electromagnetism-for-electronic-engineers-examples.pdf, Essays (university) of Engineering

Download electromagnetism-for-electronic-engineers-examples.pdf and more Essays (university) Engineering in PDF only on Docsity!

Richard Carter

Worked Examples In

Electromagnetism

Download free books at

Download free eBooks at bookboon.com 2

Richard G. Carter

Worked Examples In Electromagnetism

Download free eBooks at bookboon.com Click on the ad to read more

Worked Examples In Electromagnetism

4

Contents

Contents

Preface 6

1 Electrostatics in free space 7 1.1 Introduction 7 1.2 Summary of the methods available 7

2 Dielectric materials and capacitance 33 2.2 Summary of the methods available 33

3 Steady electric currents 57 3.1 Introduction 57 3.2 Summary of the methods available 57

4 The magnetic effects of electric currents 72 4.1 Introduction 72 4.2 Summary of the methods available 72

www.sylvania.com

We do not reinvent the wheel we reinvent light. Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.

Light is OSRAM

Download free eBooks at bookboon.com Click on the ad to read more

Worked Examples In Electromagnetism

5

Contents

5 The magnetic effects of iron 88 5.1 Introduction 88 5.2 Summary of the methods available 88

6 Electromagnetic induction 110 6.1 Introduction 110 6.2 Summary of the methods available 110

7 Transmission lines 125 7.1 Introduction 125 7.2 Summary of the methods available 125

8 Maxwell’s equations and electromagnetic waves 143 8.1 Introduction 143 8.2 Summary of the methods available 143

© Deloitte & Touche LLP and affiliated entities.

thinking.

Discover the truth at www.deloitte.ca/careers

© Deloitte & Touche LLP and affiliated entities.

Dis

eloitte & Touche LLP and affiliated entities.

thinking.

Discover the truth at www.deloitte.ca/careers

© Deloitte & Touche LLP and affiliated entities.

thinking.

Discover the truth at www.deloitte.ca/careers

Download free eBooks at bookboon.com 7

1 Electrostatics in free space

1.1 Introduction

Electrostatic problems in free space involve finding the electric fields and the potential distributions of given arrangements of electrodes. Strictly speaking ‘free space’ means vacuum but the properties of air and other gases are usually indistinguishable from those of vacuum so it is permissible to include them in this section. The chief difference is that the breakdown voltage between electrodes depends upon the gas between them and upon its pressure. The calculation of capacitance between electrodes in free space is deferred until Chapter 2.

The other problems included in this chapter involve the motion of charged particles (electrons and ions) in electric fields in vacuum. This topic remains important for certain specialised purposes including high power radio-frequency and microwave sources, particle accelerators, electron microscopes, mass spectrometers, ion implantation and electron beam welding and lithography.

1.2 Summary of the methods available

Note: This information is provided here for convenience. The equation numbers in the companion volume Electromagnetism for Electronic Engineers are indicated by square brackets.

Symbol Signifies Units İ 0 ( epsilon ) The primary electric constant 8.854 × 10-12^ F.m- Q Electric charge C q Electric line charge C.m- ı ( sigma ) Surface charge density C.m- ȡ ( rho ) Volume charge density C.m- E Electric field V.m- V Electric potential V

’ ( del ) The vector differential operator

r, x, y, z ˆ ˆ ˆ ˆ Unit vectors

Inverse square law of force between charges in free space

1 2 4 0 2 ˆ

Q Q

πε r

F = r (^) [1.1]

• Definition of the electric field of a charge in free space

(^12) 0

Q

πε r

E = r (^) [1.2]

Download free eBooks at bookboon.com 8

• Force acting on a charg0e placed in an electric field

F = Q 2 E [1.3]

• Gauss’ Theorem The flux of E out of any closed surface in free space is equal to the charge enclosed by the surface divided by ε0.
• The integral form of Gauss’ Theorem

0

S V

ρ dv ε �∫∫ E.dA = ∫∫∫ [1.5]

• The differential form of Gauss’ Theorem

0

. x^ y z

E E E

x y z

ρ ε

E   [1.9]

• Electrostatic potential difference B VB − VA = − (^) ∫ (^) A E dl ⋅ [1.13]
• Calculation of electric field from the electrostatic potential

ˆ^ V^ ˆ V^ ˆ V^ grad V V x y z

= − ∂^ + ∂^ + ∂ = − = −∇

E  x y z  [1.22]

• Poisson’s equation 2 2 2 2 (^2 2 )

V V^ V^ V

x y zx

ρ ε

∇ = ∂^ + ∂^ + ∂ = −

[1.24]

• Laplace’s equation 2 2 2 2 2 2 2 0

V V^ V^ V

x y z

∇ = ∂^ + ∂^ + ∂ =

∂ ∂ ∂ [1.27]

• The Principle of Superposition and the method of images
• The Principle of conservation of energy
• The finite difference method

Example 1.

Find the force on an electron (charge -1.602 × 10-19^ C) which is 1 nm from a perfectly conducting plane. What is the electric field acting on the electron?

