1

Physics 2

Jouko Teeriaho

Rovaniemi UAS

Main objective

• Understanding the basic phenomena and

laws of electricity, magnetism and

electromagnetic waves

Material:

* Giancoli: Physics for Engineers and Scientist

* Other material: Dr. Lewin’s video lectures for MIT students

2

Assessment criteria

• Assignments

• 2 exams

• Concepts of electrostatics

– Charge, current, electric force, induction

– Electric fields, potential, voltage

• Resistance and resistors

• Capacitance and capacitors

• Batteries

Part 1

Part 2

• Magnetic force

• Magnetic field

• Magnetic induction, coils, generators

• Transformers, 3 phase current, electric network

• AC

• Electromagnetic waves

3

Basic

Phenomena

and history

Ancient Greece:

• People observed that when one rubs amber, it starts to attract light bodies (in Greek language “electron” means amber)

• Later was found: If you rub glass with fur, it also attracts light bodies

* Two rubbed glass sticks repel each other, two amber sticks repel each other but glass and amber attract each other.

There are two kinds of electricity: +

and -

An American physicist, politician and writer Benjamin Franklin (1706-1790) suggested that there is only one kind of “electric fire”, which we call electric charge.

Negative charge is excess of electrons

positive charge is lack of electrons

History:

Amber

Benjamin Franklin 1706 -1790

US writer, politician, scientist,

inventor, civil activist, diplomat

4

Atoms a held together by electricity

+

Nucleus:

Protons:

mass mp = 1.7*10-27 kg

charge q = 1.6*10-19 C

Neutrons:

mass mn equals proton mass

charge q = 0 (neutral)

-

electrons

mass = protonmass/1830

charge q = -e =-1.6*10-19 C

Electric charge is quantized : all charges are multiples of the electron charge Q = n*e

10-12 m

10-8m

Conductors and non-conductors

Charge (typically electrons) can move freely in a conductor. Metals are good conductors.

In a non-conductor charges are not free. They are bound to atoms and cannot move along the non-conductor.

Copper is an excellent conductor

Plastic, rubber and glass are

non-conductors

5

Electric current I

* Electric current is defined as motion of electric charge

• In most cases charge is carried by electrons, sometimes ions

• Symbol of charge is Q and units 1 C = 1 Coulomb

• Smallest charge is electron charge -1.6*10-19 C

• 1 Ampere is 1 Coulomb per second

• the direction of current is opposite to the direction of electron stream

current I electrons

Q = I t charge = current x time

1 C = 1 As (“AmpereSecond”)

Three interactions of Physics

Gravitation hold the universe together

Electric force keeps the atoms together

Nuclear force keeps the nuclei of atoms together

6

Electric force

2

21

r

QQkF

1736 - 1806

The French Andre Coulomb observed, that the force between two charges is proportional to the charges and inversely proportional to the square of their distance

k = 9.0*109 Nm2/C2 = Coulombs constant

04

1

k

k can be expressed in terms of 0 , which is called vacuum permittivity

0 = 8.85*10-12 C2/Nm2

Coulomb’s law

Vector form of Coulomb’s law

rr

QQkF ˆ

2

21 q1 q2

r̂where is the unit vector from q1 to q2.

F F

* If charges have the same polarity (= sign

of charge ), the force is repulsive.

* If the charges have different polarities , the

force is attractive

7

Superposition principle

The total force due to several charges q1,q2,… on a

charge Q is the vector sum of the forces due to the individual charges.

q1

q2

+

_

+Q

F

F1

F2

)180cos(2 21

2

2

2

1 FFFFF

If is the angle between vectors F1 andF2, the sum vector is

Electric induction

Charged

rod

-

-

-

+

+

+

Induction = division of charge in conductor due to an external charge. In the picture + charge attracts free electrons, and the right side of the conductor gets a positive layer

* A charged body attracts light pieces of conductor, because of the different

distances to the induced layers in the conductor.

-q +q

+Q

8

Polarization of atoms at the presence

of external charge

symmetric atom atom polarized by induction

+ + -

external

charge

nuclei and electrons move to different directions =>

atom becomes a dipole with two electric poles

-Q

Induction appears also in non-conductors

External charge causes polarization => there will be thin charged layers on both sides of the non-conductor

A charged body attracts slightly non-conductors ( rubber balloon)

-Q + + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

- - - - - - - - - -

- - - - - - - - - -

- - - - - - - - - -

thin charged layers

9

Electroscope indicates charge

+

+ +

Metal rod Light Aluminum rod

Charged rod

+Q

Charge moves into the electroscope from the charged rod. Repulsion force makes the

Aluminum leaf rise and form an angle with the electroscope rod. This means that the

electroscope was charged.

