ece 307: electricity and magnetism fall 2012 307 chapter 8... · lorentz force law • the force on...

ECE 307: Electricity and Magnetism

Fall 2012

Instructor: J.D. Williams, Assistant Professor

Electrical and Computer Engineering

University of Alabama in Huntsville

406 Optics Building, Huntsville, Al 35899

Phone: (256) 824-2898, email: [email protected]

Course material posted on UAH Angel course management website

Textbook:

M.N.O. Sadiku, Elements of Electromagnetics 5th ed. Oxford University Press, 2009.

Optional Reading:

H.M. Shey, Div Grad Curl and all that: an informal text on vector calculus, 4th ed. Norton Press, 2005.

All figures taken from primary textbook unless otherwise cited.

mailto:[email protected]

8/17/2012 2

Chapter 8: Magnetic Forces Materials and

Devices • Topics Covered

– Forces Due to Magnetic Fields

– Magnetic Torque and Moment

– A Magnetic Dipole

– Magnetization in Materials

– Classification of Magnetic

Materials

– Magnetic Boundary Conditions

– Inductors and Inductances

– Magnetic Energy

– Magnetic Circuits

– Forces on Magnetic Materials

• Homework: 2, 5, 16, 24, 25, 32,

33, 38, 39, 40, 41, 46, 52, 53

All figures taken from primary textbook unless otherwise cited.

Lorentz Force Law

• The force on a charged particle in an electric field is simply F=qE

• However, in the presence of an electromagnetic field an additional force is imposed

from the charge displacement of velocity, u, quantified by the magnetic field, B.

• The combined force is defined by Lorentz Force Law:

• Equating the Lorentz force to Newton’s force equation, one obtains

• Requiring that the kinetic energy of a charged

particle in an electric field is therefore

8/17/2012 3

)( BuEqF

dt

udmamBuEqF

)(

2

2

1

)(

umKE

dtm

qEu

dtm

qEu

dtm

qEu

dt

udmEqF

zz

y

y

xx

dtul

dt

ldu

ii

The location of the particle can also

be found as

Where a is the acceleration of the particle is space

Forces Due to a Magnetic Field

• Similarly, one can solve for the kinetic energy of a particle in a magnetostatic field

• The kinetic energy of a charged particle

in an magnetic field is therefore

8/17/2012 4

dt

udmamBuEqF

)(

2

2

1

)()(

)()(

)()(

)(

umKE

dtm

BvBvqdt

m

Bvqu

dtm

BvBvqdt

m

Bvqu

dtm

BvBvqdt

m

Bvqu

dt

udmBvqF

xyyxzz

xzzxy

y

yzzyxx

dt

ldu

The location of the particle can also be found as

For B, u, and a in orthogonal directions,

One can deduce a coordinate system in which

0)(

ˆ)(

)(

33

212

2

11

dtm

Bvqu

uudtm

Bvqu

dtvdtm

Bvqdt

m

Bvqu

m

Bq

Cyclotron Resonance

Frequency

dtul ii

0

Velocity of a Particle with no

External Forces • Charged particles traveling at constant velocities within an electromagnetic field

have no acceleration term requiring that the total force due to the electric and

magnetic field equal zero. In this case, the velocity is the ratio of the E and B fields

8/17/2012 5

0)( BuEqF

B

Eu

quBqE

• Critical speed required to balance

the two parts of the Lorentz force

• Particles traveling at this speed

travel along a strait path

• Particles traveling slower than this

speed have travel in arc toward the

magnetic force

• Particles traveling faster than this

speed travel in an arc toward the

electric force

• Thus one can design a particle filter

Force on a Current Element

• One can also use field calculations to determine the force acting on a current element,

Idl=KdS=Jdv, due to an applied external magnetic field B.

• Assume that a copper wire carries a current density, J=vu.

• We know:

• And that: Thus we can solve for the force on the first wire:

• Requiring one to define the magnetic field, B, as the force per unit current element

8/17/2012 6

udQudvJlId v

LL

L

BdvJFBdSKF

Likewise

BlIdF

BlIdBudqFd

)(

)(

0

)(

BuqF

Ewhere

BuEqF

Force Between 2 Current Elements

• Now one can use defined the force on a current element from a magnetic field.

• However, that magnetic field must have been generated somehow. What if it was

generated by field produced from current passing through a second current

element nearby. This means that currents in neighboring wires generate magnetic

fields that generate forces on each other.

