&OLPDWH&KDQJH IB 12 %ODFN%RG\5DGLDWLRQ

Factors that affect how an object absorbs, emits (radiates), and reflects EM radiation incident on them:

1) Nature of the surface: material, shape, texture, etc.

2) Color: a) Light-colored or silvery objects: absorb little energy, reflect most energy

b) Dark objects: absorb most energy, reflect little energy

When the object is in energy absorbed = energy radiated thermal equilibrium with Pin = Pout its surroundings, Iin = Iout

An object that acts as a black-body will . . . absorb all incoming radiation, not reflect any, then radiate all of it.

%ODFNERG\UDGLDWLRQ: radiation emitted by a perfect emitter

When heated, a low-pressure gas will . . .emit a discrete spectrum

When heated, a solid will . . . emit a continuous spectrum

(PLVVLRQ6SHFWUDIRU%ODFN%RGLHV 1. Not all wavelengths of light will be emitted with

equal intensity.

2. Emitted wavelength with highest intensity (max ) is related to . . . temperature.

3. Area under curve is proportional to . . . total power radiated by body

4. As body heats up, max . . . decreases

and total power . . . increases

1

IB 12

IB 12 d) What major assumption was made in calculating the power radiated by Betelgeuse?

That it acts as a black-body

(PLVVLYLW\Hratio of power emitted by an object to the power emitted by a black-body at the same temperature.

Formula: 3H 3%% 3 H3%%

3 H $7 4V e) Compute the power radiated by Betelgeuse if its emissivity is measured to be only 0.90.

3. Calculate the power emitted by a square kilometer of ocean surface at 100C if its emissivity is 0.65.

4. Calculate the power radiated by the Earth if it is taken to be

a) a black-body at 300 K. b) at 300 K with an effective emissivity of 0.62.

3

6RODU5DGLDWLRQ IB 12

1. Calculate the power radiated by the Sun if it is taken to be a black

body at 5778 K and a mean radius of 6.96 x 108 meters.

2. What is the intensity of the solar radiation at the Suns surface?

See pg. 3 of data booklet for 1.5 x 1011 m 3. What is the intensity of the solar radiation that reaches the upper

atmosphere of Earth?

Solar constant: 1360-1370 W/m2 Rounded 1400

4. How much solar energy is incident on the Earth every second?

Take solar constant and multiply by area of disc as cross-section 1.75 x 1017 W

4

IB 12 5. What is the average intensity of the solar energy absorbed by the Earth?

Average 1.75 x 1017 W over whole surface area of Earth = 4 pi r2 340 W/m2

$OEHGRratio of total solar power scattered to total solar power incident

total scattered power D total incident power Formula:

3UHIOHFWHGD 3LQ Meaning: fraction of the total incoming solar radiation that is reflected back out into space

6. What is the albedo of a black-body? 0 What is the emissivity of a black-body? 1

7. Use the diagram at right to determine the Earths average albedo. Atmosphere, clouds, and ground

Global annual mean albedo on Earth: 0.30 = 30%

The Earths albedo varies daily and is dependent on:

1. season

2. cloud formations

3. latitude

8. How much energy is actually absorbed by the Earth each second?

0.70 x 1.75 x 1017 W = 1.23 x 1017 W

5

IB 12 9. Use the results of your prior calculations to estimate the equilibrium temperature of the Earth

and comment on your answer.

Assume black-body emiss=1 Pin=Pout Use SB T= 255 k = -18oC too cold

10. At present, the average temperature of the Earth is measured to be 288 K.

a) Calculate the average emissivity of the Earth.

Actually warmer since atmosphere absorbs some of the radiation b) Comment on why this might be. emitted by Earth surface Start to separate Earth into its parts atmosphere vs. ground

6

7KH*UHHQKRXVH(IIHFW IB 12

*UHHQKRXVH*DVHV each has natural and man-made origins

1) Water Vapor (H2O): evaporation

2) Carbon Dioxide (CO2): product of photosynthesis in plants, product of fossil fuel combustion

3) Methane (CH4): product of decay and fermentation and from livestock, component of natural gas

4) Nitrous Oxide (N2O): product of livestock, produced in some manufacturing processes

*UHHQKRXVH(IIHFW

a) Short wavelength radiation (visible and short-wave infrared) received from the Sun causes the Earths surface to warm up.

b) Earth will then emit longer wavelength radiation (long-wave infrared) which is absorbed by some gases in the atmosphere.

c) This energy is re-radiated in all directions (scattering). Some is sent out into space and some is sent back down to the ground and atmosphere.

d) The extra energy re-radiated causes additional warming of the Earths atmosphere and is known as the Greenhouse Effect.

1. What is the molecular mechanism by which greenhouse gases absorb infrared radiation?

5HVRQDQFH a transfer of energy in which a system is subject to an oscillating force that matches the natural frequency of the system resulting in a large amplitude of vibration

$SSOLFDWLRQWRWKHJUHHQKRXVHHIIHFW

The natural frequency of oscillation of the molecules of the greenhouse gases is in the infrared region (1 300 m)

2. What do the following transmittance and absorption graphs show about the atmosphere?

Sun radiates in visible

Earth radiates in infrared

Water vapor absorbs incoming solar radiation and

outgoing IR radiation

CO2 absorbs outgoing IR radiation

7

IB 12 2XWJRLQJHQHUJ\

The average albedo (reflectivity) of the Earth is about 0.3, which means that 30% of the incident solar energy is reflected back into space, while 70% is absorbed by the Earth and reradiated as infrared. The planet's albedo varies from month to month, but 0.3 is the average figure. It also varies very strongly spatially: ice sheets have a high albedo, oceans low. The contributions from geothermal and tidal power sources are so small that they are omitted from the following calculations. So 30% of the incident energy is reflected, consisting of:

x 6% reflected from the atmosphere x 20% reflected from clouds x 4% reflected from the ground (including land, water and ice)

The remaining 70% of the incident energy is DEVRUEHG:

x 51% absorbed by land and water, then emerging in the following ways: R 23% transferred back into the atmosphere as latent heat by the evaporation of water, called

latent heat flux R 7% transferred back into the atmosphere by heated rising air, called Sensible heat flux R 6% radiated directly into space R 15% transferred into the atmosphere by radiation, then reradiated into space

x 19% absorbed by the atmosphere and clouds, including: R 16% reradiated back into space

R 3% transferred to clouds, from where it is radiated back into space

When the Earth is at thermal equilibrium, the same 70% that is absorbed is UHUDGLDWHG:

x 64% by the clouds and atmosphere x 6% by the ground

Some energy balance climate models for the Earth

8

IB 12

Predict increase in planets temp using SB law like test question

9

2

IB 12

6XUIDFH+HDW&DSDFLW\&6energy required to raise the Formula: Units: temperature of a unit area of a planets surface by 1 K. 4 -&6 $'7 P . 6XUIDFHKHDWFDSDFLW\RI(DUWK CS = 4.0 x 108 J m-2 K-1 4 $& 6 '7

1. How much solar energy is needed to increase the surface temperature of one square kilometer of Earths surface by 2 K?

7HPSHUDWXUH 4 FKDQJHIRUPXOD: &6 $'7 &6 (,LQ ,RXW)'W '7' &6 3 W (,LQ ,RXW)'W'$'7 7 &6

2. If the Earth is in thermal equilibrium, it will emit as much radiation as is incident on it from the Sun (344 W/m2). Suppose a change causes the intensity of the radiation emitted by Earth to decrease 10%.

a) Suggest a mechanism by which this might happen.

Increased amounts of greenhouse gases cause more solar radiation to be trapped in atmosphere

b) Calculate the new intensity of radiation emitted by Earth. 0.90(340) = 306 W/m2

c) Calculate the amount by which Earths temperature would rise over the course of a year as a result.

(,LQ ,RXW)'W'7 &6 (340 306)(365)(24)(3600) '7

4.0 x 108

'7 2.7. 10

*OREDO:DUPLQJ IB 12 *OREDO:DUPLQJ records show that the mean temperature of Earth has been increasing in recent years.

Global mean surface temperature anomaly In specific terms, an increase of 1 or more Celsius degrees in a period of relative to 19611990 one hundred to two hundred years would be considered global warming. Over the course of a single century, an increase of even 0.4 degrees Celsius would be significant. The Intergovernmental Panel on Climate Change (IPCC), a group of over 2,500 scientists from countries across the world, convened in Paris in February, 2007 to compare and advance climate research. The scientists determined that the Earth has warmed .6 degrees Celsius between 1901 and 2000. When the timeframe is advanced by five years, from 1906 to 2006, the scientists found that the temperature increase was 0.74 degrees Celsius.

The global average surface temperature range for each year from 1861 to 2000 is shown by solid red bars, with the confidence range in the data for each

year shown by thin whisker bars. The average change over time is shown by the solid curve.

3RVVLEOHPRGHOVVXJJHVWHGWRH[SODLQJOREDOZDUPLQJ

1. changes in the composition of greenhouse gases may increases amount of solar radiation trapped in Earths atmosphere

2. increased solar flare activity may increase solar radiation

3. cyclical changes in the Earths orbit may increase solar radiation

4. volcanic activity may increases amount of solar radiation trapped in Earths atmosphere

In 2007, the IPCC report stated that:

0RVWRIWKHREVHUYHGLQFUHDVHLQJOREDOO\DYHUDJHGWHPSHUDWXUHVLQFHWKHPLGWK FHQWXU\LV YHU\OLNHO\GXHWRWKHLQFUHDVHLQDQWKURSRJHQLF>KXPDQFDXVHG@JUHHQKRXVHJDVFRQFHQWUDWLRQV. (the HQKDQFHGJUHHQKRXVHHIIHFW)

(QKDQFHG$QWKURSRJHQLF*UHHQKRXVH(IIHFW Human activities have released extra carbon dioxide into the atmosphere, thereby enhancing or amplifying the greenhouse effect.

Major cause: the burning/combustion of fossil fuels

Possible effect: rise in mean sea-level

Outcome: climate change and global warming

A column of gas and ash rising from Mount Pinatubo in the Philippines on June

12, 1991, just days before the volcanos climactic explosion on June 15.

