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