ADVANCED PHYSICS FORM SIX

ENVIROMENTAL PHYSICS

(i) Agriculture physics

- Influence of solar radiation on plant growth.

- Influence of wind, humidity, rainfall and air temperature on plant growth.

- Soil environmental component which influence plant growth.

(ii) Energy from the environment

Photovoltaic energy

Wind energy

Geothermal energy

Wave energy

(iii) Geophysics (Earth quakes)

Elastic rebound theory

Types of seismic waves

Propagation of seismic waves

Seismology

(iv) Environmental pollution

Types of pollutant in the atmosphere

Transport mechanisms of atmospheric pollutant

Nuclear waste and their disposal

Effects of pollution on visibility and optical properties of materials.

INTRODUCTION

Environmental physics is an interdisciplinary subject that integrates the physics processes in the following disciplines: the atmosphere, the biosphere, the hydrosphere, and the geosphere.

Environmental physics can be defined as the response of living organisms to their environment within the framework of the physics of environmental processes and issues.

It is structures within the relationship between the atmosphere, the oceans (hydrosphere), land (lithosphere), soils and vegetation (biosphere).

It embraces the following themes:

(i) Human environment and survival physics,

(ii) Built environment

(iii) Renewable energy

(iv) Remote sensing

(v) Weather, climate and climate change, and

(vi) Environmental health.

The environment may be defined as the medium in which any entity finds itself, For example, for a cloud its environment may be the region of the atmosphere in which it is formed.

AGRICULTURE PHYSICS

Agriculture physics is concerned with physics environment in relation to plant growth.

(a) Influence of Radiation Environment on Plant Growth

Radiation environments. Refer to radiations present in the atmosphere, commonly coming from the sun.

Components of solar radiation

The main components of solar radiation are:

(i) Visible light

(ii) Infrared radiation, and

(iii) Ultraviolet radiation.

HEATING EFFECT OF SOLAR RADIATION ON PLANTS

Positive effect

An optimum amount of heat on plant favours the process of photosynthesis. This enables a plant to make its own food and hence provide its growth.

Negative effects

(i) Excessive solar radiation (ultraviolet light) on plants leads to bleaching of green pigment (chlorophyll). This lowers the amount of food produced by photosynthesis to plant and hence a plant may die.

(ii) Excessive solar radiation on plants leads to excessive water loss in the form of water vapour commonly on plant leaves (transpiration). Hence wilting (drying) of plants may occur.

(b) Influence of Aerial Environment on Plant Growth

Aerial environments refer to the atmospheric condition resulting from a series of processes occurring in the atmosphere. These include air temperature, wind, humidity and rainfall.

WIND EFFECT ON PLANT GROWTH

Positive effects

(a) Wind acts as pollinating agent for some plants and hence favours plant productivity.

(b) Wind also favours evaporation of water from plant leaves and thus maintains water balance for proper plant growth.

Negative effects

(a) Excessive wind on environments leads to plant breaking or cutting of tree branches. This may lead to the death of plant.

(b) As the wind speed increases further, cell and Cuticular damage occurs, followed by death of plant tissue, and a gnarled appearance becomes more apparent.

(c) At low wind speeds, the effect seems to be an increase in transpiration, which results in water stress. This stress causes the plant to adapt by decreasing leaf area and internodes length, while increasing root growth and stem diameter.

(d) Strong wind may also cause shade off flowers; this lowers plant productivity.

Effect of Rainfall on Plant Growth

Positive effect

An optimum amount of rainfall on plants favours its growth. Water is a raw material for the process of photosynthesis from which plants obtain their food and hence their growth.

Negative effect

Excessive rainfall leads to water logging in soil which in turn leads to root spoil and hence the death of plant.

Effect of Humidity on Plant Growth

Positive effect

Favourable humidity on plants help plants to conserve water for various activities and in seeds helps the development of new leaves.

Negative effect

Low humidity results into a greater rate of transpiration and hence may result into plant drying.

Effect of Air Temperature of Plant Growth

Positive effect

An Optimum temperature on plants enhances enzymic activities which in turn gives favourable conditions for plant growth.

Negative effect

(a) High temperature denature enzymes commonly for photosynthesis and hence the death of plant.

(b) Low temperature inactivates the plant growth enzymes, hence low growth rate.

Wind Belts

Wind belts are seasonal strong wind moving in a specified direction in a certain region of the earth.

The global wind belts are formed by two main factors:

(i) The unequal heating of the earth by sunlight and

(ii) The earth’s spin.

Here is a simple explanation of the process

The unequal heating makes the tropical regions warmer than the Polar Regions. As a result, there is generally higher pressure at the poles and lower at the equator. So the atmosphere tries to send the cold air toward the equator at the surface and send warm air northward toward the pole at higher levels.

Unfortunately, the spin of the earth prevents this from being a direct route, and the flow in the atmosphere breaks into three zones between the equator and each pole.

These form the six global wind belts: 3 in the Northern Hemisphere (NH) and 3 in the Southern (SH). They are generally known as:

(1) The Trade winds, which blow from the northeast (NH) and southeast (SH), are, found in the sub tropic regions from about 30 degrees latitude to the equator.

(2) The Prevailing Westerlies (SW in NH in SH) which blow in the middle latitudes.

(3) The Polar Easterlies which blow from the east in the Polar Regions.

Effects of wind belts to plant

1. Wind belts because the loss of plant leaves and flowers hence lower plant productivity and growth. Loss of leaves lowers the rate of photosynthesis.

2. Wind belts sometimes cause plants to lean in direction of moving wing. This changes their direction of growth

3. Trees are broken by the strong wind.

Soil is composed of both rock particles and organic matter (humus) – the remains of plants and animals in various stages of decomposition. The humus serves as food for many living organisms. Within the soil is a large population of animals, plants. These break down the humus into soluble substances that can be absorbed by the roots of large plants.

Components of a soil

Soil is composed of:

( (a) Air, 25% by volume which supports life of soil organisms,

(b) Water, 25% which dissolves minerals so that are easily absorbed by plants,

(d) Inorganic matter (minerals), 45% by volume,

(e) Biotic organisms, micro – organisms like earth worm, centipedes, millipede, bacteria which decompose organic matter.

Types of soil

(i) Sandy soil

(ii) Silt soil,

(iii) Clay soil, and

(iv) Loamy soil (sand + silt + clay soil mixture)

Water Movement in the soil

Two forces primarily affect water movement through soils, (a) gravity and (b) capillary action.

Capillary action refers to the attraction of water into soil pores – an attraction which makes water move in soil. Capillary action involves two types of attraction – adhesion and cohesion.

Adhesion is the attraction of water to solid surfaces.

Cohesion is the attraction of water to itself.

Speed of water in a particular soil type depends on:

(i) How much water is in the soil, and

(ii) Porosity of the soil.

The movement of water in the solid is mainly due to gravity. The porosity gives a measure of how much water the soil can hold and the rate at which water flows through the soil. Large pore spaces give a faster rate and vice versa.

An experiment to study water movement in soil

An experiment to demonstrate the rate of flow of water in the soil is done using a glass tube and sand type filled in it. Water is poured into the tube and the time taken for water to reach the bottom of the tube in notes.

Unsaturated soil

Saturated soil

Soil type

Water speed

Soil type

Water speed

Sandy

Fastest

Sand

Slowest

Loamy

Moderate

Loamy

Moderate

Clay

Slowest

Clay

Highest

i. Sand soil have large pore spaces thus allows water to travel downwards through it at a fastest rate.

ii. Clay soil can hold water as has very fine pore spaces.

iii. Loamy soil allows water movement at a medium rate.

Heat transfer in the soil

Within the soil heat is transferred by a conduction process. Since soil is poor conductor of heat most of the heat from the atmosphere appears at the surface of the earth.

An optimum soil temperature favours plants growth but a high temperature can lead to the rotting of plant roots.

(d) Techniques for the Improvement of the Plant Environment

Plant environment can be improved by using wind breaks, shading and mulching.

Shading

Shading is the process of obstructing plants from excessive solar radiation.

Positive Impacts of Shading

1. Prevents excessive loss of water by plants through transpiration. This enhances plant productivity.

2. Preserve moisture in the soil and hence water supply to plant.

Mulching

Mulching is the process of covering the soil by dry leaves, grasses and or papers.

Benefits (Advantages) of Mulching

1. Improve soil moisture. Bare soil is exposed to heat, wind and compaction loses water through evaporation and is less able to absorb irrigation or rainfall. Using mulches, the soil has greater water retention, reduced evaporation, and reduced weeds. Mulch can also protect trees and shrubs from drought stress and cold injury

2. Reduce soil erosion and compaction. Mulches protect soils from wind water, traffic induced erosion and compaction that directly contribute to root stress and poor plant health.

3. Maintenance of optimal soil temperatures. Mulches have shown to lower soil temperatures in summer months. Extreme temperatures can kill fine plant roots which can cause stress and root rot. Mulches protect soils from extreme temperatures, either cold or hot.

4. Increase soil nutrition. Mulches with relatively high nitrogen content often result in higher yields, but low nitrogen mulches, such as straw, sawdust and bark, can also increase soil fertility and plant nutrition.

5. Reduction of salt and pesticide contamination. In arid landscapes, evaporating water leaves behind salt crusts. Because mulches reduce evaporation, water is left in the soil and salts are diluted. Organic mulches can actively accelerate soil desalinization and help degrade pesticides and other contaminants.

6. Improve plant establishment and growth. Mulches are used to enhance the establishment of many woody and herbaceous species. Mulches improve seed germination and seed survival, enhance root establishment, transplant survival, and increase plant performance.

7. Reduction of disease. Mulches will reduce the splashing of rain or irrigation water, which can carry spores of disease organisms to stems and leaves of plants. Populations of beneficial microbes that reduce soil pathogens can be increased with mulches. Mulches can combat disease organisms directly as well.

8. Reduction of Weeds. Using mulches for weed control is highly effective. Mulches can reduce seed germination of many weed species and reduce light, which stresses existing weeds.

9. Reduce pesticide use. Mulches reduce weeds, plant stress, and susceptibility to pests and pathogens which translates to reduced use of herbicides, insecticides, and fungicides.

Mulch Problems (disadvantages of mulching)

1. i. Acidification. Some types of mulches can increase soil acidity.

2. ii. Disease. Many mulches made from diseased plant materials can be composted or treated at temperatures that kill pathogens that can be transmitted to healthy plants.

3. iii.Pests. Many organic mulches, especially wood – based mulches, have the reputation as being “pest magnets”.

4. iv. Weed contamination. Improperly treated crop residues and composts as well as bark mulches are often carriers of weed seed. Mulch must be deep enough to suppress weeds and promote healthy soils and plants. Weed control and enhanced plant performance are directly linked to mulch depth.

v. Wind Breaks

Wind breaks are long rooted strong plants (trees) that are used to obstruct the path of wind or to slow down the wind.

Windbreaks provide many benefits to soil, water, plants, animals and man. They are an important part of the modern day agricultural landscape. Windbreaks come in many different sizes and shapes to serve many different conservation purposes.

In agriculture, wind breaks protect small growing plants from strong blowing wind Advantages of Windbreaks to Plant Environment

1. i. Control soil erosion. Windbreaks prevent wind erosion from causing loss of soil productivity. This eliminates plant roots stresses and thus favours plant growth condition.

2. ii. Increase plant yield. Windbreak research substantiates that field windbreaks improve crop yields which offsets the loss of production from the land taken out of cultivation.

3. Pesticide sprays. Windbreaks control pesticide spray drift and provide buffers to delineate property lines and protect neighbors.

EXAMPLES: SET A

Example 01

(a) What is agriculture physics? (02 marks)

(b) What are the components of a soil? How do they support the life of a plant? (06 marks)

Example 02

(a) What do you understand by the word environmental physics? (01 marks)

(b) Explain how the following climatic factors influence plant growth: air temperature, humidity, rainfall and wind. (06 marks)

Example 03

(a) What is mulching? (02 marks)

(b) Give two advantages and two disadvantages of mulching. (04 marks)

Example 04

(a) Explain two factors that primarily affect water movement in the soil (03 marks)

(b) Explain the soil environment that favours high crop yield (04 marks)

Example 05

(a) (i) Mention the components of solar radiation. (ii) How do those components affect plant growth? (04½ marks)

(b) What are wind breaks? (02 marks)

ENERGY FROM THE ENVIRONMENT

ENERGY

Energy is defined as the capacity to do work Or is defined as ability to do work.

Energy is measured in Joules (symbol J)

Types of energy according to their usefulness

(i) High grade energy

(ii) Low grade energy

i. High grade energy is the energy that is easily transformed into other forms of energy and is more suitable for doing works.

Examples are chemical and electrical energy.

ii. Low grade energy is the one that is not easily transformed into anything else.

Examples are the kinetic energy of molecules due to their randomness and the potential energy due to the forces between molecules.

ENERGY SOURCES

There are two types of energy sources, namely:

(i) Primary energy sources,

(ii) Secondary energy sources.

i. Primary energy sources

Primary energy sources are sources of energy that are used in the form in which they occur naturally.

Primary energy sources fall into two groups:

(a) Finite energy sources,

(b) Renewable sources.

a. Finite energy sources are those energy sources that last after a number of years when exploited.

Examples are coal, oil, natural gas, and nuclear fuels.

b. Renewable energy sources: these cannot be exhausted. Examples are solar energy, biofuels, hydroelectric power, wind power, wave power, tidal and geothermal power, wind power, wave power, tidal and geothermal power.

ii. Secondary energy sources

Secondary energy sources are used in the non – natural form.

SOLAR ENERGY

Nature of solar energy

The sun’s energy is produced by thermonuclear fusion.

Not all of the solar radiation arriving at the edge of the Earth’s atmosphere reaches the Earth’s surface.

About 30% is reflected back into space by atmospheric dusts and by the polar ice caps.

About 47% is absorbed during the day by the land and sea and becomes internal energy (i.e. heats the Earth). At night this is radiated back into space as infrared.

23% causes evaporation from the oceans and sea to form water vapour. This results into rain and hence hydroelectric power.

-0.2% causes convection currents in the air, creating wind power which in turn causes wave power.

-0.02% is absorbed by plants during photosynthesis and is stored in them as chemical energy. Plants are sources of biofuels

Solar constant

Solar constant is defined as the solar energy falling per second on a square meter placed normal to the sun’s rays at the edge of the Earth’s atmosphere, when the Earth is at mean distance from the sun.

Its value is about 1.35 kWm2

The amount of solar radiation received at any point on the earth’s surface depends on:

(i) The geographical location,

(ii) The season, (summer or winter)

(iii) The time of the day, the lower the sun is in the sky the greater is the atmospheric absorption.

(iv) The altitude; the greater the height above sea level the less is the absorption by the atmosphere, clouds and pollution

PHOTOVOLTAIC DEVICES (SOLAR CELLS)

A solar cell (PV, cells) is a PN junction device which converts solar energy directly into electrical energy.

How it Works

PV cells are made of at least two layers of semiconductor material. One layer has a positive charge (p – type material), the other negative (n-type material). When light enters the cell, some of the photons from the light are absorbed by the semiconductor atoms, freeing electrons from the cell’s negative layer to flow through an external circuit and back into the positive layer. This flow of electrons produces electric current.

Uses of the solar cell

1. (i)Are used to power electronics in satellite and space vehicles.

2. (ii)Are used as power supply to some calculators.

3. (iii)Are used to generate electricity for home, office and industrial uses.

Series arrangement of solar cells

Solar panel (module) is a sealed, weatherproof package containing a number of interconnected solar cells so as to increase utility of a solar cell.

When two modules are wired together in series, their voltage is doubled while the current stays constant.

When two modules are wired in parallel, their current is doubled while the voltage stays constant.

To achieve the desired voltage and current, modules are wired in series and parallel into what is called a PV array. The flexibility of the modular PV system allows designers to create solar power systems that can meet a wide variety of electrical needs, no matter how large or small.

Efficiency of a photovoltaic system

The output power of a solar cell depends on:

(i) The amount of light energy from the sun falling on a solar panel (the intensity of light).

(ii) The orientation of the solar panel. More electricity is produced if light falls perpendicular to panels.

(iii) The surface area of the panel. Large area collects more solar energy and hence greater electricity.

The best designed solar cell can generate 240 Wm-2 in bright sun light at an efficiency of about 24%.

Advantages of photovoltaic systems

1. Solar cells can produce electricity without noise or air pollution.

2. A photovoltaic system requires no fuels to purchase.

3. Panels of photovoltaic cells are used for small – scale electricity generation in remote areas where there is sufficient sun.

4. Net metering: This has the potential to help shave peak loads, which generally coincide with maximum PV power production.

5. The electricity from a PV system is controllable.

Disadvantages of photovoltaic systems

1. They require an inverter to convert the d.c output into a. c for transmission.

2. They produce electricity only when there is sunlight. Hence they need backup batteries to provide energy storage.

3. Suitable in areas which receives enough sunlight.

4. Photovoltaic large scale power generation is cost effective. This is due to large surface area of cells required for generating high power outputs and the need to convert d.c to a.c for transmission.

5. Compared to other energy sources, PV systems are an expensive way to generate electricity.

6. The available solar resource depends on two variables: The latitude at which the array is located and the average cloud cover.

WIND ENERGY

Winds are due to conventional currents in the air caused by uneven heating in the earth’s surface by the sun.

Wind energy is extracted by a device called wind turbine.

Wind speed increases with the height; it is greatest in hilly areas. It is also greater over the sea and coastal areas where there is less surface drag.

Wind turbines are also called aerogenerator or wind mills (old name)

Types of wind turbines

There are two types of wind turbines;

(i) Horizontal axis wind turbines (HAWT)

(ii) Vertical axis wind turbines (VAWT)

Horizontal axis wind turbine (HAWT)

HAWT has two or more long vertical blades rotating about a horizontal axis. Modern HAWTs usually feature rotors that resemble aircraft propellers, which operate on similar aerodynamic principles, i.e. the air flow over the airfoil shaped blades creates a lifting force that turns the rotor. The nacelle of a HAWT houses a gearbox and generator (alternator).

Advantage of HAWT

1. HAWTS can be placed on towers to take advantage of higher winds farther from the ground.

Disadvantages of HAWT

1. The alternator (generator) is paced at the top of the supporting tower.

2. Can produce power in a particular wind direction.

Vertical axis wind turbine (VAWT)

In vertical axis, the blades are long and vertical and can accept wind in any direction. The blades are propelled by the drag force on the blades as the wind flows.

Advantages of VAWT

1. It can harness wind from any direction

2. Typically operate closer to the ground, which has the advantage of allowing placement of heavy equipment, like the generator and gearbox, near ground level rather than in the nacelle.

Disadvantages of VAWT

1. Winds are lower near ground level, so for the same wind and capture area, less power will be produced compared to HAWT.

2. Time varying power output due to variation of power during a single rotation of the blade.

3. The need for guy wires to support the tower.

4. Darrieus VAWTS are not self starting like HAWTS. (More colorful picture and videos during lecture)

Power of a Wind Turbine

Consider a wind turbine with blades of length, r (area A), the wind speed is v and the air density is ρ. Assuming that the air speed is reduced to zero by the blades.

Kinetic energy of the wind, K.E =

Kinetic energy per unit volume

K.E per volume = ÷ volume =

The blades sweeps out an area A in one turn, so the volume of air passing in one second is Av.

Kinetic energy per second

= K.E per unit volume x volume per second

K.E per second = =

The available wind power is P =

Extractable power

The power extracted by the rotating blades is much less than the available wind power. This is because:

(i) The velocity of the wind is not reduced to zero at the blades

(ii) Losses due to friction at the turbine and alternator

(iii) Due to losses in both the gear train and generator.

The power actually captured by the wind turbine rotor, PR, is some fraction of the available power, defined by the coefficient of performance, Cp, which is essentially a type of power conversion efficiency:

The extractable power (electrical power output) is given by

Where ns and nb are efficiencies (power output over power input) for the generator and the gearbox.

Variations of power with wind speed

The power curve for a wind turbine shows this net power output as a function of wind speed.

i. Cut in wind speed: This is the lowest speed at which the wind turbine will start generating power.

Typical cut – in wind speeds are 3 to 5 m/s.

ii. Nominal wind speed: This is the lowest speed at which the wind turbine reaches its nominal power output.

Above this speed, higher power outputs are possible, but the rotor is controlled to maintain a constant power to limit loads and stresses on the blades.

iii. Cut – out wind speed: This is the highest wind speed which the turbine will operate at.

Above this speed, the turbine is stopped to prevent damage to the blades.

Advantages of Wind Energy

1. Wind Energy is an inexhaustible source of energy and is virtually a limitless resource.

2. Energy is generated without polluting environment

3. This source of energy has tremendous potential to generate energy on large scale.

4. Like solar energy and hydropower, wind power taps a natural physical resource,

5. Windmill generators don’t emit any emissions that can lead to acid rain or greenhouse effect.

6. Wind Energy can be used directly as mechanical energy

7. In remote areas, wind turbines can be used as great resource to generate energy

8. In combination with Solar Energy they can be used to provide reliable as well as steady supply of electricity.

9. Land around wind turbines can be used for other uses, e.g. Farming.

Disadvantages of Wind Energy

1. Wind energy requires expensive storage during peak production time.

2. It is unreliable energy source as winds are uncertain and unpredictable.

3. There is visual and aesthetic impact on region

4. Requires large open areas for setting up wind farms.

5. Noise pollution problem is usually associated with wind mills.

6. Wind energy can be harnessed only in those areas where wind is strong enough and weather is windy for most parts of the year.

7. Usually places, where wind power set-up is situated, are away from the places where demand of electricity is there. Transmission from such places increases cost of electricity.

8. The average efficiency of wind turbine is very less as compared to fossil fuel power plants. We might require many wind turbines to produce similar impact.

9. It can be a threat to wildlife. Birds do get killed or injured when they fly into turbines.

10. Maintenance cost of wind turbines is high as they have mechanical parts which undergo wear and tear over the time.

NB: Even though there are advantages of wind energy, the limitations make it extremely difficult for it to be harnessed and prove to be a setback

GEOTHERMAL ENERGY

Geothermal energy is the energy from nuclear energy changes deep in the earth, which produces hot dry rock.

Geothermal energy originates from the heat retained within the Earth since the original formation of the planet, from radioactive decay of minerals, and from solar energy absorbed at the surface.

Harnessing Geothermal Energy

Most high temperature geothermal heat is harvested in regions close to tectonic plate boundaries where volcanic activity rises close to the surface of the Earth. In these areas, ground and groundwater can be found with temperatures higher than the target temperature of the application.

Geothermal energy is extracted by using two methods:

(i) A heat pump system

(ii) Hot dry rock conversion

The heat pump system

Hot aquifers are layers of permeable (porous) rock such as sandstone or limestone at a depth of 2 – 3 km which contains hot water at temperatures of 60 – 1000C.

A shaft is drilled to aquifer and the hot water pumped up it to the surface where it is used for district space and water heating schemes or to generate electricity. A second shaft may be drilled to return the cool water to the rock.

The hot dry rock conversion

These are impermeable hot dry rocks found at depth of 5 – 6 km, have temperature of 2000C or more.

Two shafts are drilled and terminate at different levels in the hot rock about 300 m apart. The rocks near the end are fractured by explosion or by methods to reduce the resistance to the flow of cold water which is pumped under very high pressure (300 atm) down the injection shaft and emerges as steam from the top of the shallower shaft.

Uses of geothermal energy

Geothermal energy can be used for electricity production, for direct use purposes, and for home heating efficiency (through geothermal heat pumps).

Advantages of geothermal energy

1. Geothermal power plants provide steady and predictable base load power.

2. New geothermal power plants currently generate electricity at low cost.

3. Responsibly managed geothermal resources can deliver energy and provide power for decades.

4. Geothermal power plants are reliable, capable of operating about 98 percent of the time.

5. Power plants are small, require no fuel purchase and are compatible with agricultural land uses.

6. Geothermal plants produce a small amount of pollutant emissions compared to traditional fossil fuel power plants.

Disadvantages of geothermal energy

1. Many of the best potential resources are located in remote or rural areas, often of federal or state lands

2. Although costs have decreased in recent years, exploration and drilling for power production remain expensive

3. Using the best geothermal resources for electricity production may require an expansion or upgrade of the transmission system.

4. The productivity of geothermal wells may decline over time. As a result, it is crucial that

developers manage the geothermal resources efficiently.

WAVE ENERGY

Wave energy is the energy extracted from the ocean surface wave. Energy that comes from the waves in the ocean sounds like a boundless, harmless supply.

Machinery able to exploit wave power is generally known as a wave energy converter (WEC)

Wave power

Waves in the sea have kinetic energy and gravitational potential energy as the rise and fall.

Consider a sine wave of wave length λ spread over a width d the amplitude of the wave is a and the time period is T.

The power in a wave come from the change in potential energy of the water as it rotates on the circuit paths beneath the surface. It can be shown that the power carried forward by a wave is given by:

Wave Energy Flux

The mean transport rate of the wave energy through a vertical plane of unit width , parallel to a wave crest, is called wave energy flux.

From above,

Harvesting wave energy

There are two type of system:

1. i. Offshore systems in deep water more than 141 feet deep. The Salter duck method.

(a) Pumps that use bobbing motion of waves.

(b) Hoses connected to floats on surface of waves. As float rises and falls , the hose stretches and relaxes, pressurizing the water which then rotates a turbine

2. ii. Onshore systems are built along shorelines and harvest energy from braking waves.

(a)Oscillating water columns are of concrete or steel and have an opening to the sea below the waterline. It uses the water to pressurize an air column that is drawn through the turbine as waves recede.

(b)A Tapchan is a tapered water system in sea cliffs that forces waves through narrow channels and the water that spills over the walls is fed through a turbine.

(c)A Pendulor device is a rectangular box with a hinged flap over one side that is open to the sea .Waves cause the flap to swing back and forth and this powers a hydraulic pump and generator.

Advantages of Wave energy

1. Renewable: It will never run out.

2. Environment friendly: Creating power from waves creates no harmful byproducts such as gas, waste, and pollution.

3. Abundant and widely available: Another benefit to using this energy is its nearest to places that can use it.

4. Variety of ways to harness: Current gathering method range from installed power plant with hydro turbine to seafaring vessels equipped with massive structures that are laid into the sea to gather the wave energy.

5. Easily predictable: The biggest advantage of wave power as against most of the other alternative energy source is that it is easily predictable and can be used to calculate the amount that it can produce.

6. Less dependency on foreign oil cost.

7. Non damage to land.

Disadvantages of wave energy

1. Suitable to certain locations: The biggest disadvantage to getting your energy from the wave is location. Only power plants and town near the ocean will benefit direct from it.

2. Effect on marine ecosystem: Large machine have to be put near and in the water gather energy from waves .These machines disturb the seafloor, changes the habitat of near-shore creatures (like crabs and starfish) and create noise that disturb the sea life around them.

3. Wavelength: Wave power is highly dependent on wavelength i.e. wave speed, wave length, and wavelength and water density.

4. Weak performance in Rough Weather: The performance of wave power drops significantly during rough weather.

5. Noise and Visual pollution: Wave energy generators may be unpleasant for some who live close to coastal regions. They look like large machines working in the middle of the ocean and destroy the beauty of the ocean. They also generate noise pollution but the noise is often covered by the noise of waves which is much more than that of wave generators.

6. Difficult to convert wave motion into electricity efficiently.

7. Difficult to design equipment that can withstand storm damage and saltwater corrosion.

8. Total cost of electricity is not competitive with other energy sources.

9. Pollution from hydraulic fluids and oils from electrical components.

TIDAL ENERGY Tidal Power is the power of electricity generation achieved by capturing the energy contained in moving water mass due to tides.

Two types of tidal energy can be extracted: Kinetic energy of currents between ebbing and surging tides and potential energy from the difference in height between high and low tides.

Causes of Tides

Tides are caused by the gravitational pull of the moon, and to a lesser extent the sun, on the oceans. There is a high tide places near the moon and also opposite on the far side.

i. High (spring) tide: Occurs when there is full moon. The moon, sun and earth are in line the moon being between earth and sun. The pulls of the moon and sun reinforce to have extra high tides.

ii. Lowest (neap) tide: Occurs when there is half moon and the sun and moon pulls are at right angles to each other.

iii. Harnessing Tidal Energy

Tidal energy can be harnessed by building a barrage (barrier), containing water turbines and sluice gates, across the mouth of river. Large gates are opened during the incoming (flood) tide, allowing the water to pass until high tides, when they are closed.

On the outgoing tide, when a sufficient head of water has built up, small gates are opened, letting the potential energy of the trapped water drive the turbines and generate electricity.

Advantages of Tidal Energy

1. Decrease reliance on coal driven electricity so less CO2 emissions.

2. Changing technology allowing quicker construction of turbines, which in turn increases likelihood of investment with a shorter return.

3. Once constructed very little cost to run and maintain.

4. Tidal energy is renewable and sustainable.

Disadvantages of Tidal Energy

1. Intermittent energy production based around tides creates unreliable energy source.

2. High construction costs

3. Barrages can disrupt natural migratory routes for marine animals.

4. Barrages can disrupt normal boating pathways.

5. Turbines can kill up to 15% of fish in area, although technology has advanced to the point that the turbines are moving slow enough not to kill as many.

Tidal Power If the tidal height (level) is h and the estuary area is A, then the mass of water trapped being the barrier is and the centre of gravity is h/2 above the low tide level.

The maximum energy per tide is therefore

Potential Energy of tide =

Averaged over a tidal period of T (approx. 12 hours a day), this gives a mean power available of.

Average tidal power =

Note that the efficiency of the turbines (generator) will determine how much of this tidal power will be harnessed.

EXAMPLES: SET B

Example 01

The power output p of a windmill can be expressed as where A is the area swept out by the windmill blades (sails), is the density of air, v is the wind speed and k is a dimensionless constant

(a) Show that the units on both sides of this expression are the same

(b) Sketch a graph to show how the power increases with wind speed as v rises from zero to 15ms-1

Solution

(a) Units on L.H.S = Nms-1

Unit on R.H.S. = m2 (kgm-3) x (ms-1)3

= (kgms-2) ms-1=Nms-1

(b) Variation of power with speed

Example 02

The radiation received from the sun at the earth’s surface in a certain country is about 600 Wm-2 averaged over 8 hours in the absence of cloud.

(a) What area of solar panel would be needed to replace a power station of 2.0 GW output, if the solar panels used could convert solar radiation to electrical energy at an efficiency of 20%

(b) What percentage is this area of the total of the country (which is about 3 x 1011m2)?

(c) If the total power station capacity is about 140 GW, what percentage of the surface of the country would be covered by solar panels if all the power stations were replaced?

Solution

(a) Output of a solar panel

(b) Percentage area to the country

Percentage area to the country

Example 03

(a) What are aerogenerators?

(b) Estimate the maximum power available from 10m2 of solar panels and calculate the volume of water per second which must pass through if the inlet and outlet temperatures are 200C and 700C. Assume the water carries away energy at the same rate as the maximum power available. The specific heat capacity of water is 4200 Jkg-1 and solar constant is 1.4 kWm-2.

Solution

(a) Aerogenerators are devices that convert the kinetic energy of wind into electrical energy. E.g. windmill.

(b) Maximum power available from solar panel

Volume of water per second used is given by

1400 =

Example 04

A coal – fired power station has an output of 100mW. Given that its efficiency is 45%, how much coal must be supplied each day? Assume 1 tonne of coal gives 3 x 1010 of energy.

Solution

Input power of the station is given by

Total input energy in a day is

The amount of coal required is

Example 05

Calculate the energy required transport1000 tones of oil along a 100km pipeline; given that 0.05 kW hours of energy is used to shift each tone of oil along each km of pipeline. Given that 1 tonne of oil releases 4.2 x 1010 J if burned, what percentage of the total energy available from 1000 tonnes of oil is used to shift the oil along the pipeline? (Ans: 18GJ, 0.043%)

Example 06

A hydroelectric power station has efficiency of 25%. The water driving the turbines falls through a height of 300m before reaching the turbines. Calculate the volume of water that must pass through the turbines each second to give a power output of 2mW. Assume the density of water is 1000kg-3.

Solution

Power of the falling water

But,

Example 07

The solar energy flux near the Earth is 1.4W m-2. A solar power station consists of concave mirrors that focus sunlight onto a steam boiler. What must be the minimum mirror area to given an output 1 mW, assuming 100% efficiency? Why in practice, should the mirror area be greater?

Solution

Minimum mirror area is given by

The mirror area should be greater to achieve such a power output because part of the incident energy is absorbed by the mirror.

Example 08

A solar panel attached to the roof of a house is used to heat water from 50C to 400C. If the water flows through the panel at a rate of 0.012kgs-1 Calculate the heat gained per second by the water. Assume the specific heat capacity of water is 4200Jkg-1K-1. (Ans. 1764 Q)

Example 09

An aerogenerator has a power output that is proportional to (wind speed) 2 and its efficiency varies with wind speed. On a day when there is a steady wind of speed 9 ms-1, the power output is 40kW operating at an efficiency of 20%. If the wind speed on next day is 13.5 ms-1 and the efficiency increases to 25% what is the new power output?

Solution

Power output α efficiency × (wind speed)2

Example 10

Estimate the energy released from a tidal power station if 100 km3 of water raised to height of 1.5m by the tide behind a tidal barrier. What would be the mean power output of such a station if its efficiency is 25% and there are two tides per day?

Solution

The tidal power is given by

Note that the centre of gravity of water mass is at the half height up.

Mean power output is

Example 11

An open boat of width 1.0 m has a total weight of 3000N.Used near a beach, it bobs up and down through 0.5 m once every 5s. Calculate the losses of P.E. every time it drops from a crest to a through. Hence estimate the mean power available per meter of beach waves.

Solution

Loss in P.E. is given by

The mean power available per meter is 300 W

Example 12

(a) If energy is conserved, why is there energy crisis?

(b) Explain the terms high grade and low grade energy and give examples.

