elen 3371 electromagnetics fall 2008 1 lecture 3: static fields instructor: dr. gleb v. tcheslavski...

ELEN 3371 Electromagnetics Fall 2008

1

Lecture 3: Static Fields

Instructor: Dr. Gleb V. Tcheslavski

Contact:

[email protected]

Office Hours:

Room 2030

Class web site: www.ee.lamar.edu/gleb/em/Index.htm


2

1. Electrostatic Fields

1.1. Coulomb’s Law

Something known from the ancient time (here comes amber): two charged particles exert a force on each other…

1 22

0

[ ]4 R

QQF u N

R (3.2.1)

where Q1 and Q2 are charges,R –distance between particles,uR – the unit-vector

12 90

18.854 10 10 /

36F m the permittivity of free space

In this notation, negative force means attraction, positive – repelling.

Electrostatic (Coulomb’s) force:


3


1.1. Coulomb’s Law (Example)

Find the magnitude of the Coulomb force that exists between an electron and a proton in a hydrogen atom. Compare the Coulomb force and the gravitational force between the two particles. The two particles are separated approximately by 1 Ångström 1Å 10-10 m.

219

81 22 29 100

1.602 102.3 10

14 4 10 1036

C

QQF N

R

31 31

11 4722 10

9.11 10 1836 9.11 106.67 10 1.02 10

10

e pG

m mF G N

R

39: 2.27 10C

G

FRatio times

F

(3.3.1)

(3.3.2)

This is why chemical bounds are so strong!


4


1.2. Electric (electrostatic) Field

Electrostatic field due to the charge Q:

204

N V

C mR

F QE u

q R

For a system of two charges:

(3.4.1)

An “alternative definition”:

0lim q q q

q

F FdFE

dq q

What’s wrong with it?

(3.4.2)


5


1.3. Superposition

For several charges placed at different locations in space, the total electric field at the particular location would be a superposition (vector summation) of individual electric fields:

1

N

tot nn

E E

a vector sum!

(3.5.1)

(Example): find the EF at P

1 1 1

1 2 3

, , , ,

31 22 2 2

0 1 0 2 0 3

3 2 20 0 0

4 4 4

1(3 4 ) 2 3

4 5 4 4 4 3

tot P Q P Q P Q P

R R R

x y y x

E E E E

QQ Qu u u

R R R

u u u u

Q1 = +1C, Q1 = +2C, Q3 = -3C


6


1.3. Superposition (cont)

3[ / ]v

QC m

v

Volume charge density:

2[ / ]s

QC m

s

Surface charge density:

[ / ]l

QC m

l

Linear charge density:

(3.6.1)

(3.6.2)

(3.6.3)

20

10, '

4v

i R

v

if v number of volumes u dvR

(3.6.4)

There is a differential electric field directed radially from each differential charges


7


1.3. Superposition (Example)

Calculate the electric field from a finite charge uniformly distributed along a finite line.

Linear charge density: (z’)l

2 2

':

'

zR

z u uThe unit vector u

z

We assume a symmetry along z with respect to the observation point. Therefore, it will be a charge element at –z’ for every charge element at +z’. As a result, fields in z direction will cancel each other:

0iz

i

due toE symmetry (3.7.1)


8


1.3. Superposition (Example, cont)

The radial component:2 2

cos'

dE dE dE dER z

(3.8.1)

Combining (3.4.1) and (3.6.3), we arrive to:

2 20

'

4 ( ' )ldzdE

z

(3.8.2)

which, combined with (3.8.1) and integrated leads to:

3 2 22 20 0

1'

4 2'

al l

a

aE dz

az

(3.8.3)

0

22

la E

(3.8.4)


9


1.3. Superposition (Example 2)

Calculate the electric field from an infinite plane charged with s and consisting of an infinite number of parallel charged lines.

Utilize (3.8.4) and that . The linear charge density:2 2R x y l sdx

Symmetry leads to cancellation of tangent components.