• Newton’s law requires that the force, F1, acting on element 1 is equal and opposite

to the force F2 acting on element 2.

• One can calculate these interdependent forces through the following derivation.

8/17/2012 7

2

12

2202

2111

2111

4

ˆ

)(

R

aldIBd

recall

BdldIFdd

BldIFd

R

1 21 2

2

2

12

21

210

2

12

220111

2

12

220111

2

12

220111

ˆ

44

ˆ

4

ˆ

4

ˆ)(

L L

R

L L

R

L

R

R

R

aldld

II

R

aldIldIF

R

aldIldIFd

R

aldIldIFdd

onSubstituti

Biot Savart’s Law


(Example) • Consider a rectangular loop of current I1 = Il placed near an infinitely long wire

filament carrying current I2= Iw .

8/17/2012 8

aa

bIIa

a

bIIa

bIIFF

aa

bIIaadz

a

IIa

a

IldIF

adzld

aa

IB

FFor

abII

aadzII

aI

ldIF

adzld

aI

B

FFor

bz

z

L

z

b

z

z

L

z

l

l

ˆ11

2ˆ

2ˆ

2

ˆ2

ˆˆ2

ˆ2

ˆ

ˆ2

_

ˆ2

ˆˆ2

ˆ2

ˆ

ˆ2

_

00

210

0

210

0

21031

0

210

0

0

210

0

10223

2

0

101

3

0

210

00

210

0

10221

2

0

101

1

l

wlll BldIFFFFF

4321

Let’s examine the force on the sections of the loop parallel to the line charge.


(Example) (2) • Consider a rectangular loop of current I1 = Il placed near an infinitely long wire

filament carrying current I2= Iw .

8/17/2012 9 0

ˆln2

ˆ2

ˆ1

ˆ2

ˆ2

ˆ

ˆ2

_

ˆln2

ˆ2

ˆ1

ˆ2

ˆ2

ˆ

ˆ2

_

42

0

02102

21021010222

2

101

2

0

02102

21021010222

2

101

2

0

0

0

0

0

0

0

0

FF

aaII

F

adII

aadII

aI

ldIF

adld

aI

B

FFor

aaII

F

adII

aadII

aI

ldIF

adld

aI

B

FFor

z

a

z

aL

z

a

z

a

L

l

l

a

a

bIIa

a

bIIa

bIIFF

BldIFFFFFl

wlll

ˆ11

2ˆ

2ˆ

2 00

210

0

210

0

21031

4321

Let’s examine the force on the sections of the loop perpendicular to the line charge.

Magnetic Torque and Moment (show class demo here)

• Now that we have examined the force on a current carrying loop. Let’s examine the Torque

applied to it

• Torque, T, on the loop is the vector product of the force, F, and the moment arm, r.

8/17/2012

10

BmT

aISm

IBST

yielding

Slw

but

IBlwT

IBlF

Buniformforand

n

ˆ

sin

sin

___

0

sinFrT

FrT

Where we can now define a quantity m as

the magnetic dipole moment with units A/m2

which is the product of the current and area

of the loop in the direction normal the surface

area defined by the loop

• A bar magnet or small filament loop is generally referred to as a magnetic dipole

• Let us determine the magnetic field,B, of a dipole from an observation point P(r,,) due to a

circular loop carrying current, I.


Magnetic Dipole

8/17/2012 11

aar

mAB

aaa

aaIm

where

amr

ar

aIA

r

ldIA

r

rz

z

r

ˆsinˆcos24

ˆsinˆˆ

ˆ

ˆ4

ˆ4

sin

4

3

0

2

2

0

2

2

00

Note the comparison between the magnetic

dipole term and that developed for the

electrostatic dipole in Chapter 4.

)sincos2(4

cos

44

3

22

aar

pVE

r

dq

r

apV

r

o

oo

r

Comparison of Electric and

Magnetic Dipoles

8/17/2012 12

• A bar magnet or small filament loop is generally referred to as a magnetic dipole

• Assume a bar magnetic of length, l, generates a uniform magnetic field, B, and a dipole

moment, |m|=Qml


Torque and Dipole Properties

of a Bar Magnet

8/17/2012 13

ISlQ

ISBlBQT

BQF

BlQBmT

m

m

m

m

Thus, because the bar magnet represents a

magnetic dipole moment equal in magnitude

to the dipole moment of a current loop, a bar

magnet can also be taken as a magnetic

dipole

Therefore the field at a reasonable distance

away from any bar magnet is mathematically

identical to that of a dipole.

