Lecture Engineering electromagnetics: Vector analysis - Nguyễn Công Phương

Nguyễn Công Phương

Engineering Electromagnetics


Contents
I. Introduction
II. Vector Analysis
III. Coulomb’s Law & Electric Field Intensity
IV. Electric Flux Density, Gauss’ Law & Divergence
V. Energy & Potential
VI. Current & Conductors
VII. Dielectrics & Capacitance
VIII.Poisson’s & Laplace’s Equations
IX. The Steady Magnetic Field
X. Magnetic Forces & Inductance
XI. Time – Varying Fields & Maxwell’s Equations
XII. The Uniform Plane Wave
XIII.Plane Wave Reflection & Dispersion
XIV.Guided Waves & Radiation
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Introduction (1)
• The study of charges (at rest or in motion)
• Fundamental to electrical engineering
• Why study?
–
–

–
–

Electromagnetic Compatibilty
High Frequency Circuits
Communications
etc.

• Applications: communication technology, computer technology,
antenna technology, biomedical applications, military defense, etc
• Ampere, Faraday, Gauss, Lenz, Coulomb, Maxwell, …

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Introduction (2)
Electromagnetics

Electrostatics

∂q
=0
∂t

Magnetostatics

Electromagnetic Waves

∂I
=0
∂t

∂I
≠0
∂t

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Introduction (3)
• W. H. Hayt, J. A. Buck. Engineering
Electromagnetics. McGraw-Hill, 2007
• E. J. Rothwell, M. J. Cloud. Electromagnetics.
CRC Press, 2001
• N. B. Thành, N. T. Quân, L. V. Bảng. Cơ sở lý
thuyết trường điện từ. NXB Đại học & trung học
chuyên nghiệp, 1970
• />Engineering Electromagnetics - sites.google.com/site/ncpdhbkhn

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Contents
I. Introduction
II. Vector Analysis
III. Coulomb’s Law & Electric Field Intensity

IV. Electric Flux Density, Gauss’ Law & Divergence
V. Energy & Potential
VI. Current & Conductors
VII. Dielectrics & Capacitance
VIII.Poisson’s & Laplace’s Equations
IX. The Steady Magnetic Field
X. Magnetic Forces & Inductance
XI. Time – Varying Fields & Maxwell’s Equations
XII. The Uniform Plane Wave
XIII.Plane Wave Reflection & Dispersion
XIV.Guided Waves & Radiation
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Vector Analysis
1.
2.
3.
4.
5.

Scalars & Vectors
The Rectangular Coordinate System
The Dot Product & The Cross Product
The Circular Cylindrical Coordinate System
The Spherical Coordinate System

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Scalars & Vectors
• Scalar: refers to a quantity whose value may be
represented by a single (positive/negative) real number
• Ex.: distance, time, temperature, mass, …
• Scalars are in italic type, e.g. t, m, E,…
• Vector: refers to a quantity whose value may be
represented by a magnitude and a direction in space (2D,
3D, nD)
• Ex.: force, velocity, acceleration, …
• Vectors are in bold type, e.g. A
• A may be written as A
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Vector Analysis
1.
2.
3.
4.
5.

Scalars & Vectors
The Rectangular Coordinate System
The Dot Product & The Cross Product

The Circular Cylindrical Coordinate System
The Spherical Coordinate System

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The Rectangular Coordinate System (1)
z

0

y

x

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The Rectangular Coordinate System (2)
z

z = za plane

x = xa plane

0

y

x
y = ya plane
a (xa, ya, za)
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The Rectangular Coordinate System (3)
z

dS = dxdyaz
dy

dS = dydzax
0

P

dz

dV

y

dx

x

dS = dxdzay
dV = dxdydz

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The Rectangular Coordinate System (4)
z
z
r
az

ar

y

0

ax

x

y

ay

x
r=x+y+z
→ r = xax + yay + zaz = rxax + ryay + rzaz
x = xax; y = yay; z = zaz
r = rx2 + ry2 + rz2

ar =

r

rx2 + ry2 + rz2

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=
13

r
r


Ex.

The Rectangular Coordinate System (5)

Given a vector V = 5ax – 2ay + 4az, find:
a) Its components?
b) Its magnitude?
c) Its unit vector ?

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Vector Analysis
1.
2.
3.
4.
5.

Scalars & Vectors
The Rectangular Coordinate System
The Dot Product & The Cross Product
The Circular Cylindrical Coordinate System
The Spherical Coordinate System

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The Dot Product (1)
• A·B = |A||B|cosθAB
– |A|: magnitude of A
– |B|: magnitude of B
– θAB: smaller angle between A & B

• A·B = B·A

• A·B = AxBx + AyBy + AzBz
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The Dot Product (2)
B
θBa

a

B
θBa

a

B·a

(B·a)a

The scalar component
of B in the direction of
the unit vector a

The vector component
of B in the direction of
the unit vector a

Ex.: Bx = B·ax

Ex.: Bxax = (B·ax)ax

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Ex.