12 22 2 2 2

0 00 0

cos tan2 2 22 '

s s sy

sy y xE dE dx dx dx

x y yx y x y

(3.9.1)

(3.9.2)


10


1.4. Gauss’s Law

A charge Q is uniformly distributed within a sphere of radius a.

We can assume first that the charge is located at the center. Than, by (3.4.1):

204 r

QE u

a (3.10.1)

By evaluating surface integrals of both sides 204 r

QE ds u ds

a (3.10.2)

At the surface of the sphere, the unit-vector associated with the differential surface area ds points in the radial direction. Therefore, and the closed surface integral is

1r ru u 24 a


11


0

enclQE ds

(3.11.1)

1.4. Gauss’s Law (cont)

Here Qencl is the charge enclosed within the closed surface.

By using divergence theorem and volume charge density concept:

0

v

v

s v

dv

E ds Edv

0

vE

Therefore, the integral form:

Differential form:

(3.11.2)

(3.11.3)


12


1.4. Gauss’s Law (cont 2)

For the charge Q uniformly distributed within the spherical volume 34 / 3v a

The volume charge density: 343

v

Q Q

v a

(3.12.1)

0 0

v

v encl

dvQ

The total charge enclosed: (3.12.2)

a) Outside the sphere: r > a, Qencl = Q2

0 0

4v

vr

s

dvQ

E ds r E

204r

QE

r

(3.12.3)

(3.12.3)

Gauss Law


13


a) Inside the sphere: r < a

1.4. Gauss’s Law (cont 3)

33

30 0 0 0

1 4

4 33

v

enc v

dvQ Q r Q r

a a

32

0 0

4v

vr

s

dvQ r

E ds r Ea

Gauss Law

304r

QrE

a

(3.13.1)

(3.13.2)

(3.13.3)


14


1.5. Gaussian Surface

A Gaussian surface is a closed two-dimensional surface through which a flux or electric field is calculated. The surface is used in conjunction with Gauss's law (a consequence of the divergence theorem), allowing to calculate the total enclosed electric charge by exploiting a symmetry while performing a surface integral.

Commonly used are:

a)Spherical surface for•A point charge;•A uniformly distributed spherical shell of charge;•Other charge distribution with a spherical symmetryb) Cylindrical surface for•A long, straight wire with a uniformly distributed charge;•Any long, straight cylinder or cylindrical shell with uniform charge distribution.


15


1.6. Potential Energy and Electric Potential

A charged particle will gain a certain amount of potential energy as the particle is moved against an electric field.

[ ]b b

e

a a

W F dl Q E dl J (3.15.1)


16


1.6. Potential Energy and Electric Potential (cont)

Imaginary experiment: compute a total work required to bring three charged particles from - to the shaded region. No electric field exists at - and there are no friction, no gravity, and no other forces.

I

There are no forces here, therefore, no work is required! W1 = 0;


17


II

1.6. Potential Energy and Electric Potential (cont 2)

We need to overcome the Coulomb’s force, therefore, some work is required.

1 2 1 22 2 12

0 04 ( ) 4

bx

a a

QQ QQW dx Q V

x x x x

since both charges are positive

1 11 2

0 04 ( ) 4

bx

a a

Q QV dx

x x x x

V1 is an absolute electric potential caused by the charge Q1

xaxb

(3.17.1)

(3.17.2)


18



We need again to overcome the Coulomb’s force, therefore, some work is required.

1 3 2 3 1 3 2 33 3 1 3 22 2

0 0 0 04 ( ) 4 ( ) 4 4

c cx x

a b c a c b

QQ Q Q QQ Q QW dx dx QV QV

x x x x x x x x

III

Totally, for the three particles: 1 2 3 2 12 3 13 230 ( )totW W W W Q V Q V V

Or, for N particles:1 1, 10

1 1

2 24

N N Ni j

tot i ii j j i iij

QQW QV

x

1, 04

Ni

ij j i ij

QV

x

Here xij is the distance between charges i and j;

(3.18.1)

(3.18.2)

(3.18.3)

(3.18.4)


19



Note: the total work in our case is equal to the total electrostatic energy stored in the shaded region.

