Magnetic Forces, Materials and Devices
description
Transcript of Magnetic Forces, Materials and Devices
Magnetic Forces, Materials
and Devices
INEL 4151 ch 8Dr. Sandra Cruz-Pol
Electrical and Computer Engineering Dept.UPRM
http://www.treehugger.com/files/2008/10/spintronics-discover-could-lead-to-magnetic-batteries.php
Applications
MotorsTransformersMRIMore…
http://www.youtube.com/watch?v=0ajvcdfC65w
Forces due to Magnetic fieldsB=H= 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 ChargeEQFe
Analogous to the electric force:
We have magnetic force:
If the charge moving has a mass m, then:
The total force is given by:
BuQFm
me FFF
BuEQdt
udmF
Note that Fm is perpendicular to both u and B
Usually Fm < Fe
Forces on a current element
udQdt
lddQld
dt
dQlId
The current element can be expressed as:
So we can write:
BlIBuQFm
BlIdFL
m
BSKdFS
m
BdvJF
v
m
For closed path Line
I=[A] current element
Surface K=[A/m]current element
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.
2111 BldIFd
221
222 4
ˆ21
R
aldIBd Ro
I1
I2
R21
P.E. 8.4 Find the force experienced by the loop if I1=10A, I2=5A, o=20cm, a=1cm, b=30cm
I1
I2
a
b
4321 FFFFFl
z
F1
F2
F3
F4
zDivide loop into 4 segments.
o
o aIB
2
ˆ11
Since I1 is infinite long wire:
12
0
21 BldIFb
z
F1
I1I2
a
b
z
aIb
IFo
o ˆ2
121
For segment #1, Force #1
1
0
21 ˆ BadzIF z
b
z
2
ˆ11
aIB o
1222 BldIF
F2
I1I2
a
b
z
zo
oo aaII
F ˆln2
212
For segment #2, The B field at segment #2 due to current 1.
a
IadIF o
ao
o
ˆ2
ˆ 122
a
aIB
o
o
2
ˆ11
The field at segment 3:
1223 BldIF
F3
I1I2
a
b
z
aa
bIIF
o
o ˆ2
213
For segment #3, Force #3
aa
IadzIF
o
oz
bz
ˆ2
ˆ 10
23
2
ˆ11
aIB o
1224 BldIF
F4
I1I2
a
b
z
zo
oo aaII
F ˆln2
214
For segment #4, The B field at segment #4 due to current 1.
a
IadIF o
a
o
o
ˆ2
ˆ 124
The total force en the loop is
4321 FFFFFl
• The sum of all four:
• Note that 2 terms cancel out:
F1
F2
F3
F4
zI1=10A, I2=5A, o=20cm, a=1cm, b=30cm
Attracts!
Magnetic Torque and Moment
The torque in [N m]is:
• Where m is the magnetic dipole moment:
• Where S is the area of the loop and an is its unit normal. This applies if B is uniform
BmFrT
naISm ˆ
Inside a motor/generator we have many loops with currents, and the Magnetic fields from a magnet exert a torque on them.
B
z
BlIBuQFm
SIDE VIEW
B
Torque on a Current Loop in a Magnetic Field
CD Motor
Magnetic Torque and Moment
The torque in [N m]is:
BlQ
BmT
m
The Magnetic torque can also be expressed as:
naSIm ˆ)(
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 alignin the same direction.• The total magnetization [A/m]
v
mM
k
N
k
v
1
0lim
Magnetization • The magnetization current density [A/m2]
• The total magnetic density is:
• Magnetic susceptibility is:
• The relative permeability is:• Permeability is in [H/m].
MJb
)( MHB o
HM m
HB mo
)1(
HB ro
omr
)1(
o
Classification of Materials according to magnetism
Magneticr≠1
Non-magneticr=1
Ex. air, free space, most materials in their
natural state.Diamagneticr≤1
Electronic motions of spin and orbit cancel out.
lead, copper, Si, diamonds,
superconductors (r=B=0).
