Electricity

Electricity

N Bronks

Basic ideas…Electric current is when electrons start to flow around a circuit. We use an _________ to measure it and it is measured in ____.

Potential difference (also called _______) is how big the push on the electrons is. We use a ________ to measure it and it is measured in ______, a unit named after Volta.

Resistance is anything that resists an electric current. It is measured in _____.

Words: volts, amps, ohms, voltage, ammeter, voltmeter

Electrons are flowing from the negative to positive side of the

battery through the wires

Note current moves from positive to

negative, however electrons are actually

are moving in the opposite direction!

• Flow of electrons

timepoint a Passed ChargeCurrent

tQI

Current

Current and Charge

Since one Ampere flows when one coulomb of charge passes a given point in a circuit each second,

Also

Current (A) = Charge (C)

time (s)I =

Qt

or

Charge (C) = no. of electrons x charge of one electron

Amp = Coulomb

second

Example 1:

How many electrons are there in 20 Coulombs ?

No. of electrons = total charge / charge of one electron

No. of electrons = 20 / 1.6x10-19

No. of electrons = 1.25 x1020 electrons

Example 2:

The current in a circuit is 5A. What is the charge flowing in :

a. 1 second ?

b. 10 seconds ?5 Coulomb

50 Coulomb

Summary Question

1. a. The current in a certain wire is 0.35A. Calculate the charge passing a point in the wire i. in 10 s, ii in 10 min.

b. Calculate the average current in a wire through which a charge of 15C passes in i. 5s, ii 100s

Ans:

a. i. I = 0.35A t = 10s

Q = It Q = 0.35 x 10 = 3.5 C

ii. I = 0.35A t = 10 min = 600s Q = It Q = 0.35 x 600 = 210 C

b. i. Q = 15C t = 5s

I = Q/t I = 15/5 = 3A

ii. Q = 15C t = 100s

I = Q/t I = 15/100 = 0.15A

Summary Questions page 47

2. Calculate the number of electrons passing a point in the wire in 10 minutes when the current is a. 1.0µA b. 5.0A

Ans: a. I = 1.0µA t= 10min = 600s Q = It = 1x10-6 x 600 = 6x10-4 C No. of electrons = total charge / charge of one electron No. of electrons = 6x10-4 / 1.6x10-19 = 3.75x1015 electrons

I = 5A t= 10min = 600s Q = It = 5 x 600 = 3000 C No. of electrons = total charge / charge of one electron No. of electrons = 3000 / 1.6x10-19 = 1.88x1022 electrons


3. In an electron beam experiment, the beam current is 1.2mA. Calculate

a. The charge flowing along the beam each minute

b. The number of electrons that pass along the beam each minute

Ans: a. I = 1.2x10-3 A t= 1min = 60s Q = It = 1.2x10-3 x60 = 0.072 C

b. no. of electrons = total charge /charge of one electron no. of electrons = 0.072/1.6x10-19 = 4.5x1017 electrons


A certain type of rechargeable battery is capable of delivering a current of 0.2A for 4000s before its voltage drops and it needs to be recharged.

Calculate:

a. The total charge the battery can deliver before it needs to be recharged.

b. The maximum time it could be used for without being recharged if the current through it was i. 0.5A, ii. 0.1A

Ans: a. I = 0.2A t = 4000s Q = It = 0.2 x 4000 = 800Cb. i. Q = 800C I = 0.5A t = Q/I = 800/0.5 = 1600s ii. Q = 800C I = 0.1A t = Q/I = 800/0.1 = 8000s

Quiz

Current and charge quiz

1. Calculate the charge passing through a lamp in three minutes when a steady current of 0.4 A is flowing.

Q = I tQ = 0.4 x 3 x 60 = 72 Coulomb

I = Qt


2. Calculate the number of electrons flowing through a resistor when a current of 2.3 flows for 5 minutes

Q = I x tQ = 2.3 x 5 x 60 = 690 Coulombno. of electrons = 690 /1.6x10-19 = 4.31x1021

I = Qt


3. What is the current in a circuit if 2.5x1020 electrons pass a given point every 8 seconds

Charge (C) = no. of electrons x charge of one electrons

Charge (C) = 2.5x1020 x 1.6x10-19 = 40 Coulombs

I = Qt

Current = 40/8 = 5 Amps


4. How long does it take for a current of 0.3A to supply a charge of 48C?

t = Q/I t = 48/0.3 = 160 seconds

I = Qt


5. How many electrons pass a point when a current of 0.4A flows for 900 seconds?

Q = I x t = 0.4 x 900 = 360 Coulomb

I = Qt

no. of electrons = total charge / charge of one electron

no. of electrons = 360 / 1.6 x 10-19 = 2.25 x 1021


6. A torch bulb passes a current of 120 mA.How many coulombs of charge flow through the lamp in 1 minute?

Q = I x t = 120x10-3 x 60 = 7.2 Coulomb


7. A car battery is rated as 36 A h. In principle this means it could pass a current of 1 A for 36 h before it runs down. How much charge passes through the battery if it is completely run down?