Note: the total work (and the total energy) do not depend on the order, in which particles are brought.

The electrostatic energy can also be evaluated as

1[ ]

2e v

v

W Vdv J

(3.19.1)


20



We can express the electric potential difference or voltage as:

1 1a b a b

a b abV V V F dl F dl Q E dl Q E dlQ Q

[ ]

b

ab

a

J

CVV E dl or

(3.20.1)

(3.20.2)

Electric potential difference between points a and b is the work required to move the charge from point a to point b divided by that charge.


21


1.6. Potential Energy and Electric Potential (Example)

1’2’

3’4’

5’

6’

Evaluate the work (charge times potential difference) required to move a charge q from a radius b to a radius a.

The electric field is 204 r

QE u

r

The potential difference between the two spherical surfaces is

20 0 0

1 1

4 4 4

bb

ab r r

a a

Q Q QV u u dr

r r a b

The potential at r = is assumed to be 0 and is called a ground potential.The electric potential defined with respect to the ground potential is called an absolute potential.

(3.21.1)

(3.21.2)


22


1.6. Potential Energy and Electric Potential (Example, cont)

Considering the path 1-2-3-4, we notice that there are only potential differences while going 1 2 and 3 4. Therefore, these are the only paths where some work is required. When moving 2 3, the potential is constant, therefore no work is required.

A surface that has the same potential is called an equipotential surface.

If the separation between two equipotential surfaces and the voltage between them are small:

x y z

V V VdV E dl E dx E dy E dz dx dy dz

x y z

: x y z

V

m

V V VE u u u

x yThe electric fi

zeld

(3.22.1)

(3.22.2)


23



We can modify (3.22.2) as following:

E V

2

00

2 vv V

mS Vince E V

Poisson’s equationLaplace’s eqn. when v = 0

An absolute potential caused by a volume distribution that is not at the origin:

2 2 20

( ', ', ')1( , , ) ' ' '

4 ' ' '

v

v

x y zV x y z dx dy dz

x x y y z z

(3.23.1)

(3.23.2)

(3.23.3)


24



The potential energy would be

0 00

1 1( )

2 2 2 2e v

v v v v

W Vdv E Vdv VE E V dv VE ds

0

2 v

E Vdv

for R

200

2 2v

e

v

W E dvE Edv

(3.24.1)

(3.24.2)

Note that when a charged particle is moved along a closed contour, no work is required

0 0e

s

E ds EW

E dsQ

Electrostatic field is conservative and irrotational.

(3.24.3)


25


1.6. Potential Energy and Electric Potential (Example 2)

Find the potential V due to two equal charges that have opposite signs and are in the vacuum. The distance from the point of interest is much greater than the separation. The configuration is known as an electric dipole.


26


1.6. Potential Energy and Electric Potential (Example 2, cont)

Due to superposition:

0 1 0 24 4

Q QV

r r

Since r >> d; r, r1, and r2 are almost parallel.

1 2cos cos2 2

d dr r and r r

0

20

0

0

4 cos 4 cos2 2

1 cos 1 cos4

c s2

o2 4

Q QV

d dr r

Q d d

r

Qd

r r r

A vector p = Qd is a dipole moment.

(3.26.1)

(3.26.2)

(3.26.3)


27


0.002 0.004 0.006 0.008 0.01

30

210

60

240

90 270

120

300

150

330

180

0

-4 -3 -2 -1 0 1 2 3 4-0.01

-0.008

-0.006

-0.004

-0.002

0

0.002

0.004

0.006

0.008

0.01

V

Angle, rad

1.6. Potential Energy and Electric Potential (Example 2, cont)

Electric potential distribution plot of (3.26.3)

plot polar


28


1.7. On numerical integration

trapz

quad

dblquad

triplequad

Self-study

When no symmetry can be used to simplify the problem, numerical integrations are quite helpful. Numerical integration = APPROXIMATION.


29


1.8. Dielectric materials

A material can be considered as a collection of randomly (in general) oriented small electric dipoles.

If an external electric field is applied, the dipoles may orient themselves.


