11.1 – Electromagnetic induction Essential idea: The majority of electricity generated throughout the world is generated by machines that were designed to operate using the principles of electromagnetic induction. Nature of science: Experimentation: In 1831 Michael Faraday, using primitive equipment, observed a minute pulse of current in one coil of wire only when the current in a second coil of wire was switched on or off but nothing while a constant current was established. Faraday’s observation of these small transient currents led him to perform experiments that led to his law of electromagnetic induction. (1.8)

Understandings: • Electromotive force (emf) • Magnetic flux and magnetic flux linkage • Faraday’s law of induction • Lenz’s law

Applications and skills: • Describing the production of an induced emf by a changing magnetic flux and within a uniform magnetic field • Solving problems involving magnetic flux, magnetic flux linkage and Faraday’s law • Explaining Lenz’s law through the conservation of energy

Theory of knowledge: • Terminology used in electromagnetic field theory is extensive and can confuse people who are not directly involved. What effect can lack of clarity in terminology have on communicating scientific concepts to the public?

Utilization: • Applications of electromagnetic induction can be found in many places including transformers, electromagnetic braking, geophones used in seismology, and metal detectors

Aims: • Aim 2: the simple principles of electromagnetic induction are a powerful aspect of the physicist’s or technologist’s armoury when designing systems that transfer energy from one form to another

Guidance: Quantitative treatments will be expected for straight conductors moving at right angles to magnetic fields and rectangular coils moving in and out of fields and rotating in fields Qualitative treatments only will be expected for fixed coils in a changing magnetic field and ac generators

Data booklet reference f = BA cos q e = - N(Dj/Dt) e = Bvl e = BvlN

E.M.F [Electromotive force]: It provides electrical force to charges to drive the circuit. It is measured in V[volts]. A voltage generated by battery or by a magnetic force.
Force on a current in magnetic field F = BIL sin q (theta) where the units of B are T, tesla.

The flux density of a field of one tesla is therefore defined as the force per unit length on a wire carrying a current of one ampere at right angles to the field. To have an induced e.m.f, a coil must cut through the magnetic field lines or a magnet must cut through the magnetic field lines of a coil. An emf is induced in a conductor whenever magnetic field lines (magnetic flux) are crossed.

The magnitude of the induced e.m.f depending on

The rate/speed in which the magnet moves

The strength of the magnet

The number of coils of the solenoid

Dynamo Effect: The induced e.m.f as a result of a conductor moving in and out of a magnetic field. Voltage is generated whenever a wire is moved in or out of a magnetic field.

Faraday's Electromagnetic Lab** from Phet

Click to Run

Predict the direction of the magnetic field for different locations around a bar magnet and an electromagnet.

Compare and contrast bar magnets and electromagnets.

Identify the characteristics of electromagnets that are variable and what effects each variable has on the magnetic field's strength and direction.

Relate magnetic field strength to distance quantitatively and qualitatively.

Compare and contrast how both a light bulb and voltmeter can be used to show characteristics of the induced current.

Magnetic flux (ɸ) is the strength of the magnetic field cutting the field. B, flux density, and the unit area, A : When the magnetic flux passing through a surface as BA where A is the area of the surface at right angles to the field; ɸ = BAcosϴ Magnetic flux (φ) is measured in webers (Wb) where 1Wb = 1Tm2

If instead of a single loop we make a coil of N loops, the flux F through each loop is “linked” to each of the other loops in what is termed flux linkage.
Each loop produces its own emf, and the emfs from each loop add to the total emf. (Note that an emf is only produced while the flux is changing.)
Flux linkage ( Φ ) is the product of the flux Φ and the number of coils in the wire N. Φ = Nɸ

PRACTICE 1: What is the flux linkage in a coil of 15 turns and area 25cm2 in a field of strength 5T?Note: Area must be in m2 and the conversion factor is 10,000cm2 to 1m2. SOLUTION 1 from www.s-cool.co.uk : Φ = nφ = nBA = 15 x 5 x 25 x 10-4 = 0.1875 Wb.

Faraday's law
The magnitude of an induced emf is proportional to the rate of change of flux linkage (where flux linkage = Nɸ). Faraday’s law states that the emf induced in a coil is equal to the rate of change in the flux linkage in the coil. e = N DF/Dt

PRACTICE 2: Find the magnetic flux in each case. In each case the strength of the B-field is 1.5 T and the area is 0.20 m2. SOLUTION 2: Use F = BA cos ϴ (1) ϴ is 0º so F = 1.5(0.20)cos 0º = 0.30 Tm2 (2) ϴ is 30º so F = 1.5(0.20)cos 30º = 0.26 Tm2 (3) ϴ is 90º so F = 1.5(0.20)cos 90º = 0.0 Tm2 (4) ϴ is 150º so F = 1.5(0.20)cos150º = - 0.26 Tm2 (5) ϴ is 180º so F = 1.5(0.20)cos180º = - 0.30 Tm2

Lenz's law: The direction of an induced e.m.f. opposes the change causing it. German physicist Heinrich Lenz observed that the direction of the induced current in a conductor is always such as to oppose the motion which produced it. This is called Lenz's Law. The direction of current in the coil generated in a direction which opposes the movement of the approaching magnetic field.

PRACTICE 1: Watch the experiment. The maximum voltage is about 18 V. Image from https://fisitech.wordpress.com
(a) What would be the effect, if any, of reversing the magnetic so that the south pole goes in first?
(b) What would be the effect of doubling the oscillation speed of the magnet?
(c) At the original oscillation rate, what would you predict the voltage induced in a single loop to be?
(d) If there were 150 loops, what would the voltage be?
(e) Determine the direction of the induced current in the first loop of the coil while the magnet is moving right.