Dipole Moment (Example)

• Consider a small current loop L1 with a magnetic moment m1= 5 az Am2 located at the origin.

• Another loop L2 with magnetic moment m2 = 3 ay Am2 is located at P (4,-2,10).

• What is the torque on L2?

8/17/2012 14

014

5463

ˆˆˆ

5)3125(4

3

5

ˆ4

55

ˆ6

55

ˆ33

5

3125

ˆˆ4

455

ˆˆ45

1

4

02

2

1

10

3

10

1

122

aaa

T

aaam

m

aamaa

mB

BmT

r

r

r

r

5

4cos;

5

3sin,

4

3tan

5

2cos,

5

1sin

10

5tan

551034

ˆcosˆsincosˆcossin3ˆ3

ˆsinˆcos24

222

2

3

101

122

x

y

z

rr

aaaam

aar

mB

BmT

P

ry

r

• We know that all materials are made up of atoms consisting of electrons orbiting nuclei

• Each of these electrons can also be said to spin about its axis

• In certain materials these spins associated with atomic magnetic dipoles align over large

atomic distances to create magnetic domains across several thousands of atoms

• As the individual magnetic domains align, over larger and larger volumes of the material, then

the material is said to magnetize

• Magnetization M, in A/m, is the magnetic dipole moment per unit volume

• If N atoms are in a given volume, v, then the kth atom has a magnetic moment mk

Magnetization in Materials

8/17/2012 15

'1

'4

1'

'4

'

lim

0

3

2

0

1

0

dvR

MA

then

RR

R

recalling

dvaMR

Ad

dvMmd

m

M

R

N

k

k

v

• Substitution of the proper vector identities, coupled with the physics discussed in Chapter 7 will

provide sufficient analysis to develop a complete working theory between magnetization, M,

and the generation of A, J, B, and H.

Magnetization in Materials (2)

8/17/2012 16

S

b

v

b

S

n

vSv

Sv

vv

R

dSK

R

dvJA

dSR

aMdv

R

M

R

SdMdv

R

MA

SdFdvF

identitytheapply

dvR

Mdv

R

MA

R

MM

RRM

Identitytherecall

dvR

MA

'

4

'

4

ˆ

4'

'

44'

'

4

''

__

''4

''

4

''11

'

__

'1

'4

0

'

0

0

'

00

'

0

'

'

0

'

0

0

Where b in the J and K terms represents a bound current densities

• Substitution of the proper vector identities, coupled with the physics discussed in Chapter 7 will

provide sufficient analysis to develop a complete working theory between magnetization, M,

and the generation of A, J, B, and H.

Magnetization in Materials (3)

8/17/2012 17

MHB

MHJJJB

MmaterialainHowever

JB

JH

MspacefreeIn

aMK

MJ

R

dSK

R

dvJA

bf

ff

nb

b

S

b

v

b

0

0

0

0

'

0

0,___,

0,__

ˆ

'

4

'

4

Where f in the J term represents a free current density

0

00

1

1

mr

rm

m

yielding

HHHB

HM

Where is called the permeability of the

material and is measured in H/m

Henry (H) is the unit of inductance that will

be defined later in this chapter

r is called the relative permeability

m is called the magnetic susceptibility of the

medium

• In general we use the magnetic susceptibility (or relative permeability) to classify materials in

terms of their magnetic property

• A material is said to be nonmagnetic if there is no bound current density or zero

susceptibility. Otherwise it is magnetic

• Magnetic materials may be grouped into three classes: diamagnetic, paramagnetic, and

ferromagnetic

• For many practice purposes, diamagnetic and paramagnetic materials exhibit little to no

magnetic susceptibility. What magnetic properties these materials do have, follows a linear

response over a large range of applied fields

• Ferromagnetic materials kept below the Curie temperature exhibit very large nonlinear

magnetic susceptibility and are used for conventional magnetic device applications

Classification of Magnetic Materials (1)

8/17/2012 18

• Diamagnetism

– Occurs when the magnetic fields in the material due to individual electron moments

cancels each other out. Thus the permanent magnetic moment of each atom is zero.

– Such materials are very weakly affected by magnetic fields.