30


1.8. Dielectric materials (cont)

We may suggest that an external electric field causes a “thin layer of charge” of the opposite sign at either edge of the material.This charge is called a polarization charge.

The density of the polarization charge:

p P

where P is the polarization field:0

1

1lim

N

jvj

P pv

Here pj = Qdud is the dipole moment of individual dipole, N – number of atoms (dipoles)

(3.30.1)

(3.30.2)


31


1.8. Dielectric materials (cont 2)

Let us add the polarization charge density to the real charge density. The Gauss’s Law will take a form:

0

v pE

which leads tovD

where20 ( )

C

melectric displacemenD E P t flux density

(3.31.1)

(3.31.2)

(3.31.3)

The total flux that passes through the surface [ ]e

s

D ds C

(3.31.4)


32


1.8. Dielectric materials (cont 3)

Integrating (3.31.2) over a volume, leads to

enc

s

D ds Q

(3.32.1)

Dielectric materials are susceptible to polarization. Usually, polarization is linearly proportional to the applied (small) electric field. Then 0 eP E (3.32.2)

where e is the electric susceptibility

0 0(1 )e rD E E E

r is the relative dielectric constant

(3.32.3)

for linear and isotropic materials

We consider only linear materials here.


33


1.9. Capacitance

[ ]Q

C FV

(3.33.1)

A parallel-plate capacitor

Area A w z (3.33.2)

Assume A >> d


34


1.9. Capacitance (cont)

0 0 0

22

enc s s

s

Q AE ds EA E between plates

b

ab

a

V E dl Ed

0

0

,s

s

A AQC or in case of dielectric

V dC

dd

A

222

0 0

2 2 2e

v

V CVW E dv Ad

d

Stored energy:

(3.34.1)

(3.34.2)

(3.34.3)

(3.34.4)

Assumed uniform field in the capacitor and uniform distribution of charge on plates


35


1.9. Capacitance (Example)

Calculate the mutual capacitance of a coax cable with dielectric r inside…

From the Gauss’s Law:

2enc lD ds Q D L L

ln2 2

b bl l

ab

a a

bV E dl d

a

Potential difference:

The total charge: lQ L

n

2

lnl2

l

lab

LQbVa

C Lba

(3.35.1)

(3.35.2)

(3.35.3)

(3.35.4)

HW 2 is ready


36

2. Magnetostatic Fields

2.1. Electric currents

Let’s consider a wire…

The generalized Ohm’s Law specifies a current density:

2

1 A

m

VI V LRA A L A R

J E

Alternative: v drift v mJ v E E

where is the electron volume charge density, is an average electron drift velocity, is the mobility of the material.

v driftvm

1/v m R the conductivity

(3.36.1)

(3.36.2)

(3.36.3)


37


2.1. Electric currents (cont)

The total current that passes through the wire

A

I J ds

2I J aIn our case f the current is distributed uniformly in a cylindrical wire:

The power density (density of power dissipating within a conductor):

2W mp J E

The total power absorbed within the volume:

[ ]v

P pdv W

(3.37.1)

(3.37.2)

(3.37.3)

(3.37.4)


38


2.1. Electric currents (Examples)

a) Calculate the current flowing through the wire of radius a; current density:

0 zJ I ua

“skin effect”2 2

0 0

0 0

23

a

z z

aI I u d d u I

a

b) Calculate the power dissipated within a resistor with a uniform conductivity . The voltage across the resistor is V, a current passing through is I.

2

20 0 0

L a

v z

I VJ E dv d d dz

LP VI

(3.38.1)

(3.38.2)

(3.38.3)


39


2.2. Fundamentals of Magnetic Fields

Magnetic loops at the surface of the Sun, as seen with the TRACE solar spacecraft. (©TRACE operation team, Lockheed Martin)

?Magnetic monopole?