(c) Draw an energy flow diagram for a hydroelectric power station. Why does such a station have a much greater efficiency than a thermal power station?

Refer Advanced Physics by Tom Duncan fifth edition for more problems on energy.

GEOPHYSICS

Geophysics is the branch of physics which deals with the study of seismic waves and the Earth’s magnetic and gravity fields and heat flow.

Because we cannot directly observe the Earth’s interior, geophysical methods allow us to investigate the interior of the Earth by making measurements at the surface. Without studying these things, we would know nothing of the Earth’s internal structure.

STRUCTURE OF THE EARTH

Major zones of the earth

The earth is divided into two major zones, namely;

(a) Outer zone, and

(b) Inner zone.

a) Outer zone: the earth’s outer zone consists of;

(i) The hydrosphere – water bodies,

(ii) The atmosphere – gaseous envelope

(iii) The biosphere – living organisms, plant and animals

b) Inner zone: the earth’s inner zone consists of;

(i) The crust – lithosphere

(ii) The mantle – mesosphere,

(iii) The core – barysphere

Atmosphere is the envelope of gases that surround the Earth (oxygen, nitrogen, carbon dioxide, etc)

Hydrosphere is the water bodies filling the depressions in the Earth. Examples are rivers, oceans, seas, oasis,

Lithosphere is the solid outer most part of the earth.

EARTH’S LAYERS

Layers defined by composition

Layers are defined by composition because of density sorting during an early period of partial melting, Earth’s interiors not homogeneous.

 Crust – the comparatively thin outer skin that ranges from 3 kilometers at the oceanic ridges to 70 kilometers in some mountain belts. It makes up 1% of the Earth’s volume.

Continental crust (SIAL, Silicon and aluminium)

Average rock density about 2.7 g/cm3

Its density varies between 2.0 to 2.8 g/cm3

Composed of silicon and aluminium

Floats higher on the mantle forming the land masses and mountains. It is 30 to 70 km thick.

Oceanic crust (SIMA), silicon and magnesium)

Oceanic crust ranges from 3 to 15 km thick

Density vary between 3.0 to 3.1 g/cm3

Floats lower on the mantle forming the oceanic basins. It is about 8 km thick.

 Mantle – a solid rocky (silica-rich) shell that extends to a depth of about 2900 kilometers. It makes up 83% of the Earth’s volume

The mantle can further be dived into:

(i) Upper layer of mantle (Asthenosphere)

(ii) Transition layer and,

(iii) Lower layer of mantle (Mesosphere)

Upper mantle is a rigid layer of rock with average density 3.3kgm-3

Transition layer is the layer that separates upper and lower mantle.

Lower mantle plays an important role in tectonic plate movement which creates earthquakes and volcanoes.

Its density is about 5.7 kgm-3

Note: The mantle rocks are said to be in a plastic state.

The upper part of a mantle has a temperature of about 8700C. The temperature increases downwards through the mantle to about 22000C near the core.

 Core - an iron – rich sphere having a radius of 3486 kilometers making up 16% of the Earth’s volume

The core is divided into two parts:

(i) Outer core

(ii) Inner core

i. Outer core is a liquid of molten iron and nickel alloys. The Earth’s magnetic field is generated within the outer core due to convective. It is 2270 kilometers thick.

ii. Inner core is a solid iron and nickel alloys. The temperature within the inner core is higher than the outer core but the inner core is solid, this is because higher pressure in this region causes the melting point to rise. It is a sphere of radius of 1216 kilometers.

Average density is nearly 11 gcm-3and at Earth’s center.

Layers defined by physical properties

Lithosphere (sphere of rock)

Earth’s outermost layer

Consists of the crust and uppermost mantle

Relatively cook, rigid shell

Averages about 100 kilometers in thickness, but may be 250 kilometers or more thick beneath the older portions of the continents

Asthenosphere (weak sphere partially molten)

Beneath the lithosphere, in the upper mantle to a depth of about 660 kilometers

Small amount of melting in the upper portion mechanically detaches the lithosphere from the layer below allowing the lithosphere to move independently of the asthenosphere i.e. allows tectonic plate movement.

Mesosphere or lower mantle

Rigid layer between the depths of 660 kilometers and 2900 kilometers

Earth’s major boundaries

Discontinuity is the name given to any surface that separates one layer from another layer of the Earth.

The Moho (Mohorovicic discontinuity)

Discovered in 1909 by Andriaja Mohorovicic

Separates crustal materials (crust) from underlying mantle.

Gutenberg discontinuity

 Discovered in 1914 by Beno Gutenberg

 Is the boundary between the outer and inner core.

The Earth’s Structure

TEMPERATURE INSIDE THE EARTH

Earth’s temperature gradually increases with an increase in depth at a rate known as the geothermal gradient.

Temperature varies considerably from place to place

Averages between about 200C and 300C per kilometer in the crust (rate of increase is much less in the mantle and core)

The rate of heat flow within the Earth depends on:

(i) The thermal conductivity of the rock,

(ii) Temperature gradient of the rock

Sources of heat Energy within the Interior of the Earth

Major processes that have contributed to Earth’s internal heat include:

1. Heat emitted by radioactive decay of isotopes of uranium (U), thorium (Th), and potassium (K).

2. Heat released as iron crystallized to form the solid inner core.

3. Heat released by colliding particles during the formation of Earth.

4. Gravitational work done by the Earth due to its rotation through its own axis.

5. Electron motion in the core behaves like an electric current.

Heat Lost by the Earth

Heat in the earth is transferred by the process of;

(i) Convection and

(ii) Conduction

In the solid inner core and in the Earth’s crust heat is transmitted by conduction process. Rates of heat flow in the crust vary.

In the Mantle heat is transmitted by conduction process. Rates of heat flow in the crust vary.

In the Mantle heat is transmitted by convection process. There is not a large change in temperature with depth in the mantle.

Mantle must have an effective method of transmitting heat from the core outward.

Transfer of heat in the Earth by mantle convection

Uses of the Mantle

1. The mantle transfers heat by convection from the earth’s crust to the out regions of the earth and thus help it to regulate its temperature

2. The upper part of the mantle is molten, this allows tectonic plates movements.

EARTHQUAKES

An earthquake is a sudden motion or shaking of the earth caused by a sudden release of energy that has accumulated within or along edges of the earth’s tectonic plates.

Earthquakes occur within the Earth’s crust along faults that suddenly release large amounts of energy that have built up over long periods of time.

The shaking during an earthquake is caused by seismic waves.

Seismic waves are propagating vibrations that carry energy from the source of the shaking (earthquake) outward in all directions.

Seismic waves are generated when rock within the crust breaks, producing a tremendous amount of energy. The energy released moves out in all directions as waves, much like ripples radiating outward when you drop a pebble in a pond.

CAUSES OF EARTHQUAKES (SEISMIC WAVES)

The main causes of the Earthquakes and so seismic waves are:

1. Movement of tectonic plate.

2. Volcanic activity.

3. Landslide and avalanches.

4. Rebound of the crust.

5. Human activities.

Movement of tectonic plate

The Earth’s crust is made up of segment (layers) called tectonic plates which are slowly drifting in various directions. Tectonic plates may create a fault.

A boundary is a line where two tectonic plates meet.

A geologic fault is a fracture in the earth’s crust causing loss of cohesion and accompanied by displacement along the fracture.

How an earthquake is formed

Tectonic plates grind past each other, rather than slide past each other smoothly. As the plates move past each other they can become locked together due to friction. For some time, they don’t move and strain energy builds up. Stresses builds between them until fractional force holding the plates together give away. The plates move suddenly, releasing the energy and then held again. This sudden jerk is what is felt as an earthquake.

Note

(a) The Earth’s crusts near tectonic plate edges are forced to bend, compress, and stretch due to the internal forces within the earth, causing earthquakes.

(b) Nearly all earthquakes occur at plate boundaries.

Volcanic activity

Molten rock “magma” from the mantle is forced through a weak point in the Earth’s crust creating a volcanic eruption. When magma reaches the Earth’s surface it is known as “Lava”. Successive eruptions leads to the buildup of lava on the sides of the vent creating the familiar “cone – shape” of a volcanoes

Earthquakes may be created by the violent explosions which occur if there are sudden movements of the magma.

Landslides and avalanches

A landslide occurs when a large mass of land slips down a slope. An Avalanche occurs when a large mass of snow pours down a mountain side. Both of these effects can start an earthquake

Rebound of the crust

Elastic rebound theory state that "as tectonic plates move relative to each other, elastic strain energy builds up along their edges in the rocks along fault planes". Since fault planes are not usually very smooth, great amount of energy can be stored (if the rock is strong enough) as movement is restricted due to interlock along the fault. When the shearing stresses induced in the rocks on the fault planes exceed the shear strength of the rock, rupture occurs.

It follows from this that if rocks along the fault are of a certain strength, the fault is a certain length, and the plates are slipping past each other at a defined rate, it is possible to calculate the amount of time it will take to build up enough elastic strain energy to cause an earthquake and its probable magnitude.

When a fault breaks it release elastic strain energy it stored, and hence earthquake.

Human activities

Human activities such as those caused by nuclear bombs can create earthquake, together with mine actives.

EARTHQUAKE TERMS

Energy released by an earthquake moves outwards from the origin in the form of concentric waves.

Focus (Hypocenter) is the point in the Earth where seismic waves originate. Epicenter is the point on the earth’s surface vertically above the focus.

Hypocentral distance is the distance between the focus and the seismic detection station.

Epicentral distance is the distance between the epicentral and the seismic station.

S = Seismic station

E = Epicenter

ES = Epicentral distance

TYPE OF SEISMIC WAVES

i. Seismic waves are elastic waves that propagate within the earth.

There are two type of seismic waves:

1. ii. Body waves, spread outward from the focus in all directions.

2. iii. Surface waves (Long, L – waves) spread outward from the epicenter to the Earth’s surface along the crust, similar to ripples on a pond. These waves can move rock particles in a rolling motion that very few structures can withstand. These waves move slower than body waves.

BODY WAVES

There are two types of Body Waves

(1) Primary P – wave and

(2) Secondary, S – waves

1. 1. Primary Wave (P – wave): Are longitudinal (compression) wave (travels in the same direction the waves move)

Characteristics of P – waves

1. Are the fastest seismic waves (7 – 14 km/second). Arrives at recording station first, hence the name primary means first.

2. Can pass through solid, gas and liquid, hence can pass through crust, mantle and the cores.

3. Are longitudinal compression waves. The rocks that transmit the P – waves are alternately compressed and expanded.

Velocity of P – waves

The velocity of primary waves depends on the density,bulk modulus B and the shear modulus

In solid, =

In liquid =

A fluid cannot support shear stresses hence

2. Secondary Wave (S – wave): Are transverse (shear) wave (travels perpendicular to the wave movement).

Characteristics of S – waves

1. i. Slower moving (3.5 – 7 km/second) hence are detected after primary waves.

2. ii. Caused by a shearing motion

3. iii. Cannot pass through a fluid (gas or liquid) because they are transverse. Hence are unable to pass through the liquid outer core.

Velocity of S – waves

The velocity of shear waves depends on the density and the shear modulus

In solid, =

In liquid =

Note: Since the density and states of the earth layers varies, the speed of the seismic waves also vary from layer to layer, the solid part showing greater speed and the liquid ones lower speed.

Primary wave and secondary wave

Variation of speed of body waves with depth

SURFACE WAVES/LONG WAVES

Surfaces waves are produced when earthquake energy reaches the Earth’s surface.

These are the slowest moving waves, but are the most destructive for structures on earth

There are two types of L – Waves:

(i) Love long waves

(ii) Rayleigh long waves

i. Love Waves

Love waves are Transverse horizontal motion, perpendicular to the direction of propagation and generally parallel to the Earth’s surface.

They are formed by the interaction of S waves with Earth’s surface and shallow structure and are dispersive waves. The speed at which a dispersive wave travels depends on the wave’s period.

Characteristics of Love Waves

1. i. Love waves are transverse and restricted to horizontal movement (horizontally polarized).

2. ii. The amplitude of ground vibration caused by a Love wave decrease with depth. The rate of amplitude decrease with depth also depends on the period/frequency.

3. iii. Loves wave are dispersive, i.e. wave velocity is dependent on frequency; low frequency – higher velocity.

4. iv. Speed of love waves is between 2.0 and 4.4 km/s

5. v. Love waves travels within the earth’s crust only.

LOVE WAVE

Rayleigh Waves

Rayleigh waves are vertically polarized long waves. The slowest of all the seismic wave types and in some ways the most complicated.

Characteristics of Rayleigh Waves

1. Rayleigh waves are transverse and restricted to vertical movements (vertically polarized).

2. The amplitude of Rayleigh wave decreases with depth. The rate of amplitude decrease with depth depends on the period/frequency

3. Rayleigh wave are dispersive, i.e. wave velocity dependent on frequency; low frequency – high velocity

4. Speed of love waves is between 1.0 and 4.2 km/s slowest of all waves.

5. Travels within the earth’s crust only.

6. Depth of penetration of the Rayleigh waves depend frequency, with lower frequencies, penetrating greater depth.

PROPAGATION OF SEISMIC WAVES

Like all other types of waves, seismic waves may undergo,

(i) Reflection, (ii) Refraction, (iii) Dispersion, (iv) Diffraction, (v) Attenuation.

Seismic reflection: Seismic waves bounce (reflect) rock boundaries of different rock type (density).

Seismic refraction: Waves change velocity and direct (refract) when they enter a medium of different density it the one they just passed through.

Seismic Dispersion: surface waves are dispersive which means that different periods travel at different velocities. The effects of dispersion become more noticeable with increasing distance because the long travel distance spreads the energy out (it disperses to energy).

SEISMIC WAVE PATHS

By comparing the data recorded by many stations all over the world the nature, speed and the paths of the seismic waves can be determined. This information can be used to tell us about the earth’s interior such as density sand state in each layer.

L – Waves travel within the Earth’s crust only

P and S waves travel through the earth in a curve path. The waves are refracted because their speeds a constantly changing with depth due to continue increase in density. Waves are also strongly refracted the Mantle – Core boundary.

Surface waves travels through the Earth crust only

Shadow zone is the region on the Earth’s surface where no S or P waves are present.

This lies between 1050 and 1400. Only surface waves may be detected in this region.

Shadow zone occurs because:

(i) P – Waves are strongly refracted at the liquid outer core.

(ii) S – Waves can’t travel through the liquid outer core.

Seismic waves can also be used to locate the discontinuities in the earth’s crust. A change in density or crack would affect the propagation of the waves.

This alteration in the wave’s path or speed would indicate the discontinuity.

The fact that S waves do not travel through the core provides evidence for the existence of a liquid layer beneath the rocky mantle.

The change in the velocity of P waves at crust – Mantle boundary reveals the presence of Mohorovicic discontinuity

P waves passing through the inner core show increased velocity suggesting that the inner core is solid.

Both P and S – Waves slow down when they reach the asthenosphere. Because of this scientists know that the asthenosphere is partially liquid

MEASUREMENTS OF EARTHQUAKES

i. Seismology is the scientific study of earthquakes (seismic waves) and artificially produced vibrations in the earth.Seismograph is a sensitive instrument that is used to record earthquakes and seismic waves (i.e. ground movements).

ii. Seismogram is the record of ground movement drawn by a seismograph.

The arrival of seismic waves at a station

Seismograph consists of a heavy weight suspended from a frame fixed into the ground. When the earth vibrates the frame moves but the heavy weight remains stationary due to great inertia. A pen attached to weight plots the earth’s movements on a chart recorder to produce a seismogram. To obtain a complete record of the earthquake measurements must be taken in all three planes (x, y and z).

The recording of the motion caused by seismic waves can be done by using;

(a) Mechanical method, as in the drawing above.

( (b) Optical method, where light is used to write the motion on a photosensitive paper instead of using a pen.

(c) Electronic method, where a coil is fixed to the mass of the pendulum and moves in a magnetic field. This induces a voltage which is amplified so that they can be easily interpreted.

Seismometers record both the magnitude and intensity of the earthquake.

LOCATING THE EPICENTRE

Although S – waves, P – waves and surface waves all start out at the same time, they travel at different speeds. The speed of a traveling seismic wave can be used to determine the location of an earthquake epicenter.

A seismograph records the arrival time and the magnitude of horizontal and vertical movements caused by an earthquake. The arrival time between different seismic waves is used to calculate the travel time and the distance from the epicenter.

The difference in arrival time between primary waves and secondary waves is used to calculate the distance from the seismograph station to the epicenter.

It is crucial that seismic waves are recorded by three different seismograph stations in order to estimate the location of the epicenter.

(i) Locate at least 3 stations on a map that recorded the seismic waves.

(ii) Calculate the time difference between arrival of P – waves and arrival of S – waves from a seismogram. The time difference is proportional to the distance from the epicenter. Because the direction to the epicenter is unknown, the distance defines a circle around the receiving station. The radius of each circle equals that station’s distance from the earthquake epicenter.

(iii) The epicenter is where the circles intersect.

Radius = distance from epicenter

SIZE OF AN EARTHQUAKE

The size of an earthquake can be measured in terms of its intensity (Mercalli/Wood Neumann scale) or its magnitude (Richter scale).

Mercalli Intensity Scale

The Mercalli scale measures the intensity of how people and structures are affected by the seismic event. In essence, it measures damage. It is much more subjective and uses numbers ranging from 1 (no damage) to 12 (total destruction).

Degree

Explanation

Detected by a seismograph only

Felt by all, many frightened. Some heavy furniture moved, some fallen plaster, general damage small

Total damage, large cracks, waves seen moving through the ground, objects are thrown upwards.

ISOSEISMAL LINES

Intensity distribution maps can be drawn up showing the intensities of an earthquake over a region. The earthquake is most intense at the epicenter and decreases with distance.

Isoseismal lines are line joining points of equal intensity.

Richter magnitude scale

The magnitude of an earthquake is measured in terms of energy released by an earthquake. This is determined from the amplitude of the seismic wave recorded on a seismogram 100 km from the epicenter. The magnitude is equal to the logarithm of the amplitude. Therefore each successive number represents a tenfold (x10) increase in the ground motion. The Richter scale starts at 0 but has no upper limit. -However 8 represent an earthquake that causes total destruction within the region.

Magnitude

Amount of explosives (TNT) needed to release the equivalent energy, in tons

6,000

180,000

5.4 million

Intensity of an earthquake is a measure of its strength based on the changes it causes to the landscape.

EARTHQUAKE PREDICTIONS (WARNINGS)

Forecasting (predicting) earthquakes is very difficult, although there are a number of warning signs which occur before an earthquake happens.

(i) Change in the velocity of p – waves.

(ii) Electrical resistivity of the rocks decreases.

(iii) An increase in radon, emission (radon is an inert gas, radon is found to increase in soil and water samples).

(iv) Increase in fore shock (small tumors that occur just before an earthquake).

(v) Local variations in the magnetic field.

(vi) Animals begin to behave strange.

(vii) Water levels rise or fall in wells few days before earthquake.

(viii) Increase in temperature of the area few months before the occurrence of an earthquake

PRECAUTIONS

Some of the world’s populations are living in regions where there is a high risk of an earthquake. Most of these regions lie along fault lines. However a few precautions can be taken to reduce the damage caused.

(a)Build structures that can withstand the forces of an earthquake. One method is to include shock absorbers into the buildings foundations.

(b)Scientific research has shown that pumping water out of the earth reduces the stress in the crust hence preventing an earthquake. However this technique is very expensive.

(d) If you are inside a house, stay in a safe place where things will not fall on you.

EARTHQUAKE HAZARDS

Earthquake give rises to a number of hazards which pose a great risk to human life, animals, property and the environment at large. The following are some hazards:

1. Landslides and avalanches: The shaking caused by an earthquake can cause unstable hillsides, mountain slops’ and cliffs to move downwards creating landslides. Earthquakes can also trigger avalanches on snow slopes

2. Tsunamis: If an earthquake occurs under the sea or ocean, the shock waves disturb the water. The ocean floor can rise or fall causing the water to rise and fall too. This movement creates huge water waves called tsunamis that travel across the ocean.

3. Collapsing building: Buildings or structures may collapse during a strong earthquake. The collapse of the building may kill people.

4. Fire outbreak: Earthquakes can cause gas or oil pipes to break and or the collapse of electricity lines. This may set up fire.

5. Backward rivers: Tilting ground due to earthquakes can make rivers change their course.

REFLECTION SEISMOLOGY

This is the study of reflection of seismographic waves by different materials inside the earth.

Applications:

(i) Location of underground oil and water

(ii) Locate discontinuities within the earth

SEISMIC PROSPECTING

Seismic prospecting is the sending of seismic waves into the deep earth’ crust in order to study the structure of the earth or detecting oils or gases in the interior of the earth by utilizing the property of reflection and refraction of the seismic waves.

THE EARTH’S MAGNETIC FIELD

The earth has a weak magnetic field, 95% of this field is created inside the Earth’s core 5% is the result of atmospheric effects above the Earth’s surface.

Geomagnetism is science of study of the earth magnetic field, its causes and its variations.

Generation of the Earth’s magnetic field within the core

The accepted explanation for the origin of the Earth’s magnetic field within the core is given by Lemoir’s self exciting dynamo theory.

The Earth’s Outer Core consists of molten conducting metals (Iron and Nickel) which are rich in free electrons. The Earth’s rotation causes the molten metal to rotate and hence large convection currents are set up within the outer core. These currents generate a magnetic field.

Eddy currents are now generated due to a conducting material moving in a magnetic field. These Eddy currents modify the position of the Earth’s magnetic field so that it does not lie along the Earth’s axis of rotation. The present magnetic poles are situated 800km from the Earth’s axis.

Generation of magnetic field in the Atmosphere

In the Earth’s atmosphere there is a region know as the ionosphere which consists of free electrons and ions. The movement of these charges creates a magnetic field. This effect provides a small fraction of the Earth’s total magnetic field.

TERMS ASSOCIATED WITH THE EARTH’S MAGNETIC FIELD

Magnetic meridian: A vertical plane passing through the axis of a freely suspended magnetic needle.

Geographic meridian: A vertical plane passing through the geographic axis.

Magnetic equator: Is the locus of points on earth’s surface where the needle (free to rotate in a vertical plane) remains horizontal.

The Earth’s magnetic field pattern is similar to that produced by a giant bar magnet or solenoid.

Note: (i) The magnetic North pole which lies in the Northern Hemisphere behaves like a south pole or a bar magnet, i.e. the field lines are directed towards it.

(ii) The magnetic south pole which lies in the southern hemisphere behaves like a north pole of a bar magnet, i.e. the field lines are directed away from it.

ELEMENTS OF EARTH’S MAGNETISM

Angle of variation of declination, at a place is the angle between the geographic meridian and the magnetic meridian at that place.

Angle of dip or declination, at a place is the angle between the directions of intensity of the earth’s total magnetic field declinationand the horizontal direction, in the magnetic meridian at that place.

Horizontal component of Earth magnetic field It is the component of the Earth’s total magnetic field along the horizontal direction in the magnetic meridian.

By Pythagoras theorem

By trigonometric ratio

Points to note about angle of Dip

(a) At the poles,

Therefore, only horizontal component exists at the poles

(b) At the equator

At the equator only horizontal component exist.

VARIATIONS OF THE EARTH’S MAGNETIC FIELD

The Earth’s magnetic field is not constant but varies continuously with time.

(i) Short term variations (Irregular changes): The magnetic field changes daily due to variations in the magnetic field created in the ionosphere. The charged particles in this region of the atmosphere are affected by the Sun’s gravitational pull (which is stronger when the sun is directly above that area)

Also during periods of high solar activity charged particles from the solar wind are able to penetrate the magneto pause and arrange themselves under the influence of the magnetic field in a formation called Van Allen Belts. These charged particles cause further Eddy currents within the ionosphere, altering the Earth’s magnetic field strength.

Solar wind is a continuous stream of moving electrons and protons in the atmosphere which are produced from flare (eruptions) from the sun. Normally these charged particles move from west to south at 300 – 500 km/s.

Magnetic storm is a sudden worldwide disturbance of the earth’s magnetic field caused by dynamic interaction of the earth’s magnetic field and the sun. During magnetic storm, the earth’s magnetic field is unusually active.

Effects of Magnetic Storm

(a) Large storms can cause the loss of radio communication

(b) Damage satellite electronics and affect satellite operations.

(d) Induce voltage surges in electric power grids causing blackouts.

(e) Reduce the accuracy of global positioning systems.

(ii) Long term variations (Secular changes): The Earth’s magnetic field position is constantly changing, now the magnetic North pole is moving at 8 km per year, and the magnetic South Pole at 16 km per year.

Evidence from the alignment of magnetized rocks layers in the Earth’s crust show that the Earth’s magnetic field has actually reversed in direction several times during the Earth’s history (i.e. the direction of the fields have reversed causing a north acting pole to become a south acting pole.) The present polarity of the Earth’s magnetic field has not changed for 700,000 years.

VAN ALLEN BELTS

The Van Allen belts consist of two regions of highly charged particles which are trapped within the Earth’s magnetic field:

Inner Belt consists of protons and positive charged particles

Outer Belt consists of electrons and negatively charged particles.

THE ATMOSPHERE

Earth’s atmosphere is divided into five main layers, the exosphere, the thermosphere, the mesosphere, the stratosphere and the troposphere. The atmosphere thins out in each higher layer until the gases dissipate in space. There is no distinct boundary between the atmosphere and space, but an imaginary line about 110 kilometers from the surface, called the Karman line, is usually where scientists say atmosphere meets outer space.

TROPOSPHERE

The troposphere is the layer closest to Earth’s surface. It is 10 km thick and contains half of Earth’s atmosphere. Air is warmer near the ground and gets colder higher up. Nearly all of the water vapor and dust in the atmosphere are in this layer and that is why clouds are found here.

Lapse rate is the rate of fall of temperature in degrees per kilometer rise. It has an average value of 6 0C per km in the troposphere.

Tropopause is the upper boundary of the troposphere.

Importance (uses) of troposphere

1. Controls the climate and ultimately determines the quality of life in the atmosphere.

2. It supports life on earth. It contains oxygen which is used to respiration by animals.

STRATOSPHERE

The stratosphere is the second layer. It starts above the troposphere and ends about 50 km above ground.

The temperature of the stratosphere slowly increases with altitude. This temperature increase is due to the presence of Ozone layer which absorbs heat from the sun in the form of ultraviolet light.

The Ozone layer occupies the middle of stratosphere between 20 and 30 km it consists of Ozone formed by oxygen molecules dissociated and reforming into 03.

The air here is very dry, and it is about a thousand times thinner here than it is at sea level. Because of that, this is where jet aircraft and weather balloons fly.

Stratopause is the upper boundary of the stratosphere.

Importance (uses) of stratosphere

The stratosphere prevents harmful ultraviolet radiation from reaching the earth. Ozone absorbs harmful radiation from the sun. The Ozone protects plants and shield people from skin cancer and eye cataracts.

MESOSPHERE

The mesosphere starts at 50 km and extends to 80 km high. The top of the mesosphere, called the mesopause, is the coldest part of the Earth’s atmosphere with temperatures averaging about - 900C. The temperature of the mesosphere decreases with altitude (because there is no ozone to absorb heat).

This layer is hard to study. Jets and balloons don’t go high enough, and satellites and space shuttles orbit too high. Scientists do know that meteors burn up in this layer.

Importance of mesosphere

Mesosphere, thermosphere and exosphere prevent harmful radiation such as cosmic rays from reaching the earth surface.

THERMOSPHERE

The thermosphere extends from about 80 km to between 500 and 1,000 km. Temperatures increases as it approaches nearer to the sun. The heating effects of the earth no longer exist at these higher altitudes.

The thermosphere is considered part of Earth’s atmosphere (the upper atmosphere), but air density is so low that most of this layer is what is normally thought of as outer space. In fact, this is where the space shuttles flew and where the International Space Station orbits Earth.

This is also the layer where the auroras occur. Charged particles from space collide with atoms and molecules in the thermosphere, exciting them into higher states of energy. The atoms shed this excess energy by emitting photons of light, which we see as the colorful Aurora Borealis and Aurora Australis.

EXOSPHERE

The exosphere, the highest layer, is extremely thin and is where the atmosphere merges into outer space. It is composed of very widely dispersed particles of hydrogen and helium.

The upper part of the exosphere is called Magnetosphere. The motion of ions in this region is strongly constrained by the presence of the earth’s magnetic field. This is the region where satellites orbit the earth

Note:

(i)The troposphere, stratosphere, and mesosphere are collectively forms the homosphere. These layers have the same chemical composition; 78% nitrogen, 21% oxygen, 1% argon and other gasses which sum to about 0.05%. The thermosphere is excluded due to different in chemical composition.

(ii) The upper atmosphere above 90 km is called heterosphere. The atmosphere is no longer a mixture of gases but separates into layers heavier ones forming the bottom layer.

VARIATION OF TEMPERATURE WITH HEIGHT

The temperature above the Earth surface varies as shown in the graph below.

The residence time, is the mean lifetime of a gas molecule in the atmosphere

THE IONOSPHERE AND TRANSMISSION OF RADIO WAVES

The ionosphere is the region containing high concentrations of charged particles ions and electrons.

The ionosphere is created by atoms absorbing U.V radiation, gamma and X – rays.

The ionosphere extends from the lower thermosphere 55 km to 550 km above the earth’s surface.

Ionosphere layers:

Due to difference in composition of the air in the ionosphere, the ionosphere is divided into layers.

(i) The lower layer, called D layer; this layer exists only in the day time at an altitude of 55 to 90 km above the earth’s surface. Ionization in this region is relatively weak.

(ii) The next layer, E – layer: this layer is between 90 and 145 km above the earth’s surface. It has a maximum density at noon but is only weakly ionized at night.

(iii) The top layer, the F – layer: At night exists as a single layer in a region of about 145 to 400 km above the earth’s surface. During the day it splits into two layers, F1 and F2.

The Ionosphere and Communication

The ionosphere plays an important role in communication. Radio waves can be reflected off the ionosphere allowing radio communications over long distances. However this process is more successful during the night – time.

Why Transmission is better at Night?

During the day: the ionosphere extends into lower atmosphere (D layer). In this layer there is high concentration of particles and so recombination of electrons and ions due to collision is more likely to occur. The leads to the radio waves being absorbed rather than reflected. Hence distant communications are poor during the day.

During the night: The D layer disappears due to decrease in ionization of molecules but recombination of electrons and ions still occurs at a fast rate. The radio waves are then reflected by E and F layers in which recombination of electrons and ions is rare hence there is less absorption of the radio waves.

EXAMPLES: SET C

Example 01: Necta 1985 P1

(a) (i) Distinguish between P and S waves, state clearly the difference between their speeds in a medium.

(ii)Draw a schematic diagram showing how one station on the Earth’s surface can receive P or S waves from a distant source and state which waves can be refracted by the Earth’s outer core.

(b) (i) Give a summary of the origin and composition of the ionosphere.

(ii) What is the net electric charge in the ionosphere?

(iii) Show graphically how electron density changes with altitude in the ionosphere.

Answers

(a) (i) P – waves are longitudinal compression waves which can pass through solid, gas and liquid, whereas S - waves are transverse shearing waves which cannot pass thorough a fluid (gas or liquid)

The speed of P – waves in a medium is approximately twice that of the S – waves hence P – waves are faster than S – waves.

(ii) Refer the diagram for the seismic wave paths

(b) (i) Ionosphere is the upper part of the atmosphere. The ionosphere is formed due to the ionization of gaseous atoms as they absorb ultraviolet radiation from the sun, gamma and X-rays.

(ii) The net electric charge in the ionosphere is zero.

(iii) Variations of electron density in the ionosphere Electron density increases from D to F layer

Example 02: Necta 1988/1993 P1

(a) What are the factors that influence the velocities of P – and S – waves?

(b) Explain briefly the characteristics property of seismic waves which is used to locate discontinuities in the earth’s crust.

Answer

(a) The velocities of both P and S – waves are influenced by;

(i) Density of the rock material (Media),

(ii) Moduli of elasticity.

(b) Speed is the characteristic property of seismic waves that is used to locate discontinuities

Between the crust and mantle there is abrupt change of density, which shows an abrupt change in speed of both P – and S – waves, a Mohorovicic discontinuity exists here. Both P – and S waves travels across this discontinuity.

Between the mantle and the core there is the Gutenberg discontinuity only P – waves travel this discontinuity.

Example 03: Necta 1989 P1

(a) State three sources of heat energy in the interior of the earth.

(b) (i) How does temperature vary with depth of the Earth?

(ii) What are the factors that influence the flow of heat from the interior of the Earth?

Answers

(a) Refer notes

(b) (i) The temperature increases with increasing depth

(ii) The rate of heat flow (conduction) is given by

The heat flow from the interior of the earth depends on:

Thermal conductivity of the rock,

Temperature gradient of the rock

Example 04: Necta 1989 P2

(a) What do you understand by the terms?

(i) Solar wind,

(ii) Magnetopause

(iii) Magnetosphere?

(b) What are the various factors that contribute to the Earth’s magnetic field?

(c) (i) With the aid of a suitable diagram, illustrate the components of the earth’s magnetic field at a given point P in the earth’s atmosphere.

(ii) An electron whose kinetic energy is 10 eV is circulating at right angles to the earth’s magnetic field whose uniform induction is 1.0 x 10 Wbm-2. Calculate the radius of the orbit and its frequency in that orbit.

Answers

(a) (i) Solar wind is a continuous stream of fast moving charged particles in the atmosphere which are produced from flare (eruptions) from the sun:

(ii) Magnetopause is the upper boundary of the magnetosphere.

(iii) Magnetosphere is the upper most part of the exosphere consisting mainly of charged ions. These particles move under the influence of the earth’s magnetic field.

(b) Short term variations: Disturbances in the magnetosphere due to solar emissions, these charged ions travel and in the ionosphere they form ring currents which give rise to a magnetic field.

Long term variations: The molten inner core of the earth is partly ionized. The movement of this ionized core causes a magnetic field which contributes to the earth’s magnetic field.