The Dot Product (3)

Consider the vector field G = zax – 2xay + 3yaz and the point Q(4, 3, 2). Find:
a) G at Q ?
b) The scalar component of G at Q in the direction of aN = ⅓(ax + 2ay – 2az) ?
c) The vector component of G at Q in the direction of aN ?
d) The angle between G(rQ) & aN ?

a) G(rQ ) = 2a x − 2 × 4a y + 3 × 3a z = 2a x − 8a y + 9a z

1
b) G • a N = (2a x − 8a y + 9a z ) • (a x + 2a y − 2a z )
3
1
= (2 × 1 − 8 × 2 − 9 × 2) = −10.67
3

G
θ

aN
G·aN

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The Dot Product (4)

Ex.

Consider the vector field G = zax – 2xay + 3yaz and the point Q(4, 3, 2). Find:
a) G at Q ?
b) The scalar component of G at Q in the direction of aN = ⅓(ax + 2ay – 2az) ?
c) The vector component of G at Q in the direction of aN ?
d) The angle between G(rQ) & aN ?

a) G(rQ ) = 2a x − 2 × 4a y + 3 × 3a z = 2a x − 8a y + 9a z

b) G • a N = −10.67
1
c) ( G • a N )a N = ( −10.67) (a x + 2a y − 2a z )
3
= −3.55a x − 7.11a y + 7.11a z

G
θ

aN

(B·aN)aN

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The Dot Product (5)

Ex.

Consider the vector field G = zax – 2xay + 3yaz and the point Q(4, 3, 2). Find:
a) G at Q ?
b) The scalar component of G at Q in the direction of aN = ⅓(ax + 2ay – 2az) ?
c) The vector component of G at Q in the direction of aN ?
d) The angle between G(rQ) & aN ?

a) G(rQ ) = 2a x − 2 × 4a y + 3 × 3a z = 2a x − 8a y + 9a z

b) G • a N = −10.67
c) ( G • a N )a N = −3.55a x − 7.11a y + 7.11a z
d) G • a N = Ga N cosθ = −10.67
→ 22 + 82 + 92 × 1 × cos θ = −10.67

→ θ = 151o
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The Cross Product (1)
• A B = aN|A||B|sinθAB
– aN: normal (unit) vector

• B A = – (A B)
A

ax
A × B = Ax

ay
Ay

az
Az

Bx

By

Bz

θAB

B
A B

ax, ay, az : unit vectors of x, y, z axes
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Ex. 1

The Cross Product (2)

Given A = ax – 2ay + 3az and B = –4ax + 5ay – 6az . Find their cross product ?

ax
A × B = Ax

ay
Ay

Bx

By

= ax

ax a y az
az
Az = 1 −2 3
−4 5 −6
Bz

−2

3

5

−6

− ay

1

3

−4 −6

+ az

1

−2

−4

5

= −3a x − 6a y − 3a z
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Ex. 2

The Cross Product (3)

Given A = ax – 2ay + 3az and B = –4ax + 5ay – 6az . Find the angle between A & B?
Method 1:

A × B = a N A B sin θ

A×B
→ A × B = A B sin θ → sin θ =
A B

A × B = −3a x − 6a y − 3a z → A × B = 32 + 62 + 32 = 7.35

A = 12 + 22 + 32 = 3.74
B = 4 2 + 52 + 62 = 8.75

7.35
→ sin θ =
= 0.22 → θ = asin(0.22) = 12.9o
3.74 × 8.75
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The Cross Product (4)

Ex. 2

Given A = ax – 2ay + 3az and B = –4ax + 5ay – 6az . Find the angle between A & B?
Method 2:

A • B = A B cos θ → cos θ =

A•B
A B

A • B = 1( −4) − 2(5) + 3( −6) = −32
A = 12 + 22 + 32 = 3.74
B = 4 2 + 52 + 62 = 8.75

−32
→ cos θ =
= −0.97 → θ = a cos( −0.97) = 12.9o
3.74 × 8.75
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Ex. 3

The Cross Product (5)

Given A = ax – 2ay + 3az, B = –4ax + 5ay – 6az, and C = ax – ay + az. Find:
a) A ± B, B ± C, C ± A
b) A.B, B.C, C.A
c) A B, B C, C A
d) (A B).C, A.(B C)

e) What is the angle between B and C A?

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