1

Physics AElectromagnetism I

Robert Sang

Room 0.15 Science II

Email: R.Sang@sct.gu.edu.au

http://www.sct.gu.edu.au/~sctsang/

Phone: 3785 3848

Teaching Tools

• Lecturesè3 Core lectures + 1 streamed / week (all are welcome to attend both of the

streamed lectures).

èAvailable on web page

• TutorialsèSelected problems from the problem sheets. MANDATORY!

• Text Book (Halliday, Resnick and Walker) Chapters 22-31

• Griffith University Notes

2

ELECTRICITY

• Electricity is everywhere

• PlasmasèSun

èStars

• van Allen radiation belts

• Lightning

• Responsible for bonding of atoms

• Binding of electrons to the nucleus in atoms

• Electric power for industry

• Most modern appliances: TV's, microprocessors

• Biological: responsible for throught processes and movement inanimals

Charge

• Electric charge (HRW Chap 22)èElectricity is carried by charge.

èAn intrinsic characteristic of fundamental particlesmaking up those objects i.e. it automaticallyaccompanies those particles wherever they exist.

• Two types of chargeèPositive (+) eg: protons, cations

èNegative (-) eg: electrons, anions

èThose objects with equal charge of both negativeand positive charges are said to be NEUTRAL. Eg:The Earth

• Charged objects exert forces (termedelectrostatic force) on each otherèLike charges REPEL

èUnlike Charges ATTRACTîAtoms stay together! H atom

3

• Charge is Conservedè It can not be created nor destroyed

Eg:Ionisation of an atom due to a collision with an electron

He + e− → He+ + e− + e−

• Charge is Quantisedè Charge comes in integral multiples of the electronic charge, which is equalto the charge on one electron and has the unit of Coulomb (C):

-e=-1.602x10-19 C

è The charge on a proton is +e

Example: An object is said to have acquired an electron charge of -5.607x10-17-17 C,how many electrons would be on this object?

Solution: Since charge is quantised and there is 1.602x10-19 C on one electron thenumber of charges will simply be

Number of electrons =−5.607x10−17

−1.602x10−19 = 350 electrons

4

• The movement of charge (current) is a property ofthe materialèAn object that allows the movement of charge (flow) easily is called a

conductor

èAn object that does not permit easy flow of charge are termed insulators

• The earth is a conductor which is neutral

Example: A conducting sphere has a charge of -30e-, it is brought in contact withan identical conducting sphere which is neutral via a conducting wire. If the wireis now removed what will be the charge on the first sphere? If instead thecharged conducting sphere was connected to the earth what would be theresultant charge on the sphere?

-30e-

Before

-30e-

Contact

-15e- -15e-

After After

-30e-

Contact

-30e-

Before

5

Force Between Charged Objects: Coulomb'sLaw

• Charged objects exert an electrostatic force on each other

• Experimentally determined by Coulomb for point charges to be:è Proportional to product of the chargesè Inversely proportional to the square of the distance between them (Inverse Square law eg: Gravitational Force) è Points along the line joining them

F Specifies the magnitude

ˆ r 12 Specifies the direction and isa unit vector

0 Constant:permittivity of free space=8.854 x 10-12 C2N-1 m-2

ˆ r 12

Fz

x

yq1

q2 The electrostatic forceis a vector quantityF =

1

4 0

q1q2

r122

ˆ r 12

Electrostatic Force for Multiple Charges

Since the force is a vector quantity the force on a charged particle q1

due to multiple charges q2, q3, q4, …..qn is just the vector sum of theforce due to each of the charges on q1 cf:Superposition Principle

ˆ r 12

z

x

yq1

q2

q3

q4

q5

ˆ r 14

ˆ r 13

ˆ r 15

F12

F13

F14

F15

Fq1= F12 + F13 + F14 + F15 +K + F1n

Fq1=

1

4 0

q1q2

r122

ˆ r 12 +1

4 0

q1q3

r132

ˆ r 13

+1

4 0

q1q4

r142

ˆ r 14 +K +1

4 0

q1qn

r1n2

ˆ r 1n

Fq1=

1

4 0

q1qi

r1i2

ˆ r 1ii =1

n

∑

Generalised form of Coulomb's Law

6

Example 1: Determine the force on charge q1 due to q2 charge where q1 = 1.6x10-19 C and q2 = 4.8x10-19 C and the distance between themis R = 0.05m on the x axis.

x

R

+ +

q1 q2

F12 = 8.99 ×10 9[ ] 1.6 ×10−19 × 4.8 × 10−19

0.05( )2 N

= 2.76 ×10−25 N

Magnitude of the force

Direction of the force?

Since both charges are positive they must repel each other thus q2Must push q1 away ⇒ ˆ r 12 = −ˆ i

F12 =1

4 0

q1q2

r122

ˆ r 12Coulomb's Law

F12 = −2.76 ×10−25 ˆ i N⇒

F12

x

R

+ +q1

q2

+

Yq3

135oR

Example 2: Determine the force on charge q1 due to charge q2 andq3 where q1 = q3 = 1.6x10-19 C and q2 = 4.8x10-19 C and the distance between q1 and q2 is R = 0.05m on the x axis. The distance betweenq1 and q3 is R = 0.05m and q3 is located in the x-y plane at 135o tothe x axis.

Fq1=

1

4 0

q1qi

r1i2

ˆ r 1ii =1

n

∑

Fq1=

1

4 0

q1q2

r122

ˆ r 12 +1

4 0

q1q3

r132

ˆ r 13

Fq1= F12 + F13

F12 = −2.76 ×10−25 ˆ i N⇒

F13 = 8.99 ×109[ ] 1.6 ×10−19 × 1.6 ×10 −19

0.05( )2 N

= 9.23 × 10−26 N

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Direction of ?F13

x

R

+ +q1

q2

+

Yq3

135oR

F13

Fy13

Fx13

F13

Y

X

45o

45o

45o

Resolve the force into Cartesian Coords:

F13 = Fx13

ˆ i + Fy13

ˆ j

cos = adjhyp

=F y13

F13

F y13= F13 cos = F13 cos 45( ) ⇒ Fy13

= −1

2F13

ˆ j

sin = opphyp

=F x13

F13

F x13= F13 sin = F13 sin 45( ) ⇒ Fx13

=1

2F13

ˆ i

F13 = 1

2F13

ˆ i − 1

2F13

ˆ j

=1

2F13

ˆ i − ˆ j =

1

2× 9.23 ×10−26 ˆ i − ˆ j N

Total force on q1 is then given by the vector sum of the two forces

Fq1= F12 + F13

Fq1= −2.76 × 10−25 ˆ i +

1

2× 9.23 ×10−26 ˆ i − ˆ j N( )

= −2.11×10−25 ˆ i − 6.53 ×10−26 ˆ j N( )

Fq1= −2.11× 10−25( )2

+ −6.53 × 10−26( )2= 2.21 ×10−25 N( )

Direction of the force?

−2.11× 10−25 ˆ i

− 6.53 ×10−26 ˆ j θtan = opp

adj= −2.11 ×10−25

−6.53 ×10−26

⇒ = tan−1 −2.11 ×10−25

−6.53 ×10−26

= 72.83o

8

Fq1= 2.21 ×10−25(N) 72.83o from the negative y axis

x

R

+ +q1

q2

+

Yq3

135oR

Fq1 72.83o

WHY IS THIS GUY'S HAIR STANDING ON END?

1

IMPORTANT PHYSICS ATUTORIAL INFORMATION

Please ensure that you have downloaded a copy of ProblemSheet 1 prior to next week to take to your tutorial next week.

THE PROBLEM SHEETS CAN BE FOUND AT THE URL:

http://www.sct.gu.edu.au/~sctsang/

THIS IS YOUR RESPONSIBILITY!

Last Lecture: Coulomb's Law

F Specifies the magnitude

ˆ r 12 Specifies the direction and isa unit vector

0 Constant:permittivity of free space=8.854 x 10-12 C2N-1 m-2

ˆ r 12

Fz

x

yq1

q2 The electrostatic forceis a vector quantityF =

1

4 0

q1q2

r122

ˆ r 12

2

Electrostatic Force for Multiple Charges

ˆ r 12

z

x

yq1

q2

q3

q4

q5

ˆ r 14

ˆ r 13

ˆ r 15

F12

F13

F14

F15

Fq1= F12 + F13 + F14 + F15 +K + F1n

Fq1=

1

4 0

q1q2

r122

ˆ r 12 +1

4 0

q1q3

r132

ˆ r 13

+1

4 0

q1q4

r142

ˆ r 14 +K +1

4 0

q1qn

r1n2

ˆ r 1n

Fq1=

1

4 0

q1qi

r1i2

ˆ r 1ii =1

n

∑

Generalised form of Coulomb's Law

Example: Calculate the force on the point charge q1 due to the other pointcharges as shown.

q1 = -10 C

y

x

0.5m

0.5m

q3 = +10C

q2 = +10 C

Fq1=

1

4 0

q1q2

r122

ˆ r 12 +1

4 0

q1q3

r132

ˆ r 13

Fq1= F12 + F13

What are the directions of the two forces?

ˆ r 12 = ˆ i

ˆ r 12

ˆ r 13 = ˆ j

ˆ r 13 Fq1=

1

4 0

q1qi

r1i2

ˆ r 1ii =1

n

∑

Coulomb's Law:

3

⇒ Fq1= 8.99 ×109[ ] 10( ) 10( )

0.5( )2ˆ i +

1

4 0

10( ) 10( )0.5( )2

ˆ j

Fq1=

1

4 0

q1q2

r122

ˆ r 12 +1

4 0

q1q3

r132

ˆ r 13

⇒ Fq1= 3.6 ×1012 ˆ i + 3.6 ×1012 ˆ j (N)

y

x

Fy

Fx

⇒ Fq1= Fx

2 + Fy2

= 3.6 ×1012( )2+ 3.6 ×1012( )2

= 5.1× 1012 (N)

Fq1= Fx

ˆ i + Fyˆ j

tan( ) =Fy

Fx

=3.6 ×1012

3.6 ×1012= 1

⇒ = tan−1 1( ) = 45o

q1 = -10 C

y

x

0.5m

0.5m

q3 = +10C

q2 = +10 C

Fq1

Fq1= 5.1 × 1012 (N)

= 45o

4

Lecture 2Last Lecture:

F12 =1

4 0

q1q2

r122

ˆ r 12Coulomb's Lawˆ r 12

Fz

x

yq1

q2

Interesting exercise comparing Electrostatic force Fe to GravitationalForce FG between Cs+ and Cl- ions in a CsCl (cesium chloride) crystal.

Fe =1

4 0

q1q2

r122

=1

4 ×8.854 × 10−12×

1.602 × 10−19( )2

32

× 4 ×10−10

2

⇒ Fe = 1.9 ×10−9 N

The electrostatic force is 32 orders of magnitude greater than gravitational force and is responsible for bonding in the crystal

G=6.67x10-11Nm2kg-2

mCs=133 amumCl=35.5 amu1amu=1.66x10-27 kg

FG = Gm1m2

r122

= 6.67 ×10 −11 ×133 × 35.5 × 1.66 ×10−27( )2

3

2× 4 ×10−10

2

FG = 7.2 × 10−42 N

The Electric Field

• The interaction between charged bodies (eg: between two electrons) iscarried by an entity termed the electric field.

• The electric field is defined by the following:

E =F

qUnit = NC-1 = Vm-1 (working unit)

The electric field at a point in space is the electrostatic force per unitcharge felt by a positive test charge q place there.

• |E| is called the electric field strength

5

• Electric field of a point charge

Applying Coulombs law for a the force for a point charge q on a testcharge Q yields:

E =F

q=

1

4 0

Qq

r2ˆ r

q=

1

4 0

Q

r 2ˆ r

• E is a vector field hence the total field due to n point charges is simply found by summing the forces and dividing by the test point charge q:

E =1

qF1 + F2 + ... + F n( ) =

1

qF i

i =1

n

∑ = E1 + E2 + ... + En

Electric fields add vectorially!

Q

q

X

Y

F = Eq

r

r points in the direction of the force

Q would put on q

Electric Field is determined by Superposition

Q

q

X

Y

Q

QQ

Q

12

3

4

5n=5

r1 r 2r 3r 4

r 5

E =1

qF i

i =1

5

∑ = E1 + E 2 + E3 + E4 + E5

6

• Electric Field Lines

è Originally devised by Michael Faraday

è Used to pictorially represent the electric field or "lines of force"

è The direction of the field line represents the direction of force felt by atest charge. The direction of E is tangential to the field line.

è The magnitude of the electric field or the electric field strength isproportional to the number of field lines cutting a unit area perpendicularto E

è A field line begins on a positive charge and terminates on a negativecharge

è The number of field lines leaving a positive charge (or entering anegative charge) is proportional to the magnitude of the charge

• Examples of electric fields

Sphere of uniform negative charge Two equal charge positive point charges

is tangential to the field linesE

High Field Region

Weak Field Region

Zero Field

7

One positive and one negative point charge

High Field Region

Weak Field Region

Positively charged conducting Sheet

Field lines begin on positivecharge and terminate on the

negative charge

8

Example 1: Determine E at the midpoint joining two pointcharges q1=1.0x10-6 C and q2=-3.0x10-6 C, separated by 1.0cm.

q1 q2x

d d

Test charge q at x=0

Direction of E from each chargeon the test charge?

+ive +iveE1 E2

-ive

E =1

4 0

Q

r2ˆ r

E1 =1

4 0

q1

d2ˆ i E2 =

1

4 0

q2

d2ˆ i ( )

is the unit vector in the x-directionˆ i

ETOT = E1 + E2

ETOT =1

4 0

q1 + q2( )d2

ˆ i

ETOT =1

4 × 8.854 ×10−12 ×1.0 + 3.0( ) × 10−6

5.0 ×10 −3( )2ˆ i

ETOT = 1.44 ×109 NC−1 ˆ i

9

• Example 2: The Millikian oil drop experiment (1912) was used to prove that charge is quantised in units of e i.e. q=ne for n=0,±1,±2…A vertical electrostatic field holds the oil drops stationary when the gravitational and electrostatic forces are equal and opposite.

Let the oil drops radius be 1x10-6 m.If the density of oil is ρ=0.80 g/cm3,what electric field strength is requiredto stop a drop with charge q=-3e?

Millikan used X-rays to remove chargefrom the oil drops and he found that thecharge on the oil drops could only reach a minimum value and all other chargeswere multiples of that charge

- - - - - - - - - - -

- - - - - - - - - - -

+ + + + + + + +

+ + + + + + + +

-q

radius r

+

- E?

The forces are balanced when

Fe = Fg

⇒ qE = mg

⇒ E =mg

q=

× 43 1.0 ×10−6( )3( )

=m6 7 4 4 4 4 8 4 4 4 4 × 9.8

3 ×1.602 ×10−19

⇒ E = 8.5 ×103 NC−1

What forces act on the charge?

Fe

Fg

=m

V=

m4

3 r 3⇒ m = 4

3 r 3

The mass of the drop is obtained from

P1

P2

10

Example3: Deflection of Charged Particles: Determine the deflected Distance on a screen for an electron with initial velocity v0 in the x-direction

+ + + + + + + +

- - - - - - - -

e-

v0

y

x

yL1 L2

Screen

y1

y2

θE

Between the plates the electron will experience a force up due to the electric field, It therefore experiences an acceleration:

F = qE = ma

⇒ a =qE

m

⇒ a =qE

mˆ j

Hence the distance travelled vertically betweenthe plates is

s = ut + 12

at 2

y1 = 0 +1

2at 2 =

1

2

qE

m×

L1

v0

2

=qE L1( )2

2m v0( )2

Where t is given by

t =L1

v0

Note: no forces act horizontally.Thusthe velocity is constant in thisdirection

The velocity at the end of the deflection plates is given by

⇒ vy =qEL1

mv0

v = u + at

vy = 0 + at

⇒ vy =qE

mt Where the time between the plates is given by

t =L1

v0

There is no acceleration after the plates hence the velocity in the y direction isconstant, therefore the distanced travelled in this direction is

s = ut

y2 = qEL1

mv0

t

Where time travelled after the plates is given by

t =L2

v0

⇒ y2 =qEL1

mv0

×L2

v0

⇒ y2 =qEL1L2

m v0( )2

The total deflected distance is then y = y1 + y2 =qE L1( )2

2m v0( )2 +qEL1L2

m v0( )2 =qEL1

m v0( )2

1

2L1 + L2

11

Application: Cathode Ray Oscilloscope (CRO)

1

Lectures Schedule Week (Week 10)

Tuesday: Core Lecture 4

Wednesday: Physical Science Lecture 1 (Gauss's Law)

Thursday: Core Lecture 5

Friday: Biological Electromagnetism Lecture 1 (Nerves,ECG)

Problem Sheet 2 will be ready tomorrow which you willneed for next week

The slides for these lectures are on my web page:http://www.sct.gu.edu.au/~sctsang/

Last Lecture

The Electric Field

• The interaction between charged bodies (eg: between two electrons) iscarried by an entity termed the electric field.

• The electric field is defined by the following:

E =F

qUnit = NC-1 = Vm-1 (working unit)

The electric field at a point in space is the electrostatic force per unitcharge felt by a positive test charge q place there.

• |E| is called the electric field strength

2

• Electric field of a point charge

Applying Coulombs law for a the force for a point charge q on a testcharge Q yields:

E =F

q=

1

4 0

Qq

r2ˆ r

q=

1

4 0

Q

r 2ˆ r

• E is a vector field hence the total field due to n point charges is simply found by summing the forces and dividing by the test point charge q:

E =1

qF1 + F2 + ... + F n( ) =

1

qF i

i =1

n

∑ = E1 + E2 + ... + En

Q

q

X

Y

F = Eq

r

r points in the direction of the force

Q would put on q

One positive and one negative point charge

High Field Region

Weak Field RegionField lines begin on positivecharge and terminate on the

negative charge

Electric Field Lines

3

Example 1: Determine E at the midpoint joining two pointcharges q1=1.0x10-6 C and q2=-3.0x10-6 C, separated by 1.0cm.

q1 q2x

d d

Test charge q at x=0

Direction of E from each chargeon the test charge?

+ive +iveE1 E2

-ive

E =1

4 0

Q

r2ˆ r

E1 =1

4 0

q1

d2ˆ i E2 =

1

4 0

q2

d2ˆ i ( )

is the unit vector in the x-directionˆ i

ETOT = E1 + E2

ETOT =1

4 0

q1 + q2( )d2

ˆ i

ETOT =1

4 × 8.854 ×10−12 ×1.0 + 3.0( ) × 10−6

5.0 ×10 −3( )2ˆ i

ETOT = 1.44 ×109 NC−1 ˆ i

Lecture 3Electrostatic Potential

• Potential energy (PE) is the ability to do work, Electrostatic Potential is not PEbut is proportional to it.• Since the electric field exerts a force on a charge, then a free charge will move andhence alter its PE.• The electrostatic force pushes a free charge from a region of higher potential tolower potential.

Electrostatic potential (V) is the quantity, the gradient of which is equal to theelectric field and has the unit of VOLT (V), 1V=1NmC-1:

E

qLower Potential

E = −dV

dxone dimension

|E| = 10/1 =10 V/m

Note that the electrostatic potential is a scalar!

4

Relationship between Electrostatic Potential and Work (Energy)

E = −dV

dxRecall the definition of the electric field and equating:

E =F

q= −

dV

dx= −

∆V

∆x

⇒ q∆V = −F∆x ∆W = F∆x ⇒ ∆W = −q∆VWork=Force X Distance:

∆W = q∆V

Hence the energy gained by the charge due the electric field is given by

This is the work done (energy gained) on a charge q moving through a potentialdifference ∆V.

E

q

If the charge had to move against the potential it would lose this energy ∆W

Potential Difference V is called voltage in electric circuits

Energy lostby E field

Example

• Electrostatic PE is analogous to the PE of a mass raised some height in agravitational field. The greater the height raised the more the PE.

• A charge has a higher PE the more it is moved against the force of an electricfield.

• KE of the charge is converted to PE as it moves to the region of higher potential

+

Higher PotentialHigher PE

+Lower Potential

Lower PE

5

Potential of a point charge

Recall the Electric field due to a point charge Q is given by

E =1

4 0

Q

r' 2

⇒ −dV

dr=

1

4 0

Q

r' 2

⇒ dV = −1

4 0

Q

r' 2 dr

The potential at a distance r from the charge compared to infinity yields the potential:

dV (r )

V (∞ )

∫ V = −r

∞

∫1

4 0

Q

r' 2 dr'

⇒ V ∞( )

= 01 2 3 − V r( ) = −

1 × −1( )4 0

Q

r'

r

∞

=Q

4 0

1

∞−

1

r

V r( ) =Q

4 0rThe potential of a point charge Q

Exercise: Confirm that it is possible to get the electric field by taking thederivative of this function.

Example

Example: Graphical representation of the potential felt by a charge due to adipole field, the positive charge has +1C and negative -1C and 1m apart.

+

Increase in potentialto go up hill

+ +

Lower PotentialSmaller PE

+

6

Example: Determine the potential due to the arrangement of point chargesshown below where d=1.3m:

q1=+12nC, q2=+31nCq3=-24nC q4=+17nC

The potential at P is just given by the algebraic sum of the potentials from each charge and remember the potential is a scalar quantity.

V = V1 + V2 + V3 + V4 =1

4 0

q1

r1

+q2

r2

+q3

r3

+q4

r4

Recall that the potential due to a point charge Q is V r( ) =Q

4 0r

⇒ V =2

4 0dq1 + q2 + q3 + q4

By symmetry we see that each of the charges is an equal distance from P which is

r =d

2

r1

⇒ V =2

4 × × 8.85 ×10−12 × 1.3× 12( ) + 31( ) + −24( ) + 17( ) ×10−9 = 350V

Electric Current

I =dq

dt

The unit of current is the Ampere, 1A = 1 Coulomb per second C/s

• Currents are the result of an electric field exerting a force on charges that are free tomove.• Current can be due to the flow of positive or negative charge• In a conductor the flow of charge is normally due to electron movement with q=-e,as such the current flows in the opposite direction to the charge movement. This iscalled Conventional Current.• Unless otherwise specified, current (I) used in the course will be conventionalcurrent.• Current is a scalar even though we attribute a direction to it (like time).

