tesla - electromagnetic engineering

8/14/2019 Tesla - Electromagnetic Engineering

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Tom Penick [email protected] www.teicontrols.com/notes 10/22/2000 Page 1 of 13

ELECTROMAGNETIC ENGINEERING EE325

INDEX

Ampere's circuital law.....11Ampere's law ................... 6angstrom.......................... 2Avogadro's number..... ...... 2B Ampere's circuital law 11Biot-Savart law...............11Boltzmann's constant ........ 2capacitance ...................7, 8

between coaxial cylinders................................ 7

between concentricspheres .................... 7

between parallel plates. 7between two conductors7

characteristic impedance .. 2complex conjugate......... ... 1complex notation.............. 1conductance ..................... 8conductivity ..................... 8

semiconductor.............. 8

conservative field law....... 6constants.......................... 2continuity equation........... 8coordinate systems ......... .10coordinate transformations10coulomb ........................... 1Coulomb's law.................. 7cross product...................10curl.................................. 9current ............................. 8current density ................. 7D flux density ................. 6del ................................... 8divergence........................ 9

dot product....................... 9

duality of J and D............. 8E electric field................. 5electric field..................... 5electron mass ................... 2electron volt..................... 2electrostatic

force ............................ 5potential ...................... 5

electrostatics .................... 5elipse............................... 8Faraday's law ..............6, 12flux density...................... 6force

electrostatic ................. 5magnetic.....................11

Gauss' law........................ 6geometry.......................... 8grad operator.................... 8H magnetic field intensity12impedance

short-circuit ................. 2induced voltage

due to changing magneticfield........................13

due to conductor motion13Faraday's law..............12slider problem.............13

inductance.......................12J current density.............. 7

joule ................................ 2Laplacian ......................... 9Lenz's law.......................12light, speed of .................. 2line impedance................. 3

linkage............................12

magnetic energy..............12magnetic field .................11

at the center of a circularwire........................11

central axis of a solenoid...............................11

due to a finite straightconductor................11

due to an infinite straightconductor................11

magnetic field intensity...12magnetic flux..................12magnetic force.................11magnetization..................13matching transformer

inline reactive load.... 3inline resistive load... 3

mathematics..................... 8Maxwell's equations......... 6mutual inductance...........12

nabla operator .................. 8permeability..................... 2permittivity ...................... 2phase constant.................. 2Planck's constant .............. 2Poisson's equation ............ 6potential energy................ 7power

with phasor notation..... 5reactance.......................... 3reflection coefficient......... 2resistance......................... 8Rydberg constant.............. 2self-inductance................12

series stub........................ 4

shunt stub ........................ 4single-stub tuning............. 4Smith chart ...................... 4Smith charts..................... 4space derivative ............... 8sphere.............................. 8standing wave ratio .......... 4static magnetic field ........11stub length ....................... 4surface charge density...... 6time average power .......... 5vector differential equation8volume energy density...... 7wave

forward-traveling ......... 5wave equation.................. 2wavelength....................... 2We potential energy......... 7we volume energy density 7

X reactance ..................... 3Zin line impedance........... 3 electrostatic

potential ...................... 5 reflection coefficient .... 2 magnetic flux.............12 wavelength................... 2s surface charge density. 6 conductivity................. 8 del............................... 8 curl ........................... 9 divergence................ 92 Laplacian................... 9

COULOMB [C]A unit of electrical charge equal to one amp second,the charge on 6.2110

18electrons, or one joule per

volt.

COMPLEX NOTATION

)( baaejb = where b may be in radians or degrees (if noted).

COMPLEX CONJUGATES

The complex conjugate of a number is simply thatnumber with the sign changed on the imaginary part.This applies to both rectangular and polar notation.When conjugates are multiplied, the result is a scalar.

22))(( bajbajba +=+ 2))(( ABABA =

Other properties of conjugates:

*)*****()*( FEDCBAFDEABC ++=++ jBjB ee + =)*(


2/13


TRANSMISSION LINES

L REFLECTION COEFFICIENT [V/V]The reflection coefficient is a value from 1 to +1which, when multiplied by the wave voltage,determines the amount of voltage reflected at oneend of the transmission line.

0

0

ZZ

ZZ

eL

Lj

L +

==

andL

LL ZZ

+= 1

10

where: LZ is the load impedance

L is the load reflection coefficient is the reflection coefficient magnitude is the reflection coefficient phase

C

LZ =0 is the characteristic impedance

THE COMPLEX WAVE EQUATION

The complex wave equation is applicable when theexcitation is sinusoidal and the circuit is under steadystate conditions.