40


2.2. Fundamentals of Magnetic Fields (cont)

Magnetic field lines are continuous, don’t originate nor terminate at a point. There is no “magnetic monopole”…

0B ds B is a magnetic flux density, [T] = [Wb/m2]

(3.40.1)

0B

By applying the divergence theorem to (3.40.1):v

B ds Bdv

(3.40.2)

(3.40.3)


41


2.2. Fundamentals of Magnetic Fields (cont 2)

As a result, we can split a bar magnet into tiny pieces and all of them will have both north and south poles.


42



Magnetic field can be created by the electric current:

0 encB dl I electric current enclosed within a closed loop

770: 14 10 2.566 10 H mPermiability of free space

A cylindrical wire caring a current creates a magnetic field.“Right Hand Rule” (RHR).

(3.42.1)Ampere’s Law:


43



By applying Stokes’s theorem to (3.42.1), we arrive to

0

s s

B dl B ds J ds

Therefore, the differentialform of Ampere’s Law is: 0B J

An axial view of the cortically-generated magnetic field of a human listener, measured using whole-head magnetoencephalography (MEG) – from the journal “Cerebral Cortex”

It appears that even very small currents generate magnetic fields…

(3.43.1)

(3.43.2)


44


2.2. Fundamentals of Magnetic Fields (Example)

A symmetry in the system greatly simplifies evaluation of integrals in (3.42.1).

2

0

2B dl B u d u B

The right-hand side of (3.42.1) for the radius greater than a is just 0I.

0 ,2

IB a

Assuming a uniformly distribution of current within the wire, the current density inside the wire is

2 z

IJ u

a

(3.44.1)

(3.44.2)

(3.44.3)


45


2.2. Fundamentals of Magnetic Fields (Example 2)

22

20 0

enc

s r

II J ds r dr d I

a a

The total current enclosed within the circle of radius a

Therefore: 02,

2

IB

aa

(3.45.1)

(3.45.2)

We notice that at the edge of the wire, two solutions given by (3.44.1) and (3.45.2) are equal.

Can you further explain the dependence of magnetic flux on the radius?


46


2.3. Magnetic Vector Potential

A magnetic vector potential A such that:

A B

0

0A

A J

Ampere’s Law:

20A J

In the Cartesian coordinates: 0 ( ')( )

4 v

J rA r dv

R

2 2 2( ') ( ') ( ')where R x x y y z z

(3.46.1)

(3.46.2)

(3.46.3)

(3.46.4)

(3.46.5)

(3.46.6)


47


2.3. Magnetic Vector Potential (cont)

Magnetic vector potential, magnetic flux, and current element


48


2.3. Magnetic Vector Potential (cont 2)

The Magnetic flux density:

0 0( ') ( ')( ) ( )

4 4v v

J r J rB r A r dv dv

R R

( ') 0Since a a B a B and J r

0 0 02 2

( ')1( ) ( ') ( ')

4 4 4R R

v v v

u J r uB r J r dv J r dv dv

R R R

If the current is passing through a wire

02

'( )

4RI dl u

B rR

the Biot-Savart Law

(3.48.1)

(3.48.2)

(3.48.3)

(3.48.4)


49


2.3. Magnetic Vector Potential (Example)

Find the magnetic field on the axis perpendicular to the loop of current. Use the Biot-Savart Law.

We identify the terms appearing in (3.48.4):

2 2' ' , ,R zdl a d u u a u z u R R a z

2

0 02 2 3 2 2 2 3 2

( ' ) ( ) ' '( )

4 ( ) 4 ( )z za d u a u z u a d u a z d uI I

B za z a z

Due to symmetry, the terms with the unit vector u are zero.2

0 02 2 3 2 3

( )2 ( ) 2z

I a mB z u

a z R

2 the magnetic dipole momentzm I a u

(3.49.1)

(3.49.2)

(3.49.3)


50


2.3. Magnetic Vector Potential (Cont 3)

We have learned the following analytical methods to find the magnetic flux density at a point in space from a current element:

1. Application of Ampere’s Law, which requires considerable

symmetry.

2. Determination of the vector magnetic potential and the calculation

of a magnetic flux density. No symmetry is required.