SOLUTION 1:
(a) The sign of the flux would be reversed so that the meter would reverse. Thus on moving the magnet to the right the meter would deflect left.
(b) Since the Dt in DF/Dt would be cut in half, so the emf would double to about 36 V.
(c) From e = NDF/ Dt we see that
18 = 7( DF/ Dt )
2.6 V = DF/ Dt (the emf for each loop)
(d) 150 x 2.6 = 390 V
(e) Lenz’s law states that the induced current will try to oppose the flux increase.
Since the B-field is increasing right, the B-field created by the induced current will point left.
Using the right hand rule for coils, the current should flow anticlockwise as seen from the left.

Induced emf (motional emf)

e =Bvℓ

EXAMPLE 2) The emf e induced in a straight conductor of length ℓ moving at velocity v through a magnetic field of strength B:
Derive the equation.
SOLUTION 2:
Note that since v B then f = 90º: F = qvB sin f = qvB sin 90º = qvB
E = V / ℓ and that F = qE = qV / ℓ
Since qV / ℓ = F = qvB,
qV / ℓ = qvB
V = Bvℓ = e

The work done to move an unit charge from top to bottom in the wire is W = F ℓ and the force is qvB, so the work done is qvB x ℓ and thus the work done per unit charge is Bvℓ

11.2 – Power generation and transmission
Essential idea: Generation and transmission of alternating current (ac) electricity has transformed the world. Nature of science: Bias: In the late 19th century Edison was a proponent of direct current electrical energy transmission while Westinghouse and Tesla favoured alternating current transmission. The so called “battle of currents” had a significant impact on today’s society. (3.5)

Understandings: • Alternating current (ac) generators • Average power and root mean square (rms) values of current and voltage • Transformers • Diode bridges • Half-wave and full-wave rectification

Applications and skills: • Explaining the operation of a basic ac generator, including the effect of changing the generator frequency • Solving problems involving the average power in an ac circuit • Solving problems involving step-up and step-down transformers • Describing the use of transformers in ac electrical power distribution • Investigating a diode bridge rectification circuit experimentally • Qualitatively describing the effect of adding a capacitor to a diode bridge rectification circuit

Guidance: • Calculations will be restricted to ideal transformers but students should be aware of some of the reasons why real transformers are not ideal (for example: flux leakage, joule heating, eddy current heating, magnetic hysteresis) • Proof of the relationship between the peak and rms values will not be expected

International-mindedness: • The ability to maintain a reliable power grid has been the aim of all governments since the widespread use of electricity started

Theory of knowledge: • There is continued debate of the effect of electromagnetic waves on the health of humans, especially children. Is it justifiable to make use of scientific advances even if we do not know what their long-term consequences may be?

Aims: • Aim 6: experiments could include (but are not limited to): construction of a basic ac generator; investigation of variation of input and output coils on a transformer; observing Wheatstone and Wien bridge circuits • Aim 7: construction and observation of the adjustments made in very large electricity distribution systems are best carried out using computer-modelling software and websites • Aim 9: power transmission is modelled using perfectly efficient systems but no such system truly exists. Although the model is imperfect, it renders the maximum power transmission. Recognition of, and accounting for, the differences between the “perfect” system and the practical system is one of the main functions of professional scientists

Data booklet reference:

D.C.(Direct current): An electric current flowing in one direction only. / Acontinuous electric current that flows in one direction only, without substantial variation in magnitude.

A.C.(Alternating current): An electric current that reverses direction in a circuit at regular intervals. / A continuous electric current that periodically reverses direction, usually sinusoidally.

Questions 1. One design of a byclcle dynamo does not produce a big enough voltage. How could it be increased? 2. Explain why generators need slip rings and brushes? 3. What makes the generators in a power station turn around? Electromagnetic induction ac generator and motor.pptx Electromagnetic induction worksheet from Mr.Lin
PRACTICE 2:What is the effect of increasing the frequency of the generator on the induced emf?
SOLUTION 2: Increasing the frequency of a generator increases the induced emf If we decreaseDt then we will increase the induced emf but if we decrease Dt then we decrease Tthe period of rotation (time for each revolution) and thus as T decreases, f increases (f = 1 /T) The average power in an ac circuit
The RMS value is the effective value of a varying voltage or current. It is the equivalent steady DC (constant) value which gives the same effect. For example, a lamp connected to a 6V RMS AC supply will shine with the same brightness when connected to a steady 6V DC supply. (The root mean squared value extracted from the diploma in Engneering) The RMS average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 peak value

EXAMPLE 3) State that the amplitude of the sinusoidal power supply will be. The r.m.s power supply in the U.K. is 240 V.
SOLUTION 3:
As the rms value of any sinusoidal waveform taken across an interval equal to one period is 0.707 × amplitude of the waveform. Where 0.707 is an approximation of 1/ √ 2).
Vpeak = √ 2 x Vrms
Vpeak = 240 × √ 2 = 339.4 V

A basic transformer with a soft iron-core, as used for voltage transformations

Vp: Primary voltage,
Vs: Secondary voltage,
Np: Number of turns in primary coil,
Ns: Number of turns in secondary coil
Ip : Current in primary coil
Is : Current in secondary coil

For a coil of N loops rotating in a constant magnetic field we thus have the induced emf

e = NBAwsinwt w The angular frequency (measured in radians per second)
N the number of loops
B constant magnetic field
EXAMPLE 2: A 250-turn coil of wire has dimensions 2.5 cm by 3.2 cm. It is rotating at a frequency of 60 Hz in a constant magnetic field having a strength of 1.5 T
(a) What is its peak voltage e0?
(b) What is the time dependence of its emf? SOLUTION 2: w = 2pf = 2p(60) = 120p
(a) A = (0.025)(0.032) = 0.00080 m2. Then e0 = NBAw = 250(1.5)(0.00080)120p = 110 V
(b) e = NBAwsinwt =110sin(120pt) The advantages of high-voltage transmission Transmitting electricity over long distances without wasting energy is difficult. The advantage of transferring electricity at high voltages is to minimize the heat energy loss during the transmission. To transfer electricity over long distances at a given power, you either need a high voltage and low current or a low voltage and high current, since power is given by the equation: P=VI The problem with transferring electricity at a high current is that you need a very wide cable to carry the huge amount of current to reduce the resistance of the wire. This makes it more expensive that using a transformer to step-up the voltage, which in turn reduces the current flow. With a low current, the wires can be thinner, making it more cost-beneficial. When the electricity reaches to our household, transformers are used to step-down the high voltage of it.