)()( 2

2

2

zVzd

zVd=

where2

LC = =

is the phase constant

The complex wave equationabove is a second-orderordinary differential equation commonly found in theanalysis of physical systems. The general solution is:

zjzj eVeVzV ++ +=)(

wherezje and

zje + represent wave propagationin the +z and z directions respectively.

The same equation applies to current:

zjzj eIeIzI ++ +=)( and

0

)(Z

eVeVzI

zjzj ++ +=

where 0 / Z L C = is the characteristic impedanceof the line. These equations represent the voltageand current phasors.

SHORT-CIRCUIT IMPEDANCE []

( )ljZZsc = tan0

where: 0Z is the characteristic impedance

==

2LC is the phase constant

l is the length of the line [m]

CONSTANTS

Avogadros number

[molecules/mole]231002.6 =

AN

Boltzmanns constant231038.1 =k J/K51062.8 = eV/K

Elementary charge191060.1 =q C

Electron mass31

0 1011.9=m kg

Permittivity of free space12

0 1085.8= F/m

Permeability constant7

0 104= H/m

Plancks constant341063.6 =h J-s

151014.4 = cV-sRydberg constant 678,109=R cm-1

kT @ room temperature 0259.0=kT eV

Speed of light810998.2 =c m/s

1 (angstrom) 10-8 cm = 10-10m

1 m (micron) 10-4cm1 nm = 10 = 10-7cm

1 eV = 1.6 10-19J

1 V = 1 J/C 1 N/C = 1 V/m 1 J = 1 N m = 1C V

WAVELENGTH [m]

f

vp=

vp = velocity of propagation (2.998108

m/sfor a line in air)

f= frequency [Hz]


3/13


-WAVELENGTH INLINE MATCHINGTRANSFORMER resistive load

For use with a purely resistive load that does not match theline impedance. The load is matched to the line byinserting a -wavelength segment having a characteristic

impedanceZQ.

Z0

ZQ

/4

RL

LQ RZZ 0=

Z0 = characteristic impedance of the

transmission line [] = wavelength [meters]

RL = resistance of the load []ZQ = characteristic impedance of the

-wave matching segment []

-WAVELENGTH INLINE MATCHINGTRANSFORMER reactive load

For use with a reactive load. The load is matched to the

line by inserting a -wavelength segment having acharacteristic impedanceZQ at a distance lfrom the load. lis the length of transmission line required to produce thefirst voltage maximumclosest to the load. If the load isinductive, the first voltage maximum will be closer than thefirst voltage minimum, i.e. within wavelength.

0Z

0Z

/4

ZQ

l

Zin LZ

First find the reflection coefficient in order to determine the

value of . Then find the length l of the line that willconvert the load to a pure resistance, i.e. produces the first

voltage maximum. Find this resistance (Zin) using the lineimpedance formula. Then determine the impedance ZQ ofthe -wavelength segment that will match the load to theline.

0

0

ZZ

ZZe

L

Lj

L +

==

i.e. =L (radians)

=

=

42l

ljZZ

ljZZZZ

L

Lin +

+=

tan

tan

0

00

inQ ZZZ 0=

L is the load reflectioncoefficient

= phase of the reflectioncoefficient [radians]

= magnitude of thereflection coefficient []

Z0 = characteristic

impedance []= /2

= vp/f wavelength [m]Zin = impedance (resistive)of the load combined

with the l segment []ZQ = line impedance of the

-wave matching

segment []

X REACTANCE []

C

jX

C

=

LjXL

=

XC = reactance []XL = reactance []j = 1 =frequency [radians]C= capacitance [F]

L = inductance [H]

Zin LINE IMPEDANCE []

ljZZ

ljZZZZ

L

L

in ++

=tan

tan

0

00

l = distance from load [m]j = 1 =phase constant

Z0 = characteristic

impedance []ZL = load impedance []

The line impedance of a -wavelength line is the inverseof the load impedance.

Impedance is a real value when its magnitude is

maximum or minimum.

+

==11

00max ZSZZ

+

==11

00

min ZS

ZZ

Z0 = characteristic

impedance []S = standing wave ratio

= magnitude of thereflection coefficient


4/13


SMITH CHARTS

First normalize the load impedance by dividing by thecharacteristic impedance, and find this point on the chart.An inductive load will be located on the top half of thechart, a capacitive load on the bottom half.