3. Application of the Biot-Savart law. No symmetry is required.


51


2.4. Magnetic forces

If a charged particle is moving with a constant velocity v in a region that ONLY contains a magnetic field with the density B, the force that acts upon the particle is

mF q v B

Direction of the force - RHR!

F+ stands for a positively charged particle;F- represents a negatively charged one.

(3.51.1)


52


2.4. Magnetic forces (cont)

When a charged particle is going through an area with both: uniform electric field and uniform magnetic field, the force exerted on it would be the Lorentz Force:

[ ]( ) NF q E v B (3.52.1)

Recall that the work done by a charged particle moving in a field isb

a

W F dl (3.52.2)

A differential charge dQ = vdv moving at a constant velocity creates a current. If this current flows in a closed loop:

( ) ( )m vdF dQ v B v B dv J B ds dl Idl B

mF B I dl The total magnetic force:

(3.52.3)

(3.52.4)


53


2.4. Magnetic forces (cont 2)

Example: a charged particle entered a constant magnetic field will move along a circular orbit. Find the radius…

Centripetal force: , m – particle’s mass2

c cj

vF ma m

Magnetic force:mF qvB

j

mv

qB

This radius is called the Larmor radius or gyro radius. This effect is used in mass spectroscopy.

(3.53.2)

(3.53.1)

(3.53.3)


54


2.4. Magnetic forces (Example)

Evaluate the force existing between two parallel wires caring currents.

B1 will go up at the location of wire 2.From (3.52.4) force on the wire 2:

2 1 2y zF B u I dl u to the left

(3.54.1)

2 1 2( )y zF B u I dl u to the right

(3.54.2)

a)

b)


55


2.4. Magnetic forces (Example, cont)… ”Alternative approach”

Let’s re-state the force on wire 1 caused by the magnetic field generated by the current in wire 2 (from 3.52.4)

1

12 1 12

L

F I B dl

From the Biot-Savart law:

21

2

20 212 2

214R

L

u dlIB

R

21

1 2

2 10 1 212 2

214

R

L L

u dl dlI IF

R

Finally:

(3.55.1)

(3.55.2)

(3.55.3)

This is what’s called as Ampere’s force.


56


2.4. Magnetic forces (Example 2)

Consider a current-caring loop in a constant magnetic field B = B0 uz.We assume the separation between the In/Out wires to be infinitely small.

Parallel wires carry the same current in the opposite direction. Therefore, the net force will be a vector sum of all forces, which is zero!

However, there will be a torque on the loop that will make it to rotate (say, about x for simplicity).


57


2.4. Magnetic forces (Example 2, cont)

The torque on the loop is given by

1 3sin sin2 2

y yT F F

Assumptions?

1 0 3 0,where F IB x F IB x

0 sinT IB x y

Finally:

T m B magnetic moment

(3.57.1)

(3.57.2)

(3.57.3)

(3.57.4)


58


2.5. Magnetic materials

Two sources of magnetism inside an atom:1) an electron rotating around a nucleus;2) an electron spinning about its own axis.

Types of material:

1. Diamagnetic: 1) and 2) cancel each other almost completely, magnetic susceptibility m -10-5.

2. Paramagnetic: 1) and 2) do not cancel each other completely, magnetic susceptibility m 10-5.

3. Ferromagnetic: domain structure, very high m (hundreds and higher)

magnetic dipoles oriented randomly

magnetic dipoles in each domain are oriented


59


2.5. Magnetic materials: Ferromagnetics

Total magnetization (magnetic dipole moment per unit volume):

External magnetic field may change dipole orientation “permanently” - HDD.

0

1

1lim

N

jv

j

A mM mv

(3.59.1)

There is a current created inside domains (magnetization current):

m m

s s

I M dl J ds M ds

, mTherefore J M

We may modify the Ampere’s Law by adding the magnetization current:

0 0

1m

BB J J J M J M

(3.59.2)

(3.59.3)

(3.59.4)


60


(3.60.1)

2.5. Magnetic materials: Ferromagnetics (cont)

We introduce a new quantity, the Magnetic Field Intensity:

0

A mB

H M

Therefore, the Ampere’s circular law is

encH dl I (3.60.2)

: mMagnetizatio M Hn

Therefore: 0 0(1 )m rB H H H

(3.60.3)

(3.60.4)

where r is the relative permeability.