Example 05: Necta 1990 P1

(a) Define the term “isoseismal line”.

(b) Write short notes on each of the following regions of the atmosphere.

(i) Troposphere, (ii) Stratosphere, (iii) Exosphere

Answer: Refer notes

Example 06: Necta 1990 P2

(a) Explain clearly how P and S – waves were used to ascertain that the outer core of the earth is in liquid form.

(b) Giving reasons, discuss the temperature variation in atmosphere (above the earth’s surface).

Answers

(a) P – waves are longitudinal elastic, waves capable of passing through solids and liquids and S – waves are traverse elastic waves capable of a travelling through solids only.

As both waves are projected towards the surface from interior core only the P – waves are recorded. This shows that the outer core is in liquid form.

(b) From the ground level, the atmospheric temperature decreases steadily as altitude increases steadily as altitude increases up to the troposphere. Thereafter the temperature increases with altitude up to the stratosphere. The ozone of the stratosphere absorbs the incoming sun radiation hence the temperature increases. In the mesosphere there is no ozone thus there is a decrease (cooling) with increasing altitude. The heating effect of the earth ceases in the thermosphere so, the closer to the sun, the higher graph refer notes.

Example 07: Necta 1991 P2

(a) List down four physical changes that took place at a location just before onset of an earthquake at that particular location.

(b) Give brief accounts of the processes that give rise to:

(i) The earth’s magnetic field,

(ii) Volcanic eruptions

Answers

(a) Density of rocks, stresses faults and waves

(b) (i) Explain generation of the earth’s field in the atmosphere and the outer core.

(ii) The seismic or earthquakes waves result from a fracture or sudden deformation of the earth’s crust. Vast stresses do occur locally in the rocks being concentrated where the rocks are sliding over one another. In regions where pressure is reduced, pockets of molten rock called magma are formed. Once the rock has melted the pressure may force it into cracks and fissures in the surrounding solid rock. This may emerge above the surface as a lava flow or volcano.

Example 08: Necta 1992 P1

(a) What do you understand by the term ionosphere?

(b) Explain how short wave long distance transmission and reception of radio waves is more effective at night than it is during the day time.

Answer

(b) In the day time, the base of the ionosphere (D-layer) is at lower heights where the high concentration of particles allows for ionization and recombination of ions by collision. Because of this, radio waves are absorbed rather than reflected, so distance communication is poor.

During the night time, the D – layer disappear, the base of the ionosphere is higher thus the recombination of ions is rare and so less absorption of waves occurs. Obliquely transmitted waves therefore can be reflected for distant reception.

Example 09: Necta 1993 P2

(a) What is the origin of the earth’s magnetic field?

(b) The diagram below shows the structure of the Earth. Name the parts indicated by the letter A to F.

Answer

(b) A represents Mohorovicic discontinuity

B represents Gutenberg discontinuity

C represents core

D represents Mantle

E represents Epicenter

F is not clear to interpret.

Example 10: Necta 1994 P1

(a) Define the terms: angle of inclination (dip) and angle of declination (variation) as used in specifying the earth’s magnetic field at any point.

(b)The earth’s total resultant flux density BR in a certain country is found to be 5.0 x 10-5 T and the horizontal component is BH is 2.0 x 10-5 T. Calculate ;

(i) The vertical component, Bv, and

(ii) The angle of inclination in that country

Solution

(b) (i) The vertical component is given by

(ii) Angle of inclination is given by

Example 11: Necta 1994 P1

(a) (i) Name the lowest layer of the atmosphere and the lowest layer of the ionosphere.

(ii) State the importance of each of these layers.

(b) What is the ozone layer?

Answers

(a)(i) The lowest layer of the atmosphere is troposphere and the lowest layer of the ionosphere is called the D – layer.

(ii) The t troposphere supports life

The D – layer is important for communication purposes as it reflects radio waves.

(b)The ozone layer is within the stratosphere. In the ozone layer molecular oxygen (O2) is dissociated into atomic oxygen (O) which is then reformed into ozone (O3)

The ozone so formed absorbs ultra violet radiation thus protecting plants and shielding people from skin cancer and eye cataracts.

Example 12: Necta 1994 P2

(a) Illustrate the component of the earth’s magnetic field at a given point P in the earth’s atmosphere by a suitable diagram.

(b) Using a tangent galvanometer, explain how you could determine the earth’s magnetic field.

Answers

Example 13: Necta 1995 P1

(a) (i) which region of the solid earth includes the e earth’s centre?

(ii) On which region of the solid earth do the continent rests directly?

(iii) Which region of the ionosphere has the highest electron density?

(b) Briefly explain how earthquake can be detected

Answers

(a) (i) inner core (ii) crust (iii) F – region

(b) Detection of earthquake is done by recording or measuring the seismic waves generated by the earthquakes. These waves are recorded by instrument called seismograph.

Example 14: Necta 1995 P2

(a) Draw a well labeled diagram which shows the interior structure of the earth. Indicate also which part of the interior are in solid form and which are in liquid form.

(b) Name and distinguish the type of waves that are produced by an earthquake.

Answers

(a) There are four types of seismic waves:

Body waves – divided into P and S – waves

Surface waves – divided into love and Rayleigh

(b) A telecommunication problem.

Ground wave, sky wave and space waves

Example 15: Necta 1998 P1

(a) State any three magnetic components of the earth’s magnetic field

(b) The horizontal and vertical components of the earth’s magnetic field at a certain location are; 2.73 x 10-5 and 2.1 x 10-5T respectively. Determine the earth’s magnetic field at the location and its angle of inclination θ

Solution

(a) Components of the earth magnetic field are:

Vertical component (which point vertically downward)

Horizontal component which comprise lf:

Eastly component (towards geographic north pole)

Northly component (towards magnetic north pole)

(b)

Example 16: Necta 1998 P1 B

(a) What is the origin of the earth’s magnetic field?

(b) The following diagram shows the main layers forming the interior of the earth name the layers indicated by letters A to G.

Answers

(a) Refer notes

(b) A = Earth’s surface, B = Crust, C = Moho discontinuity, D = Gutenberg discontinuity, E = outer core, F = Mantle and G = inner core.

Example 17: Necta 1998 B

(a) Explain the following terms; Earthquake, Earthquake focus, Epicenter and body waves.

(b) List down three (3) sources of earthquakes,

(ii) Mention the ionosphere layers that exist during the day time

(iii) Give the reason for better reception of radio waves for high frequency signal of night than during day time.

(d) Explain briefly three different types of radio waves traveling from a transmitting station to a receiving antenna.

Answers

(a) Refer notes

(b) Refer notes

(ii) Refer Necta 1992 (b)

(d) Ground (surface wave)

Space wave

Sky waves) (refer telecommunication notes)

Example 18: nectar 2000 P1

(a) With reference to an earthquake on a certain point of the earth explain the terms ‘focus’ and ‘Epicenter’

(b) What is importance of the following layer of the atmosphere?

(i) The lowest layer

(ii) The ionosphere

(ii) Describe briefly the meaning and application of “seismic prospecting”.

Answers

(a) Refer notes

(b) (i) Importance of troposphere is supports life on earth

(ii) Ionosphere enhances communication over long distances.

(ii) Seismic prospecting is an artificial production of seismic waves purposely for searching underground fuels and oils or gases

Example 19: Necta 2001 P1

(a) (i) Define the terms “angle of declination” as used in the specification of the earth’s magnetic field at a point

(ii) The horizontal component of the earth’s magnetic field at a location was found to be 26.0 while the angle of inclination was Find the magnitude of the field and the vertical component of the field at the location

(b) (i) Define an earthquake

(ii) Distinguish between P and S waves. What factors influence their velocities?

Answers

(a) (i) Refer notes

(ii)

(b) The velocities of P and S waves are influenced by;

Density, of the media

Shear modulus, of the media, and

Bulk modulus, B of the media.

Example 20: Necta 2002 P1

(a) (i) What is the importance of ionosphere to mankind?

(ii) Explain why transmission of radio waves is better at night than at day time.

(b) (i) What is an earthquake?

(ii) Explain briefly any four (4) causes of earthquake

Example 21: Necta 2003 P2

(a) Explain the following:

(i) Earthquake (ii) Earthquake focus (iii) The epicenter.

(b) List down three sources of earthquake

(ii) State the ionosphere layer that exists during day time.

(iii) Give the reason for better waves reception for light frequencies signal at night than during the day time

Example 22: Necta 2004 P1

(a) (i) Explain the terms epicenter and focus as applied to earthquake.

(ii) State any four (4) indications that may predict the occurrence of an earthquake.

(iii) State and explain two variations of the earth magnetic field.

(iv) State one necessary precaution to be taken to people living in a region with a high risk of occurrence of earthquakes.

(b) Explain the following

(i) Solar wind (ii) Magnetopause (iii) Ionosphere.

Example 23: Necta 2005 P1

(a) Define the following terms

(i) Epicentral distance (ii) Body wave (iii) Seismograph

(b) (i) explain the meaning of reflection seismology state its application

(ii) Show how the magnetic field within the atmosphere is generated?

(ii) State their importance

Answers

(a) (i) Lowest layer of atmosphere is troposphere and that of the ionosphere is the D – layer.

Example 24: Necta 2006 P1

(a) (i) State two (2) ways by which seismic wave may be produced

(ii) What is seismic prospecting?

(b) (i) Discuss briefly the importance of the lowest layer of the atmosphere and the ionosphere.

(ii) Sketch the temperature against altitude curve for the atmosphere indicating the important atmospheric layers.

(iii)The average velocity of P – waves through the earth’s solid core is 8kms-1. If the average density of the earth’s rock is 5.5 x 103kgm-3 find the average bulk modulus of the earth’s rock.

Answer

(a) (i) Causes of an earthquake

(b) (ii) using the formula

Example 25: Necta 2007 P1

(a) (i) What are the differences between P and S waves?

(ii) Explain how the two terms of waves (P and S) can be used in studying the internal structure of the earth.

(b) Write short notes on the following terms in relation to the changes in the earth’s magnetic field; long term (secular) changes, short – period (regular) changes, and short – term (irregular) changes.

(ii) Give a summary of location, constitution and practical uses of the stratosphere, ionosphere and mesosphere.

Answers

(c) (i) Geomagnetic micro pulsation are small rapid changes in the earth’s magnetic field. They have periods between 0.2 second and 10 minutes and intensities less than 0.01% of the minimum field.

Example 26: Necta 2008 P1

(a) Define the following terms:

(i) Earthquake (ii) atmosphere

(b) Distinguish between body waves and surface waves that are produced by an earthquake.

(ii) Draw a well labeled diagram which shows the interior structure of the earth.

Example 27: Necta 2009 P1

(a) (i) What is meant by the shadow zone?

(ii) Why does the shadow zone occur?

(b) (i) Name the lowest layer of the atmosphere and the lowest layer of the ionosphere.

(ii) State the importance of each of these layers in b (i) above

(iii) Explain briefly the reason for better reception of radio waves for high frequency signals at night times than during day times.

Example 28: Necta 2010 P1

(a) (i) Explain the terms: earthquake, earthquake focus and epicenter.

(ii) Describe clearly how P and S waves are used to ascertain that the outer core of the Earth is in liquid form.

(b) (i) Define the ionosphere and give one basic use of it.

(ii) Why is the ionosphere obstacle to radio astronomy?

Example 29: Necta 2011 P1

(a) (i) Define the following terms: Geophysics, Atmosphere and Epicenter

(ii) Write down brief notes on the location, composition and importance of the following:

Troposphere, Stratosphere, Mesosphere and Thermosphere

(b) (i) Draw sketch diagram showing the working part of a Seismometer.

(ii) Explain how temperature varies with both altitude and depth of the Earth.

(iii) Write down two factors that governs heat flow from the interior of the Earth.

Example 30: Necta 2012 P1

(a) (i) Name three layers of the atmosphere

(ii) Describe any two major zones of the earth.

(b) (i) What are the factors that influence the velocities of P and S waves?

(ii) The P and S waves from an earthquake with a focus near the earth’s surface travel through the earth at nearly a constant speed of 8 km/s and 6 km/s respectively. If there is no reflection and refraction of waves how long is the delay between the arrivals of successive waves at a seismic monitoring station at 900 in the latitude from the epicenter of the earthquake?

Solution

(a) (ii) any two of core, mantle, crust, hydrosphere, atmosphere

(b) (i) the density of rock, moduli of elasticity of rock material.

(ii) Illustration (R = earth radius)

Distance travelled by the waves (distance between focus and seismic station) is

Time taken by P – waves to arrive at the station is

Time taken by the waves to arrive at the station is

The time interval between the arrival of the two waves is t = t2 – t1 = 25.1 = 18.9 = 6.2 minutes.

Example 31: Necta 2012 P1

(a) (i) What do you understand by the word environmental physics?

(ii) Briefly explain three effects of seismic waves.

(b) (i) Mention three types of environmental pollution

(ii) Explain on the following climatic factors which influence plant growth: Temperature, Relative humidity and wind.

Example 32: Necta 2013 P1

(a) (i) The main interior of the earth core is believed to be in molten form. What seismic evidence supports this belief?

(ii) Explain why the small ozone layer on the top of the stratosphere is crucial for human survival

(b) Electrical properties of the atmosphere are significantly exhibited in the ionosphere.

(i) What is the layer composed of and what you think is the origin of such constituents

(ii) Mentioned two uses of the ionosphere

Answers

(a) (i) When P and S seismic waves are sent from one side of earth to the other, only P waves can be detected on the other side. The fact that S waves do not travel through the core provides evidence for the existence of a liquid core.

(ii) Ozone absorbs harmful radiation from the sun. The Ozone projects plant and shield people from skin cancer and eye cataracts.

(b) (i) The layer is composed of free electrons and positive ions. The ionosphere is created by atoms absorbing UV radiation, gamma and x-rays.

(ii) Uses of the ionosphere

Ionosphere supports radio communication over long distances

Particles in the ionosphere absorbs U.V radiation gamma and X-rays, thus protecting people from harmful effects of these radiations

Example 33: Necta 2013 P1

(a) Briefly explain on the following types of environmental pollution:

(i) Thermal pollution

(ii) Water pollution

(b) Describe the soil temperature with regard to agriculture, physics which causes lower crop growth at a particular area

Answers

(b) High soil temperature causes the crop roots to rot, this leads to insufficient water supply to plant leaves and hence lower the growth of crop.

Lower soil temperature inactivates soil organisms. Decomposition of organic matter is lowered and hence the supply of nutrients to crop which in turn lead to lower crop growth.

TRY YOURSELF

(a) (i) What are auroras?

(ii) Define the homosphere

(b) (i) What are the factors which contribute toward volcanic eruptions?

(ii) What are the effects of volcanic eruptions?

(iii) What are lahars?

Lahars are rapidly flowing mixtures of rock debris and water that originate on the slopes of a volcano. They are also referred to as volcanic mudflows or debris flow. Volcanic eruptions may directly trigger one of more lahars by quickly melting snow and on a volcano or eject water from a crater lake. The form in a variety of at always including through intense rainfall on loose volcano rock deposits and as a consequence of debris of debris avalanches

ENVIRONMENTAL POLLUTION

Pollution is the addition of unwanted materials or pollutants into the environment.

Pollutant is any substance that does not belong in the natural system and disrupts the natural balance.

Type of Environmental pollution

(a) Air pollution (atmospheric pollution)

(b) Water pollution (hydrosphere pollution)

(d) Noise pollution

(e) Thermal pollution

ATMOSPHERIC (AIR) POLLUTION

AIR POLLUTION This is a form of environmental pollution caused by the release of gaseous materials and dust particles in the atmosphere. The main pollutants found in the air we breathe include, particulate matter, lead, ground-level ozone, heavy metals, sulphur dioxide, benzene, carbon monoxide and nitrogen dioxide

Causes of Air Pollution

Man made causes:

(i) Clearing (deforestation) and burning of vegetation. This releases carbon dioxide in the atmosphere and dust particles which may be carried by wind on bare land.

(ii) Burning of fuels: This releases green house gases in the atmosphere. Fuels are burnt in cars, power stations and industries.

(iii) Construction activities, like road, building, etc construction, can add dust particles in the atmosphere.

(iv) Automobile exhausts. Car, trains, etc burns fuels as they move his releases pollutant gases in the atmosphere.

(v) Smokes from industries also pollute the atmosphere.

(vi) Agriculture activities. The use of pesticide/insecticides pollutes the air.

(vii) Mining activities

Natural causes:

(a) Volcanic eruptions - release smoke and dust particles in the atmosphere

(b) Wind storms – carry land particles into the air

(c) Temperature inversion – the increase in temperature in the stratosphere causes high altitude particles to sink to the troposphere

WATER POLLUTION

Water Pollution is the degradation of water quality in a manner that disrupts/prevents its intended or original use.

Surface Water or Ground water may be polluted

Causes of water pollution

(i) Disposal of untreated sewage (industrial or hospital, etc) into the water bodies.

(ii) Wind may introduce dust particles into water from the land.

(iii) Agriculture activities near water bodies. Chemical used during farming may be taken to the water bodies by the rain water.

(iv) Oil spilt. The leakage of oil in under water oil pipe, leakage from boats, ships, etc pollutes the water.

(v) Fishing by using chemicals (dynamite fishing).

(vi) Volcanic activities along water bodies.

(vii) Quarrying along the coast.

LAND (SOIL) POLLUTION

Soil pollution is defined as the build – up in soils of persistent toxic compounds, chemicals, salts, radioactive materials, or disease causing agents which have adverse effects on plant growth and animal health.

A soil pollutant is any factor which deteriorates the quality, texture and mineral content of the soil or which disturbs the biological balance of the organisms in the soil.

Causes of soil pollution

(a) Chemical from industries

(b) Acid rain – this increase soil acidity

(d) Mining activities – increase rock sediment into the soil.

NOISE POLLUTION

Noise pollution is any disorganized loud sound.

Causes of noise pollution

(a) Noise from factories and workshops

(b) Thunderstorm explosion of bombs

(d) Radio on large volumes

(e) Slamming of doors

THERMAL POLLUTION

Thermal Pollution is a form of environmental pollution caused by the release of waste heat into water or air

Causes of Thermal Pollution

(a) i. Hot gases released by industries and motor vehicles warm the environment.

ii. Hot wasteful liquid from industries pumped to a river, lake, or other waterway

Effects of thermal pollution

(a) Heat introduced into water can make the water so hot that no living thing can survive in it

(b) Hot gases introduced in the atmosphere leads to green house effects.

Solutions of thermal pollution

(a)One is a cooling pond into which heated waste water is released before it enters a natural waterway. The cooling pond permits evaporation of some water, carrying heat into the air and thus releasing cooler water into the waterway

(b)The cooling tower method – either wet or dry – which also transfers heat to the air. In both types, heated water is introduced into a tower through which air is blown, and some heat is passed to the air.

PARTICULATE MATTER IN THE ATMOSPHERE (AEROSOLS)

Particulate matter (aerosol) is the general term used for a mixture of fine solid particles and liquid droplets found in the air.

Haze aerosol is frequently encountered in optical studies and includes any airborne particles that affect visibility.

Classification of Particulate

Particulates matter are classified in accordance with its formation mechanisms

(i) Primary particles (ii) Secondary particles

Primary particles are directly emitted into the atmosphere from their sources while secondary particles are formed after chemical transformation of their gaseous precursors. Chemical reactions transform primary pollutants (emitted by the sources) to secondary pollutants that are formed within the atmosphere. Ozone, sulfate aerosols, nitrates, are examples of secondary pollutants.

Particulate matters in the atmosphere are categorized as:

(i) Minerals, 72 – 91%, e.g. soil particles, hematite, mica, and talc;

(ii) Combustion products, 1 – 10%, e.g. coal and oil soot, fly ash, burned paper.

(iii) Biological materials 2 – 10% e.g. pollen, spores, starch, plant tissues and diatoms

(iv) Miscellaneous matter, trace – 8% e.g. salt, rubber, iron/steel, paint pigment and humus

Dust refers to a relatively course range of solid particles (diameter, d >1pm), produced by disintegration of minerals or from re-suspension by wind when sun blasting of soil particles may often causes comminuting.

Smokes and fumes are fine particles formed from the gas phase by condensation. In the case of fume the particles are generally from 0.01 – 1 pm diameter, and are often observed as agglomerates of smaller particles. Suspended particulate matter < 15 pm and diameter is usually defined as smoke.

Mists and fogs are liquid droplets Mists (d > 40 pm) and fogs (d = 5 – 40 pm).

Advantages of particulate matter in the atmosphere

Aerosols acts as nuclei were water vapour collects during the formation of water droplets through condensation.

Disadvantages of particulate matter in the atmosphere

(a) Cause global warming

(b) Can block the atmosphere (impair visibility)

(d) Changing the timing and location of traditional rainfall patterns

(e) Can lead to development of heart and lung diseases.

TRANSPORT MECHANISMS OF ATMOSPHERIC POLLUTANT

The transport of pollutants by the wind

The three transport processes that influence the regional dispersion are;

(a) Wind speed (shear)

(b) Directional veer (change in direction fo wind), and

Wind shear: The vertical gradient of wind speed (i.e. wind shear is responsible for lagging of low elevation pollutants behind those in the upper layers.

Directional veer: The directional veer with height causes lateral displacement of a vertically uniform puff.

The eddy motion is the vertical transport of pollutants from region of high concentration to low concentration. Eddy motions are due to random vertical and horizontal fluctuations caused by thermal and mechanical turbulence.

Both the transport speed and direction for an air parcel vary from day to day.

Stratosphere – troposphere interchange

Temperature inversion at the tropopause causes an interchange of particulate matters between Stratospheres – troposphere boundary.

EFFECTS OF POLLUTION ON VISIBILITY

Atmospheric pollution results into a reduction in visual range in the atmosphere. The reduction is visual range caused by an increase in airborne particles that affects light scattering and attenuation involves both primary and secondary aerosols, and may be experienced in rural as well as urban area.

EFFECTS OF ATMOSPHERIC POLLUTION ON THE GLOBAL ALBEDO AND CLIMATE

Increases in particulate matter in the atmosphere may:

(a) affect cloud droplet formation and precipitation,

(b) Reduce the amount of solar radiation that reaches the ground

However, controversy still remains as to whether the presence of particulate material exerts a net warming or cooling effect to enhance or offset the global warming predicted from increases in CO2 and chlorofluoro methanes in the atmosphere. In addition, considerable changes in global and surface albedo have been caused by deforestation, salinization, and desertification.

Global warming is the increase of the average temperatures near or on the surface of the earth as a result of greenhouse effect.

GLOBAL WARMING

Global warming is the increase of the average temperatures near or on the surface of the earth as a result of greenhouse effect.

Greenhouse effect

Greenhouse effect is the process in which the emission of radiation by the atmosphere warms the earth’s surface.

Greenhouse gases include carbon dioxide, methane, chlorofluorocarbons and dinitrogen oxide.

When heat from the sun reaches the earth’s surface in form of sunlight, some of it is absorbed by the earth. The rest is radiated back to the atmosphere at a long wavelength than the incoming sunlight. Some of these longer wavelengths are absorbed by the greenhouse gases in the atmosphere before they are lost out of space. The greenhouse gases reflect the heat back to the earth and warm the environment.

Sources of greenhouse gases in the atmosphere

(a) Carbon dioxide is added in the atmosphere by:

(i) Clearing and burning of vegetation

(ii) Burning of fossil fuels

(b) Methane is added in the atmosphere by:

(i) Agricultural activities;

(ii) The mining of coal and oil

(i) Combustion of fossil fuels in vehicles and power station

(ii) Use of nitrogenous fertilizer, and

(iii) The burning of vegetation and animal waste

(d) Sources of chlorofluorocarbon include fridge, air conditioners and aerosols.

Effects of Global Warming

(a) Increase in the temperature of the oceans,

(b) Rise in sea levels,

(d) Acidification of the oceans,

(e) Extreme weather events like flood, droughts, heat waves, hurricanes and tornadoes

(f) Higher or lower agriculture yields,

(g) Melting of Arctic ice and snow caps. This causes landslides, flash floods and glacial lake overflow,

(h) Extinction of some animals and plant species,

(i) Increase in the range of disease vectors (organisms that transmit disease).

Solution to Global Warming

(a) Use of cleaner alternative sources of energy such as solar and wind,

(b) Put in place energy conservation measures to reduce the use of fossil fuel,

(a) Use of cleaner alternative sources of energy such as solar and wind,

NUCLEAR WASTE AND METHODS OF DISPOSAL

Nuclear wastes are the chemical products (solid, liquid and or gases) of nuclear reactions in the nuclear reactor.

Categories of radioactive waste

For the purpose of disposal, radioactive waste is divided into the following categories:

(a) High – level waste (HLW): spent fuel (SF) not destined for reprocessing; vitrified fission product solutions from reprocessing of spent fuel.

(b) Alpha – toxic waste (STW): waste with a content of alpha – emitters exceeding a value of 20,000 Becquerel’s per gram of conditioned waste.

Nuclear Waste Disposal

(a) Deep geological repository: for spent fuel and vitrified fission product solution product solutions from reprocessing. The products are buried deep into the earth.

(b) Recycling of the nuclear waste.

ATOMIC PHYSICS

1. 1. Structure of the ATOM -Describe the Rutherford and bohr’s models of the atom. -Analyze atomic energy levels. -Discuses the hydrogen energy levels, and derives expressions for the energy levels. -Perform experiment to determine wavelength in the Balmer series of the hydrogen spectrum. 2. Quantum physics - Describe failures of classical physics. - Explain Planck’s quantum theory of blackbody radiation. - Spectral distribution of black body radiation. - Explain Einstein’s quantum theory of light. - Perform experiment to determine the Planck’s constant. - Account for the photoelectric effect phenomenon. - Deduce stopping potential threshold frequency and work function of a metal. - Explain the photo electric effect. - Deduce de Broglie wave length for electron. - Discus the wave- particle duality of electron. - Derive de Broglie’s wavelength for the electron. - Describe production and uses of x- rays. - Uses in medicine, industry and in sample analysis.

3. 3. LASER - Describe production of laser light. - Explain properties of laser light. - Distinguish types of lasers. - Discuss methods of pumping in laser production. - Identify application of laser light.

4. 4. Nuclear Physics - Describe the structure of the nucleus. * Review Rutherford experiment. - Determine half life and the decay constant (λ) of a radioactive substance. - Explain the relation of nuclear mass and binding energy. * Discuss Einstein’s mass energy equation. * Apply Einstein’s mass energy relation to determine the biding energy of nuclei. - Identify criteria for stable and unstable nucleus. * Analyze the neutron and proton ratio and plot N against Z for radioactive elements. * Establish criteria for stable and unstable nuclei - Identify uses and hazards of radioisotopes * Application * Hazards - Distinguish between fission and fusion processes * Meaning of fission and fusion * Calculate the energy released in a nuclear fission * Calculate the energy absorbed in nuclear fission * Describe the application of nuclear fission and fusion - Describe operation of a nuclease reactor * Construction and operation of nucleus reactor for safe application

THOMSON’S MODEL OF ATOM

According to Thomson an atom is a positive charged sphere in which the entire mass and positive charge of the atom is uniform distributed with negative electrons embedded in it as shown.

The number of electrons is such that their negative charge is equal to the positive charge of the atom. This atom is electrically neutral.

This model was called Thomson’s plum pudding model because the negatively charge electrons (the plums) were embedded in a sphere of uniform positive charge (the pudding).

Drawbacks of this Model 1.It could not provide stability to the atom it is because the positive and negative charges are stationary and will be drawn towards each other, thus destroying the individual negative and positive charges. 2. It could not explain the presence of discrete spectral lines emitted by hydrogen and other atoms. RUTHER FORD’S MODEL OF ATOM The salient features of this model are (i)Every atom consist of a tiny central core, called the nucleus which contains all the atom’s positive charge and most of its mass (99.9%). (iii) The electrons occupy the space outside the nucleus. Since an atom is electrically neutral the positive charge on the nucleus is equal to the negative charge on electrons surrounding the nucleus. (iv) Electrons are not stationary but revolve around the nucleus in various circular orbits as do the planets around the sun. In this way Rutherford provided stability to the atom. It is because the centripetal force required by the electrons for revolution is provided by the electrostatic force of attraction between electrons and the nucleus.

e = charge on electron

z =total number of protons in the nucleus

m=mass of the electron

r =distance of electron from the nucleus

v= linear velocity of the electron

Force of attraction between electron and the nucleus is

where Ze is a nuclear charge

The centripetal force required to keep the electron moving in circular path is

Since the atom is stable

Kinetic energy of electron

From equation (1)

Potential energy of electron

Total energy of electron

The total energy of electron in the orbit is negative hence the electron is bound to the positive nucleus

For hydrogen Atom

For hydrogen atom z= 1. Therefore K. E and P.E OF electron in hydrogen atom are

The total energy of electrons hydrogen atom is

Limitations of Rutherford’s model of atom

1. According to Maxwell’s theory of electromagnetism a charge that is accelerating radiates energy as electromagnetic waves The electron moving around the nucleus is under constant accelerating radiates energy as electromagnetic waves.

- Due to this continuous loss of energy the electrons in Rutherford’s model were bound to spiral towards the nucleus and fall into it when all of their rotational energy were radiated

- Hence Rutherford’s atomic model cannot be stable while in actual practice, an atom is stable

This shows that Rutherford’s model is not correct

1. During inward spiraling the electron’s angular frequency continuously increases

- As result electrons will radiate electromagnetic waves of all frequency i.e. the spectrum of these waves will be continuous in nature because these are continuous loss of energy.

- But this is contrary to observation experiments shows that an atom emits line spectra and each line corresponds to a particular frequency or wavelength.

Rutherford’s model failed to account for the stability of the atom. It was also unable to explain the emission of line spectra. BOHR’S MODEL OF ATOM According to Bohr's atomic model, the revolving electrons in the atom do not emit radiations under all conditions. They do so under certain conditions as expalined by him in his model. BASIC POSTULATES OF BOHR’S MODEL OF ATOM

1. The electrons revolve around the nucleus of the atom in circular orbits. The centripetal force required by electrons for revolution is provided by the electrostatic force of attraction between the electrons and the nucleus.

2. An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of

h= Plank’s constant.

Radius of orbit r

From,

Since n is a whole number only certain value of r is allowed.

Thus according to Bohr, an electron can revolve only in certain orbits of definite radii not in all these are called stable orbits (stationary orbit)

According to this postulate the angular momentum of the electron does not have continuous range i.e. the angular momentum of the revolving electron is quantized.

While revolving in stable or stationary orbits the electrons do not radiate energy inspite of their acceleration towards the centre of the orbit.

- For this reason these permitted orbits are called stable or stationary orbits.

e= charge on electron

m= mass of electron

rn= radius of the nth orbit

vn= velocity of electron in the nth orbit

Z= number of positive charge (protons) Positive charge on nucleus Ze RADIUS OF BOHR’S STATIONARY ORBITS

As the centripetal force is provided by the electrostatic force of attraction between the nucleus and electron.

According to Bohr

Consider equation

Take equation (ii) square it

Take equation (iii) equation (i)

= .

It is clear that n2, radii of the stationary orbits are in ratio 12: 22:32 ………..clearly the stationary orbits are not equally spaced.

For hydrogen atom

For hydrogen atom z= 1, so that equation become

Now = 0.53 x10-10m

e = electronic charge = (0. 53 x 10-10) metres

Thus the radii of the first, second and third stationary orbits of hydrogen atom are 0.53 Å, 2.12 Å and 4.77Å respectively.

2. VELOCITY OF ELECTRON IN BOHR’S STATIONARY ORBIT

From equation below, we have

Putting the value of into that equation

It is clear that in other words, electrons move at a lower speed in higher orbits and vice versa.

For hydrogen atom

Z =1

Then

3. FREQUENCY OF ELECTRON IN STATIONARY ORBIT

The number of revolution completed per second by the electron in a stationary orbit around the nucleus

Velocity of electron in the orbit

For hydrogen atom

Z = 1

Then,

Frequency of electron in the first orbit of hydrogen atom is n=1, r1=0.53x10-10m

Electron in first orbit of hydrogen atom will have a frequency of 6.57x 1015revolutions per second.

4. TOTAL ENERGY OF ELECTRON IN STATIONARY ORBIT

The total energy En of the electron in the nth orbit is the sum of kinetic and potential energy in the nth orbit.

- The K.E of electron in the nth orbit is

The potential energy of electron in the nth orbit is

Total energy of electron in the nth orbit is

But

Thus as n increases i.e. electron moves to higher orbit, the total energy of the electron increases i.e. total energy becomes less negative.

For hydrogen atom z=1

Thus the total energy of electron in a stationary orbit is negative which means that the electron is bound to the nucleus and it is not free to leave the atom.

We can find the total energy of electron in the various orbits of hydrogen atoms as under.

The total energy of electron increases i.e. becomes less negative as the electron goes to higher orbits

When n→∞ En =0 and the electron becomes free

Ground state/ normal state

This is the state of atom when the entire electrons in it occupies their lowest energy levels as required by their n and l values.

The energy of an atom is least i.e. largest negative value when n=1 i.e. when electron revolves in the first orbit.

The energy of hydrogen atom in the ground state is 13.6eV.

Excited state

This is the state of an atom when electrons in an atom occupy energy levels higher than those permitted by the values of n and l values.

At room temperature most of the hydrogen atoms are in the ground state

If hydrogen atom absorbs energy i.e. due to rise in temperature it may be promoted to one of the higher orbits (i.e. n=2, 3, 4…..)

The atom is said to be in the excited state. WAVE LENGTH OF EMITTED RADIATION. When an electron jumps from a higher orbit (n2) to the lower orbit (n1) the energy difference between the two orbits is released because the energy of electron in the higher orbit is more than in the lower orbit. Consider two orbits having principle quantum numbers n2 and n1 where n2>n1

Then energy of electron in the two orbits is given by

As the electron jumps from orbit n2 to n1, energy is released in the form of electromagnetic radiation.

where

f= frequency of the emitted radiation

The wavelength of the emitted radiation is given by

c=λf

This equation gives the wavelength of emitted radiation.