Draw a straight line from the center of the chart throughthe normalized load impedance point to the edge of thechart.

Anchor a compass at the center of the chart and draw anarc through the normalized load impedance point. Pointsalong this arc represent the normalized impedance atvarious points along the transmission line. Clockwisemovement along the arc represents movement from theload toward the source with one full revolution representing

1/2 wavelength as marked on the outer circle. The twopoints where the arc intersects the horizontal axis are thevoltage maxima (right) and the voltage minima (left).

Points opposite the impedance (180around the arc) areadmittance. The reason admittance is useful is becauseadmittances in parallel are simply added.

zj

Lez = 2)(

ze zj = 212 ( ) 1

( )( ) 1z

zz

=

+Z

Z

1

1

+

=L

L

LZ

0Z

ZL=Z

z = distance from load[m]

j = 1 = magnitude of the

reflection coefficient

=phase constant= reflection coefficientZ = normalized

impedance []

SINGLE-STUB TUNINGThe basic idea is to connect a line stub in parallel(shunt) or series a distance dfrom the load so

that the imaginary part of the load impedance willbe canceled.

Shunt-stub: Select dso that the

admittance Y lookingtoward the load from

a distance d is of the

form Y0 +jB. Thenthe stubsusceptance is

chosen as jB,resulting in amatched condition.

Y

Openor

short

l

Y0

0

d

Y0 YL

Series-stub: Select dso that the admittance

Zlooking toward the

load from a distance dis of the formZ0 +jX.Then the stubsusceptance is chosenas -jX, resulting in amatched condition.

LZ

l

0Z

Openor

short

0Z

d

0Z

FINDING A STUB LENGTHExample: Find the lengths of open and shorted shuntstubs to match an admittance of 1-j0.5. The admittance

of an open shunt (zero length) is Y=0; this point is

located at the left end of the Smith Chartx-axis. Weproceed clockwise around the Smith chart, i.e. awayfrom the end of the stub, to the +j0.5 arc (the valueneeded to match j0.5). The difference in the startingpoint and the end point on the wavelength scale is the

length of the stub in wavelengths. The length of ashorted-type stub is found in the same manner but

with the starting point at Y=.

rotage rnerdawoT

Admittance(short)

Admittance(open)

Shorted stub oflength .324matches anadmittanceof 1-j.5

.46.324.

47

.48

.49

.43

.44

.45

Y

1

.

0

.42

.4.4

1

.38.39

0.5

= 0

j

.06

.04

0

.01

.02

.03

.074

0

.

1

.05

Open stub oflength .074matches anadmittanceof 1-j.5

.07

0.5

0

.

5

1

.

0

.1

.08

.09

.5

1

.

0

.11.12

.33

.35.36.37

.34

2.0

.29

.3

.31

.32

5.0

.2

6

.27

.28

5

.17

2.0

2

.15.14

.13

.16

.19

.21

Y

5.0

.2

.23

.

2

5

.24

.22

=

.18

In this example, all values were in units of admittance.If we were interested in finding a stub length for aseries stub problem, the units would be in impedance.

The problem would be worked in exactly the same way.Of course in impedance, an open shunt (zero length)

would have the value Z=, representing a point at the

right end of the x-axis.

SWR STANDING WAVE RATIO [V/V]

+

===11

SWRmin

max

min

max

I

I

V

V


5/13


P(z) TIME-AVERAGE POWER ON ALOSSLESS TRANSMISSION LINE [W]

Equal to the power delivered to the load. The powerdelivered to the load is maximized under matched

conditions, i.e. = 0, otherwise part of the power isreflected back to the source. To calculate power, itmay be simpler to find the input impedance and useP = I

2R or P = IV.

( )20

2

12

)(P =+

Z

Vz

[ ]{ }1

P( ) ( ) ( ) *2

z V z I z= Re

V+

= the voltage of theforward-traveling

wave [V]Z0 = characteristic

impedance []= magnitude of the

reflection coefficientRe = "the real part"

POWER USING PHASOR NOTATION [W]

*

2

1VIS =

S = power[W]V= volts [V]

I* = complex conjugate of current [A]

V+

FORWARD-TRAVELING WAVE

( ) ( )ljLlj

Sin

in

eeZZ

VZV

+

++=

20

1

V+ = the voltage of the forward-

traveling wave [V]V0 = source voltage [V]

Zin = input impedance []ZS =source impedance []

=phase constantl=length of the line [m]L = load reflection

coefficient

ELECTROSTATICS

F ELECTROSTATIC FORCE

3

12

1221

012

)(

41

rr

rrF

= QQ 90

1094

1=

F12 = the force exerted by charge Q1onQ2. [N]r1 = vector from the origin to Q1r2 = vector from the origin to Q2.

