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Fields
OCR A2 G485 Electric fields & Magnetic fields
Transcript of Fields
Questions
Insulators can be charged by rubbing
and will then attract small, light objects.
Like charges repel and unlike charges attract.
Charging is caused by the addition or removal of electrons.
Metals contain free electrons and can only become charged if they are isolated by an insulator.
Materials that are easily charged give up their electrons easily.
Point charges & resultant electric fields
Fields
Electric field near
two positively charged
small metal spheres
Electric field around
a positively charged metal sphere
Electric field around
a negative point charge
Electric fields are represented by field lines  just like magnetic and gravitational fields.
A field line is the path followed by a small, positive test charge that is present in the field.
Electric fields are caused by an electric charge and exert an electrical force on any charged object in that field.
Fields
Shuttling ball experiment
6.1 all questions
Questions
Electric field around a positively charged object rounded at one end and sharp at the otheR
Electric field near two parallel plates,
one charged positively and the other negatively
Electric field near one positively charged
and one negatively charged small metal sphere
Static Electricity
Electric fields
electric fields are due to charges
modelling a uniformly charged sphere as a point charge at its centre
electric field lines to map electric fields
Demonstration
of electric field lines
You need to demonstrate and apply knowledge and understanding of...
The Basics:
Topic 2: Force between point charges
6.3 all questions
Worked examples: Coulomb’s Law
Questions
1. What is the force of repulsion between two electrons held one metre apart in a vacuum? What is the gravitational force of attraction between them? By what factor is the electric repulsion greater than the gravitational attraction?
2. By what factor is the electric force between two protons greater than the gravitational force between them?
3. Estimate the force that would exist between 2 students standing one metre apart if they had just 1% of the electrons in their body somehow removed, leaving them both positively charged.
Coulomb’s law questions
Find the value for absolute permittivity of free space and provide appropriate units.
k is found to have a very high value.
The practical implication of this very high value is that even small charges have strong forces acting between them.
It is the electric forces between atoms rather than gravitational forces that hold matter together.
k
Coulomb’s Law
CharlesAugustin de Coulomb (1736  1806)
The force between the two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Coulomb's Law
 Find the units for k...
And how do we get a negative E?
Electric field strength
Electric field strength
Topic 3: Electric field
6.2 Question 1 & 2
Electric fields always start on a positive charge and finish on a negative charge.
Electric field strength
F = kqQ
2
r
The strength of an electric field is defined as the force per unit charge on a small, positive test charge placed in that field.
Electric field strength is measured in NC and is a vector quantity.
Also expressed as Vm (we will see why later…)
E = F / Q
Is electric field strength a vector quantity?
E is a vector quantity and is positive as long as Q is positive.
Positive E indicates a field that radiates out away from the point charge.
If Q is negative, E is negative and this indicates a field that is directed towards the point charge.
http://www.bbc.co.uk/blogs/legacy/banggoesthetheory/2009/07/scienceputtothetest.shtml
Questions
Practical applications in xray tubes,
particle accelerators, etc.
V/d is also called the potential gradient
Electric field strength between two parallel plates
A pair of parallel plates with a battery connected across them has a uniform electric field between all but the edges of the plates
The resultant electric field strength is the vector sum of the individual electric field strengths.
Resultant electric field strength
Test charge +q on perpendicular lines from two positive point charges Q and Q .
The test charge experiences a force F = q E where E is the electric field strength due to Q and F = q E similarly.
The magnitude of the resultant force F is given by Pythagoras’ formula F = F –+ F .
Forces at right angles
to each other
Test charge +q on the line between two positive point charges Q and Q .
The test charge experiences a force F = q E where E is the electric field strength due to Q and F = q E similarly.
The two forces act in opposite directions.
The resultant force F = F– F
Forces in opposite directions
Test charge +q on the line between a negative point charge Q and a positive point charge Q .
The test charge experiences a force F = q E where E is the electric field strength due to Q and F = q E similarly.
The two forces act in the same direction.
The resultant force F = F + F
Forces in the same direction
If a test charge is in an electric field due to several point charges, each charge exerts a force on the test charge.
The resultant force F on the test charge gives the resultant electric field strength at the position of the test charge.
Electric field strength as a vector
Consider the electric field due to a point charge +Q.
The field lines radiate from the point charge because a test charge +q in the field would experience a force directly away from Q wherever it was placed.
The force on the test charge q is given by Coulomb’s Law and the electric field strength from E=F/q.
Electric field around
a positively charged metal sphere
A point charge is a convenient expression for a charged object in a situation where distances under consideration are much greater than the size of the object.
A “test” charge in an electric field is a point charge that does not alter the electric field in which it is placed.
Point charges
If Q is negative, the electric field strength will be a negative value, corresponding to the field lines pointing inwards towards Q.
Point charges
What will happen to a test charge if it is placed between two parallel plates?
a) E = V/d = 5.0 x 104 Vm
b) (i)F = 3.3 x 10 N upwards (to balance weight)
(ii) F = Eq, gives q = 6.6 x 10 C
(iii) Larger V  Larger E  Larger F  Larger a
so it will move twice as quickly!
Solution:
Treat motion in horizontal and vertical directions separately (like projectile motion).
P.E. lost as electrons go from cathode to anode = K.E. gained
Computer screen
Electrons leaving the cathode are accelerated a distance of 4cm by a uniform electric field E = 1.20 x 10 NC .
They then pass through a hole in the anode and enter a region where E = 0.
What is the speed of the electron when it
reaches the anode?
How long does it take to move from the anode
to the screen, 28cm away?
Electron gun
Electron gun or cathode ray tube
A moving charged particle passing between plates initially at right angles to the field will follow a parabolic path.
Direction of force and hence motion depends on direction of field and type of charge on particle.
Charge movement initially perpendicular to the electric field
What will happen to a test charge as it passes between two parallel plates?
A charged particle placed between a pair of oppositely charged plates experiences a force and accelerates towards the oppositely charged plate.
Charge movement in the direction of the field
Textbook page 95 question 1
Question
A beam of electrons all travelling horizontally at 1.0 x 10 ms pass between two horizontal plates 1.0 cm apart with a 100V p.d. between them.
Calculate the deflection of the beam (vertical displacement) when it has travelled a horizontal distance 2 cm between the plates.
Question
Catapult field
The magnetic field around a wire carrying a current
and a plan view of the field
A magnetic field is the region around a magnet where its effect can be felt.
Magnetic fields
What is the difference between a magnetic material
and a magnet?
How might you magnetise and demagnetise a bar of steel?
How might you investigate the magnetic field around a magnet?
Draw the magnetic field pattern you might expect to find if a north pole is brought up to a north pole.
Draw the magnetic field pattern you might expect to find
if a north pole is brought up to a south pole.
What forces act in each of the previous cases?
How is a stronger magnetic field represented with field lines?
Magnetism recap
Magnetic field around a solenoid
Magnetic fields due to current carrying conductors
Magnetic field around a straight
current carrying conductor
Magnetic field patterns
4
3
2
1
N S
S N
N S
S N
N S
N S
N S
N S
Is the magnetic field uniform anywhere?
How do you know?
S
N
8.1 Questions 1 to 3.
Magnetic field and forces 
Level 1 Questions 3, 5, 12 and 13.
Questions
Flemings left hand rule
The direction of movement, produced by the force (F), direction of magnetic field (B) and direction of current (I) flow are all perpendicular to each other.
Motor effect theory
Magnetic flux density, B, is a measure of the strength of the magnetic field.
Magnetic flux density
If the field and the wire are in the same direction, sin 0θ= 0, so F=0.
At right angles, sin 90 = 1,
so the force is maximum.
Magnetic flux density
A current carrying conductor in a magnetic field experiences a force, which causes it to move.
The motor effect
The Catapult field
(b)
(a)
The magnetic field of a currentcarrying wire superimposed on the uniform magnetic field between the poles of a horseshoe magnet;
the two fields from (a) added together to get the resultant field
Why does Bcos not contribute to the force?
F = BIL sin
Force on a wire that is not at right angles to the field
If the wire is not at right angles to the field lines, the force on
the wire is determined using the component of the magnetic
