The Steady Magnetic Field Part 2 Shoubra/Electrical... · Gauss’Law for the electric field Conservative property of the static electric field Ampere’s Circuital Law Gauss’Law

The Steady Magnetic Field

Part 2

Prepared By

Dr. Eng. Sherif Hekal

Assistant Professor – Electronics and Communications Engineering

12/12/2017 1

Agenda

• Magnetic Flux And Magnetic Flux Density

• Maxwell’s Equations for Static Fields

• Gauss’s law for magnetic field

• Magnetic Boundary Conditions

• Magnetic Forces and Inductance

• Magnetic energy density

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Magnetic Flux and Flux Density

We are already familiar with the concept of electric flux:

Coulombs

in which the electric flux density in free space is:

In a similar way, we can define the magnetic flux in units of Webers (Wb):

Webers

in which the magnetic flux density (or magnetic induction) in free space is:

and where the free space permittivity is

and where the free space permeability is

This is a defined quantity, having to do with the definition of the ampere (we will explore this later).

(1)

(2)

(3)

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Magnetic Flux and Flux Density

If the flux is evaluated through a closed surface, we have in the case of electric flux, Gauss’ Law:

If the same were to be done with magnetic flux density, we would find:

The implication is that (for our purposes) there are no magnetic charges -- specifically, no point

sources of magnetic field exist. A hint of this has already been observed, in that magnetic field lines

always close on themselves.

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Example: Magnetic Flux Within a Coaxial Line

dB

Consider a length d of coax, as shown here. The magnetic field strength between conductors is:

and so:

The magnetic flux is now the integral of B over the

flat surface between radii a and b, and of length d along z:

The result is:

The coax line thus “stores” this amount of magnetic flux in the region between conductors. This will have importance when we discuss inductance in a later lecture. 12/12/2017 5

Maxwell’s Equations for Static Fields

We may rewrite the closed surface integral of B using the divergence theorem, in which the

right hand integral is taken over the volume surrounded by the closed surface:

Because there is no isolated magnetic charge, the result is zero

This result is known as Gauss’ Law for the magnetic field in point form.

(4)

12/12/2017 6


It has just been demonstrated from Ampere’s Circuital Law that:

…..which is in fact one of Maxwell’s equations for static fields:

This is Ampere’s Circuital Law in point form.

dsJIdH enclosed

Applying the Stoke’s theory

dsJ

dsHdH

)(

(5)

12/12/2017 7


We already know that for a static electric field:

This means that:

Recall the condition for a conservative field: that is, its closed path integral is zero everywhere.

Therefore, a field is conservative if it has zero curl at all points over which the field is defined.

(applies to a static electric field)

(6)

12/12/2017 8


We have now completed the derivation of Maxwell’s equations for no time variation. In point form, these are:

Gauss’ Law for the electric field

Conservative property of the static electric field

Ampere’s Circuital Law

Gauss’ Law for the Magnetic Field

where, in free space:Significant changes in the above four equations will

occur when the fields are allowed to vary with time, as

we’ll see later.12/12/2017 9

Maxwell’s Equations in Large Scale Form

The divergence theorem and Stokes’ theorem can be applied to the previous four point formequations to yield the integral form of Maxwell’s equations for static fields:

Gauss’ Law for the electric field

Conservative property of the static electric field

Ampere’s Circuital Law

Gauss’ Law for the magnetic field

12/12/2017 10

MHB and ,To find the relationship between

s

HB /

21n̂

l

2/h

sJ

2/h

Boundary2/h

2/h

ab

c

d

1

2Region 2:

Region 1:

HB and To find normal component of at the boundary

Consider a small cylinder as 0h 0__

dsBand use

0

21

21

__

nn

nn

BB

sBsBdsB

2211 nn HH

Magnetic Boundary Conditions

(7)12/12/2017 11

s

HB /

21n̂

l

2/h

sJ

2/h

Boundary2/h

2/h

ab

c

d

11, m

22 , mRegion 2:

Region 1:

HB and To find tangential component of

Consider a closed abcd as 0h

and use encl

IdlH __

at the boundary

KHH

IlHlH

tt

enctt

21

21

K tH1 tH 2where is perpendicular to the directions of and

21ˆ

na=

x xy

z

yxz ˆˆˆ

Magnetic Boundary Conditions

2

2

1

121 or

tt

tt

BBHH If K = 0 : (8)

12/12/2017 12

Motivating the Magnetic Field Concept:

Forces Between Currents

How can we describe a force field around wire 1 that can be used to determine the force on wire 2?

Magnetic forces arise whenever we have charges in motion. Forces between current-carrying wires

present familiar examples that we can use to determine what a magnetic force field should look like:

Here are the easily-observed facts:

12/12/2017 14

Magnetic FieldThe geometry of the magnetic field is set up to correctly model forces between currents. The

magnetic field intensity, H, circulates around its source, I1, in a direction most easily determined by

the right-hand rule: Right thumb in the direction of the current, fingers curl in the direction of H

Note that in the third case (perpendicular currents), I2 is in the same direction as H, so that their cross

product (and the resulting force) is zero. The actual force computation involves a different field

quantity, B, which is related to H through B = H in free space. This will be taken up in a later

lecture. Our immediate concern is how to find H from any given current distribution.12/12/2017 15

EQFe

BQFm v

me FFF BEQF vorAlso known as Lorentz force equation.