When finding the force on one charge due to multiplecharges, the result can be found by summing theeffects of each charge separately or by converting themultiple charges to a single equivalent charge andsolving as a 2-charge problem.

E ELECTRIC FIELD

=

=n

k k

kkp

Q1

304

1

rr

rrE

( ) ldrd l = 204 1 rrRE

( )

= ldr

l

20

4

1

rrRE

Electric field from a potential:

=E refer to the NABLA notes onpage 8.

*NOTE: The l symbols couldbe replaced by a symbol forarea or volume. See WorkingWith on page 9.

Ep = electric field at pointp due to a charge Q

or charge density [V/m]

dE = an increment ofelectric field [V/m]

Q = electric charge [C]

0 = permittivity of freespace 8.85 10-12F/m

l = charge density;charge per unit

length* [C/m]dl' = a small segment of

line l*R = unit vector pointing

from r'to r , i.e. in

the direction of r- r'.

r'= vector location of thesource charge in

relation to the originr = vector location ofthe point at whichthe value of Ep isobserved

= Del, Grad, or Nablaoperator

ELECTROSTATIC POTENTIAL [V]

=

=n

k k

kQ

1041

rr

rr

= ldd l

041

ldl

= rr041

Potential due to anelectric field:

=b

aab dlE

To evaluate voltage atall points.

( ) =r

dr lE

*NOTE: The l symbolscould be replaced by asymbol for area orvolume. See WorkingWith on page 9.

= the potential [V]d = an increment of potential

[V]

ab = the potential differencebetween points a and b [V]

E = electric fielddl' = a small segment of line l*

dl = the differential vectordisplacement along the

path from a to b

0 = permittivity of free space8.85 10

-12F/m

Q = electric charge [C]l = charge density along a

line* [C/m]rk'= vector location of source

charge Qkr'= vector location of the

source charge in relationto the origin

r = vector location of

electrostatic potential in relation to the origin


6/13


MAXWELL'S EQUATIONS

Maxwell's equations govern the principles of guidingand propagation of electromagnetic energy andprovide the foundations of all electromagneticphenomena and their applications.

t

B

E = - Faraday's Law

D = Gauss' Law

t

+

D

H = J Ampere's Law*

0 B = no name law, where:E= electric field [V/m]

B = magnetic field [T]t= time[s]

D = electric flux density [C/m2] = volume charge density[C/m3]

H= magnetic field intensity [A/m]J= current density [A/m2]

*Maxwell added thet

D term to Ampere's Law.

POISSON'S EQUATION

0

2

=

s SURFACE CHARGE DENSITY [C/m2]

ns E0=

En=nE

0 = permittivity of free space 8.85 10-12

F/mEn = electric field normal to the

surface [V/m]

D FLUX DENSITY [C/m2]or ELECTRIC DISPLACEMENT PER UNIT AREA

24

r

Q

rD

ED =

Q = electric charge [C] = dielectric constant

r= 0 E = electric field [V/m]

GAUSS'S LAWThe net flux passing through a surface enclosing a chargeis equal to the charge. Careful, what this first integral reallymeans is the surface area multiplied by the perpendicularelectric field. There may not be any integration involved.

encS

Qd = sE0 == V encS QdvdsD 0 = permittivity of free space 8.85 10-12F/m

E = electric field [V/m]D = electric flux density vector [C/m2]ds = a small increment of surface S

= volume charge density [C/m3]dv = a small increment of volume VQenc = total electric charge enclosed by the Gaussian

surface [S]

The differential version of Gauss's law is:

= D or ( ) = Ediv 0

GAUSS'S LAW an example problem

Find the intensity of the electric field at distance rfrom a

straight conductor having a voltage V.

Consider a cylindrical surface of length l and radius renclosing a portion of the conductor. The electric fieldpasses through the curved surface of the cylinder but notthe ends. Gauss's law says that the electric flux passingthrough this curved surface is equal to the charge enclosed.