Conventional and electron currents

⇒ dq = Idt ⇒ dq = q∫ = I dt0

t

∫ = It

• Electric current is the net movement of charge and is analogous to the flow of waterthrough a river• Imagine a point P in an object where charges are moving past with a velocity v. If a small amount dq passes in a time dt then the current is :

dqP

v

- -

-

-- --

-- -

-

---

7

Example: A loop of copper

No net flow of charge]I=0

e-

e-e-

e-

e-

Current in opposite directionTo e- movement

• Currents are conserved (this is a result of charge conservation)

i0 = i1 + i2

Example: What is the current and the directionin the lower right hand wire

Conservation of Current iin = iout

Current in = 11A

Current out = 3A

To equate these the currentmust be out =11-3=8A

8A

8

Example: During a lightning strike ~ 104 A flow during the return stroke which takes around 100µs. How many e- pass between the ground and the charge?

- -- - - - --- - - - -- -- - - - -

- - - -

++ ++++ +++

++ +++

++++

++

+ + + + + + + + + +

Cloud

Earth

q = It

q = 104 × 100 ×10 −6 = 1C

Recall that one e- has charge 1.602x10-19 C

Number of e- =1C

1.602 × 10-19 C~ 6.2 ×1018 electrons

ICurrent Direction?

Example: Television. The image is formed by 1 (B&W) or 3 (colour) beams ofenergetic electrons striking the screen. The electrons are accelerated in an electricfield (the electron gun) E ~ 106 V/m. Say the beam current is 1.0x10-4 A (0.1 mA).How many e- are emitted per second by the electron gun?

q = it

q = 1 ×10−4 ×1 = 1 ×10−4 C

Recall that one e- has charge 1.602x10-19 C

Number of e- =1× 10-4C

1.602 × 10-19 C~ 6.2 ×1014 electrons

9

Current and Voltage in Conductors

• Current is the movement of charge• Charge movement is due to the electric fieldexerting a force on the charge: F=qE• Recall from Lecture 1 that charges moveeasily in conductors

+++

---E

Iconductor

How is E generated in a conductor? ⇒ Establish a potential difference betweentwo points in the conductor

The potential difference between two points is calledthe voltage

The electric field within a conductor is

E = V

d

V is the voltage that exist betweentwo points a distanced apart.

Note: This equation is not true in general since E may be non-uniform.

• The charges moving through a conductor with a potential difference V will have

work done on them and their energy will change by W where

∆W = q∆V

• If q increases its potential, we must do work on it and hence it loses energy• If q lowers its potential, the field does work on it and the charge gains energy

10

Example. The electron gun. An electric field isestablished between two parallel metal plates which havea small hole through which electrons enter and leave.The electrons are supplied by a filament, which is a wireheated to more than 2000 Kelvin so that it literally boils offelectrons from its surface. Each electron gains energy eVas it is accelerated between the metal plates of the gun,where V is the potential difference (voltage) between theplates.

Let d=1.0cm and V= 3000 V - typical of a CRO; TV gunsrun at 15-20kV. What is the E field and how fast will theelectrons travel?

E =V

d=

3000

0.01= 3.0 ×105 V / m

The electrons will gain kinetic energy:∆W = eV = −1.602 ×10−19 × −3000 = 4.8 × 10−16 J

The velocity of the electrons can be found since

K =1

2mv2 ⇒ v =

2K

m

⇒ v =2K

m=

2 × 4.8 ×10−16

9.11 ×10−31 = 3.3 ×107 ms-1

m = 9.11×10−31 kg

1

Lecture 4Electric Circuits

• An electric circuit conducts current from one terminal of a source though wiresand other circuit elements and back to the other terminal of the source• When referring to a circuit, voltage is interpreted as the potential difference in the circuit.• DO NOT confuse voltage with potential (V) which has the units of volts•To avoid this confusion voltage in a circuit is called the voltage drop to indicatea potential difference.

Elements of an electric circuitSources

• DC Source ⇒ source of constant potential (Battery)eg: car battery 12 V DC (Direct Current)• Current (I) flows in one direction only +ive ⇒ -ive• The flowing charge does work on the circuit elements• Within the battery the current flows from negative topositive as the battery does work on the flowing charge

Representation of aDC source in a circuit

• AC Source ⇒ source of alternating potential (alternator)eg: mains electricity supply 240 V AC• Current (I) flows alternates in both directions as each terminal alternates +ive ⇒ -ive• Simplest alternator produces a voltage output sinusoidal:v(t) =Vpeak sin( t). For the 240 V mains Vpeak=360 V

• The voltage of a non-DC source is usually specified asthe RMS value, for a sinusoid this is given by Vpeak

2

Representation of aAC source in a circuit

Circuit Elements

• Wire is assumed to be a perfect conductor andconnects other circuit elements•There is no potential drop between any two points in it.

Representation of awire in a circuit

• Switch is a device that allows current to flow when closed but blocks the current when open. Representation of a

switch in a circuit

2

• Resistor is a device that dissipates electric potential energy (ie lowers the electrostatic potential of charges flowing through it), converting it to heat.

Representation of aresistor in a circuit

• A reactor stores electrical energy; there are two types thatreact in potential or current:inductor and a capacitor. These devices will be considered further in Physics B.

capacitor

inductor

• Ammeter is a device that enables the measurement ofcurrent flowing through a circuit element

A

Representation of aammeter in a circuit

• Voltmeter is a device that measures the voltage (potentialDifference) across a circuit element

V

Representation of avoltmeter in a circuit

Connections in Circuits

• Series Connections in a Circuit: two elements are in connected in series in a circuit when the same current flows through them consecutively.

• Parallel Connections in a Circuit : two elements are connected in parallel in acircuit when the same voltage appears across them. The total current is shared between them.

3

Example: Simple electric circuit

More complex circuit: The microchip

Intel Pentium III Processor 9.5 million transistors1971 Intel 4004 microprocessor

4

Jack Kilby Nobel Laureate2000

The first microchip

5

Drug delivery via microchip

Properties of resistors and Ohm's Law

• The resistance of a conductor (R) is defined by the ratio R=V/I. A conductor ina circuit whose function is to provide a specified resistance is called a resistor.

• The unit of resistance is the ohm Ω: 1Ω = 1 VA-1

R = V

I

• Ohm's Law: The current through a device is always directly proportional to thepotential difference applied to the device:

V = IR

6

Resistivity of a material

• The resistance of a material depends on its geometric shape as well as its constitutes. The resistivity is an intrinsic property of the material and defined as:

Where J is the current density defined as J =I

A

cf: the definition of resistance=E

J Unit Ωm

Relationship between resistivity and resistance

Consider a conductor as shown:

The electric field is E =V

L The current density is J =i

A

⇒ =E

J=

V

Li

A

=V

i

A

L= R

A

L⇒ R =

L

A

• The greater the resistivity greater the resistance• The smaller the area greater the resistance • Longer the material greater the resistance • Although not specifically specified in the expression the temperature of the materialalso affects the resistance

Sometimes you will hear the term Conductivity of a material being used to describea material. The conductivity (σ) is just the reciprocal of resistivity i.e. σ=1/ρ

• Insulator (eg: Glass @ 300K) ρ ~ 1012

• Conductor (eg: Cu @ 300K) ρ ~ 2x10-8

• semiconductor (eg: Ge @ 300K) ρ ~ 0.45• Superconductor (eg: Pb @ 4K) ρ = 0

The range of resistivities of materials is great some examples are:

7

Example: Determine the current density, resistance in a aluminum wire that has a diameter of 1 mm a current of 100mA and a length of 10m, assume ρ=3x10-8 Ωm.

The current density is

J =I

A=

100 ×10−3

1× 10−3

2

2 =1.3 ×105Am -2

R =L

A= 3 ×10−8 ×

10

1×10−3

2

2 = 0.38Ω

The resistance is

The electric field is calculated from

= E

J

⇒ E = J = 3 ×10−8 ×1.3 ×105 = 0.0039Vm -1

+++

---E

Ialuminum wire

1mm

• Ohm's Law (Amended): A conducting device obeys Ohm's law when the resistanceof the device is independent of magnitude and polarity of the applied potential.

Resistor Semiconductingpn junction diodeLook at the ratio V/I

At any point eg:R=2/2x10-3

=1000 Ω

There are other materials that are called superconductors which have zero resistance

Is Ohms' Law always obeyed?

8

Power in electric circuits

• The charge flowing through a resistor does workon the resistor which will dissipate energy W:

W = qV

Consider a short time t such that charge q flows

and does work W:

W = qV Divide by t

W

t=

qV

t q

t= I

Rate of change of work = Power

⇒ P = VI Units Js-1

• This power is converted into heat

P = IV

= I × IR

= I2R

It is also worth noting that

Example 1: Calculate the current and the resistance in a 1000W hair dryer connectedto a 240 V line.

Recall that the power and current are related by

The resistance can be calculated two ways:

V = IR

⇒ R = V

I= 240

4.2≈ 57Ω

Ohm's Law:

P = IV = I2 R

⇒ R = PI2

= 10004.2 2 ≈ 57Ω

P = IV

⇒ I = P

V

⇒ I =1000

240≈ 4.2A

9

Example 2: Lightning bolt. A typical lightning bolt can transfer as much as 109J ofEnergy across a potential difference of around 5x107V in about 0.2 seconds. Determine the amount of charge transferred as well as the power.

The amount of charge can be determined by:

W = qV

⇒ q = W

V= 109

5 ×107= 20C

The current in the lightning bolt is:

I =q

t=

20

0.2= 100A

The power is then:

P =Energy

time=

109

0.2= 5 × 109 W

P = IV = 100 × 5 ×107 = 5 × 109 W

Alternative method

1

Lecture 5Electric Circuits Continued

Combinations of resistors

Resistors in Series

• A set of resistors are in series if the current through them is the same.

• The voltage drop across each one obeys Ohm's Law then the total voltageis equal to the sum of the voltage drops across each resistor.

Vtotal = IR1 + IR2 + .... + IRn = I R1 + R2 + ... + Rn( ) = IRtotal

Hence Rtotal = R1 + R2 + ... + Rn = R jj =1

n

∑ Resistors in series

V

Example 1: Circuit with a battery and a resistor

V = ir + iR = i r + R( )

r is the internalresistance

Battery

V

V

In general we will assume idealbatteries that have zero internal resistance

Rtotal = r + R

2

Example 2: Three resistors in series

V

V

Equivalent

Req = Rjj=1

3

∑ = R1 + R2 + R3

V R1 R2 R3

Resistors in Parallel

• When a voltage drop is applied to resistors connected in parallel the resistances allhave the same potential difference

I

I1 I2 I3

Recall that the current must be conserved

I = I1 + I2 + I3

Since the voltage drop across each resistor isThe same the currents I1, I2, I3 can be determinedusing Ohm's Law:

I1 = V

R1

I2 = V

R2

I3 = V

R3

I2 + I3

I2 + I3

I

⇒ I = I1 + I2 + I3 = V

R1

+ V

R2

+ V

R3

= V1

R1

+ 1

R2

+ 1

R3

⇒ I = V1

R1

+ 1

R2

+ 1

R3

cf I = V

R⇒ 1

Rtotal

= 1

R1

+ 1

R2

+ 1

R3

V Rtotal

I Equ

ival

ent

3

• Resistors that are in parallel can be replaced with an equivalent resistance Rtotal

that has the same potential drop and the same total current as the actual resistances

1

Rtotal

= 1

R1

+ 1

R2

+ 1

R3

+ ...1

Rn

= 1

R jj =1

n

∑

Series Parallel

RTotal = R jj =1

n

∑ 1

RTotal

=1

R jj =1

n

∑

Same current throughall resistors

Same potentialdifference across allresistors

Resistors in circuits summary

Example: Find the current passing through each of the resistors below:

Useful approach to solve these problems• Calculate the single resistor equivalentfor the circuit• Calculate the total current• Calculate how the current is split betweenR2 and R3.

V

R1R2 || R3

I

R2 and R3 are in parallel and can be replacedwith the equivalent resistance:1

R=

1

R2

+1

R3

⇒1

R=

R3

R2R3

+R2

R2 R3

=R2 + R3

R2R3

⇒ R2 || R3 =R2R3

R2 + R3

Two resistor in seriesRtotal = R1 + R2 || R3

V

Rtotal

I

Equivalent Circuit

⇒ Rtotal = R1 +R2 R3

R2 + R3

= 1000 +91×103 × 7500

91 ×103 + 7500= 7929Ω

4

The total current in the circuit is found from Ohm's Law:

Itotal =V

Rtotal

=12

7929= 0.0015 =1.5mA

To find the currents that pass through the resistors R2 and R3 we note that thetotal current is equal to the sum of the currents I2 and I3:

The voltage drop across each resistor is the same hence for each resistor

Divide equation (1) by I3:

Itotal = I2 + I3 (1)

VR2= VR3

⇒ I2 R2 = I3 R3 ⇒I2

I3

=R3

R2(2)

Itotal

I3

=I2

I3

+ 1

I3 =Itotal

R3

R2

+1

⇒Itotal

I3

=R3

R2

+ 1

⇒ I3 =1.5 × 10−3

7500

91×103

+ 1

=1.39 ×10−3 A =1.39mA I2 = Itotal − I3

⇒ I2 = 1.5 −1.39 = 0.11mA

We would expect I2 to be smaller since the resistance of R2 is larger than R1

Note: The currents passing through the resistors in parallel can be also determinedby recalling that the potential drop across each resistor in parallel is the same.

The potential drop across the equivalent resistance R2||R3 is

VR2 ||R3= IR2 || R3 = Itotal

R2 R3

R2 + R3

= 1.5 ×10−3 ×91000 × 7500

91000 + 7500= 10.5V

This will be the voltage drop across both R2 and R3.

It is easy to check that this is correct as the sum of the voltage drops must be equal the potential difference of the battery. This is just a statement of conservation of energy ie sum of the potential drops =0 (Kirchoff's Loop Rule) eg:

V

R1R2 || R3

I

Vbattery − VR1− VR2 ||R3

= 0

⇒ Vbattery = VR1+ VR2 || R3

= ItotalR1

1.5 ×10−3 ×1000 =1.5

1 2 3 + ItotalR2 || R3

10.51 2 4 3 4 = 12V

Now that the voltage drop across the resistors in parallel has been determined thecurrent can be found via Ohm's Law:

I2 =VR2 ||R3

R2

=10.5

91000= 0.11mA

I3 can be determined using Ohm's Law againOr using conservation of I

5

Example: Voltage divider

This type of circuit is the basis for most electronicamplitude controls eg: volume control on a radio or TV

Assume no current flows though the output terminalswhich can be accomplished via suitable electronic arrangement.

Since the resistors are in series, then the same current flows though them hence

2

I =V1

R1

=V2

R2

⇒V2

V1

=R2

R1

The voltage divides in proportion to the two resistors

The voltages drops around the circuit must sum to zero hence

Vin = V1 + V2⇒

Vin

V2

=V1

V2

+1

Note V2=Vout

⇒Vin

Vout

=R1

R2

+1 ⇒Vout

Vin

=R2

R1 + R2

Problem solving techniques for electric circuits

• Label the battery terminals

• Label the currents in each branch of the circuit. It doesn't matter if you get thedirection wrong initially as it will come out in the math.

• Apply conservation of current (Kirchoff's Current Rule)

• Apply the Potential Rule (Kirchoff's Loop Rule)

• Apply this Loop Rule in one direction only and pay attention to the signs ofpotentials.

Resistors: The sign of the potential drop is negative if the loop direction isthe same as the current direction and positive if opposite to the currentdirection. Battery: The potential will be +ive if the loop direction moves from thenegative terminal to the positive terminal. The potential difference is -ive ifthe loop direction moves from the positive terminal toward the negative.

• Solve equations generated for the required unknowns

6

Example: Find the current passing through the circuit below with R=10Ω:

+ +

R

12V 8V

To determine the current we need to know whatthe total potential drop across the circuit is thenWe can apply Ohm's Law to find I.

⇒ V12 − IR − V8 = 0

I

⇒ IR = V12 − V8

⇒ I =V12 − V8

R=

12 − 8

10= 0.4A

A

B C

D

Loop Direction

1

• Just as charge Q produces and electric field E, a current causes a Magnetic Field B.• The properties of magnetic fields make them useful in devices such as electric machines which are used in electric railway locomotives, watches, refrigerators,loudspeakers , etc.• Magnetic fields are also useful for probing the constitution of matter eg: NMR

Lecture 6Magnetism

The Magnetic Field• Recall that electric field lines radiate from a point charge and all lines start on a chargeand terminate on a charge.

• Magnetic field lines form circles or loops around currents and have no start or end

• The direction of the magnetic field is determined by the Right Hand Screw rule

2

The Right Hand Screw Rule

Grasp the element in you right hand with you thumbextended toward the the direction of the current. Yourfingers will naturally curl around which give thedirection of the magnetic field lines due to that element.B is then tangential to the field lines

Direction ConventionWhen drawing magnetic field lines

Dot = Field out of pageCross = Field into page

Example: A current loop, which is just a loop of wire carrying a current. B is strong inside and weak outside for a very simple reason: because all the B field lines are closed (and never intersect) the same number go through the loop as spread through all space outside it.

The strength of the magnetic field depends on the flux of field lines cuttingthe area.

High B field regionLow B field region

3

Permanent Magnets• Electric and Magnetic fields are described by Maxwell's Equations (Classical Physics)• Valid on the atomic scale (unusual for Classical Physics)

Example: The orbit of an electron in an atom about a nucleus

The motion of the e- inducesa current i

Spin of an electron

• Both of these are example of magnetic dipoles• In an atom the dipoles of each electrons sumvectorially to give a total magnetic dipole moment.• The magnetic properties of atoms therefore depends ontheir electron configuration cf chemical properties• In a material the total magnetic moments of theatoms sum together and if the combination of all thesemagnetic dipoles produces a magnetic field the materialis said to be Magnetic (Permanent Magnet).

• Examples of materials that are capable of being permanent magnets• Transition metals: Fe, Co and Ni• Rare Earths: Gd, Dy

• Bar Magnet• Magnetic field lines emerge from the North Pole of a permanent magnet and re-enter at the south pole.

Permanent Magnet

• If you cut a permanent magnet in two thetwo resulting pieces both have a North (N)and south(S) poles.• If you continued to do this until you onlyhad individual atoms then you still willhave north and south poles.• This implies that there is no such thing asa single magnetic pole, only N-S pairs

4

Example of attraction and repulsionin permanent bar magnet

•Since currents create magnetic fields which can exert forces on other currents,Magnetic Poles interact.•Like poles repel and unlike attract

The earth is an example of a permanent magnet

Without this magnetic field life on earth would disappear!

The poles of the earth's magnetic field reverseAbout every million years!

Magnetic Forces

Force on a moving charge• The magnetic field is is defined by the force it exerts on a moving charge:

F = qv × B

F = qvBsin( )Force on a moving charge

• This definition is analogous to the definition of the electric field as the force on a stationary charge. i.e. E=F/q

θ is the angle between the vectors v and B

A × B = A B sin( )Cross Product

• The direction of the force is always perpendicular to v and B ⇒ The magnetic force can not change the speed of a particle• The direction is found by the RH rule

• The unit of the magnetic field B is the Tesla

B =F

qvsin( ) → 1 Tesla ≡ 1 N

Cms-1 ≡ N

Cs-1m ≡

N

Am

5

RH Rule: v is swept into B with the resultantpointing in the direction of your thumb

In the case of a +ive charged particle, F is parallel tovxB

In the case of a -ive charged particle, F is anti-parallel tovxB

Recall that current is the flow of charge ie I =dq

dt

Let the charges move with a velocity v such that in a time dt they move a distance dl:

dl = vdt

Displacement vector

X by I

⇒ Idl = Ivdt ⇒ Idl =dq

dtvdt

⇒ Idl = vdq Current Element

Current Element

This definition provides a link between current and moving charge and will be usefulin this section of the course.

6

Force on a current element

• Since a magnetic field will induce a force on a moving particle we also infer thata magnetic field will induce a force on a wire carrying a current.

Consider a small parcel of charge moving with a velocity v then the force is

d F = dqv × B

⇒

Recall that the current element is given by Idl = vdq

d F = Idl × B

• The total force on a wire is found by summing (integrating) all thecurrent elements along the length of the wire.• For a wire of arbitrary shape a special type of intergralcalled a line integral is used to sum all of the current elements.• For a straight piece of wire length l the force is

F = Il × B = IlB sin( ) Force on a straight wire

The forces on moving charges and current carrying wires are the basis of many devicesEg: TV's, electric motors loudspeakers etc.

Example: The loudspeaker

7

Charged Particle in a magnetic field

Consider a (+ive) charged particle q with mass m and speed v initially in the i directionat (0,0,0) interacting with a magnetic field B =Bk in the figure below:

What is (i) The vector force on q due to B initially? (ii) The trajectory of q?

(i) The force on a moving charge in a static B field is given by

F = q v× B

⇒ F = qvi × Bk = q v B i× k( )

⇒ F = −qvBj

⇒ i × k( ) = − jDirection?

(ii) To find the trajectory recall that F is always perpendicular to B .

• B is in the k direction ⇒ F is always in the x-y plane

• Initially q will deflected in the -ive y direction, but F still must be perpendicular tov and B. Since F is perpendicular to v then speed of the particle does not change only The direction ⇒ This is a characteristic of circular motion ie the magnitude of v isconstant with the direction changing in a uniform way.

⇒Q follows a circular orbit ofconstant radius due to the magneticforce

8

The magnetic force will be balanced by the centrifugal force (this is the force thatPushes you against a door in a turning car):

Fmag = qvB Fcent =mv2

r

Proved in Mechanics B

Fmag = Fcent ⇒ qvB =mv2

r

⇒ r =mv

qBOrbit of radius of a charged particle in a perpendicular magnetic field

Applications:• Deflection of electrons in a TV• Mass Spectrometer• Particle Accelerators

Example: The Cyclotron

9

Electromagnetic Induction• This term describes the means by which we can generate electricity from mechanicalwork - the reverse of an electric motor.• This is how mains electricity is generated.