The principle of operation of a transformer An A.C. current in the primary coil produces an alternating magnetic field. The change of magnetic field within the core of wire will induced the voltage in the secondary coil.

Vp Ip = Vs Is (for 100% efficiency) POWERinput = POWERoutput Primary power supplied is equal to Secondary power supplied

Voltage(V)primary x Current(I)primary = Voltage(V)secondary x Current(I)secondary

Assuming that there is no energy loss to the surroundings and the circuit is 100% efficient.

Losses in transmission power
In addition to power losses associated with resistance of power lines (due to heat), there are also losses associated with non-ideal transformers:
1)Resistance of the windings of a transformer result in the transformer heating up.
2)Eddy currents (www.rpinceton.edu) are unwanted currents induced in the iron core. The currents are reduced by laminating the core into thin electrically insulated strips.
3)Hysteresis losses cause the iron core to heat up due to continued changes in magnetism.
4)Flux losses are caused by magnetic ‘leakage’.

Efficiency of transformers

The actual watts of power lost can be determined (in each winding) by squaring the amperes and multiplying by the resistance in ohms of the winding ( P = I2R ).

Iron losses, also known as hysteresis is the lagging of the magnetic molecules within the core, in response to the alternating magnetic flux. This lagging (or out-of-phase) condition is due to the fact that it requires power to reverse magnetic molecules; they do not reverse until the flux has attained sufficient force to reverse them.

Their reversal results in friction, and friction produces heat in the core which is a form of power loss. Hysteresis within the transformer can be reduced by making the core from special steel alloys.

The intensity of power loss in a transformer determines its efficiency.
Article taken from www.electronics-tutorials.ws

More Magnets -Sixty Symbols Published on 6 Nov 2012 including information on Eddy current

World's First Electric Generator Veritasium Published on 6 Nov 2012 Eddy current explained

Eddy current: An electric current induced in a massive conductor, such as the core of an electromagnet, transformer, etc. by an alternating magnetic field. Also called: Foucault current

Hysteresis: A lag in response exhibited by a body in reacting to changes in forces, especially magnetic forces, acting upon it.

Power losses in cables are lower when the voltage is high The major energy loss happens when current is high. Due to the flow of electrons in the wire, there is heat energy loss while transferring electrical energy in the circuit . The larger the current flowing through the wire, the bigger the energy loss as heat energy wasted to its surroundings. To reduce the energy loss, step up voltage is used in power transmission. Step-up: has more turns on the output coil than the input so the output voltage is greater than the input. These are used in power transmission (from power station to towns). Step-down: has less turns in the output than input coil so the output voltage is less than the input. These are used in electrical appliances to lessen the mains electricity before reaching the appliance

Rectification is the process of turning alternating current (AC) into direct current (DC). There are two types of rectification:
1)Half-wave rectification: A single diode will convert ac into dc. Electrical energy that is available in the negative cycle of the ac is not utilized.

2)Full-wave rectification: A diode bridge (using four diodes) utilizes all the electrical energy during a complete cycle.

Diode bridges have some very specific characteristics:

current always flows through the load (resistor) in the one direction

diodes on parallel sides of the bridge point in the same direction

the ac signal is fed to the points where opposite ends of the diodes join

EXAMPLE 2) Draw the circuit diagram for half-wave rectification

A diode bridge rectification

EXAMPLE 3) Draw the Rectifier output vs Time graph for full-wave rectification

Smoothing vlotage output
Diode bridges provide a dc but it still pulsates. To achieve a steady voltage, a smoothing device is required such as a capacitor. The capacitor is acting as a short-term store of electrical energy. The capacitor is constantly charging and discharging. The charging and discharging process of the capacitor redudces the ripple of the voltage output.

11.3 – Capacitance Essential idea: Capacitors can be used to store electrical energy for later use. Nature of science: Relationships: Examples of exponential growth and decay pervade the whole of science. It is a clear example of the way that scientists use mathematics to model reality. This topic can be used to create links between physics topics but also to uses in chemistry, biology, medicine and economics. (3.1)

Understandings: • Capacitance • Dielectric materials • Capacitors in series and parallel • Resistor-capacitor (RC) series circuits • Time constant

Applications and skills: • Describing the effect of different dielectric materials on capacitance • Solving problems involving parallel-plate capacitors • Investigating combinations of capacitors in series or parallel circuits • Determining the energy stored in a charged capacitor • Describing the nature of the exponential discharge of a capacitor • Solving problems involving the discharge of a capacitor through a fixed resistor • Solving problems involving the time constant of an RC circuit for charge, voltage and current

International-mindedness: • Lightning is a phenomenon that has fascinated physicists from Pliny through Newton to Franklin. The charged clouds form one plate of a capacitor with other clouds or Earth forming the second plate. The frequency of lightning strikes varies globally, being particularly prevalent in equatorial regions. The impact of lightning strikes is significant, with many humans and animals being killed annually and huge financial costs to industry from damage to buildings, communication and power transmission systems, and delays or the need to reroute air transport.