Q = I x t = 1 x 36 x 60 x 60 = 129600 Coulomb

H/W

• 2004 HL Q4

More basic ideas…

Another battery means more current as there is a greater push on the electrons

The extra resistance from the extra bulb means less current

Current in a series circuit

If the current here is 2 amps…

The current here will be…


And the current here will be…

In other words, the current in a series circuit is THE SAME at any

point

2A

2A

2A

Current in a parallel circuit

A PARALLEL circuit is one where the current has a “choice of routes”

Here comes the current…

And the rest will go down here…

Half of the current will go down here (assuming the bulbs are the same)…

Current in a parallel circuit

If the current here is 6 amps




And the current here will be…6A

2A

2A 2A

Voltage in a series circuit

V

V V

If the voltage across the battery is 6V…

…and these bulbs are all identical…

…what will the voltage across each bulb be? 2V

Voltmeter always in parallel

Voltage in a series circuit

V

V

If the voltage across the battery is 6V…

…what will the voltage across two bulbs be? 4V

Voltage in a parallel circuit

If the voltage across the batteries is 4V…

What is the voltage here?

And here?

V

V4V

4V

Summary

In a SERIES circuit:

Current is THE SAME at any point

Voltage SPLITS UP over each component

In a PARALLEL circuit:

Current SPLITS UP down each “strand”

Voltage is THE SAME across each”strand”

An example question:

V1

V2

6V

3A

A1

A2

V3

A3

Advantages of parallel circuits…

There are two main reasons why parallel circuits are used more commonly than series circuits:

1) Extra appliances (like bulbs) can be added without affecting the output of the others

2) When one breaks they don’t all fail

Georg Simon Ohm 1789-1854

Resistance

Resistance is anything that will RESIST a current. It is

measured in Ohms, a unit named after me.

That makes me so happy

The resistance of a component can be calculated using Ohm’s Law:

Resistance = Voltage (in V)

(in ) Current (in A)

V

RI

An example question:

V

A

What is the resistance across this bulb?

As R = volts / current = 10/2 = 5

Assuming all the bulbs are the same what is the total resistance in this circuit?

Total R = 5 + 5 + 5 = 15 Voltmeter reads 10V

Ammeter reads 2A

More examples…

12V

3A

3A

6V

4V

2A

1A

2V

What is the resistance of these bulbs?

Practice with Ohm’s Law

Ohms Volts Amps4 100 25

15 150 102 30 159 45 56 48 8

http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc

VARIATION OF CURRENT (I) WITH P.D. (V)

A

V+

6 V-

Nichrome wire

Method

1. Set up the circuit as shown and set the voltage supply at 6 V d.c.

2.Adjust the by moving the slider of the potential divider to obtain different values for the voltage V and hence for the current I.

3.Obtain at least six values for V and I using the voltmeter and the ammeter.

4.Plot a graph of V against I

Variations

(a) A METALLIC CONDUCTORWith a wire

(b) A FILAMENT BULB (c) COPPER SULFATE SOLUTION

WITH COPPER ELECTRODES(d) SEMICONDUCTOR DIODE

Done both ways with a milli-Ammeter and the a micro Ammeter

Current-voltage graphsI

VI

V

I

V

1. Resistor 3. Diode

2. BulbCurrent increases in proportion to voltage

As voltage increases the bulb gets hotter and resistance increases

A diode only lets current go in one direction

Factors affecting Resistance of a conductor

• Resistance depends on– Temperature– Material of conductor– Length – Cross-sectional area Temperature

The resistance of a metallic conductor increases as the temperature increases e.g. copperThe resistance of a semiconductor/insulator decreases as the temperature increases e.g. thermistor.

VARIATION OF THE RESISTANCE OF A METALLIC CONDUCTOR WITH

TEMPERATURE

Water Wire wound on frame

Glycerol

Heat source

10ºC

Digitalthermometer

Ω

10º C

Method1. Set up as shown.

