( ) n
n r r
 −
 =
( )
( 2)
n
n r r
 −
 =
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Electromagnetic
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10. Differentiation of Scalar Function
Prob 4: Show that is a vector perpendicular to the surface Φ(x, y, z)
= K where ‘K’ is a constant


Soln:
ˆ
ˆ ˆ
r xi yj zk
= + +
ˆ
ˆ ˆ
dr dxi dyj dzk
= + +
( , , )
x y z K
 =
( , , ) 0
d x y z
 =
0
dx dy dz
x y z
  
  
+ + =
  
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Electromagnetic
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11. Differentiation of Scalar Function
ˆ ˆ
ˆ ˆ ˆ ˆ
.( ) 0
i j k dxi dyj dzk
x y z
 
  
+ + + + =
 
  
 
. 0
dr

 =

 is perpendicular to r
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Electromagnetic
Theory

12. Differentiation of Scalar Function
Find unit normal to the surface 4
2
2
−
+
= xz
y
x
 at (-2,2,3)
)
4
2
( 2
−
+

=
 xz
y
x



k
x
j
x
i
z
xy ˆ
2
ˆ
ˆ
)
2
2
( 2
+
+
+
=


k
j
i ˆ
4
ˆ
4
ˆ
2
)
3
,
2
,
2
(
−
+
−
=

−


Prob 5:
Soln:
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13. Differentiation of Scalar Function
Downward unit normal
2
2
2
4
4
2
ˆ
4
ˆ
4
ˆ
2
ˆ
+
+
−
+
−

=
k
j
i
n
k
j
i
n ˆ
3
2
ˆ
3
2
ˆ
3
1
ˆ 
 
=
Unit normal to the surface
k
j
i
nup
ˆ
3
2
ˆ
3
2
ˆ
3
1
ˆ −
+
−
=
Upward unit normal
k
j
i
ndown
ˆ
3
2
ˆ
3
2
ˆ
3
1
ˆ +
−
=
( , , )
x y z

ˆup
n
ˆdown
n
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Electromagnetic
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14. Differentiation of Scalar Function
In the direction of
)
4
( 2
2
xz
yz
x +

=




at (1,-2,1)
Find the directional derivative of
k
j
i ˆ
2
ˆ
ˆ
2 −
+
2
2
4xz
yz
x +
=

k
zx
y
x
j
z
x
i
z
xyz ˆ
)
8
(
ˆ
ˆ
)
4
2
( 2
2
2
+
+
+
+
=


k
j ˆ
4
ˆ
2
)
1
,
2
,
1
(
+
=

−


Prob 6:
Soln:
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15. Differentiation of Scalar Function
Unit vector in the direction of k
j
i ˆ
2
ˆ
ˆ
2 −
+ is
2
2
2
2
1
2
ˆ
2
ˆ
ˆ
2
ˆ
+
+
−
+
=
k
j
i
n
k
j
i
n ˆ
3
2
ˆ
3
2
ˆ
3
1
ˆ −
+
−
=
Directional derivative 





−
+
−
+
=
 k
j
i
k
j
n ˆ
3
2
ˆ
3
2
ˆ
3
1
).
ˆ
4
ˆ
2
(
ˆ
.


3
4
3
8
3
4
ˆ
. −
=






−
=
 n




n̂
2
2
4xz
yz
x +
=

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Electromagnetic
Theory

16. Prob 7: Find the unit vector perpendicular to the surface
at point (4,2,3)
(4,2,3)
ˆ ˆ
ˆ ˆ ˆ ˆ
2 4 2 2 2 3 8 4 6
i j k i j k

 =  +  −  = + −
Differentiation of Scalar Function
2 2 2
( , , ) 11
x y z x y z
 = + − −
ˆ ˆ
ˆ ˆ ˆ ˆ
2 2 2
i j k i x j y k z
x y z
  

 
  
 = + + = + −
 
  
 