– Diamagnetic materials include Copper, Bismuth, silicon, diamond, and sodium chloride

(table salt)

– In general this effect is temperature independent. Thus, for example, there is no

technique for magnetizing copper

– Superconductors exhibit perfect diamagnetism. The effect is so strong that magnetic

fields applied across a superconductor do not penetrate more than a few atomic layers,

resulting in B=0 within the material

• Paramagnetism

– Materials whose atoms exhibit a slight non-zero magnetic moment

– Paramangetism is temperature dependent

– Most materials (air, tungsten, potassium, monell) exhibit paramagnetic effects that provide

slight magnetization in the presence of large fields at low temperatures


8/17/2012 19

• Ferromagnetism

– Occurs in atoms with a relatively large magnetic moment

– Examples: Cobalt, Iron, Nickel, various alloys based on these three

– Capable of being magnetized very strongly by a magnetic field

– Retain a considerable amount of their magnetization when removed from the field

– Lose their ferromagnetic properties and become linear paramagnetic materials (non magnetic) when the

temperature is raised above a critical temperature called the Curie temperature.

– Their magnetization is nonlinear. Thus the constitutive relation B=0rH does not hold because r

depends directly on B and cannot be represented by a single value.

– Ferromagnetic shielding

• Ferromagnetic materials can be used to “focus” and guide the flow of incident magnetic fields

• By placing a ferromagnetic material completely around a device, one can shield said device from

an external field. This shielding occurs b/c the ferromagnet acts as a magnetic waveguide, that

transmits the field around the shape of the structure and not within it.


8/17/2012 20

• Ferromagnetism - B-H Curve

– The magnetization of a ferromagnet in an external applied field, H, is presented below.

– As H is increased, the magnetic field, B, within the material increases significantly and then begins to

saturate to a value Bmax saturate as |H| approaches Hmax

– As the applied field, H, is removed, the ferromagnetic material retains some degree of its magnetization

until the point at which the applied field H is completely reversed at which time the magnetic field inside

the material saturates to the –Bmax

– The applied field is then increased again to generate the complete Hysteresis curve

– Two other defining values are indicative of every B-H magnetization (Hysteresis) curve.

• When the applied field is maxed and then again reduced to a zero value. The magnetic field within

the material remains at some positive value Br referred to as the permanent flux density.

• The value upon which B become zero under an applied H value is called the coercive field

intensity, Hc

• Materials with small coercive field intensity values are said to be soft magnetic materials and do not

retain significant magnetization upon the removal of the field

• Hard magnets (permanent magnets) have very large

coercive field intensity values


8/17/2012 21

• Magnetic boundary conditions for B and H crossing any material interface must match the

following conditions developed using Guass’s law for magnetic fields and Ampere’s circuit law

Magnetic Boundary Conditions (1)

8/17/2012 22

0 SdB

IldH

nn

nn

nn

HH

BB

yields

SBSB

2211

21

21 0

0 SdB




8/17/2012 23

0 SdB

2

2

1

121

21

122211

0

2222

tt

tt

tt

nntnnt

BBKHH

wHwHwK

h

hH

hHwH

hH

hHwHwK

IldH

0

ˆ1221

K

KaHH

yields

n

2

2

1

1

21

tt

tt

BB

HH

yields

IldH



Magnetic Boundary Conditions

(Review)

8/17/2012 24

0 SdB

IldH

2

2

1

1

21

tt

tt

BB

HH

IldH

nn

nn

HH

BB

2211

21

• These boundary conditions can be used to develop an equivalent to Snell’s law for magnetic

fields


8/17/2012 25

2112

2

2

2211

1

1

222111

tantan

sinsin

coscos

ttt

nn

BHH

B

BBBB

• We now know that closed magnetic circuit carrying current I produces a

magnetic field with flux

• We define the flux linkage between a circuit with N identical turns as

• As long as the medium the flux passes through is linear (isotropic) then then

flux linkage is proportional to the current I producing it and can be written as

Where L is a constant of proportionality called the inductance of the circuit.

A circuit that contains inductance is said to be an inductor.