Utilization: • The charge and discharge of capacitors obeys rules that have parallels in other branches of physics including radioactivity (see Physics sub-topic 7.1)

Aims: • Aim 3: the treatment of exponential growth and decay by graphical and algebraic methods offers both the visual and rigorous approach so often characteristic of science and technology • Aim 6: experiments could include (but are not limited to): investigating basic RC circuits; using a capacitor in a bridge circuit; examining other types of capacitors; verifying time constant

Guidance: • Only single parallel-plate capacitors providing a uniform electric field, in series with a load, need to be considered (edge effect will be neglected) • Problems involving the discharge of capacitors through fixed resistors need to be treated both graphically and algebraically • Problems involving the charging of a capacitor will only be treated graphically • Derivation of the charge, voltage and current equations as a function of time is not required

Data booklet reference:

Capacitance

Capacitors are devices that store charge. Capacitance is defined in terms of charge storage. The stored charge, q, is proportional to the pd across the capacitor, V, and the capacitance, C and the units are in Farads.

C = q / V where C = Capacitance (F, Farads: The units for capacitance are Coulombs per volt which are called farads and abbreviated F)
q = magnitude of charge stored on each plate (C, Coulombs)
V = Potential difference applied to the plate (V, Volts)
EXAMPLE: A 1.50 V cell is connected to a 275 mF capacitor in the circuit shown. How much charge is stored on the capacitor’s plates?
SOLUTION: Use C = q / V
q = CV
q = CV = 275 x 10-6 x 1.50 = 4.13 x 10-4 C

A battery will transport charge from one plate to the other until the voltage produced by the charge buildup is equal to the battery voltage. 1 Farad is equal to 1 coulomb per 1 volt.
Since charge cannot be added or taken away from the conductor between series capacitors, the net charge there remains zero.
You store less charge on series capacitors than you would on either one of them alone with the same voltage!

Capacitors are in parallel, they each have the cell’s voltage V.
V = V1= V2

From C = q / V we get q = CV,
thus q = CV, q1= C1V1, and q2= C2V2

Due to the conservation of charge, q = q1 + q2:
CV = C1V1 + C2V2
CV = C1V + C2V
Thus C = C1 + C2

Capacitors in series, they each have the same charge q.
Conservation of energy tells us that V = V1 + V2

From C = q / V we get V = q / C
Thus V = q / C, V1 = q1 / C1, and V2 = q2 / C2

so that
q / C = q1 / C1 + q2 / C2
1 /C = 1 / C1 + 1 / C2

EXAMPLE 5: A 1.50-V cell is connected to C1 = 275 mF and C2 = 38 mF in the circuit.
(a) What value should a single replacement capacitor have?
SOLUTION 5a: The capacitors are in parallel.
C = C1 + C2 = 275 x 10^-6 + 38 x 10^-6 = 303 x 10^-6 F

(b) How much charge has the battery placed on the capacitors?
SOLUTION 5b:
q = CV = 303 x 10^-6 x 1.50 = 455 x 10^-6 C

EXAMPLE 6: A 1.50 V cell is connected to C1 = 275 mF and C2 = 38 mF in the circuit.
(a) What value should a single replacement capacitor have?
SOLUTION 6a: The capacitors are in series.
1 / C = 1 / C1 + 1 / C2
1 / 275 x 10^-6 + 1 / 38 x 10^-6 = 29952.2 C = 1 / 29952.2 = 3.34 x 10^-5F

(b) How much charge has the battery placed on each capacitor? What are their voltages?
SOLUTION 6b: Series capacitors have the same charge
q = CV = 3.34 x 10^-5 x 1.50 = 5.01 x 10^-5C

V1 = q1 / C15.01 x 10^-5/ 275 x 10^-6 = 0.18 V
V2 = q2 / C2 = 5.01 x 10^-5 / 38 x 10^-6 = 1.32 V

Charge leakage can occur through air or vacuum. So what manufacturers do is they place a non-conductive material called a dielectric between the plates. This will both reduce arcing and the electric force, increasing the capacity of the capacitor C = e A / d The epsilon e is the permittivity of the dielectric. Recall that the permittivity of free space (and air) is e0 = 8.85 x 10-12 F m-1
q = CV and V = Ed,
q / C = V = Ed so that C = q / Ed and thus, 'The smaller the electric field is, the larger the capacitance'.

Stored Energy in Capacitors

From C = q / V , V = (1 /C) q

Image from IOP and PhysicsLab A more general approach says that in moving the charge ΔQ, the pd does not change significantly, so the energy transformed is V × ΔQ. But this is just the area of the narrow strip, so the total energy will be the triangular area under the graph. i.e. Energy stored in the capacitor = 1/2 QV = 1/2 CV2 = 1/2 Q2/C

EXAMPLE 7: A 10 mF capacitor is charged to 20 V. How much energy is stored? SOLUTION 7a: Energy stored = 1/2 CV2 = 2000 mJ b) Calculate the energy is stored at 10 V (i.e. at half the voltage). SOLUTION 7b: 500 mJ, one quarter of the previous value, since it depends on V2

When the battery is first connected there is no [charge] on the [capacitor] so the pd across the [capacitor] is zero. This means that all the potential is dropped across the resistor. The current in the circuit is proportional to the pd across the resistor so has a [maximum] value. As time progresses the charge on the [capacitor] increases so the pd across the [resistor] goes down resulting in a reduction in [current] . When fully charged the pd across the capacitor equals the [EMF] of the battery so the pd across the resistor is [zero] and no [current] flows.

Essential idea: The majority of electricity generated throughout the world is generated by machines that were designed to operate using the principles of electromagnetic induction.

