magnetism inel 4151 dr. sandra cruz-pol electrical and computer engineering dept. uprm
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Magnetism
INEL 4151Dr. Sandra Cruz-Pol
Electrical and Computer Engineering Dept.UPRM
Applications
MotorsTransformersMRIMore…
HB H= magnetic field intensity [A/m]
B = magnetic field density [Teslas]
Biot-Savart Law
• States that:
24
ˆ
R
rlIdHd
Example
34 R
RlIdHd
aI
H ˆcoscos4 12
For an infinitive line filament with current I (a1=180o and a2=0o):
a
IH ˆ
2
a2
a1
r
af
PE. 7.1 Find H at (0,0,5)
• Due to current in (figure):where a1=90o and
10A 1 y
z
x
1
(0,0,5)
aI
H ˆcoscos4 12
22
22
2
25
11cos
zyx
l
aaa
aaa
ˆ2
ˆˆ
ˆˆˆ
34 R
RlIdHd
2
ˆˆ xy aa
2
ˆˆ0
25
2
54
1 yx aaH
m
mAˆˆ30 yx aaH
Ampere’s Law
• Simpler• Analogous to Gauss Law for Coulombs• For symmetrical current distributions
Ampere’s Law SdJIldH enc
IIldH enc
adld
a
IH
2
We define an Amperian path where H is constant.
Infinitely long coaxial cable SdJIldH enc
Four cases: 1) For <a 2
2
2ˆ
a
Iadd
a
II zenc
z
2HdH
22 a
IH
Infinitely long coaxial cable SdJIldH enc
Four cases: 2) For a<<b Iadda
a
II zz
a
enc
2
02
0
ˆˆ
z
2HdH
2
IH
Infinitely long coaxial cable SdJIldH enc
Four cases: 3) For b<<b+c
2
022
ˆb
zencbcb
ddIIaddJI
z
2HdH
bcc
bIH
21
2 2
22
2
22
2 cbc
bIIIenc
Infinitely long coaxial cable SdJIldH enc
Four cases: 4) For >b+c
0 IIIenc
20 H
0H
Sheet of current distribution
0ˆ2
1
0ˆ2
1
zaK
zaKH
xy
xy
0ˆ
0ˆ
zaH
zaHH
xo
xo
bKIldH yenc
K [A/m]
ba
x
z
y
Cross section is a Line!
The H field on the Amperian path is given by:
ldHldH
1
4
4
3
3
2
2
1
bH
bHabHa
o
oo
2
))(()(0))(()(0 naKH ˆ
2
1
PE. 7.5 Sheet of currentPlane y=1 carries a current K=50 az mA/m.
Find H at (0,0,0).
naKH ˆ2
1
xyz aaaH ˆ25ˆˆ502
1
K =50
-x
y
z
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