• One can equate the inductance to the magnetic flux of the circuit as

where L is measured in units of Henrys (H) = Wb/A

• The magnetic energy (in Joules) stored by the inductor is expressed as

Inductors and Inductance

8/17/2012 26

N

SdB

LI

I

N

IL

2

2

1LIWm

• Since we know that magnetic fields produce forces on nearby current elements, and that those

magnetic fields can be generated by an isolated or coupled set of current carrying circuits, then

it is only reasonable that such circuits may induce fields and magnetization between them

• We can calculate the individual flux linkage between the two components as

• Likewise we can determine a mutual inductance between the circuits that is equal from circuit 12

as it is from circuit 21 as

• Individual inductances are

• The total magnetic energy in the circuit is

Inductors and Inductance

8/17/2012 27

1

212

S

SdB

2

121

2

1212

I

N

IM

1

11

1

111

I

N

IL

2

22

2

222

I

N

IL

2112

2

22

2

1112212

1

2

1IIMILILWWWWm

• As we eluded to before, you should think of an inductor as a conductor shaped in such a way as

to store magnetic energy

• Typical examples include toroids, solenoids, coaxial transmission lines, and parallel-wire

transmission lines

• One can determine the inductance for a given geometry using the following technique

– Choose a suitable coordinate system

– Let the inductor carry current, I

– Determine B from Biot-Savart’s or Amperes Law and calculate the magnetic flux

– Find L as a function of the flux times the number of turns over the current carried

• Mutual inductance may be calculated by a similar approach

– Determine the internal inductance, Lin for the flux generated by the first inductor

– Determine the external inductance, Lext produced by the flux external of the first inductor

– The sum of the internal and external inductance equals the individual inductances plus the

mutual inductance between the elements

• For circuit theory, we can also right the inductance as which provides a very useful equation

when quickly mapping out electronic circuits

Inductors and Inductance (2)

8/17/2012 28

1

212

S

SdB

2

121

2

1212

I

N

IM

CLext

RC

1

extext L

R

LR

Inductors and Inductance (3)

8/17/2012 29

• Calculate the self inductance per unit length of an infinitely long solenoid

Inductance Example (1)

8/17/2012 30

SnSnIl

LL

SInnl

InSBS

InHB

lengthunitperturnslNn

turnsNhassolenoid

22

2

''

'

___/

___

• Recall,

• We can derive a similar term for magnetic energy using the relation for energy as a function of

inductance

Magnetic Energy

8/17/2012 31

dvEdvDEWE

2

2

1

2

1

dvHdvBHdvwW

BBHH

v

Ww

vHW

zyxHILW

z

yx

I

yxHL

zHIwhere

I

yxH

IL

LIW

mm

m

vm

m

m

m

2

22

0

2

22

2

2

1

2

1

22

1

2

1lim

2

1

2

1

2

1

,

2

1

Assume a wire carrying current

along the z direction that

generates a field in the x-y plane

• Determine the self inductance of a coaxial cable of inner radius a and outer radius b

Inductance Example (2 part 1)

8/17/2012

32 mHa

b

l

LL

a

bl

IL

a

bIldzd

I

dzdI

dI

Idd

dzdI

dzdBd

aI

B

ba

inext

ext

al

z

enc

/ln2

ln2

ln22

21*

2

ˆ2

'

1

00

22

111

22

2

Note: this figure a as the outer conductor

radius a and b for the inner conductor

mHl

LL

l

IL

Ildzd

a

I

dzda

I

ad

I

Idd

dzda

IdzdBd

aa

IB

a

inin

in

al

z

enc

/8

8

82

2

2

ˆ2

0

'

1

00 4

3

1

4

3

2

2

111

211

21

mHa

bLLL extin /ln

4

1

2' ''

Total Inductance at b

• Alternatively we can solve the same problem form a magnetic energy perpsective

• Determine the self inductance of a coaxial cable of inner radius a and outer radius b

Inductance Example (2 part 2)

8/17/2012

33

mHa

blLLL

a

bldzdddv

I

Idv

B

IL

ldzdd

adv

a

I

Idv

B

IL

extin

ext

in

/ln4

1

2

ln24

1

22

12

2

2

848

2

2

2

22

2

2

2

2

2

42

2

42

22

2

2

1

2

dvB

dvBHLIW

aI

B

ba

aa

IB

a

m

22

1

2

1

ˆ2

ˆ2

0

22

2

21

• Determine the inductance of a two wire transmission line with wire radius a and separation

distance d


8/17/2012

34

8

ˆ2

0

1

21

Il

aa

IB

a

Method 1

a

ad

l

LL

thenad

LIa

adIl

a

adIl

a

adIldzd

Iad

a

l

z

ln4

1'

,

ln4

12

ln4

1

2

ln22

21

21

022

aI

B

ada

ˆ2

2

Same as sol. for

center wire in a

coax

• Determine the inductance of a two wire transmission line with wire radius a and separation

distance d


8/17/2012

35

mHa

adlLLL

a

adldzdddv

I

Idv

B

IL

ldzdd

adv

a

I

Idv

B

IL

dvB

dvBHLIW

aI

B

ba

aa

IB

a

extin

ext

in

m

/ln4

12

ln24

1

22

12

2

2

848

2

2

2

22

1

2

1

ˆ2

ˆ2

0

22

2

2

2

2

2

42

2

42

22

2

2

1

2

22

2

21

Method 2

• Two coaxial circular wires of radius a and b are separated by a distance h>>a,b. Find the

mutual inductance between the wires.