2 2 2
11
x y z
+ − =
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17. Unit normal to the surface at (4,2,3)
ˆ
ˆ ˆ
8 4 6
ˆ
64 16 36
i j k
n
+ −
= 
+ +
Differentiation of Scalar Function
ˆ
ˆ ˆ
8 4 6
ˆ
116
i j k
n
+ −
= 
ˆ
ˆ ˆ
8 4 6
ˆ
2 29
i j k
n
+ −
= 
4 2 3 ˆ
ˆ ˆ
ˆ
29 29 29
n i j k
=  
4 2 3 ˆ
ˆ ˆ
ˆ
29 29 29
up
n i j k
= + −
4 2 3 ˆ
ˆ ˆ
ˆ
29 29 29
down
n i j k
= − − +
( , , )
x y z

ˆup
n
ˆdown
n
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Electromagnetic
Theory

18. Differentiation of Vector Function: Divergence
( )
1 2 3
ˆ ˆ
ˆ ˆ ˆ ˆ
. .
divP P i j k iP jP kP
x y z
 
  
=  = + + + +
 
  
 
𝑷 (x, y, z) is defined and differentiable vector function at each
point (x, y, z) in a certain region of space
1 2 3
.P P P P
x y z
  
 = + +
  
1 2 3
ˆ
ˆ ˆ
P Pi P j Pk
= + +
Divergence of P Scalar function
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Electromagnetic
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19. 19
Differentiation of Vector Function: Divergence
Significance: volume density of net outward flux from a vector
field originated from a given point (source) where the point is
defined in the region of space
If divergence is negative, then inflow is signified and the
point becomes sink
. 0
P
 =
If
the vector becomes solenoidal/ also called incompressible vector field
A solenoidal vector field has zero divergence. That means that it has no sources or sinks; all
field lines form closed loops. It means that the total flux of the vector field through arbitrary
closed surface is zero.
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Electromagnetic
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20. 20
∂Ax/∂x + ∂Ay/∂y
x
y
Differentiation of Vector Function: Divergence
Visualizing Divergence
+ ∂Az/∂z
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21. Second order differentiation is called Laplacian
Scalar Differential Operator
2
.
 =  
In Cartesian coordinate
2 2 2
2
2 2 2
x y z
  
 = + +
  
In Cylindrical coordinate
2 2 2
2
2 2 2 2
1 1
r r r r z

   
 = + + +
   
In Spherical
coordinate
2
2 2
2 2 2 2 2
1 1 1
sin
sin sin
r
r r r r r

    
    
   
 = + +
   
    
   
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Electromagnetic
Theory

22. Laplacian of scalar function V= div grad V
ˆ ˆ
ˆ ˆ ˆ ˆ
.( ) .
V V V
V i j k i j k
x y z x y z
   
     
  = + + + +
   
     
   
Laplacian operator is given as
2 2 2
2
2 2 2
x y z
  
 = + +
  
Scalar Differential Operator
.( )
V V V
V
x x y y z z
 
     
   
  = + +
   
 
     
   
 
2 2 2
2
2 2 2
V V V
V
x y z
  
 = + +
  
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Electromagnetic
Theory

23. Differentiation of Vector Function: Divergence
Prob 1: Find the divergence of 𝑭 where
( )
ˆ ˆ
ˆ ˆ ˆ ˆ
. . x y z
F i j k iF jF kF
x y z
 
  
 = + + + +
 
  
 
2 ˆ ˆ
F x yzi zxj
= +
( ) ( )
2
. (0) 2xyz 0 0 2xyz
F x yz xz
x y z
  
 = + + = + + =
  
. 2xyz
F
 =
Soln:
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24. Differentiation of Vector Function: Divergence
Prob 2: 2 3 2 ˆ
ˆ ˆ
2
A x zi y zj y zk
= − +
If
at (1,1,-1)
.A

find
( )
2 3 2
ˆ ˆ
ˆ ˆ ˆ ˆ
. . 2
A i j k ix z j y z kxy z
x y z
 
  
 = + + − +
 
  
 
( ) ( ) ( )
2 3 2
. 2
A i x z y z xy z
x y z
 
  
 = + − +
 
  
 
Soln:
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Electromagnetic
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25. Differentiation of Vector Function: Divergence
( )
2 2
. 2 6
A xz y z xy
 = − +
( )
(1,1, 1)
. 2 6 1
A
−
 = − + +
(1,1, 1)
. 5
A
−
 =
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Electromagnetic
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26. Differentiation of Vector Function: Divergence
Prob 3:
Show that . 0
r
 =
Soln:
( )
ˆ ˆ
ˆ ˆ ˆ ˆ
. .
r i j k ix jy kz
x y z
 