11

(YLGHQFHOLQNLQJJOREDOZDUPLQJWRLQFUHDVHGOHYHOVRIJUHHQKRXVHJDVHV IB 12

7KH.HHOLQJ&XUYH: Named after American climate scientist Charles David Keeling, this tracks changes in the concentration of carbon dioxide (CO2) in Earths atmosphere at a research station on Mauna Loa in Hawaii. Although these concentrations experience small seasonal fluctuations, the overall trend shows that CO2 is increasing in the atmosphere.

,QWHUQDWLRQDO,FH&RUH5HVHDUFK Between 1987 and 1998, several ice cores were drilled at the Russian Antarctic base at Vostok, the deepest being more than 3600 meters below the surface. Ice core data are unique: every year the ice thaws and then freezes again, forming a new layer. Each layer traps a small quantity of the ambient air, and radioactive isotopic analysis of this trapped air can determine mean temperature variations from the current mean value and carbon dioxide concentrations. The depths of the cores obtained at Vostok means that a data record going back more than 420,000 years has been built up through painstaking analysis.

Inspect the graphical representation of the ice core data and draw a conclusion.

There is a correlation between Antarctic temperature and atmospheric concentrations of CO2

0HFKDQLVPVWKDWPD\LQFUHDVHWKHUDWHRIJOREDOZDUPLQJ 1. Global warming reduces ice and snow cover, which in turn reduces the albedo. This will result in an

increase in the overall rate of heat absorption.

2. Temperature increase reduces the solubility of CO2 in the sea and increases atmospheric concentrations.

3. Continued global warming will increase both evaporation and the atmospheres ability to hold water vapor. Water vapor is a greenhouse gas.

4. The vast stretch of permanently frozen subsoil (permafrost) that stretches across the extreme northern latitudes of North America, Europe, and Asia, also known as tundra, are thawing. This releases a significant amount of trapped CO2.

5. Deforestation results in the release of more CO2 into the atmosphere due to slash-and-burn clearing techniques, as well as reduces the number of trees available to provide carbon fixation.

Smoldering remains of a plot of deforested land in the Amazon rainforest of Brazil. Annually, it is estimated that net global deforestation accounts for about 2 gigatons of carbon emissions to the

atmosphere.

12

5LVHLQ6HDOHYHOV IB 12

Generally, as the temperature of a liquid rises, it expands. If this is applied to water, then as the average temperature of the oceans increases, they will expand and the mean sea-level will rise. This has already been happening over the last 100 years as the sea level has risen by 20 cm. This has had an effect on island nations and low-lying coastal areas that have become flooded.

&RHIILFLHQWRI9ROXPH([SDQVLRQfractional Formula: '9 Units: change in volume per degree change in temperature E 19 7' R . ' R ' 9 E9 7

1. The coefficient of volume expansion for water near 20o C is 2 x 10-4 K-1. If a lake is 1 km deep, how much deeper will it become if it heats up by 20o C? 0.4 m

Precise predictions regarding the rise in sea-levels are hard to make for such reasons as:

D $QRPDORXVH[SDQVLRQRIZDWHU: Unlike many liquids, water does not expand uniformly. From 00C to 40C, it actually contracts and then from 40C upwards it expands. Trying to calculate what happens as different bodies of water expand and contract is very difficult, but most models predict some rise in sea level.

E 0HOWLQJRILFH: Floating ice, such as the Arctic ice at the North Pole, displaces its own mass of water so when it melts it makes no difference. But melting of the ice caps and glaciers that cover land, such as in Greenland and mountainous regions throughout the world, causes water to run off into the sea and this makes the sea level rise.

Glaciers on land melting: raise sea level

Sea ice glaciers melting: dont raise sea level

13

IB 123RVVLEOHVROXWLRQVIRUUHGXFLQJWKHHQKDQFHGJUHHQKRXVHHIIHFW 3. Greater efficiency of power production.

To produce the same amount of power would require less fuel, resulting in reduced CO2 emissions.

4. Replacing the use of coal and oil with natural gas.

Gas-fired power stations are more efficient (50%) that oil and coal (30%) and produce less CO2.

5. Use of combined heating and power systems (CHP).

Using the excess heat from a power station to heat homes would result in more efficient use of fuel.

6. Increased use of renewable energy sources and nuclear power.

Replacing fossil fuel burning power stations with alternative forms such as wave power, solar power, and wind power would reduce CO2 emissions.

1. Use of hybrid vehicles

Cars that run on electricity or a combination of electricity and gasoline will reduce CO2 emissions.

2. Carbon dioxide capture and storage (carbon fixation)

A different way of reducing greenhouse gases is to remove CO2 from waste gases of power stations and store it underground.

,QWHUQDWLRQDOHIIRUWVWRUHGXFHWKHHQKDQFHGJUHHQKRXVHHIIHFW

1. ,QWHUJRYHUQPHQWDO3DQHORQ&OLPDWH&KDQJH,3&& Established in 1988 by the World Meteorological Organization and the United Nations Environment Programme, its mission is not to carry out scientific research. Hundreds of governmental scientific representatives from more than 100 countries regularly assess the up-to-date evidence from international research into global warming and human induced climate change.

2. .\RWR3URWRFRO This is an amendment to the United Nations Framework Convention on Climate Change. In 1997, the Kyoto Protocol was open for signature. Countries ratifying the treaty committed to reduce their greenhouse gases by given percentages. Although over 177 countries have ratified the protocol by 2007, some significant industrialized nations have not signed, including the United States and Australia. Some other countries such as India and China, which have ratified the protocol, are not currently required to reduce their carbon emissions.

3. $VLD3DFLILF3DUWQHUVKLSRI&OHDQ'HYHORSPHQWDQG&OLPDWH$33&'& This is a non-treaty agreement between 6 nations that account for 50% of the greenhouse emissions (Australia, China, India, Japan, Republic of Korea, and the United States.) The countries involved agreed to cooperate on the development and transfer of technology with the aim of reducing greenhouse emissions.

14

'LJLWDO7HFKQRORJ\ IB 12 'HFLPDO1XPEHUV

Symbols: 0-9

Place Values: multiples of ten

%LQDU\1XPEHUV Symbols: 0 and 1

Place Values: multiples of two

Least-significant bit (LSB)

Most-significant bit (MSB)

Example: 452 = 4 x 100 + 5 x 10 + 2 x 1

Examples: 10011 = 1 x 16 + 0 x 8 + 0 x 4 + 1 x 2 + 1 x 1 = 19

0 = 0 1 = 1 2 = 10 3 = 11 4 = 100 5 = 101 6 = 110 7 = 111 8 = 1000 9 = 1001

10 = 1010 11 = 1011 12 = 1100 13 = 1101 14 = 1110 15 = 1111 16 = 10000 17 = 10001 18 = 10010 19 = 10011 20 = 10100

'LJLWDOVLJQDO: potential difference is either High (1) or Low (0)

$QDORJXHVLJQDO: potential difference varies continuously with time

0HWKRGRIVWRUDJH 7\SLFDO LQIRUPDWLRQ 2YHUYLHZRISURFHVV $QDORJXHRU

'LJLWDO

Photocopying Text or pictures Optics and electrostatics used to fix

powder to paper Analogue

LPs (vinyl) Music or speech Sound variations stored as grooves

in vinyl Analogue

Cassette tapes Music or speech Sound variations stored in magnetic

fields on tape Analogue

Floppy disks Hard disks All forms

Bits stored as variations in magnetic fields on disk Digital

CD, DVD All forms Bits stored as series of optical bumps to be read by laser Digital

1

6WRULQJ,QIRUPDWLRQRQ&'V IB 12

A CD is a fairly simple piece of plastic, about four one-hundredths (4/100) of an inch (1.2 mm) thick. Most of a CD consists of an injection-molded piece of clear polycarbonate plastic. During manufacturing, this plastic is impressed with microscopic bumps arranged as a single, continuous, extremely long spiral track of data. Once the clear piece of polycarbonate is formed, a thin, reflective aluminum layer is sputtered onto the disc, covering the bumps. Then a thin acrylic layer is sprayed over the aluminum to protect it. The label is then printed onto the acrylic. A cross section of a complete CD (not to scale) looks like this:

The elongated bumps that make up the track are each 0.5 microns wide, a minimum of 0.83 microns long and 125 nanometers high. (A nanometer is a billionth of a meter.) Looking through the polycarbonate layer at the bumps, they look something like this: You will often read about "pits" on a CD instead of bumps. They appear as pits on the aluminum side, but on the side the laser reads from, they are bumps. The incredibly small dimensions of the bumps make the spiral track on a CD extremely long. If you could lift the data track off a CD and stretch it out into a straight line, it would be 0.5 microns wide and almost 3.5 miles (5 km) long.

Reflection from bump or pit:

Constructive interference Interpreted as 0

Reflection from edge of pit:

Destructive interference Interpreted as 1

Condition for destructive interference: height of bump/depth of pit = /4 So path difference between two light beams is /2

Example: What wavelength of laser light should be used to read the data shown encoded at right? 600 nm

2

IB 12$GYDQWDJHVRIGLJLWDOVWRUDJHRILQIRUPDWLRQRYHUDQDORJXHVWRUDJH

'LJLWDO $QDORJXH 4XDOLW\ Output can be virtually

indistinguishable from input Output can be virtually indistinguishable from input but is more liable to damage or corruption (eg. scratches on LPs)

5HSURGXFLELOLW\ Use of laser ensures that each retrieval is virtually identical since light does not damage surface

Process of retrieval often affects quality of future retrievals (eg. needle may scratch LP)

5HWULHYDOVSHHG Very high speed different section can be accessed randomly

Slow retrieval speed data needs to be retrieved in sequential order

3RUWDELOLW\RI VWRUHGGDWD

Miniaturization techniques ensure that large quantities of data can be stored in a small device (eg. flash drives)

Storage devices usually take up much more space

0DQLSXODWLRQRI GDWDHGLWLQJ

Easily achieved with little corruption of data (eg. Photoshop)

All manipulation increases possibility of data corruption

,PSOLFDWLRQVIRUVRFLHW\RIHYHULQFUHDVLQJGDWDVWRUDJHFDSDELOLWLHV

0RUDO(WKLFDO x x x

Information that is potentially problematic can easily be shared (eg. terrorism, crime) Issues concerning ownership of electronic data (eg. piracy) Privacy concerns

6RFLDO x x x

Use in documenting abuses of human rights Unequal access to the Internet Control of information and opinions

(FRQRPLF x x

Quicker access to information needed to make economic decisions (eg. price comparisons) Rise of new businesses and decline of older ones

(QYLURQPHQWDO x x

Reduction in use of paper and other materials used traditionally to stored information Recycling of electronic junk is problematic due to more dangerous materials used in their manufacture

3

'DWD&DSWXUHXVLQJ&&'V IB 12

&KDUJH&RXSOHG'HYLFH&&' a silicon chip divided into small area called pixels. Each pixel can be considered to behave as a capacitor (a device that stores charge). The CCD is used to electronically record an image focused onto its surface.