VlClQdlrEd llencrS

==== 2

000 sE

so VCdrE lr =

2

0

0 and

r

VCE l

r

02

=

Er= electric field at distance rfrom the conductor [V/m]l = length [m]r d = a small increment of the cylindrical surface S[m2]l = charge density per unit length [C/m]Cl = capacitance per unit length [F/m]V= voltage on the line [V]

CONSERVATIVE FIELD LAW

0= E 0 =S dlE E = electric field [V/m]ds = a small increment of length


7/13


COULOMB'S LAW

= D = VS dvdsD D = electric flux density vector [C/m2] = volume charge density [C/m3]ds = a small increment of surface S

We POTENTIAL ENERGY [J]The energy required to bring charge q from infinity toa distanceR from charge Q.

R

QqqWe

==4

== VVe dvdvW ED21

21

= the potential between q and Q [V]q,Q = electric charges [C] = permittivity of the material

R = distance [m] = volume charge density [C/m3]E = electric field [V/m]D=electric flux density vector [C/m2]

we VOLUME ENERGY DENSITY [J/m3]

for the Electrostatic Field

2

21

21

Ewe == ED

= the potential between q and Q [V]

= permittivity of the material

R = distance [m]E = electric field [V/m]D=electric flux density vector [C/m2]

CAPACITANCE

C CAPACITANCE [F]

=

QC

V

C ll

=

Q = total electric charge [C] = the potential between q and Q [V]Cl = capacitance per unit length [F/m]l = charge density per unit length [C/m]

V= voltage on the line [V]

C CAPACITANCE BETWEEN TWOPARALLEL SOLID CYLINDRICAL

CONDUCTORS

This also applies to a single conductor above ground,

where the height above ground is d/2.

( )ad

C

/ln

= , where d a?

or1cosh

2

Cd

a

=

C= capacitance[F/m]

= permittivity ofthe material

d= separation(center-to-center) [m]

a = conductorradius [m]

C CAPACITANCE BETWEEN PARALLELPLATES

AC

d

=

C= capacitance [F] = permittivity of the material

d= separation of the plates [m]A = area of one plate [m2]

C CAPACITANCE BETWEEN COAXIALCYLINDERS

( )

2ln /

Cb a

=

C= capacitance [F/m] = permittivity of the materialb = radius of the outer cylinder

[m]a = radius of the inner cylinder

[m]

C CAPACITANCE OF CONCENTRICSPHERES

4 abC

b a

=

C= capacitance [F/m] = permittivity of the materialb = radius of the outer sphere [m]a = radius of the inner sphere [m]

J CURRENT DENSITY

The amount of current flowing perpendicularly

through a unit area [A/m

2

]EJ =

= S dI sJ Insemiconductormaterial:

decqn vJ =

= conductivity of the material [S/m]E = electric field [V/m]I= current [A]ds = a small increment of surface Snc = the number of conduction band

electronsqe = electron charge -1.60210

-19C

vd = a small increment of surface S


8/13


CONTINUITY EQUATION

0 =

+t

J

J = current density [A/m2] EJ = = volume charge density [C/m3]

DUALITY RELATIONSHIP of J and DRESISTANCE, CAPACITANCE, CURRENT,CONDUCTIVITYWhere current enters and leaves a conductingmedium via two perfect conductors (electrodes) wehave:

=

=== Q

dddISSS

sDsEsJ

J = current density [A/m2] EJ = E = electric field [V/m]D = electric flux density vector [C/m2] ED =

As a result of this, we have the following relation,useful in finding the resistance between twoconductors:

=RC

R = resistance []C= capacitance [F] = permittivity of the material = conductivity of the material [S/m]

G CONDUCTANCE [1]

== I

RG

1

+

=

lE

sE

d

dS

R = resistance []I= current [A] = voltage potential [V] = conductivity of the material

[S/m]

SEMICONDUCTOR CONDUCTIVITY

[1]

deNq

= conductivity of the material[S/m]G = conductance [1]

q = electron charge -1.60210-19Ce = electron mobility [m2/(V-s)]

Nd

= concentration of donors, andthereby the electron concentration

in the transition region [m-3]

MATHEMATICS

WORKING WITH LINES, SURFACES, ANDVOLUMES

l(r') means "the charge density along line l as afunction of r'." This might be a value in C/m or it

could be a function. Similarly, s(r') would be thecharge density of a surface and v(r') is the

charge density of a volume.For example, a disk of radius a having a uniform

charge density of C/m2, would have a totalcharge of a2, but to find its influence on pointsalong the central axis we might considerincremental rings of the charged surface as

s(r') dr'= s2r'dr'.If dl' refers to an incremental distance along a circular

contour C, the expression is r'd, where r' is theradius and d is the incremental angle.