Consider a straight piece of wire of length l moving with velocity v perpendicular to a magnetic field B.

The charges in the wire feel the force:

F = q v× B

• The force will push (+ive) charges up the wire and (-ive) charges down the wire• This means that there will be an E-field since both ends of the wire will be at different potentials. i.e. a Potential Difference has been created.• Due to historical origins, this potential difference is called EMF (Electromotive Force) denoted with symbol ε

This potential difference is just the work done required to move the (fictitious positive)Charge from the bottom to the top of the wire:

=W

q

Where did the work come from?

⇒ Supplied by the motion of the wire through the magnetic field

Work = force x distance

W = Fl = qvBl = q

= vBl EMF induced in a moving wire

10

How to win an Igg Nobel Prize: Levitate a Frog!

B=16T

1

8 Lectures

Dr Robert Sang Room 0.15 SCII

email: R.Sang@sct.gu.edu.au

This section of the course will deal with:

Time Varying Magnetic and Electric Fields• Faradays Law• Maxwell's Equations• Time Harmonic Fields

Wave Equation (Solns of Maxwell's Eqns)

There will be two problem classes which are not assessed.Assessment: 1 AssignmentExamination in week 16?

Magnetostatics• Amperes Law• Biot-Savart Law• Magnetic Dipole• Magnetic Materials• Boundary Conditions• Magnetic Energy• Magnetic Force and Torque

Electromagnetism II SCE2311/SCE2611

Static Magnetic Fields

1.0 The Basic Equations

You have already seen the two basic equations that govern the electrostatic model:

∇ •D = ∇ × E = 0

D is the displacement vector and accounts for any polarisation of the materialE is the electric field vectorρ is the charge density

For a linear isotropic material D and E are related via the permittivity:

D = E

For a perfect vacuum then

= 0 = 8.854x10−12 ≈1

36x10−9 F / m

2

If we place a charge q in an E field then this charge will experience a force:

Fe = qE (N)

• Experiments have show that if you put a stationary charge in a B field then there is no additional force.

• If the charge is moving in the B field at some velocity v a force is experienced.

• Experimentally this was found to be proportional to the charge q, the field B and the velocity of the charge v. The force was also found to be perpendicular to both the B field direction and the velocity of motion:

Hence the total force on a charged particle in both a B and E field is given by the sum of the electrostatic force and the magnetostatic force, this sum force is called the Lorentz Force.

FTOT = Fe + FB = q(E + v × B) (N)

FB = qv × B (N)

Analogue equations for magnetostatics?

2.0 Fundamental Postulates of Magnetostatics in Free Space

• Two fundamental postulates of magnetostatics in free space have been experimentally determined to be governed by two equations:

∇ •B = 0 ∇ ×B = 0J 0 is the permeability of free space

0 = 4 × 10−7 (H / m)

J is the current density J = Nqv =I

Area (A / m2 )

N is the number of charge carriers per unit volume.

The magnetic flux density B is measured in Tesla.

Recall the vector identity: ∇ • ∇ ×A = 0

∇ • ∇ ×B( ) = 0∇• J = 0

⇒ ∇• J = 0

Is this result consistent with theelectrostatic model?

Note J is a vector

3

Recall the equation of continuity of charges

∇ •J +t

= 0

For a steady current ⇒t

= 0 and will reduce to the previous expression

Consider now the expression ∇ •B = 0

Compare this to the electrostatic equation ∇ •E =0

One can see immediately that there is no magnetic analogue to the electric charge density!

Recall the Divergence Theorem from Maths IIA

∇ •AdV = A • dSS∫

V∫

Taking the volume integral over the equation ∇ •B = 0

∇ •BdV = B• dSS∫

V∫ ⇒ B •dS = 0

S∫ S is the surface that bounds the volume

V which encloses the magnetic flux lines.

This expression indicates that the magnetic flux lines must enclose upon themselves and this is referred to as Conservation of Magnetic Flux.

This evidence points to the conclusion that Magnetic Monopoles donot exist!

Flux is defined as the volume flow rate

4

2.1 Amperes Circuital Law

Now consider the equation We can turn this into an integral equation by application of Stoke's Theorem:

∇ ×B = 0J

∇ ×A • dS = A • dlC∫

S∫

Taking the surface integral over the LHS of the equation yields:

∇ ×B • dS = B • dlC∫

By Stoke's Theorem1 2 3 S

∫

On the RHS we get 0J • dS = 0

S∫ I

Equating LHS and RHS B • dlC∫ = 0 I

where C is the contour bounding the surface S and I is the total current through S.

S

C

I

This result is know as Ampere's Circuital Law

The circulation of magnetic flux density in free space around any closed path equals µ0 times the current flowing through the surface bounded by the path.

The direction of B is found by the application of the RH Screw Rule:

Grasp the element in you right hand with you thumbextended toward the the direction of the current. Yourfingers will naturally curl around which give thedirection of the magnetic field lines due to that element.B is then tangential to the field lines

5

Differential Form Integral Form∇ •B = 0 B • dS = 0

S∫(Magnetic flux lin es must encloseupon themselves)

∇xB = µ0 J B • dlC∫ = µ0I

(Ampere's Law)

2.2 Summary of the Magnetostatic Equations

C2

r2 C1

r1b

Example:Consider an infinitely long straight conductor with cross section of radius b and carries a steady current I out of the page (see figure). Determine the magnetic flux density both inside and outside the conductor.

The solution is made by taking advantage of the cylindrical symmetry of the problem. If we align the conductor so that it points along the direction then by the RH screw rule B will be in the φ direction.

Inside the conductor we have

B • dlC∫ = 0 IB can be determined using Ampere's Law

B = B1ˆ

dl = r1d ˆ

B • dlC1

∫ = B1ˆ

0

2

∫ • r1dˆ

= B1 r d0

2

∫ = B1 r1[ ]0

2 = 2 B1 r1

Now consider the RHS. The total current through the loop C1 is given by the ratio of the cross sectional area of the loop compared to the conductor:

IC1=

r12

b2 I =r1

2

b2

I

Equating gives the magnetic flux densitywithin the conductor:

B1 = 0

2

r1b2

I for r1 ≤ b

dl

6

Outside the conductor loop C2 we again apply Ampere's law:

B • dlC2

∫ = B2ˆ

0

2

∫ • r2 d ˆ = 2 B2 r2

B = B2ˆ

dl = r2d ˆ

C2

r2 C1

r1b

The current in C2 is just the total current since the loop completely encloses the conductor hence

2 Br2 = 0 I

B = 0 I

2 r2

r2 ≥ b

Graphical Solutions

B Units of 0

2I

r in units of b

r ≤ b

B Units of 0

2I

r in units of b

r ≥ b

7

Example 2:Determine the magnetic flux density inside a closely wound toroidal coil having an air core with N turns and carrying current I. The toroid has a mean radius of b and the radius of each turn is a.

Cylindrical symmetry ensures that Bis always in the directionˆ

In the region were r<b-a there is no net current enclose by C and hence nomagnetic flux can exist so B=0

B=0

B = B ˆ

dl = rd ˆ

B • dlC∫ = 0 I

B • dlC∫ = B ˆ

0

2

∫ • rd ˆ = 2 B r

To determine the RHS of our equation we need the total current. There are N coils each with current I thus the total current = NI. Therefore the RHS is

0ITotal = 0NI

B = 0 NI

2 r

( b - a )≤ r ≤ ( b + a )Region

Equating LH and RH sides yields:

In the region were r>b-a there is no net current enclose by C and hence in this region and no magnetic flux can exist so B=0

Example 3: Determine the magnetic flux density inside an infinitely long solenoid with air core having n closely wound turns per unit length and carrying current I.

I

z

I

CL

For an infinitely long cylinder it is clear that there will be no field outside of the cylinder since there will be no current in this region.

Consider the loop C applying Ampere's law:

B • dlC∫ = 0 I

B • dlC∫ = Bz

ˆ z 0

L

∫ • dLˆ z = Bz L

The total current along the length L is nLI, thus the RHS is

RHS = 0nLI

Equating the two terms yields:

Bz = 0nI

1

Lecture 10

10.1 The Vector Potential (another method to find B)

Recall the magnetostatic equation ∇ •B = 0

Recall from Maths IIA the vector identity ∇ • ∇ × C( ) = 0

⇒ B can be written as the curl of another vector: B = ∇ ×A

The vector A is defined as the vector potential

Recall the second magnetostatic equation

∇ ×B = 0J

⇒ ∇× ∇ ×A( ) = 0 J cf vector identity: ∇ × ∇ ×C( ) = ∇ ∇ •A( ) − ∇2A

⇒ ∇ ∇ •A( ) − ∇2A = 0 J—2 is the Laplacian in Cartesian Coords is given by

∇2A = ˆ i 2 Ax2 x

+ ˆ j 2 Ay2 y

+ ˆ k 2 Az2z

There is some latitude in choosing the functional form of . The simplest is tochoose:

∇ •A

∇ •A = 0 Coulomb Gauge

⇒ ∇2A = − 0 J Vector Poisson's Equation

We compare this to the scalar Poisson's equation that you encountered previously:

∇2V = −0

Recall the solution was V =1

4 0 R∫ d

By analogy we solve for the vector potential

A = 0

4

JR∫ d [ Webber/metre (Wb/m) ]

2

The magnetic flux was given by Φ = B • dSS∫

⇒ Φ = ∇× A( )• dSS∫ Applying Stoke's Theorem: A • dS

S∫ = A • dl

C∫

⇒ Φ = A • dlC∫

• The magnetic flux can be determined from either B or A and it is it is usual to choosethe method that allows the easiest integration for a specific problem.

10.2 The Biot-Savart Law

• Allows the determination of the magnetic field associated with current carryingwires.• First proposed by Biot and Savart around 1820 based upon experimental observations. (Like most of EM until Maxwell entered the scene!)

Consider a thin wire with crosssectional area σ with a current Iflowing through it.For a small volume element dt ofwire length dl' then

I σ

dl'

dτ

Jd = J dl' = Idl'

The vector potential can be written as

A = 0

4

JR∫ d = 0

4

I

RC∫ dl' = 0 I

4

dl'R

C∫

The volume integral has been reduced to a line integral around a closed loop.Current must always flow in a closed loop from the source through the wireand back to the source.

R is the distance from the wire elementTo the field measurement point

3

The magnetic flux density can therefore be found by taking the Curl:

B = ∇ ×A = 0 I

4∇ ×

dl'R

C∫ = 0 I

4∇ ×

dl'R

C∫

We now apply the vector identity ∇ × fG( ) = f∇ ×G + ∇f( ) ×G

With f =1

RG = dl'

Thus the expression for the magnetic flux density can bewritten as

B = 0 I

4

1

R∇ ×dl' +∇

1

R

× dl'

C∫

The primed coords refer to the point on a wire while the unprimed coords due tothe grad refer to the point where the field is being measured hence ∇ ×dl' = 0

⇒ B = 0I

4∇

1

R

× dl'

C∫

We will now consider the term under the integral by considering a point (x,y,z) at a distance R from dl' .

R = RF − Rdl'

R F = xˆ i + y j + z ˆ k

Rdl' = x' ˆ i + y' ˆ j + z' ˆ k

R = x − x'( )ˆ i + y − y'( )ˆ j + z − z'( )ˆ k

By simple vector addition we infer

R = R •R = x − x'( )2 + y − y'( )2 + z − z'( )2

The term ∇1

Rcan now be evaluated, as an exercise show that it is given by:

∇1

R= −

RR3

Thus the expression for B becomesBiot-Savart Law

Sub back into expression B = 0 I

4∇

1

R

× dl'

C∫

Quite often this is written as

B = 0 I

4dl' ×

RR3

C∫ (T)

dB = 0 I

4dl' × R

R3 (T)

Vector that points fromthe origin to field point

4

Example1: A current I flows through a thin wire of length 2L. Find the magnetic flux density B at a point at a distance r along the central plane from the wire as shown.

• We first choose an appropriate coord system• In this case we choose the cylindrical coord system• Next we determine what we need to find in order todetermine B ie dl' and R.

B = 0 I

4dl' ×

RR3

C∫

dl' = dz' ˆ z

R F = rˆ r

Rdl' = z' ˆ z

R = RF − Rdl' = rˆ r − z' ˆ z

⇒ R = r 2 + z'2

⇒1

R3 =1

r2 + z' 2( )32

dl' ×R =

ˆ r ˆ ˆ z

0 0 dz'

r 0 −z'

= rdz' ˆ

B = 0 I

4dl' ×

RR3

C∫ = 0 I

4

rdz' ˆ

r2 + z' 2( )3

2_ L

L

∫ ⇒ B = 0Ir ˆ

4

dz'

r 2 + z' 2( )3

2− L

L

∫

Note that the solution to the integral is:(see handout on vectors)

1

a2 + p2( )3

2∫ dp =

1

a2

p

a2 + p2+ c

⇒ B = 0I ˆ

4

z'

r r2 + z' 2

− L

L

⇒ B = 0 IL

2 r r2 + L2ˆ

Note that we could have solved this problem by finding A using the vector PoissonEquation then taking the Curl to find B.

5

Example 2: Find the magnetic flux density at a point on the axis of a circularloop with radius b that carries a current I.

R F = zˆ z

Rdl ' = bˆ r

dl' = rd ' ˆ = bd ' ˆ

For convenience we choose the cylindricalcoord system thus

1

R3 =1

z2 + b2( )3

2

dl' ×R =

ˆ r ˆ ˆ z

0 bd ' 0

−b 0 z

= bd '. z − 0.0( )ˆ r − 0.z − −b.0 ( ) ˆ + 0 − −bbd ' ( )ˆ z

dl' ×R = bzd ' ˆ r + b2d ' ˆ z

R = RF − Rdl ' = zˆ z − bˆ r

R = R = z 2 + b2

⇒ B = 0I

4

b2ˆ z [ ]d '

b2 + z' 2( )3

20

2

∫

B = 0 I

4

b2 ˆ z

b2 + z' 2( )3

2

d '0

2

∫ = 0I

4

b2ˆ z

b2 + z'2( )3

2

'[ ]0

2

B = 0Ib2

2 b2 + z' 2( )3

2

ˆ z

B = 0 I

4

dl' ×RR3

C∫ ⇒ B = 0 I

4

bz ˆ r + b2ˆ z [ ]d '

b2 + z' 2( )3

20

2

∫

B = 0 I

4

bzˆ r [ ]d '

b2 + z' 2( )3

20

2

∫ + 0I

4

b2 ˆ z [ ]d '

b2 + z'2( )3

20

2

∫

Now applying the Biot-Savart Law:

By symmetry the first term equals zero since for every +r vector on the loop there is an opposing vector -r

ˆ r − ˆ r

6

10.2 The Magnetic Dipole (important example in magnetostatics) Consider the same circular loop of current as the last example, but now calculate the magnetic flux density B at some distance RF from the loop centre where RF>>b and off-axis wrt the centre of the loop.

• B is independent of due to

symmetry for simplicity we choose forthe field point = / 2

Recall that A = 0 I

4

dl'R

C∫

B = ∇ ×Aand

• Choose Spherical Polar Coords

It can be shown that ˆ ' = cos ' ˆ j − sin ' ˆ i

⇒ dl' = bd ' ˆ ' = b cos ' ˆ j − sin ' ˆ i ( )d '

For every Idl' there is a symmetrically located differential current element onthe other side of the y axis that will contribute an equal amount to A in the xdirection but will cancel the contribution in the y direction .

Thus the expression for dl' reduces to: dl' = −bsin ' ˆ i d '

At the the field point we are free to choose ˆ = −ˆ i

⇒ dl' = bsin ' ˆ d '

We now need to find R which can be found from the law of cosines: θ

A

B

C C2 = A2 + B2 − 2ABcos( )

⇒ R2 = RF2 + Rdl '

2 − 2RFRdl ' cosΨ

⇒ R = RF2 + b2 − 2RFbcosΨ

⇒ 1

R= 1

RF2 + b2 − 2RFbcosΨ

Applying the cosine law yields:

7

The expression for R can be simplified by the recalling the condition that b2 << RF2

⇒1

R=

1

RF2 + b2 − 2RFbcosΨ

= RF2 + b2 − 2RFbcos Ψ[ ]− 1

2 =1

RF

1+b2

RF2 −

2bcosΨRF

− 1

2

⇒1

R≈

1

RF

1 −2b cosΨ

RF

− 1

2

1 − x( )1 / 2 ≈ 1+1

2xcf

⇒1

R≈

1

RF

1 −b cosΨ

RF

is defined by the dot product of Rdl' and RFcos Ψ

Rdl ' • RF = Rdl ' RF cosΨ

Rdl ' = Rdl ' cos ' ˆ i + Rdl ' sin ' ˆ j

R F = R F cos ˆ k + RF sin ˆ j

where

x

yφ' Rdl'x

y

Rdl ' = xˆ i + yˆ j

cos '( ) = xRdl '

⇒ x = Rdl ' cos '( )

sin '( ) = y

Rdl '

⇒ y = Rdl ' sin '( )

⇒ Rdl' = Rdl' cos '( )ˆ i + Rdl ' sin '( )ˆ j

⇒ Rdl' •R F = Rdl ' cos ' ˆ i + Rdl ' sin ' ˆ j ( ) • RF cos ˆ k + RF sin ˆ j ( )⇒ Rdl ' R F sin 'sin = Rdl ' R F cosΨ

⇒ cosΨ = sin 's in

We can now sub this back into the expression for 1/R:1

R=

1

RF

1 −bsin 's in

RF

Hence the vector potential is A = 0 I

4

1

RF

1 −bsin 's in

RF

bsin '( ) ˆ d '

0

2

∫

⇒ A = 0 Ib

4 RF

sin ' d ' ˆ +0

2

∫= 0

1 2 4 4 4 3 4 4 4

0Ib2 sin

4 RF2 sin2 '

0

2

∫ d ' ˆ

⇒ A = 0Ib2 sin

4 RF2

1

2' −

1

2s in2 '

0

2

ˆ

⇒ A = 0Ib2 sin

4RF2

ˆ

8

B is found now from B = ∇ ×A

∇ ×A = 0 Ib2

4∇ ×

sin

RF2

ˆ

∇ ×A = 0 Ib2

4

1

RF2 sin

ˆ R Fˆ RF

ˆ RF sin

RF

0 0sin2

RF

B = 0 Ib2

4RF3 2cos ˆ R F + sin ˆ [ ]

We define the magnetic dipole moment such that (A.m2) then werewrite the expressions for A and B as

m = I b2 ˆ k

A = 0m × ˆ R F4 RF

2

B = 0 m4 RF

3 2cos ˆ R F + sin ˆ [ ]

[Wb/m]

[T]

Electric Dipole

V =p • ˆ R F

4 0RF2

E =p

4 0 RF3 2 c o s ˆ R F + sin ˆ [ ]

[Volts]

[V/m]

Magnetic Dipole

A = 0m × ˆ R F4 RF

2

B = 0 m4 RF

3 2cos ˆ R F + sin ˆ [ ]

[Wb/m]

[T]

1

Lecture 11

Magnetisation of Matter

• All materials consist if atoms or molecules and each one of these can be considered tohave a magnetic dipole mk. This dipole is due to the coupling of orbital and spin angular momentum of electrons as well as the spin angular momentum

mL

• In the presence of a magnetic field the dipole tries to align itself, to minimize theenergy of the interaction.

Uint eraction = −m k • Bext

Interaction energy between a dipole and an external magnetic field

Uint eraction = −m k • Bext = −m k Bext cos( ) ⇒ U is minimised for θ=0o

Proof:

mS

m k = mL + mS

• Consider a small volume of material ∆v that consists of such atoms with a magneticdipole then the number of atoms for n atoms/unit volume is

N = n∆v

We now define the magnetisation vector M for the volume which is the vector sumof the individual atomic dipoles

M = lim∆v→ 0

1

∆vmk

k =1

n∆v

∑ [A/m]

In the limit as ∆v → 0 then for a volume element dv' we find

M =dmdv'

⇒ dm = Mdv'

Recall that the vector potential for a magnetic dipole is given by

A = 0m × ˆ R F4 RF

2 ⇒ dA = 0dm × ˆ R F4 RF

2 ⇒ dA = 0M × ˆ R F4 RF

2 dv'

2

The total vector potential will just be the sum of the volume elements:

A = dAv'∫ = 0

4

M × ˆ R FRF

2

v'∫ dv'

One can show thatˆ R FRF

2 = ∇'1

RF

⇒ A = 0

4M × ∇'

1

RF

v'

∫ dv'

We now apply the vector identity: ∇ × fG( ) = f∇ ×G + ∇f( ) × G

scalar I'm a vector

f = 1

RF

G = M

with You can show that M × ∇'1

RF

=

∇' ×MRF

− ∇' ×MRF

A = 0

4

∇' ×MRF

− ∇' ×MRF

v'

∫ dv'Sub back for A ⇒

G is an arbitrary vectors' is the surface that bounds v' is the unit vector normal to the surface s'ˆ n

We now use the following relation

∇' ×Gv'∫ d v' = G × ˆ n ( )

s'∫ ds'

A = 0

4

∇' ×MRFv'

∫ dv' + 0

4

M × ˆ n RF

s'∫ ds'

⇒ ∇' ×MRF

v'∫ d v' =

M × ˆ n RF

s'∫ ds'

Hence the vector potential becomes

Define two new quantities:

J'm = ∇' ×M

J'ms = M × ˆ n

[A/m2] Volume Magnetic Current Density

[A/m2] Surface Magnetic Current Density

3

Thus A can be re-written in terms of the volume and surface magnetic current densities:

A = 0

4

J'm

RFv'∫ dv' + 0

4

J'ms

RF

s'∫ ds'

General Remarks:

• Why is this useful?• If A can be determined then B due to the material atomic dipole moments can be evaluated from B=∇xA eg: we can calculate B from a permanent magnetic• Hence we need to be able to calculate J'm and J'ms

• If a material has a uniform magnetism M then

J'm = ∇' ×M = 0 i.e. the volume magnetic current density is zero

∇' =x'

ˆ i ' +y'

ˆ j ' +z'

ˆ k '

Note: that the form of this eqn is the same as theSoln to the Vector Poisson Eqn (Lecture 10)

Example Cylindrical Bar MagneticConsider a cylindrical bar magnetic as shown with uniform axial magnetisation as labeled. Calculate B at (0,0,z).

