Transcript Magnetism - Electrical and Computer Engineering Department

Magnetic Forces,
Materials
and Devices
INEL 4151 ch 8
Dr. Sandra Cruz-Pol
Electrical and Computer Engineering Dept.
UPRM
http://www.treehugger.com/files/2008/10/spintronics-discover-could-lead-tomagnetic-batteries.php
Applications
Motors
Transformers
MRI
More…
http://www.youtube.com/watch?v=0ajvcdfC
65w
Forces due to Magnetic fields
B=mH= magnetic field density
H=magnetic field intensity
Force can be due to:
• B on moving charge, Q
• B on current
• 2 currents
B is defined as force per unit current element
Forces on a Charge
Analogous to
the electric
force:
We have
magnetic
force:


Fe  QE
The total force is given by:
  
F  Fe  Fm

 
Fm  Qu  B
Note that Fm is perpendicular to both u and B
If the charge moving has a mass m, then:


  
du
F m
 Q E uB
dt

Usually Fm < Fe

Forces on a current element
The current element
can be expressed as:

 dQ 

dl
Idl 
dl  dQ
 dQu
dt
dt
So we can write:

Fm  

   
Fm  Qu  B  Il  B
 
Idl  B
L
For closed path Line
I=[A] current element


  

Fm   KdS  B Fm   Jdv  B
S
Surface K=[A/m]
current element
v
Volume J=[A/m2]
current element
Force between two current
elements
• Each element produces a field B, which exerts
a force on the other element.
I1
R21
I2
mo I1I 2
F1 =
4p
òò
L1 L2
 

dF1  I1dl1  B2


mo I 2 dl2  aˆ R21
dB2 
2
4R21
(
dl1 ´ dl2 ´ aˆR21
R
2
21
)
P.E. 8.4 Find the force experienced by the loop if
I1=10A, I2=5A, ro=20cm, a=1cm, b=30cm
Divide loop into 4
segments.
z
z
I1
F2
I2
ro
a
F1
b
F3
r
r
    
Fl  F1  F2  F3  F4
mo I1I 2
FI1 =
4p
F4
òò
L1 L2
(
dl1 ´ dl2 ´ aˆR12
R
2
12
)
For segment #1, Force #1
Since I1 is infinite long wire:
 mo I1aˆ
B1 
2ro
z

F1  I 2
I1
b

 
dl2  B1
z 0
F1
ro
I2
b
a
r

F1  I 2
b


dz aˆ z  B1
z 0

bmo I1
 aˆ r 
F1  I 2
2ro
For segment #2,
The B field at segment #2
due to current 1.
 mo I1aˆ
B1 
2 r 
z
 

F2  I 2  dl2  B1
F2
I1
I2
ro
b
a
r
ro  a

m o I1
F2  I 2  draˆ r 
aˆ
2r
r  ro
 mo I1I 2 ro  a
aˆ z 
F2 
ln
2
ro
For segment #3, Force #3
The field at segment 3:

B1 
z
mo I1aˆ
2 r o  a 
 

F3  I 2  dl2  B1
I1
I2
b
F3
ro
a
r

F3  I 2
mo I1
z b dz aˆ z  2 ro  a  aˆ
0

mo I1I 2b
aˆr 
F3 
2 ro  a 
For segment #4,
The B field at segment #4
due to current 1.
 mo I1aˆ
B1 
2r
z
 

F4  I 2  dl2  B1
I1
I2
ro
b
a
F4
r

F4  I 2
ro
m o I1
draˆ r 
aˆ

2r
r r a
o

m o I1 I 2 r o  a
aˆ z 
F4  
ln
2
ro
The total force en the loop is
I1=10A, I2=5A, ro=20cm, a=1cm, b=30cm
• The sum of all four:
    
