1 chapter 8 the steady state magnetic field the concept of field (physical basis ?) why do forces...

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Chapter 8

The Steady State Magnetic Field

The Concept of Field (Physical Basis ?)

Why do Forces Must Exist?

Magnetic Field – Requires Current Distribution

Effect on other Currents – next chapter

Free-space Conditions

Magnetic Field - Relation to its source – more complicated

Accept Laws on “faith alone” – later proof (difficult)

Do we need faith also after the proof?

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Magnetic Field Sources

Magnetic fields are produced by electric currents, which can be macroscopic currents in wires, or microscopic currents associated with electrons in atomic orbits.

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From GSU Webpage

Magnetic Field – Concepts, Interactions and Applications

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Biot-Savart Law

dHIdL aR

4 R2

IdL R

4 R3

Magnetic Field Intensity A/m

H LI

4 R2

d a R Verified experimentally

At any point P the magnitude of the magnetic field intensity produced by a differential element is proportional to the product of the current, the magnitude of the differential length, and the sine of the angle lying between the filament and a line connecting the filament to the point P at which the field is desired; also, the magnitude of the field is inversely proportional to the square of the distance from the filament to the point P. The constant of proportionality is 1/4

Biot-Savart = Ampere’s law for the current element.

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I NK

d

The total current I within a transverse Width b, in which there is a uniform surface current density K, is Kb.

For a non-uniform surface current density, integration is necessary.

Alternate Forms H SK_x

4 R2

d aR H vJ_x

4 R2

d aR

Biot-Savart Law B-S Law expressed in terms of distributed sources

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H2

z1I

4 2

z12

3

2

d az a z1 az I

4

z1

2

z12

3

2

d a

H2I

2 a

The magnitude of the field is not a function of phi or z and it varies inversely proportional as the distance from the filament.The direction is of the magnetic field intensity vector is circumferential.

Biot-Savart Law

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Biot-Savart Law

HI

4 sin 2 sin 1 a

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Example 8.1

H2x8

4 0.3( )sin 53.1

180

1

a

H2x8

4 0.3( )sin 53.1

180

1

H2x 3.819

a must be refered to the x axis - which becomes az

H2y8

4 0.4( )1 sin 36.9

180

az

H2y8

4 0.4( )1 sin 36.9

180

H2y 2.547

H2 H2x H2y H2 6.366 az

1x 90

180 2x atan

0.4

0.3

1y atan0.3

0.4

2y 90

180

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Ampere’s Circuital Law

The magnetic field in space around an electric current is proportional to the electric current which serves as its source, just as the electric field in space is proportional to the charge which serves as its source.

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Ampere’s Circuital Law

Ampere’s Circuital Law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by the path.

We define positive current as flowing in the direction of the advance of a right-handed screw turned in the direction in which the closed path is traversed.

LH_dot_

d I

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Ampere’s Circuital Law - Example

LH_dot_

d

0

2 H

d H 0

2 1

d I

HI

2

12

HI

2 a b

a H 0 cH

I

2 a2

2 H I I

2b

2

c2

b2

HI

2

2

b2

c2

b2

H I

2

c2

2

c2

b2

Ampere’s Circuital Law - Example

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Ampere’s Circuital Law - Example

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Ampere’s Circuital Law - Example

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Ampere’s Circuital Law - Example

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CURL

The curl of a vector function is the vector product of the del operator with a vector function

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CURL

curl_H( )aN0SN

LH_dot_

d

SNlim

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CURL

Ampere’s Circuital LawSecond Equation of Maxwell

Third Equation

curl_H = x H

x H = J

x E = 0

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Illustration of Curl Calculation

CurlHy

Hzd

d zHy

d

d

axzHx

d

d xHz

d

d

ayxHy

d

d yHx

d

d

az

CurlH

ax

x

d

d

Hx

ay

y

d

d

Hy

az

z

d

d

Hz

CURL

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CURL

az

DelXHz 75DelXHzxH1y x y z( )d

d yH1x x y z( )d

d

ay

DelXHy 98DelXHyzH1x x y z( )d

d xH1z x y z( )d

d

ax

DelXHx 420DelXHxy

H1z x y z( )d

d zH1y x y z( )d

d

z 3y 2x 5Determine J at:

H1z x y z( ) 4 x2 y

2H1y x y z( ) y2 x zH1x x y x( ) y x x

2y

2

In a certain conducting region, H is defined by:

Example 1

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CURL

Example 2

H2x x y x( ) 0 H2y x y z( ) x2

z H2z x y z( ) y2 x

x 2 y 3 z 4

DelXHxy

H2z x y z( )d

d zH2y x y z( )d

d DelXHx 16 ax

DelXHyzH2x x y z( )d

d xH2z x y z( )d

d DelXHy 9 ay

DelXHzxH2y x y z( )d

d yH2x x y z( )d

d DelXHz 16 az

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Example 8.2

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Stokes’ Theorem

LH_dot_

d SDel H( )_dot_

d

S

The sum of the closed line integrals about the perimeter of every Delta S is the same as the closed line integral about the perimeter of S because of cancellation on every path.

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Example 8.3Hr r 6 r sin

H r 0

H r 18 r sin cos

segment 1

r 4 0 0.1 0

segment 2

r 4 0.1 0 0.3

segment 3

r 4 0 0.1 0.3

dL dr ar r d a

r sin d a

First tem = 0 on all segments (dr = 0)

Second term = 0 on segment 2 ( constant)

Third term = 0 on segments 1 and 3 ( = 0 or constant)

LH

d H r

d H r sin

d H r

d

since H=0

0

0.3 H r r sin

d 22.249

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Magnetic Flux and Magnetic Flux Density

B 0 H 0 4 107

H

mpermeability in free space

SB_dot_

d

SB_dot_

d 0

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The Scalar and Vector Magnetic Potentials

H Del_Vm J 0

Vma

b

LH_dot_

d

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The Scalar and Vector Magnetic Potentials

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Derivation of the Steady-Magnetic-Fields Laws