8/17/2012

36

3

22

1

1212

3

22

1

3

2

12112

3

2

11

2/322

2

1

2

2

11

2

22

4

ˆ4

ˆ4

sinˆ

4

sin

h

ba

IM

h

baIb

h

baIldA

ah

baIA

bh

abh

baIa

r

aIA

L

r

lIdA

4

0

L

r

rlIdB

3

0 '

4

Let’s recall

2/3223

22

bhr

bhr

bddl

addl

2

1

Using the equation for a magnetic dipole

aaa

aaIm

where

r

rmam

rA

rz

z

r

ˆsinˆˆ

ˆ

4ˆ

4

2

3

0

2

0

• The following relations allow one to solve magnetic field problems in a manner similar to that of

electronic circuits

• It provides a clear means of designing transformers, motors, generators, and relays using a

lumped circuit model

• The analogy between electronic and magnetic circuits is provided below

Magnetic Circuits

8/17/2012 37

• To develop our model we define a magnetomotive (mmf) force that is equivalent to voltage for

electronic circuits

• We then define a reluctance which is equivalent to magnetic resistance

• And show that the magnetic flux is equivalent to current using the following Ohmic style relation

Magnetic Circuits

8/17/2012

38

ldHNI

S

l

• Because these are often mechanical system, it is extremely useful to be able to determine the

forces generated by these circuits and those required to move components from one location to

another.

• For ease of use, we will ignore fringe fields in our calculations

• Also, by using ferromagnetic materials and applying simple magnetic boundary conditions, we

path very strong fields into specific geometric shapes. This allows one to focus the force to a

specific location and move an object in a well defined manner

• Lets examine the force required to pull a magnetic bar vertically up to an electromagnetic yoke.

Force on Magnetic Materials

8/17/2012

39

m

T

Tm

wBHB

A

Fp

SBF

FFSB

F

SdlB

dlFdW

2

1

2

2

22

2

12

0

2

2/12/1

0

2

2/1

2/12/1

0

2

0

2

Force on a single gap

Note: This diagram shows a yoke pulling a

bar magnet (keeper) at two gap locations

pressure = energy density!!!!!!

• A steel toroidal core with permeability,=r0 has a mean radius, 0, and a circular cross section

of diameter 2a. Calculate the current required to generate a flux, , in the core


(Donut Shaped Toroid)

8/17/2012

40

Method 1

Method 2

2

0

0

2

0

0

0

0

2

2

2

aNI

aNI

BdS

NI

l

NIB

r

r

r

2

0

0

2

0

0

2

0

2

0

00

2

0

0

22

2

22

aNaNNI

NIaa

NI

aS

lNI

rr

r

r

r

• A steel toroidal core with permeability,=r0 has a mean radius 0, and a rectangular cross

section, 2a *b. Calculate the current required to generate a flux, , in the core


(Toroid with Rectangular Cross-section)

8/17/2012

41

a

aIbNLIW

a

abN

I

NL

a

aNIbdNIbbd

NIBdS

NI

l

NIB

rm

r

r

a

a

r

a

a

r

r

0

0

2

02

0

0

2

0

0

0000

0

0

ln42

1

ln2

ln222

2

0

0

0

0

• For the magnetic circuit shown below, with magnetic flux density of 1.5 Wb/m2 and a relative

permeability of 50.

• Find the individual reluctances and determine the total current required to generate a particular

flux value. All branches have a cross sectional area of 0.001 cm2


(Example 2)

8/17/2012

42

20

104.7||

||

20

105

1010

1010

20

109.0

101050

1090

20

103

101050

1030

56

3516

123

143

50

8

213

21

2121

8

4

0

4

0

8

4

0

4

3

8

4

0

4

0

21

3

2

1

aT

a

r

a

r

S

l

S

l

path

andpath

path

path

A

N

SBI

SB

NI

Ta

a

Ta

16.4420400

104.710105.1 88

• A U-shaped electromagnetic is designed to lift a 400 kg mass (which includes the mass of the

keeper). The iron yoke with relative permeability of 3000 has a cross section of 40 cm2 and a

mean length of 50 cm. Each of the air gaps are 0.1mm long. Neglecting the reluctance of the

keeper, calculate the number of turns in the coil when the excitation current is 1A.


