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the magnetic flux is defined as
the product of the field magnitude by the area crossed by the field lines

ø =
 B perpendicular(component of B perpendicular to the coil)A
 BAcosθ where θ is the angle between B and the normal to the loop


if the coil has N turns in series then there is
 a flux ø through a coil of N turns and we call Nø the flux linkage
 Nø = NB (perpendicular)A

SI units of flux linkage are
webbers Wb

if the coil has N turns in series then there is a flux through a coil of N turns , and we call this
flux linkage , N thigh

flux linkage =
Nthigh = NBperpendicularA = NBAcos(angle)

provided that the magnetic flux varies in a circuit there will be
a current in the circuit no matter what method is used

generating a current in a magnetic field is called
electromagnetic induction

Faradays law
the emf in a loop of wire is proportional to the rate of change of magnetic flux through the coil

we often call the emf induced by a changing magnetic flux an
induced emf and the current it produces is called an induced current or induction current

Lenz's law
the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop

the induced current tends to maintain
the original flux through the circuit

we can write faradays law of induction mathematically
E =  N delta thigh / delta t

the minus sign indicates the
direction of the emf and is in agreement with lenz's law

motional emf 
right hand rule

to investigate the induction of emf on a moving conductor lets consider a conducting bar of length l sliding along a stationary u shaped conductor that is perpendicular to a uniform magnetic field
 the bar moves with constant speed v in the z direction
 as the bar moves the area enclosed by the loop consisting of the bar and the u shaped conductor increases
 consequently the magnetic flux through the loop increases

the emf developed around the loop is obtained from faradays law
 E = deltathigh/delta t
 E = dela BA/delta t
 E = B delta A/delta t
 E =  Bl delta x / delta t
 delta x / delta t = v
 E = Blv

if the magnetic field is uniform and the loop rotates with a constant angular speed , w , the magnetic flux through the single loop of area A may be expressed as
thigh = BperpendicularA = BAcos(angle) = BAcos(wt)

the flux is a function of time , from faradays law E =  delta thigh/delta t
E = BA delta cos(wt)/delta t

delta cos(wt)/delta t = w sin(wt) so E =
BAwsin(wt)

so generalise the result to N loops
E = NBAwsin(wt)

output of an alternator is
sinusoidal emf

when an alternator is connected to a closed circuit
it produces a sinusoidally ac current

frequency of ac current =
number of cycles per second

time period of ac current =
time for one full cycle


peak value of ac current
the maximum current that is the same value in either direction

peak to peak value
difference of the peak value in the opposite direction

the root mean square value of an ac current
the value of dc that would give the same heating effect as the ac in the same resistor


the mean power supplied to the resistor P=
 IV
 so P mean = Irms x Vrms

one of the most important applications of mutual induction and self induction takes place in a
transfomer

a transformer is a device for
increasing or decreasing ac voltage

transformer consists of
an iron core on which two coils are wound : a primary coil with Np turns and a secondary coil with Ns turns

primary coil connected to an
ac generator

ac in the primary coil establishes a
changing magnetic field in the iron core

iron is easily
magnetised

iron enhances the
magnetic field relative to that in an air coil and guides the field lines to the secondary coil

in a well designed transformer nearly all the
magnetic flux that passes through each turn in the primary coil goes through each turn in the secondary coil .

magnetic field is changing , the flux through each coil is changing consequently
emf is induced in both coils

in the secondary coil E =
Ns delta thigh/delta t

the induced emf arises from
mutila induction

in the primary coil E =
 Np delta thigh/delta t

in the primary coil the induced emf arises from
self induction

the term delta thigh/delta t is the same in both equations since the same flux penetrates each turn of both coils so Es/Ep =
Ns/Np

in a high quality transformer the resistance of the coils are
negligible

therefor the magnitudes of the emf's are
nearly equal to the terminal voltages across the coils


if Ns is greater than Np the
secondary voltage is greater than the primary voltage so we have a step up transformer

a transformer can change the voltage of secondary coil but the conservation of energy requires that
the energy delivered to secondary coil must be the same as the energy delivered to primary coil provided no energy is dissipated on heating coils or is otherwise lost

in a well designed transformer less than ... of the input energy is lost in the form of heat
1%

power is
energy per unit time

assumimg 100% efficiency energy transfer
Ps = Pp

hence
 IpVp = IsVs
 Vp/Vs = Ip/Is = Ns/Np

in fact energy is dissipated so the efficiency of the transformer can be calculated in this form

when a bar magnet is moved relative to a coil of wire connected in a circuit an
electric current is made to flow in the coil and the pd is called induced emf

direction of current depends on direction of
motion of magnet

flemmings right hand rule

induced emf can be increased by
 moving wire/magnet faster
 using a stronger magnet
 more turns in coil
 increasing area of coil
 increasing angle between magnetic field and the normal of the loop

alternate methods of inducing emf
 dynamo (spins a magnet in a coil creating a current)
 an electric motor in reverse (motor normally turns on an axle but turning axle (rotating magnet) would have reverse effect)

in a dynamo energy is transferred from
dynamo to lamp the current through the dynamo coil causes a reaction force on the coil due to the magnet . work must be done to keep the magnet spinning .