A = 0

4

J'm

RFv'∫ dv' + 0

4

J'ms

RF

s'∫ ds'

J'm = ∇' ×M = 0 since M = Cˆ k is uniform

J'ms = M × ˆ n

M = Cˆ k B = ∇ ×A

Relevant Equations:

Only the surface current density makes a contribution to the vector potential.

It makes sense to apply cylindrical symmetry to this problem and as such there arethree surfaces that we need to consider to calculate J'ms.

⇒ A = 0

4

J'ms

RF

s'∫ ds'

Surfaces S1 and S2 are parallel and antiparallel to M, for both surfacesJ'ms = 0

4

M = Cˆ k

For surface S3 dSr' = bd ' dz' ⇒ ˆ n = ˆ r

⇒ A = 0

4

J'ms

= M× ˆ n

RF

s'∫ ds' = 0Cb

4

ˆ k × ˆ r RF

0

z

∫0

2

∫ d ' dz'

⇒ A 0Cb

4

ˆ

RF0

z

∫0

2

∫ d ' dz'

The form of this integral is of the same form seen for the magnetic dipole lastlecture, hence for the small volume dv' =πb2dz'the contribution to B is

dB = 0b2C ˆ k

2 b2 + z − z'[ ]2( )3

2

The total contribution to B is just given by the sum of the volume elements:

B = dB∫ = 0b2Cdz'

2 b2 + z − z'[ ]2( )3

2z'= 0

z' = L

∫ ˆ k ⇒ B(0,0, z) = 0Cˆ k 2

z

z 2 + b2−

z − L

z − L( )2 + b2

Exercise: Show this

Magnetic Field Intensity

The presence of magnetisable materials requires modification of the magnetic flux density B due to the effect of the atomic dipole moments of the material.

∇ ×B = 0J

Recall one of the fundamental postulates for magnetostatics in free space is

We now modify this equation to include the magnetic volume current density:

1

0

∇ ×B = J + Jm ⇒1

0

∇ ×B = J + ∇ ×M ⇒ ∇×B

0

− M

= J

H =B

0

− M

⇒ ∇× H = J

We now define a new quantity, the Magnetic Field Intensity:

[A/m]

In this expression J is the volume density of free current

Note: 0 does not appear in this eqn!

5

If the magnetic material is linear and isotropic then we can write

M = mH

H =B

0

− M =B

0

− mH

then

⇒ B = 0 1 + m( )H

⇒ B = 0 r H = H

m is called the magnetic susceptibility

r is the relative permeability and is known as the absolute permeability

• Non-magnetic materials µr close to unity• Magnetisable material eg: cobalt, iron, nickel µr ranges from 50 → 10,000• Can be used to increase the magnitude of magnetic fields!

Amperes Law in Terms of the Magnetic Field Intensity

∇ ×H = J

∇ ×H( )S∫ • ˆ n ds = H • dl

C∫ = J • ˆ n ds

S∫

= I1 2 4 3 4

Applying Stoke's theorem

H • dlC∫ = I Ampere Law

6

Example The air Gap Toroid with Magnetic CoreConsider the toroid as shown with a very small gap lg with a core that comprises ofa magnetic material. A current I flows in the N turns of wire. We wish to determinethe magnetic flux density Bc in the core as well as the magnetic field intensity Hc inthe core and the Hg gap.

lgφ = 2π − δ φ = 0 o By the RH Rule B is in the

direction

ˆ

For the narrow gap we assume thatThe magnetic flux density in the gap isThe same as the core:

Bc = Bg = Bcˆ

Inside the core we have the magnetic field intensity: Hc =Bc =

Bc ˆ

But in the gap the magnetic field intensity is Hg =Bg

0

=Bc

0

ˆ

Applying Ampere's Law:

H • dlC∫ = NI = Hcrd

0

2 −

∫ ˆ

core1 2 4 3 4

+ Hgrd2 −

2

∫ ˆ

gap1 2 4 3 4

⇒ NI =Bc ˆ rd

0

2 −

∫ ˆ +Bc

0

ˆ rd2 −

2

∫ ˆ

⇒ NI =Bcr 2 −( ) +

Bcr

0

2 − 2 −[ ]( ) ⇒ NI =Bc 2 r − lg( ) +

Bclg

0

⇒ Bc = 0 NI

0 2 r − lg( ) + lg

ˆ = Bg

Hc =Bc = 0 NI

0 2 r − lg( ) + lg

ˆ

Hg =Bc

0

ˆ =NI

0 2 r − lg( ) + lg

ˆ

Hg

Hc

=0

⇒ For Large µ ⇒ Hg is Large

⇒ For Small µ ⇒ Hg is Small

The magnetic field intensity in the core is then

The magnetic field intensity in the gap is then

The ratio of the two is

7

Magnetic Materials

There are three main categories of magnetic materials which are characterised bytheir magnetic susceptibility

• Diamagnetic χm small negative number

• Paramagnetic χm small positive number

• Ferromagnetic χm large positive number

Diamagnetic Materials

• Atoms in these materials have no net magnetic moment• Due to orbital and spin angular momentum = 0• It is possible to produce a small magnetic moment by applying a perturbation

• No permanent magnetism associated with these materials• In the presence of an external magnetic field the induced magnetic moment in adirection opposite to that of the field and hence B is reduced. Therefore χm is negative with χm ~ -1x10-5.• Examples: copper, mercury, lead, gold, silver

Paramagnetic Materials

• Atoms have a small magnetic moment since spin and orbital angular momentumdo not completely cancel• Gives rise to a weak magnetic moment• Application of an external magnetic field causes diamagnetism (small) in adirection opposite to B but the permanent magnetic moments align with the field inthe direction of B.• The alignment of the permanent magnetic moment usually dominates hence thereis a net increase in B ⇒ χm positive χm ~ +1x10-5

• Examples: Aluminium, magnesium, titanium, tungsten

8

Ferromagnetic Materials

• Most important magnetic materials as they have the highest χm

• Examples:

• iron: χm~4000

• nickel: χm~250

• Cobalt: χm~600

• Mu-Metal: χm~100,000

• Materials consist of macroscopic magnetic domains (1mm-1mm in diameter)with each having a magnetic moment.

• Dividing walls are called Domain Walls and are random in orientation

• Unmagnetised materials have random domains

• Application of external magnetic fields can align the domains

• These materials can remain magnetised after the application of an externalmagnetic field.

9

The Hysteresis Curve

• Applications of small magnetic fields to ferromagnetic material will perturb thedomain walls but they can also return to the original positions after the field isremoved• Applications of large magnetic fields can distort the domain walls so that theprocess is irreversible.• The application of large fields can induce the domains to grow to encompass allother domains ⇒ Saturation ⇒ Magnetisation of the material.• The B-H curve that describes this is called a Hysteresis Curve

op

Low Field: Applying magnetic field increases the magnetisation of the materialto o to p, removal of the field returns the magnetisation to o ie unmagnetised

qr

t

Medium Field o-p: Applying magnetic field increases the magnetisation of thematerial from o-p-q. Removal of the field leaves the material magnetised atpoint r. De-magnetisation will require an external field in the opposite directionto get to point t. The field at the point t is called the coercive field Hc

q'

s

s'

High Field: It is possible to apply a magnetic field such that the magnetisationof the material can't increase further this point is called saturation.

10

Hard and Soft Magnetic Materials

• Soft magnetic materials•Thin hysteresis loops • Easy to magnetise small magnetic fields or de-magnetise• eg: iron

• Hard magnetic Materials• Wide hysteresis loops• Difficult to magnetise or de-magnetise• Often used for permanent magnets• eg: alnicoV

Curie Point Magnetic Materials

• By heating a material the atoms move around due to thermal energy• It is possible to heat a material to a point where the thermal motion dominatesover the correlation forces that define domain structures. Heating above this pointmakes the material non-magnetic.• Cooling the material results in a random domain alignment• The temperature at which this occurs is called the Curie Point• Eg: Iron Tc=770oC, Nickel Tc=358oC, Cobalt Tc=1131oC

Lecture 12

Written by R.T. Sang 1

Lecture 12

Boundary Conditions

• In this section of the course we wish to look at the effect on a magnetic field intwo mediums.• This is important as the source of the magnetic field may not be within thematerial of interest so it is important to consider the conditions that B and Hsatisfy between two media.

Normal Components

Recall that the divergence of the magnetic flux density is zero ie ∇ •B = 0

∇ •B( )dVV∫ = B • dS

S∫

Gauss's Divergence Theroem1 2 4 4 4 4 3 4 4 4 4

= B • ˆ n dSS∫ = 0 S is the surface that encloses the volume V

P

δV1

δV2

δS

Medium 2

Medium 1

Boundary Surface

ˆ n Consider two media with thearbitrary shape:

• We take a Gaussian surface(pill box) between the twomedium around a point P withsmall volumes δV1 and δV2

• is the unit vector pointing from medium 1 to medium 2ˆ n

The total volume is: V = V1 + V2

B • dSS∫ = B2 (P) • ˆ n S + B1 (P) • − ˆ n ( ) S

+ contributions from the side walls

Hence the flux at P is

In the limit that δV → 0 then the side walls give a zero contribution

⇒ B •S∫ ˆ n dS = B2(P)• ˆ n S + B1(P) • − ˆ n ( ) S = B2 (P) • ˆ n + B1(P) • − ˆ n ( )[ ] S = 0

⇒ B •S∫ ˆ n dS = B2 (P) − B1(P).[ ] • ˆ n = 0⇒ B2 ˆ n (P) − B1ˆ n (P) = 0

Lecture 12

Written by R.T. Sang 2

Hence the change in the normal component of B is unchanged betweenmedia ie ∆B ˆ n = 0

Since H is proportional to B we have: B1 = 1H1 and B2 = 2H2

1H1ˆ n = 2H2 ˆ n

Tangential Components

We now apply Ampere's Law around the closed path C around the point P andassume that the there are free surface charges on the boundary which would giverise to Js.

Consider the side view ofour two media

HC∫ • dl = H1(P) • ˆ t ∆l + H2 (P) • −ˆ t ( )∆l +Contribution from the sides 2 & 4

= JS∆l

In the limit that the contribution to the sides → 0 hence:

H1 (P) • ˆ t ∆l + H2 (P) −ˆ t ( )∆l = JS∆l

⇒ H1(P) − H2 (P)[ ]• ˆ t ∆l = JS∆l ⇒ H1(P) − H2 (P)[ ]• ˆ t = JS

⇒ The tangential components of H are discontinuous across the boundaryinterface by an amount given by the free surface current density.

Lecture 12

Written by R.T. Sang 3

Recall normal to the boundary:

Example 1: Calculate the magnitude and direction of H across a boundary oftwo linear media of permeability µ1 and µ2 respectively as shown below.Assume that there are no surface currents.

2H2 (P) − 1H1(P)[ ]• ˆ n = 0

H1 (P) • ˆ n = H1ˆ n cos = −H1 cos

H2 (P) • ˆ n = H2ˆ n cos = H2 cos

2H2 cos − 1H1 cos = 0

⇒ 1H1 cos = 2 H2 cos (1)

The tangential components:

H1(P) − H2(P)[ ]• ˆ t = JS

= 0

H1 (P) • ˆ t = H1ˆ t sin = −H1 sin

H2 (P) • ˆ t = H2ˆ t sin = H2 sin

P

Medium 1

Medium 2

ˆ n

H1

φ H2

µ1

µ 2

θ

H2 ˆ n

H1ˆ n

H1ˆ t

H2 ˆ t

⇒ H1 sin = H2 sin (2)

Dividing (2) by (1) we can determine the angleof exit:

1

1

tan =1

2

tan

= arctan 2

1

tan

Squaring (1) and (2) and summing the result we find for |H2 | :

12H1

2 cos2 + H12 sin2 = 2

2H22 cos2 + H2

2 sin2

⇒ H2 = H11

2

2

cos2 + sin2 Exercise: Confirm this result

General Remarks

• If medium 1 is non magnetic µ1~1, and medium 2 is ferrous i.e. µ2>>1, thenφ=90o. The field runs almost parallel to the interface!• If medium 2 is magnetic and medium 2 non-magnetic i.e. µ1>> µ2 then φ=0o,hence a field originating in a magnetic material will emerge in a directionperpendicular to the interface.

H2

H1

µ1

µ 2

µ1 >> µ 2

µ 2 >> µ1

H1

H2

µ1

µ2

= arctan 2

1

tan

H2 = H11

2

2

cos2 + sin2

Lecture 12

Written by R.T. Sang 4

Inductance

Consider the two closed loops with arbitrary shape as shown with loop C1

carrying a current I1. Loop C1 will clearly create a magnetic flux density B1

which some of which will link into loop C2.

C1

I1

S1

C 2

S 2S2

We now let Φ12 be the mutual flux betweenloops C1 and C2, therefore we may write:

B1 is the the magneticFlux density due to loop 1

From the Biot-Savart law we know that B1∝I1 such that

Φ12 = B1

S2

∫ • ˆ n dS2 = 0I1

4∇

1

R

× dl'l

C1

∫

S2

∫ • ˆ n dS2 = L12I1 [ Webers ]

L12 is called the mutual inductance between the two loops and is measured inHenries (H).

Φ12 = B1

S2

∫ • ˆ n dS2 [ Webers ]

For the case when there are N2 turns of loop C2, the expression is simplymodified to

Λ12 = L12I1 L12 =Λ12

I1

[H] Mutual InductanceHence

Some of the flux produced by C1 will only be linked with C1 and not with C2, thetotal flux linkage with C1 due to I1 is then defined for the general case as

Λ11 = N1Φ11

Λ12 = N2Φ12 Total flux linkage

This is called the Self Inductance. The self inductance is defined as the magnetic

flux per unit current in the loop itself. It is obvious that

L11 =Λ11

I1

[H]⇒ Λ11 = L11I1

N1Φ11 > N1Φ12

ie the flux produced by loop 1 must be less than the mutual flux

Lecture 12

Written by R.T. Sang 5

General Remarks on Inductance

• Self inductance depends on a number of factors including the shape of theloop.• If the loop encloses a linear permeable material, then the inductance isindependent of the current in the loop.• A loop of wire in an appropriate shape is called an inductor.• An inductor has the ability to store energy in the magnetic field which isanalogous to the way that a capacitor accomplishes this in an electric field.

Procedure for the Evaluation of Self Inductance

• Choose an appropriate coordinate system for the loop

• Assume a current I flows through the loop

• Find B from either Ampheres Law or the Biot-Savart Law

• Evaluate the flux linkage each turn Φ as above by integrating B over the looparea.

• Find Λ by multiplying Φ by the number of turns, and hence evaluate the selfinductance

L11 =Λ11

I1

Lecture 12

Written by R.T. Sang 6

Example 1: Find the self inductance per unit length for a very long solenoidhaving n turns per unit length carrying current I

RZ

n turns/unit length

Recall from lecture 9 that the magnetic flux density is B = 0nIˆ z

The flux is therefore given by Φ = BS∫ • dS = B • 2 rdr

0

R

∫ ˆ z

⇒ Φ = 0nI2 rdr0

R

∫ ˆ z • ˆ z = 0nI2 rdr = 0nI0

R

∫ R2

The flux linkage per unit length is

Λ11 = nΦ = 0n2 I R2

The inductance per unit length is then

L11 =Λ11

I= 0n

2 R2

# of turns per unit length

Example 2: Find the self inductance of a toroid with N turns tightly wound ona rectangular section.

I

dS

b

a

h

drI

Recall from Lecture 9 that themagnetic flux density determinedusing Ampere's law was given by:

BC1

∫ • dl = B0

2

∫ • rd( ) ˆ = 0 ITot

B = 0 ITot

2 rˆ = 0 NI

2 rˆ

The magnetic flux is then given by

Φ = BS∫ • dS = B

S∫ • ˆ n dS

⇒ Φ = 0 NI

2 rS∫ ˆ • ˆ hdr = 0 NIh

2

1

rr=b

r = a+b

∫ dr

ˆ n = ˆ dS = hdr

⇒ Φ = 0 NIh

2ln r[ ]r= b

r= a+ b = 0NIh

2ln

a + b

b

unit vector perp to surface S through which B flowsˆ n

ˆ r

Lecture 12

Written by R.T. Sang 7

⇒ Λ = NΦ = 0 N2 Ih

2ln

a + b

b

L =ΛI

= 0 N2h

2ln

a + b

b

The flux linkage is then:

The self inductance is therefore:

Note: L is not a function of I and is proportional to N2 which is generally true forwire wound inductors.

Energy Stored in Magnetic Fields

• The inductance may be used to calculate the energy stored in a magnetic field.

• If a single loop with an initial current I=0 is connected to a generator, then anemf is induced in the loop which opposes the current change. As a result work isdone on the system to reach a current I.

• For a loop of inductance L1 it can be shown that the voltage measured acrossthe inductor during the change of current is given by:

V = L1

dI

dt

The work done is therefore:

W1 = Pdt =∫ Vidti =0

i = I1

∫

= iL1

di

dtdt

i= 0

i= I1

∫

= L1 idii = 0

i = I1

∫ =1

2L1I1

2

Lecture 12

Written by R.T. Sang 8

For a linear medium recall the self inductance is

L1 =Φ1

I1

Hence the energy stored in the magnetic field is

W1 =1

2Φ1I1

For two loops whose currents are raised from 0 to I1 and I2 respectively, thenmutual inductance will also play a role. Letting the current in loop 2 be initially setto zero, then the work done in raising loop 1 to a current I1 is given by

W =1

2L1I1

2

No work is done in loop 2 at this time as the current is zero. Loop 2 is then raisedto a current I2. The work done in loop 2 is therefore

W2 =1

2L2 I2

2

Loop 1 Loop 2

I1

I1

Loop 1 Loop 2

I2

Some of the flux due to this current links loop 1 via mutual inductance whichgives rise to an induced emf in loop 1 that must be overcome to maintain thecurrent at I1. This work is given by:

W21 = V21I1dt =∫ L21

di2

dtI1dt

i 2 =0

i2 = I2

∫ = L21I1 di2i2 =0

i2 = I2

∫ = L21I1I2

I1

Loop 1 Loop 2

I2

φ12

The total work done on the system is the sum of the energies:

W = W1 + W2 + W21 =1

2L1I1

2 +1

2L2I2

2 + L21I1I2

This can be generalised for N loops as

W =1

2Ljk I jIk

k =1

N

∑j =1

N

∑

where we have used the fact that L12=L21

Lecture 12

Written by R.T. Sang 9

Finally it should be noted that

Wm =1

2ΦI where Φ = B • ˆ n

S∫ dS = ∇ × A( )

S∫ • ˆ n dS

It can be shown that the energy is also given by

Wm =1

2A • J( )

V∫ dV

where V is the volume of the loop or the linear medium in which J exists.This volume can be extended to infinity as it does not change the resultingintegral:

Wm =1

2A • J( )

V∫ dV =

1

2A •

V∫ ∇ × H( )dV

Using a vector identity this can berewritten as

Wm =1

2H • ∇ ×A( ) − ∇ • A × H( )[ ]dV

V∫ =

1

2H •B − ∇ • A × H( )[ ]dV

V∫

Applying the divergence theorem to the second term under the integral yields

Wm =1

2H • BdV

V∫ −

1

2A × H( )

S∫ • ˆ n dS

Now as the volume increases towards ∞ then where as the surfacebounding the volume is , hence the integral over the surface is of the form

A ∝1

R and H ∝

1

R2

∝ R2

lim A × H( )S∫ • ˆ n dS ∝ lim

ˆ A R

׈ H

R2

S∫ • ˆ n R2 dR ∝ lim

dR

R∫ = 0R→∞

R→∞

R→∞

Hence Wm =1

2H • BdV

V∫ =

1

2

B

• BdV

V∫ =

1

2

B2

dV

V∫ =

1

2H 2

V∫ dV

Letting the magnetic energy density wm be given such that the volume integralover all space equals the total magnetic energy:

Wm = wm

V∫ dv =

1

2H• BdV

V∫

Thus the magnetic energy density is

wm =1

2H • B =

B2

2=

2H 2 [J/m3]

Lecture 12

Written by R.T. Sang 1

Lecture 13Magnetic Force Between Current Carrying Wires

Consider a conductor which has a length dl and cross sectional area of dS.

There are N charge carriers per unit volume are moving in the conductor(assume electrons) with a velocity v in the direction dl then the magnetic forceon the element dl is given by:

dl

Conductor

v

B

dS

dFm = qv × B = −NedSdl v × B = −NedS v dl × B = Idl × B

The total force on the conductor is then

Fm = Fm∫ = I dl × BC∫ [ N ]

= I

For two loops C1 and C2 carrying currents I1 and I2 respectively, each loop willexperience the force of the other.

C1 C2

I2I1

F21

F21 = I1 dl1 × B21

C1

∫

Hence the force on C1 due to the field of C2 is givenby:

B21 can be deduced from the Biot Savart Law: B21 = 0 I2

4

dl2 × ˆ R 21

R212

C2

∫

Substitution into the expression for the force yields: F21 = 0 I1I2

4

dl1 × dl2 × ˆ R 21( )R21

2

C2

∫C1

∫ [N]

• This is known as Amperes Law of Force between current carrying loops.

• It is analogous to the Coulomb force law between two stationary charges.