Download free eBooks at bookboon.com 10

Since E is parallel to the sides of dS the flux of E through the sides is zero. Also, because the electric field within a conducting material is zero when the charges are stationary, the flux of E through the bottom of dS is zero. The flux of E through the top of dS is

d Φ = E dA (1.3)

where E is the magnitude of E (since E is normal to the top of dS). The total charge enclosed by dS is

dQ = σ dA (1.4)

By Gauss’ theorem

d^ dQ ε

Substituting in (1.5) from (1.3) and (1.4) gives

0

E^ σ ε

Note: Because a conducting surface is always an equipotential surface when the charges are stationary E must always be normal to it. If the surface is curved the electric field varies over it (1.6) shows that, locally, the charge density is always proportional to the electric field.

Example 1.

Figure 1.2 right shows a charged wire which is equidistant from a pair of earthed conducting planes which are at right angles to each other.

a) Where should image charges be placed in order to solve this problem by the method of images? b) What difference would it make if the planes were at 60° to each other? c) Could the method be used when the planes were at 50° to each other?

Fig. 1.2 A charged wire close to the intersection of two conducting planes

Download free eBooks at bookboon.com 11

Solution

a) If Cartesian co-ordinates are used to describe the positions of the wire and of its images in the plane then the image line charges are –q at (- d, d) and (d, - d) and +q at (- d, - d) as shown in fig. 1.3. b)

Fig. 1.3 Image charges for planes intersecting at 90°

c) When the planes are at 60° to each other five image charges are equally spaced on a circle as shown in fig. 1.4.

• (^) -

Fig. 1.4 Image charges for planes intersecting at 60°

d) No. The method can only be used when the angle between the planes divides an even number of times into 360°. Thus it will work for planes at angles of 1/4, 1/6, 1/8, 1/10 of 360° and so on.

Example 1.

A wire l mm in diameter is placed mid-way between two parallel conducting planes 10 mm apart. Given that the planes are earthed and the wire is at a potential of 100 V, find a set of image charges that will enable the electric field pattern to be calculated.

Download free eBooks at bookboon.com 13

Fig. 1.6 The field pattern around a set of equispaced parallel wires charged alternately positive and negative.

Example 1.

An air-spaced coaxial line has inner and outer conductors with radii a and b respectively as shown in fig.1.7. Show that the breakdown voltage of the line is highest when ln(a/b)=1.

Fig. 1.7: The arrangement of an air-spaced coaxial line

Solution

For most practical purposes the properties of air are indistinguishable from vacuum. From the symmetry of the problem we note that the electric field must everywhere be radial. The field between the conducting cylinders is identical to that of a long, uniform, line charge q placed along the axis of the system.

To find the electric field of a line charge we apply the integral form of Gauss’ equation to a Gaussian surface consisting of a cylinder of unit length whose radius is r and whose ends are normal to the line charge as shown in fig.1.8. We note that, from considerations of symmetry, the electric field must be acting radially outwards and depend only on the radius r.

Download free eBooks at bookboon.com 14

S

Fig. 1.8 A Gaussian surface for calculating the electric field strength around a line charge.

Let the radial component of the electric field at radius r be Er(r). On the curved surface of the cylinder the radial component of the electric field is constant and the flux is thus the product of the electric field and the area of the curved surface.

(^2 1) r ( ) S

�∫∫ E dA ⋅^ =^ π r^ × × E^ r (1.7)

The flux of the electric field through the ends of the cylinder is zero because the electric field is parallel to these surfaces.

We apply Gauss’ theorem [1.5] to find the relationship between the electric field, radius (r) and the unknown line charge q. Since S has unit length the total charge contained within it, which is denoted by the right-hand side of [1.5] is just q. Thus

( ) 0

2 π r Er r q ε

which can be rearranged to give

( ) 0

r 2 E r q πε r

Since the electric field is inversely proportional to r, it must be greatest when the radius is least, i.e. when r = a.

max 0

E q πε a

The potential difference between the cylinders is found from the electric field using [1.13]

( ) 0 0

(^1) ln 2 2

b^ b b a r a (^) a

V V E r dr q^ dr q^ b πε r πε a

− = − = − = − ^ 

∫ (^) ⌡ (1.11)

Download free eBooks at bookboon.com 16

Example 1.

An air-spaced transmission line consists of two parallel cylindrical conductors each 2 mm in diameter with their centres 10 mm apart as shown in fig. 1.9. Calculate the maximum potential difference which can be applied to the conductors assuming that the electrical breakdown strength of air is 3 MV m⋅ −^1.

Fig. 1.9 A cross-sectional view of a parallel-wire transmission line.

Solution

Since the diameters of the wires are small compared with their separation it is reasonable to assume that close to the surface of each wire the field pattern is determined almost entirely by that wire. The equipotential surfaces close to the wires take the form of coaxial cylinders, as may be seen in Fig. 1.10. This is equivalent to assuming that the two wires can be represented by uniform line charges ± q along their axes. Note that this approximation is only valid if the diameters of the wires are small compared with the spacing between them.