Now,

== wave number

Wave number

These are the number of waves in a unit length.

For hydrogen atom

For hydrogen atom z = 1

This gives the mathematical formula for the wavelength of radiation emitted by hydrogen atom when electron jumps from outer orbit to inner orbit.

where

RH is Rydberg constant. The value of RH can be calculated as the value of e, m, h and c are known

HOW TO CALCULATE THE RYDBERG CONSTANT USING CALCULATOR

From

Clearly, wavelength/frequency of radiation emitted from the excited atom is not continuous. They have definite value depending upon the values of , and SPECTRAL SERIES OF HYDROGEN ATOM

Bohr gave a mathematical explanation for the spectrum of hydrogen atom. The whole hydrogen spectrum can be divided into district groups of lines each group of lines is called spectral series.

The wavelength of the lines in each group can be calculated from Bohr’s formula

The following are spectral series of hydrogen atom

i) Lyman series

ii) Balmer series

iii) Paschen series

iv) Bracket series

v) Pfund series

i) Lyman series

The Lyman series is obtained when electron jump to first orbit n1=1 from outer orbits (=2, 3, 4…)

Therefore the formula for calculating the wavelength of the lines in this series is,

where

This series lies in the ultraviolet region which is the invisible region.

ii) Balmer series

Therefore the formula for calculating the wavelength of the lines in this series is

where

This series lies in the visible spectrum and was found first of all in the hydrogen series iii) Paschen series

Therefore the formula for calculating the wavelength of the lines in this series is

where

This series lies in the infrared region.

iv) Brackett series

Therefore the formula for calculated the wavelength of the lines in this series is

This series lies in the infrared region.

v) Pfund series The Pfunds series is obtained when electrons jump to fifth orbit n1 = 5 from outer orbits (n2 =6, 7, 8…..)

Therefore the formula for calculating the wavelength of the lines in this series is

where

(n2 =6, 7, 8…..)

This series also lies in the infrared region ENERGY LEVEL DIAGRAM

Energy level diagram is a diagram in which the total energies of electron in different stationary orbit of an atom represented by parallel horizontal lines drawn according to some suitable energy scale

In order to draw energy level diagram of an atom we must know the total energy of electron in different stationary orbits.

The total energy of an electron in the nth orbit of hydrogen atom is given by

By putting value of n=1, 2, 3….. we can find the total energy of electron in various stationary orbits of hydrogen atom as

Similarly we can find the total energy of electron in the higher orbits

The table below gives the total energy of electron of hydrogen atom in different stationary orbits.

The energy level diagram of hydrogen atom is shown below

Total energy of electron in a stationary orbit is represented by a horizontal line drawn to some suitable energy scale.

(i) The hydrogen atom has only one electron and this normally occupies the lowest level and has energy of -13.6eV

When the electron is in this level the atom is said to be in the ground state.At room temperature nearly all the atoms of hydrogen are in ground.

(ii) If hydrogen atom absorbs energy (due to rise in temperature )the electron may be promoted into one of the higher energy levels

The atom is now said to be in an excited state.Thus when the electron occupies other than the lowest energy level the atom is said to be in the excited state.

(iii) Once in an excited state the atom is unstable after a short time interval the electron falls back into the lowest state so that the atom is again in the ground state.

The energy that was originally impacted is emitted as electromagnetic waves.

(iv) The total energy of electron for (n=) it becomes free of atom.

The minimum energy required to free the electron from the ground state of an atom is called ionization energy

For hydrogen atom ionization energy is +13. 6eV

(v) The difference between the adjacent energy goes on decreasing as the value of n increases.

So much so that when n>10 the energy difference is almost zero this is show by closeness of energy level lines at higher levels.

(vi) Note that region is labeled continuous at energy above zero n= level, the electron is free from the atom and is at rest

Higher energy represents the translation kinetic energy of the free electron

This energy is not quantized and so all energies above n = are allowed IMPORTANT TERMS

It is desirable to discuss some important terms much used in the study of structure of atom.

(i) EXCITATION ENERGY

Excitation energy is the minimum energy required to excite an atom in the ground state to one of the higher stationary state.

Hydrogen atoms are usually in their lowest energy state where n=1

In this state (ground state) they are said to be unexcited.

However if you bombard the atoms with particles such as electron or proto collision can excite them

In other words a collision may give an atom enough energy to change it from ground state to some higher stationary state.Consider the case of hydrogen atom we know that = -13.6eV (ground state = -3.4eV (first excited state) =1.51eV (second excited state) and =0

In order to lift an electron from ground state n =1 to the first excited state n=2 energy required is E

E =-

E= -3.4 - (- 13.6)

E = 10.2eV Therefore the bombarding particle must provide an energy of 10.2eV to excite the atom from n =1 state to n=2 state

Similarly to excite the atom from n=1 state to n=3 state energy required is

E = - 1.51 – (-13.6)

E = 12.1eV

We say that first and second excitation energies of hydrogen are 10.2eV and 12.1eV respectively

(ii) EXCITATION POTENTIAL

Excitation potential is the minimum accelerating potential which provide an electron energy sufficient to jump from the ground state n=1 to one of the outer orbits

Energy required to lift a electron from ground state n=1 to n=2 state is

Hence excitation potential for the first excited state of hydrogen is 10. 2V

Similarly energy required to lift an electron from ground state n=1 to n=2

The value of excitation potential depend upon the state to which the atom is excited to which the atom is excited from the ground state

(iii) IONIZATION ENERGY

Ionization energy is the minimum energy needed to ionized an atom

Consider the case of hydrogen atom it has only one electron and this normally occupies the ground state.

The energy of the electron for n= state is zero and if the electron is lifted to this level (n=) it becomes free of hydrogen atom i.e. hydrogen atom is ionized

(iv) IONIZATION POTENTIAL

Ionization potential is the minimum accelerating potential which would provide electron energy sufficient to just remove the electron from the atom.

The ionization potential of one electron atom or ion is given by

(v) QUANTIZATION OF ENERGY

Quantization of energy is the existence of energy radiated by atoms in a specific amount which is are integral multiples of a constant (hf).

SUCCESS OF BOHR’S THEORY

The success of bohr's theory is not to be attributed so much to the mechanical picture of atom he proposed but rather to the development of mathematical explanation that agrees exactly with experimental observations. Bohr’s theory achieved the following successes.

i) MADE ATOM STABLE

Bohr’s theory made the atom stable according to this theory an electron moving in the formatted (quantum) orbits cannot lose energy even though under constant acceleration. This provided stability to the atom.

ii) INTRODUCED QUANTUM MECHANICS

Bohr’s theory introduced quantum mechanics in the realm of atom for the first time

Bohr’s explained that sub- atomic particles e.g. electrons are governed by the laws of quantum mechanics and not by classical laws of electron hydrogen as assumed by Rutherford

This completely changed our thinking and was the major step towards the discovery of the rudiment laws of the atomic world

iii) GAVE MATHEMATICAL EXPLANATION OF HYDROGEN SERIES

The hydrogen series found by various scientists were based on empirical relation but had no mathematical explanation

However these relations were easy derived by applying Bohr Theory

Further the size of hydrogen atom as calculated from this theory agreed very closely with the experimental value.

LIMITATIONS OF BOHR’S THEORY

Bohr’s simple theory of circular orbits inspire of its many successes was found inadequate to explain many phenomena observed experimentally.

This theory suffered from the following drawbacks.

(i) It could not explain the difference in the intensities of emitted radiations.

(ii) It is silent about the wave properties of electron

(iii) It could not explain experimentally observed phenomena such as Zeeman Effect, Stack effect etc.

(iv) Bohr’s model does not explain why the orbit are circular while elliptical path is also possible

(v) It could only partially explain hydrogen atom. For example this theory does not explain the fine structure of spectral lines in the hydrogen atom

WORKED EXAMPLES

1. 1. Find the radius of the first orbit of hydrogen atom. What will be the velocity of electron in the first orbit? Hence find the size of hydrogen atom

Solution

The radius of nth orbit of it atom is given by

Radius of first orbit of it atom n=1

Velocity of electron in the nth orbit of hydrogen atom is given by

Velocity of electron in the first orbit of hydrogen atom is given by

Since there is one electron in hydrogen atom the size hydrogen atom is equal to double the radius of the first orbit

Size of the atom

= 2

= 2 x 0.53Å

Size of an atom =1.06Å

2 2. (a) The hydrogen atom is stable in the ground, state why?

(b) The ionization energy of hydrogen is 13. 6eV what does it mean?

Solution

If the hydrogen atom is in the ground state (n=1) there is no state of lower energy to which a down ward transition can occur thus a hydrogen atom in the ground state is stable

a) It means that energy required to remove the single electron from the lowest energy state of hydrogen atom to becomes free electron is 13.6eV

b) Second line of Lyman series is obtained when electron jumps from third orbit =3 to the first orbit n=1

According to Bohr’s theory the wavelength of emitted radiation is given by

= x

3 3.( a) What is the meaning of negative energy of orbiting electron?

(b) What would happen if the electron in atom were stationary?

Solution a) The negative total energy means that it is bound to the nucleus. If it acquires enough energy from some external source (a collision for example) to make its total energy zero the electron is no longer bound it is free. b) If the electrons were stationary they would fall into the nucleus due to electrostatic force of attraction so atom would be unstable i.e. it would not exist c) For Paschen series we have longest wavelength line =4 This is a wavelength in the infrared part. Other lines in this series have shorter wavelength bad approach series limit of wavelength to given by This wavelength is also in the hydrogen part. This the range or centre series (820.4nm to 1875nm) is the infrared

4. a) If an electron jumps from first orbit to third orbit will it absorb energy? b) Name the series of hydrogen spectrum lying in the infrared region c) Calculate the shortest wavelength of the Balmer series d) What is the energy possessed by an electron for n=?

Solution

a) Yes it is because the energy level of third orbit is more than that of the first orbit

b) * Paschen series

* Bracket series

* P fund series Solution In Balmer series the radiation of shortest wavelength (i.e. of highest of highest energy) is emitted when electron jumps from infinity orbit = to the second orbit =2 of hydrogen atom.

5 5. a) The ionization potential of hydrogen is 13.6V what does it mean?

b) Find the longest wavelength in Lyman series

c) How much is the ionization potential of hydrogen atom?

d) The energy of the hydrogen atom in the ground state is 13.6eV. Determine the energies of those energy levels whose quantum numbers are 2 and 3.

Solution

a) The ionization energy of hydrogen is 13.6eV. Therefore, if an electron which has been accelerated from rest through a p.d of 13.6V collides with a hydrogen atom it has exactly the right amount of energy to produce ionization.

This is a common method of producing ionization and therefore the term ionization potential is often used.

b) Solution

In Lyman series the radiation of longest wavelength (i.e. lowest energy) is emitted when electron jumps from second orbit =2 to first orbit n=1 of hydrogen atom

c) The energy of hydrogen atom in the ground state is – 13.6eV. therefore its ionization energy is 13.6eV and ionization potential =13.6V

d) Solution

The energy of an electron in the nth orbit of hydrogen atom is given by

6 6. a) Name the series of hydrogen spectrum lying in the

i) Visible region

ii) Utraviolet region of electromagnetic spectrum

b) Write the empirical relation for Paschen series lines of hydrogen spectrum

c) What are the values of first and second excitation potential of hydrogen atom?

d) Calculate the radii and the energy of three lowest energy allowed orbits for the electron in Lithium ion. What is the energy of a photon that when absorbed causes an electron in Lithium ion to be excited from n=1 to n=3 state?

Solution

a) i) Balmer series

ii) Lyman series

b) The wavelength of the spectral lines in paschen series are given by

c) Excitation energy for first excited state = -3.4 – (-13.6)

=10.2eV

For second excited state

= - (1. 51 - (-13.6)

= 12.1eV

Solution

d) (I ) for a single electron atom or ion the radius of the nth orbit is given

For a single electron atom or ion the energy of electron in the nth orbit is given by

Thus the energy of n=1, 2 and 3 orbits. The photo energy must be equal to the energy needed to excite the electron

E3 – E1 =hf

(-13.6) – (-122.4) =photon’s energy Photon’s energy = 108.8eV

7. The ionization energy of hydrogen like atom is 4rydbergs

(a) What is the wavelength of radiation emitted when electron jumps from first excited state to the ground state?

(b)What is the radius of the first orbit for this atom?

(d) According to Bohr’s theory what is the angular momentum of a electron in the third orbit

Solution

The energy electron in the nth orbit of hydrogen like atom is

The energy required to excite the electron from n=1 level to n=2

If is the wavelength of the emitted radiations then, Radius of first orbit for this atom

Solution

( b) Radius of nth orbit

Angular momentum L of an electron in nth orbit is

L = n

Here n=3

Then

L= 3

7 8. The energy levels of an atom are shown in figure below.

(a)Which one of these transitions will result in the emission of photon of wavelength 275nm?

(b) An electron orbiting in hydrogen atom has energy level of 3.4eV what will be its angular momentum (c) The total energy of an electron in the first excited state of hydrogen atom is about – 3. 4eV what is the wavelength? solution (a) Energy of emitted photon E

Therefore photon of wavelength 275nm will be emitted for transition B

Solution

(d) K.E of electron = -(total energy of electron)

K.E of electron =3.4eV

ii) P.E of electron = 2xtotal energy

P.E of electron = -6.8eV

10. (a) How many lines can be drawn the energy level diagram of hydrogen atom?

(b) Use Bohr's model to determine the ionization energy of the He ion also calculate the minimum wavelength a photo must have to cause ionization

(d) i) In neon atom the energies of the 3s and 3p states are respectively 16.70eV and 18.70eV. What wavelength corresponds to 3p -3s transitions in neon atom?

ii) The wavelength of the first member of the Balmer series in hydrogen spectrum is 6563Å. Calculate the wavelength of first member of Lyman series in the same spectrum

PLANCK’S QUANTUM THEORY OF BLACK BODY RADIATION

The findings in the black body radiation led Max plank in 1901 to postulate that radiant energy is quantized i.e. it is radiated in form of energy packets.

BASIC POSTULATES OF THE PLANK’S THEORY

1. Any radiation is associated with energy.

2. Radiant energy is emitted or absorbed in small packets known as quanta.

3. The energy associated with a quantum is proportional to the frequency f of the radiation

4. The energy is absorbed or emitted only in whole number of quanta

Black Body Radiation

A blackbody is a substance that absorbs all light fall on it and does not reflect any light.

It is not easy to get a black body however a sealed metal box with a very small hole on it is very close to a black body.

From law of physics it follows that a good absorber of radiation also is a good radiator. A black body is supposed to be the best radiator.

When a black body is heated it emits light. The colour of light emitted changes from red to yellow then to white as the temperature is increased.

The change in colour with temperature shows that the frequency changes with temperature.

This in contradiction with the classical wave theory since in the classical wave theory energy is uniformly distributed over the wave form when heating the black body the colour of radiation should stay the same

Only the intensity is supposed to increase with temperature.

DISTRIBUTION OF ENERGY IN THE SPECTRUM OF A BLACK BODY

Lamer and Pringshein investigated the distribution of energy amongst the different wavelength of a thermal spectrum of a back body radiation

The results obtained by Lamer and Pringshein are shown in figure below

Results:

1. At a given temperature the energy is not uniformly distributed in the radiation spectrum of a hot body

2. At a given temperature the intensity of radiations increases with increased in wavelength and at a particular wavelength λ its value is maximum with further increase in wavelength the intensity of heat radiations decreased

3. With increase in temperature wave length increases, wavelength emission of energy takes place.

The points on the dotted line represent wavelength at various temperatures

4. For all wavelength an increase in temperature causes an increase in the energy emission The area under each curve represents the total energy emitted for the complete spectrum at a particular temperature

5. This area increases in temperature of the body. It is found that the area is directly proportional to fourth power of the temperature of the body

This represents Stefan’s Boltzmann’s law.

Plank’s constant

Plank's constant is a fundamental constant equal to the ratio of the quantum energy to the frequency of the radiation.

ELECTRON EMISSION

This is the liberation of electron from the surface of a substance.

For electron emission metals are used because they have many free electrons.

If a piece of metal is investigated at room temperature the random motion of free electrons is as shown in figure below

However these electrons are free only to the extent that they may transfer from one atom to another within the metal but they cannot leave the metal surface to provide electron emission. It is because the free electron that start at the surface of metal find behind them positive nuclei pulling them back and none pulling forward. This at the surface of a metal a free electron encounters force that prevents it to leave the metal

In other words the metallic surface offers a barrier to free electrons and is known as surface barriers

However if sufficient external energy is given to the free electron its kinetic energy is increased and thus electron will cross over the surface barrier to leave the metal. WORK FUNCTION OF THE METAL

This is the additional energy required by an electron to overcome the surface barrier of the metal.

This is the minimum energy required by an electron to just escape from the metal surface.

The work function depends on

i) Nature of metal

ii) Conditions of the metal surface

It is measured by a smaller unit of energy called electron volt. (eV) because this is the conventional unit of energy i.e. joule is very large for computations in atomic and nuclear physics.

Electron volt.

One electron volt is the amount of energy acquired by an electron when it is accelerated through a potential difference of IV

Since potential difference V

Work done = QV

For an electron

The electron volt is the kinetic energy gained by an electron being accelerated by a potential difference of one volt.

The work function of pure metal varies roughly from V to V as shown in table below

Metal

Work Function Wo (eV)

2.14

2.30

2.75

3.20

4.17

4.25

4.28

4.49

4.65

4.70

5.15

5.65

It is clear from the table above that the work function of platinum is the highest while it is lowest for Cesium.

It is desirable that metal used for electron emission should have low work function so that a small amount of energy is required to cause emission of electrons

PHOTOELECTRIC EFFECT

Photoelctric effect is the phenomenon of emission of electron from a metallic surface when radiation of suitable frequency falls on it or is the phenomenon where electromagnetic radiation of certain frequency when incident on certain material liberates electron from the surface of the material.

Photo emission

Photo emission is the emission of electron by electromagnetic radiation.

Photo electrons

These are emitted or rejected electrons from the surface of the cathode.

Photo electric effect is a general phenomenon exhibited by all substances but is most easily observed with metals.

When radiation of suitable frequency the threshold frequency is incident on a metallic surface electrons are emitted from the metal surface.

The threshold frequency is different for metals.

Certain alkali, metals e.g. sodium potassium, calcium show photo electric effect when visible light falls on them.

However, metal like zinc, calcium, magnesium etc show photo electric effect to ultra violet light.

Threshold Frequency

The threshold frequency is the minimum frequency of the incident radiation which is just sufficient to eject photo electron from surface of a metal Or is the minimum frequency of radiation below which no photo electron emission occurs.

It is denoted by

Illuminating a metal surface with light of frequency less than will not cause ejection of photo electrons, no matter now great is the intensity of radiation.

But illumination with a frequency greater than causes emission of photo electrons even if the radiation intensity is very small

Threshold wavelength

The threshold wavelength is the maximum wavelength of the incident radiation at which photo electric emission occurs.

It is the wavelength corresponding to threshold frequency

If the wavelength of the incident radiation is greater than threshold wavelength then there will be no photo electric emission.

Photoelectric current is the photo-electrons emitted per second

EXPERIMENTAL STUDY OF PHOTO ELECTRIC EFFECT

Figure below shows the experimental set up for studying the photoelectric effect.

The arrangement consists of an evacuated glass or quartz tube inclusive a photosensitive cathode C and metallic A.

A transparent window W is sealed onto the glass tube which can be covered with different filters to obtain the desired frequency.

The anode and cathode are connected to a battery through a potential divided by which potential difference between anode and cathode can be changed.

The reversing switch RS tends to make anode positive or negative with respect to cathode.

The P.D between anode and cathode is measured by the voltmeter V while photoelectric current is indicated by the micro ammeter.

1. 1.Effect of intensity of light on photo electric current

The anode A is maintained at positive potential with respect to cathode C and a radiation of suitable frequency (above threshold frequency) is incident on cathode C.

As a result photo electric current is set up.

Keeping the frequency of incident radiation and accelerating potential fixed the intensity of the incident radiation is changed in steps.

For each value of intensity of radiations the corresponding value of photo electron current is noted.

If we plot a graph between intensity of radiation and photoelectric current it is found to be a straight line passing through the origin O as shown in figure below

This shows that photoelectric current is directly proportional to the intensity of incident radiation

The intensity of radiation can be changed by changing the distance between cathode C and the source of radiation.

Effect of potential of anode with respect to cathode on photoelectric current

We keep the anode at some positive accelerating potential with respect to cathode C and illuminate the cathode with radiation of fixed frequency f above threshold frequency and fixed intensity I

If we increase the positive potential on anode gradually, it is found that photo electric current also increases a stage comes when the photo electric current becomes maximum.

If we increase the positive potential on anode further the photo electric current does not increase.

This maximum value of photo electric current is called saturation current and corresponds to the photo electrons emitted by the cathode reach the anode A

Now saturation current is of higher value as shown in figure below

This is expected because the greater the intensity of incident radiation the greater is the photo electric current

Stopping potential

Stopping potential is the minimum retarding potential at which photoelectric current becomes zero Or is the potential difference when no electrons are able to reach the anode.

It is also known as stopping voltage or cut-off potential.

Stopping potential is a measure of the maximum kinetic energy of the photo electrons.

Since potential difference V

For stopping Voltage Vo

For an electron

At V0, even the photo electrons having maximum kinetic energy K.E max (i.e. fastest photo electrons) cannot reach the anode A.

Therefore, the stopping potential V0 is a measure of the maximum kinetic energy K.E max of the photo electrons.

eV0 is the work done by the retarding force to stop the photo electron with maximum kinetic energy and is therefore equal to K.Emax.

At V0, it is found that the photoelectric current cannot be obtained even if we increase the intensity of radiation. It is same for different intensities I1, I2 and I3 of incident radiation.

3. Effect of frequency of incident radiation on stopping potential.

We now study the relation between the frequency f of the incident radiation and the stopping potential V0.

For this purpose, we take the radiations of different frequencies but of the same intensity.

For one frequency say f1, of the incident radiation, we plot the graph between photoelectric current and potential of anode A with respect to cathode C at a constant intensity of incident radiation

Keeping the intensity of incident radiation the same, we repeat the experiment for frequency f2 of the incident radiation.

The following is the resulting graph

Observations from the graphs

1. The value of stopping potential is different from radiation of different frequencies

2. The value of stopping potential is move in low higher frequency. This implies that the value of maximum kinetic energy depend on the frequency of incident radiation.

The greater the frequency of incident radiation, the greater is the kinetic energy of emitted photo electrons.

3. The value of saturation current depends on the intensity of incident radiation but is independent of the frequency of incident radiation.

If we draw a graph between the frequency of incident radiation (f) and the stopping potential (V0) at constant intensity of radiation, it will be a straight line AB as shown in figure below

From the graph

At fo, stopping potential V0 = 0. It means that at fo, the photo electric current is just zero (i.e. photo electrons and emitted with zero velocity) and there is no retarding potential.

V0 = 0

This limiting frequency fo is called threshold frequency for the cathode material.

It is a minimum frequency of the incident radiation which is just sufficient to eject photo electrons (i.e. with zero velocity) from the surface of a metal.

Stopping potential is directly proportional to the frequency of incident radiation. V0 α f

The greater the frequency of incident radiation, the higher is the stopping potential and vice versa.

Experiments show that photo electric emission is an instantaneous process.

As soon as light of suitable frequency (equal to or greater than fo) is incident on the surface of the metal, photo electrons are emitted from the metal surface.

The time delay is less than 10-9 second

LAWS OF PHOTOELECTRIC EMISSION

The above experimental study of photoelectric effect leads to the following laws of photoelectric emission.

i. For a given metal, there exists a certain minimum frequency of incident radiation below which no emission of photo-electrons takes place. This cut off frequency is called threshold frequency fo.

ii. For a given metal and frequency of incident radiation (>fo) the photo electric current is directly proportional to the intensity of incident radiation.

iii. Above the fo, the maximum kinetic energy of the emitted photo-electron is independent of the intensity of the incident radiation but depends only upon the frequency of the incident radiation.

iv. The photoelectric emission is an instantaneous process.

The above laws of photoelectric emission cannot be explained on the basis of light or radiation. This gave death blow to the wave theory of light or radiation.

FAILURE OF WAVE THEORY/CLASSICAL PHYSICS TO EXPLAIN PHOTOELECTRIC EFFECT

The wave theory of radiation failed to explain photoelectric effect. This will become clear from the following discussion.

I. According to wave theory of radiation the greater the intensity of the wave the greater the energy of the wave.

So wave theory does explain why the number of emitted photoelectrons increase as the intensity of radiation is increased.

But it fails to explain the experimentally observed fact that the velocity or kinetic energy of the emitted photoelectron is independent of the intensity of incident radiation.

According to the wave theory, an increasing in the intensity of radiation should increase the kinetic energy of the emitted photoelectrons but it is contrary to the experimentally observed fact.

II. According to wave theory, intensity of radiation is independent of it is frequency it depends upon the amplitude of electric field vector.

Therefore, an increase in the frequency of radiation should not affect the velocity or kinetic energy of the emitted electrons.

But it is observed experimentally that if the frequency of the incident radiation is increased, the kinetic energy of the emitted electrons also increases.

III. According to the wave theory, electrons should always be emitted from a metal by radiation of any frequency if the incident been is strong enough.

However experiments show that no matter how great is the intensity of the incident radiation; no electrons are emitted from the metallic surface if the frequency of radiation is less than a particular value i.e. threshold frequency.

IV. According to the wave theory the energy of radiation is spread continuously over the wave fronts of the radiation.

Therefore, a single electron in the metal will intercept only a small fraction of the wave’s energy.

Consequently considerable time should be needed for an electron to absorb enough energy from the wave to escape the metal surface. But experiment show that electron are emitted as soon as radiation of suitable frequency falls on the metallic surface.

In other words photoelectric emission is instantaneous there is no delay.

The above discussion is a convincing proof of the inability of the wave theory to explain the photoelectric effect.

EINSTEIN QUANTUM THEORY OF LIGHT

Einstein explained photoelectric effect on the basis of Planck’s quantum theory.

According to Einstein light radiation consist of tiny packets of energy called quanta.

Photon

Photon is the single quantum of light radiation which travels with the speed of light.

The energy of a photon is given by E. E=hf

where

f –frequency of light radiation

h – Plank’s constant.

Further, Einstein assumed that one photon of suitable frequency (=fo or >fo) can eject only one photoelectron from the metal surface.

He suggested that the energy of a single photon cannot be shared among the free electron in the metal.

Only one electron can absorb the energy of a single photon.

EINSTEIN’S PHOTOELECTRIC EQUATION

According to Einstein, one photoelectron is emitted from a metal surface if one photon of suitable frequency is incident on the metal.

Suppose a photon of suitable frequency f (< than the fo for the metal) is incident on the metal.

The energy hf of the photon is spent in two ways

I. A part of photon energy is used in liberating the least tight bound electron from the metal surface which is equal to the work function W0 of the metal.

II. The rest of the energy of photon appears as maximum kinetic energy of the emitted electron.

Einstein summarized this idea in what is called the Einstein photoelectric equation.

Photon = work function + maximum kinetic energy

The above equation is known as Einstein’s photoelectric equation

If the frequency of the incident radiation is fo then the emitted photoelectron will have zero velocity.

Maximum kinetic Energy of emitted photoelectrons is

If f < fo, then from above equation K.Emax is negative which is impossible therefore, photoelectrons emission cannot occur if the frequency of incident radiation is less than fo.

If f > fo, then equation above K.Emax α f. This means that max kinetic energy of photoelectrons depends only on the frequency (f) of the incident radiation. NUCLEAR PHYSICS

RADIOACTIVITY

This is the emission of radiations from heavily elements such as uranium whose nuclei are unstable.

Radiations emitted are called alpha , Beta (β) particles and gamma rays.

RATE OF DISINTEGRATION

The number of atoms of radioactive elements disintegrating per second is directly proportional to the number of atom present at that instant.

Where λ = decay constant

The negative sign (-) indicates that N decreases as time (t) increases.

If N0 is the number of atoms at time t= 0 and N is the number of atoms at time t. then:-

HALF LIFE

The half life time of a radioactive element is the times taken for the atoms disintegrate to half their initial number.

NECTA 1994/1/19

Draw a graph of

SOLUTION:

This is the graph of radioactive decay in time.

X-RAYS

X-rays are produced by bombarding a target of heavy metal with high energy electron.

NECTA 1984/2/8

The emission of X-rays may be regarded as the inverse of photo electric effect.Explain

SOLUTION

X-Rays which are waves are produced by bombarding a hard metal with electrons(particles)where as in photoelectric effect electron(particles) are liberated from ametal surface by incident radiation(waves)

MODERN X-RAY TUBE

Electron are obtained from the filament by thermionic emission and are accelerated to the anode having small target of high melting point such as tangstem.

X-RAY QUANTITY

Refers to the intensity of X-rays which increases with the number of electrons limiting the target.This depends on the cathode temperature controlled by the heating current

X-RAY QUALITY

Refer to the penetrating power of X-ray and determined by velocity whith which electrons strikes the target.

SOFT X-RAY: Are those which can penetrate soft objects such as flesh

HARD X-RAY Are those which can vibrate much more solid material

PROPERTIES OF X-RAY

1.They travel in a straight line.

2.They readily penetrate matter.

3.They affect photographic plates

4.They are not deflected by electric or magnetic field,because they have no charge

They are wave of wavelength 10A

X-RAY SPECTRA

CONTINUOUS SPECTRUM

Explained by electromagnetic theory.

It is due to electrical interaction between the bombarding electrons and the nuclei of the target atom

An electron approaching a positively charged nucleus is accelerated and according to electromagnetic theory accelerated motion is accompanied by emission of radiations in this case x-ray

DISCRETE SPECTRA

These are explained by quantum theory.They are produced when the incident electron interact with electron close to nucleus of the target atom.The bombarding electrons knock off electron in their orbits and the created gaps are filled by electrons in the high energy levels,when they moves in these gaps they produce x-ray photon.

The cut off wavelength or short wave limit(λ min)

This is the minimum wavelength of x-radiations which corresponds to the maximum energy of the X-rays produced by electrons which have given out all their K.Æ on a single encounter with the target nucleus

ENERGY OF X – RAY:

Energy of an electron striking the atom of the target is eV where e = electronic charge.

V = p.d across the X-ray tube.

If a direct collision is made with a target atom and the energy is absorbed then on quantum theory X-rays produced hence a maximum energy hV

Therefore

eV = 1/2mV2 = hV

eV = hc/ λ minimum

λmin = hc/eV

NECTA 1989/1/18

Calculate the wavelength of most energetic X-rays produced by a tube operating at 1.0 x 105V.

USES OF X – RAYS

1.in investigation of suspected bone fractures

2.Detection of lung tuber culosis (feased tissue is denser than a heatthy one).

3.Treatment of cancer (cancer cells are killed by X-rays)

4.Casting and welding joints can be inspected for internal imperfection.

5.It is used in the study of crystal structure refer to Bragg’s law.

NUCLEAR FISSION FUSSION

Nuclear fission is a splitting disintegrating of heavy nucleus such as uranium into two other lighter nuclei (with several neutrons)

Example:

Nuclear fusion

This is the combination of light nuclei to form a heavier nucleus.

Example:

NECTA 1990/2/8

(1)Both fission and fusion are source of energy

Explain.

SOLUTION:

Fission-During disintegration of heavy nucleus such as uranium, energy is released which is

transformed into K.E of fragments.

Fussion-When light nuclei are fused together a large amount of energy is released in term of heat.(eg

in the sun)

2T1

ELECTROMAGNETISM

This is the production of a magnetic field by current flowing in a conductor. The magnetic effect of current was discovered by Ousted in 1820. The verified magnetic effect of current by the following simple experiment. Figure below shows a conducting wire AB Above a magnetic needle parallel to it. So long as there is no current in the wire, the magnetic needle remains parallel to the wire i.e. there is no deflection in the magnetic needle. As soon as the current flows through the wire AB, the needle is deflected. Magnetic needle When the current in wire AB is Reversed the needle is deflected in the opposite direction This Deflection is a convincing proof of the existence of a magnetic field around a current carrying conductor. On increasing the current in the wire AB the deflection of the needle is increased and vice versa. This shows that magnetic field strength increases with the increase in current and vice versa It is clear from Worsted’s experiment that current carrying conductor produces a magnetic field around it. The larger the value of current in the conductor the stronger is the magnetic field and vice versa. Magnetic field Is the region around a magnet where magnet effect can be experienced. OR Is the space around a current carrying conducting (magnet) where magnetic effects can be experienced. The Direction of a field at a point is taken to be the direction in which a north magnetic pole would move more under the influence of field if it were placed at that point. The magnetic field is represented by magnetic lines of force which form closed loops. The magnetic field disappears as soon as the current is switched off or charges stop morning. Magnetic flux Φ is a measure of the number of magnetic field lines passing through the region. The unit of magnetic flux is the Weber (Wb) The flux through an area A on figure below the normal to which lies at angle ðœƒ to a field of flux density B Is a quantity which measures the strength of the magnetic field It is sometimes called magnetic It is a vector quantity The SI unit of Magnetic flux density is Tesla (T) or Wb/m2 Magnetic flux density is simply called magnetic field B B = θ/A FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD Consider a positive charge +Q moving in a uniform magnetic field with a velocity Let the Angle between and be θ as shown It has been found experimentally the magnetic field exerts a force F on the charge. The magnitude F of this force depends on the following factors (i) F α θ (ii)F α B (iii) Combining the factors we get Where K is a constant of proportionality The unit of B is so defined that K = 1 Equation (a) can be written in a vector form as:- F = the force of the particle (N) B = the magnitude of the magnetic flue density of the field T Q = the charge on the particle V= the magnitude of the velocity of the particle Definition of From F = BQVsinâ¡θ If V = 1, Q = 1, θ= 90 then F = Sin90 F=B Magnetic field ( ) at a point in space is equal to the force experienced by a unit charge moving with a unit velocity perpendicular to the direction of magnetic field at that point Right Hand Grip Rule

Grip the wire using the right hand with the thumb pointing in the direction of the current the other fingers unit point in the direction of the field.