• Exactly, the same arguments can be applied for the force on loop C2:

B12 = 0 I1

4

dl1 × ˆ R 12

R122

C2

∫C1

∫ ⇒ F12 = 0 I1I2

4

dl2 × dl1 × ˆ R 12( )R12

2

C2

∫C1

∫

Lecture 12

Written by R.T. Sang 2

It is interesting and important to note that Newton's third law of force (i.e.action and reaction are equal and opposite) does NOT hold for the elementsdF12 since

dF12 ≠ −dF21

Proof:

dF21 = 0 I1I2

4

dl1 × dl2 × ˆ R 21( )R21

2 Replacing ˆ R 21 = − ˆ R 12

dF21 = − 0I1I2

4

dl1 × dl2 × ˆ R 12( )R21

2Changing the order of the cross product yields:

dF21 = 0 I1I2

4

dl2 × dl1 × ˆ R 12( )R21

2Comparing to dF12 = 0 I1I2

4

dl2 × dl1 × ˆ R 12( )R12

2

Hence dF12 = dF21

Since the forces are not equal and opposite, Newton's third law is notapplicable to the elemental forces.

What about the total force?

For F12 we can expand the vector triple cross product using the vectoridentity:

dl1 × dl2 × ˆ R 21( ) = dl2 dl1 • ˆ R 21( ) − ˆ R 21 dl1 • dl2( )

Substituting into the integral reveals:

F21 = 0 I1I2

4

dl2 dl1 • ˆ R 21( ) − ˆ R 21 dl1 • dl2( )R21

2C2

∫C1

∫

= 0I1I2

4

dl2 dl1 • ˆ R 21( )R21

2

C2

∫C1

∫ − 0I1I2

4

ˆ R 21 dl1 •dl2( )R21

2

C2

∫C1

∫

= 0I1I2

4dl2

dl1 • ˆ R 21

R212

C1

∫C2

∫ − 0I1I2

4

ˆ R 21 dl1 •dl2( )R21

2C2

∫C1

∫

Applying the relationship: ∇1

1

R21

= −

ˆ R 21

R212

Lecture 12

Written by R.T. Sang 3

⇒ F21 = − 0 I1I2

4dl2 dl1

C1

∫C2

∫ • −∇1

1

R21

− 0 I1I2

4

ˆ R 21 dl1 • dl2( )R21

2C2

∫C1

∫

= − 0I1I2

4dl2 x1

ˆ i + y1ˆ j + z1

ˆ k ( ) •x1

ˆ i +y1

ˆ j +z1

ˆ k

1

R21

+

ˆ R 21 dl1 • dl2( )R21

2

C2

∫C1

∫C1

∫C2

∫

= − 0I1I2

4dl2 d

1

R21

+

ˆ R 21 dl1 •dl2( )R21

2C2

∫C1

∫C1

∫C2

∫

The first term is zero since the line integral has symmetric limits, hence

⇒ F21 = − 0 I1I2

4

ˆ R 21 dl1 • dl2( )R21

2

C2

∫C1

∫Recall that F21 was also equal to

F21 = 0I1I2

4

dl2 × dl1 × ˆ R 12( )R12

2 = 0I1I2

4

dl1 dl2 • ˆ R 12( ) − ˆ R 12 dl2 • dl1( )R12

2

C1

∫C2

∫C1

∫C2

∫

Thus

F21 = − 0I1I2

4dl1 d

C2

∫1

R21

C1

∫ +ˆ R 12 dl2 •dl1( )

R122

C1

∫C2

∫

Using the same procedure as before we find

F21 = − 0I1I2

4

ˆ R 12 dl2 •dl1( )R12

2 =C2

∫C1

∫ 0 I1I2

4

ˆ R 21 dl1 • dl2( )R21

2

C2

∫C1

∫ = −F12

Provided that: ˆ R 12 = − ˆ R 21; R122 = R21

2 ; dl2 •dl1( ) = dl1 • dl2( )

Hence Newton's third law holds for the total force as we would expect!

Lecture 12

Written by R.T. Sang 4

Example 1: The force between two parallel conducting thin wires

Consider the two wires as shown, which carrycurrents I1 and I2 respectively. The wires are assumedto be infinitely long allowing cylindrical symmetryabout the wire to be used. Let F12 be the force onwire 2 from the current on wire 1 and B12 be themagnetic flux density at wire 2 due to wire 1. Then:

⇒ F12 = −I2ˆ k × B12

O

z

x

y

r

I1

I2

B12 direction? F12 direction?

B12

F12

F12 = I2 dl2 × B12

C2

∫

B12 can be found from Ampere's Law:

B12 • dl1

C∫ = 0 I1 = B12 • ˆ ( )

0

2

∫ rd = rB12ˆ d

0

2

∫ = 0I1 ⇒ B12 = 0I1

2 rˆ = − 0 I1

2 rˆ i

Where we have substituted since that is the direction of B12 at wire 2ˆ = −ˆ i

Substituting this result into the expression for the force onthe wire yields

F12 = − I2ˆ k × B12 = −I2

ˆ k × − 0 I1

2 rˆ i =

ˆ i ˆ j ˆ k

0 0 − I2

− 0 I1

2 r0 0

⇒ F12 = 0I1I2

2 rˆ j

The force on wire 2 due to the current inwire 1 therefore tends to push the wiresapart if the currents flow in opposingdirections. The converse is true if theyare running in the same direction wherethe force is attractive.

O

z

x

y

r

I1

I2

B12

F12

F21

B21

⇒ F12 = 0I1I2

2 rˆ j

We can immediately infer the force on wire 1 since forthe total force F21=-F12:

Lecture 12

Written by R.T. Sang 5

Magnetic Torque

• The B⊥ component (out of the page)

• Force on the loop from the varioussides tends to expand the loop which⇒ Zero Net Force

B⊥

Fm

Fm

Fm

Fm

Fm

Fm

Fm

Fm I

I

Resolving the forces on the current loop:

Consider a circular loop carrying current Iimmersed in a constant flux density B asshown. The flux vector B can be resolvedinto 2 components, one in the plane of theloop and one perpendicular to the plane:B=B⊥+B||.

z

I

I

B

B⊥

B||

y

z

x

T

b

b

φ

φ

dl1

dl2

r1

r2

O

B||

I

• B|| component case

• Force on element I dl1 at angle φproduces a force dF1 out of the paper.

• Force on element I dl2 at an angle -φproduces a force dF2 into the paper.

• From symmetry the total force on theloop is zero.

• Although Net force is zero, a torque Texists due to the B|| component of themagnetic flux density ⇒ the loop willspin.

•The differential torque around the axisshown produced at dl1 and dl2 is given by

dT = r1 × dF1 + r2 × dF2 = r1ˆ j × dF1

ˆ k + r2 −ˆ j ( ) × dF1 − ˆ k ( )

x

F-y

Fy

dl = bd= bsin dF1ˆ i + bsin dF2

ˆ i = 2bsin IdlB|| sin( )ˆ i = 2bsin Ibd B|| sin( )ˆ i = 2IB||b

2 sin2 ˆ i

Lecture 12

Written by R.T. Sang 6

The torque is therefore given by the sum of the elements dT:

T = dT∫ = 2IB||b2 sin2

0∫ d ˆ i

= 2IB||b2 1

2−

1

2sin2

0

ˆ i

= b2I( )B||ˆ i [Nm]

Recall the definition of the magnetic dipole moment: m = b2 Iˆ n

The unit vector points perpendicular to the surface as the right hand follows thedirection of the current. Hence the torque can be rewritten as

T = m × B [Nm]

• This is applicable to any shape of planar current loop• This torque is applicable to microscopic and well as macroscopic particles• Responsible for the alignment of magnetic moments in materials and hencemagnetisation of materials• This equation does not hold if B is non-uniform over the loop since there willunbalanced forces around the loop

Example: Torque on a Rectangular loop. Theloop is in the x-y plane as shown, in Cartesiancoords B is given by

1

2

3

4a

b

y

xO

B||

IB = Bx

ˆ i + Byˆ j + Bz

ˆ k

Hence the parallel and perpendicularcomponents of the field are given by:

B|| = Bxˆ i + By

ˆ j B⊥ = Bzˆ k

The force due to B⊥ is the sum of the forces due to each currentelement:

FB⊥= Sum of the forces for each side of the loop∑

= FB⊥( )1

+ FB ⊥( )2

+ FB⊥( )3

+ FB⊥( )4

= IbBz −ˆ j ( ) + IaBz − ˆ i ( ) + IbBzˆ j ( ) + IaBz

ˆ i ( ) = 0

As we would expect hence there is no Torque due to the forceson this loop since this force tries to stretch the loop.

Lecture 12

Written by R.T. Sang 7

The total force due to B|| is given by the sum of the force due to each side ofthe loop:

FB ||= FB ||

( )1+ FB ||

( )2

+ FB ||( )

3+ FB ||

( )4

where each term is given by

FB ||( )

1= I dx

−b

2

b

2

∫ ˆ i × B|| = I dx−

b

2

b

2

∫ ˆ i × Bxˆ i + By

ˆ j ( ) = IbByˆ k

Hence again we see for the total force∑ FB||= 0

1

2

3

4a

b

y

xO

B||

I

FB ||( )

3= IbBy − ˆ k ( )

FB ||( )

4= IaBx − ˆ k ( )

FB ||( )

2= I dy

−a

2

a

2

∫ −ˆ j ( ) × B|| = I dy−

a

2

a

2

∫ −ˆ j ( ) × Bxˆ i + By

ˆ j ( ) = IaBxˆ k

Although the net forces are zero; a torque T exists due to these forces:

T1 = r1 × FB ||( )

1=

a

2ˆ j × IbBy

ˆ k =a

2bIBy

ˆ i

1

2

3

4a

b

y

xO

B||

I

T3 = r3 × FB ||( )

3=

a

2−ˆ j ( ) × IbBy − ˆ k ( ) =

a

2bIBy

ˆ i

T4 = r4 × FB ||( )

4=

b

2−ˆ i ( ) × IaBx − ˆ k ( ) =

b

2aIBx −ˆ j ( )

T2 = r2 × FB ||( )

2=

b

2ˆ i × IaBx

ˆ k =b

2aIBx −ˆ j ( )

The total torque is therefore:

T = T1 + T2 + T3 + T4

= abIByˆ i − abIBx

ˆ j

= IS Byˆ i − Bx

ˆ j ( )= m × Bx

ˆ i + Byˆ j ( )

= m × Bxˆ i + By

ˆ j + Bzˆ k ( )

= m × B

m = IS − ˆ k ( ) = Iab − ˆ k ( )where

Lecture 12

Written by R.T. Sang 8

The Principle of Virtual DisplacementAlternative method for determination of magnetic forces and torques.

Constant Flux Conditions

Let a loop be immersed in a magnetic field such that the flux passing throughthe loop is φ. If the loop is moved through a small displacement dl (a virtualdisplacement) such that there is no change in flux linkage (i.e. no induced emfby the source), the work done by the system on the magnetic field results in adecrease in magnetic energy which is given by:

FΦ = −∇Wm [N]

FΦ represents the force under constant flux conditions.

If the circuit is constrained to rotate about an axis (say the z axis) by anamount dφ, then the a torque results on the loop which in terms of thechange in energy is given by

TΦ( )z

= −Wm [Nm]

where (Tφ)z represents the torque around the z-axis under constant fluxconditions.

Constant Current Conditions

For the case where the source is connected to a constant current supplywhich counteracts any induced emf from the change in φ due to thedisplacement dl, the force on the loop will be then given by:

FI = +∇Wm [N]

where FI represents the force under constant current conditions.

Once again if the loop is constrained to rotate around an axis (say the zaxis) by an amount dφ, then the a torque results on the loop which in termsof the change in energy is given by

TI( )z

= +Wm [N]

where (TΙ)z represents the torque around the z-axis under constantcurrent conditions.

Lecture 12

Written by R.T. Sang 9

Forces and Torques in Terms of Mutual Inductance

Recall from the last lecture that for two circuitscarrying currents I1 and I2 of inductances L1 and L2

coupled via a flux linkage giving rise to a mutualinductance L12 then the magnetic energy is givenby:

I1

Loop 1 Loop 2

I2

φ12

W21 =1

2L1I1

2 + L21I1I2 +1

2L2I2

2

Hence if one of the circuits is given a virtual displacement, while thecurrents are held constant only the mutual inductance L12 will change,therefore the force is

FI = +∇Wm = I1I2∇L21 [N]

If one of the loops is constrained to rotate then there will be a torque:

TI( )z

= +Wm = I1I2

L21 [Nm]

Example: Force between two Magnetic Dipoles:Consider the two sets of coils that carry currents I1

and I2 having the number of turns N1 and N2

respectively separated by a distance d along the zaxis.

x

z

yO

R

I2Dipole 2

Dipole 1

I1

P

d

N1

N 2

θ

b1

b2

Let z=d+|dl| >>b1,b2 where dl will be the virtualdisplacement distance. Recall from lecture 10 thatthe for a single loop magnetic dipole that the vectorpotential is given by:

A ≅ 0 Ib2 sin

4R2ˆ

At the point P in this example which consists of N1 loops then the vectorpotential is

A12 ≅ 0 N1I1b12 sin

4R2ˆ where sin =

b2

R

= 0N1I1b12

4R2

b2

R

ˆ

= 0N1I1b12b2

4R3ˆ = 0N1I1b1

2b2

4 z2 + b22( )

3

2

ˆ

Lecture 12

Written by R.T. Sang 10

The flux coupling circuit 1 to circuit 2 is given by:

Φ12 = B • dS2

S2

∫ = ∇ ×A( )• dS2

S2

∫ = AC2

∫ • dl2

Therefore

Φ12 = AC2

∫ • dl2 = 0 N1I1b12b2

4 z 2 + b22( )

3

2

ˆ C2

∫ • b2d ˆ

= 0N1I1b12b2

2

4 z2 + b22( )

3

2

d0

2

∫ = 0 N1I1 b12( )b2

2

2 z2 + b22( )

3

2

= 0 m1 b22

2 z2 + b22( )

3

2

The mutual inductance is then

L12 =N2Φ12

I1

= 0 m1 N2b22

2I1 z2 + b22( )

3

2

The force on the dipole using the principle of virtual displacement due withdisplacement dl=dz is given by

FI = I1I2∇L21 = I1I2

L21

z

z =d

ˆ k

= 0I1I2 m1 N2b22

2I1

ˆ k z2 + b2

2( )− 3

2

z

z =d

= −3d 0 m1 N2 I2 b2

2( )2 d2 + b2

2( )5

2

ˆ k = −3d 0 m1 m2

2 d2 + b22( )

5

2

ˆ k

We have assumed that d>>b2 as such thus the expression for theforce reduces to

FI ≅ −3 0 m1 m2

2 d4ˆ k

The negative sign indicates that the force is attractive between thedipoles, when the current is in the same direction.

It should also be noted that the force drops off rather rapidly as afactor of R-4 where R is the distance between the dipoles.

Lecture 12

Written by R.T. Sang 1

Lecture 14Time Varying Electric and Magnetic Fields

Summary of the Electro-magnetostatic Models

Electrostatic Model Magnetostatic Model∇ ×E = 0 ∇ •B = 0∇ •D = ρ ∇ ×H = JD = εE

H =1

µB

The equations of Electrostatic and Magnetostatic models can betreated independently.

Even if an electric field induces a steady current which creates a magneticfield the independence still holds.

The equations are no longer independent when the fields are time varying

Faraday's Law

In 1831 Faraday experimentally discovered that a time rate of change ofmagnetic field induced a current in a loop (an hence an electric field). Thisdiscovery can be written as Faraday's Law of Electromagnetic Induction:

∇ ×E = −Bt

Taking the surface integral over this expression allows us to determine theintegral form of Faraday's law:

∇ ×E( )S∫ • ˆ n dS = −

Bt

• ˆ n dSS∫

LHS = ∇ ×E( )S∫ • ˆ n dS = E • dl

C∫

Stokes's Theorem 1 2 4 4 4 4 3 4 4 4 4

= V

RHS = − Bt

• ˆ n dSS∫ = −

tB• ˆ n dS

S∫

=Φ1 2 4 3 4

= − Φt

V = −Φt

[Volts]

Equating both yields:

Remarks

• It has been assumed that the loopof contour C binding the surface S isstationary.

• V is known as the induced emf inthe loop and Φ is the magnetic fluxthrough the loop.

• Note that the emf induces a currentin a direction to oppose the changingflux (hence the negative sign) knownas Lenz's Law.

Lecture 12

Written by R.T. Sang 2

Example: A circular loop of N turns emersed in a magnetic field with

B =B0

rsin( t) ˆ k

Let the loop be positioned at the originand have radius a as shown. Then the fluxis

Φ = B • ˆ n dSS∫ =

B0

rsin( t)

r = 0

a

∫=0

2

∫ ˆ k • ˆ k rdrd

= B0 sin( t) d drr = 0

r = a

∫=0

= 2

∫ = 2 aB0 sin( t)

Hence the induced emf for N turns according to Faraday's law is:

V = −NΦt

= −Nt

2 aB0 sin( t)( )= −2 NaB0 cos( t)

= 2 NaB0 sin t −2

The emf voltage is out of time phase with the magnetic flux by 90o

Application of Faraday's Law of Induction: Transformers

Transformers are used to convert AC voltages, AC currents and impedancesbetween two parts of a circuit by making direct use of Faraday's Law ofInduction.

Consider the transformer as shown which is composed of a ferrous core ofsquare cross section area S and permeability µ.

It can be shown that for a magnetic circuit such as the one above that:

N j I j = RkΦk

k∑

j∑ where Rk is called the relutance of the magnetic

circuit and is given by R k =

l

kS

l is the length of the circuit, S is the cross sectionalarea and k is the permeability of the core

Around a closed path in a magnetic circuit the algebraic sum of ampere-turns is equal to the algebraic sum of the product of the reluctances andthe fluxes.

Lecture 12

Written by R.T. Sang 3

Applying this expression to the transformer yields:

N1I1(t) − N2I2 (t) = RΦ =

l

SΦ

(a) Consider the case of an Ideal Transformer: → ∞

N1I1(t) − N2I2 (t) = 0 ⇒I1 (t)

I2 (t)=

N2

N1

Also we have from Faraday's Law: V1(t) = −N1

Φ(t)

t,V2 (t) = −N2

Φ(t)

t

⇒V1(t)

V2(t)=

N1

N2

For a load impedance ZL, the effective load asseen by the AC supply is given by:

Z1( )eff

=V1(t)

I1(t)=

V2 (t)N1

N2

I2(t)N2

N1

=N1

N2

2V2(t)

I2 (t)=

N1

N2

2

ZL

(b) Real Transformer: In this case the flux linking the primary and secondarycircuits is given by

Λ1 = N1Φ(t) =SN1

lN1I1(t ) − N2I2 (t)( ) =

S

lN1

2 I1(t) − N1N2 I2 (t)( )

Λ2 = N2Φ(t) =SN2

lN1I1(t) − N2 I2( t)( ) =

S

lN1N2 I1( t) − N2

2 I2(t)( )

Applying the magnetic circuit equation to determine the flux, hence theflux linkage is:

Applying Faraday's law enables the induced emf to be determined:

V1(t) = − Λ1(t)t

= Sl

N12 dI1(t)

dt− N1N2

dI2( t)dt

= L1

dI1(t)dt

− L12

dI2( t)dt

V2 (t) = −Λ2 (t)

t=

S

lN1N2

dI1(t)

dt− N2

2 dI2 (t)

dt

= L12

dI1 (t)

dt− L2

dI2 (t)

dt

L1 = Sl

N12

L2 =S

lN2

2

L12 =S

lN1 N2

where

L1 and L2 are the self inductances of theprimary and secondary circuits

L12 is the mutual inductance of the primaryand secondary circuits.

This results is the same as for long solenoidscoupled together.

Lecture 12

Written by R.T. Sang 4

For an ideal transformer (no flux leakage beyond the transformer)thenL12 = L1L2

For a real transformer, some flux will leak out and as such the mutualinductance is

L12 = k L1L2

where k<1 is called the coefficient of coupling.

Further conditions that effect the operation of real transformers are:

• Leakage flux• Finite resistance of the windings• Capacitance of the windings (mainly high frequency circuits)• Hysteresis in the magnetic core• Eddy currents in the core due to finite conductivity⇒ Power loss due toheating• Nonlinearity of the magnetic core (saturation effects)

Although many of these effects can be addressed, an accurate descriptionof a transformer's characteristics can be very complex due to these effects.

Moving Conductor in a Static Field

For a conductor moving with a velocity v in a static magnetic field there is aforce on the charged particles given by Fm = qv × B

For an observer moving with this conductor, this will appear as a force Fm

moving the charges which can be interpreted as due to an inducedElectric Field E' along the conductor.

Hence the potential difference measure will be:

V21 = E' •dl1

2

∫ = Fm

q• dl

1

2

∫ = v × B( ) • dl1

2

∫ [Volts]

If the moving conductor part of a closed circuit C the flux cutting emfgenerated will be given by:

V ' = v × B( )• dlC∫ [V]

Lecture 12

Written by R.T. Sang 5

Example 2: The Faraday Disk Generator

V

1

2

3 bφ4

5

6

ωB = B0

ˆ k Consider the spinning conducting disk ofradius b as shown which is immersed in auniform magnetic field with . A voltmeter measures the emf induced between thedisk edge and centre when the disk spins withangular velocity ω.

B = B0ˆ k

For the closed loop 123456 the only portion that moves in the uniformmagnetic field is contribution 3 to 5, hence

V = v × B( )• dlC∫ = v × B( ) •

1→6

6→1

∫ dl = 0 + v × B( )•3

5

∫ dl + 0

= r ˆ × B0ˆ k ( )•

3

5

∫ dl = r B0ˆ r •

3

5

∫ dl = r B0ˆ r •

3

5

∫ dr ˆ r ( )

= B0 rr =b

r =0

∫ dr

∴V = −B0b

2

2 (Volts)

Moving Conductor in a Time Varying Field

This is the most general case for conductors in magnetic fields.