61


2.5. Magnetic materials: Ferromagnetics (cont 2)

Example: a magnetic flux density B = 0.05 T appears in a material with r = 50.Find the magnetic susceptibility and the magnetic field intensity.

1 50 1 49m r 7

0

0.05796 [ ]

50 4 10r

BH A m

Hysteresis

Magnetic flux density B exhibits nonlinear dependence on the magnetic field intensity H.

Because of hysteresis, magnetic materials “remember” the magnitude and direction of magnetic flux density. They can be used as memory elements.

saturation

saturation


62


2.6. Magnetic circuits

Just like electrical circuits, we can build magnetic circuits where magnetic flux “flows”.

L – mean length of the iron region

g – length of the gap.

m – the reluctance.

Assumptions:1) the gap is very small;2) the cross-sectional area of the gap is identical to the cross-sectional area of the magnetic material.


63


2.6. Magnetic circuits (cont)

The Ampere’s circular law leads to ( ) encH L g I NI

where I is the current flowing through the N turns of a wire.

0 0 0 0m

r r

B B L gL g NI

A A

Here A is the cross-sectional area of the iron

mF NI is the magnetomotive force [A-turns]

, ,0 0

m iron m gapr

L g

A A is the reluctance

(3.63.1)

(3.63.2)

(3.63.3)

(3.63.4)

The Hopkinson’s Law (aka Ohm’s):m m mF (3.63.5)


64


2.6. Magnetic circuits (Examples)


65


2.7. Inductance (an ability to create magnetic flux)

When j = k – self –inductance; otherwise – mutual inductance.

where is a magnetic flux linkage.

jjk

k

H Wb ALI

Ex. 1: Let us consider a solenoid of the length d, cross-section area A, and having N turns.It may also have a core made from a magnetic material.z is the solenoid’s axis.

(3.65.1)


66


2.7. Inductance (cont)

The magnetic flux density at the center of the solenoid is:

z

NIB

d

The total magnetic flux: m

NIA

d

The magnetic flux linkage:2

m

N IN A

d

Therefore, the self-inductance of a solenoid is

2NL A

d

(3.66.1)

(3.66.2)

(3.66.3)

(3.66.4)


67


2.7. Inductance (cont 2)

Example 2: a self-inductance of a coaxial cable

Here, the magnetic flux linkage equals to the total magnetic flux.

What is the main difference as compared to a solenoid?

2

IB

r

mfd:

0

ln2 2

z b

m

z r a

I I bdr dz z

r a

ln2

bL z

a

Therefore:

(3.67.1)

(3.67.2)

(3.67.3)


68



Example 3: a mutual inductance between two circular solenoids, whose individual lengths are d and areas S1 and S2, separated by x; x << d

x

First coil:

0 1 11

N IB

d

,1 1 1m B S

Assuming that the magnetic flux has the same value in the second solenoid:

2 2 1 1N B S

Therefore, the mutual inductance:0 1 2 12

1

N N SM

I d

(3.68.1)

(3.68.2)

(3.68.3)

(3.68.4)


69



Example 4: consider a transformer with N1 and N2 turns.

Alternative formulas: 12 1 2 12M N N P

1 2M k L L

correction coefficient

permeance of the space occupied by the flux

coefficient of coupling

(3.69.1)

(3.69.2)


70



Solenoids, transformers, etc. can store magnetic energy:2

0 0 0 0

' ' '' 2

t t t t

m

dI LIW p dt IV dt I L dt L I dI

dt

221

2z

m

B dNW A

d N

for a solenoid:

The total magnetic energy stored within a volume

2

2m

v

JB

W dv

(3.70.1)

(3.70.2)

(3.70.3)

Your homework 3 is available through “My Lamar”

elen 3371 electromagnetics fall 2008 1 lecture 3: static fields instructor: dr. gleb v. tcheslavski...

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