field perpendicular to the wire, Bsin .
8.1 Questions 1 to 3.
Magnetic field and forces 
Level 1 Questions 3, 5, 12 and 13.
8.2
Questions
The force on a charged particle moving in a magnetic field
Force on a charged particle in a magnetic field
Assuming that the particle (of charge q) is moving perpendicular to the field B, with a velocity v, show that the force experienced by the particle is given by:
All charged particles moving across the lines of a magnetic field are acted on by a force due to the field.
Positive charges are pushed in the opposite direction to negative charges.
Moving charges in a magnetic field
θ
θ
When a currentcarrying conductor is spun in the plane of the magnetic field,
the direction and magnitude of force that it feels will vary.
θ
Currentcarrying conductor perpendicular to the plane of the field 
Bainbridge mass spectrometer
What is the speed of the ions that get through?
The radius of the ion as it travels through and out of the mass spectrometer depends upon its mass, velocity and charge.
When the ions are produced, they will have a spread of velocities (Gaussian distribution). A velocity selector is used to ensure only a certain velocity is selected.
So, in actual fact, r is dependent on m/q only.
A Negative particle traveling in a vacuum and acted upon by both an electric and a magnetic field
An electric field and a magnetic field are used to "select" ions with particular speeds. When ions are produced they have a range of speeds.
Using a magnet and a pair of parallel plates at spacing, d, and voltage, V, only ions with the right velocity will travel straight through the two fields.
Velocity selector
...used to analyse the type of atoms present in a sample.
Atoms of the sample are ionised and directed, in a narrow beam and at constant velocity, into a uniform magnetic field.
Each ion, in the magnetic field, traverses a circular path of radius determined by its mass and charge.
Therefore each ion is deflected by a different amount and can be separated.
Mass spectrometer
8.3 charged particles in circular orbit
Questions
Motion of charged particle
In A magnetic field
206
82
238
92
A spectrometer used to collect the nuclides U and Pb from a small sample of rock to date it.
Conductors contain positive and negative charges
Charges experience a force given by Fleming's left hand rule
Positive charges in one direction (but are fixed in their lattice positions), negative charges in the other (and are free to move)
Charge separation occurs. “Battery” is formed, i.e. emf induced.
Why is an EMF induced?
Moving the wire faster,
Using a stronger magnet,
Making the wire into a coil and pushing the magnet in or out of the coil.
The induced EMF can be increased by:
Generating an electric current
A wire being moved between the poles of a horseshoe magnet in order to produce an electric current by electromagnetic induction
When a conductor is moved at right angles to a magnetic field an emf is induced in the conductor.
If the conductor is part of a complete circuit, a current will flow.
EM induction
A traditional electric doorbell
Deflection of electrons in a magnetic field
William Gladstone, the Chancellor of the Exchequer asked about the practical worth of electricity, to which Faraday replied:
EM induction
When asked,
“What is the use of electricity?”
he replied,
“What is the use of a baby?”
EM induction was discovered by Michael Faraday in 1831.
Why is it necessary to be careful about whether each side of the coil is wound clockwise or anticlockwise?
A large electromagnet with two separate coils.
For a moving conductor of length, l,
in a magnetic field,
Where v is the speed of the conductor.
Faraday’s law
Why is an emf induced in the coil as the magnet first enters it?
Why is no EMF induced while the magnet is entirely inside the coil?
Why is one peak positive and the other negative?
Why is the 2nd peak narrower and higher?
Faraday’s law demonstration
Faraday’s law demonstration
The magnitude of the induced emf is equal to the rate at which magnetic flux is cut.
For a coil: The induced emf is equal to the rate of change of flux linkage.
Faraday’s Law of
9 Electromagnetic Induction: 1 to 4 and 6.
9.1 Questions
Questions
Be careful how the angle is defined!
Flux linkage
Magnetic flux density, B, is the number of magnetic field lines per unit area.
Unit: Tesla or Weber m
Magnetic flux, , is the strength of the magnetic field multiplied by the area over which it is acting. = BA.
Unit Weber (Wb).
Flux linkage through a coil of N turns is N .
Unit Weber turns (Wb turns).
Some key terms:
Magnetic flux
(a) If the loop of area A is at right angles to the field of flux density,
B, the flux through the loop is BA
(b) If the loop has been rotated through an angle, θ , from its original position, the component of area perpendicular to the field is A cos θ, so the flux through the loop is BA cos θ.
As the magnet is moved towards the solenoid, the current induced in the solenoid causes a magnetic field which repels the magnet.
As a result, a force has to be exerted on the magnet to keep it moving towards the solenoid.
In this way, energy is transferred from whatever system is pushing the magnet towards the solenoid to the electrical circuit, through the medium of the magnetic field.
Why does this happen?
The induced EMF always acts in a direction to oppose the change causing it.
Lenz’s law
When the magnet is pulled away from the solenoid, the current flows in the opposite direction and the magnet experiences an attractive force, requiring a further energy transfer from the “pulling system” to the electrical circuit.
The induced current flows in such a direction as to oppose the change causing it.
Lenz’s Law is an example of the conservation of energy.
The induced current sets up a force on the magnet which the mover of the magnet must overcome.
The work done in overcoming this force provides the electrical energy of the current.
This energy is dissipated as heat in the coil.
Calculate the maximum e.m.f. induced in a coil with 500 turns of area 1.0 x 10 m , rotating at 20 radians per second in a field of flux density 0.20 T.
Question
The peak emf induced can be increased by increasing:
N
B
the size of the coil
the speed of rotation.
Graph showing the variation of current with time in an alternating current.
The direct current that provides the same heating effect is also shown.
Flux linkage in a spinning coil
EMF v. time for an a.c. generator
The a.c. generator – top view
(induced emf in a rotating coil)
A simplified diagram of an a.c. generator
A local mains transformer with an input at 11 000 V and an output at 230 V
You can never get more power out of a transformer than you put into it.
Transformers are very efficient  almost 100%!
The equation of turns ratio assumes 100% efficiency.
This is true as long as resistance of the coils is negligible  such as when thick copper coils are used.
However, this is expensive  a balance must be found between cost and efficiency.
The efficiency of a transformer
When there is an a.c. in the primary coil, the iron core becomes magnetised and then remagnetised in the opposite direction 50 times a second.
This rapid change in magnetic flux means an equally fast change in magnetic flux through the secondary coil.
According to Faraday’s law, the more rapidly the flux changes and the more coils on the secondary coil, the larger the induced e.m.f.
The usual arrangement of the coils in a transformer;
Another possible arrangement of the coils in a transformer;
The circuit symbol for a transformer.
In a typical mains electrical distribution system, a generator will have an output voltage of 6000V.
This will be stepped up at the power station to 275 000V for transmission.
A large electricity substation, will step down the voltage, usually in stages, to 11 000V.
A small transformer unit, somewhere close to homes, will further step down the voltage to 230V.
Power stations are linked by a national grid, which is also connected to many large substations for area distribution.
Mains distribution
Transformers change electrical supplies from low voltage, high current, to high voltage, low current, or vice versa.
They are over 99% efficient and need no maintenance as there are no moving parts.
The transformer can cope with varying output without any need for manual adjustment. It automatically adjusts to different demands for power.
Assuming 100% efficiency does not produce highly accurate answers – answers obtained are accurate to 2 significant figures.
Two key facts about transformers
The simplest transformer consists of two coils of insulated copper wire wound on top of one another on a core of easily magnetised iron.
One of the coils is connected to the a.c. input and is called the primary coil.
The second is connected to the output and is called the secondary coil.
In order to get as strong a magnetic field as possible, the core is usually a complete loop of iron.
Transformer structure
Economic cost vs. physics efficiency.
Cooling tubes containing oil are used where a convection current is set up when the transformer gets hot.
The oil is cooled by the air outside.
For a 100% efficient transformer:
The number of turns in the primary coil also has to be considered, so the ratio of the output to input e.m.f. equals the turns ratio.
Transformers can be used to increase or decrease the e.m.f. supplied to the primary coil.
Those that increase the e.m.f. are called stepup transformers.
Those that decrease are called stepdown transformers.
Use very high voltages for transmission (up to 450 000V) and to use 230V for domestic use.
To change from high voltage to low voltage efficiently requires a transformer.
Transformers will only work on a.c., so all mains electrical supplies must be a.c.
2
1
1
1
2
1
2
1
1
1
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
QUESTION:
5
1
+