2. Use the thermometer to note the temperature of the glycerol, which is also the temperature of the coil.

3. Record the resistance of the coil of wire using the ohmmeter.

4. Heat the beaker.5. For each 10 C rise in temperature record

the resistance and temperature using the ohmmeter and the thermometer.

6. Plot a graph of resistance against temperature.

Graph and Precautions

Precautions - Heat the water slowly so temperature does

not rise at end of experiment-Wait until glycerol is the same temperature as

water before taking a reading.

R

LengthResistance of a uniform conductor is directly proportional to its length.

i.e. R L


Cross-sectional areaResistance of a uniform conductor is inversely proportional to its cross-sectional area.

i.e. R 1 A


• MaterialThe material also affects the resistance of a

conductor by a fixed amount for different materials. This is known as resistivity ().

R = L = Resistivity A Unit: ohm meter m

RESISTIVITY OF THE MATERIAL OF A WIRE

Micrometer

Metre stick

l

Bench clamp

Stand

Nichrome wire Crocodile clips

Method1. Note the resistance of the leads when the crocodile

clips are connected together. Could also be precaution.2. Stretch the wire enough to remove any kinks or ‘slack’ in

the wire.3.Read the resistance of the leads plus the resistance of wire

between the crocodile clips from the ohmmeter. Subtract the resistance of the leads to get R.

4.Measure the length l of the wire between the crocodile clips, with the metre stick.

5.Increase the distance between the crocodile clips. Measure the new values of R and l and tabulate the results.

6.Make a note of the zero error on the micrometer. Find the average value of the diameter d.

1. Calculate the resistivity where A =

2. Calculate the average value.

Precautions Ensure wire is straight and has no kinks like ....Take the diameter of the wire at different angles

,Al

Rñ

4

2d

ρ

H/W

• 2004 HL Q4

Resistors in series and Parallel

321 IIIIT

V1

I

I1

V

I2

IT

R1R1

R2 R3

R2

321 VVVVT


321 IRIRIRIRT

V1

I

I1

V

I2

IT

R1R1

R2 R3

R2

321 RRRRT

321 VVVVT


321 R

V

R

V

R

V

R

V

T

V1

I

I1

V

I2

IT

R1R1

R2 R3

R2

321

1111

RRRRT

321 IIIIT

H/W

• 2005 HL Q9

Wheatstone BridgeUses

– Temperature control– Fail-Safe Device (switch

circuit off)– Measure an unknown

resistance

– R1 = R3 (When it’s balanced R2 R4 Galvanometer reads

zero)

Metre Bridge R1 = R2 (|AB|)

|BC|

I

r 1

r2

r 4

r3

A C

B

D

Effects of an Electric Current

•Heat•Chemical•Magnetic

Chemical Effects of an Electric Current

• Electrolysis is the chemical effect of an electric current

• Voltameter consists of electrodes, an electrolyte and a container

• Inactive electrodes are electrodes that don’t take part in the chemical reaction e.g. platinum in H2SO4

• Active electrodes are electrodes that take part in the chemical reaction e.g. copper in CuSO4

Chemical Effects• Ion is an atom or molecule

that has lost or gained 1 or more electrons

• Charge Carriers in an electrolyte are + and – ions

Uses Electroplating to make metal look better, prevent corrosion Purifying metals Making electrolytic capacitors

Current-voltage graphs

I

V

I

V

1. Active Electrodes

2. Inert Electrodes

e.g. Copper in Copper Sulphate

e.g. Platinum in Water

Current Carriers

Medium Carrier

Solid (Metal) Electrons

Liquid (Electrolyte) Ions

Gas Electrons and Ions

Resistance in Semiconductors

2) Thermistor – resistance DECREASES when temperature INCREASES – Due to more charge carriers being liberated by heat

1) Normal conductor like metal resistance increases as vibrating atoms slow the flow of electrons

Resistance

Temperature

Resistance

Temperature

Fuse – Safety device

Fuses are designed to melt when too large a current tries to pass through them to protect devices.

Prevent Fires

Modern fuse boxes contain MCB (Miniature circuit breakers) that trip when too much current flows to protect the circuit

2A5A

Which Fuse

• A i-pod charge uses 200W and is plugged into the mains at 230v. What fuse is in the plug?

