Chapter One: Vector Analysis The use of vectors and vector analysis ...

Electromagnetic Theorem
(Dr. Omed Ghareb Abdullah) University of Sulaimani –College of Science – Physics Department
Solution:
(a). since the vector is in cylindrical coordinate, and we have the following matrix to
convert it to Cartesian coordinate:
A
A
A
x
y
z
cos
sin
0
sin
cos
0
0
0
1
A
A
A
z
A
A
A
x
y
z
cos
sin
0
sin
cos
0
0
0
1
cos
0
2
z sin
A
A
A
x
y
z
cos
2
sin
cos
0 0
0
0 z
A
x
0 0
2
sin
2
5cos
( 53.13)
1.8
at point (3,‐4,0) : tan
A
y
z 0
Therefore, the vector in Cartesian coordinate is:
2
2
A cos aˆ
cos
sin
aˆ
z sin
aˆ
x
y
x
2
1
y
y
x
2
tan
9 16
5
1
5sin( 53.13)cos(
53.13)
2.3
z
A 1.8
aˆ
x
4
53.13
3
2.3aˆ
y
0
A
z
5
0
sin( 53.13)
0
(b). since the vector is in cylindrical coordinate, and we have the following matrix to
convert it to spherical coordinate:
A
A
A
r
sin
cos
0
0
0
1
cos
sin
0
A
A
A
z
A
A
A
r
sin
cos
0
0
0
1
cos
sin
0
cos
0
2
z sin
The spherical point of ( 3, ‐4, 0) is given by :
A
r
sin
cos
z
2
cos
sin
A
r
x
2
cos
tan
y
1
1
z
r
y
x
cos
cos
z
2
2
z
cos
tan
2
1
1
sin
sin
9 16
0 5
0
90
5
4
53.13
3
A
0
0 0 0
A r
A
5sin 90 cos( 53.13)
5
0
cos90 sin( 53.13)
5cos90
cos( 53.13)
5
0
sin 90 sin( 53.13)
A
0
A (
sin
cos
z
A 3aˆ
0 aˆ
0 aˆ
z
2
cos
sin)
aˆ
A 3aˆ
3
0
( cos
cos
z
2
sin
sin)
aˆ
0 aˆ
z
55