A conductor moving in an electric and magnetic field E and B then anobserver moving with the conductor will see an apparent electric field E'given by:

E'= E + v × B

Hence for a loop of contour C and surface S moving with velocity v:

E'C∫ • dl = E

C∫ • dl + v × B( ) • dl

C∫ = −

Bt

•S∫ ˆ n dS + v × B( ) • dl

C∫ [V]

General form of Faraday's Law for moving conductors in timevarying magnetic fields.

• The line integral on the left represents the emf induced in the movingframe of reference

• The first term on the right is the emf due to the variation of B and thesecond term is due to the motion emf of the conductor moving in B

Lecture 12

Written by R.T. Sang 6

Consider a circuit with contour C evolvingfrom C1 at time t to C2 at time t+∆timmersed in a time varying B field asshown in the figure above. The motion ofthe circuit may include distortions,translations and rotations. The time rate ofchange of magnetic flux is then given by:

v∆tdl

dS3

S1

S2

C2

C1

dΦdt

=d

dtB(t)

S( t)∫ • ˆ n dS = lim

∆t →0

1

∆tB(t + ∆t)

S( t +∆ t )∫ • ˆ n dS − B(t)

S(t )∫ • ˆ n dS

= lim∆t →0

1

∆tB(t + ∆t)

S2

∫ • ˆ n 2 dS2 − B(t)S1

∫ • ˆ n 1dS1

B(t)can be expanded using a Taylor's series: B(t + ∆t) = B(t) +B(t)

t∆t + H.O.T.

Substituting back into our expression for the time varying flux yields:

dΦdt

=d

dtB(t)

S( t)∫ • ˆ n dS = lim

∆t →0

1

∆t

B(t)S2

∫ • ˆ n 2 dS2 + B(t)t

∆tS2

∫ • ˆ n 2dS2 − B(t)S1

∫ • ˆ n 1dS1

+H.O.T.

now in the limit ∆t→0, then as such our expression may berewritten as

lim∆ t→ 0

S2( ) → S

d

dtB(t)

S( t)∫ • ˆ n dS ≅

B(t)

tS∫ • ˆ n dS + lim

∆ t→ 0

1

∆tB(t)

S2

∫ • ˆ n 2dS2 − B(t)S1

∫ • ˆ n 1dS1

In going from C1 to C2 the circuit covers the region that is bounded by thesurfaces S1, S2 and S3 (the surface S3 is the area swept out by the contour intime ∆t. an element of the side surface is . Applying thedivergence theorem to our expression reveals:

dS3 = dl × v∆t

∇ •BdVV∫ = 0 = B

S1

∫ (t) • − ˆ n 1dS1( ) + B(t)S2

∫ • ˆ n 2dS2 + B(t)S3

∫ • ˆ n 3dS3

∴0 = B(t)S2

∫ • ˆ n 2dS2 − BS1

∫ (t) • ˆ n 1dS1 + B(t)C∫ • dl × v∆t( )

∴0 = B(t)S2

∫ • ˆ n 2dS2 − BS1

∫ (t) • ˆ n 1dS1 + ∆t B(t)C∫ • dl × v( )

∴0 = B(t)S2

∫ • ˆ n 2dS2 − BS1

∫ (t) • ˆ n 1dS1 + ∆t v × B(t)( )• dlC∫ (V.I.)

⇒ B(t)S2

∫ • ˆ n 2dS2 − BS1

∫ (t) • ˆ n 1dS1 = − ∆t v × B( t)( )• dlC∫

Lecture 12

Written by R.T. Sang 7

Hence

dΦdt

=d

dtB(t)

S( t)∫ • ˆ n dS ≅

B(t)

tS∫ • ˆ n dS + lim

∆ t→ 0

1

∆tB(t)

S2

∫ • ˆ n 2dS2 − B(t)S1

∫ • ˆ n 1dS1

∴dΦdt

=B(t)

tS∫ • ˆ n dS + lim

∆t →0

1

∆t− ∆t v × B(t)( )• dl

C∫

∴ dΦdt

= B(t)

tS∫ • ˆ n dS − v × B(t)( ) • dl

C∫ = − E' •dl

C∫

Letting be the induced emf in the moving circuit yields:V ' = E' •dlC∫

V ' = −dΦdt

[V]

Faraday's Law of Induction applies equally well to a moving circuitas it does to a stationary circuit.

Example 3: A rectangular loop in a time varying B field

z

x

y

a

b

ˆ n

α

α

dl

ω

V

B = B0 sin ωt( )ˆ j

B = B0 sin ωt( )ˆ j α

ω

ˆ n

α = ωt

y

View from the +x direction

v

v

z

1

2

3

4

Let a rectangularconducting loop along and b wide beimmersed in a Bfield whereB = B0 sin t( )ˆ j

If the loop is at rest (v=0) at an angle α wrt z-axis then the induced emf in theloop Vrest is

Vrest = − B(t)t

S∫ • ˆ n dS + v × B(t)( ) • dl

C∫ = − B(t)

tS∫ • ˆ n dS

Vrest = −t

S∫ B0 sin t( )ˆ j • ˆ n dS = −

tS∫ B0 sin t( ) ˆ j ˆ n

=1

cos dS

Vrest = − B0 cos t( )cos dSS∫

= ab

∴Vrest = − (ab)B0 cos t( )cos

Lecture 12

Written by R.T. Sang 8

Vmove = −B( t)

tS∫ • ˆ n dS + v × B(t)( ) • dl

C∫

Vmove = − (ab)B0 cos t( )cos + v × B(t)( )• dlC∫

We now let the loop rotate with constant angular velocity , and letα=0 at t=0. Hence and the induced emf in the loop is given by:

=d

dt= t

⇒ Vmove = − (ab)B0 cos t( )cos + v × B( ) • dl12

1

2

∫ + v × B( ) • dl23

2

3

∫

+ v × B( ) • dl34

3

4

∫ + v × B( ) •dl41

4

1

∫Consider each integral in turn:

v × B( ) •dl12

1

2

∫ = B0 sin t( ) b

2

ˆ n × ˆ j

=sin ˆ i

• dx

1

2

∫ ˆ i ( ) =b

2B0 sin t( )sin dx

0

a

∫

v × B( ) •dl23

2

3

∫ = 0 Since velocity component = 0

v × B( ) •dl34

3

4

∫ = B0 sin t( ) b

2

− ˆ n ( ) × ˆ j

=− sin ˆ i 1 2 4 3 4

• dx3

4

∫ ˆ i ( ) = −b

2B0 sin t( )sin dx

a

0

∫

v × B( ) •dl41

4

1

∫ = 0 Since velocity component = 0

Therefore

Vmove = − (ab)B0 cos t( )cos +b

2B0 sin t( )sin dx

0

a

∫ −b

2B0 sin t( )sin dx

a

0

∫⇒ Vmove = − (ab)B0 cos t( )cos + ab B0 sin t( )sin

⇒ Vmove = − B0(ab)cos t( )cos + B0(ab)sin t( )sin

Recall that , hence= t

Vmove = − B0 (ab)cos2 t( ) + B0(ab)sin 2 t( )⇒ Vmove = B0 (ab) sin2 t( ) − cos2 t( )[ ]∴Vmove = − B0 (ab) co s 2 t( ) (Volts)

Lecture 12

Written by R.T. Sang 9

V ' = −dΦdt

The induced emf can be calculated more easily, directly from the flux since :

Φ = B(t) •S∫ ˆ n dS

= B0 sin t( )ˆ j • ˆ n cosdS

S∫

= B0 sin t( )cos dx + dz0

b

∫0

a

∫

∴Φ = (ab)B0 sin t( )cos = (ab)B0 sin t( )cos t( )

Vmove = −dΦdt

= −d

dt(ab)B0 sin t( )cos t( )

= −(ab)B0 cos2 t( ) − sin2 t( )[ ]∴Vmove = − B0 (ab) c o s2 t( ) (Volts)

Hence

This is the same results as before that both methods are equallyviable.

z

x

y

a

b

ˆ n

α

α

dl

ω

V

B = B0 sin ωt( )ˆ j

B = B0 sin ωt( )ˆ j α

ω

ˆ n

α = ωt

y

View from the +x direction

v

v

z

1

2

3

4

Lecture 12

Written by R.T. Sang 1

Lecture 15

Maxwell's Equations

Electrodynamic Model Magnetostatic Model

∇ × E = −∂B∂t

∇ •B = 0

∇ •D = ρ ∇ × H = JD = εE

H =1µ

B

The equations for electromagnetostatics given in the last lecture can now bewritten to include magnetic induction. Hence we have

A complete set of equations must also satisfy charge conservation :

∇ •J +t

= 0

J is the free current density and ρ the free charge density. One equation inthe above table does not satisfy this expression and as such an additionalterm must be added.

Proof: Consider the expression for the magnetic field intensity:

∇ ×H = J

Taking the divergence of this expression yields

∇ • ∇ ×H( )= 0 Vector Identity1 2 4 3 4 = ∇ •J

⇒ ∇• J = 0

∇ •J +t

= 0 ⇒ ∇• J = −t

But according to the equation for charge continuity:

Hence to make the above equation satisfy the equation of charge continuitywe need to add a term to it:

⇒ ∇× H = J +Dt

Lecture 12

Written by R.T. Sang 2

This set of equations are called Maxwell's equations. It is straightforward to show that these equations then satisfy the conservation ofcharge:

∇ • ∇ ×H( )

= 0 V.I.1 2 4 3 4 = ∇• J +

Dt

⇒ 0 = ∇ • J +

Dt

⇒ 0 = ∇ •J + ∇ • Dt

⇒ 0 = ∇ •J + ∇ •D( )=6 7 8

t

∴0 = ∇• J +t

The full set of Electromagnetic Equations

Electrodynamic Model Magnetodynamic Model

∇ ×E = −Bt

∇ •B = 0

∇ •D = ρ∇ ×H = J +

Dt

D = εEH =

1µ

B

+Dt

The term has the same units as the current density (A/m2) and is calledthe Displacement Current Density after Maxwell.

Dt

It should be noted that ρ is the volume density of free charges and J is thedensity of free currents, which can include currents of the form(these are called convection currents) and conduction currents .

J v = vJ c = E

Maxwell's equations are Vector equations. In 3-D spacethere are 12 partial differential equations that need to besolved simultaneously.

Lecture 12

Written by R.T. Sang 3

Integral form of Maxwell's equationsDifferential

FormIntegral Form Significance

∇ × E = −∂B∂t

E • dlC∫ = −

∂Φ∂t

∇ •D = ρ DS∫ • ˆ n dS = Q

∇ × H = J +∂D∂t

H • dlC∫ = I +

∂D∂t

S∫ • ˆ n dS

∇ •B = 0 BS∫ • ˆ n dS = 0

D = εEH =

1µ

B

Faraday's Law

Ampere's Law

Gauss's Law

No MagneticMonopoles

Coupling Equations

The integral form of these equations is obtained using Gauss'sDivergence theorem and Stokes theorem (this is left as an exercise).

Note that in all cases, Maxwell's equations reduce to theelectromagnetostatic equations when the time dependenceis removed.

Potential Functions

Recall that , hence in Maxwell's equation:∇ •B = 0

∇ ×E = −Bt

= −t

∇ ×A( )

⇒ ∇× E + At

= 0

Now for a scalar function β we have the identity so byanalogy we deduce for the expression above that

∇ × ∇ = 0

E +At

= −∇V

where we have chosen the scalar potential function to be consistent withthe electrostatic model. Hence

E = −∇V −At

[V / m]

Lecture 12

Written by R.T. Sang 4

Consider the Maxwell equation: ∇ ×H = J +Dt

Recall that H =1

B =1

∇ × A( )

∇ × ∇ ×A( ) = J +Dt

⇒ ∇× ∇ ×A( ) = J +Et

⇒ ∇× ∇ ×A( ) = J +t

−∇V −At

⇒ ∇ ∇ •A( ) − ∇2A

From V.I.1 2 4 4 3 4 4 = J +

t−∇V −

At

⇒ ∇2 A −2At2 = − J + ∇

V

t+ ∇ •A

=0 Lorentz Gauge1 2 4 4 4 3 4 4 4

where in the last line wehave used the LorentzGauge which is defined for

and can be shown to be avalid solution of Maxwell'sequations.

V

t+ ∇ •A = 0

⇒ ∇2 A −2At2 = − J

This is the non-homogeneous wave equation forA it can also be derived for the scalar potential Vin the Lorentz gauge.

The expression for the two potential functions are given by

∇2V −2V

t 2= −

∇2A −2At 2 = − J

The Lorentz Gauge Decouples the Wave Equations for V and A.

Lecture 12

Written by R.T. Sang 5

Boundary Conditions for Electromagnetics

• It is essential to understand boundary conditions in EM theory since thesolution of the partial D.E.'s defined by Maxwell's equations containconstants evaluated at any interface.

• It is usual to use the integral form of Maxwell's equations close to theboundary to evaluate these conditions recall the derivation of theboundary conditions for the magnetostatics case (see Lecture 12).

P

δV1

δV2

δS

Medium 2

Medium 1

Boundary Surface

ˆ n

Normal Components Tangential Components

E1t = E2t [V / m]

ˆ n 2 × H1 − H2( ) = Js [A / m]

ˆ n 2 • D1 − D2( ) = s [C/m2 ]

B1n = B2n [T]

For a boundary between media 1 and 2 letting be the outward normalto the boundary from 2 to 1:

ˆ n 2

The General Case

• The tangential component of E is continuous across the boundary• The tangential component of B is discontinuous across the boundary byan amount given by the surface current density• The normal component of D is discontinuous across the boundary by anamount given by the surface charge density• The normal component of B is continuous across the boundary

Lecture 12

Written by R.T. Sang 6

Boundary Conditions for Lossless Linear Media

Assume permittivity ε1 and ε2 with permeability µ1 and µ2. Since there is noloss in the media, there can be no conduction and as such σ1 and σ2 =0. Thisimplies that ρs=0 and Js=0. Hence:

E1t = E2t ⇒ D1t

D2t

= 1

2

H1t = H2t ⇒B1t

B2t

= 1

2

D1n = E2 n ⇒D1n

D2n

= 2

1

B1n = B2n ⇒H1n

H2n

= 2

1

• The tangential component of E is continuous across the boundary• The tangential component of H is continuous across the boundary• The normal component of D is continuous across the boundary• The normal component of B is continuous across the boundary

Boundary Conditions for a Dielectric to Perfect Conductor Interface

Assume that medium 1 is a lossless dielectric and medium 2 a perfectconductor. Hence the conductivity in the medium 1 is σ1=0 whereas inmedium 2 σ2=∞ and a surface current Js and surface charge density ρs canoccur.

In medium 2, σ2=∞ implies that E2 and D2 within the medium must be zerosince Ohm's Law would imply infinite current density which isimpossible.

For time varying fields, Maxwell's equations couples the E and B fieldstogether. Hence if E2=0 and D2=0 then B2=0 and H2=0

Lecture 12

Written by R.T. Sang 7

E1t = E2t = 0

ˆ n 2 × H1 − 0( ) = Js ⇒ ˆ n 2 × H1 = J s ⇒ ˆ n 2 × B1 = 1Js ⇒ B1t = 1 J s

ˆ n 2 • D1 − 0( ) = s ⇒ ˆ n 2 • D1 = s ⇒ ˆ n 2 • E1 = s

1

⇒ E1n = s

1

B1n = B2n = 0

• The tangential component of E is zero on both sides of the boundary• The tangential component of H is zero in the conductor and isgiven by the surface current on the boundary• The normal component of D is zero in the conductor and is givenby the surface charge density on the boundary• The normal component of B is zero on both sides of the boundary

For a perfect conductor, the electric field must beperpendicular to the boundary surface.

This is usually a good approximation to make for many other non-perfectconductors such as gold, silver, copper, aluminium etc, where theconductivity is high. Only superconductors approach infinite conductivityin reality.

Potential Wave Equation Solutions

Recall that in the Lorentz Gauge the two equations for the potentialfunctions were:

∇2V −2V

t 2= −

∇2A −2At 2 = − J

z

x

y

V'

∆V'

V(R), A(R)

R

R'

Charge Distribution ρV

Ri

For a distribution of charges as shown,the solution to the time independentexpressions are given by Poisson's Scalarand Vector equations:

∇2V = − v (Ri) ⇒ V(R) =1

4v ' (Ri)

R'V '∫ dV'

∇2A = − J(Ri ) ⇒ A(R) =4

J(Ri )

R'V '∫ dV'

Lecture 12

Written by R.T. Sang 8

For a time varying distribution ofcharges it is necessary to accountfor the finite propagation time thatany variation of ρ and J at thesource makes to the field potentialsat the field point. This time isdetermined by R' and the velocityof propagation . The solution tothe time dependent equations isgiven by

v p

z

x

y

V'

∆V'

V(R), A(R)

R

R'

Charge Distribution ρV

Ri

V(R,t' +R'

vp

) =1

4v' (R i ,t ')

R'V'∫ dV ' ⇒ V(R,t) ≅

1

4

v' (Ri ,t −R

vp

)

RV '∫ dV ' [V / m]

A(R,t' +R'

v p

) =4

J(Ri ,t' )

R'V '∫ dV ' ⇒ A(R,t) ≅

4

J(Ri ,t −R

vp

)

RV '∫ dV ' [A / m]

In these expressions R' has been replaced by R for the far field point since thisis where the propagation time becomes of significance.

• The first equation indicates that the scalar potential at a distance R from thesource at time t depends on the value of the charge density at an earlier time

i.e. it takes time for the effect of ρ to be felt at a distance R.

• As a result V(R,t) is called the retarded scalar potential. Similarly A(R,t) iscalled the retarded vector potential.

• It takes time for electromagnetic waves to travel and for the effects oftime varying charges and currents to be felt at distant points.

t −R

vp

R

v p

z

x

y

V'

∆V'

V(R), A(R)

R

R'

Charge Distribution ρV

Ri

Lecture 12

Written by R.T. Sang 9

Source Free Solutions to Maxwell's Equations

• For a solution to Maxwell's equations in a region of space far from anysources of charge or current, we can make the assumption that ρ=0 andJ=0. In this case we are interested in how the electromagnetic wavepropagates in a region such as this rather than knowing about its origin.

• Assume that the propagation occurs in a non-conducting, linear, isotropic,homogeneous medium with permittivity ε and permeability µ. ThenMaxwell's equations are given by:

∇ ×E = −Bt

∇ •D = 0 ⇒ ∇• E = 0

∇ ×H =Dt

=Et

∇ •B = 0 ⇒ ∇• H = 0

∇ × ∇ ×E = −∇ ×B( )

t

Taking the curl of the differential form of Faraday's Law

⇒ ∇ × ∇ ×E = −∇ ×H( )

t

⇒ ∇ × ∇ ×E = −t

Et

⇒ ∇ ∇ •E( ) − ∇2EV . I .

1 2 4 4 3 4 4 = −2 Et2

⇒ −∇2E = −2Et2

⇒ ∇2 E −2Et2 = 0

⇒ ∇2 E −1

vp2

2Et2 = 0 vp =

1 is the velocity of the electric fieldpropagation through the medium.

Similarly you can show ∇2H −1

vp2

2Ht2 = 0

These expressions are the Homogeneous Vector Wave Equation Solutionsto Maxwell's Equations in a linear isotropic non-conducting medium.

Lecture 12

Written by R.T. Sang 10

DifferentialForm

Integral Form Significance

∇ × E = −∂B∂t

E • dlC∫ = −

∂Φ∂t

∇ •D = ρ DS∫ • ˆ n dS = Q

∇ × H = J +∂D∂t

H • dlC∫ = I +

∂D∂t

S∫ • ˆ n dS

∇ •B = 0 BS∫ • ˆ n dS = 0

D = εEH =

1µ

B

Faraday's Law

Ampere's Law

Gauss's Law

No MagneticMonopoles

Coupling Equations

Lecture 16

Written by R.T. Sang 1

Lecture 16Time Harmonic Fields

• The study of the solutions of Maxwell's equations for sinusoidal(harmonic) fields ie E=E0cos(ωt).

• By Fourier's theorem all other fields can be generated as a summation ofsinusoidal components.

• Since Maxwell's equations are linear differential equations the principleof superposition then allows a solution to be found for an arbitrary field

• It is useful to introduce phasors to describe the harmonic field.

• Noting a sinusoidal quantity requires an amplitude, frequency and a phasefor a complete definition then

F(r,t) = F0 cos k •r ± t +( )

j = −1Note:

⇒ F(r,t) = Re F(r)e± j t( )We now define the phasor F(r) such that F(r) = F0e

j k •r +( )

e± j = cos( ) ± j sin( )where we have used the relation

F(r) is called a vector phasor and is independent of frequency and in generalis complex in form and gives magnitude, direction and phase information

Taking the time derivative of F(r,t):

F(r,t )

t=

tF0 cos k •r ± t +( ) = m F0 sin k• r ± t +( )

= F0 cos k• r ± t + ±2

⇒F(r,t)

t= ± j Re F(r)e± j t( )

= Re F0ej k•r + ±

2

e j t

= ± j Re F0ej k•r +( )e j t( )

Hence the time derivative is represented by the multiplication of thephasor by . It is possible to also show that the time integral isrepresented by the division of the phasor by .

j

j

Lecture 16

Written by R.T. Sang 2

Derivation of the Time Harmonic Maxwell's Equations using Phasors

Time harmonic EM fields travelling in a linear isotropic and homogeneousmedium of permittivity ε and permeability µ may be written in phasor for as

E(r,t) = E(x,y,z,t) = Re E(r)e j t( ) = Re E(x, y, z)e j t( )H(r,t) = H(x, y, z,t) = Re H(r)e j t( ) = Re H(x, y, z)e j t( )

∇ ×E(r,t) = −H(r,t)

t

Hence the first Maxwell equation may be written:

⇒ Re ∇ ×E(r)e j t( ) = Re − j H(r)e j t( )

⇒ ∇× Re E(r)e j t( ) = −Re j H(r)e j t( )

⇒ ∇× E(r) = − j H(r)

F(r,t )

t= j Re F(r)e± j t( )

It is possible to remove the condition of the Re part as the curl, div andgrad operators do not produce cross terms between real and imaginaryparts of the phasor hence the exp terms can be cancelled.

