Tải bản đầy đủ (.ppt) (12 trang)

Section 1 3 TRƯỜNG ĐIỆN TỪ

Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (93.09 KB, 12 trang )

Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
to supplement “Elements of Engineering
Electromagnetics, Sixth Edition”
by

Nannapaneni Narayana Rao

Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, Urbana, Illinois, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India


1.3
Cylindrical and Spherical
Coordinate Systems


1.3-3

Cylindrical and Spherical Coordinate Systems
Cylindrical (r, , z)
z

Spherical (r, , )
z

az

a



90Þ

x

x  r
y

z

ar
a

 r

ar
y


x

Only az is uniform.

All three unit

ar and a are
nonuniform.

vectors are
nonuniform.


90Þ

a
y


1.3-4

= r cos 
= r sin 
=z

x
y
z
D1.7 (a)

x = r sin  cos 
y = r sin  sin 
z = r cos 

(2, 5/6, 3) in cylindrical coordinates
z

3


2


x

5/6

y

x 2 cos 5 6 – 3 
 3  1 2
y 2 sin 5 61

z 3


1.3-5

(b) (4, 4 3, – 1) in cylindrical coordinates
z
1

4
4/3

y

x

x 4 cos 4 3 – 2


 4  12 4

y 4 sin 4 3 – 2 3 
z – 1


1.3-6

(c) (4, 2 3,  6) in spherical coordinates
z
2/3
x

y

/6
4

2

x 4 sin
cos 3 

3
6

2

y 4 sin
sin  3  9  3  4 4
3
6




z 4 cos – 2
3



(d)



1.3-7



8,  4,  3 in spherical coordinates.
z
/4
x

/3

8
y



x  8 sin cos 1 


4
3




y  8 sin sin  3 1  3  4  8
4
3



z  8 cos 2

4



1.3-8

Conversion of vectors between coordinate systems
a rc  cos  sin  0 a x 
  
a 
a

–sin

cos


0
   
 y 

0
1 
 0

a z 
 
a z 


a rs  sin  cos  sin  sin 
  
a  cos  cos  cos  sin 
  
cos 

a 
  – sin 

a
az
ay

ax arc

a


az


ars
arc
a

cos  a x 
 
– sin  a y 
 
0 a z 


1.3-9

P1.18

A

= ar at (2, /6, 2)

B

= a at (1, /3, 0)

C

= i at (3, /4, 3/2)


A

z
/6



x

/2

y



A sin a y  cos a z
6
6
1
3
 ay 
az
2
2
1 3
 1
4 4


1.3-10

z

 3

1

y

B

x

1 3
 1
4 4

z
C

3

/4
3/2

x



B sin a x – cos a z
6

6
1
3
 ax –
az
2
2

y

C a x


1.3-11

1
 1

3
3
(a) A  B  a y 
az    ax  
az 
2
2  2
2 

3

4


1
3 
(b) A  C  a y 
az   ax
2
2 

0


1.3-12

1
3 
(c) B  C  ax 
a

a

z
x
2

2


1

2

(d) A B • C C • A B
1
= 0
1
2
–

0
1
2

0
3
2
3
0 –
2
3
4



×