  
 = + + + +
 
  
 
.r x y z
x y z
 
  
 = + +
 
  
 
. 1 1 1
r
 = + + . 3
r
 =
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Electromagnetic
Theory

27. Differentiation of Vector Function: Divergence
Prob 4: 2 1
0
r
 
 =
 
 
Show that
Soln:
ˆ
ˆ ˆ
r xi yj zk
= + +
( )
1
2 2 2 2
1
x y z
r
−
= + +
( )
2 2 2 1
2 2 2 2 2
2 2 2
1
x y z
r x y z
−
   
  
 = + + + +
   
    
 
 
Arpan Deyasi
Electromagnetic
Theory

28. Differentiation of Vector Function: Divergence
( )
2 1
2 2 2 2 2
2
1
.........
x y z
r x
−
   

 = + + +
   
  
 
 
( )
1
2 2 2 2 2
1
.....
x y z
r x x
−
   
 
 = + + +
   
   
 
 
( )
3
2 2 2 2 2
1
.....
x x y z
r x
−
    
 = − + + +
   
    
 
Arpan Deyasi
Electromagnetic
Theory

29. Differentiation of Vector Function: Divergence
( ) ( )
5 3
2 2 2 2 2 2 2 2
2 2
1
3 .....
x x y z x y z
r
− −
 
 = + + − + + +
 
 
 
( ) ( ) ( )
2 2 2 2 2 2 2 2 2
2
5 5 5
2 2 2 2 2 2 2 2 2
2 2 2
1 2 2 2
x y z y z x z x y
r x y z x y z x y z
  − − − − − −
 = + +
 
 
  + + + + + +
2 1
0
r
 
 =
 
 
 
Arpan Deyasi
Electromagnetic
Theory

30. Differentiation of Vector Function: Divergence
3
. 0
r
r
 
 =
 
 
 
Show that
Soln:
Prob 5:
&
3
1
r
 = A r
=
Let
( )
3
. .
r
A
r

 
 = 
 
 
 
( )
3 3
3
. . ( )( . )
r
r r r r
r
− −
 
 =  + 
 
 
 
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Electromagnetic
Theory

31. Differentiation of Vector Function: Divergence
5 3
3
. 3 . 3( )
r
r r r r
r
− −
 
 = − +
 
 
 
5 2 3
3
. 3 3( )
r
r r r
r
− −
 
 = − +
 
 
 
3
. 0
r
r
 
 =
 
 
 
Arpan Deyasi
Electromagnetic
Theory

32. Differentiation of Vector Function: Divergence
Prob 6:
Determine the constant ‘p’ so that the vector
ˆ
ˆ ˆ
( 3 ) ( 2 ) ( )
A x y i y z j x pz k
= + + − + + is solenoidal
Soln:
ˆ ˆ
ˆ ˆ ˆ ˆ
. . ( 3 ) ( 2 ) ( )
A i j k x y i y z j x pz k
x y z
 
    
 = + + + + − + +
   
  
 
. ( 3 ) ( 2 ) ( )
A x y y z x pz
x y z
 
  
 = + + − + +
 
  
 
Arpan Deyasi
Electromagnetic
Theory

33. For solenoidal
Differentiation of Vector Function: Divergence
. 0
A
 =
( 3 ) ( 2 ) ( ) 0
x y y z x pz
x y z
 
  
+ + − + + =
 
  
 
( )
1 1 0
p
+ + =
2
p = −
Arpan Deyasi
Electromagnetic
Theory

34. Prob 7: Find the Laplacian of the following scalar field sin2 cos
x
V e x y
=
2 2 2
2
2 2 2
V V V
V
x y z
  
 = + +
  
2
2
sin2 cos
x
V
e x y
x x x
  
 
=  
  
 
Differentiation of Vector Function: Divergence
Soln:
( )
 