- For an electron (negatively charged) entering the magnetic field as shown below

The Direction of positive charge will be exactly opposite. Applying Right hand Grip Rule it is clear that Direction of force on the electron will be vertically upward

For a positively charged particle, it will be vertically downward

Direction of magnetic field means from N –pole to S-pole.

SOME CASES OF MAGNETIC FORCE F

Consider an electric charge Q moving with a velocity V through a magnetic field B. then the magnetic force F on the charge is given by

F = BQV

(i) When = 0o or 1800

F = BQV

F= BQVSin00 or F = BQV

F= 0

Hence a charged particle moving parallel(or Anti parallel) to the direction of magnetic field experiences no force

(ii) When =900

F = BQV

F = BQV =1

F = BQV

Hence a force experienced by charged particle is maximum when it is moving perpendicular to the direction of magnetic field.

(iii) When V=O, the charge particle is at rest.

F = BQV

F=BQ(0)

∴F=O

If a charged particle is at rest in a magnetic field it experiences no force.

(iv) When Q = O

F = BQV F = 0

Hence electrically neutral particle (eg neutron) moving in a magnetic field experiences no force.

The magnetic force F acts perpendicular to velocity V (as well as B)

This means that a uniform magnetic field can neither speed up nor slow down a moving charged particle; it can charge only the Direction of V and not magnitude of V

Since the magnitude of V does not charge the magnetic force does not change the kinetic energy of the charged particle. UNITS AND DIMENSIONS OF

The SI unit of magnetic field B is Tesla

Now

F = BQV

If Q = 1C, V =1m/s, Q= 900 F= 1N

B = 1T

Hence the strength of magnetic field at a point is 1T if a charge of 1C when moving with a velocity of 1m/s at right angles to the magnetic field, experiences a force of 1N at that points.

Magnetic field of earth at surface is about 10 – 4T. On the other hand, strong electromagnets can produce magnetic fields of the order of 2T.

Dimensions of

Worked Examples 1. A proton is moving northwards with a velocity of m/s in a magnetic field of 0.1Tdirected eastwards. Find the force on the proton. Charge on proton = 1.6 x 10 -19C. Solution

F = BQV

B= 0.1T

V= m/s F=0.1 X 1.6 X 10-19 X 5 X 106X Sin 90

Q = 1.6 X 10-19C

= 900

2. An electron experiences the greatest force as it travel at 3.9 x105 m/s in a magnetic field when it is moving westward. The force is upward and is of magnitude N what is the magnitude and direction of the magnetic field. Solution

The conditions of the problem suggest that the electron is moving at right angles

To the direction of the magnetic field

F = BQV, F = 8.7 x 10 -13N

Q= 1.6 X10 -19C

V=3.9X105m/s

B = 13.14T

By right hand rule per cross product, the direction of the magnetic field is towards northward.

3. An α - particle of mass 6.65 x 10-27 kg is travelling at right angles to a magnetic field with a speed of 6x105m/s. The strength of the magnetic field is 0.2T.calculate the force on the - particle and its acceleration. Solution Force on α – particle F = BQV

M = 6.65 X10-27Kg

V = 6 x 105m/s

B = 0.2T

= 900

F = BQV

= (0.2 x 2x 1.6x10-19) x x Sin90Ëš

Acceleration of α – particle

F= mÉ‘

É‘= =

4. A copper wire has 1.0 x 1029 free electrons per cubic meter, a cross sectional area of 2mm2 and carries a current of 5A . The wire is placed at right angle to a uniform magnetic field of strength 0.15T. Calculate the force the acting on each electron.

Solution

I = neA

Drift velocity =

n= 1x1029m-3 e = 1.6x10-19c A= 2mm2 = 2x10-6m2

I = 5A

Force on each electron F= BQ Sin

Q= 1.6 x 10-19c

B= 0.15T

Q=900

BIOT –SAVART LAW

The Biot – Savart law states that the magnitude of magnetic flux density dB at a point P which is at a distance r from a very short length dl of a conductor carrying a current I is given by.

where is the Angle between the short length dl and the line joining it to point P

K is a constant of proportionality its value depends on the medium in which the conductor is situated and the system of units adopted.

For free space vacuum or air

This equation is known as Biot –Savart Law and gives the magnitude of the magnetic field at a point due to small current element

Current element

Is the product of current (I) and length of very small segment () of the current carrying conductor.

Current element =

Current element produces magnetic field just as a stationary charge produces an electric field the current element is a vector.

Its Direction is Tangent to the element and acts in the direction of current flow in the conductor

Biot -Savart law holds strictly per steady currents

Direction of B

The direction of is perpendicular to the plane containing and by right hand rule for the cross product the field is directed inward.

Special cases

(i) When = 00 or 1800

i.e Point P lies on the axis of the conductor

Hence there is no magnetic field at any point on the thin current carrying conductor minimum value.

(ii) When = 900

When point P lies at a perpendicular position w .r. t current element

Hence magnetic field due to a current element is maximum in a plane passing through the element and perpendicular to its axis. Important point about Biot – Savant law

(i) Biot – Savant law is valid per symmetrical current distributions. (ii) Biot – Savant law cannot be proved experimentally because it is not possible to have a current carrying conductor of length dl

(iii) Like coulomb’s law in electrostatics, Biot- Savant law also obeys inverse square law

(iv) The Direction of dB is perpendicular to the plane containing and

(v) This law is also called Laplace’s law and inverse square law’

BIOT – SAVART LAW VERSUS COULOMB’S LAW IN ELECTROSTATICS

According to coulomb’s law in electrostatics, the eclectic field due to a charge element dQ at a distance r is given by

According to Biot – Savart law the magnetic field due to a current element at a distance r is given by

From the above two equations we note the following points of Similarities and Dissimilarities.

Similarities

(i) Both laws obey inverse square law

(ii) Both the fields(magnetic field and Electro static field) obey superposition principles

(iii)Both the fields are long range fields.

Dissimilarities

(i) The Electric field is produced by a scalar source i.e. Electric charge. However the magnetic field is product by a vector source i.e. current

(ii) The Direction the Electric field is along the displacement vector i.e. The line joining the source and field point. However the direction of magnetic field is perpendicular to the plane containing current element and displacement vector

(iii) In Biot –Savant law the magnitude of magnetic field dB α Sin Where is the Angle between current element and displacement vector However there is no angle dependence in coulomb’s law for electrostatics

MAGNETIC FIELD AT THE CENTER OF CURRENT CARRYING CIRCULAR COIL

Consider a circular coil of radius r and carrying current I in the Direction shown in figure

Suppose the loop lies in the plane of paper it is desired to find the magnetic field at the centre O of the coil

Suppose the entire circular coil is divided into a large number of current elements each of length

According to Biot – Savant law, the magnetic field at the centre O of the coil due to current element is given by

……………

The direction of dB is perpendicular to the plane of the coil and is Directed inwards

Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the centre O can be found by integrating equation…………(i)

L- Total length of the coil = 2r

If the coil has N turns each carrying current in the same direction then contribution of all turn are added up.

MAGNETIC FIELD DUE TO INFINITELY LONG CONDUCTOR

The flux density dB at P due to the start length dl given by equation as

From the figure (A)

r =

= a cot

= -a

Substituting for and gives

The total flux density B at P is the sum of the flux densities of all the short lengths and can be found by letting d→O and integrating over the whole length of the conductor.

The limits of the integration are and 0 because these are values of ðœƒ at the ends of the conductor

FLUX DENSITY AT ANY POINT ON THE AXIS OF A PLANE CIRCULAR

Circular coil with its plane perpendicular to that of the paper

The flux density dB at p due to the short length dl of the coil at X, where X is in the plane of the paper, is given by equation as

By symmetry, when all the short lengths are taken into account the components of magnitude sum to zero.

Each short length produces a component of magnitude Sin α parallel to the axis and all those components are in the direction shown

The total flux density is therefore in the direction of Sin α and its magnitude B is given by

The radius vector XP of each small length is perpendicular to it, so that =900 and there pore Sin = 1

Since,

= 2(the circumference of the coil)

, But =

For a coil of N Turns

When S= r

Also from the figure

AMPERE’S CIRCUITAL LAW

States that the line integral of magnetic field around any closed path in vacuum/air is equal to times the total current (I) enclosed by that path I = current enclosed by that path.

Ampere’s law is an alternative to Biot - Savart law but it is useful for calculating magnetic field only in situations with considerable symmetry.

This law is true for steady currents only.

In order to use law it is necessary to choose a path for which it is possible to determine the value of the line integral

It is because there are many situations where there is no such path that the law is of limited use.

Hence the application of ampere law (i) Magnetic field due to constraining conductor carrying current (ii)Magnetic field due to solenoid carrying current (iii)Magnetic field due toroid

MAGNETIC FIELD DUE TO STRAIGHT CONDUCTOR CARRYING CURRENT

Consider a long straight conductor carrying current I in the direction as shown in the figure below It is desired to find the magnetic field at a point p at a perpendicular distance r for the conductors

Applying Ampere’s circuital law to this closed path

SOLENOID

Is a long coil of wire consisting of closely packed loops

Is a cylindrical coil having many numbers of turns

By long solenoid we mean that the length of the Solenoid is very large as compared to its Diameter.

Figure below shows the magnetic field lines due to an air cored solenoid carrying current

Inside the solenoid the magnetic field is uniform and parallel to the solenoid axis.

Outside solenoid the magnetic field is very small as compared to the field inside and may be assumed zero.

It is because the same no of field line that are concentrated inside the solenoid spread out into very faster space outside

Magnetic flux density due to an Axis of an in finely long Solenoid

Consider the magnetic flux density at P due to a section of the solenoid of length

n = number of turns per unit length.

N= number of turns the section can be treated as a plane circular coil of N turns in which case dB is given by

Since dx is small, the section can be treated as a plane circular coil or N turns in which case dB is given by

From the figure

Also

Substituting for and dx gives,

The flux densities at P due to every section of the Solenoid are all in the same direction and therefore the total flux density B can be found by letting dB→o and integrate over the whole length of the solenoid.

The limits of integration are and 0 because these values of β at the end of the solenoid.

If the Solenoid is Iron-cored of relatively permeability magnitude of magnetic field inside the Solenoid is

From

At points near the ends of an air cored Solenoid, the magnitude of magnetic field is

The magnetic field outside a solenoid is zero

Also in a current carrying long solenoid the magnetic field produced does not depend upon radius of the Solenoid. TOROID Toroid is a solenoid that bent into the form of the closed ring. The magnitude field B has a constant magnitude every where inside the toroid while it is zero in the open space interior and exterior to the toroid. If any closed path is inside the inner edge of the toroid then ther is no current enclosed. Therefore, by Ampere's circular law B=0. Magnetic field due to toroid Consider the diagram below Let r = mean radius of toroid I = Current through toroid n = number of turns permit length B = magnitude of magnetic field inside the toroid Then

FORCE ON A CURRENT CARRYING CONDUCTOR PLACED IN A MAGNETIC FIELD We know that a moving charge in a magnetic field experiences a force

Now electric current in a conductor is due to the drifting of the force electrons in a definite direction in the conductor

When such a current carrying conductor is placed in a uniform magnetic field, each free electron experiences a force.

Since the free electrons are constrained in the conductor, the conductor itself experiences a force.

Hence a current carrying conductor placed in magnetic field experiences a force F.

Consider a conductor of the length L and area of cross- section a placed at an angle ðœƒ to the direction of uniform of magnetic field B.

- is the angle between the plane of the conductor. The magnetic force experienced by the moving charge in a conductor is F = BQV Sin

For steady current I =

Q =I t

The velocity for direct current is constant

F = B I t

F= Force on the conductor (N)

B= Magnitude of the magnetic flux density of the field (T)

I = Current in the Conductor (A)

L= length of the conductor (M)

The current in the conductor I

Special cases

Thus if current carrying conductor is placed parallel to the direction of the magnetic field of the conductor will experience no force.

ii.)

F = BIL

Hence current carrying conductor will experience maximum force when it placed at right angles to the direction of the field.

One Tesla

Is the magnetic flux density of a field in which a force of IN acts on a 1M length of a conductor which is carrying a current of IA and is perpendicular to the field.

B = Tesla

The Direction of the force

Experiment shows that the force is always perpendicular to the plane which contains both the current and the external field at the site of the conductor

The direction of the force can be found by using Fleming’s left hand rule

Fleming’s left hand rule

States that if the first and the second fingers and the thumb of the left hand are placed comfortably at right angles to each other, with the first finger pointing in the direction of the current then thumb points in the direction of the force i.e. Direction in which Motion takes place If the conductor is free to move.

Maxwell’s Corkscrew rule

States that if a right handed corkscrew is turned so that its point travels along the direction, the direction of rotation of corkscrew gives the direction of the magnetic field.

FORCE BETWEEN TWO PARALLEL CONDUCTORS CARRYING CURRENTS

When two parallel current carrying conductors are close together, they exert force on each other.

It is because one current carrying conductor is placed in the magnetic field of the other

If currents are in the same direction the conductor attract each other and If currents are in the opposite directions conductors repel each other

Thus like currents attract, unlike currents repel.

Consider two infinitely long straight parallel conductors X and Y carrying currents I1 and I2 respectively in the same direction.

Suppose the conductors are separated by a distance rin the plane of the paper.

As each conductor is in the magnetic field produced by the other, therefore each conductor experiences a force

The current carrying conductor Y is placed in the magnetic field produced by conductor X

Therefore force act on the conductor Y. The magnitude of the magnetic field at any point P on the conductor Y due to current I, in the conductor X is

By right hand grip rule ; the direction of B is perpendicular to the place of the paper and is directed inwards.

Now conductor Y carrying current is placed in the magnetic field produced by conductor X

Therefore force per unit length of conductor Y will experience a force given by

According to FLHR, force on conductor Y acts in the place of the paper perpendicular to Y and is directed towards to the conductor X.

Similarly, the Force on conductor X per unit length is = ByI1L

But

Hence when two long parallel conductors carry currents in the same direction they attract each other. The force of attraction per unit length is

This shows that the attraction between two parallel straight conductors carrying currents in the same direction in terms of magnetic field lines of conductors

It is clear that in the space between X and Y the two fields are in opposition and hence they tend to cancel each other

However in the space outside X and Y the two fields assist each other. Hence resultant field distribution will be

If two straight current carrying conductors of unequal length are held parallel to each other then force on the long conductor is due to the magnetic field of the short conductor

I1 = Current through short conductor

l = Length of short conductor

I2 = Current through long conductor

L = Length of long conductor

If r is the separation distance between these parallel conductors

Force on Long conductor = force on short conductor

Force on each conductor is the same in magnitude but opposite in direction (Newton's third law)

DEFINITION OF AMPERE

Force between two current currying conductors per unit length

If And r =1m then

Ampere

Is that steady current which when it is flowing in each of two infinitely long, straight parallel conductors which have negligible areas of cross – section and are 1m apart in a vacuum, causes each conducts to exert a force of N on each mete of the other.

WORKED EXAMPLES

1. The plane of a circular coil is horizontal it has 20 turns each of 8cm radius A current of 1A flows through it which appears to be clockwise from a point vertically above it. Find the Magnitude of the magnetic field at the centre of the coil.

Solution The magnitude of the magnetic field at the centre of the coil carrying current is given by, As the currents appears to be clockwise from appoint vertically above the coil the direction of the field will be vertically downward (By R.H.G.R)

2. A wire placed along the South-North direction carries currents of 5A from South to North. Find the magnetic field due to a 1cm piece of wire at a point 200cm North-East from the place. Solution

By RHGR, The field is vertically vertical downwards

3. A coil of radius 10cm and having 20 turns carries a current of 12A in a clockwise direction when seen from east. The coil is in North – South plane. Find the magnetic field at the centre of the coil.

Solution The magnitude of the magnetic field at the centre of the coil

The electron of hydrogen atom moves along a circular path of radius 0.5 x 10-10 with the uniform speed of 4 x 106 m/s. Calculate the magnetic field produced by electron at the centre ( e= 1.6 x 10-9c)

Number the revolution made by the electron in 1 second is

Current =

I = 1.27 X 1016 X 1.6 X10-19

I = 2.04 X 10-3A

Magnetic field produced by the electron at the centre is

5. A circular coil of 100 turns has a radius of 10cm and carries a current of 5A Determine the magnetic field

(i) At the centre of the coil

(ii) At a point on the axis of the coil at a distance of 5cm from the centre of the coil. Solution

(i) Magnetic field at the centre of the coil is

= 4 x 10-7 TA -1

N = 100 turns

I = 5A

r = 10x10-2m

B = 4 x 10-7x 100 x S

2 X 0.1

B= 3.14 X10-3 T

The magnetic field of the centre of the coil = 3.14 X10-3 T

(ii) Magnetic field on the axis of the coil at a distance X from the centre is

= 4 x 10-7 TA -1

N = 100 turns

I = 5A

r = 10 x 10-2

x = 0.05m

6. An electric current I is flowing in a circular wire of radius at what dose from the centre on the axis of circular wire will the magnetic field be 1/8th of its value at the centre?

Solution Magnetic field B at the centre of the circular coil is Suppose at a distance X from the centre on the axis of the circular coil the magnetic field is

7. In Bohr’s model of hydrogen atom the electron circulates around nucleus on a path of radius 0.51Å at a frequency of 6.8x is rev/second calculate the magnetic field induction at the centre of the orbit. Solution

The circulating electron is equivalent to circular current loop carrying current I given by

I = 1.6

I = 1.1A

Magnetic field at the centre due to this current is

= 14T

8. A long straight wire carries a current of 50A. An electron moving at 107ms is 5cm from the wire

Find the Magnetic field acting on the electron velocity is directed

(i) Towards the wire

(ii) Parallel to the wire

(iii) Perpendicular to the directions defined by I and ii

Solution The magnetic field produced by current carrying long wire at a distance r

The field is directed downward perpendicular to the plane of the paper

( i) The velocity V1 is towards the wire. The angle between VI and B is 900 force on electron

F= BQV

F = 2x 10-4 x 1.6x10-19x107x Sin 900 F = 3.2 x 10-16 N

(ii) When the electron is moving is moving parallel to the wire ,angle between V2 and B is again 90Ëš Therefore, force is again 3.2x10-16N

(iii) When the electron is moving perpendicular to the directions defined by (i) and (ii) the angle between V and B is O

F = O

9. A solenoid has a length of 1 .23 m and inner diameter 4cm it has five layers of windings of 850 turns each and carries a current of 5.57A. what is the magnitude of the magnetic field at the centre of the solenoid

Solution The magnitude of the magnetic field at the centre of a solenoid is given by

But

10. A to void has a core ( non - ferromagnetic) of inner radius 20cm and over radius 25cm around which 1500 turns of a wire are wound. If current in the wire is 2A

Calculate the magnetic field

(i) Inside the to void

(ii) Outside the to void

Solution

( i) The magnitude of the magnetic field inside the toroid is given by 2 B = 0.003T

(ii)The magnetic field outside the toroid is Zero. It is all inside the toroid.

11. A solenoid 1.5m long and 4cm in diameter possess 10 turnscm. A current of 5A is flowing through it. Calculate the magnetic induction

(i) Inside and

(ii) At one end on the axis of the solenoid

Solution n = = =

(i) Inside the solenoid , the magnetic induction is given by

B =

B = 4

B =

(ii) At the end of the solenoid the magnetic induction is given by

12. (a) How will the magnetic field intensity at the centre of a circular loop carrying current change, if the current through the coil is doubled and the radius of the coil is halved?

(b) A long wire first bent in to a circular coil of one turn and then into a circular

coil of smaller radius having n turns, if the same current passes in both the cases, find the ratio of magnetic fields produced at the centers in the two cases.

If the ratio of their radii is 1:2 and ratio of flux densities at O due to A and B is 1:3, find the value of

Solution (a) Magnetic field at the centre of circular coil

( b)Suppose r is the radius of one turn coil and the r1 is the radius of n-turn coil. Then

First case Second case

Solution C. Magnetic field at the centre of circular coil

13. A helium nucleus makes a full rotation in a circle of radius 0.8m in two seconds. Find the value of magnetic field at the centre of the circle.

Solution

The charge on helium nucleus

Q= e

Q= 1.6 X10-19c

Current produced I =

I = 2 x 1.6 x 10-19

I =1.6 x10-19A Magnetic field at the centre of the circle orbit of the helium is,

B= 1.256 x 10-25T

14. A soft Iron ring has a mean diameter of 0.20m and an area of cross section 5x10-4m2 it is uniformly wound with 2000turns carrying a current of 2A and the magnetic flux in the iron is 8x 10-3Wb. What is the relative permeability of iron?

Solution

Length of ring l

l = 2

l = 2 x 0.10m

Number of turns per unit length n

n = =

If M is the absolute permeability of iron, then magnetic flux density of iron ring is

B =

Magnetic flux

Magnetic flux = BA

Relative permeability of Iron μr

15. Two flat circular coils are made of two identical wires each of length 20cm one coil has number of turns 4 and the other 2. If the some current flows though the wire in which will magnetic field at the centre will be greater?

Solution

For the first coil

For second coil

Therefore, magnetic field will be greater in coil with 4 turns

16. A plat circular coil of 120 turns has a radius of 18cm and carries currents of 3A. What is the magnitude of magnetic field at a point on the axis of the coil at a distance from the centre equal to the radius of the coil?

Solution Number of turns n = 120

Radius of the coil r = 0.18 m

Axial distance x = 0.18m

Current in coil I = 3A

B = (4 x 10-7) x 120 x3 x0.182

2(0.182 + 0.182) 3/2 B= 4.4 x 10-4T

17. A current of 5A is flowing upward in a long vertical wire. This wire is placed in a uniform northward magnetic field of 0.02T. How much force and in which direction will this field exert on 0.06 length of the wire?

Solution

B = 0.02T

I = 5A

L = 0.06

= 900 F= 0.02 X 5 X 0.06Sin900 F = 0.006N

By Fleming’s Left hand rules the force is directed towards West

18. A straight wire of mass 200g and length 1.5m carries a current of 2A. It is suspend in mind air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field? solution

M = 200 X 10-3 kg

I = 2A

l = 1.5m

B =? F=BIL

Mg = BIL

B = Mg = 200 x 10-3 x 9.8

IL 1.5 X 2

B = 0.65T

19. Two long horizontal wires are kept parallel at a distance of 0.2cm apart in a vertical plane . both the wires have equal currents in the same direction the lower wire has a mass of 0.05kg/m if the lower wire appears weightless what is the current in each wire ?

Solution

Let I amperes be the current in each wire the lower wire is acted upon by two forces.

Since the lower wire appears weightless the two forces were equal over 1m length of the wire 10-4I2 = 0.49

20. The horizontal component of the earth magnetic field at a certain place is 3 x 10-5 and the direction of the field is from the geographic south to the geographic North A very long straight conductor is carrying a steady current of 1A. what is the force per unit length on it when it is placed on a horizontal table and the direction of the current is

(a) East to West

(b) South to North

Solution

(a) When current is flowing from east to west 900

Force on the conductor per unit

(b) When current is flowing from south to north = 00

Force on the conductor per unit length

21. A horizontal straight wire 5cm long of mass 1.2gm-1 placed perpendicular to a uniform magnetic field of 0.6T if resistance of the wire is 3.85cm-1 calculate the P.d that has to be applied between the ends of the wire to make it just self supporting

Solution

The current (i) in the wire is to be in such a direction that magnetic force acts on it vertically upward. To make the wire self supporting its weight should be equal to the upward magnetic force.

Resistance of the wire

R = 0.05 x 3.8

= 0.19

Required P. (I) V = IR

V = 19.6X10-3 X 0.19

V = 3.7 X 10-3V

22. A conductor of length 2m carrying current of 2A is held parallel to an infinitely long conductor carrying current of 10A at a distance of 100mm. find the force on small conductor

Solution

II = 2A

I2 = 10A

r = 100 x 10-3m

l = 2m

Force on unit length of short conductor by the long conductor is give by

Force on length l = 2m of short conductor by the long conductor is

The force will be attractive if the direction of current is the same in two conduction and it will be repulsive if the conductors carry current in the opposite directions.

23. In the figure below, determine the position between two wire which experience zero resultant force due to charge Q placed at that point

Solution The force unit length acting in each wire of the parallel wire is given by

Let be the force per unit length in the wire carrying a current of 14A

Since F1 and F2 have the same magnitude but they are acting in opposite direction for resultant force to be zero

Assume that the charge Q is placed at a distance X from the wire carrying the

The charge Q is placed 4m from the either wire.

CLASSIFICATION OF MAGNETIC MATERIALS All substances are affected by magnetic field some attain weak magnetic properties and some acquire strong magnetic properties and some acquire strong magnetic properties.

The magnetic properties of the substances are explained on the basis of modern atomic theory.

The atoms that make up any substance contain electrons that orbit around the central nucleus.

Since the electrons are charged they constitute an electric current and therefore produce magnetic field .

Thus an atom behave as a magnetic dipole and possesses magnetic dipole moment.

The magnetic properties of a substance depend upon the magnetic moments of its atoms.

IMPORTANT TERMS USED IN MAGNETISM The following terms are used in describing the magnetic properties of the materials: (i) Magnetic flux density (B)

Is a measure of the number of magnetic field lines passing per unit area of the material.

The greater the number of magnetic field lines passing per unit are of the material

(ii) Magnetic permeability

Is a measure of its conductivity for magnetic field lines

The greater the permeability of the material the greater is its conductivity for the magnetic field line and vice versa

Since magnetic field strength B is the magnetic field lines passing per unit area of the material, it is a measure of magnetic permeability of the material.

Suppose magnetic flux density in air or vacuum is. If vacuum/air is replaced by a material, suppose the magnetic flux density in the material becomes B

Then ratio B/ called the relative permeability . of the material.

(i) Relative permeability. Is the ratio of magnetic flux density B in that material to the magnetic flux density that would be if the material were replaced by vacuum/ air .

Clearly is a pure number and its value per vacuum/air is 1

Relative permeability of a material may also be defined as the ratio of absolute permeability of the material to absolute permeability of vacuum/air.

(ii) Magnetizing force/ Magnetic intensity

Is the number of ampere - turns flowing per unit length of the toroid.

The SI Unit of magnetizing force H is Ampere – turns per meter (AT/m)

Consider a toroid with n turns per unit length carrying a current I. if the absolute permeability of toroid material is M, then magnetic flux density B in the material is

The quantity is called magnetizing force or magnetic intensity

Therefore, the ratio in a material I is from

; B=

Thus if the some magnetizing force is applied to two identical air cored and iron cored toroid, then magnetic flux density produced inside the toroid is

(iii) Intensity of magnetization () is the magnetic moment developed per unit volume of the material.

When a magnetic material is subjected to a magnetizing force , the material is magnetized

Intensity of magnetization is the measure of the extent to which the material is a magnetized and depends upon the nature of the material

where:

= magnetic moment developed in the material

V= volume of the material

If m is the pole strength developed,

is the area of X – section of the material and 2l is the magnetic length. Then

Hence Intensity of magnetization of a material may be defined as the pole strength developed per unit area of cross – section of the material.

Thus the SI unit of I is Am-1 which is the same as the SI unit of H

Magnetic susceptibility is the ratio of intensity of magnetic on I developed in the material to the applied magnetizing force H

The magnetic susceptibility of a material indicates how easily the material can be magnetized.

The unit of I is the same as that of H so that is a number

Since I is magnetic moment per volume is also called volume susceptibility of the material .

Consider a current carrying toroid having core material of relative permeability

The total magnetic flux density B in the material is given by

Where

= magnetic flux density due to current in the coils.

= magnetic flux density due to the material (Magnetization of the material) .....................(i)

.....................(ii)

Here I is the intensity of magnetization induced in the toroid material

B = +

Now,

Equation (iii) give the relation between relative permeability (μr ) and magnetic susceptibility (Xm).

CLASSIFICATION OF MAGNETIC MATERIALS

All materials or substances are affected by the external magnetic field. Some attain weak magnetic properties and acquire strong magnetic properties. On the basis of their behavior in external magnetic field , the various substance classified into the following three categories (i)Diamagnetic materials (ii)Paramagnetic materials (iii)Ferromagnetic materials

(i) DIAMAGNETIC MATERIAL When a diamagnetic substance is placed in a magnetic field in the magnetic field lines prefer to passs through the surrounding air rather than through the substance. Diamagnetic materials are materials which can not be affected by the magnetic field.

They are repelled by magnetic field e.g. lead, silver, copper, zinc, water, gold bismuth etc.

These substances when placed in a magnetic field are weakly magnetized in a direction opposite to that of the applied field. PROPERTIES OF DIAMAGNETIC MATERIALS

1. A diamagnetic substance is feebly repelled by a strong magnet.

2. The magnetic susceptibility () of a diamagnetic substance has a small negative value.

3. The relative permeability () of a diamagnetic substances is slightly less than 1

4. When a rod of diamagnetic substances is suspended freely in a uniform magnetic field, the rod comes to rest with its axis perpendicular to the direction of the applied field. See figure below

This gives the relation between relative permeability and magnetic susceptibility of the material.

(ii)PARAMAGNETIC MATERIALS

Are materials which when placed in a magnetic field are weakly magnetized in the direction of the applied field

The paramagnetic substances include the Aluminum antimony , copper sulphate, Crown grass etc

Since the weak induced magnetic field is in the direction of the applied field, the resultant magnetic field in the paramagnetic substance is slightly more than the external field

Hence the magnetic susceptibility of a paramagnetic substance is positive having

It clear that the relative permeability for such substances will be slightly more than 1 = 1 +

Paramagnetic substance loses its magnetism as soon as the external magnetic field is removed BEHAVIOR OF PARAMAGNETIC SUBSTANCES IN AN EXTERNAL MAGNETIC FIELD

When a paramagnetic substance is placed in an external magnetic field the dipoles are partially aligned in the direction of the applied field.

Therefore the substance is feebly magnetized in the direction of the applied magnetic field. This result into a weak attractive force on the substances.

In the absence of the external magnetic field the dipoles of the paramagnetic substances are randomly oriented and therefore the net magnetic moment of the substance is zero.

Hence the substance does not exhibit Para - magnetism PROPERTIES OF PARAMAGNETIC SUBSTANCES

1. The relative permeability of a paramagnetic substance is always more than 1

The result field B inside a paramagnetic substance is more than the external field Bo

2. The magnetic susceptibility of the paramagnetic substance has small positive value

It is because and

3. The magnetic susceptibility of a paramagnetic material varies inversely as the absolute temperature

Paramagnetism is quite sensitive to temperature. The lower the temperature the stronger is the paramagnetism and vice versa

4. A paramagnetic substance is feebly attracted by the strong magnet. It is because a paramagnetic substance develops weak magnetization in the direction of the applied external magnetic field

5. When a paramagnetic substance is placed in a magnetic field, the magnetic field lines of force prefer to pass through the substance rather than through air.

Therefore the resultant field B inside the substance is more than the external field Bo

FERROMAGNETIC MATERIALS

Are the materials which when placed in a magnetic field are strongly magnetic in the direction of the applied field.

Ferromagnetic substances includes

· Iron

· Cobalt

· Nickel

· Fe2O3

· Gadolinium

Since the strong induced magnetic field is in the direction of the applied magnetic field, the resultant magnetic field inside the ferromagnetic substance is very large compared to external field

It is clear that ferromagnetism is very stronger form of magnetism. When external field (magnetic field) is removed some ferromagnetic substances retain magnetism

PROPERTIES OF FERROMAGNETIC SUBSTANCES 1. The relative permeability () of the ferromagnetic substance is very large

Now

The resultant field B inside a ferromagnetic substance is very large as compared to the external filed Bo

2. The magnetic susceptibility () of a ferromagnetic substance is positive has a very high value.

It is because = 1 and 1 for this reason, ferromagnetic substance can be magnetized easily and strongly.

3. A ferromagnetic substance is strongly attracted by a magnet

4. When a rod of ferromagnetic substance is suspended in a uniform magnetic field, it quickly aligns itself in the direction of the field.

5. They retain their magnetization even when their magnetizing force is removed.

6. When a ferromagnetic substance is placed in a magnetic field the magnetic field lines tend to crowd into the substance

DOMAIN

Is the region of the space over which the magnetic dipole movements of the atoms are aligned in the same direction.

(i) In the absence of the external magnetic field the domain of the ferromagnetic materials are randomly oriented as shown below.

In other words, within the domain all the magnetic moments are aligned in the same direction but different domains are oriented randomly in different direction.

The result is that one domain cancels the effect of the other so that the net magnetic moment in the material is zero.

Therefore a ferromagnetic material does not exhibit magnetism in the normal state

(ii) When a ferromagnetic substance is placed in an external magnetic field a net magnetic moment develops the substance.

This can occur in two ways

(a) By displacement of boundaries of the domains i.e. the domains that already happen to be aligned with the applied field may grow in size whereas those oriented opposite to the external field reduce in size.

(b) By the rotation of the domains i.e. the domains may rotate so that their magnetic moments are more or less aligned in the direction of the magnetic field.

The result is that there is net magnetic moment in the material in the direction of the applied field.

Since the degree of alignment is very large even for a small external magnetic field the magnetic field produced in ferromagnetic material is often much greater than the external field.

CURIE TEMPERATURE Is the temperature at which the ferromagnetic substance becomes paramagnetic

It is also known as Curie point of the substance

Ferromagnetism decreases with the increases in temperature

When a ferromagnetic substance is heated magnetization decreases because random thermal motions tend to destroy the alignment of the domains

At sufficiently high temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic.

In a ferromagnetic substance the atom appear to be grouped magnetically into what are called domains.