+
+

+

Motion in electric fields
Question:
In an (old fashioned) computer, electrons are accelerated through 25kV.
With what speed to they hit the fluorescent screen?
2
1
Solution:
Horizontal direction: –
constant velocity, so use: speed = distance/time
Vertical direction:
constant acceleration, so use: F = eV/d and F = ma
Gives s = 3.5 x 10 m vertical displacement of 3.5 mm
Comparison between electric and gravitational fields
A lowvoltage supply powers the heater, which heats the cathode, which throws off some of its electrons as its atoms vibrate more vigorously.
A highvoltage power supply attracts electrons from the cathode to the anode.
Some pass through the hole in the anode and produce an electron beam, seen as a visible spot when it hits the fluorescent screen.
Electron gun
4
1
19
Textbook p 95 qu 1
N
S
N
S
S
S
N
N
N
N
N
N
S
S
S
S
Two magnetic fields from two different sources.
The field lines are shown as solid lines for one and dotted lines for the other. The direction of each field is given by the arrows on the lines.
uniform field
+
radial field
=
Force on a current carrying
conductor in a magnetic field
F = B I L
The strength of the magnetic field, B
The size of the current, I
The length of the conductor, L
The size of the force is proportional to:
B is measured in Tesla.
How would you define a B of 1 Tesla?
1 Tesla is the magnetic flux density when a wire of length 1m and carrying a current of 1 A at right angles to the field experiences a force of 1 N.
If the field and the wire are at an angle, ,
to each other,
the perpendicular component of force
is given by F sin ,
whilst the parallel component is F cos .
Any currentcarrying wire that is parallel to a field will not experience a force.
cos 90 = 0
sin 90 = 1
F
B
I
Why is there a force?
Record data.
Plot two graphs.
Find B in both cases.
Magn
Field
Patt
The
Motor
Effect
etic
erns
Mov
Cha
Mass
Spect
rometer
&
Velocity
Sele
ctor
Elec
tro
magn
etic
Ind
uct
ion
Fara
day's
Law
Lenz's
Law
Alt
ern
ator
Tran
sfor
mers
“Nothing is too wonderful to be true if it be consistent with the laws of nature.”
“Why, sir, there is every probability that you will soon be able to tax it!”
Faraday
Michael
Hint: Analysis is similar to force on a current carrying wire.
F = Bqv
According to Fleming’s LHR, the electron moves into the page.
(a)
(b)
(flux through a coil of N turns)
emf = Blv
flux cut / time
wire moves down through field
induced emf
flux cut / time induced emf
Faraday’s law concentrates on the emf produced by electromagnetic induction.
uniform field
What would the resultant magnetic field look like?
+

connected to power supply
moveable rod
parallel rods held in place
N
S
motion
variable resistor
ammeter
power supply
birds eye view
yoke magnets
toppan balance
currentcarrying wire
Investigating the force
on a currentcarrying wire in a magnetic field
change the number of magnets to vary length of wire in field
change the current
ing
rges
B
F
currentcarrying wire
currentcarrying wire
B
F
BIL
force always BIL,
but direction of the force changes
What does this mean?
What would happen if the magnet was attracted to the solenoid when pushed towards it?
The plates and the magnet are aligned so that each ion passing through is acted on by an electric field force
F elec = QV/d and a magnetic field force F mag = BQv.
Only ions moving at a speed such that F elec = F mag
(equal and opposite forces) get through undeflected.
v = Electric field strength/Magnetic flux density
What happens to a charged particle moving through a magnetic AND an electric field?
Both fields will exert an individual force on the particle. Dependent on the fields' directions, the particle's deflection can be used to identify the particle's mass and velocity.
r = mv / Bq
Flux linkage = BAN sin t (for a rotating coil)
Therefore emf = BAN cos t
For max. emf, cos t = 1,
so max emf = BAN
3
2
Answer: 2V
High voltage is lethal in a house, but 230V can only be transmitted with cables that have a much lower resistance, requiring thick, costly cables.
What's the solution?
N
S

equivalent to current
Region of magnetic field B out of the screen
Q
Force = QE
Force = BQv
Electric field
What would happen to an electron entering the fields?
It travels straight through with v = E/B
TAP 4061 from http://www.tap.iop.org/fields/electrical/406/page_46863.html
Draw the field lines for the different examples:
1
1
2
1
1
7
3
9.2 Questions
Questions
Taken from TAP worksheet.
2
Electromagnetic Induction
from the side:
from above:
motion
spinning allows the wire to cut the field lines constantly
the spinning wire only cuts through the field lines at certain angles

+
A positive particle traveling in a vacuum and acted upon by both an electric and a magnetic field
Region of magnetic field B down / into the screen
+Q
Force = BQv
Force = QE
Electric field