2
2
cos sin2 2 cos2
x x
V
y e x e x
x x
 
= +
 
( )
2
2
cos sin2 2 cos2 2 cos2 4 sin2
x x x x
V
y e x e x e x e x
x

= + + −

Arpan Deyasi
Electromagnetic
Theory

35. Differentiation of Vector Function: Divergence
( )
2
2
cos 4 cos2 3 sin2
x x
V
y e x e x
x

= −

2
2
sin2 cos
x
V
e x y
y y y
 
  
=  
  
 
( )
 
2
2
sin2 sin
x
V
e x y
y y
 
= −
 
2
2
sin2 ( sin )
x
V
e x y
y y
 
= −
 
Arpan Deyasi
Electromagnetic
Theory

36. Differentiation of Vector Function: Divergence
2
2
sin2 cos
x
V
e x y
y

= −

2
2
sin2 cos
x
V
e x y
z z z
  
 
=  
  
 
( )
( )
2
cos 4 cos2 3 sin2 sin2 cos
(cos ) 4cos2 3sin2 sin2
x x x
x
V y e x e x e x y
y e x x x
 = − −
= − −
 
2
2
0 0
V
z z
 
= =
 
Collaborating
( )
2
4(cos ) cos2 sin2
x
V y e x x
 = −
Arpan Deyasi
Electromagnetic
Theory

37. Differentiation of Vector Function: Divergence
( )
1 2 3
ˆ ˆ
ˆ ˆ ˆ ˆ
. .
P i j k iP jP kP
x y z
 
  
 = + + + +
 
  
 
1 2 3
.P P P P
x y z
  
 = + +
  
( )
1 2 3
ˆ ˆ
ˆ ˆ ˆ ˆ
. .
P iP jP kP i j k
x y z
 
  
 = + + + +
 
  
 
1 2 3
.
P P P P
x y z
  
 = + +
  
. .
P P
  
Arpan Deyasi
Electromagnetic
Theory

38. Differentiation of Vector Function: Curl
𝑷 (x, y, z) is defined and differentiable vector function at each
point (x, y, z) in a certain region of space
1 2 3
ˆ
ˆ ˆ
P Pi P j Pk
= + +
( )
1 2 3
ˆ ˆ
ˆ ˆ ˆ ˆ
( )
CurlP P i j k iP jP kP
x y z
 
  
=  = + +  + +
 
  
 
1 2 3
ˆ
ˆ ˆ
i j k
P
x y z
P P P
  
 =
  
Arpan Deyasi
Electromagnetic
Theory

39. Differentiation of Vector Function: Curl
3 2 1 3 2 1
ˆ
ˆ ˆ
P i P P j P P k P P
y z z x x y
 
   
     
 
 = − + − + −
 
 
   
     
 
   
 
Significance: Magnitude of rotation of the vector field originated from the
point in the given region of space
Curl of P Vector function
Arpan Deyasi
Electromagnetic
Theory

40. 0
P
 =
If
the vector becomes irrotational
Differentiation of Vector Function: Curl
Rotation of the vector field stops. So the field becomes conservative
Arpan Deyasi
Electromagnetic
Theory

41. 41
.
+∂Ay/∂x + ∂Ax/∂y
.
.
x
y
Differentiation of Vector Function: Curl
Visualizing Curl - one component
- ∂Ax/∂y
+∂Ay/∂x
Arpan Deyasi
Electromagnetic
Theory

42. Differentiation of Vector Function: Curl
Visualizing Curl
x
y z
.
.
-(∂Ax/∂y)
(∂Ay/∂x)
+
-(∂Ay/∂z)
(∂Az/∂y)
+
-(∂Az/∂x)
(∂Ax/∂z)
.
.
Arpan Deyasi
Electromagnetic
Theory

43. Differentiation of Vector Function: Curl
Prob 1: If
( ) ( ) ( )
3 2 2 ˆ
ˆ ˆ
2 2 3 2
A xy z i x y j xz k
= + + + + −
Then show that 𝑨 is irrotational vector
( ) ( ) ( )
3 2 2
ˆ
ˆ ˆ
2 2 3 2
i j k
A
x y z
xy z x y xz
  