This occurs because the magnetic dipole moments of atoms of a paramagnetic substance exert strong force on their neighbor so that over a small region of space the moments are aligned with each other even with no external field.

Above Curie temperature these forces disappear and ferromagnetic substances become paramagnetic. HYSTERESIS

Is the phenomenon of lagging of flux density (B) behind the magnetic force (H) in ferromagnetic materials subjected to cycles of magnetization.

When a ferromagnetic substance e.g. iron is subjected to cycle of magnetization (i.e. it is magnetized first in one direction and then in the other) it is found that flux density B in the materials lags behind the applied magnetizing force H.

This phenomenon is known as Hysteresis.

If a piece of ferromagnetic material is subjected to one cycle of magnetization the result B-H curve is a closed loop a b c d e f a called Hysteresis loop.

B always lags behind H, Thus at point b, H is zero but flux density B has a finite positive value ob similarly at point e, H is zero but flux density B has a finite negative value X .

HYSTERESIS LOOP

Consider an Iron cored toroid carrying current I

If N is the total number of turn and l the length of toroid, then magnetizing force is

The value of H can be changed by varying current in the coil

Consider that when the Iron cored toroid is subjected to a cycle of magnetization the resultant B- H curve traces a loop a b c d e f a called hysteresis loop

(i) To start with the toroid is unmagnetised and its situation is represented by point O in graph

As H is increased ( by increasing current I),B increases along and reaches its saturation value at a this stage

(i) all the domains are aligned

(ii) If now H is gradually reduced by decreasing current in the toroid it is found that curve follows the path instead of

At point b, H = O but flux density in the material has a finite value of +Br called residual flux density

REMANENCE

Is the flux density left behind in the sample after the removal of the magnetizing force (H). It is also called Residual magnetism or retentively.

B lags behind H. This effect is called Hysteresis

(iii) In order to reduce flux density in the material to zero, it is necessary to apply H in the reverse direction

This can be done by reversing the current in the toroid

When H is gradually increased in the reverse direction the curve follows the path

At point C, B =O and H = -HC, the value of H needed to wipe out residual magnetism is called coercive force it.

COERCITIVITY OF THE SAMPLE

Is the value of reverse magnetizing force required to wipe out the residual magnetism in the sample

Now H is further increased in the reversed direction until point d is reached where the sample is saturated in the revision direction ().

If the H is now reduced to zero, point e is reached and the sample again retain magnetic flux density ()

The remaining part of the loop is obtained by increasing current to produce H in the original direction .

The hysteresis loop results became the domains do not become completely unaligned when H is made zero.

The area enclosed by the hysteresis loop represents loss in energy

This energy appears in the material as heat.

HYSTERESIS LOSS

This is the loss of energy in the form of heat when a ferromagnetic material is subjected to cycles of magnetization.

Hysteresis loss is present in all those electrical machines whose iron parts are subjected to cycles of magnetization.

The obvious effect of hysteresis loss is the rise in temperature of the machine.

HYSTERESIS LOOP

Is the loop traced by the resultant B-H curve when the Iron – cored toroid is subjected to a cycle of magnetization.

The shape and size of hysteresis loop largely depends upon the nature of the material

The choice of a ferromagnetic material per a particular application often depends upon the shape and size of the hysteresis loop

(i) The smaller the hysteresis loop area of a ferromagnetic material the smaller is the hysteresis loss

The hysteresis loop per silicon steel has a very small area.

For this reason, silicon steel is widely used per making transformer cores and rotating machines which are subjected to rapid reversals of magnetization

(ii) The hysteresis loop per hard steel indicates that this material has high retentivity and Coercivity.

Therefore hard steel is quite suitable for making permanent magnets.

But due to the large area of the loop there is a greater hysteresis loss

For this reason, hard steel is not suitable for the construction of electrical machines

(iii)The hysteresis loop for wrought iron shows that this material has fairly good residual magnetism and Coercivity.

Hence it is suitable for marking cores of electromagnets

APPLICATIONS OF FERROMAGNETIC MATERIALS Ferromagnetic material (E.g. iron, steel nickel, cobalt etc) are widely used in a number of applications

The choice of ferromagnetic material for a particular for a particular application depends upon its magnetic properties such as

(i) Retentivity

(ii)Coercivity

(iii) Area of the hysteresis loop

Ferromagnetic materials are classified as being either

(i) Soft (soft iron)

(ii) Hard (steel)

Figure below shows the hysteresis loop for soft and hard ferromagnetic materials

The table below gives the magnetic properties of hard and soft ferromagnetic materials

Magnetic property

Soft

Hard

Hysteresis loop

narrow

Large area

Retentivity

High

Coercivity

low

high

(i) PERMANENT MAGNETS

The permanent magnets are made for hard ferromagnetic materials (steel, cobalt, carbon steel)

Since these materials have high relativity the magnet is quite strong

Due to their high Coercivity, they are unlikely to be demagnetized by stray magnetic field

(ii) TEMPORARY MAGNETS / ELECTROMAGNETIC The Electromagnets are made from soft ferromagnetic materials e.g. soft iron.

Since these materials have low coercively they can be easily demagnetized.

Due to high saturation flux density they make strong magnets

(iii) TRANSFORMER CORES

The transformer cores are made from soft ferromagnetic materials

When a transformer is in use, its core is taken through many cycles of magnetization

Energy is dissipated in the core in the form of heat during each cycle. The energy dissipated is known as hysteresis loss. And is proportional to the area of hysteresis loop

Since the soft Ferromagnetic materials have narrow hysteresis loop (smaller loop areas) they are used for making transformer cores.

WORKED EXAMPLE 1. (a) How does a permanent magnet attract an unmagnetised iron object?

(b) Show that the unit of magnetizing force is Nm-2T-I or Jm-Iwb-I

(d) An Iron ring has a cross - sectional area of 400and a mean diameter of 25cm. It is wound with 500trns. If the relative permeability of iron is 5000 find; (i) The magnetizing force

(ii) The magnetic flux density set up in the ring. The coil resistance is 474 and the supply voltage is 240V Solution

( a) The magnet’s field cause a slight alignment of the domains in the unmagnetised iron object so that the object becomes a temporary magnet with its north pole facing the south pole of the permanent magnet and viceversa. Therefore attraction results.

(b) The SI units of magnetizing (H) are Ampere/M(Am-I )

Now

Also

( c) The soft Iron has very small residual magnetism and coercive force. Therefore the material loses magnetism as soon as the magnetizing force is removed for this reason electromagnets are made of soft iron

(d) Current through the coil I

Mean length of the ring l

l = 2

l = 2 x (12.5 x 10-2)

l = 0.785m

(i) Magnetizing force H

(ii) Magnetic flux density B

From

B = 2.02wb/m2

2. (a) Diamagnetic is a property of a material. Discuss

(b) What is the magnetic susceptibility and permeability of a perfectly diamagnetic Substance?

(d) The core of a toroid having 3000 turns has inner and outer radii of 11cm and 12cm respectively. The magnetic flux density in the core per a current of 0.70A is 2.5T

What is the relative permeability of the core?

Solution

(a) Diamagnetism is a natural reaction to the applied magnetic field. Therefore it is present in all materials but is weaker even than paramagnetism. As the result, diamagnetism is overwhelmed by paramagnetic and ferromagnetic effects in materials that display these other forms of magnetism

Solution

B = H + I

B = (H+I)

For perfectly diamagnetic substance

B = O

(H+I) = O

H +I = O

I = -H

Susceptibility

Also

= 1 +

= 1 – 1

= O

(c) The induced magnetic moment in atoms of a diamagnetic substance is not affected by the thermal motion of the atoms. For this reason, diamagnetic is independent of temperature.

(d) Solution

Mean radius of toroid

The magnetic flux density in the toroid is

3 .(a) (i) What is a non Magnetic material?

(ii) Can there be a material which is non -magnetic?

(b) What do you mean by the greater susceptibility of a material?

Calculate its magnetic permeability. Given susceptibility of iron are 599.

(d) Which material is used to make permanent magnets and why?

Solution

(a) (i) A non –magnetic material is that which is not affected even by strong magnetic fields.

(ii) No, every material is at least diamagnetic.

(b) From

For a given H, α Thus the greater value of the susceptibility of the material the greater will be its intensity of magnetization i.e. more easily can be magnetized .Thus greater value of its susceptibility for iron means that it can be easily magnetized.

= 1 +

But

= (1+)

=4-7(1+599)

=7.54 x 10-4TA-4M

(d) Steel it is because has high coactivity. This ensures the stay of magnetism in steel for a longer period

4. (a) A toroid of mean circumference 50cm has 500 turns and carries a current of 0.15A

(i) Determine the magnetizing force and magnetic flux density if the toroid has an air core

(ii) Determine the magnetic flux density and intensity of magnetization if the core is filled with iron of relative permeability 5000

(b) Why do magnetic lines of force prefer to

Solution

(a) (i) given

L = 50cm = 50 x 10-2 m

N = 500 turns

I = 0.15A

Magnetizing force H

Magnetic flux density

(b) It is because permeability of Iron (Ferromagnetic material ) is very high as compared to that of air .

Therefore, have no Units

5. (i) Figure below shows the variation of intensity of magnetization (I) versus the applied magnetic field intensity (H) for two magnetic material A and B

(a) Identify the materials A and B

(b) For the material A, plot the variation of I with temperature.

(ii) The relative permeability of a material is

(i) 0.999

(ii) 1.001 Solution

The slope I-H graph gives the magnetic susceptibility of the material

For material A the slope is positive and has a small value. Therefore, material A is paramagnetic.

For material B, the slope is position and has a large value. Therefore , material B is ferromagnetic

(a) The intensity of magnetization of a paramagnetic material A is inversely proportional to the absolute temperature therefore I – T graph for material A will be as shown in figure below .

(ii) (i) Diamagnetic material

(ii) Paramagnetic material

6. (a) Graph below shows the variation of intensity of magnetization(I) versus the applied

Magnetic field intensity (H) for two magnetic material A and B

(i) identify the materials A and B

(ii) draw the variation of susceptibility with temperature for material B

(b)A magnetizing for of 360 produce a magnetic flux density of 0.6T in a ferromagnetic material. Calculate

(i) Permeability (ii) susceptibility of the material

Solutions

(a) for material A the susceptibility ( = slope of I – H graph) is small and positive therefore material A is paramagnetic and for material B, Susceptibility is small and Negative. Therefore , material B is Diamagnetic

iii) The susceptibility of a diamagnetic material B is independent of temperature therefore – T graph for material B will be shown in graph below.

(b) (i) Permeability of material

= 1.67 X 10-3 A -1Tm

(ii) Susceptibility of the material

= (1 +)

= 1328.62 Am -1

6. (a) What is Magnetic solution?

(b) Name two materials which have

(i) Position susceptibility

(ii) Negative susceptibility

(c) Obtain the earth’s magnetization , assuming that the earth’s field can be approximated by a giant bar magnet of magnetic moment 8.0 x 10 22 A? Radius of earth = 6400km

(d) A bar magnet has Coercivity of 4x103 A/m it is desired to demagnetize it by inserting it inside a solenoid 12cm long and having 60turns. With current should be sent through the solenoid

Solution

(a) Magnetic saturation

Is the maximum magnetization that can be obtained in the material when all the domains of a ferromagnetic material are in the direction of the applied magnetic filed

(b) (i) paramagnetic material e.g. Aluminum and Antimony

(ii) Diamagnetic materials e.g. Copper and Zinc

Magnetization of the earth is given by

The bar magnetic has a Coercivity of 4 i.e. it needs a magnetic intensity H = 4 to get magnetized

7. (a) Define hysteresis loop (b)What does the area of hysteresis loop indicate?

(d) Why is soft iron preferred in making the core of a transformer?

Solution

(a) hysteresis loop

Is the resulting B – H curve (Closed loop) obtained when a ferromagnetic material is subjected to one circle of the magnetization

(b) the area of hysteresis loop is a measure of energy wasted in a sample when it is taken through s complete of magnetization

(c) The hysteresis loop of a material tells us about hysteresis loss retentively and Coercivity. This knowledge helps us in selecting materials for making electromagnetic permanent magnets cores of transformer

(d) The area of hysteresis loop for soft iron is small. Therefore energy dissipated in the core for cycle magnetization is small. For this reason, the core of a transformer is made of soft iron

8. (a) state curie law

(b) Give the graph between I and B/T

(d) Copper Sulphate is paramagnetic with a susceptibility of 1.68x10-4 at 293K. What is the susceptibility of copper at 77.4K if it fellows curie law?

Solution

(a) Curie law

States that intensity of magnetization (I) if a paramagnetic substance is directly proportional to the external magnetic field (B) and inversely proportional to the absolute temperature(T) of the substance.

Combining these two factors, we have

where

C is a constant of proportionality and is called curie constant

This law is physically reasonable As B increase the alignment of magnetic moments increases and therefore I increases

If the temperature is increased the thermal motions will make alignment difficult thus decreasing I

The curve law is found to hold good so long as does not become too large

Since

(b) The temperature of molten iron 7700C is above the Curie temperature i.e. On malting the iron becomes paramagnetic. Therefore it loses its magnetism

Solution

According to curie law the susceptibility depends inversely on the temperature

10. (a) A solenoid 0.6m long is wound with 1800 turns of copper wire. An iron rod having a relative permeability of 500 is placed along the axis of the solenoid. What are the magnetic intensity H and field B when a current of 0.9A flows through the wire? What is the intensity of magnetization I in the iron? Find the average magnetic moment per iron atom. Density of iron is 7850 Kg/m3.

(b) An Iron sample having mass 8.4Kg is repeadly taken over cycle of magnetization at a frequency of 50cyles per second it is found that energy equal to 3.2 x J is dissipated as heat in the sample in 30Minutes if the density of the iron is 7200kg/m3 find the energy dissipated per unit volume per cycle in the iron sample.

(c) A domain in ferromagnetic iron is in the form of a cube of side length 1μM. Estimate the number of Iron atom in the domain and the maximum possibility dipole moment and magnetization of the domain and the maximum possible dipole moment and magnetization of the domain. The molecular mass of Iron is 55g/mole and its density is 7.9g/cm2 assume that each Iron atom has a dipole moment of 9.27 x 10-24 Am2

Solution

H = 2700A/m

I = 1.35 X 106A/m

One Kilo mole (55.85Kg) of Iron has 6.2 x1026 atoms. Therefore, number of atoms in 1 m3 of Iron

Average magnetic moments per iron atoms

=1.59 x 10-23 Am2

Solution

f =50 HZ

cycle

Length of cubic domain l

=1M = 10-6M

Volume of Domain

V = 3

V = (10-6)3

V = 10-18M3

Mass of domain

=Volume X Density

=10-12cm x7.9g

= 7.9 x10 -12g/cm

It is given that 55g of Iron contain 6.023 x1023 Iron atoms (Avogadro’s no)

Number of atoms in the domain is

N = 6.023 X 1023 X 7.9 X10-12

N = 8.65 X 1010atoms

The maximum possible dipole moment is achieved per the case when all the atomic domains are perfect aligned (This condition is unrealistic)

= (8.65 x1010) x (9.27 x10-24)

= 8 x 10-13 AM2

Maximum intensity of magnetization of the Domain is

= =

= 8 x105 Am-1

NUMERICAL PROBLEMS

1. (1)The magnetic moment of a magnet (10cm x 2cm x 1cm) is 1AM? What is the intensity of magnetization?

I = 5 X A/m

2. (2)An iron rod of cross sectional area 4 is placed with its length parallel to a magnetic flied of intensity 1600 A/M the flux through the rod is 4 x 10-4Wb what is the permeability of the material of the rod?

μ = 0.625 x10-3 Wb A -1 m-1

3. (3) A toroid winding carrying a current of 5A is wound with 300turns/miter of core. The core is Iron which has a magnetic permeability of 5000Mo under the given conditions

Find (i) the magnetic intensity H

(ii) Flux density B

(iii) Intensity of magnetization I

i) 1500AT/m

ii) 9.43T

iii) 7.5 X106A/m

4. (4)A specimen of Iron is uniformly magnetized by a magnetizing field of 500 A/m. if magnetic induction in the specimen is 0.2Wb/m2, find the relative permeability and susceptibility

Xm = 317.5

Mr = 318.5

5. Consider a toroid of 1000 turns and mean radius 25cm. what is the B- field in the toroid if there is a current of 2A?

What will be the field when the toroid is filled with Iron per which μ = 100H/m?

= 1 .6 x 10 -3

B = 0.16T

6. An Iron of volume 10-4m3 and relative permeability 1000 is placed inside a long solenoid wound with storms/cm. if a current of 0.5A is passed through the solenoid, find the magnetic moment of the rod.

M = 25Am2

7. The flux through a certain toroid clangs from 0.65m Wb to 0.91M Wb when Air core is replaced by another material. What are

i) The relative permeability

ii) Absolute permeability of the material

= 1. 4

μ= 5.6 x10-7H/m

8. Answer the following Questions

a) Why does a paramagnetic sample display greater magnetization (per the same magnetizing field) when cooled?

b) Why is diamagnetism, in contrast almost independent of temperature?

c) Distinguish between a soft and a hard magnetic material, giving an example of soft magnetic materials are those which can easily be magnetized but do not retain their magnetism (retentively )

An example of soft magnetic material

Is soft Iron i.e. Iron in a reasonably pure state. It is otherwise known as wrought iron

Hard magnetic material

Are those which are difficultly to magnetic but once magnetized, can retain the magnetism per long

These are usually used making permanent magnetic

An example of hand magnetic material is steel which consists of iron and a small % of carbon

MOTION OF CHARGED PARTICLE IN UNFORM MAGNETIC FILED

Consider a charged particle of charge +Q and mass M moving with a velocity V in the plane of the paper.

Suppose this charged particle enters a uniform magnetic filed B which is perpendicular to the plane of the paper and directed outward

Clearly the entry of the charged particle is at right angles to the magnetic field

The force i.e. magnetic force Fm on the charged particle is given by

Fm = BQV

The magnetic force Fm acts at right angle to the plane containing V and B

On entering the magnetic field at M the charged particle experiences a force of magnitude and is deflected in the direction shown

This force is at right angle to the direction of motion of the charge particle and therefore, cannot change the speed of charge particle it only charge its direction of motion

A moment later, then the particle reaches point N the magnitude of force Fm acting on it is the same as it was at M but the direction of force is different (Fm is still perpendicular to V )

Thus the force is perpendicular to the direction of motion of the charged particle at all times and has a constant magnitude

The magnetic force does not change the speed or kinetic energy of the charge particle it only charges the direction of the charged particle

When the moving charged particle is inside the uniform magnetic field, it moves along a circular path.

When the initial velocity of the particle is parallel to the magnetic field

= 00

From

Fm = BQV

Fm = 0

Thus in this case the magnetic field does not exert any force on the charge particle

Therefore the charged particle will continue to move parallel to the magnetic field then = 1800

Therefore, the particle will continue to the move in the original direction.

When the initial velocity of the particle is perpendicular to the magnetic field =900

From

Fm = BQV Max. Value Fm = BQV

PARAMETERS OF MOTION

A force of constant magnitude always acts perpendicular to the direction of motion of the charged particle.

Therefore , provides the necessary centripetal force to more the charged particle in a circular path in the circle of radius r perpendicular to the field

i) RADIUS OF PATH

The acceleration of a particle moving along a circular path of radius r is given by

For a given charge mass and magnetic field r V. this means that fast particles move in large circles and slow ones in small circles.

ii) TIME PERIOD

The time taken by the charged particle to complete one circular revolution in the magnetic field is its Time period T

From

Thus Time period of the charged particle is independent of the speed (V) and the radius of the path

It only Depends on the magnitude of B and charge to mass ratio of the particle .

FREQUENCY

The number of circular revolutions made by the charged particle in one second is its frequency f

f =

f = 1

There Frequency of the charged particle is also independent of speed (V) and radius (r) of the path

ANGULAR FREQUENCY

From

=2πf

But

Then

Again Angular frequency of the charged particle is independent of the speed (V) and radius (r) of the path..

Since T, f and of a charged particle moving in a magnetic field are independent of its speed (V) and the radius (r) of the path.

In fact all the charged particles with same Q/M and moving in a uniform magnetic field B will have the same value of T, f and w

MOTION OF CHARGED PARTICLE ENTERING UNIFORM MAGNETIC FIELD AT AN ANGLE

Suppose the charged particle moving with velocity V enters a uniform magnetic field B making an angle to the direction of the field

Diagram

The velocity V can be resolved into two rectangular components

i) V1 = VActing in the direction of the field

ii) V2 = Vsin acting perpendicular to the direction

The perpendicular component V2 moves the charged particle in a circular party while the horizontal componentmoves it in the direction of the magnetic field

In other words, the charged particle corers circular path as well as linear path. Consequently the charged particle will follow a helical path.

The charged particle rotates in a circle at speed V2 while moving in the direction of the field with a speed VI

PARAMETERS OF MOTION The perpendicular component of velocity V2 determines the parameters of the circular motions while the horizontal component of velocity VI decides the pitch of helix

(i) Radius of path

From

(ii) T, f and w

Since time period (T), frequency (f) and Angular frequency(ω) of a charged particle moving in a uniform magnetic field are independent of speed V and radius (r) of the path, these values remain the same

(iii) Pitch of helix (d) It is the linear distance covered by charged particle when it completes one circular revolution Or It is the linear distance covered by charged particle during time T

d = T

d = VT.

d = V

The following points may be noted about the behavior of charged particle in a Uniform magnetic field

(i) If a charged particle is at rest V=0 in a magnetic field, it experiences no force

From

= BQV

= BQ = 0

(ii) If a moving charged particle enters a uniform magnetic field at right angles to the field it describes a circular path

(iii) If a moving charged particle enters a uniform magnetic field. Making an angle to the direction of the field it describes a helical path

(iv) A moving charged particle in a magnetic field experience maximum force when angle between V and B is 900

(v) Since magnetic force does not change the speed of a charged particle it means that K.E of the charged particle remains constant in the magnetic field.

(vi) Since magnetic force (Fm) is perpendicular to V, it does not work. Therefore work done by the magnetic force on the charged particle is zero

WORKED EXAMPLE

1. An electron and a proton moving in a circular path at 3x106 Ms-1 in a uniform magnetic field of magnitude 2 x 10-4T. Find the radius of the path

Solution

2. An electron and a proton moving with the same speed enter the same magnetic field region at right angles to the direct of the field. For which of the two particles will the radius of circular path be smaller?

Solution

From

Since the mass of electron is less than that of the proton the radius of the circular path of electron will be smaller.

3. (a) What will be the path of a charged particle moving along the direction of a uniform magnetic field?

(b) A moving charged particle enters a magnetic

Solution

(a) When a charged particle moves along the direction of a uniform magnetic field ,it experiences no force = 00 therefore the charged particle will more along its original straight path

(b) Helical path since the velocity of the charged particle can be resolved into two rectangular components one along the field, and the other perpendicular to the field. The velocity component perpendicular to the field causes the charged particle to more in a circular path while the velocity component along the field cause it to more it in the direction of the field. The combination of these two motions course the charged particle to move in a helical path.

4. A become of α-particles and of proton of the same velocity V, entries a uniform magnetic field at right angles to the field lines. The particle describes circular paths. What is the ratio of radii of the two paths?

Solution

Radius of α-particle path

Of proton path

= ……………………….

Take equation (i) equation (ii)

Therefore, radius of x - particle path is twice that of proton’s path

Note

α – particle

Charge = 2e

Mass = mass of helium nucleus

5. A proton with charge – mass ratio of 108 CKg -1 is moving in a circular orbit in a uniform magnetic field of 0.5T. calculate the frequency of revolution

f = BE

2πM

F = B. e

2π M

F = 0.5 . 108

2π

6. (a) What happens when a charged particle is projected perpendicular to a uniform magnetic field ?

(b) A beam of protons moving with a velocity of 4 x 105 M5-1 enters a uniform field of 0.3T at an angle of 600 to the direction of a magnetic field find

(i) The radius of helical path of proton beam

(ii) Pitch of helix

Given mass of proton = 1.67 x 10 -27 Kg charge on proton = 1.6 x 10 -19C

Solution

(a) When a charged particle is projected perpendicular to a uniform magnetic field

(i) Its path is circular in plane perpendicular to B and V

(ii) Its speed and Kinetic energy remain the same

(iii) The magnitude of force remains the same Fm = BQV. Only the direction of velocity of the particle charges.

(iv) The force acting on the particle is independent of the radius of the circular path

(v) The time period of revolution of the particle is independent if V and r

(b) (i) Solution

r = 1.2 x10-2M

(ii) Pitch of helix of d

d = 4.37 x10-2M

7. (a) A particle of charge ðœƒ moves in a circular path of radius r in a uniform is P = BQr

(b) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0KV enters a region of magnetic field of 0.15 determine the trajectory of the electron if the field

(i) is traverse to its initial velocity

(ii) makes an angle of 30 with the initial velocity

c) A proton a deuteron and an α – particle whose kinetic energies are same enter perpendicularly to a uniform magnetic field. Compare the radii of their circular paths

(a) Solution

The magnetic force Fm provides the necessary centripetal force Fc

Fm = Fc

BQ =

Momentum P = BQr

(b) Solution

When an electron (e) is accelerated through a p.d of V, it acquires energy eV. IF Vis the velocity gained by electron, then

V =

V=9X10-31

(i) Force on the electron due to transverse field is

Fm = BQr ðœƒ = 900

Since magnitude of Fm is constant and Fm is perpendicular to both V and B the electron will move in a circle of radius r. The necessary centripetal force is provided by Fm

r = 10-3M

(ii) When electrons enters the magnetic field making an angle ðœƒ = 300 with the field

r = 0.5 x10-3M

Let 1, 2 and 3 be the suffix force proton, deuteron and α - particle respectively

K .E, = K.E2 = K.E3

1 MIV1 = 1 M2V2= 1 M3V3

2 2 2

If MI then M2 = 2M and M3 = 4M

MV2I = 2MV22 = 4MV23

VI = 2 = 2V3

V2 = VI

Also

Radius of the path r

r = MV

If QI = Q, Then Q2 = Q and Q3 = 2Q

rI = MIVI = MVI

BQI BQ

r2 = M2V2 = 2M. V1

BQ2 BQ

rI: r2 : r3 = MVI : MVI : MVI

BQ BQ BQ

rI : r2 : r3 = 1 : : 1

NUMERICAL PROBLEMS 1. What is the radius of the path of an electron (mass 9 x 10-31)Kg and charge(1.6 x 10-16C) moving at a speed of 3 x107m/s in a magnitude field of 6 x 10-4T perpendicular to it? What its frequency? Calculate its energy in KeV(1Ev=1.6x10-19 J)

r= 0.28m

f= 1.7 x10-7 Hz

E = 2.53KeV

2. An electron after being accelerated through a p.d of 100V enters a uniform magnetic field of 0.004T perpendicular to its direction of motion. Find the radius of the path described by the electron

r = 8.4 10-3m

3. An α-particle is describing a circle of radius of 0.45m in a field of magnetic Induction 1.2Wb/m2. Find its speed, frequency of rotation and kinetic energy. What potential difference will be required which will accelerate the particle so as to this much energy to it? The mass of α-particle is 6.8 x10-27Kg and its charge is 3.2 x 10-19 C

V=2.6 X107 m/s f= 9.2 x106 sec-1

MAGNETIC TORQUE ON RECTANGULAR COIL IN UNIFORM FIELD Consider a rectangular conductor ABCD having length L and N-turns carrying a current I and placed in a magnetic field between N and S-pole of bar magnets.

Since the velocities of the electrons in the sides side AB and BC are perpendicular to the magnetic induction B then these sides will experience the maximum magnetic forces equal to F.

F= BIL

The direction of the magnetic force in these two sides is given by Fleming’s Left Hand Rule as shown in the figure below.

The two parallel and equal forces will constitute the turning of the rectangular called the magnetic torque.

Mathematically

The magnetic torque or couple is given by

= Total force x Perpendicular distance

τ=BIL x b

τ=BI (L x b)

Lb = cross-sectional area=A

For the rectangular coil of N- turns

ELECTROMAGNETIC MOMENT(M) This is the magnetic torque acting on the coil when it is parallel to a uniform field whose flux density is one tesla.It is the property of the coil is defined as the couple required to hold the coil at right angles to field.

Thus, in equation τ=m when B=1I from

Consider a rectangular coil ABCD placed at an angle to the magnetic field of flux density B

The perpendicular distance between the parallel force is and not b. Then the magnetic force or compile is given

Special cases

(i) When

Plane of the loop is parallel to the direction of magnetic field

Thus, the Torque on a current loop is maximum when the plane of the loop is parallel to the direction of magnetic field is given by

(ii) when ðœƒ = 900

Plane of the loop is perpendicular to the direction of magnetic field

Thus , the torque on a current loop is minimum (zero) when the plane of the loop is perpendicular

(iii) When B = 1T

when B = 1T, then the magnetic torque is numerical equal to magnetic moment

From

B = 1T

MAGNETIC TORQUE AT AN ANGLE α BETWEEN THE AXIS OF THE COIL AND NORMAL TO THE PLANE OF THE COIL

We can also express torque in another useful form. If normal to the plane of coil makes an angle α with the direction of the magnetic field

From

τ = BANI cos (90 – α)

τ = ANIBSin α

Since M = IAN

When a current carrying coil is placed in a uniform magnetic field, torque acts on it which tends to rotate the coil so that the plane of the coil is perpendicular to the direction of magnetic field.

WORK DONE BY TORQUE If the magnetic torque displaces the coil through the small angular displacement

The work done by the torque is given by

The total work done is obtained by integration the above equation within limits

If the magnetic field displaces the coil through small angular displacement the work done by torque is given by

dw = τdα

dw = mbSinαdα

WORKED EXAMPLES 1. A vertical rectangular coil of sides 5cm by 2cm has 10 turns and carries a current of 2A. Calculate the torque on the coil when it is placed in a uniform horizontal magnetic field of 0.1T with its plane.

(a) Parallel to the field

(b) Perpendicular to the field

Solution

The area of the coil

A= (5 x 10 -2) x (2 x10-2)

A= 10-3 m2

(a) From

A=10-3

I = 2A

N=10

B=0.1T

τ= 0.1 X10-3 X10 X2

τ = 2 X10-3NM

(b)

τ=BANI

τ= 0

(c)

τ = BANICos ðœƒ

τ= 0.1 x10-3 x10x2xCos 600

τ= 10 -3NM

2. 2.Given a uniform magnetic field of 100T in East to West direction and a 44cm long wire with a current carrying capacity of at most 10A. what is the shape and orientation of the loop made of this wire which yields maximum turning effect on the loop?

Solution

A current carrying planar loop will experience maximum together if its area to the direction of the magnetic field for a given perimeter , a circle has the maximum area.

If 44cm wire is bent into a circular

2πr = 44

πr = 22

22r = 22

r = 1

R = 7cm

Area of loop = πr2

= πx72

= 154cm2 = 154x10-4m2

Magnetic toque τ

τ = BANI

τ = (154 X10-4) X 100 X100

τ = 150 T

3. A circular coil of wire of 50 turns and radius 0.05 carries current of 1A the wire is suspended vertically in a uniform magnetic field of 1.5T. the direction of magnetic field is parallel to the plane of the coil

(a) Calculate the Toque on the coil

(b) Would your answer charged if the circular coil is replaced by a plane coil of some irregular shape that has the same area (all other particulars are unaltered? )

(a) Solution

B = 1.5T

A = πr2 = π (0.05) M2

A = 7.85 X 10-3

N = 50

I = 1A

τ = 1.5 X (7.85 X10-3) X50X1

τ = 0.589NM

(b) Since torque on the loop is independent of its shape provide area (A) remains the same the magnitude of the torque will remain unaltered.

4. A circular coil of 20turns and radius 10 cm is placed in a uniform magnetic field of 0.2T normal to the coil. If current in the coil is 5A find.

(i) Total torque on the coil

(ii) Total force on the coil

(iii) Average force on each electron in the coil due to the magnetic field. The coil is made of copper wire of cross-sectional area 10-5m2 and force of electron density in the wire is 1029m-3

Solution

(i) The toque on the coil is given by

Since

τ = 0

(ii) The net force on a planar current loop in a uniform magnetic field is always zero

(iii) Magnetic force on each electron

F = BeVd

F = Be.

F =

F= 10-24N

MOVING COIL METERS

A galvanometer deflects or measures small amount of current passing through it and it gives the direction to which that current is flowing.

In these instruments a rectangular of fine insulated copper wire is suspended in an strong magnetic field as shown in the figure below. The field is set up between soft iron poler pierces Ns attacked to a powerful permanent magnet.

(i) Millimeter

The magnet field is radial to the core and pole pieces over the region which the coil can swing. In this case the deflected coil always comes to rest with the plane parallel to the field in which it then situated.

The moving coil galvanometer ha s has hair spring and jewel bearings. The coil is around in the rigid but light aluminium frame which also comes to carries a pivot. The current is led in and out of the springs.

Aluminum pointer P shows the deflection of coil, it is balanced by counter weight Q.

THEORY OF MOVING COIL GALVANOMETERS

The rectangular coil is situated in the radia field B when the current is passed into it the coil rotate to an angle Q which depend on the length of the spring.

No matter where the coil comes to rest, the field B in which it is situated always along the plane of the coil because the field is radial.

It is shown that the torque T of the coil is given by T = B A N I. ie.

In equilibrium the deflecting torque T is equal to the opposing torque due to the elastic forces in the springs. The opposing torque

Torque = CQ

Where C is the constant of the spring.

Example 1

A galvanometer coil has a coil has 100 turns which each turn having an area of 2.5cm2. If the coil is in the radial field of 2.0 10-9Nm per degree what current is needed to give a deflection of 600?

Example 2

A moving coil galvanometer with a coil of 15 turns and an area of 0.02m2 is suspended by a torsion wire which has restoring constant of 9.00 10-6Nm per degree of twist if the current of â„¦MA is passed through the coil whose plane is parallel to a uniform magnetic field of 0.03T. What will be the deflection of the coil?