,PDJHFDSWXUH Incident light causes charge to build up within each pixel due to the photoelectric effect. An electrode then measures the potential difference developed across each pixel and converts it into a digital signal. The position of the pixel is also stored.

,PDJHUHWULHYDOSince both the location and potential difference of each pixel are recorded, all the information needed to store and reconstruct the image is saved. Each p.d. can be converted to a digital signal and the digital signals can be converted to an image. The intensity of the light at each pixel, and thus the image, can be reconstructed.

&DSDFLWDQFH: the ratio of charge to potential difference

Formula: C = Q/V Units: C/V or farad (F) Type: Scalar

1. The potential difference measured across a 100 pF pixel is 25 mV. Determine the charge and number of electrons stored in the pixel.

4XDQWXP(IILFLHQF\ ratio of the number of photoelectrons emitted to the number of photons incident on the pixel

2. If the quantum efficiency of the pixel described above is 90%, how many photons were incident on it?

4

IB 12 0DJQLILFDWLRQratio of the length of the image on the CCD to the length of the object

3. A digital camera is used to photograph an object that is 3.0 x 10-1 m2 in area. The image that is focused onto the CCD is 4.5 x 10-3 m2. What is the magnification of the camera?

5HVROXWLRQ: the ability to distinguish between two sources of light

Applied to CCD: two points on an object may be just resolved on a CCD if the images of the points are at least two pixels apart

4. The CCD of a digital camera has a square image collection area that measures 25 mm on each side and a resolution of 5.0 megapixels. An object that is photographed by the camera has an area of 4.6 x 10-3 m2. The image formed on the CCD has an area of 1.0 x 10-4 m2.

a) Calculate the magnification.

b) Estimate the length of a pixel on the CCD.

c) Two small dots on the object are separated by a distance of 0.20 mm. Deduce whether the images of the dots will be resolved.

5

IB 12

(IIHFWRQTXDOLW\RISURFHVVHGLPDJH

4XDQWXP HIILFLHQF\ The greater the quantum efficiency, the greater the sensitivity of the CCD.

0DJQLILFDWLRQ A greater magnification means that more pixels are used for a given section of the image. The image will be more detailed.

5HVROXWLRQ The greater the resolution, the greater the clarity of the image and the amount of detail recorded.

Some advantages of using CCDs compared with the use of film:

a) Each photo does not require film and is thus cheaper and uses less resources/produces less waste.

b) Traditional film has a quantum efficiency of less than 10% while CCDs can have quantum efficiencies of over 90%. This means that very faint images can be photographed with CCDs.

c) The image is digital and can be stored and edited more easily. d) Images can be viewed immediately with no processing time delay. e) Storage, archiving, and retrieving a large number of photos is easy and efficient.

CCDs are used for image capturing in a large range of the electromagnetic spectrum (not just visible light).

'LJLWDO FDPHUDV

Very convenient to take and share photographs, but image quality can be less than that of traditional film unless the camera is of high quality (more expensive).

9LGHR FDPHUDV

Digitized images are usually better quality than analogue images stored on magnetic videotape and are easier to store and transport. It is possible to continuously record video without interruption during playback. Searches are faster and easier to perform. Digital storage is fast and utilizes re-usable media, an advantage for security cameras.

7HOHVFRSHV Sensitivity of CCDs is better than traditional film and allows for detailed analysis over a range of frequencies. CCDs also allow for remote operation of telescopes, both ground-based and in orbit, like the Hubble space telescope.

0HGLFDO ;UD\ LPDJLQJ

Digital X-rays have better contrast and can be processed, allowing for enhancements and detailed study. Information can be quickly shared between hospitals and more easily stored and retrieved.

6

(OHFWULF&LUFXLWV IB 12 In the electric circuit shown below, energy is transferred from the battery to the light bulb by charges

that move through a conducting wire because of a potential difference set up in the wire by the battery.

The circuit shown contains a typical 9-volt battery.

a) What is the emf of the circuit?

b) How much energy does one coulomb of charge carry around the circuit?

Schematic

mark potentials at each spot

c) How much energy do two coulombs of charge carry around the circuit?

d) How much energy does each coulomb of charge have at point B?

e) How much energy does each coulomb of charge have at point C?

f) What is VB? What is VC?

g) What is VBC? What is VCD? What is VDA?

(OHFWULF&XUUHQW

Formula: Units: Type: I = q/t A (ampere) = C/s Scalar

Unofficial definition: rate of flow of electric charge

2IILFLDO'HILQLWLRQRI2QH$PSHUH$RIFXUUHQWDIXQGDPHQWDOXQLW

One ampere is the amount of current flowing in each of two infinitely-long parallel wires of negligible cross-sectional area separated by a distance of one meter in a vacuum that results in a force of exactly 2 x 10-7 N per meter of length of each wire.

6KRUWIRUP Current is defined in terms of the force per unit length between parallel current-carrying conductors.

Closed circuit: complete pathway Open circuit: incomplete pathway for current Short circuit: circuit with little to no for current break in circuit infinite resistance resistance extremely high current

overheating sketch

sketch

1

5HVLVWDQFH IB 12 5HVLVWDQFH: ratio of potential difference applied across a piece of material to the current through the material

Formula:

R = V / I

Units:

ohm () = V/A

Type:

scalar

For a wire conductor: Formula:

A short fat cold wire is the best conductor R = L/A

A long hot skinny wire has the most resistance

Power: energy per unit time Unit: W = J/s Type: scalar

Mechanical Power:

P = W/t = F s/t = F v

Electrical Power: P = E/t = (q V)/t

P = I V

Alternate Formulas: Substitute V = IR

P = I (IR) = I2 R

P = (V/R)V = V2/ R

Meters in a circuit

Schematic diagram

Ammeter: measures current

Placement: Must be placed in series to allow current to flow through it

Circuit must be broken to insert ammeter

Ideal ammeter: has zero resistance so it will not affect current flowing through it

Voltmeter: measures potential difference

Placement: Must be placed in parallel to measure potential difference between two points circuit does not to be broken

Ideal voltmeter: has infinite resistance so it will not allow any current to flow through it and disrupt circuit

2

6HULHVDQG3DUDOOHO&LUFXLWV IB 12

&KDUDFWHULVWLF 6HULHV&LUFXLW 3DUDOOHO&LUFXLW Number of pathways for current one More than one

Current Same everywhere same for all devices Current splits shared among devices

Potential Difference (Voltage)

Voltage shared among devices voltage splits Same for all devices

Overall resistance high low

Power low high

Formulas:

Voltage Ratio 6HULHV &LUFXLWV

VT = V1 + V2 +

IT = I1 = I2 =

RT = R1 + R2 +

PT = P1 + P2 + . . .

Current Ratio

VT = V1 = V2 = . . .

IT = I1 + I2 + . . .

1/RT = 1/R1 + 1/R2 + . . .

PT = P1 + P2 + . . .

3DUDOOHO &LUFXLWV

Power Ratio Power Ratio

3

$QDO\]LQJ&LUFXLWV IB 12 Determine the current through and the voltage drop across each resistor in each circuit below.

1. 2.

3. 4.

3RWHQWLDO'LYLGHU Resistors in series act as a potential (voltage) divider. They split the potential of the source between them.

5. A 20device requires 40 V to operate properly but no 40 V source is available. In each case below, determine the value of added resistor R1 that will reduce the voltage of the source to the necessary 40V for device R2.

$ % & '

6. A mini light bulb is rated for 0.60 W at 200 mA and is placed in series with a variable resistor. Only a 9.0 volt battery is available to power it. To what value should the variable resistor be set to power the bulb correctly?

Bulb needs only 3 V

Bulb has resistance of 15 at rated power

Added resistance should be 30 ohms

4

7KH8VHRI6HQVRUVLQ&LUFXLWV IB 12

1. /LJKW'HSHQGHQW5HVLVWRU/'5 or /LJKW6HQVRU: A photo-conductive cell made of

semiconducting material whose resistance decreases as the intensity of the incident light increases.

$XWRPDWLFOLJKWVZLWFK Describe how the LDR activates the light switch.

As ambient light decreases, resistance of LDR increases

Potential difference across LDR increases

Switch needs minimum PD to turn on

When light intensity drops to desired level, PD is high enough to turn on switch

2. 1HJDWLYH7HPSHUDWXUH&RHIILFLHQW17&7KHUPLVWRU or 7HPSHUDWXUH6HQVRU: A sensor made of semiconducting material whose resistance decreases as its temperature increases.

)LUHDODUPDescribe how the NTC thermistor activates the fire alarm.

As external temperature increases, resistance of NTC decreases

Potential difference across R2 increases

Switch needs minimum PD to turn on

When temperature increases to desired level, PD is high enough to turn on switch

3. 6WUDLQ*DXJH or)RUFH6HQVRU: A long thin metal wire whose resistance increases as it is stretched since it becomes longer and thinner.

Describe how the strain gauge can measure the strain put on a section of an airplane body.

As strain increases, resistance of strain gauge increases

Potential difference across R2 decreases

Voltmeter can read change in voltage which can be used to determine amount of strain on part

5

&RPELQDWLRQ6HULHV3DUDOOHO&LUFXLWV IB 12

1. Determine the current through and the voltage drop across each resistor.

2. The battery has an emf of 12 V and negligible internal resistance and the voltmeter has an internal resistance of 20 k. Determine the reading on the voltmeter.

