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E & EP outcome 3 Nat Dip/cert notes yr1
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OUTCOME 3. Understand the principles andproperties of magnetism.
Magnetism.
In this section we must consider the relationshipsbetween magnetism (magnets) and electricity(conductors carrying a current).
We have all played with simple bar magnets astoys, or as simple fridge magnets. The basic barmagnet has a NORTH and SOUTH pole, thesepoles allow a freely suspended bar magnet to alignitself with the N & S poles of the earth. This isperhaps the oldest and one of the most importantuses of magnetism by man, and is known as acompass. This was used as one of the mostimportant navigation tools, and is still used today.
Only a small number of materials can bemagnetised, and not all of these will staymagnetised and so produce a permanent magnet.The materials commonly used are steels(permanent) and Iron or Ferrite (non-permanent),others such as nickel alloys and rare earthminerals are used in special applications (cost).
The lines of force around a magnet are known asLINES OF FLUX, and show the path of a, “freeNorth pole”. A free N poles does not really exist butwe can imaging the path it would take if we couldproduce one. This path would be toward the South
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pole of the magnet, and away from the North polethis is due to:-,
LIKE POLES (N & N or S & S) REPEL &
UNLIKE POLES (N & S) ATTRACT.
Let us consider a simple bar magnet,
We can see in this drawing that the lines of fluxflow around the magnet with an indication of theirdirection (away from the north pole towards thesouth).
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N S
Questions 12.Complete the following diagrams by sketching inthe lines of flux and their direction:-,
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N S NS
a)
NS N S
b)
S NSN
c)
Magnetic fields are also produced by a conductor(wire) carrying a current (amperes). This can also,repel and attract, other current carrying conductorsand/or magnets. This is the bases of many of ourelectrical and electronic devies/machines etc.
We can use the right hand screw rule to help usvisualise the fields around a conductor. This saysthat if we look along a conductor in the directionthe current flows then the field rotates in aclockwise direction. In a similar manner asscrewing a screw into something.E.G.
In this diagram wesee that the conductor is marked X if it flows intothe paper and - if it flows out towards you.Complete the diagram for the current flowing out ofthe paper by sketching in the lines of flux and theirdirection.
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DIR
EC
TIO
N O
F I
X -
I Flows into paper I Flows Out
Questions 13.IF we place a conductor in to a magnetic flux thenwe find that a force is caused on the conductor,depending on the current flowing, direction,proximity of the N & S poles of the magnetic fluxetc.Research and sketch a typical conductor in a barmagnets field diagram, showing the N & S poles,lines of flux, direction of current and the forceproduced.
An aid to us in determining the motion, field andcurrents direction is FLEMING’S LEFT-HANDRULE.With the aid of sketches describe this rule and howit is used.
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To allow us to use this relationship betweenmagnets and electricity we must be able to quantifyas many of the variables as possible. We have anumber of very complex relationships which usesome very complex and advanced maths (fieldtheory, calculus etc.), however, we can use somemore basic relationships at this level ofunderstanding.
Let us consider the force F produced,
F BIlwhere,
F ForceNB Flux DensityTeslaTI CurrentAmpsl Length of Conductorm
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Let us consider a simple conductor suspendedwithin a magnetic flux,
In the above example we have:-,
I 10A,B 1.2T & l 0.2mWe can calculate the force, F on the conductor,
F BIlF 1.2 10 0.2F 2.4N
If we consider the direction of the movement thenusing Fleming’s left-hand rule, we find that theconductor will move down.
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N SX
1.2T
I=10A
length =200mm
Flux Density is a measure of the intensity of amagnetic field at a particular place, It is found fromthe total magnetic flux divided by the area that it isspread across.
Flux density B Total FluxArea of Field
where,Total flux WeberWb
Area of field m2Example,
Calculate the force, F acting on a conductor oflength 400mm, in a magnetic flux of 200Wb. Ifthe conductor is carrying a current of 12A and themagnets pole faces are 120mm x 100mm.
B A A 0.12 0.1
A 0.012m2
B 200106
0.012 0.0167T
F BIl 0.0167120.4
F 0.08016N
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Questions 14.1) Calculate the force acting on conductor if, itcarries a current of 8amps, has a length of 0.5mand is subjected to a magnetic flux density of15mT.
2) Calculate the flux density if a magnet of area0.01m2, produces a magnetic flux of 25mWb.