Electric field E

Q q

Positive test charge

F

P

Electric field at point P is defined q

FE

Electric field = electric force / charge

Unit = 1 N/C , which equals 1 V/m as later shown

10

Field of a point charge Q

Q

E=F/q

2

0

2 4

1

r

Q

r

QkE

r

k = Coulomb constant = 9.0*109 Nm2/C2

Q = electric charge

r = distance of P from Q

P

Field lines describe electric field

+Q -Q

Field lines go away from positive charge

Field lines come towards a negative charge

11

Two charge system field lines

Dipole field Two positive charges

Field lines start from positive pole

and bend towards the negative

pole

Field lines bend as if they would

“repel” each other

Homogenous electric field

Charged cloud + + + + + + + + + + + + + + + + + + + + + + + + +

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

Electric field E

Friction of the motion of air layers charges the cloud. Induction attracts an equal, opposite

charge to the ground. Field lines are parallel and the field is uniform.

Breakdown voltage of air is 3 MV/m. When the field exceeds it, electrons are removed from

their host atoms and lightning starts.

Ground

12

Conductor in an electric field

1) Inside the conductor electric field becomes zero

2) Field lines penetrate the surface in 90 o angle

Charge distribution in a conductor

+ +

+

+

+ +

+ +

It can be shown that the charge density and electric field is greatest at the edges of a conductor.

The sharper the edge, the bigger field and charge density

E

E = 0

13

Figure: conductors in electric field

Electrostatic Induction:

Electrons move inside the

conductor to new positions:

1. The electric field inside the

conductor becomes zero

2. Field lines penetrate the

conductor surface in straight

angles.

Faraday cage

Faraday cage is made of metal.

Inside the cage electric field = 0

regardless of outside fields.

Faraday cage is a shield

against static fields and low

frequency alternating fields (Long Wave radio is not heard

inside)

High frequency GSM field can

penetrate into the cage

Car is a good shelter during

lightning of a thunderstorm

Link: MIT-video on Faraday cage

14

Non conductor in an E-field

-

-

-

-

-

+

+

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

E is weakened in the nonconductor due to atomic polarization

Field is weakened by factor which is called the dielectric

constant or relative permittivity of the of the material

Eint Eext

For paper = 3 , rubber = 7, glass = 7 , water = 80

extE

E int

Motion due to the electric force

According to Newtons law F = ma = qE =>

Em

qa

Positive charge ”falls” in the direction of

field lines

The trajectory is a parabola as in the free fall under gravity

trajectory E

F = q E

15

+ + + + + + +

Uniform field near a charged plane

dQ = dA = 2 x dx

02

E

where = Q/A = charge per unit area (“surface charge density”)

x

x

d r2=d2+x2

Derivation by integration

Result: Field near a charged plane

Uniform field between two charged planes

+ + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - - - -Q

+Q

A

E

Field between two charged plates = 2 x field of a single charged plate:

0

E

where = Q/A = charge per unit

area (“surface charge density”)

Link: MIT-video on uniform electric field

16

Potential V

Voltage U

Definition of potential V

Potential of point P = work per unit charge needed to bring a positive test charge from infinity to P

q

WV unit J/C = 1 V (Volt)

Voltage means difference of potentials of points A and B.

Voltage is thus the potential energy difference per unit charge

U = VB - VA

17

Gravitational potential V and electric potential V

m

h

gravitational field = g

force F = m g

q

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

x

E = electric field

force F = q E

+ + + + + + + + + + + + +

W = m g h W = q E x

V=W/m = g h V=W/q = E x

In gravitational field g Potential is defined as V = W/m (energy per unit mass)

In electric field E Potential is defined as V = W/q (energy per unit charge)

gravity electricity

m

h

gravity

pot. energy pot.energy

Voltage between charged plates

+ + + + + + + + + + + + + + + + + + + + +

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

E

d Voltage U = ?

Force on a positive test charge q is F = qE,

Work needed in moving the charge is W = F*d = qEd =>

The work is called potential energy and is qEd.

Voltage U = W/d = Ed

+

F=qE

dEU In uniform field E the voltage = the field

times the distance in the direction of the field

18

Potential of a point charge

2r

QkE

r

Qqdr

r

QqdrqEW

rP

P

0

2

0 4.

4.

Q V = 0

r

Work needed to bring +q :n to point P is

Division by q gives

r

QrV

04)(

Potential of a point P at distance r from point charge Q

Piste P

Potential on the surface and inside the sphere is 90kV

Outside the sphere potential decreases proportional to 1/r

Figure shows electric potential inside and outside of a charged metal sphere with Q = 1 Coulomb and r = 0.10 m

90 kV

Link: MIT-video on a light tube in a field of a charged sphere

19

Energy principle in

electric field:

Application: TV - tube

TV - tube

When electrons are accelerated using voltage U, the potential energy is transformed to kinetic energy

2

21 mvqU

TV tube formula

20

In TV –tube electrons are accelerated using 3.0 kV voltage. Calculate their final speed when hitting the screen.