+
N.B. q/m is known as the specific charge
F = Bqv sin
A charged particle moving at an angle, θ, to the magnetic field will experience a force,
Why the negative symbol?
The negative symbol can be explained by Lenz's Law.
A charged object is normally modelled as a point charge...
positive as long as Q is positive
Equipment:
Logbook ML
Microammeter
Connecting cables
Helmholtz coil
Open Datalogging Insight and pick COM3
Go to: Set up  Timespan
Select: 0.01 seconds
"Start" to collect data.
Go to: Edit  Copy Data to export to excel.
When a wire is moved between opposite poles, cutting through a uniform field, a voltage is induced across its ends. The induced voltage pushes current around the circuit through the meter.
A wire contains many free electrons because it is made of metal. When the wire is moved across the field, the field pushes the electrons along the wire.
Why does this happen?
If the wire stops moving, no voltage is induced in the wire.
The induced voltage can be made larger by using a stronger magnet, using more wire or moving the wire faster.
If the wire is moved along the lines of the field, no voltage is produced.
N
coil
coil
N
coil
coil
coil
coil
N
coil
S
S
S
S
N
How could we use what we know to find the magnetic flux density
of the Earth?
Observe and explain
why the swinger comes to a rest so quickly...
Coulomb’s law
the concept of electric fields as being one of a number of forms of field giving rise to a force
You need to demonstrate and apply knowledge and understanding of...
6.1 all questions
Questions
Questions
Textbook p427
Q 2, 4 & 5
Questions
You need to demonstrate and apply knowledge and understanding of...
electric field strength
Textbook page 91
All questions
Questions
A charged object is normally considered to be a point charge.
Electric field strength
of a point charge
Relate E = F/Q to Coulomb’s Law to derive the electric field strength of a point charge...
Fields are regions where a force is exerted on an object at a distance.
Some key points:
The direction of the field is shown by an arrow (+ to )
Electric fields lines are at right angles to the surface of a conductor
A uniform field is shown by equally spaced parallel lines
Stronger fields are represented by more densely packed field lines.
Uniform field here!
analogous to Newton's law of gravitation  but for charged particles
Any two point charges exert an electrostatic (electrical) force on each other that is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
where k is a constant
absolute permittivity of free space
electric field strength for a point charge
similarities and differences between the gravitational field of a point mass and the electric field of a point charge
You need to demonstrate and apply knowledge and understanding of...
Questions
Textbook p 431
Q 3, 4, 6 & 7
Questions
Investigating Coulomb's law
0.03 g
top pan balance
+
+
what will happen?
+
The electric field strength is directly proportional to the charge and obeys the inverse square law of the distance.
E =
F
q
E =
kQ
r
2
electric field strength for a point charge
Questions
Textbook p 431
Q 5
Questions
6.2 questions 3 & 4
Questions
Textbook page 93
question 1
Questions
Electric potential & energy
motion of charged particles in a uniform electric field
You need to demonstrate and apply knowledge and understanding of...
electric potential at a point as the work done in bringing unit positive charge from infinity to the point; electric potential is zero at infinity
electric potential at a distance from a point charge; changes in electric potential
force–distance graph for a point or spherical charge; work done is area under graph
electric potential energy a distance from a point charge
You need to demonstrate and apply knowledge and understanding of...
uniform electric field strength
parallel plate capacitor; permittivity
You need to demonstrate and apply knowledge and understanding of...
Uniform electric fields
oppositely charged plates
small test charge
+
experiences constant force
The charge gains energy as it moves from positive plate to negative plate.
F = E/Q
RECALL: p.d. = work done per unit charge
and work done on the charge is Fxd
W = Fd
VQ = (EQ)d
E = V/d
V = W/Q
and
but this equation only works for parallel plates
This is really useful as we can determine the electric field strength between parallel plates using just a voltmeter and a ruler.
E = V/d
NC = Vm
1
1
1
1
cancel Q
EQUATE
Textbook p 435
Questions
Wider CONTEXT:
Parallel Plate Capacitor
For plates in a vacuum (or air), experiments show that capacitance is directly proportional to the area and inversely proportional to the separation. It will also depend upon the insulator that is used.
A simple capacitor can store charge by separating said charge on two plates with an insulator in between.
The constant of proportionality in this relationship is the permittivity of free space, , as used in Coulomb's law.
C =
3
A
0
r
3
0
When an insulator other than a vacuum is used, we use the permittivity for the insulator such that:
C =
3
A
r
permittivity of insulator
About E=V/d: Q 2&3
About capacitors: Q 46
oppositely charged plates
Electron