3) Calculate the force on a conductor if its length is65mm, and it carries a current of 8amps, when it issubjected to a magnetic flux of 120Wb from amagnet whose poles face is 120 x 200mm.
4) Calculate the current required to produce aforce of 4.5N on a 0.6m length conductor, if theflux density is 0.3T.
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DC Motors.
If we wrap the conductor around a former toproduce a coil and pivot this coil within a magneticflux then we can produce a rotational movement.E.g.
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X ..N SF
F
d
Torque = Force, F X Dist, d
From simple mechanics we know that:
TorqueNm ForceN distmWe know that:
ForceN BIlWe can combine these expressions to give:
T BIldIn real DC motors we use an armature, whichconsists of the spindle, former, and coil. The use ofa coil to produce the current carrying conductorallows a large length of conductor to be used in amore compact design. The former used iscommonly manufactured from Iron or ferrite toconcentrate the field around the conductor. Thearmature also carries the commutator, this is a setof split rings used to carry the conductors currentto he correct end of the wire. This comes intocontact with the brushes which supply the currentto the motor.
In very large powerful motors the magnetic fluxmaybe produced by an elecrto-magnet. A solenoidcan be manufactured that produces a highermagnetic flux in a smaller area then a permanentmagnet.
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Example.
Calculate the force acting on the side of the coil,and the maximum torque that the armature canproduce.
F BIlF 300 103 20103 30103 500
F 9102 or 0.09N
T F d (sides)T 0.09 15 103 2
T 2.7 103Nm.
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mag
netic
fiel
d
15mm
30mm
500 turnsFlux Density = 300mT
Current = 20mA
Questions 15.1) Sketch a simple DC motor showing how theconnections are made using the commutator.
2) Calculate the torque produced by a motor whichhas a coils consisting of 800 turns, carrying4.5amps. If the magnetic flux density is 2T, and thecoils dimensions are:Armature to coil centre line 30mm.Magnetic pole face length 120mm.
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This affect as well as being used to give us amotor can be used as a method of measuringphysical quantities. This application gives us thesimple analogue measuring instruments,multimeters with analogue displays commonly usea moving coil. This can be used to measurevoltage, resistance and current, by using differentwiring and resistors we can extend the range. Thesimplest of these instruments use a coil supportedin a magnetic field, the electricity to be measured isfed into the coil, where it produces a torque. Thistorque tries to turn the coil, with the attachedneedle, but acts against a spring to give a readingon a display. The display can be calibrated to showdifferent scales and/or physical quantities.
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magnet
needle
coil soft iron core
Induction & generation.
To generate or produce electricity, we require verysimilar equipment to a motor. We require thecurrent carrying conductor, the magnetic field etc.But instead of putting current in and getting torqueout, we put torque in and get current out.
Whenever, relative movement occurs between amagnetic field and a conductor then anelectro-motive-force (emf) is induced in theconductor. This emf is a voltage that is producedor induced in the conductor, by the process knownas electro-magnetic-induction.
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N
S
V emf
relative movement
coil N turns
We can calculate the emf induced using thefollowing relationship:
emf Nddt Volts
where,N NO of turnsddt rate of change of flux
When we consider these variables we find that theamount of emf induced depends on, the length ofthe conductor (i.e. No of turns etc.), speed of theconductor passing through the magnetic field (dt),and the size of the magnetic flux.
These variables can be brought together to give usthe expression:
E BLv sinwhere,
B flux density TL length of conductor mv relative velocity of magnet & conductor m/s anglebetween conductor & mag flux
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Example.
A closed conductor of length 25 cm cuts themagnetic flux field of 1.5T, at an angle of 60O, witha velocity of 20 m/s. If the conductor carries acurrent of 13A calculate the induced emf, and themaximum emf that could be produced by thismachine.
Emf.emf BLv sinemf 1.5 25 102 20 sin60emf 6.5volts.
The maximum emf that the machine can producewould occur when the armature allows the coils tocut the lines of magnetic flux at right-angles (90O).
Max. emf.emfmax BLv sin90emfmax 1.525102 20 1emfmax 7.5volts
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Questions 16.A number of Laws and rules are used to give usthe directions of current and the size of the emf induced etc:
Fleming’s Right Hand Rule.Lenz’s Law.Faraday’s Law.
Research these Laws and rule, using sketcheswhere required, describe these laws and rule inconnection with the generation of current.
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Mutual Inductance and the transformer.
