Magnetism

INEL 4151Dr. Sandra Cruz-Pol

Electrical and Computer Engineering Dept.UPRM ch 7

http://www.treehugger.com/files/2008/10/spintronics-discover-could-lead-to-magnetic-batteries.php

http://videos.howstuffworks.com/hsw/11955-magnetism-introduction-to-magnetism-video.htm

Applications

MotorsTransformersMRIMore…

http://videos.howstuffworks.com/hsw/18034-electricity-and-magnetism-magnetic-levitation-video.htm

HB H= magnetic field intensity [A/m]

B= magnetic field density [Teslas]

In free space the permeability is:H/m 104 7 o

Magnetic FieldBiot-Savart Law

• States that:

Example

aI

H ˆcoscos4 12

For an infinite line filament with current I (1=180o and 2=0o):

a

IH ˆ

2

2

1

a

PE. 7.1 Find H at (0,0,5)

• Due to current in (figure):where 1=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

2

ˆˆ xy aa

2

ˆˆ0

25

2

54

1 yx aaH

m

mAˆˆ30 yx aaH

Circular loop• Defined by• Apply Biot-Savart:

• Only z-component of H survives due to symmetry:

0,922 zyx

dl

R

z

y

dHz

dH

xzadahd

h

d

aaa

Rld

ˆˆ

0

00

ˆˆˆ

2

z

2

02/322

2

4

ˆ

h

adIH z

2/322

2

2

ˆ

h

aI z

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

The H field is given by:

2

4

1

3

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 mA/m

-x

y

z

Toroidal inductors can have higher Q factors and higher inductance than similarly constructed solenoid coils. This is due largely to the smaller number of turns required when the core provides a closed magnetic path. The magnetic flux in a toroid is largely confined to the core, preventing its energy from being absorbed by nearby objects, making toroidal cores essentially self-shielding.

A toroidA circular ring-shaped magnetic core of iron powder, ferrite, or

other material around which wire [N- loops] is coiled to make an inductor. Toroidal coils are used in a broad range of applications, such as high-frequency coils and transformers.

NIH

SdJIldH enc

2

elsewhere0

core theinside2 l

NINIH

Fields stay inside core, no interference.

Magnetic Flux Density, B• The magnetic flux is defined as:

which flows through a surface S.• The total flux thru a closed surface in a

magnetic field is:

[Wb] S

SdB

0S

SdB

0 vS

dvBSdB

0 B

Monopole doesn’t exist.

vD

Maxwell’s Equations for Static Fields

Differential formDifferential form Integral FormIntegral FormGaussGauss’’ss Law for Law for EE field.field.

GaussGauss’’ss Law for Law for HH field. Nonexistence field. Nonexistence of monopole of monopole

FaradayFaraday’’ss Law; E Law; E field is conserved.field is conserved.

AmpereAmpere’’ss Law Law

vD

0 B

0 E

JH

v

v

s

dvSdD

0s

SdB

0 ldEL

sL

SdJldH

Magnetic Scalar and Vector Potentials, Vm & A

When J=0, the curl of H is =0, then recalling the vector identity:

• We can define a Magnetic Scalar Potential as:

• The magnetic Vector Potential A is defined:

0J if

mVH

VH 0

AB

The magnetic vector potential, A, is

AB

L

Roo R

alIdHB

24

ˆ

L

o

R

IdlA

4

It can be shown that:

2

11

RR

Substituting into equation for Magnetic Flux:

SS

SdASdB

L

ldA

L

ldA

The magnetic vector potential A is used in antenna theory.

This is another way of finding magnetic flux.

P.E. 7.7 A current distribution causes a magnetic vector potential of:

Find :• B at (-1,2,5)Answer:• Flux thru surface z=1, 0≤x≤1, -1≤y ≤4

Answer :

zxyzyxyxyxA ˆ4ˆˆ 22

AB

SS

ldASdB

[Wb] 20

[T] ˆ3ˆ40ˆ20 zyxB