Nature of science:Experimentation: In 1831 Michael Faraday, using primitive equipment, observed a minute pulse of current in one coil of wire only when the current in a second coil of wire was switched on or off but nothing while a constant current was established. Faraday’s observation of these small transient currents led him to perform experiments that led to his law of electromagnetic induction. (1.8)

Understandings:• Electromotive force (emf)

• Magnetic flux and magnetic flux linkage

• Faraday’s law of induction

• Lenz’s law

Applications and skills:• Describing the production of an induced emf by a changing magnetic flux and within a uniform magnetic field

• Solving problems involving magnetic flux, magnetic flux linkage and Faraday’s law

• Explaining Lenz’s law through the conservation of energy

Theory of knowledge:• Terminology used in electromagnetic field theory is extensive and can confuse people who are not directly involved. What effect can lack of clarity in terminology have on communicating scientific concepts to the public?

Utilization:• Applications of electromagnetic induction can be found in many places including transformers, electromagnetic braking, geophones used in seismology, and metal detectors

Aims:• Aim 2: the simple principles of electromagnetic induction are a powerful aspect of the physicist’s or technologist’s armoury when designing systems that transfer energy from one form to another

Guidance:Quantitative treatments will be expected for straight conductors moving at right angles to magnetic fields and rectangular coils moving in and out of fields and rotating in fields

Qualitative treatments only will be expected for fixed coils in a changing magnetic field and ac generators

Data booklet referencef = BA cos q

e = - N(Dj/Dt)

e = Bvl

e = BvlN

E.M.F [Electromotive force]:It provides electrical force to charges to drive the circuit. It is measured in

V[volts]. A voltage generated by battery or by a magnetic force.Force on a current in magnetic field F = BIL sin q (theta) where the units of B are T, tesla.

The flux density of a field of one tesla is therefore defined as the force per unit length on a wire carrying a current of one ampere at right angles to the field.

To have an induced e.m.f, a coil must cut through the magnetic field lines or a magnet must cut through the magnetic field lines of a coil. An emf is induced in a conductor whenever magnetic field lines (magnetic flux) are crossed.

The magnitude of the induced e.m.f depending on

Dynamo Effect:The induced e.m.f as a result of a conductor moving in and out of a magnetic field. Voltage is generated whenever a wire is moved in or out of a magnetic field.

Faraday's Electromagnetic Lab** from PhetMagnetic flux (ɸ) is the strength of the magnetic field cutting the field. B, flux density, and the unit area, A :

When the magnetic flux passing through a surface as BA where A is the area of the surface at right angles to the field;

ɸ = BAcosϴMagnetic flux (φ) is measured in webers (Wb) where 1Wb = 1Tm2

Image from www.schoolphysics.co.uk

If instead of a single loop we make a coil of N loops, the flux F through each loop is “linked” to each of the other loops in what is termed flux linkage.

Each loop produces its own emf, and the emfs from each loop add to the total emf. (Note that an emf is only produced while the flux is changing.)

Flux linkage ( Φ ) is the product of the flux Φ and the number of coils in the wire N.

Φ = Nɸ

What is the flux linkage in a coil of 15 turns and area 25cm2 in a field of strength 5T?Area must be in m2 and the conversion factor is 10,000cm2 to 1m2.Note:SOLUTION 1 from www.s-cool.co.uk :

Φ = nφ = nBA = 15 x 5 x 25 x 10-4 = 0.1875 Wb.

Faraday's lawThe magnitude of an induced emf is proportional to the rate of change of flux linkage (where flux linkage = Nɸ).

Faraday’s law states that the emf induced in a coil is equal to the rate of change in the flux linkage in the coil.

e = N DF/Dt

PRACTICE 2:

Find the magnetic flux in each case. In each case the strength of the B-field is 1.5 T and the area is 0.20 m2.

SOLUTION 2: Use F = BA cos

ϴ(1)

ϴ is0º so F = 1.5(0.20)cos 0º = 0.30 Tm2(2)

ϴ is30º so F = 1.5(0.20)cos 30º = 0.26 Tm2(3)

ϴ is90º so F = 1.5(0.20)cos 90º = 0.0 Tm2(4)

ϴ is150º so F = 1.5(0.20)cos150º = - 0.26 Tm2(5)

ϴ is180º so F = 1.5(0.20)cos180º = - 0.30 Tm2Lenz's law: The direction of an induced e.m.f. opposes the change causing it.German physicist Heinrich Lenz observed that the direction of the induced current in a conductor is always such as to oppose the motion which produced it. This is called Lenz's Law.

The direction of current in the coil generated in a direction which opposes the movement of the approaching magnetic field.Direction of the induced e.m.f from faculty.www.edu

Reason for Opposing, Cause of Induced Current in Lenz's Law? from www.electrical4u.com

Image from www.physchem.co.za

Lenz's law worksheet from TES prepared by robcowen robcowen

Lenz's law lab worksheet from myslu.stlawu.edu

Faraday’s Law and Lenz’s Law can be combined to relate the emf generated in a coil of N turns with a rate of change of flux:

ε =- N (Δɸ/Δt)Flux and induced EMF graphs Dave McBain Published on 31 Jul 2014 youtube.com

Lenz's law and Faraday's law practice questions from spot.pcc.edu

Induced emf questions form www.s-cool.co.uk and physics.bu.edu

PRACTICE 1: Watch the experiment. The maximum voltage is about 18 V.

Image from https://fisitech.wordpress.com

(a) What would be the effect, if any, of reversing the magnetic so that the south pole goes in first?

(b) What would be the effect of doubling the oscillation speed of the magnet?

(c) At the original oscillation rate, what would you predict the voltage induced in a single loop to be?