CURRENT SENSITIVITY.

Therefore the greater the sensitivity is obtained with a stronger field B atom value of c that is a week springs and greater value of N and the value of A. However the size and number of turns of a mound increase the resistance of the meter which is not desirable

VOLTAGE SENSITIVITY

If the resistance of a moving coil meter is R the p.d.v across its terminals where a current I flows it. It is given by V = IR--------------- (1)

But we have BAIN = C ------------ (2)

Hence the voltage sensitivity depends on the resistance R of the meter.

Unlike the current sensitivity.

CONVERTING A MILLIAMETER TO AN AMMETER.

Moving coil meters give full scale deflection for current smaller than those generally met in laboratory. In order to measure the current of the order of an ampere or more we connect a low resistance s called the shunt across. The terminals of moving coil meter

The shunt turns most of the current to be measured I away from the coil.

Suppose the coil of meter has a resistance r of 20â„¦ and full deflected by the current of 5MA. If we want to convert it so that its full scale deflection is 5A; Then the shunt s must be connected which will extra current that is (5- 0.005) A or 4.995A.

Potential difference across the shunt p.d across the coil ie.

Example

A milli ammeter has a full scale reading of 0.8MA and resistance of 75 ohms. What is the value of a single resistor which mould a current it into an ameter capable of reading 15amps at full scale

SOLUTION:

CONVERTING A MILLIAMETER TO VOLTIAMETER

Suppose we have a moving coil meter which requires 5MA for full scale deflection and also let suppose that the resistance of its coil r is 20 â„¦

When this milliameter is full deflected the p.d across it is given by u = rI

= (20â„¦) (510-3)

=0.1u

If the coil resistance is constant the instrument can be used as a voltimeter giving a full scale deflection for p.d of 0.1 or 100mv so this milliameter can possess two scales for current and for voltage as shown in (1) above.

The p.d to be measured in the laboratory is usually greater than 100mv. Therefore to measure such a p.d we put resistor R in series of with the coil as shown in fig (2) above:

For example if we wish to measure to measure up 10v is applied between terminals CD then the scale current of 5MA flows through the moving coil. That is

V = (R + r) I

10 =(R + 20) 5 10-3

R =2000-20

=1980â„¦

The resistance r is called a multiplier.

Many voltmeters contain a series of multiplier contains a series of multiplier of different resistances which can be chosen by a switch…. and socket arrangement shown in fig (3) above

Example

A voltmeter whose range is 0-200v has resistance R of 1500 per volt.(fsd) what resistance should be converted in series with it to give a range of 0-2000v.

Solution

Given

But when the resistance in series they have the same current through it

New solution:

Total resistance for this voltmeter is

=1500 200 =30 10-4â„¦

Total resistance for the new scale is

=15002000 = 30 105 â„¦

Extra resistance required is

(30-3) 105 =27 105 â„¦

MAGNITUDES FOR CURRENT CARRYING CONDUCTORS

Laws of Biot and Savant.

It state that the flux density dB= at point P due to a small element dl of a conductor carrying

Where r is the distance from the point P to the element is the angle formed it to P.

B: FLUX DENSITY = INDUCTION OR MAGNETIC INDUCTION

MAGNETIC FIELD

Equation (1) can be written as

dB=KIdlsinx

Where K is the constant of proportionality and it depends on the medium in which the

B AT THE CENTRE OF A NARROW CIRCULAR COIL

Suppose the coil is in air has a radius of r carries a steady current I and it is considered to consist of current element of length dl. Each element is at the distance r from the centre o and it is at the

Example

A coil of wire with 15 turns of radius 6.0cm, has a current of 3.5A flowing through it. What is the magnetic flux density at the center of the coil?

Example

What is the magnitude of the flux density produced the center of a coil of radius 5cm carrying current of 4A in air.

Example

A circular coil of radius 6cm consisting of 5 turns carries a current supplied from 2v accumulator of negligible internal resistance. If the coil has a total resistance of 2â„¦. Calculate the magnetic field induced at the centre

B DUE TO A LONG STRAIGHT WIRE AT A DISTANCE d SIDE THE WIRE

Consider a very long wire YN carrying a current I. Take P to be a point outside the wire but also this point is considered to be very near to this wire.

THE HALL EFFECT Is the phenomenon where by e .m. f or voltage is set up transversely or across a current carrying conductor when a perpendicular magnetic field is applied

Consider a piece of conducting material in a magnetic field of flux density B

Suppose that the field is directed (perpendicularly) into the paper and that there is a current flowing from right to left.If the material is a metal the current is carried by electrons moving from left to right

Consider the situation of one of these electrons and suppose that it has a velocity V

The electron feels a force F which by Fleming’s left hand rule, is directed downwards .

Thus in addition to the electron flow from left to right electrons are urged away from face Y and towards face X. Anegative charged builds up on X, leaving a positive charged on Y so that a potential difference is established between X and Y. The buildup of charge continues until the potential difference becomes so large that it prevents any further increase . This maximum, potential difference is called the Hall voltage

Hall voltage

Is the potential difference created across a current carrying metal strip when the strip is placed in a magnetic field perpendicular to the current flow in the strip.

Actually, the magnetic field does not have to be totally perpendicular to the strip the magnetic field only needs to have a component that is perpendicular

The flow ceases when the e .m .f reaches a particular VH called Hall voltage

MAGNITUDE HALL VOLTAGE

Suppose VH is the magnitude of the Hall voltage and d is the width of the slab (the separation of x and y). Then the Electric field strength E set up across the slab is numerical equal to the potential gradient.

E =

let Fv be the force exerted on an electron by the P.d between X and Y. Therefore when the buildup of charged on X and Y has ceased

F = Fv

BeV = eE

BV = E

BV= VH=BVd.......................(i)

Where

E = The strength of the uniform electric field between X and Y due to the Hall voltage

VH = Hall voltage

d = The separation of X and Y

HALL VOLTAGE Is the potential difference created across a current carrying metal strip when the strip is placed in a magnetic field perpendicular to the current flow in the strip

Actually the magnetic field does not have to be totally perpendicular to the strip the magnetic field only needs to have a component that is perpendicular The flow ceases when the e .m. f reaches a particular value VH called Hall voltage. It has been shown that the current I in a material is given by I = neAV

Where

n = the number of electron per unit value

e = the charge on each electrons

v = the drift velocity of the electrons

A = the cross – sectional area of the material

V = ----------------- (ii)

Sub equation (ii) into equation (i)

From

In figure A = dt and therefore

ELECTROMAGNETIC INDUCTION An electric current create magnetic field, the reverse effects of producing electricity by magnetism was discovered by Faraday and is called electromagnetic induction Induced can be generated in two ways (a) By relative moment ( The generator effect ) if the bar magnet is moved in and out of a stationary galvanometer or small current is recorded during the motion but not at other time movement of the coil towards or away from the stationary magnet has the same results (figure above) relative motion between the magnet and coil is necessary, the direction of the of induced current depends on the direction of the relative motion. And magnitude of current is produced increase with i. the speed of the motion ii. The number of turns in the coil iii. The strength of the magnet used (b) By changing a magnetic field (transformer effects) In this case two coils are arranged one inside the other (Figure below) to galvanometer

Rheostat the other called secondary is connected to galvanometer, switching the current on or off in the primary causes impulse of and current to be induced in the secondary. Varying the primary current quietly altering the value of rheostat has the same effect. Electromagnetic induction thus they occurs only when there is only change in primary current and also in magnetic fields it induces LAWS OF ELECTROMAGNETIC INDUCTION While the magnitude of the induced EMF is given by Faraday law. Its direction can be predicted by Lenz's Law LENZ'S LAW The direction of induced is such that it tends to oppose the flux change which causing it and does oppose it if induced current flows Faraday or Newman’s laws The induced is directly proportional to the rate of change of the flux through the the coil. If E = induced then NOTE I). The minus sign express Lenz's Law II). NÉ¸ is the flux linkage in the coil INDUCED EMF IN A MOVING ROD

Area swept in 1 second

AB is a wire which can be moved by a force F in a contact with a smooth metal rails PQ and RS. A magnetic field of flux density B acts downwards perpendicular to the plane of the system. As the wire AB cuts the flux density the is produced by the current I and is in opposition to the motion

Therefore

F= BIL ..............................................................i Where l is the distance between two rails And I = .....................................................................ii Where is the resistance of the wire If the wire is moving with a speed V then F' = F ...................iii

F’ = ……………………………4

Power = = = Force x velocity

= ……………………… 5

Also power = = = …………….6

Equating equation 5 and 6

I.e. E = BLV (This is the induced in a moving coil)

INDUCED EMF IN A ROTATING COIL

Consider a coil of an area A and its normal makes an angle of with the magnetic field BY

The flux linkage with the coil of n turns is expressed as

N = ………………………………………………………1

The induced emf is given by

E = = - = = E = since

If the maximum value of emf is denoted by o

Then

E = Eo sinwt where Eo = NABw

A gain w =

Therefore

2. Eo =

Exercise 1

The magnetic flux QB through the loop perpendicular to the plane of the coil and directed into the paper as shown in the diagram is varying according to the equation QB = 8t2 +5t +5 where QB is measured in millimebers and t in seconds

i. What is the magnitude of induced in the loop when

ii. What is the direction of the current through R?

Solution

E =

E = 16t + 5

E = 53Mv

Exercise 2

What is the maximum induced in a coil of 500turns, each with an area of , which makes 50reflections per second in a uniform magnetic field of flux density 0.04T?

Solution

B = 0.04T

2.5Volts

INDUCED EMF IN ROTATING DISC – DYNAMO

Consider a copper disc which rotates between poles of magnets. Connections are made to its circle and the circumference. An induced emf is obtained between the Centre of the disc and one edge. We assume that magnetic field is uniform over the radius xy

The radius continuously cuts the magnetic flux between the poles of the magnet. For this straight conductor, the velocity at the end of x is zero and that at the other end y where w is the angular velocity of the disk

Average velocity of is

An induced in straight conductor is given by

In this case

…………………………………………………….i)

Since ………………………………………...ii)

If the disc has the radius r1 and an axle at the Centre of radius r2 the area swept out by a rotating radius of the metal disc is - = - in this case the induced would be

-f

The direction of the E is given by Fleming’s right hand rule

As the disc rotates clockwise the radius moves to the left at the same time as the radius moves to right

If the magnetic field covers the whole disk, induced in the two radii would be in opposite direction. So the resultant emf between yz would be zero. The emf between the Centre and the rim of the disc is the maximum which can be obtained

Qn.

A circular metal disc with a radius of 10cm rotates at 10revolutions per seconds. If the disc is in a uniform magnetic field of 0.02T at a right angle to the plane of the disc. What will be the induced between the Centre and the rim of the disc?

Solution

B = 0.02T

SELF INDUCTANCE (L)

An induced emf appear in the coil if the current in that coil is changed is called self-induction and produced is called self-induced

For a given coil produced no magnetic materials nearly the flux linkage proportional to the current I

Where L is a constant proportionality which is called self-inductance of a coil

From Faraday’s law in such a coil the induced

Substitute i) in ii)

Hence the unit of inductance. A special name the Henry has been given to this combination of units

Two coils A and B have 200 and 800turns respectively. A current 2Amperes in A produces a magnetic flux of in each turn of A, compute:

i. Mutual inductance

ii. Magnetic flux through A when there is a current of 4.0 Ampere in B

iii. The induced when the current in A changes 3A to 1A in 0.2seconds

SELF INDUCTANCE (L) FOR THE COIL

The induced,

(By integrating the equation we have)

Therefore

The self-inductance may be defined as the flux linkage per unit current, when is in wabers and I is in amperes then L is in henry:

Magnetic flux density for a long coil is given by with an iron core with a relative permeability of

The flux density is given by since

Thus the flux linkage

(Unit for L is Henry)

ENERGY STORED IN AN INDUCTOR

Because of of the self-induction that act when the current in the coil change, electrical energy must be supplied in setting up the current against the .

If L is the self-inductance of the inductor then the back across it is given by

…………………………i)

Hence rate at which work is done against the backward emf.

Power = EI…………………………ii)

Substitute equation i) into ii)

Then equation ii) becomes

The work done to bring the current from zero to a steady state value Io is

Therefore

MUTUAL INDUCTANCE (M)

The may be induced by in one circuit by changing current in another. This phenomenon is often called mutual induction and the pairs of circuits which shows it are said to have mutual inductance

The mutual inductance m between the two circuits is defined by the following equation

Induced in B by changing = M (rate of change of current in A) i.e.

The unit of mutual inductance is Henry the same as that of self-inductance

MUTUAL INDUCTION

Since the rate of change in flux in B then

QUANTITY OF ELECTRICITY INDUCED

Consider a close circuit of total resistance R Ohms which has a total flux linkage with magnetic field B. if the flux linkage starts to change

Induced, but current

Flux linkage will not change at a steady rate and a current will not be constant. But throughout it changes. Its charge is being carried round the circuit. If a time t seconds is taken to reach a new constant value the charge carried round the circuit in that time is

Where is the number of linkage at t=o and is the number of linkage time t

Thus

AC THEORY

When a battery is connected to a circuit the current flows steadily in one direction, this is called a Direct current (d.c). The use of Direct currents is limited to a few applications e.g charging of batteries, electroplating etc. Most of electrical energy is generated and used in the form of alternating current due to many reasons including,

i) Alternating voltages can be changed in value very easily by means of transformers.

ii) A.c motors are simpler in construction and cheaper than d.c motors.

ALTERNATING VOLTAGE AND CURRENT

i) Alternating voltage

An alternating voltage is one whose magnitude changes with time and direction reverses periodically. The instantaneous value (i.e value at any time t) of an alternating voltage is given by,

where,

E = Value of the Alternating voltage at time t

E0 = Maximum value of the Alternating voltage

ω = Angular frequency of supply From ,where f is the frequency of the alternating voltage, If T is the time period of alternating voltage then

The voltage varies from zero to a positive peak (+E0) then back via zero to negative peak (-E0) and so on. In time period T, the wave completely cycle.

ii) (ii)Alternating current

This is one whose magnitude changes with time and direction reverses periodically.

The Instantaneous value ( value at any time t) of sinusoidally varying alternating current is given by

where

I = value of alternating current at time t

I0 = maximum value (Amplitude) of alternating current.

ω = Angular frequency of supply.

Figure below shows the waveform of alternating current.

Current varies sinusoidally with time. The current increases gradually from zero to a positive peak (+Io), then back via zero to a negative peak (-Io) and so on.

In time period T, the wave completes a cycle.

An Alternating voltage and current also be represented as a cosine function of time.

Both these representations give the same result as is given by the one containing sine functions.

MEASUREMENT OF ALTERNATING CURRENT

Since the average value of sinusoidal alternating current is zero, an ordinary (DC) ammeter or galvanometer will not show any deflection when connected in an AC circuit.

Due to inertia it will not be possible for the needle to oscillate with the frequency of the current.

Therefore to measure AC we use hot wire instruments because the heating effect of current is independent of the direction of current.

MEAN OR AVERAGE VALUE OF ALTERNATING CURRENT

The mean value or average value of alternating current over one complete circle is zero.

It is because the area of positive half cycle is exactly equal to the area of the negative half cycle.

However, we can find the average or mean value of alternating current over any half cycle.

Half Cycle Average Value of a.c This is that value of steady current (d.c) which would send the same amount of charge through a circuit for half the time period of a.c as it sent by the a.c through the same circuit in the same time.It is represented by Im or The instantaneous value of alternating current is given by, Suppose current I remains constant for a small time. Then small amount of charge sent by alternating current in a small time is given by

If Q is the total charge sent by the positive half cycle of a.c (

If the Im is the half cycle average means values of positive half cycle of a.c then by definition

From equation (i) and (ii)

Hence half cycle average value of a.c is 0.637 times the peak value of a.c For positive ½ cycle Im +0.637 I0

For negative half cycle Im -0.637 I0

Obviously, average value of a.c over a complete cycle is zero.

MEAN OR AVERAGE VALUE OF ALTERNATIVE VOLTAGE Half Cycle Average value of alternating

This is that value of steady, (which would send the same amount charge through a circuit for half the time period of alternating as is sent by the alternating through the same circuit in the same time.It is denoted by The instantaneous value of alternating e. m. f is given by Suppose this alternating is applied to a circuit of resistance R. Then by ohm’s law the instantaneous value of alternating current is

If this current remains constant a small time, then small amount of charge send by alternating in small time is given by

If Q is the total charge sent by positive half cycle of a alternating then

If is the half cycle average or mean value of the positive half cycle of alternating then by definition

From equation (i) and equation (ii)

Therefore, half cycle average value of alternating is 0.637 times the peak value of alternating . For positive half cycle

For negative half cycle

A d.c voltmeter or ammeter reads average (or d.c) value. Therefore, they can be used to measure alternating voltage on current.

It is because the average value of alternating voltage or current over a complete cycle is zero.

We use a.c meters to measure alternating voltage/current.

ROOT MEAN SQUARE VALUE OF ALTERNATING CURRENT

The average value cannot be used to specify an alternating current (or voltage).

It is because its value is zero over one cycle and cannot be used for power calculations.

Therefore we must search for more suitable criterion to measure the effectiveness of an alternating current or voltage.

The obvious choice would be to measure it in terms of direct current that would do work (or produce heat) at the same average rate as a.c under similar conditions.

This equivalent direct current is called the root mean square () or effective value of alternating current.

Effective or value of Alternating Current

The root mean square (r.m.s) of alternating current is that steady current (d.c) which when flowing through a given resistance for given time produces the same amount of heat as produced by the alternating current when flowing through the same resistance for the same time.

- It is also called virtual value of a.c

- It is denoted by called

For example, when we say that or effective value of an alternating current is 5A, it means that the alternating current will do the work (or produce heat) at the same rate as 5A direct current under similar conditions.

RELATION BETWEEN R.M.S. VALUE AND PEAK VALUE OF A.C

Let the alternating current be represented by

If this alternating current flows through a resistance R for a small time, then small amount of heat produced is given.

In one complete cycle for time 0 to T of alternating current the total amount of heat produced in R is given by

If Irms is the virtual or value of the alternating current, then heat produced in R in the same time (0 to T) is given by

From equation (i) and (ii), we have

Hence the value of effective value or virtual value of alternating current is 0.707 times the peak value of alternating current.

The value is the same whether calculated for half cycle or full cycle.

Alternative Method Let the alternating current be represented by

If this current is passed through a resistance R, then power delivered at any instant is given by

Because the current is squared, power is always positive since the value of varies

between 0 and 1

Average power delivered

If is the virtual or value of the alternating current then by definition

Power delivery P

From equation (I) and (ii)

ROOT MEAN SQUARE VALUE OF ALTERNATING E.M.F

The root mean square (r.m.s) value of alternating voltage is that steady voltage (d.c voltage when applied to a given resistance for a given time produces the same amount of heat as is produced by the alternative when applied to the same resistance for the same time.

It is also called virtual value of alternating e.m.f and is donated by of or The instantaneous value of alternating is given by

When this alternating voltage is applied to a resistance R, small amount of heat produced in a small time is

But

Then,

In one complete cycle for time 0 to T the total amount of heat produced in resistance R is

If Erms is the value of the alternating, then heat produced in the same resistance R for the same time T is

From the equation (i) and equation (ii)

Hence the value of alternating is 0.707 times the peak value of the alternating

Therefore, value is the same whether calculated for half cycle or full cycle.

Importance of r.m.s values

An alternating voltage or current always specified in terms of values.

Thus an alternating current of 10A is the one which has the same heat effect as 10A d.c under similar conditions. The following points may be noted carefully.

i) The domestic a.c supply is 230V, 50H. It is the or effectively value. It means that the alternative voltage available has the same effect as 230 V d.c under similar conditions.

The equation of this alternative voltage is

ii) When we say that alternating current in a circuit is 5A, we specifying the value.

- It means that the alternating current flowing in the circuit has the same heating effect as 5A d.c under similar conditions.

iii) A.C ammeters and voltmeters record values of current and voltage respectively.

The alternating voltage/current can be measured by utilizing the heating effect of electric current.

Such Instruments are called hot wire instruments and measure the value of the voltage/current since r.m.s value is the same for half cycle or complete cycle.

WORKED EXAMPLES

1. An a.c main supply is given to be 220V what is the average during a positive half cycle?

Solution

Average during positive half is given by 2. An alternating current I is given by

Find

(i) The maximum value

(ii) Frequency

(iii) Time period

(iv) The instantaneous value when t= 3ms

Solution

Comparing the given equation of the alternating current with the standard form

(i) Maximum value,

(ii) Frequency f

f = 50 HZ

(iii) Time period T

(iv)

When

3. Calculate the value of the current shown in figure below

Solution

4. An a.c voltmeter records 50V when connected across the terminals of sinusoidal power source with frequency 50Hz. Write down the equation for the instantaneous voltage provided by the source.

Solution

An a.c voltmeter records values

Now

Here

5. An alternating voltage of 50 Hz has maximum value of 200 volts. At what time measured from a positive maximum value will the instantaneous voltage be 141.4 volts.

Solution

This equation is valid when time is measured from the instant the voltage is zeroi.e point O. Since the time is measured from the positive maximum value at point A.

The above equation is modified to

Let the value of voltage become 141.4 volts t second after passing then the maximum positive value.

6. An alternating current of frequency 60 Hz has a maximum value of 120 A.

i) Write down the equation for the instantaneous value.

ii) Recording time from the instant the current is zero and becoming positive. Find the instantaneous value after 1/360 second. iii) Time taken to reach 96 A for the first time.

Solution

i) The instantaneous value of alternating current is given by

ii) Since point 0 has been taken as the reference the current equation is

When

iii) To reach the current 96A for the first time

7. Find the r.m.s value of the current after current I shown in figure below

Solution

Consider the current variation over one complete cycle.

A.C CIRCUIT

An A.C circuit is the closed path followed by alternating current. When a sinusoidal alternating voltage is applied in a circuit, the resulting alternating current is also sinusoidal and has the same frequency as that of applied voltage.

However there is generally a phase difference between the applied voltage and the resulting current.

As we shall see, this phase difference is introduced due to the presence of inductance (L) and capacitance (C) in circuit.

While discussing A.C circuits our main points of interest are;

(i) Phase difference between the applied voltage and circuit current (ii) Phasor diagram. It is the diagram representation of the phase difference between the applied voltage and the result circuit current. (iii) Wave diagram (iv) Power consumed.

A.C CIRCUIT CONTAINING RESISTANCE ONLY

When an alternating voltage is applied across a pure resistance, then from electrons current flow in one direction for the first half cycle of the supply and then flow in the opposite direction during the next half cycle, thus constitute alternate current in the circuit.

Consider a pure resistor of resistance R connected across an alternating source of

Suppose the instantaneous value of the alternating is given by

If I is the circuit current at that instant, then by ohm’s law

The value of I will be maximum Io when

Therefore equation (ii) becomes

From

Since sin t = 1 then I = Io

1) Phase Angle

It is clear from equation (i) and (iii) the applied and circuit current are in phase with each other they pass through their zero values at the same instant and attain their peak value both positive and negative peaks at the same instant.

This is indicated in the wave diagram shown in figure below.

The diagram shown in figure below also reveals that current is in phase with the applied voltage.

Hence in an a.c circuit, current through R is in phase with voltage across R.

This means that current in R varies in step with voltage across R. If voltage across R is maximum current in R is also maximum, if voltage across R is zero, current in R is also zero and so on.

2) Power Absorbed

In a.c circuit, voltage and current vary from instant to instant. Therefore power at any instant is equal to the product of voltage and current at that instant.

Instantaneous power P

Since power varies from instant to instant the average power over a complete cycle is to be considered.

This is found by integrating equation...........(iv) with respect to time for 1 cycle and dividing by the time of 1 cycle. The time per one cycle is T. Average power P

Therefore, average power absorbed by a resistor in an a.c circuit is equal to the product of virtual voltage ( Erms) across it and virtual current ( Irms) through it.

Obviously, this power is supplied by the source of alternating .

Since

WORKED EXAMPLES 1. An a.c circuit consists of a pure resistance of 10Ω and is connected across an a.c supply of 230V, 50 Hz.

Calculate

i) Circuit current

ii) Power dissipated and

iii) Equations for voltage and current

Solution

Erms = 230V R = 10Ω f = 50HZ

i) Circuit current

ii) Power dissipated P

= 230 x 23

∴ P = 5290 W

iii)Equations for voltage and current

= x 230 = 325. 27V

= 2 x 23

∴ = 32.52A

ω = 2

= 2

ω = 314

∴ The equations of voltage and current

E = 325.27 sin 314t and I = 3.52 sin 314t

2. In a pure resistive circuit, the instantaneous voltage and current are given

E = 250 sin 314t

I = 10 sin 314t

Determine

i) Peak power

ii) Average power

Solution

In a pure resistive circuit

i) Peak power =

= 250 x 10

∴ Peak power = 2500W

ii) Average power P

P =

∴ P = 1250 W

3. Calculate the resistance and peak current in a 1000 W hair dryer connected to 120V, 60Hz supply. What happens if it is connected to 240V line?

Solution

Peak current Io

Resistance of hair dryer R When connected to 240V line, the average power delivered would be

This would undoubtedly melt the heating element or the coils of the motor.

4. A voltage E = 60sin 314t is applied across a 20 Ω resistor. What will;

i) An a.c ammeter

ii) Ordinary moving coil ammeter in series with resistor read?

Solution

i) E = 60 sin 314t

An a.c ammeter will read the r.m.s value.

= =

Therefore a.c a meter will read 2.12A

ii) An ordinary moving coil ammeter will read average value of alternating current. Since the average value of a.c over one cycle is zero, this meter will record zero reading.

5. What is the peak value of an alternating current which produces three times the heat per second as a direct current of 2A in a resistor R?

Solution

Heat per second by 2A

= R

H = R

Three times heat per second

3 = 12R

If Iv is the r .m .s value of the a.c heat per second in R

12R = R

= 12

Peak value is given by,

∴ Peak value = 24 = 4.9A

6. An a.c voltage of 4V peak (maximum) is connected to a 100Ω resistor R

a) What is the phase of the current and voltage?

b) Calculate the current in R in mA

c) What is the power in R in mW

Solution

a) The current and voltage are in phase

b) Current is R

c) Power in R

P = R

P = 0.0282 x 100

P = 0.078W

P = 78mW

A.C CIRCUIT CONTAINING INDUCTANCE

When alternating voltage is applied across a pure inductive coil a back (E) is induced in the coil due to its self inductance.

E = -

The negative sign indicates that induced e. m .f opposes the change in current.

In order to maintain the flow of current the applied voltage must be equal and opposite to induced voltage at every instant.

Consider a pure inductor of inductance L connected across an alternating source of

E = Eosin ωt

Suppose the instantaneous value of the alternating e. m. f is given by

Esin wt ………. ( ) If I is the current in the circuit and is the rate of change of current at that instant, then e. m. f induced in L is given by

As applied voltage is equal and opposite to induce e .m .f at every instant

From, Equation (i) and Equation (ii)

Integrating both sides we get

The value of I will be maximum Io when, Sin = 1

Then,

Substituting the value of = in equation (iii)

I = sin (wt –

……………….. (iv)

I) Phase Angle

It is clear from equation (i ) and (iv) that circuit current lags behind the applied voltage by (/2) radians or 900.

This fact is also indicated in the wave diagram.

The phase diagram in figure below also reveals the fact that lags behind by 90º Hence in an a.c circuit current through L lags behind the voltage across L by 90º

This means that when voltage across L is zero, current through L is maximum and vice versa.

From

E = L

Now, is maximum when circuit current is zero and is zero when circuit current is maximum. INDUCTIVE REACTANCE

Inductive reactance is the opposition in which an inductor offers to current flow. It is denoted by XL

Inductance not only causes the current to lag behind the voltage but it also limits the magnitude of current in the circuit.

We have seen above that,

Io =

ωL =

Clearly the opposition of inductance to current flow is ωL. This quantity ωL is called inductive reactance XL of the inductor.

Inductive reactance XL

a) From XL =

But = and Io =

Then, XL =

b) For d.c

f = 0

so that,

XL = 2

Therefore a pure inductance offers zero opposition to d.c c) XL = 2

Therefore,the greater f,the greater is XL and vice versa.

d) We can show that the units of XL are that of ohm

XL = ωL = x Henry = and

XL =

III) Average Power consumed

E = sin ωt

I = sin (ωt- I = - cos ωt

Instantaneous power P

P = EI

P = (sin ωt) (-cos ωt)

P = - sin ωt cos ωt

P = Sin 2ωt

Average power P is equal to average of power over one cycle.

P = Sin 2ωtdt P = 0

Hence average power absorbed by pure inductor is zero

During one quarter cycle of alternating source of e .m .f. energy is stored in the magnetic field of the inductor this energy is supplied by the source.

During the next quarter cycle the stored energy is returned to the source. For this reason average power absorbed by a pure inductor over a complete cycle is zero.

NUMERICAL EXAMPLES

1. A pure inductive coil allows a current of 10A to flow from a voltage of 230V and frequency 60Hz supply.

Find

i) Inductive reactance

ii) Inductance of the coil

iii) Power consumed

Write down equations for voltage and current:

Solution

EV = 230V

IV = 10A

f =50Hz

i) Inductive reactance XL

XL = =

ii) Inductance of the coil L

From

XL =2

L = 0.073 H

iii) Power absorbed = 0

Also

= 230 x = 10 x

= 325.27V = 14.14A

ω = 2

ω = 314

Since in pure Inductive circuit current lags behind the applied voltage by radians. The equation for voltage and current are,

E = 325 .27 sin 314t, I = 14.14 sin (314t)

2. Calculate the frequency at which the inductive reactance of 0.7H inductor is 220Ω

Solution

XL = 220 Ω

L = 0.7H

f =?

f =

f = 50 Hz

3. A coil has self inductance of 1.4H. The current through the coil varies sinusoidally with amplitude of 2A and frequency 50 Hz

Calculate

i) Potential difference across the coil

ii) r .m .s value of P.d across the coil.

Solution

(i) P.d across the coil

E = L

E = L cos t,

E = L 2 cos2

E = 2

E = 880 cos 100

ii) r.m.s value of potential different across the coil

= 622.2V

4. How much inductance should be connected to 200V, 50 Hz a.c supply so that a maximum current of 0.9A flows through it?

Solution

= 200V

Io = 0.9A

f = 50 Hz

Peak value of voltage

Inductive reactance L XL =

XL =

XL = 314.27 Ω

Inductance L, L =

L = 1 H

5. An Inductor of 2H and negligible resistance is connected to 12V, 50Hz supply. Find the circuit current, what current flows when the inductance is changed to 6H?

Solution

* For the First case XL

XL = 2

XL = 628 Ω

Circuit current

= 0.019A

* For the second case XL’

XL’ = 2

XL’ = 1884 Ω

Circuit current

= 0.0063A

A.C CIRCUIT CONTAINING CAPACITANCE ONLY

When an alternating voltage is applied to a capacitor, the capacitor is charged first in one direction and then in the opposite direction. The result is that electrons move to and fro round the circuit connecting the plates, thus constituting alternating current.

Consider a capacitor of capacitance C connected across an alternating source of e .m. f .

Suppose the instantaneous value of the alternating e.m.f E is given by

E = sin ωt …….. (i)

If I is the current in the circuit and Q is the charge on the capacitor at this instant, then the Potential difference across the capacitor VC

At every instant the applied e.m.f E must be equal to the potential difference across the capacitor.

E =

Q = C sin ωt

From, I =

I =

I = Cωcos ωt

I = Cω sin (ωt + 2 The value of I0 will be maximum Io when sin (wt+) = 1

Substituting the value of Io in equation (ii)

1) Phase Angle

It is clear from equation (I) and (ii) that circuit current leads the applied voltage by π/2 radians or 90Ëš.

This fact is also indicated in the wave diagram.

It also reveals that Iv leads Ev by 90°, hence in a.c circuit current in capacity leads the voltage. This means that when voltage across capacitor is zero, current in capacitor is maximum and vice versa. When P.d across capacitor is maximum, the capacitor is fully charged, i.e circuit current is zero. Since the rate at which a sinusoidally varying p.d falls is greater as it reaches zero, the current has its maximum value when P.d across capacitor is zero. Hence, current and voltage are out of phase by 90°.

CAPACITIVE RESISTANCE

Capacitive resistance is the opposition which a capacitor offers to current flow. It is denoted by XC.

Capacitance not only causes the voltage to lag behind the current but it also limits the magnitude of current in the circuit.

We have seen above that

Then

Clearly, the opposition offered by capacitance to current flow is 1/ωC

The quantity 1/ωC is called capacity reactance XC of the capacitor

XC will be in Ω if f is in Hz and C in Farad

a) From

Then

b) For d.c

f = 0 XC = =

XC = XC = ∞

Therefore a pure capacitance offers infinite opposition to d.c. In other words, a capacitor blocks d.c.

c. From XC =

Therefore the greater the f the smaller is XC and vice versa

d) From

Therefore the units XC are Ohm

(iii) AVERAGE POWER ABSORBED

From

E = sin ωt

I = sin (ωt+2)

I = cos ωt

Instantaneous power P

P = EI

P = Sin ωt. Cos ωt

P = (sinωtcosωt)

P = Sin 2ωt

∴ Average power P = Average of P Over one cycle

P =Sin 2ωtdt

P = 0

Hence average power absorbed by pure capacitance is zero. During one quarter cycle of the alternating source of energy is stored in the electric field of the capacitor.

This energy is supplied by the source during the next quarter cycle, the stored energy is returned to the source

WORKED EXAMPLES

1. A 318μF capacitor is connected 230V, 50Hz supply. Determine

i) The capacitive reactance

ii) r.m.s value of circuit current

iii) Equations for voltage and current

Solution

C = 318μF = 318 x 10-6 F

f = 50 Hz

i) Capacitive reactance XC

XC =

XC = 10 Ω ii) r.m.s value of current

= =

= 23A

iii ) = . = √2 x 23

= 230 = 32.53A

= 325.27 V

ω= 2

ω = 2

ω= 314

E0 = 325.27sin 314t and I = 32.53 sin314t

2. A coil has an inductance of 1H

a) At what frequency will it have a reactance of 3142Ω?

b) What should be the capacitance of a condenser which has the same reactance at that frequency?