 =
  
+ + −
Soln:
Arpan Deyasi
Electromagnetic
Theory

44. Differentiation of Vector Function: Curl
( ) ( )
( ) ( ) ( ) ( )
2 2
2 3 2 3
ˆ 3 2 2
ˆ
ˆ 3 2 2 2 2
A i xz x y
y z
j xz xy z k x y xy z
x z x y
 
 
 = − − +
 
 
 
 
   
 
− − − + + + − +
   
   
   
Then 𝑨 is irrotational vector
( ) ( ) ( )
2 2 ˆ
ˆ ˆ
0 0 3 3 2 2
A i j z z k x x
 = − − − + −
0
A
 =
Arpan Deyasi
Electromagnetic
Theory

45. Prob 2: If
Differentiation of Vector Function: Curl
2 ˆ
ˆ ˆ
2 2
A x yi zxj yzk
= − + find ( )
P
 
Soln:
2
ˆ
ˆ ˆ
2 2
i j k
P
x y z
x y zx yz
  
 =
  
−
2
ˆ
ˆ(2 2 ) ( 2 )
P i x z k x z
 
 = + − +
 
Arpan Deyasi
Electromagnetic
Theory

46. Differentiation of Vector Function: Curl
2
ˆ
ˆ ˆ
( )
(2 2 ) 0 ( 2 )
i j k
P
x y z
x z x z
  
  =
  
+ − +
ˆ
( ) (2 2)
P x i
  = +
Arpan Deyasi
Electromagnetic
Theory

47. Prob 3: Determine the relation between linear velocity and angular velocity
Differentiation of Vector Function: Curl
Say a rigid body is rotated about an axis through O with angular
velocity 𝝎. Then the velocity 𝒗 of the point P(x,y,z) is given by
v r

= 
( ) ( ) ( )
ˆ
ˆ ˆ
ˆ
ˆ ˆ
x y z y z z x x y
i j k
v i z y j x z k y x
x y z
        
= = − + − + −
Soln:
Arpan Deyasi
Electromagnetic
Theory

48. Differentiation of Vector Function: Curl
( ) ( ) ( )
ˆ
ˆ ˆ
y z z x x y
i j k
v
x y z
z y x z y x
     
  
 =
  
− − −
( ) ( )
( ) ( )
( ) ( )
ˆ
ˆ
ˆ
x y x z
y z x y
x z y z
v i y x z x
y z
j z y y x
z x
k z x z y
x y
   
   
   
 
 
 = − − −
 
 
 
 
 
+ − − −
 
 
 
 
 
+ − − −
 
 
 
Arpan Deyasi
Electromagnetic
Theory

49. Differentiation of Vector Function: Curl
( ) ( ) ( )
ˆ
ˆ ˆ
ˆ
ˆ ˆ
2
2
x x y y z z
x y z
v
i j k
i j k
     
  


= + − − − + +
 
= + +
 
=
( )
1
2
v

 = 
Arpan Deyasi
Electromagnetic
Theory

50. Differentiation of Vector Function: Curl
Prob 4: Show that Curl Grad Φ = 0
Soln:
( ) ˆ
ˆ ˆ
i j k
x y z
  

 
  
  =  + +
 
  
 
( )
ˆ
ˆ ˆ
i j k
x y z
x y z

  
  
   =
  
  
  
Arpan Deyasi
Electromagnetic
Theory

51. Differentiation of Vector Function: Curl
( )
ˆ
ˆ
ˆ
i
y z z y
j k
z x x z x y y x
 

   
 
 
 
   
 
− +
 
 
   
   
   
 
 
   =  
 
 
       
 
     
 
− + −
 
     
 
 
 
       
     
   
 
 
( )
2 2 2 2 2 2
ˆ
ˆ ˆ
i j k
y z z y z x x z x y y x
     

 
     
     
   = − + − + −
 
     
           
     
 
( ) 0

   =
Arpan Deyasi
Electromagnetic
Theory

52. Differentiation of Vector Function: Curl
Prob 5: Calculate ( )
. A r
  if 𝑨 is irrotational vector
Soln:
ˆ
ˆ ˆ
r xi yj zk
= + + 1 2 3
ˆ
ˆ ˆ
A Ai A j A k
= + +
Let
1 2 3
ˆ
ˆ ˆ
i j k
A r A A A
x y z
 =
2 3 3 1 1 2
ˆ
ˆ ˆ
( ) ( ) ( )
A r i zA yA j xA zA k yA xA
 