(d) If there were 150 loops, what would the voltage be?

(e) Determine the direction of the induced current in the first loop of the coil while the magnet is moving right.

SOLUTION 1:

(a) The sign of the flux would be reversed so that the meter would reverse. Thus on moving the magnet to the right the meter would deflect left.

(b) Since the Dt in DF/Dt would be cut in half, so the emf would double to about 36 V.

(c) From e = NDF/ Dt we see that

18 = 7( DF/ Dt )

2.6 V = DF/ Dt (the emf for each loop)

(d) 150 x 2.6 = 390 V

(e) Lenz’s law states that the induced current will try to oppose the flux increase.

Since the B-field is increasing right, the B-field created by the induced current will point left.

Using the right hand rule for coils, the current should flow anticlockwise as seen from the left.

## Induced emf (motional emf)

## e

EXAMPLE 2) The emf e induced in a straight conductor of length ℓ moving at velocity v through a magnetic field of strength B:=BvℓDerive the equation.

SOLUTION 2:

Note that since v B then f = 90º:

F = qvB sin f = qvB sin 90º = qvB

E = V / ℓ and that F = qE = qV / ℓ

Since qV / ℓ = F = qvB,

qV / ℓ = qvB

V = Bvℓ = e

11.2 – Power generation and transmission

Essential idea: Generation and transmission of alternating current (ac) electricity has transformed the world.

Nature of science:Bias: In the late 19th century Edison was a proponent of direct current electrical energy transmission while Westinghouse and Tesla favoured alternating current transmission. The so called “battle of currents” had a significant impact on today’s society. (3.5)

Understandings:• Alternating current (ac) generators

• Average power and root mean square (rms) values of current and voltage

• Transformers

• Diode bridges

• Half-wave and full-wave rectification

Applications and skills:• Explaining the operation of a basic ac generator, including the effect of changing the generator frequency

• Solving problems involving the average power in an ac circuit

• Solving problems involving step-up and step-down transformers

• Describing the use of transformers in ac electrical power distribution

• Investigating a diode bridge rectification circuit experimentally

• Qualitatively describing the effect of adding a capacitor to a diode bridge rectification circuit

Guidance:• Calculations will be restricted to ideal transformers but students should be aware of some of the reasons why real transformers are not ideal (for example: flux leakage, joule heating, eddy current heating, magnetic hysteresis)

• Proof of the relationship between the peak and rms values will not be expected

International-mindedness:• The ability to maintain a reliable power grid has been the aim of all governments since the widespread use of electricity started

Theory of knowledge:• There is continued debate of the effect of electromagnetic waves on the health of humans, especially children. Is it justifiable to make use of scientific advances even if we do not know what their long-term consequences may be?

Aims:• Aim 6: experiments could include (but are not limited to): construction of a basic ac generator; investigation of variation of input and output coils on a transformer; observing Wheatstone and Wien bridge circuits

• Aim 7: construction and observation of the adjustments made in very large electricity distribution systems are best carried out using computer-modelling software and websites

• Aim 9: power transmission is modelled using perfectly efficient systems but no such system truly exists. Although the model is imperfect, it renders the maximum power transmission. Recognition of, and accounting for, the differences between the “perfect” system and the practical system is one of the main functions of professional scientists

Data booklet reference:D.C.(Direct current):

An electric current flowing in one direction only. / A continuous electric current that flows in one direction only, without substantial variation in magnitude.

A.C.(Alternating current):

An electric current that reverses direction in a circuit at regular intervals. / A continuous electric current that periodically reverses direction, usually sinusoidally.

## AC and DC

A rotating-coil generator and the use of slip ringsImage of

a rotating-coil generatorfrom resources.teachnet.ieWhile the coil rotating in the magnetic field, the slip rings and brushes allow the coil to rotate freely.

Generators from schoolphysics.co.uk

generators_.doc

Questions

1. One design of a byclcle dynamo does not produce a big enough voltage. How could it be increased?

2. Explain why generators need slip rings and brushes?

3. What makes the generators in a power station turn around?

Electromagnetic induction ac generator and motor.pptx

Electromagnetic induction worksheet from Mr.Lin

PRACTICE 2:What is the effect of increasing the frequency of the generator on the induced emf?

SOLUTION 2: Increasing the frequency of a generator increases the induced emf

If we decreaseDt then we will increase the induced emf but if we decrease Dt then we decrease Tthe period of rotation (time for each revolution) and thus as T decreases, f increases (f = 1 /T)

The average power in an ac circuit

The RMS value is the effective value of a varying voltage or current. It is the equivalent steady DC (constant) value which gives the same effect. For example, a lamp connected to a 6V RMS AC supply will shine with the same brightness when connected to a steady 6V DC supply. (The root mean squared value extracted from the diploma in Engneering)

The RMS average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 peak value

Image taken from Institude of Physics

EXAMPLE 3) State that the amplitude of the sinusoidal power supply will be. The r.m.s power supply in the U.K. is 240 V.

SOLUTION 3:

As the rms value of any sinusoidal waveform taken across an interval equal to one period is 0.707 × amplitude of the waveform. Where 0.707 is an approximation of 1/ √ 2).

Vpeak = √ 2 x Vrms

Vpeak = 240 × √ 2 = 339.4 V

A basic transformer with a soft iron-core, as used for voltage transformations

Image from www.polytechnichub.com

(Ip / Is)

where,

Vs: Secondary voltage,

Np: Number of turns in primary coil,

Ns: Number of turns in secondary coil

Ip : Current in primary coil

Is : Current in secondary coil

e = NBAwsinwtw The angular frequency (measured in radians per second)

N the number of loops

B constant magnetic field

EXAMPLE 2: A 250-turn coil of wire has dimensions 2.5 cm by 3.2 cm. It is rotating at a frequency of 60 Hz in a constant magnetic field having a strength of 1.5 T

(a) What is its peak voltage e0?