GEOMETRYSPHEREArea 24 rA =

Volume 3

34

rV =

ELLIPSEArea ABA = Circumference

22

22ba

L+

NABLA, DEL OR GRAD OPERATOR[+ m-1]

Compare the operation to taking the time

derivative. Where /tmeans to take the derivativewith respect to time and introduces a s-1 component tothe units of the result, the operation means to takethe derivative with respect to distance (in 3dimensions) and introduces a m-1 component to theunits of the result. terms may be called spacederivativesand an equation which contains the operator may be called a vector differential

equation. In other words A is how fast A changesas you move through space.in rectangularcoordinates:

A A A

x y z x y z

= + +

A

in cylindrical

coordinates:

1

A A A

r zr r z

= + + A in sphericalcoordinates:

1 1 sin

A A Ar

r r r

= + +

A


9/13


2 THE LAPLACIAN [+ m-2]in rectangularcoordinates:

0 2222 =++= zyx AAA zyxA

022

2

2

2

22 =

+

+

zyx

in spherical andcylindricalcoordinates:

( )

( ) ( )AA

AAA

curlcurldivgrad

2

=

for example, theLaplacian of electro-static potential:

022

2

2

2

22 =

+

+

=zyx

DIVERGENCE [+ m-1]The del operator followed by the dot product operatoris read as"the divergence of" and is an operation

performed on a vector. In rectangular coordinates, means the sum of the partial derivatives of themagnitudes in thex,y, andz directions with respect tothex,y, andz variables. The result is a scalar, and afactor of m

-1is contributed to the units of the result.

For example, in this form of Gauss' law, where D is a

density per unit area, D becomes a density per unitvolume.

div yx zDD D

x y z

= = + + =

D D

D = electric flux density vector D = E [C/m2] = source charge density[C/m3]

In the electrostatic context, the divergence of D is thetotal outward flux per unit volume due to a sourcechar e. The diver ence of vector D is:in rectangularcoordinates:

z

D

y

D

x

D zyx

+

+

=Ddiv

in cylindricalcoordinates: ( )

z

DD

rrD

rr

zr

+

+

= 11

div D

in spherical coordinates:

( ) ( )

+

+

=

D

r

D

rr

Dr

r

r

sin1sin

sin11

div2

2D

CURL [+ m-1]The circulation around an enclosed area. The curl ofvector B isin rectangular coordinates:

curl

y yx xz zB BB BB B

x y z y z z x x y

= =

+ +

B B

in cylindrical coordinates:

( )

curl

1 1 z r z rrBB B B B B

r zr z z r r r

= =

+ +

B B

in spherical coordinates:

( )

( ) ( )

sin1curl

sin

1 1 1 sinr r

B Br

r

rB rBB Br r r r

= = +

+

B B

The divergence of a curl is always zero:

( ) 0 = H

DOT PRODUCT [= units2]The dot product is a scalar value.

( ) ( )zzyyxxzyxzyx BABABABBBAAA ++=++++= zyxzyxBA

ABcos = BABA

0 =yx , 1 =xx ( ) yzyx BBBB =++= yzyxyB

B

A

AB

Projection of Balong :

( )aaB

B

B

The dot product of 90 vectors is zero.The dot product is commutative and distributive:

ABBA = ( ) CABACBA +=+


10/13


CROSS PRODUCT

) )( ) ( ) ( )xyyxzxxzyzzy

zyxzyx

BABABABABABA

BBBAAA

++=

++++=

zyx

zyxzyxBA

ABsin = BAnBA

where n is the unit vector normal toboth A and B (thumb of right-hand rule).

BAAB = zyx = zxy = 0= xx

=z r = r z The cross product is distributive:

( ) CABACBA +=+ Also, we have:

( ) ( ) ( ) = A B C A C B A B C

n

AB

A

B

COORDINATE SYSTEMS

Cartesian or Rectangular Coordinates:

( , , ) x y z xx yy zz= + +r x is a unit vector

222 zyx ++=r

Spherical Coordinates:

),,( rP ris distance from center is angle from vertical is the CCW angle from the x-axis

r, , and are unit vectores and are functions ofpositiontheir orientation depends on where theyare located.