• P=I.V• 200=I.230• I = 200/230 = 0.87A is current used• So the most the fuse should be is a

1A

Other safety devices…1) Insulation and double insulation

2) Residual Current Circuit Breaker

In some parts of Europe they have no earth wire just two layer of insulating material the sign is

An RCCB (RCB) detects any difference in current between the live and neutral connectors and the earth it switches off the current when needed. They can also be easily reset.

http://www.picsearch.com/info.cgi?q=wire%20insulation&cid=928494693268

Electrical Safety• A combination of fuse and Earth

A.C. Supply

That Hurts!

The fuse will melt to prevent electrocution and the electricity is carried to earth

The casing touches the bare wire and it becomes live

Fuse on live wire !!

Wiring a plug

Earth wire

Neutral wire

Insulation

Live wire

Fuse

1.

2.

3.

4.

5.

6. Cable grip

Capacitors• A device for storing

charge.• A pair of metal plates are

separated by a narrow gap -

-----

-

+

+++++

- - -

electrons

capacitor charge

charged capacitor

capacitor discharge

electrons

Charge & Discharge

Capacitor Construction• Two metal plates• Separated by

insulating material• ‘Sandwich’

construction• ‘Swiss roll’ structure• Capacitance set by...

d

AC

Uses of Capacitors

• Storing charge for quick release – Camera Flash

• Charging and discharging at fixed intervals – Hazard Lights

• Smoothing rectified current – See Semiconductors

Smoothing

• Add capacitor

http://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svg

variable capacitor

smoothing capacitors

Parallel Plate Capacitors• The size of the capacitor depends on1. The Distance the plates are apart d

-

-

-

+

+

+

d

Parallel Plate Capacitors

2 /.The area of overlap A

-

-

-

+

+

+

A

Parallel Plate Capacitors• 3/.The material between ()

-

-

-

+

+

+

High material

Called a

DIELECTRIC

--

--

++

++

Finding Capacitance

A

s

V

Vecapacitanc

Equations

C

d

A=

For the parallel plate capacitor

Distance in meters

Area In m2

Permitivity inFm-1

CapacitanceIn Farads

Example 1

0

C0.01m

0.04m2

=

The common area of the plates of an air capacitor is 400cm2 if the distance between the plates is 1cm and ε0=8.5x10-12Fm-1.

C

d

A=

8.5x10-12Fm-

1x=3.4x10-11F.

Capacitance experiment on the internet

http://www.absorblearning.com/media/attachment.action?quick=14a&att=2889

http://www.absorblearning.com/media/attachment.action?quick=14a&att=2889

Equations

C

V

Q=

Capacitance on any conductor

Potential Difference in volts

Charge in Coulombs


Placing a charge of 35μC on a conductor raises it's potential by 100 V. Calculate the capacitance of the conductor.

Info Q = 35μC and V = 100V find C=?

Using Q=VC or C = Q/V

= 35 x 10-6/100

= 35 x 10-8 Farads

Equations

C½Work Done

(V)2=

Energy stored on a capacitor

Voltage Squared


Energy Stored

Example 3Find the capacitance and energy stored of a

parallel plate capacitor with 2mm between the plates and 150cm2 overlap area and a dielectric of relative Permittivity of 3. The potential across the plates is 150V.

A = 150cm2=0.015m2, d = 2x10-3m,

ε = 3xε0 = 27x10-12Fm-1

As C = ε0A/d = 27x10-12 x 0.015/0.002 = 2.025x10-9 F

Energy stored = ½ C V2 = ½ x 2.025x10-9x (150)2

= 2.28x10-5 Joules

Types of BatteriesType of Battery

Contains Uses

Wet cell rechargeable

Lead and acid Cars, industry

Dry cell rechargeable

Nickel, cadmium, lithium

Mobile phones, power tools

Dry cell non-rechargeable

Zinc, carbon, manganese, lithium

Torches, clocks, hearing aids

Why use rechargeable batteries?

• Long long-term expense

• Can be used many times

• Less energy to produce

Why use standard batteries?

• No need for charger

• Less expensive

• Rechargeables contain carcinogens

http://www.softgun.ch/shop/catalog/product_info.php?cPath=33_79&products_id=117&osCsid=bb0137b2664b3d2e9f3f706ddfdd136c

There are 2 types of currents:

• Direct Current (DC) – Where electrons flow in the same direction in a wire.

There are 2 types of currents:

• Alternating Current (AC) – electrons flow in different directions in a wire

DC and AC

DC stands for “Direct Current” – the current only flows in one direction:

AC stands for “Alternating Current” – the current changes direction 50 times every second (frequency = 50Hz)

Find Root Mean Square of voltage by

Vrms= Vpeak/ √2

1/50th s

240V

V

V

Time

T

The National Grid

If electricity companies transmitted electricity at 240 volts through overhead power lines there would be too much energy lost by the time electricity reached our homes.

