prof. d. wilton ece dept. notes 25 ece 2317 applied electricity and magnetism notes prepared by the...

Prof. D. WiltonECE Dept.

Notes 25

ECE 2317 ECE 2317 Applied Electricity and MagnetismApplied Electricity and Magnetism

Notes prepared by the EM group,

University of Houston.

Magnetic FieldMagnetic Field

r

Lorentz Force Law:

N Sq

In general, (with both E and B present):

F q v B

F q E v B

This experimental law defines the magnetic flux density vector B.

The units of B are Webers/m2 or Tesla [T].

v

Magnetic Gauss LawMagnetic Gauss Law

x

y

z

B

S (closed surface)

N

Magnetic pole (not possible) !

0S

B n ds

N

S

S

No net flux out !

Magnetic Gauss Law: Differential FormMagnetic Gauss Law: Differential Form

0S

B n dS

0B

Apply the divergence theorem: 0V

B dV

Since this applies for any volume,

B field flux lines either close on themselves or at infinity!

Ampere’s LawAmpere’s Law

y

x Iron filings

0

70

2

4 10 H/m

IB

I

(exact value)

Ampere’s Law (cont.)Ampere’s Law (cont.)

Note: the definition of one Amp is as follows:

72 10 T

at 1 m

B

0

70

2

1 A2 10 T

2 1 m

IB

1 [A] current produces:

So

70 4 10 H/m


Define:

0

0

1H B

B H

A/m2

IH

The units of H are [A/m].

H is called the “magnetic field”

(for single infinite wire)


y

x

I

2

0

2

0

2

22 2

C C

C

H dr H d d z dz

H d

Id

I Id

I

C

The current is inside a closed path.

2 12 2

0

CC C

H dI

rI

The current is outside a closed path.


y

x

I

C

2C1C

Iencl

Hence encl

C

H dr I

“Right-Hand Rule”

Although this law was “derived” for an infinite wire of current, the assumption is made that it holds for any shape current.

This is an experimental law.

Ampere’s Law


C

Amperes’ Law: Differential FormAmperes’ Law: Differential Form

C

S

n

encl

C

encl

S

S S

H dr I

H n dS I

H n dS J n dS

H n S J n S

H J

(from Stokes’ theorem)

S is a small planar surface

Since the unit normal is arbitrary,

Maxwell’s Equations (Statics)Maxwell’s Equations (Statics)

H J

0E

vD

0B

electric Gauss law

magnetic Gauss law

Faraday’s law

Ampere’s law

In contrast to electric fields, the sources for magnetic fields produce a circulation and are vectors! !

Maxwell’s Equations (Dynamics)Maxwell’s Equations (Dynamics)

vD

0B

electric Gauss law

magnetic Gauss law

Faraday’s law

Ampere’s law

BE

t

D

H Jt

Displacement CurrentDisplacement Current

DH J

t

Ampere’s law:

“displacement current”

(This term was added by Maxwell.)

1" " sI Q

J z z zA A t t

D z Dz

t t

h

A

insulator

I

z

Q+ + + + + + + + + + +

The current density vector that exists inside the lower plate.

Ampere’s Law: Finding Ampere’s Law: Finding HH

encl

C

H dr I

1) The “Amperian path” C must be a closed path.

2) The sign of Iencl is from the RH rule.

3) Pick C in the direction of H (as much as possible).

4) Wherever there is a component of H along path C , it must be constant. => Symmetry!

ExampleExample

Calculate H

An infinite line current along the z axis.

First solve for H .

“Amperian path”

y

x

I

C

z

r

Example (cont.)Example (cont.)

2

0

2

encl

C

encl

C

H dr I

H d I I

H d I

H I

2

IH

y

x

I

C

z

r


I

S

3) H = 0

h

0

2 0S

B n dS

B h

Magnetic Gauss law:

0

0

encl

C

z

H dr I

H h

2) Hz = 0

I

C

h

Hz dz = 0

H d cancels

=


A/m2

IH

y

x

I

z

r

ExampleExamplecoaxial cable

< a2

2A/mz

IJ

a

b < < c 22 2

A/mz

IJ

c b

The outer jacket of the coax has a thickness of t = c-b

Note: the permittivity of the material inside the coax does not matter here.

C

c

a

b

x

y

ra

b

I I

z

c


2

0

encl

C

encl

H d I

H d I

encl

C

H dr I

This formula holds for any radius, as long as we get Iencl correct.

H H

2enclI

H

C

c

a

b

x

y

r

The other components are zero, as in the wire example.


< a

a < < b

b < < c

2 22

Aencl z

II J

a

enclI I

2 2

2 22 2

2 2

2 2

Bencl zI I J b

II b

c b

cI

c b

> c 0enclI I I

Note: There is no magnetic field outside of the coax (“shielding property”).

C

c

a

b

x

y

r


< a

a < < b

b < < c

22

IH

a

2

IH

2 2

2 22

I cH

c b

> c 0H

Note: There is no magnetic field outside of the coax (“shielding property”).

C

c

a

b

x

y

r

ExampleExample

Solenoid

z

n = N/L= # turns/meter

0 r

I

aCalculate H

< a

h

Find Hz

C

encl

C

z encl

z

H dr I

H h I I nh

H nI

L


> a

h

Hz

C

0

0

0

encl

z

z

I

H h

H

A/m ,

0,zH nI a

a


The other components of the magnetic field are zero:

2) H = 0 from

1) H = 0 since

0S

B n ds

0enclI C

S

2 enclH I

2 0B h


Summary:

A/m ,

0,

H z nI a

a

0 ,rB H a

z

n = N/L = # turns/meter

0 r

I

a

L

ExampleExample

y

x

z

Infinite sheet of current

A/ms szJ z JCalculate H

y

x

top view+

-


By symmetry:

0zH (superposition of line currents)

(magnetic Gauss Law)

x xH y H y

0yH

Hx-

Hx+

y

x

xH x H

Note: No contribution from the left and right edges (the edges are perpendicular to the field).encl

C

H dr I 2

2

ˆ ˆ( )

w

x xwbot

H dr H x x dx H w

2

2

ˆ ˆ( )

w

x xwtop

H dr H x x dx H w

Example (cont.)Example (cont.)w

-

y

x+

C

2y

Note: The magnetic field does not depend on y.

1

2

x x sz

x x sz

x sz

H w H w J w

H H J

H J

, 02

, 02

sz

sz

JH x y

JH x y


ExampleExample

A/m

A/m

bots

tops

IJ z

w

IJ z

w

Find H everywhere

h

w

I

Ix

y

z w >> h


A/m

A/m

botsz

topsz

IJ

w

IJ

w

y

x

h

Two parallel sheets of opposite surface current


y

x

“bot” “top”

h

magnetic field due due to bottom sheet

magnetic field due to top sheet

I

-IInfinite current sheet approximation of finite width current sheets


2 2 , 02

0, otherwise

bot top

botbot sz

H H H

JH H x y h

H

[A/m], 0

0, otherwise

IH x y h

w

H

A/mbotsz

IJ

w

Hence

prof. d. wilton ece dept. notes 25 ece 2317 applied electricity and magnetism notes prepared by the...

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