Cylindrical Coordinates:

),,( zr C ris distance from the vertical (z) axis is the CCW angle from the x-axis

z is the vertical distance from origin

COORDINATE TRANSFORMATIONS

Rectangular to Cylindrical:

To obtain: ( , , ) r zr z rA A zA = + +A

22yxA

r+= cos sinr x y= +

x

y1tan = sin cosx y = +

zz = z z= Cylindrical to Rectangular:

To obtain: ( , , ) x y z xx yy zz= + +r

= cosrx cos cosx r=

= sinry sin cosr y = + zz = z z=

Rectangular to Spherical:

To obtain: ( , , ) rr rA A A = + + A

222

zyxAr ++= sin cos sin sin cosr x y z= + +

222

1cos

zyx

z

++=

cos cos cos sin sin x y z = +

x

y1tan = sin cosx y = +

Spherical to Rectangular:

To obtain: ( , , ) x y z xx yy zz= + +r

= cossinrx sin cos cos cos sinx r=

= sinsinry sin sin cos sin cosy r= + +

= cosrz cos sinz r=


11/13


THE STATIC MAGNETIC FIELD

F F12 MAGNETIC FORCE [N/m]due to a conductor

If the current in the two wires travels in oppositedirections, the force will be attractive.

d

II

= 2

21012

xF

F12 = the force exerted by conductor 1

carrying current Ion conductor 2.

[N/m]0= permeability constant 410-7

[H/m]I= current [A]d= distance between conductors [m]

BP BIOT-SAVART LAWDetermines the B field vector at any point P identified

by the position vector r, due to a differential current

elementI dl' located at vector r'.

20

4

'

R

dI

d P

=Rl

B

( )

=

CP

dI3

0

'

''4 rr

rrlB

''

rr

rrR

=

BP = magnetic field vector

[T]

0= permeability constant410-7 [H/m]

Idl'= current element [A]R = unit vector pointing

from the current

element to point P

R= distance between thecurrent element and

point P [m]

B AMPERE'S CIRCUITAL LAWAmpere's law is a consequence of the Biot-Savart

law and serves the same purpose as Gauss's law.Ampere's law states that the line integral of B around

any closed contour is equal to 0 times the total netcurrentIpassing through the surface S enclosed by

the contour C. This law is useful in solvingmagnetostatic problems having some degree ofsymmetry.

I

ddSC

0

0

=

= sJlB

B = magnetic field vector, equal to

B times the appropriate unit

vector [T]0= permeability constant 410-7

[H/m]

dl= an increment of the line whichis the perimeter of contour C

[m]J = current density [A/m2] EJ = ds= an increment of surface [m2]

B MAGNETIC FIELD [T or A/m]due to an infinite straight conductor

May also be applied to the magnetic field close to aconductor of finite length.

02P

I

r

=

B

BP = magnetic field vector [T]0= permeability constant 410-7 [H/m]

I= current [A]r= perpendicular distance from the

conductor [m]

B MAGNETIC FIELD [T]due to a finite straight conductor at a point

perpendicular to the midpoint

0

2 2

2

P

Ia

r r a

= +

B

a

r

I

BP = magnetic field vector [T]0= permeability constant

410-7 [H/m]I= current [A]a= half the length of the

conductor [m]r= perpendicular distance

from the conductor [m]

B MAGNETIC FIELD [T]at the center of a circular wire of Nturns

a

NIB

ctr 2 0

= z

B = magnetic field [T]0= permeability const. 410-7 [H/m]

N= number of turns of the coil

I= current [A]a = radius [m]

B MAGNETIC FIELD [T]along the central axis of a solenoid

( )( )

( )

( )

( )

+

++

+=2222

0

2/

2/

2/

2/2

lza

lz

lza

lz

l

NIzB z

and at the center of the coil:l

NIBctr

0 z

B = magnetic field [T]0= permeability constant

410-7 [H/m]N= number of turns

I= current [A]

l= length of the solenoid [m]z= distance from center of

the coil [m]a = coil radius [m]


12/13


H MAGNETIC FIELD INTENSITY [A/m]The magnetic field intensity vector is directlyanalogous to the electric flux density vectorD in

electrostatics in that both D and H are medium-independent and are directly related to their sources.

MB

H

0

t+=DJH

H = magnetic field [A/m]B = magnetic field vector [T]0= permeability const. 410-7 [H/m]M = magnetization [A/m]J = current density [A/m2] EJ = D = electric flux density vector

[C/m2]

, , (,lambda) MAGNETIC FLUX, LINKAGE

Flux linkage is the ability of a closed circuit to storemagnetic energy. It depends, in part, on thephysical layout of the conductors. It is the totalmagnetic field due to circuit #1 passing through thearea enclosed by the conductors of circuit #2. The

text seemed to describe as the flux due to one turn

and as the flux due to all of the turns of the coil, butwas not consistent so be careful.