This is explained by JOULES LAW

Power stationStep up

transformerStep down

transformerHomes

The National Grid

Power Transmitted is = P = V.I

JOULES LAW gives us the power turned into heat

Power Lost = I2R

So if we have a high voltage we only need a small current. We loss much less energy

Power stationStep up

transformerStep down

transformerHomes

Power loss in Transmission lines

A power company wants to send 100000w of power by a line with a resistance of 12 ohms. If it uses 100A as the current

Power transmitted = V . I 100000 = V . 100 So V=1000Volts

But the loss is from Joules law = I2R= (100)2.12 = 120000watts

Power loss in Transmission lines

If we want the same power but use only 1A as the current

Power transmitted = V . I 100000 = V . 1

So V=100000Volts But the loss is from Joules law = I2R

= (1)2.12 = 12watts

10000 times less!

Joules law

Heating coil Lagging

Calorimeter Water

A

LidDigitalthermometer

10°C

Method1. Put sufficient water in a calorimeter

to cover the heating coil. Set up the circuit as shown.

2. Note the temperature. 3. Switch on the power and simultaneously

start the stopwatch. Allow a current of 0.5 A to flow for five minutes. Make sure the current stays constant throughout; adjust the rheostat if necessary.

4. Note the current, using the ammeter. 5. Note the time for which the current

flowed. 6. Stir and note the highest temperature.

Calculate the change in temperature ∆.

Calculation and GraphRepeat the above procedure for increasing values of current I, taking care not to exceed the current rating marked on the rheostat or the power supply. Take at least six readings. Plot a graph of ∆(Y-axis) against I 2 (X-axis).

A straight-line graph through the origin verifies that ∆ I 2 i.e. Joule’s law.

Electrical Power lost as Heat P I2 is Joules lawThe power lost (Rate at which heat is produced) is

proportional to the square of the current.

∆

I2

H/W

• 2006 HL Q 4

Experiment to Show shape of Electric Field

• The electrodes connected to high voltage source is placed in the shallow glass dish containing a mixture of semolina and castor oil. The semolina aligns itself along the lines of the electric field.

http://lecturedemo.ph.unimelb.edu.au/var/lecdem/storage/images/electrostatics/el_5_visualisation_of_electric_field_lines/efiled_visualisation_pic_4/5289-1-eng-AU/efiled_visualisation_pic_4_imagelarge.jpg

http://lecturedemo.ph.unimelb.edu.au/var/lecdem/storage/images/electrostatics/el_5_visualisation_of_electric_field_lines/efiled_visualisation_pic_3/5286-1-eng-AU/efiled_visualisation_pic_3_imagelarge.jpg

The Electroscope

+

+

++

- - - -

The electroscope detects charge The Gold leaf and post repel each other

H/W

• 2006 HL Q9

Electric and Magnetic Fields• Electric Field- region of space where a

charged particle feels a electrostatic force.

• Magnetic field – region where a magnet feels a force other than gravity.

• Field lines are the path a positive charge or north pole would travel

Coulomb's Law

• Force between two charged bodies

Force = f Q1.Q2

d2

Q1 Q2d

Put this as a sentence to get a law!

Coulomb Calculations

• We replace the proportional with a equals and a constant to get an equation

Force =f Q1.Q2

d2

Force = f = Q1.Q2

4d2

= permitivity as in capacitors

Coulomb's Law Calculations

• Force between these bodies

Force = f = Q1.Q2

4d2

2C 4mCd=2m

= 3.4 x 10-11



Force = f = 2 x 0.004

4 x3.4 x 10-11x 22

2C 4mCd=2m



Force = f = 7.49 x 10-15 N

2C 4mCd=2m



2C 4mCd=2m

Electric Field Strength = E = F/q

Electric Field Strength =

E = 7.49 x 10-15 N /2C

= 3.75 x 10-15 N /C

Precipitator

• Carbon and ash - can be removed from waste gases with the use of electrostatic precipitators

Precipitator

• Dirt particles are charged then made to stick to oppositely charged plates

Photocopier

• Charging:• Exposure: • Developing:• Transfer: • Fusing:• Cleaning:

Potential Difference (V)

Potential difference is the work done per unit charge to transfer a charge from one point to another (also Voltage)

i.e V = W Q

Potential Difference (V)