Force on a Moving Point Charge

The force on a moving particle arising from combined electric and magnetic fields is obtained easily

by superposition,

In an electric field, the force on a charged particle is given by

A charged particle in motion in a magnetic field of flux density B is found experimentally to

experience a force whose magnitude is

Where v is the velocity, B is the flux density

(9)

(10)

(11)

12/12/2017 16

EQ

BQ v

Charge

Condition Field

Stationary -

Moving

E

EQ

B Field

BEQ v

Combination

E Band

EQ

Force on charge in the influence of fields:


• The electric force is usually in the direction of the electric field while,

the magnetic force is perpendicular to the magnetic field

• The electric force acts on a charged particle whether or not it is

moving, while the magnetic force acts moving charged particle only

• The electric force expends energy in displacing a charged particle,

while the magnetic one does no work when the particle is displaced

because it is perpendicular to the velocity

v

v

v

12/12/2017 17


4.

12/12/2017 18

Force on a Differential Current Element

The force on a charged particle moving through a steady magnetic field may be written as the

differential force exerted on a differential element of charge,

BdQFd v

dvdQ v and vvJ

dvBJFd

We saw in Chapter 8, part1 that J dν may be interpreted as a differential current element; that is,

and thus the Lorentz force equation may be applied to a differential current filament,

BIdFd 12/12/2017 19


dBI

BIdF

-

One simple result is obtained by applying (12) to a straight conductor in a uniform magnetic field,

(12)

The magnitude of the force is given by the familiar equation

BLIF

sinBILF

where θ is the angle between the vectors representing the direction of the current flow and the

direction of the magnetic flux density.

(13)

12/12/2017 20


Ex. 8.3: A square conductor current loop

is located in z = 0 plane with the edges

given by the coordinates (1,0,0), (1,2,0),

(3,0,0) and (3,2,0) carrying a current of 2

mA in anti clockwise direction. A

filamentary current carrying conductor of

infinite length along the y axis carrying a

current of 15 A in the –y direction. Find

the force on the square loop.

I1

I2

12/12/2017 21

Solution:

(1,0,0)

(1,2,0)

(3,0,0)

(3,2,0)

x

y

z

2 mA

15 A

T ˆ103

104 ∴A/m ˆ

2

15ˆ

26

1

7-

101

11

zx

HHB

zx

zx

IH

-

Field created in the square loop due to

filamentary current :


I1I2

zxy

aa Rl

ˆˆˆ

ˆˆˆ

12/12/2017 22

Hence: __

121

__

2 dlBIBdlIF

1

3

0

2

3

1

2

0

63

ˆ1

ˆˆ

ˆ

ˆ3

ˆˆ

ˆ103102

x y

x y

ydyz

xdxx

z

ydyz

xdxx

zF

nN ˆ8

ˆ2ˆ3

1lnˆ

3

2ˆ3ln106

ˆˆlnˆ3

1ˆln106

9

0

2

1

3

2

0

3

1

9

x

xyxy

xyyxxyyxF

(1,0,0)

(1,2,0)

(3,0,0)

(3,2,0)

x

y

z

2 mA15 AI1 I2


𝑎𝑥

𝑎𝑦𝑎𝑧

12/12/2017 23


5.

12/12/2017 24

P1(x1,y1,z1)

P2(x2,y2,z2)

__

11 dlI

2

__

2 dlI

I1 I2

12ˆ

RaR12

x

y

z

Force Between Differential Current Elements

12/12/2017 25

The magnetic field at point P2 due to the filamentary current I1dl1 :

2

12

11

24

ˆ x 12

R

aldIHd

R

(A/m)

BldIFd x (N)

2

12

11

2224

ˆ x x 12

R

aldIldIFdd

Ro

222212

11

222 x4

ˆ x x

1

12 BldIR

aldIldIFd

l

Ro

We have :

where is the force due to I2dl2 and due to the magnetic field of wire l2Fd


12/12/2017 26

Integrate:

2 1

12

2

12

11

2224

ˆ x x

l l

Ro

R

aldIdlIF

2 1

12

22

12

1212 x

xˆ

4l l

Ro ldR

ldaIIF

2s

22 dsBJF s

dvBJFv

222

For surface current :

For volume current :


12/12/2017 27


6.

12/12/2017 28

Magnetic Force between Two current Elements

By inspection of the figure we see that ρ = y and a = -ax. Inserting this in the

above equation and considering that dL2 = dzaz, we have

Now let us consider a second line of current parallel to the first.