In the last section we considered inductance, thisinductance was caused by the relative movementbetween a conductor and a magnetic flux. We canproduce a similar affect by changing the magneticflux in a coil instead of a permanent magnet. Thisis known as mutual induction and occurs betweentwo conductors (normally coils) in close proximityto each other. When one has a changing level ofcurrent this will induce a current in the other.
This phenomena is used in transformers to allowus to have a magnetic couple between circuits, andto transform the current/voltage ratio.
In should be noted that we require an AC powersource, not the DC supply we have beenconsidering.
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Varying Flux
coil BCoil AI
E=-M dIdt
Transformers.
Above is a simple diagram of a basic transformer,a consists of an iron core (ferrite) which is normallymade up from a number of laminates. Wrappedaround this are two coils the supply coil known asthe primary, and the load coil which is known asthe secondary coil.
The transformer is used to provide a method ofchanging voltage in AC circuits. This can on alarge scale the sub-stations used to convert the400 +kVolts from the UK national grid, to 230 voltdomestic supply. On a smaller scale the batterycharger used to charge mobile phones at 6 or 9volts is another application of the simpletransformer. The core is made from a material which canmagnetise easily but also, will not become apermanent magnet. Also, we manufacture this ininsulated laminates, this reduces losses due toeddy-currents within the core material.
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R
core
load
supp
ly
primary coilseconday coil
The coils are commonly wound on top of eachother this reduces the physical size of thecomponent and increases the efficiency of thetransformer by the concentration of the magneticflux.
The ratios of the number of turns in each of thecoils controls the voltage/current ratio, between theprimary and secondary coils.
Basic Transformer Equation.
VPVS NP
NS IS
IP
where,VP&VS primary & secondary VoltagesNP &NS primary & secondary NO of turnsIP &IS primary & secondary currents
In the above expression we notice that when the voltage increases the current will decrease, andvisa versa.
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Example.
In the above transformer we are given the primaryvoltage and number of turns. Also, we have beengiven the secondary coils voltage and the expectedoutput current.
Calculate the primary current and the secondarycoils number of turns.
VP 230v , NP 400turns, VS 12v , IS 0.4A
VPVS NP
NS
23012 400
Ns
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PRIMARY SECONDARY23
0vol
ts
400 Turns
12 V
olts
I=400mA
Ns 40012230
Ns 20.9TurnsVPVS IS
IP
23012 0.4
IP
IP 0.412230
IP 0.021A or 21mANote The voltage has been decreased from primary tosecondary (230v to 12v).The current has increased from primary tosecondary (0.021A to 0.4A).If we consider the power using the simplifiedexpression:
Power V IPowerP 230 0.021 4.83wattPowerS 12 0.4 4.8watt
We can see that the power does not change allthat has changed is the V/I ratio or form theelectrical power takes.
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Questions 17.1) Calculate the secondary voltage & secondarycurrent of a transformer which has an input voltageof 220 volts and an input current of 0.2 amps. If theratio of turns is given as Primary 1000 turns,Secondary 150 turns.
2) Calculate the required input current (primary) ina transformer if it is used to drop 240 volts to 14volts, with an output current (secondary) of 150mA.
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3) Calculate the secondary voltage and secondarycurrent of a transformer that has 600 turns in itsprimary, which is supplied with 230 volts at 12amps. If it has 48 turns in its secondary coil.
4) List 3 piece of equipment or electricalappliances that use some sort of transformer,either internal or external. Giving the primary andsecondary voltages.
Secondary VPrimary VEquipment/Applience
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Self-Inductance.
As well as mutual inductance found between twocoils, a back emf also occurs when the fluxchanges within a coil. This effect is known asself-inductance and gives rise to an importantcomponent the induction coil. This componentfinds many uses including control circuits for light,radio transmission etc.
E LdIdt
where,E back emf voltsL INDUCTANCE henryHdIdt rate of change of current
A circuit is said to have an Inductance L of henry ifone volt is induced by a rate of change of currentdI/dt of one ampere per second.
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Example.
A coil has a self-inductance of 25 mH and issubject to a rate of change of current of 600 A/s.What is the value of the back emf.
E LdIdt
E 25 103 600
E 15volts
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Questions 18.1) Calculate the rate of change of current requiredto produce a back emf of 10 volts, from a coil whohas a self-inductance of 60 mH.
2) Calculate the size of inductor required (L), toproduce a back emf of 150 volts, if the currentsrate of change is 2000As-1.
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