(Example 3a)

8/17/2012

43

16215

110001.011.1

65

6

48

105

004.0

0001.02

48

106

004.03000

50.02

/11.1

22

22

2

1

2

1

0

0

6

00

6

00

20

0

2

0

2

_,

22

N

lBlHNI

NINI

S

l

S

l

NI

mWbS

mgB

mgSB

F

FdySdyB

dWdW

ydFdvBdvHW

aaaa

gy

g

g

gy

g

gy

g

g

g

r

y

gy

a

a

gapairmm

v Lv

m

y

N

iy

yLi

x

WF

NyL

NI

S

ty

S

l

S

l

NI

titi

NL

tLiW

S

mgB

mgSB

x

WF

iml

i

kgy

g

kgy

g

g

g

rk

kk

ry

y

y

i

aa

am

a

aml

2

2

000

2

0

0

2

2

1)(

2

1

)(

)(2,,

)()(

)(2

1

22

• Yoke pulling a keeper from both ends


(Example 3b)

8/17/2012

44

FdydyS

SdyB

dWdW

ydFdvBdvBHW

gapairmm

v Lv

m

00

2

_,

2

222

2

1

2

1

mgltxl

tiSN

mdt

dv

vdt

dx

mgSB

ltxl

tiSN

x

titxL

x

WF

ltxl

SNNtxL

NI

S

tx

S

l

S

l

NI

titi

NL

tLiW

S

mgB

mgSB

x

WF

kryrkryyrk

arkry

a

kryrkryyrk

arkryaml

kryrkryyrk

rkry

i

kgy

g

kgy

g

g

g

rk

kk

ry

y

y

i

aa

am

a

aml

2

2222

0

2

0

2

2

2222

0

22

0

22

000

2

0

0

2

)(2

)(1

22

)(2

)()())((

2

1

)(2))((

)(2,,

)()(

)(2

1

22

• Yoke pulling a keeper by gravitational force


(Example 3c)

8/17/2012

45

• Yoke pulling a spring loaded keeper from both ends


(Example 3d)

8/17/2012

46

xkxkxkltxl

tiSN

mdt

dv

vdt

dx

xkxkxkSB

ltxl

tiSN

x

titxL

x

WF

ltxl

SNNtxL

NI

S

tx

S

l

S

l

NI

titi

NL

tLiW

S

mgB

xkxkxkSB

x

WF

sss

kryrkryyrk

arkry

sssa

kryrkryyrk

arkryaml

kryrkryyrk

rkry

i

kgy

g

kgy

g

g

g

rk

kk

ry

y

y

i

aa

am

a

sssam

l

212

2222

0

2

21

0

2

2

2222

0

22

0

22

000

2

0

21

0

2

)(2

)(1

22

)(2

)()())((

2

1

)(2))((

)(2,,

)()(

)(2

1

22

• The system is driven by voltage control and not steady state current


(Example 3e)

8/17/2012 47

vdt

dx

xkxkxkltxl

tiSN

mdt

dv

tutviltxl

tiSNtRi

xLdt

tdi

isequationsofsetcompletetheThus

tutviltxl

tiSNtRi

xLdt

tdi

dt

dx

dt

xdLi

dt

tditRitu

dt

tixLdtRi

dt

dtRitu

xkxkxkSB

ltxl

tiSN

x

titxL

x

WF

sss

kryrkryyrk

arkry

aa

kryrkryyrk

arkry

aa

aa

kryrkryyrk

arkry

aa

aa

aa

aaaa

sssa

kryrkryyrk

arkryaml

212

2222

0

2

2

2222

0

2

2

2222

0

2

21

0

2

2

2222

0

22

)(2

)(1

)()()(2

)(2)(

)(

1)(

:______

)()()(2

)(2)(

)(

1)(

)()()()(

)(*)()()()(

22

)(2

)()())((

2

1

ece 307: electricity and magnetism fall 2012 307 chapter 8... · lorentz force law • the force on...

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