Magnetic Forces, Materials
and devices
INEL 4151Dr. 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://videos.howstuffworks.com/hsw/18034-electricity-and-magnetism-magnetic-levitation-video.htm
Forces due to Magnetic fieldsEQFe
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
BuEQdtudmF
Forces on a current element
udQdtlddQld
dtdQlId
The current element can be expressed as:
So we can write:
BlIBuQFm
BlIdFL
m
BSKdFS
m
BdvJF
vm
Line
current element
surfacecurrent element
volumecurrent 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
RaldI
Bd Ro
I1
I2
R21
221
21211
21
1 2
ˆ4 R
aldldIIF R
L L
o
P.E. 8.4 Find the force experienced by the loop if I1=10A, I2=5A, ro=20cm, a=1cm, b=30cm
I1
I2
rro a
b
4321 FFFFFl
2
12
12212
12
1 2
ˆ4 R
aldldIIF R
L L
oI
z
F1
F2
r
F3
F4
zDivide loop into 4 segments.
o
o aIB
r
2ˆ1
1
Since I1 is infinite long wire:
120
21 BldIFb
z
F1
r
I1I2
roa
b
z
rr aIbIF
o
o ˆ2
121
For segment #1, Force #1
10
21 ˆ BadzIF z
b
z
r
2ˆ1
1
aIB o
1222 BldIF
F2
r
I1I2
roa
b
z
zo
oo aaIIF ˆln2
212 r
r
For segment #2, The B field at segment #2 due to current 1.
r
r
rr rr aIadIF o
ao
o
ˆ2
ˆ 122
aaI
Bo
o
r
2ˆ1
1
The field at segment 3:
1223 BldIF
F3
r
I1I2
roa
b
z
rr a
abIIF
o
o ˆ2
213
For segment #3, Force #3
r a
aIadzIF
o
oz
bz
ˆ2
ˆ 10
23
r
2ˆ1
1
aIB o
1224 BldIF
F4
r
I1I2
roa
b
z
zo
oo aaIIF ˆln2
214 r
r
For segment #4, The B field at segment #4 due to current 1.
r
r
rr rr aIadIF o
a
o
o
ˆ2
ˆ 124
The total force en the loop is
4321 FFFFFl
rrrr a
aIbIF
ooo
oloop ˆ11
21
2
• The sum of all four:
• Note that 2 terms cancel out:
F1
F2
r
F3
F4
zI1=10A, I2=5A, ro=20cm, a=1cm, b=30cm
r aN ˆ5
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 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 Dipoles, m
24ˆ
ramA ro
Current loop Magnet
naSIm ˆ)(
Where the magnetic moment is:
rAB
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, many materials in their
natural state.Diamagnetic
r≤1Electronic motions of spin
and orbit cancel out.lead, copper, Si,
diamonds, superconductors.
Are weakly affected by B Fields.
Paramagneticr≥1
Temperature dependent.Not many uses except in
masers
Ferromagneticr>>1
Iron, Ni,Co, 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.
HB
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 122211hHhHwHhHhHwH
wKIdlH
nntnnt
l
KHH tt 21
q1
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 magnetric material with r=10
• Assume boundary is current free.
21 /ˆ2.ˆ4.ˆ1.0 mWbzyxB
P.E. 8.7 B-field in a magnetic material.
• In a region with r=4.6• Find H and M and susceptability
2/ˆ10 mmWbzeB y
rm 1
BH
HM m
6.3m
zyaeH 1730
A/m6228 zyaeM
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 turn, the total Flux Linkage is
• This is proportional to the current • The energy stored in
the inductor is
• So we can define the inductance as:
NI
NL
LI
2
21 LIWm
When more than 1 inductor
1
212S
SdB
212
121
2
1212 M
IN
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
NI
L
2
22
2
222 I
NI
L
21122
222
11
1221
21
21 IIMILIL
WWWWm
See table 8.3 in textbook with formulas for inductance of common elements like coaxial cable, two-wire line, etc.
Magnetic Energy
• The energy in a magnetostatic filed in a linear medium is:
• Similar to E field
22
21
21 LIdvHdvHBWm
dvEdvEDWe2
21
21
P.E. 8.10 solenoidA long solenoid with 2x2 cm cross section has
iron core (permeability is 1000x ) and 4000 turns per meter. If carries current of .5A, find:
• Self inductance per m• Energy per m stored in its field
SnlLL 2'
2'21' ILWm
InlIN
lH
lB
lBS
l
'
nl
Nl '
mH
cmo
/042.8240001000 22
mJ /005.1
)5)(.042.8(21 2
Example: Calculate self-inductance of coaxial cable
We’ll find it in two parts:
extLLL int
21int 2 aIHBr
r 22
IHBext
dzdaIdSBd rr
2int1 2
2
2
111 ad
IIdd enc
r
dzdaa
Ia l
z
rrr
r2
2
20 0
1 2
8
Il
8
1' lI
Lin
8inL
Example: (cont.) coaxialWe’ll find it in two parts:
extLLL int
21int 2 aIHBr
r 22
IHBext
dzdIdSBd ext rr
22
dzdIdd rr
222
dzdIb
a
l
z
rr
r 202
abIl ln
2
abl
ILext ln
22'
[H/m]ln41
2
abLLL extin
Example: Find inductance for a 2-wire transmission line
• Lin is same as before:
• We’ll find Lext from the definition of energy:
2
21 LIWm 2
2IWL m
dvBdvHBWm2
21
21
dvBWm2
21
8inL
dzddIWad
a
l
zextm rr
r
r22
222
00, 2
)(2 int extLLL
aadlLext
ln
2
aadlL
ln41
Magnetic Circuits
• Manetomotive force
• Reluctance
• Like V=IR
• Ex: magnetic relays, motors, generators, transformers, toroids
• Table 8.4 presents analogy between magnetic and electric circuits
dlHNIF
Sl
R
RF
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.
dlHNIFHlNI
31312600
3.25.1
ooIBlN
Force on Magnetic materials
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
dlSBdWFdl
om )2(
21 2
SBFo
2
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
mgSBFo
2
Esto nos da el B en el air gap.
dlHNIF Sl
R