experiences constant force
F = E/Q
A p.d. of just 1.5 V from a cell can accelerate an electron up to 700 km/s.
Even greater voltages can be used to accelerate particles close to the speed of light. Protons reach energies of around 400 MeV in a linear particle accelerator.
The acceleration of the particle will be constant due to the constant electrostatic force.
We can analyse the motion of a charged particle moving at right angles to an electric field in much the way we dealt with projectile motion in mechanics.
The particle's motion parallel to the plates is at a constant velocity and unaffected by the electrostatic force. Indeed, there are no other forces acting on the particle (in most cases), so the acceleration = 0.
The particle's motion perpendicular to the plate is at a constant acceleration as provided by the constant electrostatic force of the parallel plates.
a = F/m = EQ/m
u = 0
a = 0
u = v
t = L/v
where L = horizontal distance moved parallel to plates
use suvat to solve
Textbook p 438
Questions
Q 24
Textbook p 442
Questions
About V: Q 14 & 6
About capacitors: Q 5
Textbook p 4435
MORE Questions
Q 1, 2 & 5
About capacitors: Q 3 & 4
Questions
6.4 questions 13
Electric potential energy
Charges repel each other so you have to do work to decrease the separation between the charges.
+
+
More force is needed to push the charges together as the separation decreases. All the work done is stored as electric potential energy  this energy is recoverable, just let go!
force
0
0
area = work done
F is inversely proportional to r
2
r
separation
r
+q
+Q
The force between the two positive charges varies with their separation.
The total work done to bring the particles from infinity to a separation r is the total area under
the graph.
It is the same as the electric potential energy.
W =
kQq
r
work done = electric potential energy
If one of the charges is negative, then the value for W will also be negative  this represents the amount of external energy required to completely separate the charged particles to infinity.
This is what happens when atomic or molecular bonds are broken.
Electric potential
The electric potential, V, at a point is defined as the work done per unit charge in bringing a positive charge from infinity to that point. If the test charge is q, the equation for V can be determined by dividing the electric potential energy W by q.
V = =
W
q
kQ
r
The unit for electric potential is unit JC or volts.
1
Electric potential difference (p.d.) is the work done per unit charge between two points around the particle of charge Q.
+Q
A
B
r
A
r
B
The electric p.d. between point A and B is the difference in the potentials at these two points.
Placing a voltmeter between A and B would show the work done per unit charge  if a voltmeter could be placed in empty space around charged spheres.
V =
kQ
r
A
V =
kQ
r
B
Wider CONTEXT:
Parallel Plate Capacitor
A capacitor stores charge, as does an isolated charged sphere of radius, R.
The capacitance of a charged sphere is the ratio of the charge it stores, Q, to the potential, V, at its surface.
C =
Q
V
R
k
V =
kQ
r
far away from other charged objects
=
Imagine the Earth as a huge capacitor floating is space.
If R = 6400km,
calculate
Earth's capacitance.
7.1 x 10 F
4
C =
Equally spaced parallel lines
Magnetic field lines tell us the strength and the direction of the field
 N to S  the direction that
a free North pole would move.
created by the electrons moving within the wire
magnetic field created by electrons moving around iron nuclei
the iron atoms act as tiny magnets, all alligned in the same direction
direction determined by right hand grip rule
Questions
Textbook p 447
Q 3 & 4
Questions
Questions
Textbook p 451
Q 3 to 5
Questions
Questions
Textbook p 456
Q 2 to 5
Questions
7
q V = ½ m v
v = 9.4 x 10 ms
PAG 11.3
• To determine the magnetic field strength of a magnet
• To demonstrate investigative skills
• To research, plan and implement a practical activity
• To make quantitative measurements
• To relate observations to research or theory
AIM:
a) Calculate the electric field strength between the plates.
b) A tiny sphere of weight 3.3 x 10 N has acquired a charge so that it is held in equilibrium midway between the plates by the electric field.
14
(i) State the magnitude and direction of the electric force on the sphere.
(ii) Calculate the charge on the sphere.
(iii) Describe the motion
of the sphere if V doubles.
You need to demonstrate and apply knowledge and understanding of...
magnetic fields are due to moving charges or permanent magnets
magnetic field lines to map magnetic fields
magnetic field patterns for a long straight current
carrying conductor, a flat coil and a long solenoid
You need to demonstrate and apply knowledge and understanding of...
Fleming’s lefthand rule
force on a currentcarrying conductor
magnetic flux density; the unit tesla
techniques and procedures used to determine the uniform magnetic flux density between the poles of a magnet using a currentcarrying wire and digital balance
You need to demonstrate and apply knowledge and understanding of...
force on a charged particle travelling at right angles to a uniform magnetic field; F = BQv
charged particles moving in a uniform magnetic field; circular orbits of charged particles in a uniform magnetic field
You need to demonstrate and apply knowledge and understanding of...
charged particles moving in a region occupied by both electric and magnetic fields; velocity selector
You need to demonstrate and apply knowledge and understanding of...
Faraday’s law of electromagnetic induction
You need to demonstrate and apply knowledge and understanding of...
Lenz’s law
emf =  rate of change of magnetic flux linkage
techniques and procedures used to investigate magnetic flux using search coils
You need to demonstrate and apply knowledge and understanding of...
magnetic flux, the unit weber
magnetic flux linkage
You need to demonstrate and apply knowledge and understanding of...
simple a.c. generator
You need to demonstrate and apply knowledge and understanding of...
simple laminated ironcored transformer
techniques and procedures used to investigate transformers
Full transcriptInsulators can be charged by rubbing
and will then attract small, light objects.
Like charges repel and unlike charges attract.
Charging is caused by the addition or removal of electrons.
Metals contain free electrons and can only become charged if they are isolated by an insulator.
Materials that are easily charged give up their electrons easily.
Point charges & resultant electric fields
Fields
Electric field near
two positively charged
small metal spheres
Electric field around
a positively charged metal sphere
Electric field around
a negative point charge
Electric fields are represented by field lines  just like magnetic and gravitational fields.
A field line is the path followed by a small, positive test charge that is present in the field.
Electric fields are caused by an electric charge and exert an electrical force on any charged object in that field.
Fields
Shuttling ball experiment
6.1 all questions
Questions
Electric field around a positively charged object rounded at one end and sharp at the otheR
Electric field near two parallel plates,
one charged positively and the other negatively
Electric field near one positively charged
and one negatively charged small metal sphere
Static Electricity
Electric fields
electric fields are due to charges
modelling a uniformly charged sphere as a point charge at its centre
electric field lines to map electric fields
Demonstration
of electric field lines
You need to demonstrate and apply knowledge and understanding of...
The Basics:
Topic 2: Force between point charges
6.3 all questions
Worked examples: Coulomb’s Law
Questions
1. What is the force of repulsion between two electrons held one metre apart in a vacuum? What is the gravitational force of attraction between them? By what factor is the electric repulsion greater than the gravitational attraction?
2. By what factor is the electric force between two protons greater than the gravitational force between them?
3. Estimate the force that would exist between 2 students standing one metre apart if they had just 1% of the electrons in their body somehow removed, leaving them both positively charged.
Coulomb’s law questions
Find the value for absolute permittivity of free space and provide appropriate units.
k is found to have a very high value.
The practical implication of this very high value is that even small charges have strong forces acting between them.
It is the electric forces between atoms rather than gravitational forces that hold matter together.
k
Coulomb’s Law
CharlesAugustin de Coulomb (1736  1806)
The force between the two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Coulomb's Law
 Find the units for k...
And how do we get a negative E?
Electric field strength
Electric field strength
Topic 3: Electric field
6.2 Question 1 & 2
Electric fields always start on a positive charge and finish on a negative charge.
Electric field strength
F = kqQ
2
r
The strength of an electric field is defined as the force per unit charge on a small, positive test charge placed in that field.
Electric field strength is measured in NC and is a vector quantity.
Also expressed as Vm (we will see why later…)
E = F / Q
Is electric field strength a vector quantity?
E is a vector quantity and is positive as long as Q is positive.
Positive E indicates a field that radiates out away from the point charge.
If Q is negative, E is negative and this indicates a field that is directed towards the point charge.
http://www.bbc.co.uk/blogs/legacy/banggoesthetheory/2009/07/scienceputtothetest.shtml
Questions
Practical applications in xray tubes,
particle accelerators, etc.
V/d is also called the potential gradient
Electric field strength between two parallel plates
A pair of parallel plates with a battery connected across them has a uniform electric field between all but the edges of the plates
The resultant electric field strength is the vector sum of the individual electric field strengths.
Resultant electric field strength
Test charge +q on perpendicular lines from two positive point charges Q and Q .
The test charge experiences a force F = q E where E is the electric field strength due to Q and F = q E similarly.
The magnitude of the resultant force F is given by Pythagoras’ formula F = F –+ F .
Forces at right angles
to each other
Test charge +q on the line between two positive point charges Q and Q .
The test charge experiences a force F = q E where E is the electric field strength due to Q and F = q E similarly.
The two forces act in opposite directions.
The resultant force F = F– F
Forces in opposite directions
Test charge +q on the line between a negative point charge Q and a positive point charge Q .
The test charge experiences a force F = q E where E is the electric field strength due to Q and F = q E similarly.
The two forces act in the same direction.
The resultant force F = F + F
Forces in the same direction
If a test charge is in an electric field due to several point charges, each charge exerts a force on the test charge.
The resultant force F on the test charge gives the resultant electric field strength at the position of the test charge.
Electric field strength as a vector
Consider the electric field due to a point charge +Q.
The field lines radiate from the point charge because a test charge +q in the field would experience a force directly away from Q wherever it was placed.
The force on the test charge q is given by Coulomb’s Law and the electric field strength from E=F/q.
Electric field around
a positively charged metal sphere
A point charge is a convenient expression for a charged object in a situation where distances under consideration are much greater than the size of the object.
A “test” charge in an electric field is a point charge that does not alter the electric field in which it is placed.
Point charges
If Q is negative, the electric field strength will be a negative value, corresponding to the field lines pointing inwards towards Q.
Point charges
What will happen to a test charge if it is placed between two parallel plates?
a) E = V/d = 5.0 x 104 Vm
b) (i)F = 3.3 x 10 N upwards (to balance weight)
(ii) F = Eq, gives q = 6.6 x 10 C
(iii) Larger V  Larger E  Larger F  Larger a
so it will move twice as quickly!
Solution:
Treat motion in horizontal and vertical directions separately (like projectile motion).
P.E. lost as electrons go from cathode to anode = K.E. gained
Computer screen
Electrons leaving the cathode are accelerated a distance of 4cm by a uniform electric field E = 1.20 x 10 NC .
They then pass through a hole in the anode and enter a region where E = 0.
What is the speed of the electron when it
reaches the anode?
How long does it take to move from the anode