Fl  F1  F2  F3  F4
z
F2
F1
F3
r
F4
• Note that 2 terms cancel out:
bmo I1 é 1
1 ù
Floop = -I 2
ê ú aˆr
2p êë ro ( ro + a) úû
Attracts!
é
ù
(0.3)mo10 1
1
Floop = -5
ê ú aˆr = -5 [m N] aˆr
2p (0.2) êë.2 (.2 +.1) úû
Magnetic Torque and Moment
Inside a motor/generator we have many loops with currents, and
the Magnetic fields from a magnet exert a torque on them.
The torque in [N m]is:
    
T  r  F  m B

   
Fm  Qu  B  Il  B
z
• Where m is the magnetic
dipole moment:

m  IS aˆn
• Where S is the area of the
loop and an is its unit normal.
This applies if B is uniform
B
SIDE VIEW
B
Torque on a Current Loop in a Magnetic Field
CD Motor
Magnetic Torque and Moment
The Magnetic torque can also be expressed as:
The torque in [N m]is:
  
T  m B
 
 Qml  B

m  I (S )aˆn
Magnetization
(similar to Polarization for E)
• Atoms have e- orbiting and spinning
– Each have a magnetic dipole associated to it
• Most materials have random orientation of their
magnetic dipoles if NO external B-field is applied.
• When a B field is applied, they try to align
in the same direction.
• The total magnetization [A/m]
N

M  lim
v 0

k 1

mk
v
Magnetization
• The magnetization current density [A/m2]


Jb    M
• The total magnetic density is:

 
B  mo (H  M )
• Magnetic susceptibility is:


M  m H
mo


B  mo (1  m )H


B  mo mr H
• The relative permeability is:
m
mr  (1   m ) 
• Permeability is in [H/m].
mo
Classification of Materials
according to magnetism
Non-magnetic
mr=1
Ex. air, free space, most
materials in their
natural state.
Magnetic
mr≠1
Diamagnetic
mr≤1
Electronic motions of spin
and orbit cancel out.
lead, copper, Si,
diamonds,
superconductors
(mr=B=0).
Are weakly affected by B
Fields.
Paramagnetic
mr≥1
(air, platinum, tungsten,
platinum )
Temperature dependent.
Not many uses except in
masers
Ferromagnetic
mr>>1
Iron, Ni,Co, steel, alloys
Loose properties if heated
above Curie T (770C)
Nonlinear:
mr varies


B  mo H


B  mo (1  m )H
B-H or Magnetization curve
• When an H-field is applied to ferromagnetic material,
it’s B increases until saturation.
B
m
H
•
But when H is decreased,
B doesn’t follow the same
curve.
Hysteresis Loop
• Some ferrites, have almost
rectangular B-H curves, ideal
for digital computers for
storing information.
• The area of the loop gives
the energy loss per volume
during one cycle in the form
of heat.
• Tall-narrow loops are
desirable for electric
generators, motors,
transformers to minimize the
hysteresis losses.
Magnetic B.C.
• We’ll use Gauss Law
&
• Ampere’s Circuit law
 B  dS  0
 H  dl  I
B2n
B.C.: Two magnetic media
• Consider the figure below:
 B  dS 0
S
B1n  B2 n
B1n S  B2 n S  0
is continuous.
B1n
m1H1n  m2 H 2n
B1
S
B1t
B2
B2t
m1
m2
h
B.C.: Two magnetic media
• Consider the figure below:
 H  dl  I  K  w
l
h
h
h
h
 H1t w  H1n
 H 2n
 H 2t w  H 2 n
 H1n
2
2
2
2
H1n
H1
H1t  H 2t  K
q1
m1
H2n m2
a
at boundary
b
K
H1t
if no current
h
H2
H2t
d
w
c
H1t  H 2t
P.E. 8.8 Find B2
• Region 1 described
by 3x+4y≥10, is free
space
• Region 2 described
by 3x+4y≤10 is
magnetic material
with mr=10
Assume boundary is
current free.
B1 = 0.1xˆ +.4 yˆ +.2 zˆ Teslas
OJO: Error en
Solucionario!
B
Teslas
B2 ==-1.052
0.24 xˆxˆ++1.26
3yˆ + y2ˆ +zˆ 2 zˆTeslas
2
P.E. 8.7 B-field in a magnetic
material.
• In a region with mr=4.6
• Find H and susceptability
 m 1  mr
H