Electron’s charge and mass are 1.6*10-19 C and 9.11*10-31 kg

Electron being very light is easy accelerated. If the accelerating voltage is tens of hundreds kiloVolts like in X-ray tube, we cannot use classical formula ½ mv2 for kinetic energy. Then the equation must be replaced by

2

21 mvqU sm

m

Uqv /

1011.9

3000106.12231

19

=>

= 3.24621×107 = 32 Mm/s ( 10% of the speed of light)

22

2

2

1mc

mcqU

c

v

where c = the speed of light 300 Mm/s

In this Einstein’s formula m = the mass at rest, m/(1-v2/c2) is

the increased kinetic mass in the speed of v.

2

21

r

QQkF

q

FE

2

21 mvqU

Q = I t

dEU

q

WV

pot

U = VB - VA

Summary of formulas of electrostatics

Charge and current

Electric force

Electric field

Electric potential

Voltage between points A and B

Voltage in a uniform field

TV tube formula

2r

QkE Electric field of a point charge

Electric potential around a point charge r

QkV

UqW Potential energy difference

1

DC circuits

• Resistance, resistors

• Batteries

• DC circuits

Direct Current = DC

In TV tube the electric potential energy is transformed into kinetic energy

according to q U = ½ m v2

Resistance

wire

heat

+ -

In a wire the electrons collide with atoms and all the electric potential energy is

transformed into thermal energy

Created heating power

UIt

qU

t

EP

pot

2

Ohm’s law

Denote l = length. A = cross area of a wire, U = voltage and I = current in the wire.

It can be shown that voltage and current are linearly dependent : U = RI. Coefficient R depends on length l, cross area A, density of free electrons n, and mean time between collisions T.

RIIA

lI

A

l

nTeU

2

1

Parameters n, T are properties of the material. They determine a quantity .

= resistivity of the material , unit m (Ohmmeter)

Resistance unit

1 = 1 Ohm

e = electron charge

Ohm’s law

U = R I Ohm’s law

A

lR

= resistivity of material

l = length of wire

A = cross area of wire

Resistance R depend on following parameters:

Copper Cu 1.7 ( 10-8 m)

Iron Fe 9.7

Silver Ag 1.6

Tin Sn 11

Carbon C 1000

Table of resistivities

of common conductors

3

Resistors

Resistors are common components in electronics

Heating resistors are used in grills, electric heaters.

Combining P = U I and U = RI we get three different formulas for

power consumed in a resistor

R

URIUIP

22

Power

consumption

W = P t Energy consumed

in time t

linkki

Examples

Ex 1: Calculate the resistance of 100 km long 2.0 cm thick Copper wire

4.5)(*

100000107.1

2

202.0

8

A

lR

Ex 2: How much current uses a 1200 W car heater, which uses 220 V AC.

P = U I => I = P / U = 1200 W / 220 V = 5.5 A

Ex 3: Calculate the cost when 600 W car heater is used 12 h per day during January (unit price of electricity is 10 cnt/kWh)

Energy used = 0.6 kW * 31*12 h = 223 kWh

Cost = 22 Euro

4

More examples

242000

)220( 22

W

V

P

UR

Ex 4: Calculate a) resistance of a 6.0 kW electric sauna b) resistance of 2000 W electric cooker?

Ex 5: How many 12 Ampere fuses is needed for the 6 kW sauna, if the 220 V AC is used.

1.86000

)220( 222

W

V

P

UR

R

UPSauna

Cooker

Current I = P / U = 6000W / 220V = 27.3 A

=> Sauna needs 3 fuses

Resistor symbol 10

Resistors in series

R2 R1

U I

U1 U2

R I = R1 I + R2 I

R = R1 + R2 + …

or

Total voltage U is divided to U1 and U2.

U = U1 + U2

By Ohm’s law this becomes

where R = total resistance of the system

Dividing I we get

2

1

2

1

R

R

U

U

Total resistance

Note! Voltage between resistors is divided in the

ration of resistances

5

Parallel resistors

21

111

RRR

21 R

U

R

U

R

U

R1 R2

U

I I1 I2

Ohm’s law gives

Here the voltage in both resistors is the

same. Current I is divided in I1 and I2. By

conservation of current

I = I1 + I2

From which

Total resistance R of parallel

resistors

Current is divided inversely proportional to the

resistances

1

2

2

1

R

R

I

I

Examples

5.1)( 1

81

51

31

Ex6: Calculate total resistances of following connections

10

10

5

3

8

5)( 1

101

101

4 5

3

Upper branch is 5 + 4 = 9 Ohm.

Total resistance

R = (1/3+ 1/9)-1 = 2.25 Ohm

a)

b)

c)

6

Ex7: An old cooker works with three 50 Ohm resistors, which can be combined in 7 ways to give different powers. Draw all possible combinations, calculate their resistances and heating powers, if U = 220 V

50

50

50

50 50

50 50 50

50

50

50

50 50

50

50 50

50

50

100

25

150

17

75

33

968 W

484 W

1936 W

323 W

2847 W

645 W

1467 W

figure resistance power P = U2/R

(2202/50)

Resistance depends on temperature

• Light bulbs resistance becomes 10 times large when its hot.