3. A cell with negligible internal resistance is connected to three resistors as shown. Compare the currents in each part of the circuit.

6

IB 12 4. Determine the current through and the voltage drop across each resistor.

5. A battery with emf ( and negligible internal resistance is connected in a circuit with three identical light bulbs.

a) Determine the reading on the voltmeter when the switch is open and when it is closed.

b) State what effect closing the switch has on the current through each bulb and the brightness of each bulb.

7

2KPV/DZ IB 12

5HVLVWDQFH: ratio of potential difference applied across a piece of material Formula: to the current through the material R = V / I

Relationship:2KPV/DZ: for a conductor at constant temperature, the current flowing V Ithrough it is proportional to the potential difference across it

2KPLF'HYLFH a device that obeys Ohms law for a wide range of potential differences

Meaning: a device with constant resistance Example: resistor

1. On the axes at right, sketch the ,9FKDUDFWHULVWLFV for a resistor.

Resistance:

a) R = V/I at any point

b) related to slope of graph

(Reciprocal = resistance)

2. A resistor is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it.

a) Calculate the resistance of the resistor.

R = V/I R = 7.5

b) If the voltage is doubled, what is the new current?

V = IR for resistor Resistance is constant so double current

8

IB 12 1RQ2KPLF'HYLFH a device that does not obey Ohms law

Meaning: resistance is not constant Example: filament lamp

1. On the axes at right, sketch the ,9FKDUDFWHULVWLFV for a filament lamp. Resistance:

a) R = V/I at any point

b) as current increases, wire filament heats up and resistance increases

c) Resistance is NOT related to the slope

d) except in initial linear region

2. A flashlight bulb is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it.

a) Calculate the resistance of the bulb.

R = V/I R = 7.5

b) If the voltage is doubled, what is the new current? V = IR for bulb but resistance is not 7.5 ohms any more R increases with T so less than double current

3. Discuss how the resistance varies with increasing potential difference for devices X, Y, and Z.

X: resistance increases - ratio V/I increases

Y: resistance is constant ratio V/I is constant

Z: resistance decreases ratio V/I decreases

9

8VLQJD3RWHQWLRPHWHUWR0HDVXUH,9&KDUDFWHULVWLFV IB 12 3RWHQWLRPHWHU a type of variable resistor with three contact points

Common use: as a potential divider to measure the I-V characteristics of a device

The schematic shows how a potentiometer can be used as a potential divider to measure the I-V characteristics of a filament lamp. It is placed in parallel with the lamp and the slider (center contact point) effectively splits the potentiometer into two separate resistors AB and BC. By moving the slider, the ratio of the voltage drops across the resistors AB and BC is varied.

Redraw the schematic with an ammeter and a voltmeter correctly placed to measure the I-V characteristics of the filament lamp.

Comment on the circuit characteristics as the slider is moved from A to B to C.

Slider at A:

Slider at B:

Slider at C:

10

IB 12,QWHUQDO5HVLVWDQFHRI%DWWHULHV A 6 volt battery is connected to a variable resistor and the current in the circuit and potential difference across the terminals of the battery are measured over a wide range of values of the resistor. The results are shown in the table.

5HVLVWDQFH 3UHGLFWHG &XUUHQW$ $FWXDO

&XUUHQW$ 9ROWDJHDFURVV EDWWHU\9

2000 0.003 0.003 6.00 200 0.03 0.03 5.99 20 0.3 0.29 5.85 2 3 2.4 4.80

0.2 30 8.8 1.71 0.02 300 11.5 0.23 0.002 3000 12.0 0.02 0.0002 30000 12.0 0.00

Why does the current seem to be limited to a maximum of 12.0 amperes and why does the voltage across the battery not remain constant at 6.0 volts?

The battery has some internal resistance. As the external resistance decreases, more and more of the energy supplied by the battery is used up inside the battery.

(OHFWURPRWLYHIRUFHHPI total energy per unit charge supplied by the battery Symbol: RU(

Units: V = J/C

7HUPLQDO9ROWDJH9WHUP potential difference across the terminals of the battery

Ideal Behavior: Vterm always equals emf since no internal resistance

Real Behavior:

1) Think of battery as internal E and tiny internal resistor r

2) Vterm only equals the emf when no current is flowing

3) E is split between R and r

4) When R>>r, Vterm emf

5) As R decreases, Vr increases and VR decreases 11

IB 12

5HODWLRQVKLSEHWZHHQHPIDQGWHUPLQDOYROWDJH

Treat internal resistance as a series resistor

= I RT = I (R + r)

= IR + Ir

Note that in the absence of internal resistance, = Vterm

1. A resistor is connected to a 12 V source and a switch. With the switch open, a voltmeter reads the potential difference across the battery as 12 V yet with the switch closed, the voltmeter reads only 9.6 V and an ammeter reads 0.40 A for the current through the resistor. Sketch an appropriate circuit diagram and calculate the internal resistance of the source.

2. Discuss the expected I-V characteristics for this battery and how they can be experimentally determined.

R can be adjusted from 0 to its max value

A graph of Vterm vs. I can be drawn

Specific equation of graph can be compared to math model to derive internal resistance

Emf = Vterm + Ir

Vterm = -Ir + emf so slope = -r and y-intercept = emf

12

(OHFWURPDJQHWLVP IB 12 Direction of magnetic field lines: the direction that the North pole of a small test compass would point if Magnetic Field placed in the field (N to S) around a Bar

Magnet

7KXPEdirection of conventional current

)LQJHUWLSV direction of magnetic field tangent to circle

What is the cause of magnetic fields? Moving electric charges

Therefore: current in a wire will produce a magnetic field

The Right Hand Rule for the Magnetic Field around a Wire

a) head-on view b) side view c) side view

Draw dots and crosses Draw concentric circles Draw concentric with increasing spacing and circles around wire. arrows in circular fashion.

Alternate Right Hand Rule for Loops )LQJHUWLSV direction of current

7KXPEpoints North Note that a wire bar

magnetloop acts like a:

Component fields Resultant field Your turn

Solenoid: coil of wire many loops Draw the magnetic field around this solenoid. Use alternate RHR to find North.

magnet 1 barNote that a solenoid

acts like a:

0DJQHWLF)RUFHRQD:LUH IB 12 If a wire with current flowing

Two magnetic fields around wire and from through it is placed in an external external magnet will either attract or repel magnetic field, it will experience

a force. Why?

The Right Hand Rule for the Magnetic )ODW+DQG thumb and fingers at right angles Force on a Current-Carrying Conductor in a Magnetic Field )LQJHUV external B field north to south

7KXPE current

3DOP force palm pushes Maximum force occurs when: current is perpendicular to B field

No force occurs when: current is parallel to B field

Use the right hand rule for forces to confirm the direction of the force in each case.

Magnetic field strength0DJQLWXGHRI ) %,/VLQ Magnetic field intensityWKHPDJQHWLF Magnetic flux density

Where is angle between currentIRUFHRQDZLUH and B field Units: Tesla (T)

'HILQLWLRQRIPDJQLWXGH The ratio of the magnetic force on a wire to the product of the current RIPDJQHWLFILHOG in the wire, the length of the wire and the sine of the angle between

the current and the magnetic field B = F / (IL sin )

Find the magnitude and direction of the force on the wire segment confined to the gap between the two magnets as shown when the switch is closed. The strength of the magnetic field in the gap is 1.9 T.

0.62 N up

2

0DJQHWLF)RUFHRQD0RYLQJ&KDUJHG3DUWLFOH IB 12 Why is there a magnetic force on a Moving charged particle creates its own magnetic charged particle as it moves through a field two magnetic fields interact magnetic field?

The Hand Rule for the )ODW+DQG thumb and fingers at )LQJHUV external B field Magnetic Force on a Charge right angles north to south

5LJKW+DQGpositive charge 7KXPE velocity

/HIW+DQGnegative charge 3DOP force palm pushes

Maximum force occurs when: velocity is perpendicular to B field

No force occurs when: velocity is parallel to B field

moving in a Magnetic Field

Find the direction of the magnetic force on each particle below as each enters the magnetic field shown.

0DJQLWXGHRIWKHPDJQHWLFIRUFH ) TY%VLQ RQDPRYLQJFKDUJHGSDUWLFOH

Where is angle between v and B

'HILQLWLRQRIPDJQLWXGH The ratio of the force on a charged particle moving in a magnetic field to the RIPDJQHWLFILHOG product of the particles charge, velocity and sine of the angle between the

direction of the magnetic field and velocity. B = F / (qv sin )

A proton in a particle accelerator has a speed of 5.0 106 m/s. The proton encounters a magnetic field whose magnitude is 0.40T and whose direction makes an angle of T = 30.0 with respect to the proton's velocity. Find the magnitude of the magnetic force on the proton and the protons acceleration. How would these change if the particle was an electron?

a) 1.6 x 10-13 N

b) 9.6 x 1013 m/s2

c) 1.8 x 1017 m/s2, same force but opposite direction

3

0RWLRQRID&KDUJHG3DUWLFOHLQD0DJQHWLF)LHOG IB 12 1. A charged particle will follow a circular path in a magnetic

field since the magnetic force is always perpendicular to the velocity.

2. The magnetic force does no work on the particle since the magnetic force is always perpendicular to the motion.

3. The particle accelerates centripetally but maintains a constant speed since the magnetic force does no work on it.

5DGLXVRI&LUFXODU3DWK

F = m ac FB = m v2 /r

Q v B = m v2 /r

r = mv/ qB

a) Sketch the paths of a slow and a fast

moving proton at constant speed.

c) How would the radius of the path change if the particle were an alpha particle?

b) Sketch the path of a proton that is slowing

down and one that is speeding up.

&RPSDULQJ(OHFWULFDQG0DJQHWLF)LHOGVDQG)RUFHV

(OHFWULF)LHOG 0DJQHWLF)LHOG

draw paths for stationary, parallel and perp charges

4

(OHFWULF)LHOGVDQG0DJQHWLF)LHOGV IB 12 1. A proton is released from rest near

the positive plate and leaves through a small hole in the negative plate where it enters a region of constant magnetic field of magnitude 0.10T. The electric potential difference between the plates is 2100 V.

a) Describe the motion of the proton while in the electric field

constant acceleration in a straight line

b) Describe the motion of the proton while in the magnetic field

constant acceleration and constant speed circular path

c) Find the speed of the proton as it enters the magnetic field.