B
m

y
2
B  10e zˆ mWb / m
 m  3.6
-y
H =173e az mA/m
How to make traffic light go Green
when driving a bike or motorcycle
• http://www.wikihow.com/Trigg
• Stop directly on top of
er-Green-Traffic-Lights
induction loop on the
street
• Attach neodymium
magnets to the vehicle
• Move on top of loop
• Push crossing button • http://www.labreform.org/educatio
n/loops.html
• Video detectors
 
   B  dS [Wb]
S
• If flux passes thru N
turns, the total Flux
Linkage is   N
• This is proportional
to the current I
Inductors
• So we can define
the inductance as:
  LI
• then:
N
L
I
Units of Inductance
When more than 1 inductor
12


  B2  dS
S1
M 12 
12
I2
N112

 M 21
I2
*Don’t confuse the Magnetization vector, M, with the mutual inductance!
L1 
11
I1
Self -inductance
N11

I1
L2 
22
I2
N 2 2

I2
• The total energy in the magnetic field is the sum of the
energies:
Wm  W1  W2  W12
1
1
2
2
 L1 I1  L2 I 2  M 12 I1 I 2
2
2
• The positive is taken if currents I1 and I2 flow such that the
magnetic fields of the two circuits strengthen each other.
See table 8.3 in textbook with formulas for inductance of common
elements like coaxial cable, two-wire line, etc.
P.E. 8.10 solenoid
A long solenoid with 2 x 2 cm cross section has
iron core (permeability is 1000x ) and 4000
turns per meter. If carries current of 0.5A,
find:
• Self inductance per meter
B = mH =
m IN
l
Y = BS
= m In
L
2
L' = = NY / Il = m n S
l
Note L is Independent of current
 1000mo 4000 2cm
2
l = NY
 8.042H / m
2
Magnetic Circuits
• Magnetomotive force
In units of ampere-turns
F  NI   H  dl
• Reluctance
• Like V=IR
l
R
mS
F  R
• Ex: magnetic relays,
motors, generators,
transformers,
toroids
• Table 8.4 presents
analogy between
magnetic and
electric circuits
V  IR
Magnetic Circuits
Ex. Find current in coil needed to produced
flux density of 1.5T in air gap.
Assume mr=50 and all cross sectional area is 10 cm2
8.42 A cobalt ring (mr=600) has mean
radius of 30cm.
• If a coil wound on the ring carries 12A, calculate the N
required to establish an average magnetic flux density of 1.5
Teslas in the ring.
F  NI   H  dl
NI  Hl
radius of 30cm
1.5  2 .3
N

 313 vueltas
mmo I 600mo12
Bl
Force on Magnetic materials
1
2
Wm 
B
dv

2m
Relay
• N turns, current I
• B.C. B1n=B2n (ignore
fringing)
• Total energy change
is used to displace
bar a distance dl.
• S=cross sectional
area of core
 1 B2

 Fdl  dWm  
(2S )dl
 2 mo

F 
B2
mo
S
P.E. 8.16 U-shape electromagnet
F 
B2
mo
• Will lift 400kg of mass(including
keeper + weight)
• Iron yoke has mr=3,000
• Cross section =40cm2
• Air gap are 0.1mm long
• Length of iron=50cm
• Current is 1A
Find number of turns, N
Find force across one air gap
S  mg
Esto nos da el B en el air
gap.
F  NI   H  dl
l
R
mS
MagLev
magnetic Levitation
• Floating one
• Use diamagnetic
magnet over
materials, which
another
repel and are
repelled by strong H • Regular train
fields.
187 mph
• Maglev train
• Superconductors
are diamagnetic.
312 mph