• Cold air resistance is high, but hot air resistance lower

• Resistivity of pure water is high but resistivity of salt water is low

7

Dropping resistors

Ex.8 Car battery voltage is 12 V. The voltage of the fan engine can be adjusted by using a dropping resistor in series with the fan. Assuming that the fan’s resistance is 200 Ohm, what should be the resistance of the dropping resistor in order to get a) 6 V b) 4 V operating voltage for the fan.

Principle: In serial coupling the ratio of resistances equals the ratio of voltages.

R 200

12V

6V 6V

a) R/200 = 6/6

=> R = 200 Ohm

R 200

12V

b) R/200 = 8/4

=> R = 400 Ohm

4V 8V

2

1

2

1

U

U

R

R

Shunt resistors

Ex9. Switching with different measurement ranges in Ammeters is implemented with shunt resistors parallel to the ammeter. Calculate the size of the shunt resistor, if we want to increase the measured range from 0 – 1.0 A to 0 – 10 A. Ammeter’s resistance = Rm.

Principle: In parellel coupling the current divides inversely proportional to resistances

R =?

Rm

I

1/10 I

9/10 I

Voltage over both resistances is the same

=> 1/10 I*Rm = 9/10 I*R => R = 1/9*Rm

Result: Shunt resistance must be 1/9 of the resistance of the ammeter.

8

Batteries

• Batteries are chemical pumps , which move charge to higher potential.

• They are based on the electrochemical series. Battery EMF equals the difference of the normal potential difference of metals : f.e Cu – Zn pair voltage is 0.76 V – (– 0.34 V) = 1.1 V.

The battery voltage, when current is not taken from the battery is called electromotive force E (EMF)

Most car batteries are lead batteries. For each cell E = 2.1V. There are 6 cells in series giving 12.6 V

K, Li, Ba, Ca ,Na ,Mg ,Al, Zn, Cr, Fe, Co, Ni, Sn, Pb, H, Cu, Hg, Ag, Au, Pt

Part of the electrochemical series:

+ -

E

RECHARGEABLE BATTERIES

Type Cell

voltage Duration (discharge- charge cycles)

Energiatiheys

Wh/kg

Käyttö

Pb-Acid

lyijyakku

2.1 V 500-800 30-40 Vehicles

NiCd 1.2 V 1500 40-60 Forbidden

(poison)

NiMH 1.2 V 500-1000 30 – 80 Laptops

Li-ion

(LiFePO4)

3.3 V 2000+ 80-120 Electric

Vehicles

Li-ion

(Coboltti)

3.7 V 1200 150-250

Phones

explosive?

9

Terminal voltage

When current is taken (battery is loaded) , the voltage between its terminals is less than E due to the internal resistance of the battery (the battery gets warmer in use)

U = E - Rc I

E = EMF of battery

I = current

Rc = internal resistance

U = terminal voltage

Terminal

voltage:

In words: Terminal voltage = EMF – voltage loss in the internal resistance

Battery banks

Serial coupling gives bigger EMF:

E = E1 + E2 + E3

In car battery there are 6 – 8 cells in

series.

In parallel coupling the EMF does not

increase, but internal resistance is

halved, charge and energy is doubled

if batteries are identical

bulb

Batteries in series

Parallel batteries

10

Ex10. When 0.20 A is taken from the battery , terminal voltage is 4.41 V. When 0.55 A is taken, terminal voltage is 4.38 V. Calculate E and Rs.

U = E - Rs I The two ( I , U) pairs are set into equation

resulting in a pair of linear equations of unknowns E and Rs

4.41 = E - Rs* 0.2

4.38 = E – Rs* 0.55 Elimination of E by subtraction of equations

4.41 - 4.38 = - Rs*+0.2 + Rs* 0.55 =>

0.03 = 0.35 Rs

=> Rs = 0.086 = n. 9 m

E is obtained by substituting Rs back to equation 1

E = 4.41 + 0.086*0.2 = 4.43

Ex11.

Car battery holds text: 60Ah, 12V.

a) Determine its charge and energy

b) The owner has forgotten lights (90 W) on. How long the battery energy lasts ? 60Ah, 12 V

a) Charge is 60 Ah = 60 A*3600h = 216 000 C

Energy W = QU = 216000C*12V = 2592000 J = 2.6 MJ

More straightforward way is to calculate current from

I = P/U = 90/12 A = 7.5 A and then

time t = Q/I = 60 Ah/7.5A = 8 h

b) Battery lasts time t = W/P = 2592000/90 s = 28 800 s

= 480 min = 8 h

11

DC circuits

- Consists of resistors and batteries. They can have one or several loops

Solving DC circuit means solving currents in all its branches.

In practise this means solving a group of linear equations obtained from Kirchoff’s rules.

Kirchoff’s laws

1. Law : The sum of currents arriving to a node = the sum of currents leaving it

2. Law : In every closed loop the sum of potential changes = 0 ( sum of battery EMF’s

and potential losses in resistors = 0)

I1

I2

I3

I1 = I2 + I3

0iii IRE

Solved examples Esim12: Light bulb (R = 4.0 ) is in series with a battery with E= 4.50 V and Ri = 0.10 .