Use conserv. Of energy a) 6.3 x 105 m/s

d) Find the radius of the circular path of the proton in the magnetic field.

b) 6.6 x 10-2 m

2. A 9HORFLW\6HOHFWRU is a device for measuring the velocity of a charged particle. The device operates by applying electric and magnetic forces to the particle in such a way that these forces balance.

a) Determine the magnitude and direction of an electric field that will apply and electric force to balance the magnetic force on the proton.

perp to v and B from bottom to top of page

F = 0 FB Fe = 0

FB = Fe

qvB = Eq

E = vB

b) What is the resulting speed and trajectory of the proton?

v = E/B in a straight line

5

7KH0DVV6SHFWURPHWHU IB 12 A PDVVVSHFWURPHWHU is a device used to measure the masses of isotopes. Isotopes of the same element have the same charge and chemical properties so they cannot be separated by using chemical reactions but have different masses and so can be separated by a magnetic field. A common type of mass spectrometer is known as the %DLQEULGJHPDVVVSHFWURPHWHU and its main parts are shown below.

Ion Source: source of charged isotopes same charge different mass

Velocity selector: so all ions have the same speed

Magnetic deflection chamber: radius is proportional to mass

1. A singly charged ion with mass 2.18 x 10-26 kg moves without deflection through a region of crossed magnetic and electric fields then is injected into a region containing only a magnetic field, as shown in the diagram, where it is deflected until it hits a photographic plate. The electric field between the plates of the velocity selector is 950 V/m and the magnetic field in both regions is 0.930 T. Determine the sign of the charge and calculate where the ion lands on the photographic plate.

Sign: could be either velocity selector magnetic chamber in velocity selector v = E/B Only positive in v = 1.0 x 103 m/s r = mv/qB deflection chamber

r = 1.5 x 10-4 m

d = 3.0 x 10-4 m

2. A hydrogen atom and a deuterium atom (an isotope of Hydrogen = 0.20 mhydrogen) move out of the velocity selector and into the region of a constant 0.10 T magnetic field at point S, as Deuterium = 0.41 m shown below. Each has a speed of 1.0 x 106 m/s. Calculate where they each hit the photographic plate at P.

6

(OHFWURPDJQHWLF,QGXFWLRQ IB 12 In 1819, Hans Christian Oersted discovered that a magnetic compass experiences a force in the vicinity of an electric current that is, that electric currents produce magnetic fields.

Because nature is often symmetric, this led many scientists to believe that magnetic fields could also produce electric currents, a concept known as HOHFWURPDJQHWLFLQGXFWLRQ.

Free electrons in the wire are charged particles moving through Why does moving a wire through a magnetic field so there is a qvB force on them causing them to a magnetic field induce a current move resulting in a current.in the wire?

'HULYDWLRQRIIRUPXODIRU(0)LQGXFHGLQDPRYLQJZLUH A straight conductor is moved at constant velocity perpendicular to a uniform magnetic field.

1. Electrons in the moving conductor experience a downward magnetic force and migrate to the lower end of the conductor, leaving a net positive charge at the upper end.

2. As a result of this charge separation, an electric field is built up in the conductor.

3. Charge builds up until the downward magnetic force is balanced by the upward electric force due to the electric field. At this point, the charges stop flowing and are in equilibrium.

4. Because of this charge separation, a potential difference is set up across the conductor.

FB = qvB

LHR for electrons to show direction of force

E

FB = Fe qvB = Eq E = vB

V = E d = E V= vB %Y

If the conductor is connected to a complete circuit, the induced emf will produce an induced current.

is equivalent to $PRXQWRI&XUUHQW 'LUHFWLRQRI&XUUHQW The amount of induced The direction of the induced emf current in the circuit is and induced current can be found

given by from the right hand rule for forces to find the force on a positive

charge in the conductor. %Y

= I R

I = Bv/R

7

, IB 127ZR2SSRVLQJ)RUFHV

, The magnetic force acts to oppose the)% applied force, like drag Y

or friction.

At a constant speed, Fapp = FB = BI

An applied force (Fapp) in the direction of the velocity induces an The induced current now

emf which causes current to be generates a magnetic field around the moving bar that causes a pushed upwards. magnetic force (FB) on itself.

TY% 3DOPSXVKHVFXUUHQWXS

)% 3DOPSXVKHVEDUEDFN )% %,O

Suppose a rod is moving at a constant speed of 5.0 m/s in a direction perpendicular to a 0.80-T magnetic field as shown. The rod has a length of 1.6m and negligible electrical resistance. The rails also have negligible resistance. The light bulb, however, has a resistance of 96 :. Find:

a) the emf produced by the motion of the rod 6.4 V

(b) the magnitude and direction of the induced current in the circuit

0.067 A CCW

c) the electrical power delivered to the bulb 0.43 W

d) the energy used by the bulb in 60.0 s. 26 J

e) How much external force is applied to keep the rod moving at this constant speed? 0.086 N

f) How much work is done by the applied force in 60.0 seconds? 26 J

g) What happens to this work? Converted to electrical energy

8

0DJQHWLF)OX[ IB 12 6\PERO: 0DJQHWLF)OX[

Number of field lines 8QLWVWeber (Wb) = T m2

0DJQHWLF)OX['HQVLW\ ILHOGVWUHQJWKLQWHQVLW\

)RUPXOD: B = / A 6\PERO: B

number of field lines per unit area or = B A 8QLWV

Wb/m2 = T (tesla)

$QJOH'HSHQGHQFHRI)OX[ What is the amount of magnetic flux if the field lines are not perpendicular to the cross-sectional area? Only the perpendicular component of the magnetic field contributes to the magnetic flux.

1RUPDOOLQH: line perpendicular to plane of cross-sectional area

)RUPXOD: = (B cos ) A = B A cos

$QJOH: = angle between normal line and field lines

0DJQHWLF)OX[ product of the magnetic field strength and a cross-sectional area and the cosine of the angle between the magnetic field and the QRUPDO to the area

)RUPXOD %$FRV 8QLWV7P

0DJQHWLFIOX[OLQNDJHPDJQHWLFIOX[OLQNLQJDFRLOproduct of magnetic flux through a coil of wire and the number of turns of the wire

)RUPXOD1 1%$FRV 8QLWV7P

1. A single loop of wire whose cross-sectional area is 0.50 m2 is located in a 0.20 T magnetic field as shown. Calculate the flux through the loop in each case.

a) 0.10 T m2

b) 0.050 T m2

c) 0

2. If the coil of wire in the above example consisted of 50 turns of the wire, calculate the amount of flux linking the coil in each case.

a) 5.0 Tm2 b) 2.5 T m2 c) 0

9

(0),QGXFHGE\D7LPH&KDQJLQJ)OX[ IB 12

Moving a magnet towards a coil will Holding the magnet stationary will not Moving the magnet away from the coil increase the magnetic flux linking change the amount of magnetic flux will decrease the magnetic flux linking the coil and will induce an emf and linking the coil and so will not induce the coil and will induce an emf and a

a current in a certain direction. an emf or current. current in the opposite direction.

Methods of inducing an EMF by a time-changing flux

2. Move magnet or coil 2. Rotate coil 1. Vary magnetic field

)DUDGD\V/DZ an induced emf is proportional to the rate of change of the flux linkage

)RUPXOD = - N (/t)

1. A coil of area 0.030 m2 with 300 turns of wire rotates as shown in 0.10 second in a magnetic field of constant 0.25 T strength.

a) What is the magnitude of the induced emf? 11.3 V

b) What is the magnitude of the induced emf if the coil were stationary at 00 but the field strength changed from 0.25 T to 0.60 T in 0.10 second? 22.5 V

10

IB 12 2. A 50 turn coil of wire of area 0.20 m2 is perpendicular to a magnetic field that varies

with time as shown by the graph.

a) Determine the emf induced in the coil during each time interval. 3 V, 0 V, -1.5 V

b) Sketch a graph of the induced emf vs. time. Emf = - derivative of flux

/HQ]V/DZ)LQGLQJWKH'LUHFWLRQRIWKH,QGXFHGHPI /HQ]V/DZ The direction of an induced emf is such that it produces a magnetic field whose flux opposes the flux change that induced it.

(An emf will be induced so as to keep the net flux constant.)

a) Original flux change an increasing b) Induced flux opposes increasing c) Result - two magnetic fields flux induces an emf and current. flux by pointing in opposite direction acting to keep net flux constant.

thus current is in direction shown.

1. If the magnetic field linking this coil is decreasing 2. The diagrams show a conducting ring that is placed in a uniform with time, in which direction is the induced current? magnetic field. Deduce the direction of the induced current in each

case if there is

(a) an increasing B field (b) a decreasing B field 11

IB 12 3. If the current in the wire is increasing, in which 4. If the wire loop moves away from a steady current

direction will there be an induced current in the in the straight wire, in which direction will there be rectangular wire loop? an induced current in the loop?

5. A conducting loop moves at a constant speed into and through a uniform magnetic field as shown in the diagram. Indicate the direction of the induced current. Graph the flux through the loop and the induced emf as a function of time.

6. If a clockwise current through the primary 7. Determine the direction of the 8. Determine the direction(s) of coil is increasing with time, what effect current in the solenoid in the induced current as the will this have on the secondary coil? each case. magnet falls through the loop.

12

$OWHUQDWLQJ&XUUHQW*HQHUDWRUV IB 12 Basic Operation:

1. coil of wire is turned by mechanical means in an external magnetic field

2. emf and current are induced in coil as coil cut flux lines

3. current varies in magnitude and direction as flux linkage changes current and emf variations are sinusoidal

4. brushes and rings maintain contact with external circuit without getting tangled

5RWDWLRQRID&RLOLQD8QLIRUP0DJQHWLF)LHOGLQGXFHVDQ(0)

3RVLWLRQAs the coil rotates, the flux 0D[LPXP(0)DQG&XUUHQW

linking it changes 1. sides of coil cut field lines perpendicularly

2. plane of coil is parallel to field lines

3. normal to coil is perpendicular to field lines (900)

3RVLWLRQ 0LQLPXP(0)DQG&XUUHQW 1. sides of coil do not cut field lines perpendicularly

(move parallel to them)

2. plane of coil is perpendicular to field lines

3. normal to coil is parallel to field lines (00)

Sketch a graph of the induced emf for a coil with: Mark when the coil is in Sketch the graph of the

positions 1 and 2. induced current. twice the frequency of rotation. half the frequency of rotation.