energy from the coil to the lamp is equal to
the work done on the coil to keep it spinning assuming no wastage

the rate of transfer of energy from source of emf to the other components of the circuit is equal to
the product of induced emf and current

this is because
 emf is energy transferred from the source per unit charge that passes through the source
 E = V/Q
 current is the charge flow per second Q = It
 E = V/It
 V/It = EQ/It
 VI = EQ x 1/Qt
 VI = E/t

when a north magnet is pushed towards the coil the induced emf causes
 current to flow which creates a magnetic field , the end of the coil nearest to the magnet becomes a north pole which repels the magnet . the quicker you push the magnet into the coil the larger the induced emf , the larger the force opposing the motion
 can't be south because that would attract bar magnet and this would increase kinetic energy and electrical energy of magnet which is impossible

the induced emf is equal to the
rate of change of magnetic flux linkage

when a coil cuts the field lines an
emf is induced

the greater the rate of cutting lines the
greater the emf

when the plane of the coil is parallel to the field lines the flu linkage is
0 the coil doesn't experience magnetic force so no emf induced

rate of change of flux linkage at this point is
max

when the plane of the coil is perpendicular to the field the flux linkage is
maximum but the rate of change of flux linkage is zero

when the plane of the coil is again parallel to the field the rate of change of flux linkage is
again maximum but the sides of the coil are moving through the field in the opposite direction thus an alternating emf is induced

emf is the rate of change of
flux linkage

fixed coil in changing magnetic field
 the B field of the solenoid passes through the small coil
 if the current in the solenoid changes an emf is induced in the small coil
 this is because the magnetic field through the coil changes so the flux linkage through it changes , causing an induced emf
 Nthigh = BAN
 Ndeltathigh = delta BAN
 emf = N deltathigh/delta t = ANdeltaB/delta t

because B is proportional to I in the solenoid the magnitude of the emf is
proportional to the rate of change of current in the solenoid
 F = BIL
 Emf = ANdletaB/deltat

consider a rectangular coil of N turns , length l and a width w moving into a uniform magnetic field of flux density B at constant speed v . suppose the coil enters the field at time t=0
time taken to enter field completely =
coil width / speed = w/v

during this time the flux linkage Nthigh increases from 0 to

therefor the rate of change of flux per second
N deltathigh/delta t = BNlwv/w = BNlv

when the coil is completely in the field
the flux linkage doesn't change so induced emf = 0

a battery powered or hybrid electric vehicle contains an alternator that can be used as
an electric motor

when the alternator is used as an electric motor its driven by the
batteries

when the brakes are applied the alternator is
used to generate electricity which is used to recharge the battery

som kinetic energy is transferred to
electrical energy in the battery

the induced current through the alternator coil creates
a magnetic field that acts against the magnetic field of the alternator , so the alternator experiences a braking force which helps slow the vehicle down

the fuel consumption of a hybrid car is ls less than that of a petrol vehicle this is because
some of the kinetic energy is converted to chemical energy in its battery when the vehicle brakes . the battery supplies this energy to the motor when it takes over from the the petrol engines at low speeds

generators
produce current


motor
current is put in and spins coil producing kinetic energy

regenerative braking uses a
motor first then a generator


B is
magnetic flux density


if got a magnetic flux density vs t graph to find emf
 E = delta thigh/delta t x N
 E = delta BAN/delta t
 so E = gradient x AN

if got a magnetic flux density vs t graph to draw emf graph
 E = delta NAB/delta t
 draw a delta B delta t graph since N and A are constant

a power station alternator has three sets of coils at
120 degrees to one another

each coil produces an alternating emf
120 degrees out of phase with each of the other two emf's

the coils are called the
stators because they are stationary so they don't need slip ring connectors and an electromagnet called the rotor spins between them

the electromagnet is supplied with a current from
a generator

so the turning rotor
induces an emf in each set of stator coils

the three phases are distributed via
transformers and power lines to factories and local sub stations

a local sub station supplies
mains electricity to local premises , a third on each of the three phases this is why your home can sometimes suffer a blackout when other homes nearby still have electricity . This happens when a fault in the local sub station cuts out one phase but not the others

back emf
an emf is induced in the spinning coil of an electric motor because the flux through the coil changes

the induced emf is referred to as back emf because
it acts against the pd applied to the motor in accordance with lens's law

at any instant
 V  emf = IR
 where I is current through motor and R is circuit resistance

emf is proportional to
v so the current changes as the motor speed changes

at high speeds
low current because induced emf is high

at low speeds
high current because induced emf is low

IV  emfI =
 I^{2}R  electrical power wasted due to circuit resistance

IV is
electrical power supplied by source

Emf is
electrical power transferred to mechanical power

prove T is 2pi/Bq
 F = Bqv
 Bqv = mv^{2}/r
 Bq = mv/r
 Bqr/m = v
 v = s/t
 Bqr/m = 2pir/T
 BqrT = 2pirm
 T = 2pim/Bq

alpha particle electric charge
+2

beta particle electric charge
1

angle in F = Bqvsinangle
angle between the field and direction of the particles motion

prove E = blv
 E = delta thigh/delta t
 E = delta BA/delta t
 E = B deltaA/delta t
 A = ls
 v = s/delta t
 v x delta t = s
 E = Blv delta t / delta t

units of magnetic flux (thigh)