Calculate current and power consumption in the light bulb.

X

I 4.0

0.1

4.5V

Kirchoff’s 2. law = > the sum of EMF’s + potential losses in resistors = 0

a) 4.5 – 0.1 I - 4.0 I = 0

current I = 4.5/4.1 = 1.098 A

b) Power

Light bulb P = R I2 = 4.0*1.0982 = 4.82W = 4.8 W

Internal resistance: P = RI2 = 0.1*1.0982 W= 0.12 W

12

Ex13. Solve currents in following circuits.

5 10

6

3

I1

I2

I3

Currents are denoted by I1, I2 and

I3, assumed directions are marked

with arrows. Solution gives the

final directions. 12V

6V

4V

Kirchoff 1 => I1 = I2 + I3

+12 – 5 I1 – 3 I3 + 6 – 6 I1 = 0 (left loop)

+12 – 5 I1 –10 I2 - 4 – 6 I1 = 0 (big loop)

Normal form of the group : I1 - I2 - I3 = 0

-11 I1 - 3 I3 = -18

-11 I1 - 10 I2 = -8

1 1 1

11 0 3

11 10 0

1

.

0

18

8

Out[3]=

1.17919

0.49711

1.6763

=

Matrix solution

Ans. I1 = 1.2 A, I2 = -0.5 A (opposite to the

assumed direction). I3 = 1.7 A

Kirchoff 2 =>

Measuring current and voltage

V

A

= voltmeter

= ammeter

Symbols

X 4.5V

X = bulb

A

V

I

Ammeter is always in series with other components, volt meter is parallel

Ammeters resistance is very small => it does not change the current

Voltmeters resistance is very large => it does not change the voltage

13

Capacitance

and capacitors

Capacitance

Metal surfaces can store charge when its potential is raised or lowered with respect to the environment.

This property is called capacitance.

V

QC

Capacitance = charge / potential

Unit 1 C/V = 1 Farad = 1 F

+ + + + + + + + + V

The bigger surface area, the bigger capacitance

14

Metal sphere capacitance

RV

QC 04

Q

R

Potential on the sphere is

Capacitance is thus

R

QV

04

Sphere’s capacitance is proportional to the radius.

Capacitors

• They store charge

• They are used in electrical devices:

• Radio , TV, phones, …

• Camera flash light

• Rectifiers, which transform AC to DC

• Low frequency filters

• Ultracapacitors in electric buses /vehicles

symbol

15

Plate capacitors

d

A

d

AC 0

voltage

U

capacitance : d

A

d

A

U

QC

Mostly there is insulator filling between plates

= dielectric constant of the insulator material

= Q/A = surface charge dens.

= 0 = permittivity of filling

E = / = electric field

U = E d = voltage

Energy density of the E-field

C

QCUQUW

22

2

1

2

1

2

1

AdEA

dAEW 2

21

222

21

Capacitor energy in three forms

For plate capacitor Q = A= EA ja C= A/d , whence

where Ad = V = volume between plates

Energy density of electric field

2

21 Ew unit : J/m3

Energy is usually = QU, but the

average charge during charging the

capacitor is ½ Q, => capacitor

energy = ½ QU

16

Cylindrical capacitor

)ln(

2

1

2

0

R

R

lC

Capacitance

l = length

r1 and r2 = radii of inner and

outer cylinder

0 = vacuum permittivity

= dielectric constant of

insulator

Cylindrical capacitors

Discharge through a resistor

R U C

charge discharge

I

Capacitor charge is Q = CU.

During the discharge the sum of voltages = 0

Q/C + R I = 0 where current I = Q’(t))

Charge and current drop exponentially. Product = CR is called time constant of the circuit and it gives the speed of discharge.

Solving this differential equation gives a mathematical discharge law:

RC

t

oeQQ

0.5 1.0 1.5 2.0 2.5 3.0

0.2

0.4

0.6

0.8

1.0

17

Capacitor connections

21

111

CCC

serial

21 CCC

parallel

Usage of capacitors:

* When we want a repeatedly identical current flow, capacitor discharge gives that

* examples: flash light, car ignition circuit

C1 C2

C2

C1

Other examples

LCf

2

1

We tune the radio receiver with an variable

capacitor.

In this old fashioned variable capacitor the

area A is varied by turning the knob

vrt. kaava

symbol

Variable capacitor

*) formula for received frequency

d

A

d

AC 0

18

Flash light

Large (in picture 185 F) capacitor is charged with

1,5 V battery and transformer circuit to high

voltage.

Capacitor discharge causes the flash

The energy released in the flash

light is given by W = ½ C U2

In the example 185F capacitor with 100

Volt gives energy

W = ½ 185*10-6*1002J = 1.9 J

Ultracapacitors

In city busses power source can be

ultracapacitor, which is charged at bus

stops.

Capacitance: 3000 F

life time: 10 years

Charging time : 1 - 10 sec

Range : 4 -5 blocks

Benefits:

low costs ( can save 200 000 € / year

good for environment

Ultracapacitor

buses in China.