Applying the identical procedure to all of Maxwell's equations then yields thephasor form of Maxwell's Equations as summarised below:

∇ ×E(r) = − j H(r)

∇ ×H(r) = J + j E(r)

∇ •D(r) = ⇒ ∇• E(r) =

∇ •B(r) = 0 ⇒ ∇ •H(r) = 0

You should satisfy yourself that you can derive these equations!

Lecture 16

Written by R.T. Sang 3

The Scalar and Vector Potentials in Phasor Form

Applying a similar procedure as to the Maxwell Equation case:

∇2V(r,t) −2V(r,t)

t2 = −

∇2V(r) + k2V(r) = −

⇒ ∇2V (r) + 2V(r) = −

⇒ ∇2V (r) − j( )2V(r) = −

k = =vp

=2 f

f=

2⇒ vp =

1

The wavenumber is related to the speed of the EM wave (vp) and itswavelength (λ) and frequency (f) by

You can show for the vector potential ∇2A(r) + k 2A(r) = − J

k =Define the wavenumber ⇒ Non-HomogeneousHelmholtz Waveequations

Source free EM fields in Simple Media

Consider a homogeneous non-conducting medium (ie σ=0) ofpermittivity ε and permeability µ with no sources ⇒ ρ=0 and J=0.

Maxwell's equations in phasor form for these conditions are

∇ ×E(r) = − j H(r)

∇ ×H(r) = + j E(r)

∇ •E(r) = 0

∇ •H(r) = 0

The homogeneous vector Helmholtz wave equations can be

derived as we have done for Maxwell's equations to give

∇2E(r) + k2E(r) = 0

∇2H(r) + k2H(r) = 0

You should convince yourself that you canderive these expressions from Maxwell'sEquations

Lecture 16

Written by R.T. Sang 4

For a medium that is conducting (σ≠0) a current density canflow and as such Maxwell's Eqns become

J = E

∇ ×E(r) = − j H(r) ∇ •E(r) = 0 ∇ •H(r) = 0

∇ ×H(r) = J + j E(r)

εc is defined as the complex permittivity.

It can be shown that the imaginary term of the permittivity results inpower losses of the EM wave in the medium.

We may write εc as c = R − j I

whereI =

= E(r) + j E(r) = + j( )E(r)

= j − j +

E(r) = j cE(r)

Replacing ε by εc in Maxwell's equations, then the vector Helmholtz waveequations can be derived to yield:

∇2E(r) + k2E(r) = 0

∇2H(r) + k2H(r) = 0

These equations look unaltered but now the wave number is given by

k = c

⇒ k = R − j I( )

k = R − j

These expressions allow us to define whether a medium is a goodconductor or or an insulator:

Good Conductor >>

Good Insulator <<

Lecture 16

Written by R.T. Sang 5

Example. Conduction of seawater at 1kHz and 500Thz

Seawater has a conductivity of σ~4 Siemens/metre and has a realpermittivity εR~80ε0 F/m. At 1kHz (very low frequency radio waves) weget

R = 2 f R = 160 f 0 ≈ 4.5x10−6 <<Hence at these frequenciesseawater is a good conductorforthe EM wave.

For a frequency of 500Thz (light waves with λ=600nm) then

R = 2 f R = 160 f 0 ≈ 2.2x10+6 >>

Hence for visible light seawater is a good insulator (a dielectric).

Plane Wave Solutions to Maxwell's Equations

• An important set of solutions to Maxwell's equations are the so called PlaneWave Solutions.

• A uniform plane wave can be considered as a solution with E having thesame direction, same magnitude and same phase over planes perpendicular tothe propagation direction (similarly for B).

• There are also no componets of E and B in the direction of propagation of thewave.

• E ⊥ B

Lecture 16

Written by R.T. Sang 6

Recall the Helmholtz Equations in a source free vacuum are:

∇2E(r) + k2E(r) = 0

∇2H(r) + k2H(r) = 0where k0 =

c

c is the speed of propagation of the EM wave in vacuum which is aconstant ~ 3x108 m/s.

Consider a plane wave propagating in the z direction characterised by Ex

uniform in magnitude and phase. Helmholtz's equation in E can be written inCartesian coordinates as:

∇2E(r) + k2E(r) = 0 ⇒d2Ex

dz 2 + k02 = 0⇒

2

x2 +2

y2 +2

z2 + k02

Ex

ˆ i = 0

where

since the wave is uniform over the xy plane. The solution to this secondorder liner differential equation is given by

2 Ex

x 2 =2Ex

y2 = 0

Ex (z) = E0+e− jk 0z + E0

−e+ jk0 z

This solution can be confirmed by substitution back into the de. Thisexpression reduces to for plane waves travelling in the +zdirection.

Ex (z) = E0+e− jk 0z

The full solution is found when boundary conditions are introduced intothe system.

The time dependent solution for waves travelling in the +z direction isgiven by

Ex (z, t) = Re Ex+(z)e j t[ ] = Re E0

+e− jk0z e j t[ ] = E0+ cos t − k0 z( ) (V / m)

This is just the familiar wave solution that you have seen in first yearyielding a travelling wave in the z direction, with a phase velocity given by

vp =k0

=1

0 0

= c

Lecture 16

Written by R.T. Sang 7

The H field for this wave can be determined from Maxwell's equations:

∇ ×E(r) = − j H(r)

⇒

ˆ i ˆ j ˆ k

x y zEx

+(z ) 0 0

= − j 0 Hx+ˆ i + Hy

+ˆ j + Hz+ ˆ k ( )

⇒ 0ˆ i +Ex

+(z)

zˆ j −

Ex+(z)

yˆ k = − j 0 Hx

+ˆ i + Hy+ˆ j + Hz

+ˆ k ( )

⇒Ex

+ (z)

zˆ j = − j 0 Hx

+ˆ i + Hy+ˆ j + Hz

+ ˆ k ( )Evaluating the individual components yields

Hx+ = Hz

+ = 0

Hy+ = −

1

j 0 zEx

+ (z)( )

As such only the component is non zero. We can further evaluate H:ˆ j

Hy+ = −

1

j 0 zEx

+ (z)( ) = −1

j 0 zE0

+e− jk 0z( )

⇒ Hy+ = −

− jk0

j 0

E0+e− jk0z( )

⇒ Hy+ =

k0

0

Ex+(z)

Hy+ =

1

0

Ex+(z) (A / m)

where η0 is defined as the IntrinsicImpedance of Free Space and is givenby

0 =1

k0

0 =

c

0 = 0c = 0

0 0

= 0

0

≈ 377Ω

Notice that the units are given in Ohms and hence the impedance terminology.

Hence for a plane wave with , the H field is given byE = Ex± (z,t)ˆ i

H = Hy±(z,t)ˆ j =

Ex± (z,t)

0

ˆ j

Lecture 16

Written by R.T. Sang 8

Example: Calculate E and H for a sinusoidal plane wave travelling in alossless medium with =4 0, = 0 and =0, where the frequency of the EM

wave is f=100MHz. The amplitude of the E vector is 100 V at t=0 and

z=0.25m.

Since the wave is travelling in a medium other than vacuum it is necessary tocalculate its wavelength and speed from the medium permitivity andpermeability. Hence

vp = 1 = 1

0 4 0

= 1

2 0 0

= c

2

k =vp

=2

c=

2 × 2 ×108

3 ×108 =4

3 (rad / m)

∴E(z,t ) = Ex(z,t)ˆ i = 10−4 cos 2 ×108t −4

3z +

ˆ i

Now at E(z = 0.25,t = 0) = 10−4 cos −4

3

1

4+

ˆ i = 10−4 ˆ i To be consistent then ⇒ =

3

∴E(z,t ) = 10−4 cos 2 ×108t −4

3z +

3

ˆ i

∴E(z,t ) = 10−4 cos 2 ×108t −4

3z −

1

4

i (V / m)

The magnetic field is now given by:

H(z,t) = Hy(z,t)ˆ j

⇒ H(z,t) =Ex(z,t) ˆ j = Ex(z,t)ˆ j = 2 0

0

Ex (z,t)ˆ j

∴H(z,t) ≈2 ×10−4

377cos 2 ×108 t −

4

3z −

1

4

j (A / m)

Lecture 16

Written by R.T. Sang 9

Transverse Electromagnetic Waves

We have just demonstrated that the propagation of a plane EM wave in the zdirection is characterised by both an electric field E vector with a H vectorthat also lies in the x-y plane which are given by

E = Exˆ i H = Hy

ˆ j

We can therefore write the phasor form of the wave aswhere E0 is a constant vector (in this case the x direction).

E(z) = E0e− jkz

This is a case of Transverse Electromagnetic (TEM) Wave propagation.

A more general form is given by

E(x,y,z) = E0e− jk xxe

− jk y ye− jk z z where kx

2 + ky2 + kz

2 = 2

Is required to satisfy Helmholtz's equation. If we define a wavenumbervector:

k = kxˆ i + ky

ˆ j + kzˆ k = k ˆ n = k ˆ n

Defining a radius vector:

R = xˆ i + yˆ j + z ˆ k

Then E can be re-written as

E(x,y,z) = E0e− jk x xe

− jk y ye− jk z z = E 0e

− jk•R = E0e− jk ˆ n •R (V / m)

The E0 vector lies in the plane perpendicular as shown in the figure

Plane of constant phase

ˆ n

R

z

x

y

P

ˆ n

Lecture 16

Written by R.T. Sang 10

Applying Maxwell's equations we have for H.

H(R) = −1

j∇ ×E(R)

It maybe shown that

H(R) =1

jˆ n × E(R) =

1 ˆ n × E0( )e− jk ˆ n •R

where =k

= (Ω) is the intrinsic impedance of themedium

Clearly from this result both E and H are perpendicular to the direction ofpropagation and E ⊥ H.ˆ n

This is the General form of a TEM Wave

Example: Given a magnetic TEM Field H(R) find the associated E(R).

From the above we have

H(R) =1 ˆ n × E(R)

ˆ n × H(R) =1 ˆ n × ˆ n × E(R)( )

⇒ ˆ n × H(R) = ˆ n • E(R)( ) − ˆ n • ˆ n E(R)

⇒ ˆ n × H(R) = −E(R)

∴E(R) = − ˆ n × H(R)

Taking the cross product of both sides by :ˆ n

ˆ n ⊥E(R)

Lecture 17

Written by R.T. Sang 1

Lecture 17Plane Waves in a Lossy (Conducting) Media

Recall from the last lecture that the Helmholtz equation can be written for thephasor E(r) as

where

∇2E(r) + kc2 E(r) = 0

kc = c = R − j I( ) = R 1− j I

R

⇒ kc = R 1 − jR

Let = jkc = + j

= jkc = j R 1− jR

= + j

Note that if σ=0 (perfect dielectric) then and α=0. Helmholtz'sequation can be written as

kc =

∇2E(r) − 2E(r) = 0

This has a solution for a linearly polarised plane wave propagatingin the z direction given by:

E = Exˆ i

E = Exˆ i = E0e

− ze− j zˆ i

It can be shown that both α and β are positive. Hence the term attenuatesthe wave as z increases through the medium, with α being the so calledattenuation constant. The term is a phase term with β being the phaseconstant.

e− z

e− j z

Lecture 17

Written by R.T. Sang 2

Low Loss Dielectrics

In this case conduction is poor. The material is an imperfect insulator andTherefore we may write

<< 1

= + j ≅ j R 1− j I

2 R

+1

8I

2

2 R2

⇒ = I

2 R

, = R 1 +1

8I2

2 R2

[rad / m]

The attenuation constant is approximately proportional to the frequencywhereas the phase constant varies only slightly from that of a perfectdielectric.

The intrinsic impedance of the imperfect dielectric medium is also complexsince

=c

=R

1− j I

R

− 1

2

≈R

1 + j I

2 R

[Ω]

Since Ex and Hy intensities are directly related to each other via the intrinsicimpedance it is clear they are no longer in time phase as is the case for aperfect dielectric.

Finally the phase velocity of the wave is given by:

vp = ≅1

R

1+1

8I2

2 R2

-1

[m / s]

=1

R

1− 18

I2

2 R2

[m / s]

Lecture 17

Written by R.T. Sang 3

Good Conductors

In this case we find that . Thus>> 1

= j R 1 − jR

1

2

≅ j R − jR

1

2

= j R j R

1

2

⇒ = =2

= f

= jj

=1+ j

2

The attenuation constant and the phase constant are approximately equaland increase with both and .f

The intrinsic impedance of a good conductor is complex since

=c

=j

⇒ =1 + j

2

= 1+ j( ) 2 f

= ( 1 +j) [Ω]

The electric field intensity therefore leads the magnetic field intensity by45o.

Finally the phase velocity in the conductor is given by:

vp = ≅2 f

f=

4 f=

2 [m / s]

Lecture 17

Written by R.T. Sang 4

Skin Depth of Penetration

The attenuation of the wave in a conductor is given by where .It is usual to define the skin depth as the depth where the amplitude of the wavereduces to a value e-1. Hence

e− z = f

=1

=1

f [m]

since α=β (for a conductor) then skin depth can also be written as

=1

=2

[m]

The skin depth of various conductors with waves of different frequencies aregiven below:

Material σ(S/m) f=50Hz f=1MHz f=1GHz f=500THzSilver 6.17x107 9.06mm 64µm 2.0 µm 2.9nmGold 4.10x107 11.1mm 66µm 2.1 µm 3.5nm

Copper 5.80x107 9.35mm 79µm 2.5 µm 3.0nmAl 3.54x107 12.0mm 84µm 2.7 µm 3.8nm

It should be noted that = 0 for most conductors.

Example: 10kHz plane waves in seawater with ε=80ε0 and σ=4S/m. Find wherethe field is 1% of the surface intensity and evaluate E and H. Given that at z=0the E field is given by .E = E0 cos 2 ×104 t( )ˆ i To solve this problem we need to find the appropriate quantities α,β,η,vp,δ.

To determine whether the sea water medium is a conductor or aninsulator at this frequency we look at the ratio

R

=4

2 × 104 × 80 × 8.84 × 10−12 = 22,500 >> 1 ⇒ seawater is a good conductor.

= f = 0.398 ⇒ = 0.398 [rad / m]

⇒ = ( 1 +j) = (1+ j)0.398

4≅

2

10e

j4 (Ω)

⇒ vp =2

=1.58x105 (m / s)

⇒ =1

= 2.51 (m) ∴2

= 2.51 ⇒ = 15.8m

Lecture 17

Written by R.T. Sang 5

Hence for the intensity of the field to be 1% the depth into the water z1% isdetermined by

0.01= e− z1% ⇒ 0.01 = e−0.398z1%

⇒ ln(0.01) = −0.398z1%

⇒ z1% = 11.6m

To evaluate the E and H fields we have

E(z,t) = Re(Exej tˆ i ) = Re E0e

− ze j t − z( )ˆ i ( )⇒ E(z,t) = E0e

−0.398z cos 2 ×104 t − 0.398z( )ˆ i

To calculate H we need to use the phasor notation. Hence:

= Re10

2

E0e−0.398z e− j 0.398z

ej

4

e j 2 ×10 4 tˆ j

H(z,t) = ReEx e j tˆ j

⇒ H(z,t) =10

2E0e

−0.398 z cos 2 ×104t − 0.398z −4

ˆ j

= Re10

2E0e

−0.398z ej 2 ×10 4 t −0.398 z−

4

ˆ j

The Poynting Vector

Clearly an EM wave propagating from a source (such as an antenna) mustcarry power for it to be receivable at the receiver (such as a radio or TV). ThePoynting Vector gives a description of this power transfer.

Consider Maxwell's Equations for a medium of permeability µ andpermittivity ε and conductivity σ:

∇ ×E = −H( )t

∇ ×H = E +E( )t

∇ • E × H( ) = H • ∇ × E( ) − E • ∇ ×H( )

= −H •H( )t

− E • E +

E( )t

Then:

= −1

2 tH •H( ) − E • E −

1

2 tE • E( )

= −t

1

2H 2 +

1

2E2

− E2

Lecture 17

Written by R.T. Sang 6

Gauss's Theorem

⇒ E × H( )S∫ • ˆ n dS = −

t

1

2H 2 +

1

2E 2

V∫ dV − E2

V∫ dV

∴ ∇ • E × H( )V∫ dV = −

t

1

2H 2 +

1

2E 2

V∫ dV − E2

V∫ dV

Integrating over the volume:

• The first and second term on the RHS are equivalent to the time rate ofchange of the stored energy in the field.

• The last term on the RHS is the ohmic power dissipated in the medium dueto the finite conductivity.

• Hence the RHS is the rate of decrease of energy stored in the field and mustequal the power leaving the volume V through the surface S.

• The vector therefore represents the power flow per unit area and iscalled the Poynting vector.

E × H

P = E × H [W / m2 ]

• Notice that the power is orthogonal to both E and H. Also note that if σ=0there is no ohmic losses in the system.

• In a static condition only the ohmic losses occur.

• Taking the negative of both sides of the above expression yields Poynting'sTheorem:

− E × H( )S∫ • ˆ n dS =

t

1

2H2 +

1

2E2

V∫ dV + E2

V∫ dV

= − P • ˆ n dSS∫

Poynting's Theorem:The total power flowing into a closed surface at any instant equals the sum ofthe rate of increase of stored energy in the EM field together with the ohmicpower dissipated in the total volume.

Lecture 17

Written by R.T. Sang 7

Example: The Poynting vector on the surface of a long wire radius b andconductivity σ carrying current I. For the wire let

J =I

b2ˆ k

L

dz

z

bP

E

H

P P

P

P

P PP

P

y

x

H

rφ

Top View

dS

We can determine E directlysince

J = E ⇒ E =J

=I

b2ˆ k

Recall from lecture 9 that for a longstraight wire that the B field at radiusr from the wire was

B =rI

2 b2ˆ ⇒ H(b) =

bI

2 b2ˆ =

I

2 bˆ

Hence the power is given by

P = E × H = I

b2

I

2 bˆ k × ˆ ( ) = I2

2 2b3− ˆ r ( )

⇒ P = −I2

2 2b3ˆ r

To confirm this solution in the steady state we integrate P over the surface:

− P • ˆ n dSS∫ = − −

I 2

2 2b3ˆ r

S∫ • ˆ n dS

What we would expect for a conductor!

=I2

2 2b3

0

2

∫ bd dz ˆ r • ˆ r ( )0

L

∫

=I2

2 2b3 2 bL( )

= I2 L

b2

⇒ − P • ˆ n dSS∫ = I2 R

Lecture 17

Written by R.T. Sang 8

Average Power Density in the EM Wave

• The Poynting Vector gives the instantaneous power in the EM wave.• Often it is useful to consider the average power in the EM wave, sincemany detectors only average quantities (they usually can not respond at thefrequency of the EM wave).• Using the phasor notation developed previously, we may write for a linearlypolarised plane wave in a medium of permeability µ and permittivity ε withconductivity σ:

E(z,t) = Re(Ex(z)e j tˆ i ) = Re E0e− + j( )z

e j tˆ i ( )

=E0e

− z

cos t − z − n( )ˆ j

= E0e− z cos t − z( )ˆ i

H(z,t) = Re(Ex(z)

e j tˆ j ) = ReE0e

− + j( )z

e j ne j tˆ j

= ReE0e

− + j( )z

ej t − n( )ˆ j

Hence the instantaneous Poynting Vector is given by:

P(z,t) = E(z,t) × H(z,t) = Re(E(z )e j tˆ i ) × Re(H(z)e j tˆ j )

=E0

2e−2 z

cos t − z( )cos t − z − n( )ˆ k

=E0

2e−2 z 1

2cos n( ) + cos 2 t − 2 z − n( )

ˆ k [W / m2 ]

= E0e− z cos t − z( )ˆ i × E0e

− z

cos t − z − n( )ˆ j

cos A( )cos B( ) = 12

cos A + B( ) + cos A − B( ) where A = t − z B = t − z − n

We have applied the trig identity:

Lecture 17

Written by R.T. Sang 9

Pav (z) =1

TP(z,t)dt =

20

T

∫ P(z,t)dt0

2

∫

=E0

2e−2 z

2cos n( ) +

1

4sin 4 − 2 z − n( ) − sin −2 z − n( )( )

ˆ k

=E0

2e−2 z

2cos n( )ˆ k [W / m2 ]

=E0

2e−2 z

2 2cos n( )dt +

2cos 2 t − 2 z − n( )dt

0

2

∫0

2

∫

ˆ k

=E0

2e−2 z

2cos n( ) +

2

1

2sin 2 t − 2 z − n( )

0

2

ˆ k

To calculate the average power in the EM wave we need to time integrateover one complete cycle of the wave :T = 2

sin A + B( ) = sin A( )cos B( ) + cos A( )sin B( )where A = 4 B = −2 z − n

It should be noted that in phasor notation the average power is

Pav (z) =1

2Re E(z) × H*(z)[ ] (W / m2)

E(z) × H*(z) = E0e− ze− j z ˆ i ( ) ×

E0e− ze+ j z + n( )

ˆ j

Proof:

=E0

2e−2 z e+ j n

k

∴Pav (z) =1

2ReE(z) × H*(z)[ ] = Re

E02e−2 ze+ j n

2

k

= E02e−2 z

2cos n( ) ˆ k

END OFCOURSE

Physics A SCE1301NR.T. Sang 2001

Page 1 of 10

Physical Science Context Lecture 1Gauss's Law

Gauss's law is a generalised reformulation of Coulomb's law which allows one tosimplify the determination of an electric field. It is especially useful in symmetricalsituations which greatly reduce the amount of work required to calculate an electricfield.

Central to Gauss's law is the hypothetical closed surface called a Gaussian surfacewhich can be of any shape that we choose to make it but it is usually chosen tosimplify the calculation for the problem. Quite often these surfaces are spheres orcylinders or some other symmetrical body but the important aspect as the surfacemust be closed, ie there must be a clear distinction between the inner surface and anouter surface of the object.

Gauss's law relates the electric fields at a point on a (closed) Gaussian surface to thenet charge enclosed by that surface.