Are weakly affected by B Fields.
Paramagneticr≥1
(air, platinum, tungsten, platinum )
Temperature dependent.Not many uses except in
masers
Ferromagneticr>>1
Iron, Ni,Co, steel, alloysLoose properties if heated
above Curie T (770C)Nonlinear:r varies
HB mo
)1( HB o
B-H or Magnetization curve• When an H-field is applied to ferromagnetic material, it’s B increases until saturation.
• But when H is decreased, B doesn’t follow the same
curve.
H
B
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
0dSB
IdlH
B.C.: Two magnetic media• Consider the figure below:
B1
B2
B1t
B1n
B2t
B2n
h
0
0
21
SBSB
dSB
nn
S
S
2
1
.continuous is21 nn BB
nn HH 2211
B.C.: Two magnetic media• Consider the figure below:
H1
H2
H1t
H1n
H2t
H2n
a b
cd w
h
2222 122211
hH
hHwH
hH
hHwH
wKIdlH
nntnnt
l
KHH tt 21
1
2
1 tt HH 21
boundaryat
current no if
K
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 r=10
Assume boundary is current free.
OJO: Error en Solucionario!
P.E. 8.7 B-field in a magnetic material.
• In a region with r=4.6
• Find H and susceptability
2/ˆ10 mmWbzeB y
rm 1
B
H
6.3m
A/m
How to make traffic light go Green
when driving a bike or motorcycle• Stop directly on top of
induction loop on the street
• Attach neodymium magnets to the vehicle
• Move on top of loop• Push crossing button• Video detectors
• http://www.wikihow.com/Trigger-Green-Traffic-Lights
• http://www.labreform.org/education/loops.html
Inductors
• If flux passes thru N turns, the total Flux Linkage is
• This is proportional to the current I
• So we can define the inductance as:
• then:
N
I
NL
LI
[Wb] S
SdB
Units of Inductance
When more than 1 inductor
1
212
S
SdB
212
121
2
1212 M
I
N
IM
*Don’t confuse the Magnetization vector, M, with the mutual inductance!
Self -inductance
• The total energy in the magnetic field is the sum of the energies:
• The positive is taken if currents I1 and I2 flow such that the magnetic fields of the two circuits strengthen each other.
1
11
1
111 I
N
IL
2
22
2
222 I
N
IL
21122
222
11
1221
2
1
2
1IIMILIL
WWWWm
See table 8.3 in textbook with formulas for inductance of common elements like coaxial cable, two-wire line, etc.
P.E. 8.10 solenoidA 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
mH
cmo
/042.8
240001000 22
Note L is Independent of current
Magnetic Circuits
• Magnetomotive force
• Reluctance
• Like V=IR
• Ex: magnetic relays, motors, generators, transformers, toroids
• Table 8.4 presents analogy between magnetic and electric circuits
dlHNIF
S
l
R
RF IRV
In units of ampere-turns
Magnetic Circuits
Ex. Find current in coil needed to produced flux density of 1.5T in air gap.
Assume r=50 and all cross sectional area is 10 cm2
8.42 A cobalt ring (r=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.
dlHNIF
HlNI
vueltas31312600
3.25.1
ooI
BlN
radius of 30cm
Force on Magnetic materialsRelay• 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
dlS
BdWFdl
om )2(
2
1 2
SB
Fo
2
dvBWm2
2
1
P.E. 8.16 U-shape electromagnet
• Will lift 400kg of mass(including keeper + weight)
• Iron yoke has r=3,000
• Cross section =40cm2
• Air gap are 0.1mm long• Length of iron=50cm• Current is 1AFind number of turns, NFind force across one air gap
mgSB
Fo
2
Esto nos da el B en el air gap.
dlHNIF S
l
R
MagLevmagnetic Levitation
• Use diamagnetic materials, which repel and are repelled by strong H fields.
• Superconductors are diamagnetic.
• Floating one magnet over another
• Maglev train312 mph
• Regular train 187 mph