19

Breakdown voltages of capacitor fillings

Capacitor’s have a maximal voltage which is determined by the breakdown voltage

values of the insulator material between the plates.

Breakdown voltages:

Air 1 – 3 MV/m

Paper 50 MV/m

Oil 16 MV/m

Bakelite 15 MV/m

Paraffin 10 MV/m

Oil filled capacitor, where the width of

the gap between plates is 0.10 mm

Maximal possible voltage is

U = Emax*d = 16*106 V/m *0.0001m

= 1600 V

1

Magnetism

Magnetic force

Magnetic field B

History

• Already 500 eKr Greeks knew, that certain stones attract iron. These stones were magnetite FeO2 and they were found in province of Magnesia in Greece

• 1100 the Chinese made a compass from magnetite needles

• In 13th century was found that magnets have two poles P and S , P and S attracting each other but P and P and S and S repelling each other.

• It is not possible to isolate magnetic poles, electric poles can be isolated.

• 1500 Gilbert found out that the Earth is a big magnet and he made first magnetic maps of Earth

2

• Today magnetic poles are called “North”and “South” . The “N” of a magnet turns to the geographic North (which is thus magnetic

“South”)

• 1819 the Danish Ørsted noticed, that current affects magnetic needles

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3

Other magnetic phenomena

• Magnetic field affects moving charges

• Between two currents exists a magnetic force

• Changes of magnetic field induce currents and EMF into wires

• Oscillating charges send EM waves

Magnetic force

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Nikolai Tesla was an american engineer who invented first AC motor.

4

Magnetic force properties

1. Magnetic force is always perpendicular to the velocity => no work is done, kinetic energy is conserved

2. If the moving charge enters magnetic field in straight angle � =90o, its trajectory is a circle

3. If � � 90o , the velocity component in direction of magnetic field is conserved and orbit is spiral

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Northern lights

Corona discharge of the Sun emit

charged particles (Solar wind)

In the atmosphere these particles travel

along spiral trajectories following the

magnetic field lines to the poles.

They ionize the atoms of atmosphere and

visible light is emitted

6

Magnetic bottles

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DC motors

DC motors are based on the magnetic force acting on current loops

7

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8

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9

DC -motor animation

http://www.youtube.com/watch?v=Xi7o8cMPI0E&feature=fvw

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11

Origin of magnetic field

Biot- Savart law

Ampere’s law

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12

Biot- Savart’s law (1820)

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r

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13

Field in the middle of a loop

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I

r

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⋅

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14

Ampere’s laws applications

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15

Summary of laws:

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16

Magnetic Induction

Magnetic flux

Lenz law

History

Faraday invented induction in his

experiment 1831.

Turning current on or off in the primary coil makes the ammeter pointer swing.

But when the current has been on for a while, ammeter shows zero.

Conclusion is that, inductive current is seen only when the magnetic field is changing, but not when the magnetic field is constant.

Modern society with electric energy is based on induction.

17

Magnetic flux �

Magnetic flux through the loop is defined as

αcosBA=Φ

Unit of flux is 1 Tm2 = 1 Wb = 1 Weber

= quantity needed to understand magnetic induction

Induction law

dt

dE

Φ−=

If the magnetic flux through a coil is changed, a voltage called EMF (EMF = electromotive force) resisting the change is induced. Its magnitude is

In a coil of N windings the induced EMF is

dt

dNE

Φ−=

Magnetic flux can be changed in three ways:

1) Field B is changes

2) Loop area is changed

3) Angle � of the loop is changed

18

Increasing B causing induction

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Decreasing field causing induction

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19

Rotating magnet induces AC

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20

Generator of hydroelectric power plant

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tee ωsinˆ= where peak voltage is ê = �NBA

21

Bicycle dynamo

In a dynamo the rotating magnet is called the anchor and the coil is called the stator.

Eddy currents

N

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22

Moving eddy current

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Magnetic train

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23

Next topics

• Magnetic materials• Self induction• Coils , inductance• Transformers• Three phase currents• Electromagnetic waves

Magnetic materials

In a medium the magnetic field B usually becomes stronger ( para- and ferromagnetism) , in some cases weaker (diamagnetism).

The vacuum permeability �0 gets in formulas a multiplier coefficient � , which is called the relative permeability of the material.

l

NIB 0

κµ=F.e a coil with an iron core has a

magnetic field given by

B0 B= � B0

24

Diamagnetic Paramagnetic Ferromagnetic materials

materials materials

Au, Cu, Pb, N2, Si

� < 1

* Magnetic field slightly smaller than in the air

* The magnetic dipole moments of atoms turn against the external field for reasons that can be explained only through quantum theory

Al, Ca, O2, Pt

� > 1

*Magnetic field slightly greater than in the air

•The atoms are small magnets, because the electron orbits are current loops

• The atoms turn to the direction of outside field

Fe, Ni, Co

� = 400 - 2000

* Magnetic field much greater than in the air

•Small magnetic domains turn in external field to the same direction=> becomes a permanent magnet

• Above Curie –temperature (Iron 1043K) is not magnetized

Often is written � = 1 + � where

� is called the susceptibility of the material

Self induction

2 4 6 8 10

1

2

3

4

5

6

time

current

LE

1

2

dtdILRIE +=

X

R

Self induction: When the switch is turned to position 1, the increase of current changes the magnetic flux in the coil => The coil reacts to the change with an induced EMF, which tries to prevent the change.