to the screen, 28cm away?
Electron gun
Electron gun or cathode ray tube
A moving charged particle passing between plates initially at right angles to the field will follow a parabolic path.
Direction of force and hence motion depends on direction of field and type of charge on particle.
Charge movement initially perpendicular to the electric field
What will happen to a test charge as it passes between two parallel plates?
A charged particle placed between a pair of oppositely charged plates experiences a force and accelerates towards the oppositely charged plate.
Charge movement in the direction of the field
Textbook page 95 question 1
Question
A beam of electrons all travelling horizontally at 1.0 x 10 ms pass between two horizontal plates 1.0 cm apart with a 100V p.d. between them.
Calculate the deflection of the beam (vertical displacement) when it has travelled a horizontal distance 2 cm between the plates.
Question
Catapult field
The magnetic field around a wire carrying a current
and a plan view of the field
A magnetic field is the region around a magnet where its effect can be felt.
Magnetic fields
What is the difference between a magnetic material
and a magnet?
How might you magnetise and demagnetise a bar of steel?
How might you investigate the magnetic field around a magnet?
Draw the magnetic field pattern you might expect to find if a north pole is brought up to a north pole.
Draw the magnetic field pattern you might expect to find
if a north pole is brought up to a south pole.
What forces act in each of the previous cases?
How is a stronger magnetic field represented with field lines?
Magnetism recap
Magnetic field around a solenoid
Magnetic fields due to current carrying conductors
Magnetic field around a straight
current carrying conductor
Magnetic field patterns
4
3
2
1
N S
S N
N S
S N
N S
N S
N S
N S
Is the magnetic field uniform anywhere?
How do you know?
S
N
8.1 Questions 1 to 3.
Magnetic field and forces 
Level 1 Questions 3, 5, 12 and 13.
Questions
Flemings left hand rule
The direction of movement, produced by the force (F), direction of magnetic field (B) and direction of current (I) flow are all perpendicular to each other.
Motor effect theory
Magnetic flux density, B, is a measure of the strength of the magnetic field.
Magnetic flux density
If the field and the wire are in the same direction, sin 0θ= 0, so F=0.
At right angles, sin 90 = 1,
so the force is maximum.
Magnetic flux density
A current carrying conductor in a magnetic field experiences a force, which causes it to move.
The motor effect
The Catapult field
(b)
(a)
The magnetic field of a currentcarrying wire superimposed on the uniform magnetic field between the poles of a horseshoe magnet;
the two fields from (a) added together to get the resultant field
Why does Bcos not contribute to the force?
F = BIL sin
Force on a wire that is not at right angles to the field
If the wire is not at right angles to the field lines, the force on
the wire is determined using the component of the magnetic
field perpendicular to the wire, Bsin .
8.1 Questions 1 to 3.
Magnetic field and forces 
Level 1 Questions 3, 5, 12 and 13.
8.2
Questions
The force on a charged particle moving in a magnetic field
Force on a charged particle in a magnetic field
Assuming that the particle (of charge q) is moving perpendicular to the field B, with a velocity v, show that the force experienced by the particle is given by:
All charged particles moving across the lines of a magnetic field are acted on by a force due to the field.
Positive charges are pushed in the opposite direction to negative charges.
Moving charges in a magnetic field
θ
θ
When a currentcarrying conductor is spun in the plane of the magnetic field,
the direction and magnitude of force that it feels will vary.
θ
Currentcarrying conductor perpendicular to the plane of the field 
Bainbridge mass spectrometer
What is the speed of the ions that get through?
The radius of the ion as it travels through and out of the mass spectrometer depends upon its mass, velocity and charge.
When the ions are produced, they will have a spread of velocities (Gaussian distribution). A velocity selector is used to ensure only a certain velocity is selected.
So, in actual fact, r is dependent on m/q only.
A Negative particle traveling in a vacuum and acted upon by both an electric and a magnetic field
An electric field and a magnetic field are used to "select" ions with particular speeds. When ions are produced they have a range of speeds.
Using a magnet and a pair of parallel plates at spacing, d, and voltage, V, only ions with the right velocity will travel straight through the two fields.
Velocity selector
...used to analyse the type of atoms present in a sample.
Atoms of the sample are ionised and directed, in a narrow beam and at constant velocity, into a uniform magnetic field.
Each ion, in the magnetic field, traverses a circular path of radius determined by its mass and charge.
Therefore each ion is deflected by a different amount and can be separated.
Mass spectrometer
8.3 charged particles in circular orbit
Questions
Motion of charged particle
In A magnetic field
206
82
238
92
A spectrometer used to collect the nuclides U and Pb from a small sample of rock to date it.
Conductors contain positive and negative charges
Charges experience a force given by Fleming's left hand rule
Positive charges in one direction (but are fixed in their lattice positions), negative charges in the other (and are free to move)
Charge separation occurs. “Battery” is formed, i.e. emf induced.
Why is an EMF induced?
Moving the wire faster,
Using a stronger magnet,
Making the wire into a coil and pushing the magnet in or out of the coil.
The induced EMF can be increased by:
Generating an electric current
A wire being moved between the poles of a horseshoe magnet in order to produce an electric current by electromagnetic induction
When a conductor is moved at right angles to a magnetic field an emf is induced in the conductor.
If the conductor is part of a complete circuit, a current will flow.
EM induction
A traditional electric doorbell
Deflection of electrons in a magnetic field
William Gladstone, the Chancellor of the Exchequer asked about the practical worth of electricity, to which Faraday replied:
EM induction
When asked,
“What is the use of electricity?”
he replied,
“What is the use of a baby?”
EM induction was discovered by Michael Faraday in 1831.
Why is it necessary to be careful about whether each side of the coil is wound clockwise or anticlockwise?
A large electromagnet with two separate coils.
For a moving conductor of length, l,
in a magnetic field,
Where v is the speed of the conductor.
Faraday’s law
Why is an emf induced in the coil as the magnet first enters it?
Why is no EMF induced while the magnet is entirely inside the coil?
Why is one peak positive and the other negative?
Why is the 2nd peak narrower and higher?
Faraday’s law demonstration
Faraday’s law demonstration
The magnitude of the induced emf is equal to the rate at which magnetic flux is cut.
For a coil: The induced emf is equal to the rate of change of flux linkage.
Faraday’s Law of
9 Electromagnetic Induction: 1 to 4 and 6.
9.1 Questions
Questions
Be careful how the angle is defined!
Flux linkage
Magnetic flux density, B, is the number of magnetic field lines per unit area.
Unit: Tesla or Weber m
Magnetic flux, , is the strength of the magnetic field multiplied by the area over which it is acting. = BA.
Unit Weber (Wb).
Flux linkage through a coil of N turns is N .
Unit Weber turns (Wb turns).
Some key terms:
Magnetic flux
(a) If the loop of area A is at right angles to the field of flux density,
B, the flux through the loop is BA
(b) If the loop has been rotated through an angle, θ , from its original position, the component of area perpendicular to the field is A cos θ, so the flux through the loop is BA cos θ.
As the magnet is moved towards the solenoid, the current induced in the solenoid causes a magnetic field which repels the magnet.
As a result, a force has to be exerted on the magnet to keep it moving towards the solenoid.
In this way, energy is transferred from whatever system is pushing the magnet towards the solenoid to the electrical circuit, through the medium of the magnetic field.
Why does this happen?
The induced EMF always acts in a direction to oppose the change causing it.
Lenz’s law
When the magnet is pulled away from the solenoid, the current flows in the opposite direction and the magnet experiences an attractive force, requiring a further energy transfer from the “pulling system” to the electrical circuit.
The induced current flows in such a direction as to oppose the change causing it.
Lenz’s Law is an example of the conservation of energy.
The induced current sets up a force on the magnet which the mover of the magnet must overcome.
The work done in overcoming this force provides the electrical energy of the current.
This energy is dissipated as heat in the coil.
Calculate the maximum e.m.f. induced in a coil with 500 turns of area 1.0 x 10 m , rotating at 20 radians per second in a field of flux density 0.20 T.
Question
The peak emf induced can be increased by increasing:
N
B
the size of the coil
the speed of rotation.
Graph showing the variation of current with time in an alternating current.
The direct current that provides the same heating effect is also shown.
Flux linkage in a spinning coil
EMF v. time for an a.c. generator
The a.c. generator – top view
(induced emf in a rotating coil)
A simplified diagram of an a.c. generator
A local mains transformer with an input at 11 000 V and an output at 230 V
You can never get more power out of a transformer than you put into it.
Transformers are very efficient  almost 100%!
The equation of turns ratio assumes 100% efficiency.
This is true as long as resistance of the coils is negligible  such as when thick copper coils are used.
However, this is expensive  a balance must be found between cost and efficiency.
The efficiency of a transformer
When there is an a.c. in the primary coil, the iron core becomes magnetised and then remagnetised in the opposite direction 50 times a second.
This rapid change in magnetic flux means an equally fast change in magnetic flux through the secondary coil.
According to Faraday’s law, the more rapidly the flux changes and the more coils on the secondary coil, the larger the induced e.m.f.
The usual arrangement of the coils in a transformer;
Another possible arrangement of the coils in a transformer;
The circuit symbol for a transformer.
In a typical mains electrical distribution system, a generator will have an output voltage of 6000V.
This will be stepped up at the power station to 275 000V for transmission.
A large electricity substation, will step down the voltage, usually in stages, to 11 000V.
A small transformer unit, somewhere close to homes, will further step down the voltage to 230V.
Power stations are linked by a national grid, which is also connected to many large substations for area distribution.
Mains distribution
Transformers change electrical supplies from low voltage, high current, to high voltage, low current, or vice versa.
They are over 99% efficient and need no maintenance as there are no moving parts.
The transformer can cope with varying output without any need for manual adjustment. It automatically adjusts to different demands for power.
Assuming 100% efficiency does not produce highly accurate answers – answers obtained are accurate to 2 significant figures.
Two key facts about transformers
The simplest transformer consists of two coils of insulated copper wire wound on top of one another on a core of easily magnetised iron.
One of the coils is connected to the a.c. input and is called the primary coil.
The second is connected to the output and is called the secondary coil.
In order to get as strong a magnetic field as possible, the core is usually a complete loop of iron.
Transformer structure
Economic cost vs. physics efficiency.
Cooling tubes containing oil are used where a convection current is set up when the transformer gets hot.
The oil is cooled by the air outside.
For a 100% efficient transformer:
The number of turns in the primary coil also has to be considered, so the ratio of the output to input e.m.f. equals the turns ratio.
Transformers can be used to increase or decrease the e.m.f. supplied to the primary coil.
Those that increase the e.m.f. are called stepup transformers.
Those that decrease are called stepdown transformers.
Use very high voltages for transmission (up to 450 000V) and to use 230V for domestic use.
To change from high voltage to low voltage efficiently requires a transformer.
Transformers will only work on a.c., so all mains electrical supplies must be a.c.
2
1
1
1
2
1
2
1
1
1
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
QUESTION:
5
1
+