13

$OWHUQDWLQJ&XUUHQW IB 12 The output of an AC generator is an emf V0 = peak/ maximum voltage I0 = peak/ maximum current

that varies sinusoidally with time.

V = Vo sin t (where = 2f)

I = V/R so

I = (Vo/R) sin t = Io sin t

The power output of an AC generator P = I V

P = (Io sin t)(Vo sin t)

P = Io Vo sin2 t

Pav = Io Vo

= (Io/ rad 2)(Vo / rad2)

= Irms Vrms

Maximum Power

Pmax = Io Vo

Pmax = 2 Pav

Average Power RMS Values

Irms = Io / rad 2

Vrms = Vo / rad 2

Root-Mean-Squared values (RMS):

The rms value of an alternating current (or voltage) is that value of the direct current (or voltage) that dissipates power in a resistor at the same rate.

1. In the USA, most household voltage is stated as 120 V at 60 Hz. This is the root-mean-square voltage and the frequency of the AC voltage. Calculate the maximum voltage and mark Vo, Vrms, on the graph.

Vo = 170 V

2. In Europe, the mains electricity is rated at 230 V. What is the peak household voltage in Europe?

14

IB 12

Rating: rms values are given as the AC values to be used in calculations, as if they were DC values

Formula: R = V0/I0 = Vrms/Irms

1. A stereo receiver applies an AC voltage of 34 V to a speaker. The speaker behaves approximately as if it has a resistance of 8.0 :, as the circuit figure indicates. Determine a) the maximum voltage, b) the rms current,

c) the average power for this circuit.

a) 48 V b) 4.25 A c) 145 W

2. A 100 W light bulb is designed to operate from a 120 VAC mains. Determine:

a) the maximum power of the light bulb

b) the maximum current drawn by the bulb

a) 200 W b) 1.2 A

3. A maximum alternating voltage of 170 V is applied across a 50 resistor. Determine:

a) the maximum current through the resistor

b) the average power dissipated by the resistor

a) 3.4 A b) 289 W 15

7KH7UDQVIRUPHU IB 12

According to Michael Faradays original experiment that first produced electromagnetic induction, an emf and current were only induced in the secondary coil when the switch in the primary coil was being opened or closed, that is, when the current in the primary coil was changing (increasing or decreasing).

No emf or current was induced in the secondary coil while the switch was stationary in the open or closed position, that is, when the current was steady or off.

Therefore, emf can only be induced in the secondary coil when the magnetic field from the current in the primary coil is building up or dying down, that is, while the magnetic flux is changing.

7UDQVIRUPHUa device that increases or decreases AC voltage.

6WUXFWXUHDQGRSHUDWLRQRIDWUDQVIRUPHU NP, then VS > VP and voltage increases from primary to secondary

6WHS'RZQ7UDQVIRUPHU: If NS < NP, then VS < VP and voltage decreases from primary to secondary

16

IB 12

How can the voltage increase or decrease without violating the conservation of energy principle?

The power input at the primary equals the power output at the secondary. (This assumes 100% efficiency and such a transformer is termed an LGHDOWUDQVIRUPHU.)

,GHDO7UDQVIRUPHU)RUPXOD

PP = PS

VP IP = VS IS

VP/ VS = IS / IP

Voltage and current in inverse ratio

1. A 120 VAC wall outlet is used to run a small electronic appliance with a resistance of 2.0 , as shown in the diagram.

a) Is the transformer a step-up or step-down transformer? Cite evidence for your answer.

b) How much voltage does the device need? c) If the current in the primary coil is 150 mA, how much current does the device use? Assume an ideal transformer.

a) down b) 6 V c) 3 A

Real Transformers

Ps < PP

eff = Ps / PP

Modern transformers are up to 99% efficient

Reasons for power losses in real transformers

1. resistance of wires in P and S coils causes heating of coils

2. not all flux from P coil is linked to S coil

3. core warms up as result of cycles of flux changes (hysteresis)

4. small currents are induced in core (eddy currents) reduce by lamination

17

IB 12 2. The figure shows a step-down transformer used to light a

filament lamp with a resistance of 4.0 under operating conditions. The secondary coil has an effective resistance of 0.2 and the primary current is 150 mA. Calculate:

a) the reading on the voltmeter with switch S open 12 V

d) the power taken from the mains supply 36 W

b) the current in the secondary coil with switch S closed 2.86 A

e) the efficiency of the transformer 95%

c) the power dissipated in the lamp and the secondary coil 3.27 W and 1.6 W

+HDOWKDQG6DIHW\&RQFHUQVDVVRFLDWHGZLWK+LJK9ROWDJH3RZHU/LQHV 1. Extra-low-frequency electromagnetic fields, such as those produced by electrical appliances and power

lines, induce currents within a human body.

Just as AC can induce emfs and currents in secondary coils, so to can they be induced in the human

body since it is a conducting medium

Changing magnetic field induces current in human body

2. Current research suggests that low-frequency fields do not harm genetic material.

f = 60 Hz individual photons of this frequency do not have enough energy to cause ionization in the body childhood leukemia clusters are suspected to have a link to living near overhead power cables

3. The risks attached to the inducing of current in the human body are not well-understood.

Risks are likely to be dependent on current density, frequency, and length of exposure

18

3RZHU7UDQVPLVVLRQ IB 12

Power loss in transmission lines

When current flows through a wire, some energy is lost to the surroundings as the wire heats up due to the collisions between the free electrons in the current and the lattice ions of the wire. This is known as -RXOH KHDWLQJor UHVLVWLYHKHDWLQJ. Since the energy lost per second, or power loss, is proportional to the square of the current (P = I2 R), this energy loss is also know as I2R loss.

Methods of reducing I2R loss in power transmission lines

1. Reduce resistance: thicker cables low resistivity material

Constraints: lengths are fixed, thicker cables are heavier and more expensive

2. Increase voltage: step voltage up to very high levels

Constraints: high voltages are dangerous must be stepped back down for household use

For economic reasons, there is no ideal value of voltage for electrical transmission. Typical values are shown below.

1. AC power is generated at a power plant at 12,000 V and then stepped up to 240,000 V by step-up transformers.

2. The high-voltage, low-current power is sent via high-voltage transmission lines long distances.

3. In local neighborhoods, the voltage is stepped-down (and current is stepped-up) to 8000 V at substations.

4. This voltage is stepped-down even further at transformers on utility poles on residential streets.

An average of 120 kW of power is delivered to a suburb from a power plant that is 10 km away. The transmission lines have a total resistance of 0.40 . Calculate the power loss if the transmission voltage is

a) 240 V a) 240,000 V

I = 500 A P = 100 kW I = 0.50 A P = 0.10 W 19

(OHFWURVWDWLFV IB 12 1) electric charge: 2 types of electric charge: positive and negative

2) charging by friction: transfer of electrons from one object to another

3) positive object: lack of electrons negative object: excess of electrons

4) Types of materials:

a) &RQGXFWRUV: materials in which electric charges move freely (e.g. metals, graphite)

b) ,QVXODWRUV: materials in which electric charges do not move freely (e.g. plastic, rubber, dry wood, glass, ceramic)

c) 6HPLFRQGXFWRUV: materials with electrical properties between those of conductors and insulators (e.g. silicon)

d) 6XSHUFRQGXFWRUV: materials in which electrical charges move without resistance (e.g. some ceramics at very low temperatures)

3URSHUWLHVRI$WRPLF 3DUWLFOHV

H = elementary unit of charge (magnitude of charge on electron)

H = 1.60 x 10-19 C

Particle Mass Electric Charge

Electron PH [ NJ T H T [&

Proton PS [ NJ T H T [&

Neutron PQ [ NJ T T &

1. A balloon has gained 2500 electrons after being rubbed with wool. What is the charge on the balloon? What is the charge on the wool?

q = -4.0 x 10-16 C q = +4.0 x 10-16 C

2.8125 x 1013 e2. A rubber rod acquires a charge of -4.5 C. How many excess electrons does this represent?

&RQVHUYDWLRQRI(OHFWULF&KDUJH The total electric charge of an isolated system remains constant.

1

(OHFWULF)RUFH(OHFWURVWDWLF)RUFH&RXORPE)RUFH IB 12

&RXORPEV/DZ The electric force between two point charges is directly proportional to the product of the two charges and inversely proportional to square of the distance between them, and directed along the line joining the two charges.

&RXORPE)RUFH

T T k = Coulomb constant (electrostatic constant))H N 1 2 U2 2 C -2k = 8.99 x 109 Nm

NOTE: +-F denotes direction of force not sign of charge

Point charge: a charged object that acts as if all its charge is concentrated at a single point

Alternate formula for 1 T T1 2) k = 1/ 40Coulomb force: H 24SH0 U 0 = permittivity of free space T T -21 2) = 8.85 x 10-12 C2 N-1 mH U 24SH0

Use the Coulomb force to estimate the speed of the electron in a hydrogen atom.

2

7KH3ULQFLSOHRI6XSHUSRVLWLRQ IB 12 The net electric force acting on a charged particle is the vector sum of all the electric forces acting on it.

1. Determine the net electrostatic force on charge q1, as shown below.

2. Where can a third charge of +1.0 C be placed so that the 3. Three point charges of net force acting on it is zero? -2.0 C are arranged

as shown. Determine the magnitude and direction of the net force on charge T.