The coil in front of the light causes a delay, when the light is turned on.

25

time

current

LE

1

2

dtdILRIE +=

X

R

When the switch is turned to position 2, the current decreases rapidly and the coil opposes the change with an induced EMF, which tries to preserve the current.

The coil in front of the light causes a delay, when the light is turned on.

1 2 3 4 5

0.2

0.4

0.6

0.8

1.0

;���������>

>����� �

l

NIB 0

µ=The field inside the coil is

The magnetic flux l

NAIBA 0

µ==Φ

If the current through the coil is changing with rate dI/dt , according to Lenz law there appears an induced EMF that opposes the change:

dt

dI

l

AN

dt

dN

2

0µ

−=Φ

−=Ε

where L is called coil inductance

dt

dILE −=

Unit of inductance is 1 Vs/A = 1 �s = 1 Henry = 1 H

=>

l

ANL

2

0µ

=

26

Coil energy

The energy stored in the magnetic field of the coil

2

2

1LIW =

R

Car

battery

switch

resistance of the circuit

spark plug

secondary coil

CAR IGNITION SYSTEM

When the switch is turned on, the current will reach the value E/R in time, which is given by L/R (time constant)

When the current is cut off, the time constant is very low, because the resistance in L/R will be extremely large. So dI/dt will be very large and also the induced voltage in the primary coil will be very large: maybe 100 V or more.

This induced peek is however magnified with a transformer ( 1:10 or 1:20) to

10 kV – 20 kV peek, which will create a spark in the plug.

see the end of MIT lecture #24 (end)

Problem: Assume that the primary coil has an inductance L = 2 H and the total resistance of the primary circuit is 6 ohms.

Calculate the energy of the spark.

27

Relay

A small control current in coil terminals will close the main circuit.

(Example: garage doors, alarm systems )

Microphone

sound

The sound waves cause vibrations in the paper foil and the magnet fixed in it.

The vibrations of the magnet cause induced EMF in the coil, which can be amplified.

A microphone modulates 9V DC. A capacitor is used to filter the DC off.

out

micro-

phone

28

Thief alarm systems

There are magnets fixed to the products. The magnet causes an induction current in the coils of the gate and causes an alarm.

A gate with induction coils

Electromagnets

29

Metal detectors

An alternating current generates a time-varying magnetic field around the coil.

This field induces eddy currents in a nearby metal object which, in turn, generate a time-varying magnetic field of their own.

These fields induce a voltage in the receive coil which, when amplified, reveal the presence of the metal object or target

sending coilreceiving coil

Metal object with induced eddy current and induced field

Transformers

Np windings Ns windings

Up

Us

Current Is

Current Ip

(made of isolated plates of iron to prevent eddy currents)

30

l

IN

l

INB sspp 00 µµ

==The same magnetic field is going through both coils. Hence

sspp ININ =

Assuming that the transformer is ideal ( the energy and power is not lost), we have also UpIp = UsIs.

So we get

p

s

s

p

p

s

N

N

I

I

U

U==Formulas of an ideal transformer.

A transformer is an AC circuit component that changes the voltage and current keeping the transferred power P = UI constant.

When the voltage increases , current decreases, and vice versa.

”transform ratio”

Example (transfer without transformers)

GeneratorE = ?

500 m wires R1 = 2 ohm

There is an engine

U=220V, P=10 kW

inside the building

Questions:

a) How much over 220V generator has to produce

b) Calculate the voltage loss in transfer

c) Calculate the power loss in transfer

d) Calculate transfer efficiency

Solution:

The current through the engine I = P/U = 10000/220 = 45.5 Amperes

The voltage loss in transfer = R1*I =2*45.5 = 91 Volts

The Generator has to produce 220 V + 91 V = 311 Volts

The power loss in wires = RI2 = 2*45.52 = 4.1 kW

Transfer efficiency = 10kW /(10+4.1) kW = 71 %

31

Example (transfer with transformers)

GeneratorE = ?

500 m wires R1 = 2 ohm

There is an engine

U=220V, P=10 kW

inside the building

Solution:

The current through the engine I = P/U = 10000/220 = 45.5 Amperes

The current in the wires I1 = 1/20 * 45.5 A = 0.455 A

The voltage loss in transfer = R1*I1 =2*0.455 =0. 91 Volts

The Generator has to produce 220 V + 0.91 V = 221 Volts

The power loss in wires = RI2 = 2*0.4552 = 0.41 W

Transfer efficiency = 10000W /(10000.41) W = nearly 100 %

M1 M2

1:2020:1

In practice the transformers do heat => there is loss of energy inside transformers.