 = − + − + −
 
Arpan Deyasi
Electromagnetic
Theory

53. Differentiation of Vector Function: Curl
2 3 3 1 1 2
.( ) ( ) ( ) ( )
A r zA yA xA zA yA xA
x y z
 
  
  = − + − + −
 
  
 
2 3
3 1 1 2
.( )
z A y A
x x
A r
x A z A y A x A
y y z z
 
 
 
− +
 
 
 
 
 
  =
 
 
   
 
− + −
 
 
 
   
 
 
 
Arpan Deyasi
Electromagnetic
Theory

54. Differentiation of Vector Function: Curl
2 1
1 3 3 2
.( )
z A A
x y
A r
y A A x A A
z x y z
 
 
 
− +
 
 
 
 
 
  =
 
 
   
 
− + −
 
   
   
 
 
 
 
2 1
1 3 3 2
ˆ
ˆ
ˆ ˆ
.( ) ( ).
ˆ ˆ
k A A
x y
A r xi yj zk
j A A i A A
z x y z
 
 
 
− +
 
 
 
 
 
  = + +
 
 
   
 
− + −
 
   
   
 
 
 
 
Arpan Deyasi
Electromagnetic
Theory

55. Differentiation of Vector Function: Curl
.( ) .( )
A r r A
  = 
𝑨 is irrotational vector 0
A
 =
.( ) 0
A r
  =
Arpan Deyasi
Electromagnetic
Theory

56. Differentiation of Vector Function: Curl
Prob 6: Show that Div Curl 𝑨 = 0
Soln:
1 2 3
ˆ
ˆ ˆ
i j k
A
x y z
A A A
  
 =
  
3 2 1 3 2 1
ˆ
ˆ ˆ
A i A A j A A k A A
y z z x x y
 
   
     
 
 = − + − + −
 
 
   
     
 
   
 
Arpan Deyasi
Electromagnetic
Theory

57. Differentiation of Vector Function: Curl
3 2
1 3 2 1
.( )
A A
x y z
A
A A A A
y z x z x y
 
 
  
− +
 
 
  
 
 
  =
 
 
     
 
− + −
 
   
     
 
 
 
 
2 2
3 2
2 2 2 2
1 3 2 1
.( )
A A
x y x z
A
A A A A
y z y x z x z y
 
 
 
− +
 
 
   
 
 
  =
 
   
   
 
− + −
   
       
 
   
 
Arpan Deyasi
Electromagnetic
Theory

58. Differentiation of Vector Function: Curl
2 2 2
3 2 1
2 2 2
3 2 1
.( )
A A A
x y x z y z
A
A A A
y x z x z y
 
  
− +
 
     
 
  =
 
  
− + −
 
     
 
.( ) 0
A
  =
Arpan Deyasi
Electromagnetic
Theory

59. Differentiation of Vector Function: Curl
Prob 7: Evaluate Curl Curl 𝑨
Soln:
1 2 3
ˆ
ˆ ˆ
i j k
A
x y z
A A A
  
 =
  
3 2 1 3 2 1
ˆ
ˆ ˆ
A i A A j A A k A A
y z z x x y
 
   
     
 
 = − + − + −
 
 
   
     
 
   
 
Arpan Deyasi
Electromagnetic
Theory

60. Differentiation of Vector Function: Curl
3 2 1 3 2 1
ˆ
ˆ ˆ
( )
i j k
A
x y z
A A A A A A
y z z x x y
  
  =
  
   
     
 
− − −
 
   
     
 
   
Arpan Deyasi
Electromagnetic
Theory

61. Differentiation of Vector Function: Curl
2 1 1 3
3 2 2 1
1 3 3 2
ˆ
ˆ
( )
ˆ
i A A A A
y x y z z x
A j A A A A
z y z x x y
k A A A A
x z x y y z
 
 
 
     
 
− − −
 
 
 
 
     
 
 
 
 
 
 
   
     
 
  = + − − −
 
   
 
     
   
