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