Transfer of electric energy

Principal lines : 400 kV or 220 kV

Major lines: 110 kV

Local lines: 20 kV or 10 kV

Consumer lines: 220 V

Finland

32

Three phase current

Most alternating-current (AC) generation and transmission, and a good part of use, take place through three-phase circuits. If you want to understand electric power, you must know something about three-phase.

Benefits of three phase current1) If the loads of all phases are equal (as it

should be the case), the current in the fourth (neutral wire) is zero, which decreases the transfer losses in wires.

2) The system gives two different voltages: A) the phase voltage V between one face and the neutral wire

B) higher voltage between two phases, which is �3 V

3) Three phase system makes it possible to build three phase motors(*), which are based on the circulating magnetic field that three phase voltage creates.

(*)look MIT video #18

33

Electromagnetic waves

1864 Maxwell predicted the existence of electromagnetic waves when he derived the following wave equation from the Maxwell equations:

2

2

22

21

x

E

ct

E

∂

∂=

∂

∂ the speed of light in vacuum

c = 1/�(0�0) = 300Mm/s

(one dimensional wave equation)

A German physicist Heinrich Hertz made 1887 an experiment that proved the existence of electromagnetic waves.

The solution is any function of form E = f (x – c t). The solution represents a wave traveling at the speed of c.

Electric field and magnetic field oscillate perpendicular to each other. The changing electric field creates always a magnetic field and vice versa.

34

Spectrum of EM waves

Gamma rays

X - rays

UV

Visible light

IR

Microwaves

Radio waves

9/5/2014

1

AC circuits

1. Resistance, capacitance and inductance in AC 2. Impedance Z, phase 3. Power in AC circuits

Apparent power S Real power P Reactive power Q

Resistance and AC

If voltage has a form u = û sin(t) , then the current i = î sin(t) has the same phase

iRu ˆˆ Ohm’s law holds for peak values as well as for instantaneus values

Power consumption P = ½ û î = ½ R î2

or using effective values P = U I = R I2

Power p=u i = û î sin2( t) is all the time positive,

9/5/2014

2

Capacitance and AC

If current has form i = î sin(t) then voltage phase is 90o behind u = -û cos(t)

iC

iXu ˆ1ˆˆ

(analogy to Ohm’s law)

Real power consumption = 0

X = reactance

Power oscillates around zero. Average power = 0. It is called ”reactive power”. Phase = -90o

Inductance (coil) and AC

If current has form i = î sin(t) then voltage is 90o ahead u = û cos(t)

iLiXu ˆˆˆ

(analogy to Ohm’s law)

X = L (reactance)

Power oscillates around zero. Average power = 0. It is called ”reactive power”. Phase = +90o

Real power consumption = 0

9/5/2014

3

All three together in AC circuit

If current has form i = î sin(t) , then voltage is u = û sin( t + )

iZu ˆˆ Z = impedance

2222 )1

( XRC

LRZ

R

X

R

CL

1

tan

phase

Power curve visits negative side, which is a sign of existence of reactive power

Power in AC circuits

2cos RIUIP

UIS

sinUIQ

Real power

Apparent power

Reactive power

P

Q

S

P2 + Q2 = S2

In electrical network, occurence of phase difference and reactive power is bad . Consumer does not get as much power as he should. For network companies reactive power currents mean additional costs. Finally the customer pays also for that. Cheap ”energy saving light bulbs” cause reactive power.

CLX

1

Big industrial or even school buildings should by a ”compensation equipment ”, which can vary capacitance in order to eliminate reactive power.

9/5/2014

4

Paying for reactive power

As mentioned ”energy saving bulbs and LED has incresed reactive power, which leads to additional cost for energy companies.

* Reactive power is seen already in electricity consumption meters * Normal consumer does not pay yet for reactive power • Companies, which consume > 100 A already pay 4-5 Euro / month per kVAR

COMPENSATION EQUIPMENT

Vattenfall: ” For companies who pay for reactive power and buy compensation equipment, the investment pays itself back already in one year”

EU: Less than 25 W bulbs do not need an inbuilt compensation circuit

Example: AC power supply: 24 V , 60 Hz is in series with 50 Ohmresistance, 4 mikroFarad capacitor and 2.0 mH coil. Calculate a) reactance X b) impedance Z c) phase d) current e) apparent power f) real power g) reactive power. h) If frequency could be varied, which would be the resonance frequency?

OhmC

LX 4.6621

f = 60.0 Hz, u = 24.0 V, R = 50 , C=4.0.10-6 F, L = 0.002H, = 2 f =377.0 s-1

OhmXRZ 3.66422

reactance

impedance

phase 7.85)(tan 1

Z

X

current mAAZUI 36036.0/

apparent power real power reactive power

S = UI = 0.87 W P = UI cos = 0.065 W Q = UI sin = -0.86 W

Resonance frequency, when X = 0 :

HzHzLC

f 1779100.4002.02

1

2

1

6