 
 
 
 
     
 
 
+ − − −
 
   
 
     
 
 
 
 
 
Arpan Deyasi
Electromagnetic
Theory

62. Differentiation of Vector Function: Curl
2 2 2 2
2 1 1 3
2 2
2 2 2 2
3 2 2 1
2 2
2 2 2 2
1 3 3 2
2 2
ˆ
ˆ
( )
ˆ
i A A A A
y x y z z x
A j A A A A
z y z x x y
k A A A A
x z x y y z
 
 
 
   
− − +
 
 
 
     
 
 
 
 
 
 
   
 
  = + − − +
 
 
 
     
 
 
 
 
 
 
   
+ − − +
 
 
 
     
 
 
 
 
Arpan Deyasi
Electromagnetic
Theory

63. Differentiation of Vector Function: Curl
2 2 2 2 2 2
2 3 1 1 1 1
2 2 2
2 2 2 2 2 2
3 1 2 2 2 2
2 2 2
2 2 2 2 2 2
1 2 3 3 3 3
2 2 2
ˆ
ˆ
ˆ
i A A A A A A
y x z x x x x y z
LHS j A A A A A A
z y x y y y z y x
k A A A A A A
x z y z z z x y z
  
 
     
+ + − − −
 
 
        
 
 
 
 
     
= + + + − − −
 
 
        
 
 
 
 
     
+ + + − − −
 
 
        
 
 


 
 
 
 
 
 
 
 
 

Arpan Deyasi
Electromagnetic
Theory

64. Differentiation of Vector Function: Curl
2 2 2
1 2 3
2 2 2
1 2 3
1 2 3
1 2 3
ˆ
ˆ ˆ
( )
ˆ
ˆ
ˆ
Ai A j A k
x y z
i A A A
x x y z
LHS
j A A A
y x y z
k A A A
z x y z
 
 
  
− + + + +
 
 
  
 
 
 
 
 
   
 
+ +
 
 
   
 
 
 
=  
 
 
   
 
+ + +
 
 
 
   
 
 
 
 
 
 
   
+ + +
 
 
 
   
 
 
 
 
Arpan Deyasi
Electromagnetic
Theory

65. Differentiation of Vector Function: Curl
2
1 2 3
( )
A A A A A
x y z
 
 
  
  = − +  + +
 
 
  
 
 
2
( ) ( . )
A A A
 
  = − +  
 
Arpan Deyasi
Electromagnetic
Theory

66. Differentiation of Vector Function: Curl
Prob 8: I𝐟 𝑨 is irrotational, then evaluate the constants where
Soln:
( ) ( ) ( ) ˆ
ˆ ˆ
2 8 3 4
A x y z i x y z j x y z k
  
= + + + − + + − −
ˆ
ˆ ˆ
( 2 ) ( 8 ) (3 4 )
i j k
A
x y z
x y z x y z x y z
  
  
 =
  
+ + − + − −
Arpan Deyasi
Electromagnetic
Theory

67. Differentiation of Vector Function: Curl
ˆ
ˆ ˆ
( 1) ( 3) ( 2)
A i j k
  
 = − − + − + −
As 𝑨 is irrotational
0
A
 =
ˆ
ˆ ˆ
( 1) ( 3) ( 2) 0
i j k
  
− − + − + − =
1
3
2



= −
=
=
Arpan Deyasi
Electromagnetic
Theory

68. Differentiation of Vector Function: Curl
1 2 3
ˆ
ˆ ˆ
i j k
P
x y z
P P P
  
 =
  
3 2
1 3 2 1
ˆ
ˆ
ˆ
i P P
y z
P
j P P k P P
z x x y
 
 
 
− +
 
 
 
 
 
 =
 
 
   
 
− + −
 
   
   
 
 
 
 
1 2 3
ˆ
ˆ ˆ
i j k
P P P P
x y z
 =
  
  
2 3
3 1 1 2
ˆ
ˆ
ˆ
i P P
z y
P
j P P k P P
x z y x
 
 
 
− +
 
 
 
 
 
 =
 
 
   
 
− + −
 
   
   
 
 
 
 
P P
  
Arpan Deyasi
Electromagnetic
Theory