FluxWe define the flux as the volume flow rate of something through an area. Forexample the flux of air particles passing though a square loop as in figure 1 below:

Figure 1

Physics A SCE1301NR.T. Sang 2001

Page 2 of 10

The number of particles passing through the loop or the flux of these particles throughthe loop depends on the velocity of the particles as well as the area of the loop. Fromfigure 1(a) the flux is simply

Φ = vA

In the general case where the trajectories of the particles make an angle θ with respectto the surface of the loop as in figure 1(b) we have

Φ = v cos( )A

This can be written more generally as a vector quantity:

Φ = v • A

A is know as an area vector with a direction perpendicular to the surface as shown infigure 1(c).

Flux of an Electric Field

We now define an arbitrary closed surface (Gaussian surface) as shown in figure 2,that is immersed in an electric field.

Figure 2

Physics A SCE1301NR.T. Sang 2001

Page 3 of 10

This surface can be broken up into a series of smaller surfaces ∆A each with area ∆Aand a vector associated with it ∆A with the direction of the vector perpendicular to thesurface. The flux of the electric field lines passing through the small surface ∆A isgiven by

Φ∆A = E• ∆A

The sign of the scalar product depends on the product directions and may be positive,negative or zero (ie E parallel to ∆A). The total flux will be just the sum of each of thefluxes due to each ∆Ai:

ΦA = E • ∆Aii

∑

In the limit, as the each of the surfaces tend to zero, then the summation becomes acontinuous integral:

ΦA = lim∆Ai → 0

E • ∆Ai =i

∑ E •d A∫ Electric Flux through a Gaussian Surface

The electric flux F is proportional to the net number of electric field lines passingthrough the surface.

The integral defining the flux is called a surface integral and the loop symbol tells usthat the integral is over the entire surface. These types of integrals will be consideredin-depth in Maths IIA and EMII next year.

Example: Calculate the electric flux though a cylindrical Gaussian surface in auniform E field as shown

The flux is given by

ΦA = E •d A∫The integral is over the entire surface which can be broken into three surfaces, two ateach end of the cylinder (a and c) and one surface that is perpendicular to the electricfield (surface b) as shown above. Hence

Physics A SCE1301NR.T. Sang 2001

Page 4 of 10

ΦA = E •d A∫ = Ea∫ • d Aa + E

b∫ • d Ab + E

c∫ • dA c

where each integral is given by

E •d Aa

a∫ = E cos 180( )dAa∫ = − EdAa = −EA∫ a

E

b∫ •d Ab = E cos 90( )dAb

E⊥dA b

1 2 4 4 3 4 4 ∫ = 0

E •d Ac

c∫ = E cos 0( )dAc∫ = EdAc = EA∫ c

Hence the flux is

⇒ Φ = −EAa + EAc

A a = Ac

1 2 4 3 4 = 0

This is what we would expect since the E field lines passing through the left cap exitat the right of the surface ⇒ Net Flux over the entire surface = Zero

Gauss's Law

This law relates the net flux of an electric field through a closed surface to the netcharge enclosed by that surface:

Φ =qenc

0

where qenc is the total net charge enclosed by the surface. Recall that the flux is relatedto the electric field by

ΦA = E •d A∫Hence Gauss's law can be rewritten as

E •d A∫ =qenc

0

Gauss's Law

This equation is valid for an electric field in a vacuum (and you can assume for ourpurposes air). E is the total electric field due to charges inside and outside theGaussian surface.

If qenc is positive then the flux is outward while the flux will flow into the surface ifthe enclosed charge is negative. The charges outside the surface do not count towardsthe flux, only those bounded by the surface, and the location of the charges within theGaussian surface doesn't matter. This law is completely equivalent to Coulomb's lawas will be shown soon.

Physics A SCE1301NR.T. Sang 2001

Page 5 of 10

Example: Consider two point charges in the figure below which have charges that areequal in magnitude but opposite in sign with four different Gaussian surfaces.

Surface S1: The E field flux due to the enclosed charges is outward since qenc>0.

Surface S2: The E field flux due to the enclosed charges is inward since qenc<0.

Surface S3: The E field flux =0 since there are no enclosed charges hence the samenumber of field lines enter the surface that exit the surface.

Surface S4: The E field flux = 0 since the net charge enclosed by the surface qenc=0.

Physics A SCE1301NR.T. Sang 2001

Page 6 of 10

Example 2: Determine the net flux through the Gaussian surface below with eachpoint charge have a charge of 3nC.

The charges outside of the surface do not contribute to the flux, only those enclosedby the surface. The net charge enclosed within the surface is:

qenc = 4 × q = 4 × 3 ×10− 9 = 12nC

Hence by Gauss's law the flux is

Φ =qenc

0

⇒ Φ =12 × 10−9

8.85 × 10−12 CC2

Nm2

UnitAnalysis

1 2 4 3 4

=1356 Nm2

C

S q

q

q q

q q

q

q

Physics A SCE1301NR.T. Sang 2001

Page 7 of 10

The Equivalence of Gauss's law and Coulomb's law

If these two laws are equivalent as stated earlier, then it should be possible to deriveone from the other. Consider a point charge as shown in the figure below which issurrounded by a spherical Gaussian surface

The flux though the surface is given by Gauss's law:

E •d A∫ =qenc

0

The symmetry of this situation dictates that the E will be perpendicular to everysurface of the sphere, hence

⇒ dA || E ⇒ E •∫ dA = EdA =∫qenc

0

E is a constant over the spherical surface and hence can be taken outside the integral:

⇒ EdA = E dA∫ =∫q

0

⇒ EdA = E4 r2 =∫q

0

Hence the electric field within the Gaussian surface is

⇒ E =q

4 0r2

Comparing this to the electric field of a point charge determined using Coulomb'slaw:

cf: Coulomb's Law

⇒ E =F

q=

qq

4 0r2q

=q

4 0r2

Gauss's Law is equivalent to Coulomb's law

Physics A SCE1301NR.T. Sang 2001

Page 8 of 10

Using Gauss's Law to Calculate Electric fields

As we saw in the last section Gauss's law can be used to determine the electric fieldwithin a Gaussian surface. As another example we will determine the electric field ofa very long uniformly charged rod with charge density λ as shown in the figurebelow:

To solve this problem we first draw an appropriate Gaussian surface about theconductor, in this case a cylindrical surface makes the most sense. The flux is thengiven by

Φ = E • d A∫ =qenc

0

⇒ Φ = E • dA 1

A1

∫ + E • d A2

A2

∫ + E • d A3

A3

∫It is obvious from the figure that

E • d A1

A1

∫ = − E • dA 2

A2

∫hence the expression for the flux reduces to

Φ = E • dA 3

A3

∫

⇒ Φ = E cos( ) dA3

A3

∫ = EA3 cos( )= 0

= EA3

The area of surface A3 is given by A3 = 2 rh substituting into the above expressionfor the flux yields

⇒ Φ = E 2 rh( )

Applying Gauss's Law: Φ =qenc

0

we need to determine the charge enclosed by the

surface which will just be the charge density times the distance h : qenc = h hence

Φ = E(2 rh) =h

0

⇒ E =0 2 r

Electric field for a line of charge infinitely long.

A1

A2

A3

Physics A SCE1301NR.T. Sang 2001

Page 9 of 10

The Electric Field of a Charged Isolated Conductor

If an excess of charge is placed on an isolated conductor, that amount of excesscharge will move entirely to the surface of the conductor. None of the excess chargewill be found within the body of the conductor. Hence

E = zero inside the conductor

This result is due to an electrostatic equilibrium from the charges repelling each other.If an E field did exist within the conductor then current would flow, which is clearlynot what we observe in an isolated conductor. The electric field must be perpendicularto the surface as if there were parallel components of the electric field then thecharges would move until there were no parallel forces on them which occur when theelectric field is entirely perpendicular to the surface.

Example: Consider a point charge in a spherical conducting shell as shown below

The arrows indicate the direction of the electric field. The positive charge in thecentre of the conducting ring induces the inner surface of the conductor to be negativeand the outer part of the ring positive. Within the ring the field lines emanate from thepositive charge to the inner face of the conducting ring. The electric field inside thewall of the conductor is zero, since the charges in the conductor are symmetricallyspread around the conductor and the E field in this region therefore cancels. Hence

Good Conductor+

-

-

-

-

-

-

+

+

+

+

+

+

Physics A SCE1301NR.T. Sang 2001

Page 10 of 10

there are no field lines in this region. Outside the conductor the electric field linesemanate perpendicular to the surface of the conductor in the radial direction as if theconductor was not there.

Example 2: E field shielding: Consider a hollow conducting box between twoparallel plates as shown. What is the electric field direction in the centre of theconducting box?

The electric field within the conducting box will be zero since the electrons willdistribute themselves on the outer surface of the box such the electric fields willcancel inside the conductor. It is easy to show this using Gauss's Law. If we draw aGaussian surface of the same shape as the inner surface of the conductor, then sincethe enclosed charge is zero, the electric field must also be zero.

This has a useful application for the shielding of equipment from electric fields and isthe reason why your car is a reasonably safe place in a storm (if you can see the roadwhile you are driving!).

+

+

+

+

+

-

-

-

-

-

E

+

+

+

-

--

1

SCE2311/SCE2611: EMIIR.T. Sang 1999

Necessary Vector Analysis Knowledge

In this part of the course vectors will be written in bold eg: A will mean the vector A.

Unit vectors will be written as ˆ φ , the norm of a unit vector =1 eg: ˆ φ =1

Vector Mag nitude:

The magnitude or the norm of a vector is defined as A

x

y

z

A

In Cartesian coords we define the vector A = Axˆ i + A y

ˆ j + Azˆ k

The norm is defined as

A = A x

2+ Ay

2+ Az

2

Dot Product : This product yields a scalar

A • B = A B cosθ

In Cartesian coords

A • B = Ax Bx + A yBy + AzBz

Cross Product: This yields another vector

A × B = ˆ n A B sin θ

where ˆ n is a unit vector that points perpendicular to the vectors A and B .

2

SCE2311/SCE2611: EMIIR.T. Sang 1999

In Cartesian coords

A × B =

ˆ i ˆ j ˆ k

A x Ay A z

Bx By Bz

= ˆ i AyBz − A zBy( ) − ˆ j A xBz − AzBx( ) + ˆ k A yBx − A xBy( )

It should also be noted that: A × B = −B × A

Gradient Operator

The gradient is defined as ∇ it is also called del and is a differential operator that is a vector.In Cartesian coords it is defined as:

∇ ≡∂∂x

ˆ i +∂∂y

ˆ j +∂∂z

ˆ k

Divergence of a Vector

The divergence of an vector is defined as the dot product between the gradient operator andthe vector. Thus the result of this operator results in a scalar. In Cartesian coords thedivergence of an arbitrary vector A is

∇• A =∂Ax

∂x+

∂A y

∂y+

∂Az

∂z

Divergence Theorem:

This allows the conversion of a volume integral to a surface integral. The volume integralof ∇• A over any volume V is related to the flux of a vector field by

∇• AdVV∫ = A •dS

S∫

Where S is the surface that bounds the volume V.

Curl of a Vector

The curl of a vector is defined as the cross product between the gradient operator and avector: The action of the curl produces a new vector. The curl describes the circulation ofthe vector field. In Cartesian coords the curl is defined as

∇× A =

ˆ i ˆ j ˆ k ∂∂x

∂∂y

∂∂z

A x Ay A z

= ˆ i ∂∂y

Az −∂∂z

Ay

− ˆ j

∂∂x

Az −∂∂z

Ax

+

ˆ k ∂∂x

A y −∂∂y

Ax

3

SCE2311/SCE2611: EMIIR.T. Sang 1999

Stokes's Theorem

This theorem allows the conversion of a surface integral to a line integral:

∇× A( ) • dSS∫ = A

C∫ •dl

where C is the contour bounding the surface S.

The Laplacian

The Laplacian is defined as

∇• ∇A( ) = ∇2A =∂2

∂x2 +∂2

∂y2 +∂2

∂z2

A

The Laplacian of a vector is a vector, while the Laplacian of a scalar is a scalar.

Important Vector Identities:

The following identities are used extensively in this section of the course:

1) ∇× A + B( ) = ∇× A + ∇ × B

2) ∇• A + B( ) = ∇ • A +∇• B

3) ∇• ∇× A( ) = 0

4) ∇× ∇f( ) = 0

5) ∇2A = ∇ ∇• A( ) − ∇ × ∇× A( )6) ∇• fA( ) = f∇• A + A •∇f

7) ∇× fA( ) = f∇× A + ∇f( ) × A

where A and B are arbitrary vectors and f is a scalar.

Orthogonal Coordinate Systems:

In this EM course it is often required that you perform line, surface and volume integrals.In each case we need to express the differential length change corresponding to the changein one of the coordinates. In some cases the change in coordinate may not be length, it maybe for example the change in an angle and as such a conversion factor is required to convertthis into a change in length. We will always use orthogonal coordinate systems whichhave coordinates that are mutually perpendicular. In this course we will use three:

1) Cartesian Coordinates2) Cylindrical Coordinates3) Spherical Coordinates

In the electromag problems encountered in this course we choose a coordinate system thatsuits the geometry of the problem.

4

SCE2311/SCE2611: EMIIR.T. Sang 1999

Let say we have some general coordinates ui (i=1,2,3) then the differential change in lengthto each of the general coordinates are:

dl i = h idui

where hi is a scaling factor and is called the metric coefficient and may be just a constantor functions of u1, u2, u3.

hi is defined as

h i =∂A

∂u i

The arbitrary differential length change with coords ui (i=1,2,3) is then defined as

dl = ˆ a u1h1du1( ) + ˆ a u2

h2du2( ) + ˆ a u3h3du3( )

The differential volume change is

dv = h1h2h3du1du2du3

For objects defined in a orthogonal coord system there will be three differential surfaces.The differential surface is defined as

dS = ˆ n ds

where ˆ n is the unit vector pointing perpendicular to the surface S.In terms of the differential length for this arbitrary coord system we have:

dS1 = dl2dl2 = h2 h3du2du3ˆ a u1

dS2 = dl1dl3 = h1h3du1du3ˆ a u2

dS3 = dl1dl2 = h1h2du1du2ˆ a u3

In terms of the generalised coordinate system the grad operator, divergence, curl andLaplacian are defined as:

∇f =1

h1

∂f

∂u1

ˆ a u1+

1

h2

∂f

∂u2

ˆ a u2+

1

h3

∂f

∂u3

ˆ a u3

∇• A =1

h1h2 h3

∂∂u1

h2h3A1( ) +∂

∂u2

h1h3A2( ) +∂

∂u3

h1h2A3( )

5

SCE2311/SCE2611: EMIIR.T. Sang 1999

∇× A =1

h1h2h3

ˆ a u1h1

ˆ a u2h2

ˆ a u3h3

∂∂u1

∂∂u2

∂∂u3

h1A1 h2A2 h3A2

∇2f =1

h1h2 h3

∂∂u1

h2h3

h1

∂f

∂u1

+

∂∂u2

h3h1

h2

∂f

∂u2

+

∂∂u3

h1h2

h3

∂f

∂u3

Where A is a vector and f is a scalar.

Example: Cylindrical Coords

The relationship between the x, y and z coords with r, φ and z are:

x = rcos φy = rsinφz = z

We define the arbitrary vector A in Cartesian coords as

A = xˆ i + y j + zˆ k = rcos φˆ i + rsinφˆ j + zˆ k

We want to find dl, ds, dv, grad, div, curl and the Laplacian in cylindrical coordinates r, φand z.

6

SCE2311/SCE2611: EMIIR.T. Sang 1999

We will use the approach given on the previous page, hence let

u1 = r

u2 =φu3 = z

du1 = dr

du2 = dφdu3 = dz

ˆ a u1= ˆ r

ˆ a u 2= ˆ φ

ˆ a u3= ˆ z

We now need to find the scaling factors:

h r =∂A

∂r=

∂rcos φˆ i + rsin φ j + zˆ k

∂r

∂A

∂r= ∂

∂rrcos φˆ i + rsin φ j + z ˆ k = cosφˆ i + sinφˆ j

∂A

∂r= cosφ 2 + sinφ 2 = 1

h r = 1

and

hφ = ∂A

∂φ= ∂rcos φˆ i + rsinφˆ j + zˆ k

∂φ

∂A

∂φ=

∂∂φ

rcos φˆ i + rsin φ j + zˆ k = −rsin φˆ i + rcos φˆ j

∂A

∂φ= −rsinφ 2 + rcosφ 2 = r2 cos2 φ + cos2 φ( ) = r2

hφ = r

finally

h z =∂A

∂z=

∂rcosφˆ i + rsin φ j + z ˆ k

∂z

∂A

∂z= ∂

∂zrcos φˆ i + rsin φ j + z ˆ k = 1ˆ k

∂A

∂z= 12 = 1

h z = 1

From our change of basis formulas the elemental change in length is

7

SCE2311/SCE2611: EMIIR.T. Sang 1999

dl = ˆ a u1h1du1( ) + ˆ a u 2

h2du2( ) + ˆ a u3h3du3( )

⇒ dl = ˆ r h rdr( ) + ˆ φ hφdφ( ) + ˆ z hzdz( )⇒ dl = ˆ r 1dr( ) + ˆ φ rdφ( ) + ˆ z 1dz( )⇒ dl = drˆ r + rdφˆ φ +dzˆ z

The change in volume is

dv = h1h2h3du1du2du3

⇒ dv = rdrdφdz

The change in the surfaces are

dS1 = h2h3du2du3ˆ a u1

⇒ dSr = hφh zdφdzˆ r

⇒ dSr = rdφdzˆ r

dS2 = h1h3du1du3ˆ a u 2

⇒ dSφ = h rh zdrdzˆ φ

⇒ dSφ = drdzˆ φ

dS3 = dl1dl2 = h1h2du1du2ˆ a u3

⇒ dSz = h rhφ drdφˆ z

⇒ dSz = rdrdφˆ z

The del operator is

∇• A = 1h1h2 h3

∂∂u1

h2h3A1( ) + ∂∂u2

h1h3A2( ) + ∂∂u3

h1h2A3( )

⇒ ∇• A =1

h rhφh z

∂∂r

hφh zAr( ) +∂∂φ

h rh zAφ( ) +∂∂z

h rhφA z( )

⇒ ∇• A =1

r

∂∂r

rAr( ) +∂∂φ

Aφ( ) +∂∂z

rAz( )

⇒ ∇• A = 1r

∂∂r

rAr( ) + 1r

∂∂φ

Aφ + ∂∂z

Az

You can show that the grad and curl operators are

∇f =∂∂r

fˆ r +1

r

∂∂φ

fˆ φ +∂∂z

fˆ z

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∇× A =1

r

ˆ r ˆ φ r ˆ z ∂∂r

∂∂φ

∂∂z

A r rAφ A z

= ˆ r 1

r

∂Az

∂φ−

∂Aφ

∂z

+ ˆ φ

∂A r

∂z−

∂Az

∂r

+

ˆ z 1

r

∂ rAφ( )∂r

−∂A r

∂φ

Trigonometric Substitution Integration Revisited

In this section of the course you will often come into contact with integrals that containfactors such as:

a2 − x2( ) , a2 + x2( ) , x2 − a2( )where x would be the variable that you are trying to integrate and a is some constant.Integrals that involve forms of these functions can be difficult to integrate when they appear

under radical powers eg: 1

x2 − a2( )3

2

unless you make the appropriate substitution.

Trigonometric substitutions enable these types of integrals to be solved in a general sort ofway. The substitutions are based on trig square identities (sometimes called Pythagoreanidentities) which are as follows:

cos2 θ + sin2 θ =1 ⇒1 − sin2 θ = cos2 θ

1 + tan2 θ = sec2 θ⇒ sec2 θ − 1 = tan2 θ

The substitutions themselves consist of writing either x = asinθ , x = atan θ , x = asecθ

The trigonometric functions have been defined wrt the right angle triangle below:

y

x

r

θ

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SCE2311/SCE2611: EMIIR.T. Sang 1999

sinθ =y

r cosecθ =

r

y cosecθ =

1

sin θ

cosθ =x

r sec θ =

r

x secθ =

1

cosθ

tanθ =x

y cotθ =

y

x cotθ =

1

tanθ tanθ =

sinθcosθ

The choice of substitution depends on what turns up in the integrand. This choice can besummarised by the following table:

Integral involves Substitution(i) a2 − x2( ) x = asinθ(ii) a2 + x2( ) x = atan θ(iii) x2 − a2( ) x = asecθ

The purpose of the substitution is to change the factors, a2 − x2( ) , a2 + x2( ) , x2 − a2( ) thatappear in the integral to squared trigonometric functions.

Example: For the factor a2 − x2( ) if we make x = asinθ then

a2 − x2( ) = a2 − a2 sin2 θ( ) = a2 1 − sin2 θ( ) = a2 cos2 θ

Example of an i ntegration involving a trigonometric substitution

Evaluate:

1

a2 + x2( )3

2∫ dx since the integral involves a2 + x2( ) we make the substitution :

x = atan θ thus dx

dθ= asec2 θ⇒ dx = asec 2 θdθ

Making our substitution yields the integral:

1

a2 + x2( )3

2∫ dx =

asec2 θdθ

a3 1+ tan2 θ( )3

2∫ =

sec2 θdθ

a2 sec2 θ( )3

2∫ =

1

a2

dθsecθ∫

From above we recognise that cosθ =1

secθ , as such the integral simply reduces to

1

a2

dθsecθ∫ =

1

a2 cosθ∫ dθ =1

a2 sin θ + C

It is preferable to put the solution in terms of x and a which is easily done by Pythagoras'stheorem:

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SCE2311/SCE2611: EMIIR.T. Sang 1999

x = atan θ⇒ tanθ =x

a thus the right angle triangle for this is given by

a

x

θ

x2 + a2

⇒ sinθ =adj

hyp=

x

a2 + x2

Substituting back into the solution for our integral then yields:

1

a2 + x2( )3

2∫ dx =

1

a2

x

a2 + x2+ C