-
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 =
- I2R - 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 = mv2/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)
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