The force dF12 from the magnetic field of line 1 acting on a differential section of line 2 is

12 2 2 1d I d F L B

The magnetic flux density B1 for an infinite length line of current is recalled from equation

to be

1

12

oI

B a

a = -ax

ρ = y

1 1

12 2 2 1 2 22 2

o o

z z x

I II d I dz I dz

F a a a aL B -

1 2122

o

y

I I

ydz

F a

12/12/2017 29

To find the total force on a length L of line 2 from the field of line 1, we must integrate dF12 from +L to

0. We are integrating in this direction to account for the direction of the current.

0

1 2

12

1 2

2

2

o

L

o

I Idz

y

I I L

y

y

y

F a

a

This gives us a repulsive force.

Had we instead been seeking F21, the magnetic force acting on line 1 from the

field of line 2, we would have found F21 = -F12.

Conclusion:

1) Two parallel lines with current in opposite directions experience a force of

repulsion.

2) For a pair of parallel lines with current in the same direction, a force of

attraction would result.

a = -ax

ρ = y


12/12/2017 30

In the more general case where the two lines are not parallel, or not straight, we could use the Law

of Biot-Savart to find B1 and arrive at

This equation is known as Ampere’s Law of Force between a pair of current carrying circuits and is

analogous to Coulomb’s law of force between a pair of charges.

2

12

121212

012

)(

4 R

adLdLIIF


12/12/2017 31

Example: Find force per meter between two parallel infinite conductor carrying

current, I Ampere in opposite direction and separated at a distance d meter.

Solution:

2B at position conductor 2

d

Ix

r

IHB

c

2

ˆ

2

ˆ101

0202

(N/m) 2

ˆ2

ˆ

2

ˆ)ˆ(

2

ˆ

2

0210

1

0

102

102

1

0

22

d

Iy

d

IIy

d

IxdzzI

d

IxdIF

Hence:d

z

2B

21FI1

I2

I1 = I2 = I

x

y


12/12/2017 32

+

_

VL

Coil

I

Magnetic

flux

From circuit theory the induced potential across a wire wound coil such as solenoid or a toroid :

dt

dILVL

where L is the inductance of the coil, I is the time varying current flowing through the coil – inductor.

Self Inductance

Inductance is the last of the three familiar parameters from

circuit theory that we are defining in more general terms.

We can define the inductance (or self-inductance) as

the ratio of the total flux linkages to the current which

they link,

I

N

IL

where (lambda) is the total flux linkage of the inductor

Henry

12/12/2017 33

2

2

1CVWE

In a capacitor, the energy is stored in the electric field :

In an inductor, the energy is stored in the magnetic field,

as suggested in the diagram :

+

_

VL

Coil

I

Magnetic

flux

switch

Idtdt

dILIdtVW

tt

t

tt

t

Lm 00

00

)( 2

1 2

0

0

JouleLIdILI

tt

t

Self Inductance

12/12/2017 34

Example: Obtain the expression for self inductance per meter of the coaxial cable when the

current flow is restricted to the surface of the inner conductor and the inner surface of the outer

conductor as shown in the diagram.

Solution:

a

b

The Φ will exist only between a and b and

will link all the current I

a

b

I

dzdr

r

I

I

dzdrH

IIL

b

a

c

c

b

a

c

ln2

2

1

0

1

0

Self Inductance

12/12/2017 35

Example: Obtain the self inductance of the long solenoid shown in the diagram.

Solution: Assume all the flux links all N turns and that does not vary over the cross section area of the solenoid.

B

N turns

22

22

2

have We

al

IN

Nal

NINaH

HB

NaBN

l

aN

IL

22

Self Inductance

12/12/2017 36

I

SNb

NI

I

Nac

B

I

N

IL

2

4

2

Example: Obtain the self inductance of the toroid shown in the diagram.

a

cb

0

avel Bads ˆ __

Mean path

Cross sectional area

S

Feromagnetic core

N turns

I

Solution:

area sectional cross toroidal-

radiusmean - where

2

2

S

b

b

SNL

Self Inductance

12/12/2017 37

Henry I

L

We have :

Joule 2

1

2

1

2

1 22 III

LIWm

BSNINIWm2

1

2

1Consider a toroidal ring : The energy in the magnetic field :

Multiplying the numerator and denominator by 2b : bSb

NIBWm

2

22

1

a

cb

IS

N turns

toroid theof volume theis )2( and 2

where VbSHb

NI

Magnetic Energy Density

12/12/2017 38

BHVWm2

1Hence :

32 2

1

2

1 JmHBHV

Ww mm

HBwm 2

1In vector form :


12/12/2017 39

Example: Derive the expression for stored magnetic energy in a coaxial cable with the length l

and the radius of the inner conductor a and the inner radius of the outer conductor is b. The

permeability of the dielectric is .

(J) ln4

21

8

1

82

1

2

2

22

2

22

22

a

blI

drrlr

IW

dvr

IdvHW

r

IH

b

a

m

vvm

Solution:

a

b


12/12/2017 40

Documents

The Steady Magnetic Field Part 2 Shoubra/Electrical... · Gauss’Law for the electric field Conservative property of the static electric field Ampere’s Circuital Law Gauss’Law