Solution

a) L = 1H

XL = 3142Ω

f = ?

XL = 2

f = = :

f= 500Hz

b) XC = 3142 , f=500 Hz

C =? XC =

C =

C = 0.11

3. A 50 capacitor is connected to a 230V, 50Hz supply. Determine

i) The maximum charge on the capacitor

ii) The maximum energy stored in the capacitor

Solution The charge and energy in capacitor will be maximum when p.d across the capacitor is maximum.

i) Maximum charge on the capacitor

Q = C

Q = (50x) x (230 x)

Q = 16.26 x C

ii) Maximum energy stored in the capacitor U

U = C

U = 1/2 x (50x) x (230 x 2

U = 2.65J

4. The Instantaneous current in a pure inductance of 5H is given be

I = 10sin (314t- amperes

A capacitor is connected in parallel with the inductor. What should be the capacitance of the capacitor to receive the same amount of energy as inductance at the same terminal voltage? Solution The current flowing through pure inductor is

I = 10 Sin (314t)

= 10A

ω = 314s-1

Maximum energy stored in the inductor

UL = 250J ………. (i)

Now

= ωL

= 314 x 5 x 10

= 15700v

Max energy stored in the capacitor of capacitance C

= ……….(ii)

Equate the equation (I and (ii)

= = 250

C =

C = 2.03 x F

C = 2.03

A.C CIRCUIT CONTAINING R AND L IN SERIES

Consider a resistor of resistance R ohms connected in series with pure inductor of L Henry.

Let

= r.m.s value of applied alternating e.m.f

=r.m.s value of the circuit current

VR = R when VR is in phase with

VL = where leads by 90º

Taking current as the reference phasor, the phasor diagram of the circuit can be drawn as shown in figure.

The voltage drop VR is in phases with current and is represented in magnitude and direction by the phase OA.

The voltage drop VL leads the current by 90º and is represented in magnitude and direction by the phase AB.

The applied voltage is the phasor sums of these two voltage drops

= +

1) Phase Angle

It is clear from the phasor diagram that circuit current lags behind the applied voltage by Φº.

Therefore we arrive at a very important conclusion that in an inductive circuit current lags behind the voltage.

NUMERICAL EXAMPLE

1. Three impedance are connected in series across a 200V, 50Hz supply. The first impedance is a 10â„¦ resistor and the second is a coil or 15 â„¦ inductive reactance and 5 â„¦ resistance while the third consists of a 15 â„¦ resistor in series with a 25â„¦ capacitor

Calculate

i) Circuit current

ii) Circuit phase angle

iii) Circuit power factor

iv) Power consumed

Solution

i) Total circuit Resistance

R = 10 +5 + 15

R = 30 â„¦

Total Circuit reactance

X = XL - XC

X = 15 – 25

X = -10 â„¦ (capacitive)

Circuit impedance Z

Z = 31. 6 â„¦

Circuit current IV

IV = E v Z

I V = 200 31.6

IV = 6. 33 A

ii) Circuit phase Angle

iii) Circuit power factor

Power factor = COS θ

= COS 18 .26º

Power factor = 0. 949

iv) Power consumed P

P = EV IV Cos ∅

P = (200 x 6.33) x 0. 949

P = 1201. 4 W

Alt

P = Iv2 R

2. A 230V, 50Hz supply is applied to a coil of 0.06 H inductance and 2.5 Ω resistance connected in series with 6.8 μF capacitor

Calculate

i. Circuit impedance

ii. Circuit current

iii. Phase angle between EV and Iv

iv. Power factor

v. Power consumed

Solution

i) Inductive reactance XL

XL = 2ΠfL

XL = 2Πx 50 x 0.60

XL = 18 .85 Ω

Capacitive reactance XC

XC = 468 Ω

Circuit Impedance

Z = 449. 2 Ω

ii) Circuit current IV

IV = 0.152 A

iii) Phase angle between EV and IV

iv) Power factor

Cos Φ = 0.0056

v) Power consumed P

P = EVIS Cos Φ

P = 230.x 0.512 x 0.0056

P = 0.66 W

3. A resistance R, and inductance L = 0.01H and a capacitance C are connected in series. When an alternating voltage E = 400Sin (3000t- 200) is applied to the series combination, the current flowing is 10√2 Sin (3000t - 650). Find the value of R and C

Solution

The circuit current lags behind the applied voltage by θ

θ = 650 – 20

θ = 450

This implies that the circuit is inductive i.e.

XL > XC

The net circuit reactance X

X = XL –XC

Now

XL = ωL

XL = 3000 x 0.01

XL = 30 Ω

Also

From

Circuit Impedance Z

Z = 28.3 Ω

Now

Z2 = R2 + X2

Z = R2 + R2

Z2 = 2R2

Z = R √2

R = 20Ω

Now

X = XL – XC

20 = 30 – XC

XC= 10Ω

From

C = 33.3 x 10 -6 F

4. A series RLC circuit is connected to an a.c (220V, 50 H) as shown in the figure below If the reading of the three volt meter V1 V 2 and V3 are 65V, 415V and 204V respectively.

Calculate

i. The current in the circuit

ii. The value of inductor L

iii. The value of capacitor C

Solution Here voltmeters are considered ideal i.e. having infinite resistance.

Therefore, it is a series RLC circuit

i) Circuit current IV

IV = 0. 65 A

ii) Inductive reactance XL

XL = 318. 85 Ω

Inductance L

L = 1H

iii) Capacitive reactance XC

XC = 638. 46Ω

Capacitance C

C = 5 x 10-6 F

5. A coil of resistance 8Ω and inductance 0.03H is connected to an a.c supply of 240V, 50 Hz.

Calculate

i) The current the power and power factor.

ii) The value of a capacitance which when connected in series with the above coil causes no change in the value of current and power taken from the supply Solution i) Reactance of the coil XL

XL=9.42 Ω

Impedance of the coil Z

Z = 12. 46 Ω

Circuit Current IV

IV = 19.42 A

Power consumed P= (19.42)2 x 8

P= 3017 W

Power factor Cos Φ

Cos Φ = 0.65 lag ii) To maintain the same current and power, the impedance of the circuit should remain unchanged. Thus the value of capacitance in the series circuit should be such so as to cause the current to lead by the same angles as it previously lagged.

This can be achieved if the series capacitor has a capacitive reactance equal to twice the inductive reactance.

XC = 2XL

XC = 2 x 9 .42

XC = 18.84 Φ

Now

C = 169 x 10-9 F

RESONANCE IN R – L – C SERIES A.C CIRCUIT The R- L – C series A.C circuit is said to be in electrical resonance when the circuit power factor is unity

XL = XC

This is called series resonance.

Resonant frequency

This is the frequency at which the reactance of the coil has the same magnitude and so giving the current in a circuit its maximum value. XL = XC

where

XC = capacitive reactance

XL= inductive reactance

The frequency at which resonance occurs is called the resonant frequency

The resonance in R – L – C series circuit can be achieved by changing the supply frequency because XL and XC are frequency dependent

At a certain frequency, called the resonant frequency fo, XL becomes equal to XC and resonance occurs.

At resonance

XL = XC

If L and C are in Henry and farad respectively, then ƒo will be in Hz EFFECTS OF SERIES RESONANCE When series resonance occurs, the effect on the circuit is the same as though neither inductance nor capacitance is present. The current under this condition is dependent solely on the resistance of the circuit and voltage across it.

(i)The impedance of the circuit is minimum and equal to the resistance of the circuit.

From

When Z=ZR, XL = XC

ZR = R

At series resonance

(ii) The circuit current is maximum as it is limited by the resistance of the circuit alone.

(iii) Since at series resonance the current flowing in the circuit is very large, the voltage drops across L and C are also very large.

In fact, these drops are much greater than the applied voltage.

However, voltage drop across L – C combination as a whole will be zero because these drops are equal in magnitude but 1800 out of phase with each other.

RESONANCE CURVE Resonance curve is the curve between the circuit current and the supply frequency.

Figure below shows the resonance curve of a typical R-L-C series circuit. Current reaches the maximum value at the resonant frequency fo, falling off rapidly on either side at that point. It is because if the frequency is below fo, XC > XL and the net reactance is no longer zero. Net reactance X X = XC - XL If the frequency is above fo, then XL > XC and the net reactance is again not zero.

Net reactance X X = XL - XC

In both cases, the circuit impedance will be more than the impedance ZR at resonance.

The result is that magnitude of circuit current decreases rapidly as the frequency changes from the resonant frequency.

The effect of resistance in the circuit. The smaller the resistance, the greater is the current at resonance and sharper the resonance curve.

On other hand, the greater the resistance, the lower is the resonant peak. Q – FACTOR OF SERIES RESONANT CIRCUIT At series resonance, the p.d across L or C (the two voltage drop being equal and opposite) builds up to a value many times greater than the applied voltage EV.

The voltage magnification produced by series resonance is termed as Q – Factor of the series resonant circuit.

Q – Factor of a resonant. R-L circuit

This is the ration of voltage across L or C to the applied voltage.

This is the ratio of power stored to power dissipated in the circuit reactance and resistance respectively.

Where

X = Capacitive or inductive reactance at resonance

R = Series resistance

The Q – factor of a series resonant circuit can also be expressed in terms of L and C.

From,

Also

The value of Q-factor depends entirely upon the design of the coil i.e. R – L part of the R – L – C circuit because resistance arises in this rather than. With a well designed coil, the quality factor can be 200 or more.

PHYSICAL MEANING OF Q – FACTOR The Q- Factor of series a.c circuit indicates how many times the p.d across L or C is greater than the applied voltage at resonance.

For example, consider on R-L- C series circuit connected to 240v a.c source.

If Q – factor of the coil is 20, then voltage across L or C will be,

VC = VL = QVR = 20 x 240

VC = VL = 4800 V Q – FACTOR AND RESONANCE CURVE At series resonance the circuit current is maximum (IR = EV/R) and is limited by circuit resistance only.

The smaller the circuit resistance, the greater is the circuit current and sharper will be the resonance curve. Smaller circuit resistance means large value of Q – Factor. Therefore, the greater the Q – Factor of resonant R-L-C circuit, the sharper is the resonance curve.

APPLICATION OF SERIES RESONANCE One important applicant of series resonance is to tune radio and TV receivers.

The input signal comes from the antenna and induces a voltage E in L of the series resonant circuit. VC = QE

The voltage across the capacitor becomes VC.

where:

Q = Quality factor of the circuit

As the value of Q is generally large, the original signal received by the antenna increases many times in value and appears across C.

The value of VC is much more than that could have been obtained by direct transformer ratio.

Thus amplifier receives a greatly increased signal.

Bandwidth of a series resonant circuit.

This is the range of frequencies over which circuit current is equal to or greater than 70.7% of maximum current. (IR, current at resonance).

The two frequencies are cut off frequency f1 and upper off frequency f2.

Bandwidth (BW) = f2 - f1

It can be shown by using mathematical equation:-

The circuit has a general appearance of a parallel circuit but actually it is a series circuit. It is because no separate volt is applied to L but instead a voltage E is induced in it which is co as a voltage in series with LC.

Q – Factor of a resonant Circuit

Is the ration of resonant frequency fo to the bandwidth of the circuit RESONANCE IN PARALLEL A.C CIRCUIT

A parallel a.c circuit containing reactive element (L and C) is said to be in electrical resonance when the circuit power factor is unity. This is called parallel resonance

Consider a pure inductor of inductance L connected in parallel with a capacitor of capacitance across an a .c source of voltage EV ( r. m . s)

The circuit will be in resonance when the circuit power factor is unity. This means that wattles component of the circuit current should be zero.

IL - IC=0

The resonance in a parallel a .c circuit can be achieved by changing the supply frequency because XL and XC are frequency dependent .

At a certain frequency called resonant frequency f0, IC become equal to IL and resonance occur. At resonance

I C = IL

From

f0 will be in Hz if L is in Henry and C is in farad

EFFECTS OF PARALLEL RESONANCE

1) The circuit power factor becomes unity. This implies that the circuit act as a resistor 2) The impedance (resistive) of the circuit becomes maximum. 3) The circuit current is minimum the small current IV flowing in the circuit is only the amount needed to supply the resistance losses. RESONANCE CURVE This is the curve between the circuit current and the supply frequency. Figure below shows the resonance curve of a parallel a.c circuit. The circuit current IV is minimum at parallel resonance.

As the frequency changes from resonance, the circuit current increases rapidly.

This action can be explained as follows. For frequencies other than the resonance, the reactive currents (IL and IC) in the two branches of the circuit are not equal. The resultant reactive current must be supplied by the a.c source . As the difference of the reactive currents in the two branches increase with the amount of deviation from the resonant frequency, the circuits current will also increase.

ADVANTAGE OF A.C OVER D.C

(i)Alternating voltages can be stepped up or stepped down efficiently by a transformer. This permits the transmission of electric power at high voltages to achieve economy and distribute the power at utilization voltages. (ii)A.C motors are cheaper and simpler in construction than the d.c motors. (iii) A. C can be easily converted into d.c by rectifiers. (iv)Alternating current can be controlled with a choke coil without any appreciable loss of electrical energy. (v) The switch gear (e.g. switches, circuit breakers) for a.c system is cheaper than the d.c system.

DISADVANTAGE OF A.C OVER D.C

(i) For the same voltage (same value of voltage), a.c is more dangerous than d.c. (ii) The shock of a.c is attractive whereas that of d.c is repulsive. (iii) A.C cannot be used for some processes e.g. electroplating, charging of batteries etc. (iv) A.C is transmitted more over the surface of the conductor than from inside. This is called skin effect. To avoid skin effect, a.c is transmitted over several fine insulated wires instead of a single thick wire.

WORKED EXAMPLES

1. A coil of resistance 100Ω and inductance 100μH is connected in series with a 100 pF capacitor. The circuit is connected to a 10V variable frequency source. Calculate: (i) Resonant Frequency (ii) Current at resonance (iii) Voltage across L and C at resonance.

Solution:

Capacitance C = 100pF = 100 x 10-12 F Inductance L = 100μH = 100 x 10-6 H (i) Resonant frequency f0 f0 = 1 . 59 x 106 HZ (ii) Current at resonance IR IR = 0.1 A (iii) Voltage across L VL = Ir XL VL = 100V (iv) Voltage across C VC = 100V

At series resonance , the voltage across L or C is much greater than the applied voltage.

2. A circuit, having a resistance of 4Ω and inductance of 0.5 H and a variable capacitance in series, is connected across a 100V, 50 Hz supply. Calculate;

(i) The capacitance to give resonance

(ii) The voltage across inductance and capacitance. Solution

(i) At series resonance

XL = XC

C = 20.26 x 10-6F

(ii) Current at resonance Ir.

Ir = 25 A

P.d across L

VL = Ir XL

VL = 3927V

P.d across C

VC = Ir XC

VC = 3927 V

3. A series R – L- C circuit consists of a 100 Ω resistor, an inductor of 0.318 H and a capacitor of unknown value. When this circuit is energized by √2 x 230 sin 314t volts are supply the current is found to be √2 x2.3 sin 314t find. (i) The value of capacitor in microfarad.

(ii) Voltage across indicator

(iii) Total power consumed.

Solution

From, EO = √2 EV

E = √2 x 230 Sin 314t

E=Ev Sin ωt

Ev = √2 x 230

√2 x230 = √2 x EV EV = 230V.

Circuit current IV

Iv = 2.3 A

Also further phase angle is zero

ω = 314 S-1

f = 50 Hz

i) At resistance

XL = XC

C = 31.4 x 10 -6 F

C = 31.4 x μF

ii) Voltage across inductor VL

VL = 230 V iii) Total power consumed

P= 2.32 x 100

P = 529 W 4. A coil of inductance 8 μH is connected to a capacitor of capacitor of capacitance 0.02 μF. To what wavelength is this circuit is tuned?

Solution

L = 8 x 10 -6 H

C = 0.02 x 10 -6F

Resonant frequency f0

f0 = 3.98 x 105 Hz

If C = 3. 108 m /s is the velocity of the e.m wave, then wavelength λ λ= 7.54 X 102 m

5. a ) A Sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to A series L.C. R circuit in which R = 3 Ω , L = 25.48 μH and C = 796 μF . Find

i) The impendence of the circuit.

ii) The phase difference between voltage across the source and the current

iii) The power dissipated in the circuit

iv) Power factor.

b) Suppose the frequency of the source in the previous example can be varied.

i) What is the frequency of the source at which resonance occurs?

ii) Calculate the impedance the current and the power dissipate at the resonant condition.

Solution

a) E0 = 283 V

f= 50 Hz

R = 3Ω

L = 25.48 x 10- 3 H

C = 796 x 10-6 F

XL = 8Ω

XC = 4Ω i) Circuit impedance Z

Z = √25

Z = 5 Ω

ii) The phase difference between voltage and circuit current iii) Circuit current IV

IV = 40 A iv) Power dissipated in the circuit

P = IV R

P = 402 x 3

P = 4800 W

Power factor = Cos 53.1º

b) i) the frequency at which resonance occur f0 = 35.4 HZ ii) The impedance z at resonance is equal to R

Z = 3Ω

The circuit Current IV

IV = 66 .7 A

Power dissipate at resonance is

P = (66.7) x 3

P = 13350 W

L - C IN PARALLEL CIRCUIT.

Consider inductor of inductance L Henry connected in parallel with capacitor of capacitance C. Figure below shows the two component of current.

IL lags by 90º on EV but IC leads by 90º on E

CASE 1

If IC is greater than IL at the particular frequency

Since I leads by 90º on E in this case we say that the circuit in net capacitive.

CASE 2

If IL is greater than IC

I = IL – IC

Since I lags by 90º on E in this case, the circuit is Net inductive.

CASE 3

At the Resonance

XC = XL

ELECTRONICS

1. Conductors

Posses free electrons

Metals are all good conductors due to having low resistance to the flow of current.

2. Insulators

They do not have free electrons for conduction. They have high resistance to the flow of current.

All non metals are bad conductors. Eg. dry wood, paper and air.

3. Semiconductors

These are class of materials whose conductivity is between that of good conductors and insulators

Silicon and Germanium are examples of semiconductors elements widely used in electronic industry.

INTRINSIC SEMICONDUCTORS

These are pure semiconductors.

EXTRINSIC SEMICONDUCTORS

These are impure semi conductors material.

DOPING

Is a process of introducing a tiny amount of impurity into a semiconductors material to form extrinsic semiconductors shells.

N-SEMI CONDUCTORS -Silicon and germanium atoms are tetravalent

-They have four electrons in their outermost shell. -When a doner atom with fine electrons in its outer most shell (ie Arsenic) is added to a silicon crystal, the fifth electrons becomes a free change carriers since there is production of large number of negative charge carriers(electrons) the impure semiconductors is called N-Semiconductors

P-SEMICONDUCTORS

A-P- Semiconductor is made by adding a trivalent atom (an acceptor) such as B or on to pure semi conductor such as germanium. Since there is a production of large number of holes (positive charges) the impure semiconductor is called P- Semiconductor.

P-N=JUNCTION/DIODE:

This is formed when P and N semiconductors are melted to form a junction between them

The marrow region at the P-n junction which contains the negative and positive charge is called depletion layer.

A barrier dip is a p.d which oppose more diffusion of charges across the junction.

This is produced when the flow of +ve and -ve Charges ceases

P – N JUNCTION AS A RECTIFIER:

FORWARD BIAS. Is said to be forward biased when its P- semiconductor is connected to the +ve terminals of the battery and its N- Semi conductors is connected to the -ve terminal at the battery.

In this case electrons and holes flow across the P-n junction. This happen because the +ve pole of the battery repel the +ve charge and –Ve pole rel the –ve charges.

REVERSE BIAS

A-P-N junction is said to be reverse biased when its P. Semiconductor is connected to the negative pole junction of a battery and N. Semiconductor is connected to the +ve [p;e pf the battery in this case only a very small a current flows.

P-N JUNCTION AS RECTIFIERS

The graph shows that P-N junction acts as a rectifier, it has low resistance in one direction of P.d (ie + v) and higher resistance in the opposite direction of P.d (-v)

RECTIFIER CIRCUITS:

HALF-WAVE RECTIFIER CIRCUIT

A rectifier is a circuit which allow the flow of current I P.d in one direction only:

FULL –WAVE RECTIFIERS CIRCUITS:

a) Using centre –tapped transformer.

b) Using bridge circuits

On one half of a cycle when P is +v relative to Q only diode D1 conducts

On one other half the same cycle only the diode D2 conducts.

In both cases the current gees through resistor RL in the same direction.

The large capacitor C is used for stabilizing the marying d.c voltage.

B) TRANSISTORS

Transistor is a component which amplifies current. It is made from three layers of P and n. Semiconductors. The layers are called the emitter (E) base (B) an collector (C)

There are two types of transistors.

I. n.p.n transistor

II. p.n.p transistor

Formation of a transistor

A transistor is formed by putting the doped semiconductors together in such a way that two junction are formed.

The pnp transistor (bipolar transistor).

-Bipolar means n p n and p n p transistor as they have two opposite polarity of doped semiconductors and voltages across terminals

P n p n p n

Transistor configuration

There are 3 basic configurations

1. Common emitter configuration

2. Common base configuration

3. Common collector configuration

1. Common emitter configuration (n p n)

Under thus configuration the transistor has both voltage gain and current gain.

To get volt you need a resistance RL

= I

Ring is used for injecting only a small current for great amplification on E by C

Current gain

= very large

But

Also

is the reflection of

Common base configuration (PnP)

Under this configuration the transistor has voltage gain but no current gain

Earthing puts the common line at p.d=0

Common collector configuration

Under this configuration the transistor has current gain but no voltage gain

=Amplification factor

For common emitter

Vo = IC R2

Vi = IBRB

From, IE = IB + IC

Common emitter characteristic curve

The circuit above is for investigating the variation of current with voltage in the input and output circuits. OUTPUT CHARACTERISTICS IC-VCE with IB constant . The results are plotted below. The knee of the curves shown corresponds to a low P.d(0.2) the output for higher P.d the output IC varies linearly with VCE for a given value of base current IB. The linear part of the characteristic is the one used in the audio frequency (a.f) amplifiercircuits so that the output is undistorted.

INPUT CHARACTERISTICS

IB-VBE with VCE constant The results are as follows:- The input characteristics is non-linear

TRANSFER CHARACTERISTICS IC-IB with VCE constant

The results are as shown below:- The output current IC varies linearly with the input current IB. The current transfer ratio or current gain is given by In the figure below

Questions

1. An npn transistor has a current gain (Beta) value of 200. Calculate the base current required to switch a resister of 4µA.

2. An npn transistor has a dc base bias voltage of and an input base resistor of 100kΩ.What will be the of base current into the transistor

(The transistor is a silicon type)

Data

From Kirchhoff law

For silicon

=10v

= 100kΩ

=0.6v (wasted voltage)

Solution

--=0

Example

Given the circuit below, determine

The transistor has =150

The transistor of silicon type

Solution

Second Kirchhoff’s law in the input circuit

× = 0

=0.6v

= 3.4V

=5+ ( )

=5+ (10000×4.4×10-4) (6.6)

=2.8v

Example

1. A common emitter amplifier has = 1.2kÎ and supply Voltage of V=12v. Calculate the maximum collector current following throughout resistor when switched fully on (saturation assume.Also find with a voltage drop of 1v across it, the transistor silicon.

Solution

Quiescent point:

It’s a point when the current flow is smooth i.e. not being clicked (excess) and transistor functions.

Saturation point:

If =0 transistor is in the cutoff region, there is a small current collector leakage, CEO

Normally is neglected so that =

In cutoff both the base emitter and base collector junction are reverse based.

When base emitter becomes forward based. is increase, then IC also increases when decreases as a result.

When reaches its saturation value BC junction becomes forward based and can increase no further even with continued increase in

At the point of saturation () not longer valid)

VCE(Sat) for a transistor occurs somewhere below the knees of the collector curve.

The saturation value for (Sat) is usually a few tenth of volt for silicon transistors.

The DC load line, the cutoff and saturation can be illustrated by the load line.

Between the cutoff point and the saturation point is where the transistor is active and as most active at the quiescent point.

Self biasing /fixed bias

Outer loop

10 = 9.4×10-3×100 +

= 9.06V

Common emitter amplifier circuit

Faithful amplification- is the application or the output that is not distorted.

Question

a) Pd across base resistor

Consider loop (L), from Kirchhoff’s law

3V=0 but =0.7

= 2.3V

b) From Ohms law

= 1.53

= ) + (

= (20- 3) + (2.3-1.836)

= (17+0.463)

=17.463v

d) Find

From Kirchhoff’s law

Given

--=0

β=20

=20v

20v-(1.5×1.224-=0

=18.164v

2. Q= =

= +

= 25V

= 47mA = 4.7×10 -2A

=0.3659Ω

Question

For the circuit above the transistor has a current gain =80 the collector supply voltage = 40 . The required biased conditions are = 0.7V and = 1mA. Determine the suitable values for resistors,,& ,

R2 = 10RE, VE = 1

VE = IV.

Given v

= 1×10 -3A

=80

=0.0000125

== 0.0010A

=1kΩ

=110

= 1.7 but

+ =10

Operational Amplifier (Op amp)

An operational amplifier (op amp) is an electronic device consist of a large number i.e. twenty and above.

It has 3 terminals two input terminals and one output terminal.

The op amp can perform electronically mathematically of such as additional, subtraction, multiplication, differentiation, integration

Properties of an op- amp

i) It has got a very high voltage gain called the open loop gain which typically is 105 for dc and low frequency but decrease with frequency.

ii) It has a very high input resistance typically 10, it draw a minute current from the signal source.

iii) It has a very low output resistanceR0, typically 100Î.

Description

It has one output and two inputs and one non inverting ( and one inverting (-).

Its operation must convenient from a dual balanced power supply giving its equal +ve and –ve voltage (+Vs, or,-Vs)

Inverting amplifier

=(-)

Some of the output goes back to the input .This red called the amplification from A0 to A

But

I1, = I2

Example

(i)Find the closed loop gain of the inverting amplifier

From

= -10

ii) Supposed the voltage gain is to be increased to 40 and the current of remains the same .What are the values of the resistors required to gain this

NON INVERTING AMPLIFIER

The fraction β fed back via Rf

LOGIC GATES (Non inverting)

LOGIC GATES

SYMBOL

i. NOT GATE(INVERTER)

he It has only one input and one output.

ii. OR GATE This can have many number of inputs but only one input. It gives high output if either of the inputs is high or all inputs are high.

TRUTH TABLE FOR OR GATE

iii. AND GATE It can have many number of inputs but only only one output. It gives high output when both input are high.

TRUTH TABLE FOR AND GATE

iv. NOR GATE This is equivalent to OR gate followed by NOT gate.All outputs of OR gate are inverted

TRUTH TABLE FOR NOR GATE

v. NAND GATE This is the AND gate followed by NOT gate . This is widely used gate . In this case the outputs of AND gate re inverted.

TRUTH TABLE FOR NAND GATE

All logic gates described can be connected together to form different function (i)They are used to control traffic light (ii)They are used in communication system (iii)They are used in arithmetic and data processing

Questions

1. Find the expression for Y and form the truth table of the following diagram.

Solution

2. From the logic circuit below form the Boolean expression and draw the truth table

Solution

Truth Table

Laws of Boolean algebra

T1: Commutative law

a) A+B=B+A

b) AB=BA

T2: Associative law

a) (A+B) +C=A+ (B+C)

b) (AB) C=A (BC)

T3: Distributive law

a) A (B+C) =AB+AC

b) A+BC= (A+B) (A+C)

T4: Identity law

a) A+A+=A

b) AA=A

T6: Redundancy law

a) A+AB=A

b) A (A+B) =A

T7: a) 0+A=A

b) 0A=0

T8: a) I+A=I

b) 1A=A

T11: De Morgan’s theorem Example

1) Prove that

Algebraically

=A (I+B) +B

Question

For a lift (L), these are the conditions:

i) The lift door must be closed giving d=1

ii) The appropriate floor button (B) must be pressed B

L= Bd but not L=B+ d

A boiler shut down solenoid (s) will operate if the temperature T reaches 50 and the circulating pump P ise turned off or if the pilot light L goes out.

CURRENT ELECTRICITY

An electric current is set up when a net charge Q passes through any section of the conductor in time t.

The charge assumed to be constant: i.e If the value of the rate flow of charge is not constant the current varies with time and is given by the equation.

VELOCITY OF CURRENT CARRIES OR DRIFT VELOCITY (V).

The conduction of electricity in metal is due to free electrons. Free electron have the thermal energy and wonder randomly through the metal from atom to atom and hence collide.

The amplitude of vibration are then increased and the temperature of the metal rise.

On the average the electron drift in the direction of the drift in the mean velocity called drift velocity

Consider a metal wire of cross sectional area A and of length L.

Let “n” be the number of electrons per unit volume and “e” is the charge of an electron

The volume occupied by electron is AL

The total number of electrons is nAL

The total charge (Q) in the wire is nALe........ (1) If I is the current flowing through this wire then, I = Q/t ---------------------(3) From equation 5

Examples

Electrons each having a charge passing through point x towards a point y in 0.1 sec. What is the current and its direction?

2. A conductor with a cross section of and electric current of 1.2A. If the number of electrons be calculate the electron velocity.(Take charge of an electron=

Q3 A uniform copper wire of diameter 0.051mm carries a current 10A. what is the drift velocity of electrons in the wire, assuming the charge of electrons is

1.6 X 10-19 C

RESISTANCE AND RESISTIVITY:

Ohm had shown that by using wires of different lengths and diameters.

Resistance of a wire R is proportional to its length

and inversely proportional to its cross section area. A i.e

From (1) and (2) we have

Where is the constant for the material of the wire and it is called. Resistivity.

The resistance of pure metal increases with temperature.

Example

A copper coil has a resistance of at what its resistance at ?

Take the temperature coefficient of copper is

Solution

HEAT AND ELECTRICAL POWER

Consider an electrical device k and let potential difference between A and B be V.

If Q is the charge passing through the ends in a time t then Q = It........... (1)

Total energy given out by the conductor is W = VQ ..................(2)

Substitute eg (1) into eq (2) we get

W= VIt ...... (3)

but workdone = electrical heat developed (H)

hence W= H =VIt ............(4)

Now V= IR so the eqn (4) becomes

KIRCHHOFF’S LAWS

Some circuits cannot be broken down into set of series and parallel combination of conductors. It is then necessary to make use of generalized form of circuits laws already considered. These laws are known as Kirchhoff’s law

Kirchhoff’s first law:

The algebraic sum of current at any junction in the net work is zero

i.e. ∑ I = 0

A current is given a positive sign if it flows towards a junction and negative sign if it flows away from the junction.

From figure I above we have

Kirchhoff’s second law:

The algebraic sum of emf’s around the loop is equal to the algebraic sum of all potential differences in that circuit i.e

RULES FOR FINDING POTENTIAL DIFFERENCE(pd) IN A LOOP

If the resistance is traversed in the direction of the current,the chage in potential is -IR in the opposite direction is +IR

If a set of emf is traversed in the direction of emf the change in potential is +ve(+E) in the opposite direction it is(-E)

The choice of the direction of the traverse depend on yourself

Solution

Applying Kirchhoff’s laws in the clockwise direction we have

Qn. Determine the magnitude of

The minus sign tells that the current is in of the site direction (from the one assumed)

Qn. On the basis of Kirchhoff’s laws determine the numerical values for the current

Take the loop B E D C in clockwise direction.

POTENTIAL DIVIDER

The resistance box in series are often used to provide known fraction of a given potential difference

(iii) With load.

A resistor with a sliding contact can similarly be used as shown in figure (ii) to provide continuous variable potential difference from zero to a full supply by value V0.

This is a convenient way of controlling the voltage applied to a load such as lump. The resistor of the load R3 however acts in parallel with resistor R1 hence equation (1) is no longer true.

Qn.

What is the final potential difference between A and B in the direction circuit below (I) in the circuit as shown (II) if an additional 5000Ω resistor where connected from A to B

Solution

(i) Final p.d between

OHM’S LAW FOR COMPLETE CIRCUIT

There exist certain device such as battery and electronic generators which are able to maintain a potential balance between two points to which they are attached to devices are called a set of electromotive force put the normally have some internal resistances. When the current flow through the battery setsup a p.d Ir across the internal resistance. The resistor of resistance R which is connected to the battery is called the load.

OUT PUT AND AN EFFICIENCY

power generated by the source the current

WHEATSTONE BRIDGE:

- Wheat stone bridge is a circuit that enable resistance to be measured accurately:

Four resistors P, Q , R and S are joined as shown in the figure above

If P is the known resistor Q must be known as well as R and S or their ratio one or more of Q, R and S are adjusted until there is no deflection on G, the bridge is then said to be balanced.

At balance point no current through G so p.d across BD=0 and p.d across AB=p.d across AD. Also current through R = current through Current through p=current through

Qn.

A wire of length 1.1 meter and Radius is connected across the right gap of the metre bridge when a resistance of 45 Ω is introduced in the left gap from a resistance box connected across it. The balance point is obtained 0.6m from left side. Calculate the specific resistance (resistivity) of the material of the wire

Qn.

A and resistor are connected in series combination is connected across a 100v supply of negligible internal resistance as shown in the figure below:

(i) What is the output voltage of the load resistance

(ii) What current resistance through the load resistor in (i) above.

I=21.15.

Solution

Search This Blog

Jdan.com

ADVANCED PHYSICS FORM SIX

Comments

Post a Comment

Popular posts from this blog

EDUCATION PSYCHOLOGY

FOLLOW THIS BLOG FOR MORE GOSPLE SINGLES