V = W Q

Unit Volt V or J C-1

Volt is the p.d. between two points if one joule of work is done bringing one coulomb from one point to the other

Potential at a point is the p.d. between a point and the Earth, where the Earth is at zero potential

Current in a Magnetic Field

N S N S

Current in a Magnetic Field

N S

Force

CurrentMagnetic Field

A conductor carrying a current in a magnetic field will always feel a force

The force is perpendicular to the current and the field. – This is THE MOTOR EFFECT

Fleming’s Left Hand Rule

I used my left hand to show the direction the wire would move

The Size of the Force

Force = F = B.I.lWhere B = Magnetic Field Density in Tesla (T)

I= Current in Amps (A)…………………………… L = length if the conductor in metres…

Example What is the force acting on a conductor of length 80cm carrying a current of 3A in a 4.5T magnetic field?

Using Force = F = B.I.l = 4.5x3x0.8

= 10.8N

Two Parallel Wires• Wires also produce magnetic fields

when a current flows

Attraction

Two Parallel Wires• The fields act like magnets when the

current flows

Repulsion

The Ampere• Basic unit of electricity

F=2x10-

7N/m

1m

The current flowing is 1A when the force between two infinitely long conductors 1m apart in a vacuum is 2x10-7N Per metre of length.

Demo

• OHP and coils and compass

Moving Charge• When any charged particle moves it is like a

small current of electricity• It feels the same force• The crosses show a magnetic field into the

screen

e-Velocity

Force

e -

VelocityForce

e -

Velocity

Forcee-

e-

Moving Charge• A positive will move the other way

e-Velocity

Force

+

All charged particles moving in magnetic fields always have a force at right angles to their velocity so follow a circular path due to FLH Rule

See particles motion

http://www.absorblearning.com/media/attachment.action?quick=17d&att=3111

Force 0n a Particle

Force = F = B.q.vWhere B = Magnetic Field Density in Tesla (T)

q=charge on the particle (C)

v=velocity of the particle…

Example What is the force acting on a particle travelling at 80m/s carrying a charge of 0.1C in a 10T magnetic field?

Using Force = F = B.q.v= 10x.1x80

= 80N

Demo

• CRT and magnet

Inductionis where changes in the current flow in a circuit are caused by changes in an external field.

N

Moving Magnet

Circuit turning off and on

Electromagnetic induction

The direction of the induced current is reversed if…

1) The magnet is moved in the opposite direction

2) The other pole is inserted first

The size of the induced current can be increased by:

1) Increasing the speed of movement

2) Increasing the magnet strength

3) Increasing the number of turns on the coil

Demo

• Coils and spot galvo• Internet

http://phet.colorado.edu/en/simulation/faraday



Generators (dynamos)

Induced current can be increased in 4 ways:

1) Increasing the speed of movement

2) Increasing the magnetic field strength

3) Increasing the number of turns on the coil

4) Increasing the area of the coil

Electric motor

http://www.absorblearning.com/media/attachment.action?quick=7k&att=536

Faraday’s LawBasically

1. More turns (N) more EMF

2. Faster movement more EMF

Rate of change of FLUX DENSITY is proportional to induced EMF

Induced EMF = E = - Nd ( =B.A)

dt

Lenz’s LawThe induced EMF always opposes the current/Motion

You get ought for nought

A version of Newton III and of energy conversion

The induction always tries to stop the motion or change in the field.

Aluminum Ring

The ring moves away as the induced current is preventing more induction

Mutual induction

• Induction in a second circuit caused by changes in a first circuit

• Main use in a transformer• As the current changes the

field changes giving a EMF in the second circuit.

TransformersThis how A.C. changes voltage up or down

V In

V Out

Turns 2

Turns 1=

Self Induction

• property whereby an electromotive force (EMF) is induced in a circuit by a variation of current in the circuit its self

D.C. SourceCurrent Back EMF

Another example on LENZ’S LAW

Flux Density• Magnetic flux, represented by the

Greek letter Φ (phi), total magnetism produced by an object. The SI unit of magnetic flux is the Weber

• Magnetic field (B) is the flux through a square meter (the unit of magnetic field is the Weber per square meter, or Tesla.)

As the flux expands the density through any square meter decreases

http://en.wikipedia.org/wiki/Phi

http://en.wikipedia.org/wiki/SI

http://en.wikipedia.org/wiki/Unit_of_measurement

http://en.wikipedia.org/wiki/Weber_%28unit%29

http://en.wikipedia.org/wiki/Tesla_%28unit%29

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