D = 2/3 m

3

(OHFWULF)LHOG IB 12

Electric field: a region in space surrounding a charged object in which a second charged object experiences an electric force

Test charge: a small positive charge used to test an electric field

(OHFWULF)LHOG'LDJUDPV 1. Positively charged sphere 2. Positive point charge 3. Negative point charge

Radial Field: field lines are extensions of radii

5. Two positive charges 6. Two negative charges 7. Two unlike charges

8. Oppositely charged parallel plates

Uniform Field: field has same intensity at all spots

Edge Effect: bowing of field lines at edges

3URSHUWLHVRI(OHFWULF)LHOG/LQHV

1. Never cross

2. Show the direction of force on a small positive test charge

3. Out of positive, into negative

4. Direction of electric field is tangent to the field lines

5. Density of field lines is proportional to field strength (density = intensity)

6. Perpendicular to surface

47. Most intense near sharp points

(OHFWULF)LHOG6WUHQJWK IB 12 (OHFWULF)LHOG6WUHQJWK,QWHQVLW\electric force exerted per unit charge on a small positive test charge

Electric Field:)( H T Units: N/C

Electric Force:

)H (T Units: N

Electric Field for a Point Charge:

4TN U 2 4 1 4( N 2 2T U 4SH0 U

3RLQW&KDUJH 6SKHULFDO &RQGXFWRU

1. a) Find the magnitude and direction of the electric field at 2. a) Find the magnitude and direction of the gravitational field at a spot 0.028 meter away from a sphere whose charge an altitude of 100 km above the surface of the Earth. is +3.54 microcoulombs and whose radius is 0.60 centimeters.

b) Find the magnitude and direction of the electric force b) Find the magnitude and direction of the gravitational force acting on a -7.02 nC charge placed at this spot. exerted on a 6.0 kg bowling ball placed at this spot.

c) Find the electric field strength at the surface of the sphere. c) Find the gravitational field strength at the surface of the Earth.

5

3. a) Find the magnitude and direction of the net electric field halfway between the two charges shown below. IB 12

b) Determine the electric force on a proton placed at this spot.

4. Two charged objects, $ and %, each contribute as follows to the net electric field at point 3: ($ = 3.00 N/C directed to the right, and (% = downward. What is the net electric field at 3?

E = 3.61 N/C Theta =33.70

5. a) Two positive point charges, T1 = +16 PC and T2 = +4.0 PC, are separated in a vacuum by a distance of 3.0 m. Find the spot on the line between the charges where the net electric field is zero.

6

6. A proton is released from rest near the positive plate. The distance between the plates is IB 12 3.0 mm and the strength of the electric field is 4.0 x 103 N/C.

a) Describe the motion of the proton.

constant acceleration in a straight line

b) Write an expression for the acceleration of the proton.

c) Find the time it takes the proton to reach the negative plate. d) Find the speed of the proton when it reaches the negative plate.

7. A particle is shot with an initial speed through the two parallel plates as shown.

a) Sketch and describe the path it will take if it is a proton, an electron, or a neutron.

b) Which particle will experience a greater force?

c) Which particle will experience a greater acceleration?

d) Which particle will experience a greater displacement?

8. In the figure, an electron enters the lower left side of a parallel plate capacitor and exits at the upper right side. The initial speed of the electron is 5.50106 m/s. The plates are 3.50 cm long and are separated by 0.450 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

7

Base level where EP = 0

(OHFWULF3RWHQWLDO(QHUJ\ IB 12

(OHFWULF3RWHQWLDO(QHUJ\(3 the work done in bringing a small positive test charge in from infinity to that point in the electric field

Derivation for Point Charges EP = 0(Work done by field)

(3 : )V cos T (3 (T'V

U N4 Electric Potential Energy due to a point charge (3 f ( V2 T)GV Formula: Units: Type: N4T U N4T J scalar V(3 (3f U N4T(3 U 8

High amount of EP

Low amount of EP

*UDYLWDWLRQDO3RWHQWLDO(QHUJ\(3 Reason for EP:

1. Test object has mass (test mass = m)

2. Test mass is in a gravitational field (g) caused by larger object (M)

3. Larger object exerts a gravitational force on test mass (Fg = mg)

4. Test mass has tendency to move to base level due to force

5. Work done moving object between two positions is path independent.

Gravitational potential energy: EP = mgh W = EP = mg h

High amount of EP

Low amount of EP

Base level where EP = 0

(OHFWULF3RWHQWLDO(QHUJ\(3 Reason for EP:

1. Test object has charge (test charge = +q)

2. Test charge is in an electric field caused by larger object (Q)

3. Larger object exerts an electric force on test charge (FE = Eq)

4. Test charge has tendency to move to base level due to force

5. Work done moving object between two positions is path independent.

Electric potential energy: EP = Eq h W = EP = Eq h

IB 12 (OHFWULF3RWHQWLDO9 work done per unit charge moving a small positive test charge in from infinity to a point in an electric field.

(OHFWULF3RWHQWLDOGXHWRDSRLQWFKDUJH N4T Units: Type:

Formula: (3 U9 ( T9 J/C scalarT T 3 = volts(V) N4 49 U 4SH0U

$% Higher potential Lower potential

Zero potential

$%

Lower potential Higher potential Zero potential

1. a) Calculate the potential at a point 2.50 cm away from a +4.8 C charge.

b) How much potential energy will an electron have if it is at this spot?

9

3. What is the potential where a proton is placed 0.96 m from a -1.2 nC charge?

IB 123RLQW&KDUJHV

4 4

(OHFWULF)RUFH (OHFWULF)LHOG Two objects needed One object needed property interaction between the two of that one object

Magnitude: F = Eq Magnitude: E = F/q

F = kQq/r2 E = kQ/r2

Units: N Units: N/C

Type: vector Type: vector

Direction: likes repel, Direction: away from unlikes attract positive, towards negative

Sign: dont use when Sign: dont use when calculating check frame of calculating check frame of reference reference

F = 0 where E = 0

4 4

(OHFWULF3RWHQWLDO(QHUJ\ (OHFWULF 3RWHQWLDO

Two objects needed quantity One object needed property of possessed by the system the field

Magnitude: EP = qV Magnitude: V = EP/q

P = kQq/r V = kQ/r

Units: J Units: J/C

Type: scalar (+/-) Type: scalar (+/-) E Sign: use signs of Q and q Sign: use sign of Q

EP = 0 where V = 0

10

IB 12

1. a) Calculate the net electric field at each spot (A and B):

b) Calculate the net electric force on a proton placed at each spot.

2. a) Calculate the net electric potential at each spot (A and B):

b) Calculate the electric potential energy of a proton placed at each spot.

11

(OHFWULF3RWHQWLDODQG&RQGXFWRUV IB 12 *UDSKVIRUDVSKHULFDOFRQGXFWRU

Value at surface = kQ/r2

For a hollow or solid conductor,

1. all the charge resides on the outside surface

2. the electric field is zero everywhere within

3. the external electric field acts as if all the charge is concentrated at the center

4. the electric potential is constant ( 0) El

ectri

c Fi

eld

Stre

ngth

Elec

tric

Pote

ntia

l

everywhere within and equal to the radius radius potential at the surface Distance Distance

A spherical conducting surface whose radius is 0.75 m has a net charge of +4.8 C.

a) What is the electric field at the center of the sphere?

b) What is the electric field at the surface of the sphere?

c) What is the electric field at a distance of 0.75 m from the surface of the sphere?

d) What is the electric potential at the surface of the sphere?

e) What is the electric potential at the center of the sphere?

f) What is the electric potential at a distance of 0.75 m from the surface of the sphere?

12

(TXLSRWHQWLDO6XUIDFHV IB 12 (TXLSRWHQWLDOVXUIDFH: a surface on which the electric potential is the same everywhere

1. Locate points that are at the same electric potential around each of the point charges shown.

2. Sketch in the electric field lines for each point charge.

3. What is the relationship between the electric field lines and the equipotential surfaces?

Perpendicular Field lines point in direction

of decreasing potential

(OHFWULF3RWHQWLDO*UDGLHQW Formula: '9( The electric field strength is the negative of the '[ electric potential gradient.

Units: N/c or V/m

For each electric field shown, sketch in equipotential surfaces.

Sketch in equipotential surfaces for the two configurations of point charges below.

13 http://wps.aw.com/aw_young_physics_11/0,8076,898593-,00.html http://www.surendranath.org/Applets.html

(OHFWULF3RWHQWLDO'LIIHUHQFH IB 12

(OHFWULF3RWHQWLDO'LIIHUHQFH9work done per unit charge moving a small positive test charge between two points in an electric field

Formula: :'9 T '( '( 3 T 9' Units: 3'9

J/C = V T

+LJKDQG/RZ3RWHQWLDO 1. a) Which plate is at a higher electric potential? positive

b) Which plate is at a lower electric potential? negative

c) What is the electric potential of each plate? Arbitrary relative to base level

d) What is the potential difference between the plates? Not arbitrary depends on charge, distance between, strength of electric field, geometry of plates, etc. Mark plates with example potentials, as well as

spots within field Mark ground mark equipotentials

e) Where will: a proton have the most electric potential energy?

an electron? a neutron? an alpha particle? Not arbitrary

2. An electron is released from rest near the negative plate and allowed to accelerate until it hits the positive plate. The distance between the plates is 2.00 cm and the potential difference between them is 100. volts.

a) Calculate how fast the electron strikes the positive plate. Eo = Ef Ee = EK

qV = mv2 v = sqrt (2qV/m)

v = sqrt(2(1.6 x 10-19)(100 V) /(9.11 x 10-31))

v = 5.9 x 106 m/s

b) Calculate the strength of the electric field.

Formula:

qV = mv2 Ve = mv2

Formula:

E = -V/x E = V/d

14

7KH(OHFWURQYROW IB 12

(OHFWURQYROW: energy gained by an electron moving through a potential difference of one volt

Derivation: Ee = qV Ee = (1e)(1 V) = 1 eV

Ee = (1.6 x 10-19 C)( 1 V) Ee = 1.6 x 10-19 J

Therefore: 1 eV = 1.60 x 10-19 J

1. How much energy is gained by a proton moving through a potential difference of 150. V?

150 eV or 150(1.60 x 10-19) = 2.4 x 10-17 J

2. A charged particle has 5.4 x 10-16 J of energy. How many electronvolts of energy is this?

Factor-label (5.4 x 10-16 J) (1 eV/1.6 x 10-19) = 3375 eV

3. An electron gains 200 eV accelerating from rest in a uniform electric field of 150 N/C. Calculate the final speed of the electron.