=2

2112 S

dsB

12112 = N

= S dN sB

12= the magnetic flux passingthrough coil 2 that is produced

by a current in coil 1 [Wb]= total flux linkage [Wb]B = magnetic field vector [T]


ds= an increment of surface [m2]

LENZ'S LAW

Induced voltage causes current to flow in the directionthat produces a magnetic flux which opposes the fluxthat induced the voltage in the first place. This law isuseful in checking or determining the sign or polarityof a result.

L INDUCTANCE [H]Inductance is the ability of a conductor configurationto "link magnetic flux", i.e. store magnetic energy.Two methods of calculating inductance are givenbelow.

I

L

=

2

2

I

WL m=

= flux linkage [Wb]I= current [A]Wm = energy stored in a magnetic field

[J]

L11 SELF-INDUCTANCE [H]When a current in coil 1 induces a current in coil 2,the induced current in coil 2 induces a current back incoil 1. This is self-inductance.

1

112

1

1

11111

I

N

I

NL

=

=


11= the total flux linked by a singleturn of coil 1 [Wb]

I1 = current in coil 1 [A]

11 = the magnetic flux produced bya single turn of coil 1 and linkedby a single turn of coil 1 [Wb]

L12 MUTUAL INDUCTANCE [H]The mutual inductance between two coils.

1

1212

1

12212

I

NN

I

NL

=

=

Neumann formula:

=1 2 '

4

2121012

C C

ddNNL

rr

ll

N= number of turns ofthe coil

= flux linkage [Wb]I= current [A]= magnetic flux [Wb]r = vector to the point

of observationr' = vector to source

Wm MAGNETIC ENERGY [J]Energy stored in a magnetic field [Joules].

= Vm dvBW '21 2

0

Wm = energy stored in a magnetic

field [J]0= permeability constant

410-7 [H/m]B = magnetic field [T]

FARADAY'S LAW

When the magnetic flux enclosed by a loop of wirechanges with time, a current is produced in the loop.The variation of the magnetic flux can result from atime-varying magnetic field, a coil in motion, or both.

t

= BE E = the curl of the electric fieldB = magnetic field vector [T]

Another way of expressing Faraday's law is that achanging magnetic field induces an electric field.

ind C Sd

V d ddt= =

E l B s

where S is the surface

enclosed by contourC.

(see also Induced Voltage below)


13/13


Vind INDUCED VOLTAGE

The voltage induced in a coil due to a changingmagnetic field is equal to the number of turns in thecoil times the rate at which the magnetic field ischanging (could be a change in field strength or coilarea normal to the field).

dt

dNV

ind

=

= Cind dV lE


= the magnetic flux produced by

a single turn of the coil [Wb]

Vind INDUCED VOLTAGE DUE TOMOTION

When conductors move in the presence of magneticfields, an induced voltage is produced even if themagnetic fields do not vary in time. For the voltageproduced due to both a changing magnetic field and aconductor in motion:

( ) indS C

V d d

t

= +

Bs v B l

B = magnetic field vector [T]v = velocity vector of the conductor [m/s]ds = increment of the surface normal to the magnetic field

vector [m2]dl= incremental length of conductor [m]

INDUCED VOLTAGE SLIDER PROBLEM

A frictionless conducting bar moves to the right atvelocity v produces a currentI.

R

I

0B v

d

h

An expanding magnetic field area having a staticmagnetic field directed into the page produces aCCW current.

0indV B hv=

0mag B Ih=F x 2 dE I R

v=

Vind = induced voltage [V]B0 = static magnetic field [T]h = distance between the conductor rails

[T]v = velocity of the conductor [m/s]Fmag = magnetic force opposing slider

[N]

x = unit vector in the direction againstconductor movement [m/s]

I= current [A]E= energy produced [J or W/s]R = circuit resistance []d= distance the conductor moves

[m]

M MAGNETIZATION [A/m]The induced magnetic dipole moment per unitvolume.

e

e

m

aNq

4

22 BM =

or

0

=B

M m

where

e

e

mm

aNq

40

22 =


qe = electron charge -1.60210

-19C

a = orbit radius of an electron [m]B = magnetic field vector [T]

0= permeability constant 410-7

[H/m]

me = who knows?

m = magnetic susceptibility

tesla - electromagnetic engineering

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