+
+

+

Motion in electric fields
Question:
In an (old fashioned) computer, electrons are accelerated through 25kV.
With what speed to they hit the fluorescent screen?
2
1
Solution:
Horizontal direction: –
constant velocity, so use: speed = distance/time
Vertical direction:
constant acceleration, so use: F = eV/d and F = ma
Gives s = 3.5 x 10 m vertical displacement of 3.5 mm
Comparison between electric and gravitational fields
A lowvoltage supply powers the heater, which heats the cathode, which throws off some of its electrons as its atoms vibrate more vigorously.
A highvoltage power supply attracts electrons from the cathode to the anode.
Some pass through the hole in the anode and produce an electron beam, seen as a visible spot when it hits the fluorescent screen.
Electron gun
4
1
19
Textbook p 95 qu 1
N
S
N
S
S
S
N
N
N
N
N
N
S
S
S
S
Two magnetic fields from two different sources.
The field lines are shown as solid lines for one and dotted lines for the other. The direction of each field is given by the arrows on the lines.
uniform field
+
radial field
=
Force on a current carrying
conductor in a magnetic field
F = B I L
The strength of the magnetic field, B
The size of the current, I
The length of the conductor, L
The size of the force is proportional to:
B is measured in Tesla.
How would you define a B of 1 Tesla?
1 Tesla is the magnetic flux density when a wire of length 1m and carrying a current of 1 A at right angles to the field experiences a force of 1 N.
If the field and the wire are at an angle, ,
to each other,
the perpendicular component of force
is given by F sin ,
whilst the parallel component is F cos .
Any currentcarrying wire that is parallel to a field will not experience a force.
cos 90 = 0
sin 90 = 1
F
B
I
Why is there a force?
Record data.
Plot two graphs.
Find B in both cases.
Magn
Field
Patt
The
Motor
Effect
etic
erns
Mov
Cha
Mass
Spect
rometer
&
Velocity
Sele
ctor
Elec
tro
magn
etic
Ind
uct
ion
Fara
day's
Law
Lenz's
Law
Alt
ern
ator
Tran
sfor
mers
“Nothing is too wonderful to be true if it be consistent with the laws of nature.”
“Why, sir, there is every probability that you will soon be able to tax it!”
Faraday
Michael
Hint: Analysis is similar to force on a current carrying wire.
F = Bqv
According to Fleming’s LHR, the electron moves into the page.
(a)
(b)
(flux through a coil of N turns)
emf = Blv
flux cut / time
wire moves down through field
induced emf
flux cut / time induced emf
Faraday’s law concentrates on the emf produced by electromagnetic induction.
uniform field
What would the resultant magnetic field look like?
+