4. In Rutherfords famous scattering experiments (which led to the planetary model of the atom), alpha particles were fired toward a gold nucleus with charge +79H. An alpha particle, initially very far from the gold nucleus, is fired at 2.00 107 m/s directly toward the gold nucleus. Assume the gold nucleus remains stationary. How close does the alpha particle get to the gold nucleus before turning around? (the distance of closest approach)

2.74 x 10-14 m

15

(QHUJ\DQG3RZHU IB 12 3RZHU*HQHUDWLRQLQDW\SLFDOHOHFWULFDOSRZHUSODQW

a) Some fuel is used (coal, natural gas, oil, uranium) to release thermal energy which is used to boil water to make steam.

b) *HQHUDWRU'\QDPR - Steam turns turbines attached to coils of wire which turn in a magnetic field inducing an alternating potential difference.

c) Potential difference is stepped up by transformers in order to reduce I2R loss of power in transmission lines then stepped down for consumer use.

What are the energy transformations that take place?

Chemical (nuclear) energy in fuel

thermal energy in steam

rotational mechanical energy/kinetic energy in turbine

electrical

energy in

generator

Thermal energy hot Thermal energy Thermal energy friction gases go out radiation and convection in the generator chimney/stack from boiler

Degraded energy: energy transferred to the surroundings that is no longer available to do useful work cant be converted into other forms

1

IB 12 Why does the generation of electrical power involve the degradation of energy?

1. Thermal energy can be completely 2. A continuous conversion of thermal energy converted to work in a single process. into work requires a cyclical process.

Example: isothermal expansion Q = U + W U = 0 so Q = W

6HFRQG/DZRI7KHUPRG\QDPLFV

1) The total entropy of the universe is increasing.

2) No cyclical process (engine) is ever 100% efficient. Some energy is transferred out of the system (lost to the surroundings) as unusable energy (degraded energy).

6DQNH\GLDJUDPVHQHUJ\IORZGLDJUDPV: used to keep track of energy transfers and transformations

1) Thickness of arrow is proportional to amount of energy.

2) Degraded energy points away from main flow of energy.

3) Total energy in = total energy out.

)XHO 7\SLFDO (IILFLHQF\ Coal 30-35%

Natural Gas 50%

Oil 30-35%

Output 30%Fuel boiler dynamo

Input Electrical energy 100% Chemical energy

10% Thermal energy Friction in dynamo

20% Thermal energy

40% boiler Thermal energy exhaust gas out chimney 2

)XHOV IB 12

Fuel: source of energy (in a useful form)

How does a fuel work? A fuel releases energy by changing its chemical (or nuclear) structure. Chemical (or nuclear) bonds are broken reducing the fuels internal potential energy but increasing the kinetic energy of the substances particles which is seen macroscopically as an increase in the temperature of the substance. It is this thermal energy that is used to heat the water that will change to steam to turn the generators turbines.

Fossil fuels: coal, oil, natural gas, peat

Origins of fossil fuels: organic matter decomposed under conditions of high temperature and pressure over millions of years

Non-renewable fuels: rate of production of fuel is much smaller than rate of usage so fuel will be run out - limited supply

Renewable fuels: resource that cannot be used up or is replaced at same rate as being used

7\SHRIIXHO 5HQHZDEOH" &2 HPLVVLRQV" Fossil fuels No Yes

Nuclear No No

Hydroelectric Yes No

Wind Yes No

Solar Yes No

Wave Yes No

This histogram shows the relative proportions of world use of the different types of energy sources, though it will

vary from country to country.

NOTE: In most instances, the prime energy source for world energy is . . . the Sun.

Exceptions: nuclear, tidal (Moon)

3

)RVVLO)XHO3RZHU3URGXFWLRQ IB 12

Historical and geographical reasons for the widespread use of fossil fuels:

1. industrialization led to a higher rate of energy usage (Industrial Revolution)

2. industries developed near large deposits of fossil fuels (coal towns)

Transportation and storage considerations:

1. Natural gas is usually transported and stored in pipelines.

$GYDQWDJHV cost effective

'LVDGYDQWDJHV unsightly, susceptible to leaks, explosions, terrorist activities, political instability (withholding use of pipelines or terminals for political reasons)

2. Many oil refineries are located near the sea close to large cities. Oil is transported via ships, trucks, and pipelines.

$GYDQWDJHV workforce and infrastructure in place, easy access to shipping

'LVDGYDQWDJHV oil spills and leakage, hurricanes, terrorist activities

3. Power stations using coal and steel mills are usually located near coal mines.

$GYDQWDJHV minimizes shipping costs

'LVDGYDQWDJHV environmental impact (strip mining), mine cave-ins

8VHRIIRVVLOIXHOVIRUJHQHUDWLQJHOHFWULFLW\ $GYDQWDJHV 'LVDGYDQWDJHV

1. high energy density 1. combustion produces pollution, especially SO2 (acid rain)

2. relatively easy to transport 2. combustion produces greenhouse gases (CO2)

3. cheap compared to other sources 3. extraction (mining, drilling) damages environment

4. power stations can be built anywhere 4. nonrenewable

5. can be used in the home 5. coal-fired plants need large amounts of fuel

4

IB 12 (QHUJ\GHQVLW\RIDIXHOthe ratio of the energy released from the fuel to the mass of the fuel consumed

Formula: De = E/m Units: J/kg

Use: to compare different types of fuels

How is choice of fuel influenced by energy density?

Fuels with higher energy density cost less to transport and store

1. An oil-fired power station produces 1000 MW of power.

a) How much energy will the power station produce in one day?

(3 W ( 3W ( (1000 x 10 6:)(24 3600) [ ( 8.6 10 13 -[

b) Estimate how much oil the power station needs each day.

useful out ( (HII RXW ' total in (LQ H P [ 13 - 2.46 1014 -8.6 10 - [41.9 106.35 [ NJ P(LQ

14 P 5.9 10[ 6 NJ (LQ 2.46 10[ -

)XHO (QHUJ\'HQVLW\ 0-NJ

Fusion fuel 300,000,000 Uranium-235 90,000,000 Natural gas 53.6 Gasoline (Petrol) 46.9 Diesel 45.8 Biodiesel 42.2 Crude oil 41.9 Coal 32.5 Sugar 17.0 Wood 17.0 Cow dung 15.5 Household waste 10

5

IB 12 2. A 250 MW coal-fired power plant burns coal with an energy density of 35 MJ/kg. Water enters

the cooling tower at a temperature of 350 K and leaves at a temperature of 293 K and the water flows through the cooling tower at a rate of 4200 kg/s.

a) Calculate the thermal energy removed from the water in the cooling towers each second.

4 PF '7 (3 3 - W4 (4200 NJ [ )(350 293) 9)(4.19 10 NJ. 1.0 10 -[3

4 1.0 109 - 1V[ 3 [ 9 10000:1.0 10 :

b) Assuming the only significant loss of energy is this thermal energy of the water, calculate the energy produced by the combustion of coal each second.

(LQ (RXW (LQ 1000 MJ + 250 MJ ELQ 12500-

c) Calculate the mass of coal burned each second.

useful out 3RXW (HII 'H total in 3LQ P 2500: 6 1250 106 -- [35 10

12500: NJ P HII .20 P 36NJ HII [

6

1XFOHDU(QHUJ\ IB 12 Most common source: fissioning of uranium-235 with conversion of some mass into energy

Process:

a) unstable uranium nucleus is bombarded with a neutron and splits into two smaller nuclei and some neutrons

Why use neutrons? Neutral, not repelled by nucleus

b) rest mass of products is less than reactants so some matter is converted into energy

Form of energy: KE of products (thermal energy) 235 1 141 92 18 Qo %D .U 3 Q92 0 56 36 0

c) released neutrons strike other uranium nuclei causing further fissions

1) A particular nuclear reactor uses uranium-235 as its fuel source. When a nucleus of uranium-235 absorbs a neutron, the following reaction can take place:

235 1 144 90 18 Qo ;H 6U [ Q 92 0 54 38 0 a) How many neutrons are produced in the reaction? 2

b) Use the information to show that the energy released in the reaction is approximately 180 MeV.

235 5 2rest mas of 92 8 2.1895 x 10 MeV c rest mas of 10Q 939.56 MeV c 2

144 5 2rest mas of 54 ;H 1.3408 x 10 MeV c 90 5 2rest mas of 38 6U 8.3749 x 10 MeV c

7

IB 12 2. The energy released by one atom of carbon-12 during combustion is approximately 4 eV. The

energy released by one atom of uranium-235 during fission is approximately 180 MeV.

a) Based on this information, determine the ratio of the energy density of uranium-235 to that

of carbon-12. (Then, check your answer with the given table of energy densities.)

1 1PDVV= x molar mass PDVV= x molar mass1$ 1$ 1 1 mass = x .235 kg mass = x .012 kg23 236.02 10[ 6.02 10[

25 26mass=3.90x10 NJ mass=1.99x10 NJ

1.60 10 -4H9 [ 19 196 19 [ -180 10[ H9 1.60 10 [ - 11 6.40 10 2.88 10 - 1 1H9 [ 1 1H9

11 6.40 1019 -[[ 'H 3.22 10[ 7 -/ NJ 2.88 10 - 13 26' 7.38 10[ -/ NJ 1.99 10 NJ [ H 3.90 1025 NJ[ 32.20-/ NJ

7.38 10[ 7 0- / NJ

'H U-235 7.38 107 6[ 2.3 10['H C-12 32.2

b) Based on your answer above, suggest one advantage of uranium-235 compared with fossil fuels.

Higher energy density implies that uranium will produce more energy per kilogram less fuel needed to produce the same amount of energy

8

1XFOHDU)XHODQG5HDFWRUV IB 12

Naturally Occurring Isotopes of Uranium:

1) 8UDQLXP most abundant (99.3%) but not used for fuel since it has a very small probability of fissioning when it captures a neutron.

2) 8UDQLXP rare (0.3%) but used for fuel since it has a much greater probability of fissioning when captures a neutron but must be a low-energy neutron (thermal neutron).

7KHUPDO1HXWURQ low-energy neutron (1eV) that favors fission reactions energy comparable to gas particles at normal temperatures

)XHO(QULFKPHQW p