EKT 241: ELECTROMAGNETIC THEORY
SEM 2 2010/2011
ANSWER FOR TUTORIAL 1: VECTOR ANALYSIS
At point P 3,4,5 , express the vector that extends from P to Q2,0,1 in:
(a) Cartesian coordinates.
(b) Cylindrical coordinates
(c) Spherical coordinates
(d) Show that each of these vectors has the same magnitude
1.
Solution:
(a)
Cartesian coordinates:
R PQ  Q  P  xˆ 5  yˆ 4  zˆ 6 ,
then
(b)
R PQ  25  16  36  8.8
Cylindrical coordinates:
  3  53.1 , and z  5
at P, r  5,   tan 1 4
0
Now,
R PQ  R PQ  r̂  xˆ5  yˆ 4  zˆ6  r̂  5 cos   4 sin   6.20
r
R PQ

 R PQ  ˆ  xˆ5  yˆ 4  zˆ6  ˆ  5 sin   4 cos   1.60
Thus,
R PQ  rˆ6.20  ˆ1.60  zˆ6
R PQ  6.20 2  1.60 2  6 2  8.8
and
(c)
Spherical coordinates:
R  9  16  25  7.07,  cos 1 ( 5
at P,
R PQ
R PQ
R

7.07
)  45 0 ,
  tan 1 ( 4  3)  53.10
 R PQ  R̂   xˆ 5  yˆ 4  zˆ 6   R̂
 5 sin  cos   4 sin  sin   6 cos   0.14
 R  ˆ  xˆ5  yˆ 4  zˆ 6  ˆ
PQ
 5 cos  cos   4 cos  sin   (6) sin   8.62
R PQ

 R PQ  ˆ  xˆ5  yˆ 4  zˆ6  ˆ
 5 sin   4 cos   1.60
Thus, R PQ  Rˆ 0.14  ˆ8.62  ˆ1.60
And
(d)
R PQ  0.14 2  8.62 2  1.60 2  8.8
Each does, as shown above.
-------------------------------------------------------------------------------------------------------------------------------------2.
Find the gradient of each following:
(a) V2  V0 e 2  sin 3 in cylindrical coordinates.
(b) V3  V0 (a ) cos 2 in spherical coordinates.
r
Solution
a.
In cylindrical coordinates;
 
1 

V2   rˆ
 ˆ
 zˆ V0 e 2 r sin 3
 
z 
 

3V0 e 2 r cos 3
2r
ˆ
ˆ
 r 2V0 e sin 3  
r

3 cos 3 

 V0 e  2 r  rˆ 2 sin 3  ˆ
r 

b.
In spherical coordinates;
 
1 
1
  a
V0   cos 2
V3   Rˆ
 ˆ
 ˆ
R 
R sin     R 
 R
ˆ V0 a cos 2  θˆ 2V0 a sin 2
 R
R2
R2


V0 a ˆ
R cos 2  θˆ 2 sin 2
R2

------------------------------------------------------------------------------------------------------------------------------------3.
Given
A   cos a    sin a   3 za z , find   A at (2,0,3) in cylindrical
coordinates.
Solution
Solution:
Change the notation of

into
r.
A  r cos a r  r sin a  3za z
1 
1 A Az
rAr 

r r
r 
z
1  2
1 
r sin     3z

r cos 
r r
r 
z
 2 cos  cos  3
A


Therefore,
  A  2, 0, 3   2  1  3  6
------------------------------------------------------------------------------------------------------------------------------------
4.
Evaluate the line integral of
shown in the Figure 1.
E  xˆ x  yˆ y along the segment P1 to P2 of the circular path
Figure 1
--------------------------------------------------------------------------------------------------------------------------
5.
Transform the vector A  3aˆ x  2aˆ y  aˆ z into spherical vector and evaluate it at the
point P1  (2,  1, 1) .
Solution
For the vector A  3a x  2a y  a z ,
Ax  3 , Ay  2 , Az  1
AR  Ax sin  cos   Ay sin  sin   Az cos 
 3 sin  cos   2 sin  sin   cos 
A  Ax cos  cos   Ay cos  sin   Az sin 
 3 cos  cos   2 cos  sin   sin 
A   Ax sin   Ay cos 
 3 sin   2 cos 
Hence, the vector A in spherical coordinates:
A  (3sin  cos   2 sin  sin   cos )a R
 (3 cos  cos   2 cos  sin   sin  )a θ  (3 sin   2 cos  )a φ
Point P1 in spherical coordinates:
R
x2  y2  z2
 2 2  (1) 2  12
 6

x2  y2
z
 2 2  (1) 2
 tan 1 

1


  65.9


 1
  26.6
 2 
 y
x
  tan 1    tan  1
Hence, at point P1;
A  3.67a R  0.59aθ  0.45aφ
---------------------------------------------------------------------------------------------------------------------------------6.
Find
at (3, π/6, 0) in spherical coordinates for the vector field
Solution
-----------------------------------------------------------------------------------------------------------------------------------------7.
A section of a sphere is described by 0  R  2, 0    90, and 30    90. Find:
(a) the surface area of the spherical section.
(b) the enclosed volume
ˆ 3R 2 , evaluate both sides of the divergence theorem for the
For the vector field D  R
region enclosed between the spherical shells defined by R=1 and R=2.
8.
Solution
 D  dS
S
 D  dS

S  outer
2


 
 D  dS

S inner

2
2
 Rˆ 3R  R sin ddRˆ
0 0
2

2

  Rˆ 3R
0 0
r 2
2


  sin dd  3(1 )  sin dd
 3(2 4 )
4
0 0
0 0
 48 cos 0    3 cos  0  0

2
0

2
 192  12
 180
For the right hand side of the divergence theorem,
D 
1  2
R AR
R 2 R

1 
( R 2 (3R 2 ))
2
R R
1 
 2
(3R 4 ))
R R
1
 2 (12 R 3 )
R
 12 R


2
ˆ)
 R 2 sin dd (R
r 1
Thus, the right hand side of the divergence theorem,
   Ddv
v


2
R 1
2
 
0
2
0
 12  R dR
3
R 1
12 R ( R 2 sin ddRd  )

2
 sin d  d
0
0
2
 R4 

2
 12   - cos 0  0
 4 1
 45(2)( 2 )
 180
Thus, left hand side of divergence theorem = right hand side. Hence, divergence
theorem has been verified.
--------------------------------------------------------------------------------------------------------------------9.
r
For the vector field, E  rˆ10e  zˆ 3 z , verify the divergence theorem for the cylindrical
region enclosed by r =2, z =0 and z = 4.
10.
ˆ  rˆr cos   φˆ sin   by evaluating:
Verify Stokes’s theorem for the vector field B
 B  dl over the semicircular contour shown in Figure 2, and
(b)    B  dS over the surface of the semicircle.
(a)
C
S
Figure 2