Fig. 1.10 The field pattern around a parallel-wire transmission line

Download free eBooks at bookboon.com 17

The electric field of either wire is then given by Equation (1.9) (for r ≥1 mm) with the appropriate sign for q. Since the strength of the electric field of each line charge is inversely proportional to the distance from the charge, the greatest electric field must occur on the plane passing through the axes of the two conductors. Using the notation of Fig. 1.9 and Equation (1.9) the electric field on the x axis between the wires is found by superimposing the fields of the two wires.

x (^2 0) ( 12 ) 2 0 ( 12 ) E q^ q πε d x πε d x

It is easy to show that this expression is a maximum on the inner surfaces of the wires (as might be expected from Fig. 1.10), that is, when x = ± ( 12 d − a ). The maximum permissible charge is therefore given by ( ) max 2 0 max

q E a d^ a

d

= πε − (1.16)

The potential at points on the x axis between the wires is found from (1.15) using [1.13]

( ) ( ) ( )

(^12) 2 0 12 12 2 0 ln^12

V x q^ q^ q^ dx q^ d^ x C πε d x d x πε d x

  ^ 

where C is a constant of integration. It is convenient to choose C = 0 so that the potential is zero at the origin.

The maximum permissible potential at A is obtained by substituting the maximum charge from (1.16) into (1.17) and setting x = (^) ( 12 d − a )to give

( ) A max ln V E a d^ a^ d^ a d a

= −^ ^ − 

The potential at B is -VA so the maximum potential difference between the wires is 2VA. Substituting the numbers gives the maximum voltage between the wires as 5.9 kV.

When the wires are not thin compared with their separation the method of solution is similar but, as can be seen from the equipotentials in Fig. 1.10, the equivalent line charges are no longer located at the centres of the wires.

Example 1.

A metal sphere of radius 10 mm is placed with its centre 100 mm from a flat earthed sheet of metal. Assuming that the breakdown strength of air is 3 MV.m -1, calculate the maximum voltage which can be applied to the electrode without breakdown occurring. What is then the ratio of the maximum to the mean surface-charge density on the sphere?

Download free eBooks at bookboon.com 19

The first step is to use Gauss’ theorem to find the electric field at a distance r from a point charge Q. The problem has spherical symmetry and therefore the electric field must be constant on the surface of a sphere of radius r centred on the charge and directed radially outwards. The surface area of a sphere of radius r is 4 π r^2 so that from [1.5]

(^2) ( ) 0

4 π r Er r Q

so that

( ) (^2) (^40) r E r Q πε r

Next we use [1.13] to find the potential at a distance r from the charge.

( ) ( ) (^2) (^4 0 ) r V r E r dr Q^ dr Q C

πε r πε r

∫ (^) ⌡

where C is a constant of integration. Now

r = d − x (1.22)

so that the potential due to the first charge is

( )

V x Q C

πε d x

Similarly the potential due to the other charge is

( )

V x Q C

πε d x

Superimposing the potentials of the two charges gives

( )

V x Q^^1 1 Q x

πε d x d x πε d x d x

= ^ − = ⋅

 −^ +^  −^ +

At the surface of the first sphere x^ =^ d^ −^ a and

( ) (^2 0) ( 2 )

V d a Q^ d^ a

πε a d a

The electric field at the surface of the sphere is found by superimposing the fields of the two charges using (1.20)

( ) 0 2 (^ )^2

E d a Q πε a (^) d a

Download free eBooks at bookboon.com 20

Eliminating Q between (1.27) and (1.28) gives the relationship between the breakdown field and the breakdown voltage

( ) (^) ( )

1 max max 2 2

V E d^ a a d a a (^) d a

− = − ⋅ + − (^) −

  ^ 

  ^ 

Substituting the numerical values of the quantities we find that the maximum voltage is 28.3 kV.

From example 1.2 we know that the maximum surface charge density is

12 6 6 2 σ (^) max ε 0 E max (^) 8.854 10 3 10 26.6 10 C m = = × −^ × × = × −^ ⋅ − (1.30)

The total charge on the sphere can be computed from (1.28)

( )

1 max 0 2 2

Q E

a (^) d a

− = + −

so the average charge density is

( ) ( )

1 2 1 max 4 02 12 1 2 0 max^12 av (^4 2 ) E E a a a (^) d a d a

σ^ πε ε

− − = + = + − −

and the ratio of peak to average charge density is

( )

max^2 1 2 2 1. av

a d a

σ σ

Example 1.

An electron starts with zero velocity from a cathode which is at a potential of -10 kV and then moves into a region of space where the potential is zero. Find its velocity.

Solution

The principle of conservation of energy requires that the sum of the kinetic energy and the potential energy of the electron must be constant. Thus

mv + qV = (1.34)

where q is the charge on the electron and V is the potential relative to the cathode. The charge to mass ratio of an electron q / m = −1.759 × 1011 C kg. −^1 and the region of zero potential has a potential relative to the cathode +10 kV so that, rearranging (1.34) we obtain

v^2^ qV 2 1.759 1011 104 59.3 106 m.s- m

= = × × × = × (1.35)