(b) What is the time dependence of its emf?

SOLUTION 2:

w = 2pf = 2p(60) = 120p

(a) A = (0.025)(0.032) = 0.00080 m2. Then

e0 = NBAw = 250(1.5)(0.00080)120p = 110 V

(b) e = NBAwsinwt =110sin(120pt)

The advantages of high-voltage transmission

Transmitting electricity over long distances without wasting energy is difficult. The advantage of transferring electricity at high voltages is to minimize the heat energy loss during the transmission.

To transfer electricity over long distances at a given power, you either need a

high voltage and low currentor alow voltage and high current, since power is given by the equation: P=VIThe problem with transferring electricity at a high current is that you need a very wide cable to carry the huge amount of current to reduce the resistance of the wire. This makes it more expensive that using a transformer to step-up the voltage, which in turn reduces the current flow. With a low current, the wires can be thinner, making it more cost-beneficial. When the electricity reaches to our household, transformers are used to step-down the high voltage of it.

The principle of operation of a transformer

An A.C. current in the primary coil produces an alternating magnetic field. The change of magnetic field within the core of wire will induced the voltage in the secondary coil.

Vp Ip = Vs Is (for 100% efficiency)

POWERinput = POWERoutput

Primary power supplied is equal to Secondary power supplied

Assuming that there is no energy loss to the surroundings and the circuit is 100% efficient.Voltage(V)primary x Current(I)primary = Voltage(V)secondary x Current(I)secondaryLosses in transmission power

In addition to power losses associated with resistance of power lines (due to heat), there are also losses associated with non-ideal transformers:

1)Resistance of the windings of a transformer result in the transformer heating up.

2)Eddy currents (www.rpinceton.edu) are unwanted currents induced in the iron core. The currents are reduced by laminating the core into thin electrically insulated strips.

3)Hysteresis losses cause the iron core to heat up due to continued changes in magnetism.

4)Flux losses are caused by magnetic ‘leakage’.

## Efficiency of transformers

The actual watts of power lost can be determined (in each winding) by squaring the amperes and multiplying by the resistance in ohms of the winding ( P = I2R ).Iron losses, also known as hysteresis is the lagging of the magnetic molecules within the core, in response to the alternating magnetic flux. This lagging (or out-of-phase) condition is due to the fact that it requires power to reverse magnetic molecules; they do not reverse until the flux has attained sufficient force to reverse them.

Their reversal results in friction, and friction produces heat in the core which is a form of power loss. Hysteresis within the transformer can be reduced by making the core from special steel alloys.

The intensity of power loss in a transformer determines its efficiency.

Article taken from www.electronics-tutorials.ws

Power losses in cables are lower when the voltage is high

The major energy loss happens when current is high. Due to the flow of electrons in the wire, there is heat energy loss while transferring electrical energy in the circuit . The larger the current flowing through the wire, the bigger the energy loss as heat energy wasted to its surroundings. To reduce the energy loss, step up voltage is used in power transmission.

Step-up: has more turns on the output coil than the input so the output voltage is greater than the input. These are used in power transmission (from power station to towns).Step-down: has less turns in the output than input coil so the output voltage is less than the input. These are used in electrical appliances to lessen the mains electricity before reaching the applianceRectificationis the process of turning alternating current (AC) into direct current (DC). There are two types of rectification:1)Half-wave rectification: A single diode will convert ac into dc. Electrical energy that is available in the negative cycle of the ac is not utilized.

2)Full-wave rectification: A diode bridge (using four diodes) utilizes all the electrical energy during a complete cycle.

EXAMPLE 2) Draw the circuit diagram for half-wave rectification## A diode bridge rectification

EXAMPLE 3) Draw the Rectifier output vs Time graph for full-wave rectificationSmoothing vlotage outputDiode bridges provide a dc but it still pulsates. To achieve a steady voltage, a smoothing device is required such as a capacitor. The capacitor is acting as a short-term store of electrical energy. The capacitor is constantly charging and discharging. The charging and discharging process of the capacitor redudces the ripple of the voltage output.

11.3 – Capacitance

Essential idea: Capacitors can be used to store electrical energy for later use.

Nature of science:Relationships: Examples of exponential growth and decay pervade the whole of science. It is a clear example of the way that scientists use mathematics to model reality.

This topic can be used to create links between physics topics but also to uses in chemistry, biology, medicine and economics. (3.1)

Understandings:• Capacitance

• Dielectric materials

• Capacitors in series and parallel

• Resistor-capacitor (RC) series circuits

• Time constant

Applications and skills:• Describing the effect of different dielectric materials on capacitance

• Solving problems involving parallel-plate capacitors

• Investigating combinations of capacitors in series or parallel circuits

• Determining the energy stored in a charged capacitor

• Describing the nature of the exponential discharge of a capacitor

• Solving problems involving the discharge of a capacitor through a fixed resistor

• Solving problems involving the time constant of an RC circuit for charge, voltage and current

International-mindedness:• Lightning is a phenomenon that has fascinated physicists from Pliny through Newton to Franklin. The charged clouds form one plate of a capacitor with other clouds or Earth forming the second plate. The frequency of lightning strikes varies globally, being particularly prevalent in equatorial regions. The impact of lightning strikes is significant, with many humans and animals being killed annually and huge financial costs to industry from damage to buildings, communication and power transmission systems, and delays or the need to reroute air transport.

Utilization:• The charge and discharge of capacitors obeys rules that have parallels in other branches of physics including radioactivity (see Physics sub-topic 7.1)

Aims:• Aim 3: the treatment of exponential growth and decay by graphical and algebraic methods offers both the visual and rigorous approach so often characteristic of science and technology

• Aim 6: experiments could include (but are not limited to): investigating basic RC circuits; using a capacitor in a bridge circuit; examining other types of capacitors; verifying time constant

Guidance:• Only single parallel-plate capacitors providing a uniform electric field, in series with a load, need to be considered (edge effect will be neglected)

• Problems involving the discharge of capacitors through fixed resistors need to be treated both graphically and algebraically

• Problems involving the charging of a capacitor will only be treated graphically

• Derivation of the charge, voltage and current equations as a function of time is not required

Data booklet reference:

Capacitors are devices that store charge. Capacitance is defined in terms of charge storage.CapacitanceThe stored charge, q, is proportional to the pd across the capacitor, V, and the capacitance, C and the units are in Farads.

C= q / Vwhere

C= Capacitance (F, Farads: The units for capacitance are Coulombs per volt which are called farads and abbreviated F)q = magnitude of charge stored on each plate (C, Coulombs)

V = Potential difference applied to the plate (V, Volts)

EXAMPLE: A 1.50 V cell is connected to a 275 mF capacitor in the circuit shown. How much charge is stored on the capacitor’s plates?

SOLUTION: Use

C= q / Vq =

CVq =

CV = 275 x 10-6 x 1.50 = 4.13 x 10-4 CA battery will transport charge from one plate to the other until the voltage produced by the charge buildup is equal to the battery voltage. 1 Farad is equal to 1 coulomb per 1 volt.

Since charge cannot be added or taken away from the conductor between series capacitors, the net charge there remains zero.

You store less charge on series capacitors than you would on either one of them alone with the same voltage!

Capacitors are in parallel, they each have the cell’s voltage V.

V = V1= V2

From C = q / V we get q = CV,

thus q = CV, q1= C1V1, and q2= C2V2

Due to the conservation of charge, q = q1 + q2:

CV = C1V1 + C2V2

CV = C1V + C2V

Thus C = C1 + C2

Capacitors in series, they each have the same charge q.

Conservation of energy tells us that V = V1 + V2

From

C= q / V we get V = q /CThus V = q /

C, V1 = q1 /C1, and V2 = q2 /C2so that

q /

C= q1 /C1 + q2 /C21

/C= 1 /C1 + 1 /C2(a) What value should a single replacement capacitor have?

SOLUTION 5a: The capacitors are in parallel.

C = C1 + C2 = 275 x 10^-6 + 38 x 10^-6 = 303 x 10^-6 F

(b) How much charge has the battery placed on the capacitors?

SOLUTION 5b:

q = CV = 303 x 10^-6 x 1.50 = 455 x 10^-6 C

EXAMPLE 6: A 1.50 V cell is connected to

C1 = 275 mF andC2 = 38 mF in the circuit.(a) What value should a single replacement capacitor have?

SOLUTION 6a: The capacitors are in series.

1 /

C= 1 /C1 + 1 /C21 / 275 x 10^-6 + 1 / 38 x 10^-6 = 29952.2

C= 1 / 29952.2 = 3.34 x 10^-5F(b) How much charge has the battery placed on each capacitor? What are their voltages?

SOLUTION 6b: Series capacitors have the same charge

q =

CV = 3.34 x 10^-5 x 1.50 = 5.01 x 10^-5CV1 = q1 / C15.01 x 10^-5/ 275 x 10^-6 = 0.18 V

V2 = q2 / C2 = 5.01 x 10^-5 / 38 x 10^-6 = 1.32 V

More practice questions on capacitors from www.kitronik.co.uk

Charge leakage can occur through air or vacuum. So what manufacturers do is they place a non-conductive material called a dielectric between the plates. This will both reduce arcing and the electric force, increasing the capacity of the capacitor C = e A / d

The epsilon e is the permittivity of the dielectric. Recall that the permittivity of free space (and air) is e0 = 8.85 x 10-12 F m-1

q =

CV and V = Ed,q / C = V = Ed so that C = q / Ed and thus, '

The smaller the electric field is, the larger the capacitance'.## Stored Energy in Capacitors

From C= q / V , V = (1/C) qImage from IOP and PhysicsLab

A more general approach says that in moving the charge ΔQ, the pd does not change significantly, so the energy transformed is V × ΔQ. But this is just the area of the narrow strip, so the total energy will be the triangular area under the graph.

i.e. Energy stored in the capacitor = 1/2 QV = 1/2 CV2 = 1/2 Q2/C

EXAMPLE 7:

A 10 mF capacitor is charged to 20 V. How much energy is stored?

SOLUTION 7a:

Energy stored = 1/2 CV2 = 2000 mJ

b) Calculate the energy is stored at 10 V (i.e. at half the voltage).

SOLUTION 7b:

500 mJ, one quarter of the previous value, since it depends on V2

When the battery is first connected there is no

[charge]on the[capacitor]so the pd across the [capacitor] is zero. This means that all the potential is dropped across the resistor. The current in the circuit is proportional to the pd across the resistor so has a [maximum]value. As time progresses the charge on the[capacitor]increases so the pd across the[resistor]goes down resulting in a reduction in[current]. When fully charged the pd across the capacitor equals the[EMF]of the battery so the pd across the resistor is[zero]and no[current]flows.Charging and discharging a capacitor from schoolphysics.co.uk, Capacitor Current Charge Stored from www.s-cool.co.uk

The charging voltage across the capacitor PHY 124 - AC circuits at Stony Brook Physics Laboratory Manuals,

DC Circuits Containing Resistors and Capacitors from philschatz.com

Supplementary Notes for PHY 316 ELECTRICITY & MAGNETISM from bolvan.ph.utexas.edu