connected to power supply
moveable rod
parallel rods held in place
N
S
motion
variable resistor
ammeter
power supply
birds eye view
yoke magnets
toppan balance
currentcarrying wire
Investigating the force
on a currentcarrying wire in a magnetic field
change the number of magnets to vary length of wire in field
change the current
ing
rges
B
F
currentcarrying wire
currentcarrying wire
B
F
BIL
force always BIL,
but direction of the force changes
What does this mean?
What would happen if the magnet was attracted to the solenoid when pushed towards it?
The plates and the magnet are aligned so that each ion passing through is acted on by an electric field force
F elec = QV/d and a magnetic field force F mag = BQv.
Only ions moving at a speed such that F elec = F mag
(equal and opposite forces) get through undeflected.
v = Electric field strength/Magnetic flux density
What happens to a charged particle moving through a magnetic AND an electric field?
Both fields will exert an individual force on the particle. Dependent on the fields' directions, the particle's deflection can be used to identify the particle's mass and velocity.
r = mv / Bq
Flux linkage = BAN sin t (for a rotating coil)
Therefore emf = BAN cos t
For max. emf, cos t = 1,
so max emf = BAN
3
2
Answer: 2V
High voltage is lethal in a house, but 230V can only be transmitted with cables that have a much lower resistance, requiring thick, costly cables.
What's the solution?
N
S

equivalent to current
Region of magnetic field B out of the screen
Q
Force = QE
Force = BQv
Electric field
What would happen to an electron entering the fields?
It travels straight through with v = E/B
TAP 4061 from http://www.tap.iop.org/fields/electrical/406/page_46863.html
Draw the field lines for the different examples:
1
1
2
1
1
7
3
9.2 Questions
Questions
Taken from TAP worksheet.
2
Electromagnetic Induction
from the side:
from above:
motion
spinning allows the wire to cut the field lines constantly
the spinning wire only cuts through the field lines at certain angles

+
A positive particle traveling in a vacuum and acted upon by both an electric and a magnetic field
Region of magnetic field B down / into the screen
+Q
Force = BQv
Force = QE
Electric field

+
N.B. q/m is known as the specific charge
F = Bqv sin
A charged particle moving at an angle, θ, to the magnetic field will experience a force,
Why the negative symbol?
The negative symbol can be explained by Lenz's Law.
A charged object is normally modelled as a point charge...
positive as long as Q is positive
Equipment:
Logbook ML
Microammeter
Connecting cables
Helmholtz coil
Open Datalogging Insight and pick COM3
Go to: Set up  Timespan
Select: 0.01 seconds
"Start" to collect data.
Go to: Edit  Copy Data to export to excel.
When a wire is moved between opposite poles, cutting through a uniform field, a voltage is induced across its ends. The induced voltage pushes current around the circuit through the meter.
A wire contains many free electrons because it is made of metal. When the wire is moved across the field, the field pushes the electrons along the wire.
Why does this happen?
If the wire stops moving, no voltage is induced in the wire.
The induced voltage can be made larger by using a stronger magnet, using more wire or moving the wire faster.
If the wire is moved along the lines of the field, no voltage is produced.
N
coil
coil
N
coil
coil
coil
coil
N
coil
S
S
S
S
N
How could we use what we know to find the magnetic flux density
of the Earth?
Observe and explain
why the swinger comes to a rest so quickly...
Coulomb’s law
the concept of electric fields as being one of a number of forms of field giving rise to a force
You need to demonstrate and apply knowledge and understanding of...
6.1 all questions
Questions
Questions
Textbook p427
Q 2, 4 & 5
Questions
You need to demonstrate and apply knowledge and understanding of...
electric field strength
Textbook page 91
All questions
Questions
A charged object is normally considered to be a point charge.
Electric field strength
of a point charge
Relate E = F/Q to Coulomb’s Law to derive the electric field strength of a point charge...
Fields are regions where a force is exerted on an object at a distance.
Some key points:
The direction of the field is shown by an arrow (+ to )
Electric fields lines are at right angles to the surface of a conductor
A uniform field is shown by equally spaced parallel lines
Stronger fields are represented by more densely packed field lines.
Uniform field here!
analogous to Newton's law of gravitation  but for charged particles
Any two point charges exert an electrostatic (electrical) force on each other that is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
where k is a constant
absolute permittivity of free space
electric field strength for a point charge
similarities and differences between the gravitational field of a point mass and the electric field of a point charge
You need to demonstrate and apply knowledge and understanding of...
Questions
Textbook p 431
Q 3, 4, 6 & 7
Questions
Investigating Coulomb's law
0.03 g
top pan balance
+
+
what will happen?
+
The electric field strength is directly proportional to the charge and obeys the inverse square law of the distance.
E =
F
q
E =
kQ
r
2
electric field strength for a point charge
Questions
Textbook p 431
Q 5
Questions
6.2 questions 3 & 4
Questions
Textbook page 93
question 1
Questions
Electric potential & energy
motion of charged particles in a uniform electric field
You need to demonstrate and apply knowledge and understanding of...
electric potential at a point as the work done in bringing unit positive charge from infinity to the point; electric potential is zero at infinity
electric potential at a distance from a point charge; changes in electric potential
force–distance graph for a point or spherical charge; work done is area under graph
electric potential energy a distance from a point charge
You need to demonstrate and apply knowledge and understanding of...
uniform electric field strength
parallel plate capacitor; permittivity
You need to demonstrate and apply knowledge and understanding of...
Uniform electric fields
oppositely charged plates
small test charge
+
experiences constant force
The charge gains energy as it moves from positive plate to negative plate.
F = E/Q
RECALL: p.d. = work done per unit charge
and work done on the charge is Fxd
W = Fd
VQ = (EQ)d
E = V/d
V = W/Q
and
but this equation only works for parallel plates
This is really useful as we can determine the electric field strength between parallel plates using just a voltmeter and a ruler.
E = V/d
NC = Vm
1
1
1
1
cancel Q
EQUATE
Textbook p 435
Questions
Wider CONTEXT:
Parallel Plate Capacitor
For plates in a vacuum (or air), experiments show that capacitance is directly proportional to the area and inversely proportional to the separation. It will also depend upon the insulator that is used.
A simple capacitor can store charge by separating said charge on two plates with an insulator in between.
The constant of proportionality in this relationship is the permittivity of free space, , as used in Coulomb's law.
C =
3
A
0
r
3
0
When an insulator other than a vacuum is used, we use the permittivity for the insulator such that:
C =
3
A
r
permittivity of insulator
About E=V/d: Q 2&3
About capacitors: Q 46
oppositely charged plates
Electron

experiences constant force
F = E/Q
A p.d. of just 1.5 V from a cell can accelerate an electron up to 700 km/s.
Even greater voltages can be used to accelerate particles close to the speed of light. Protons reach energies of around 400 MeV in a linear particle accelerator.
The acceleration of the particle will be constant due to the constant electrostatic force.
We can analyse the motion of a charged particle moving at right angles to an electric field in much the way we dealt with projectile motion in mechanics.
The particle's motion parallel to the plates is at a constant velocity and unaffected by the electrostatic force. Indeed, there are no other forces acting on the particle (in most cases), so the acceleration = 0.
The particle's motion perpendicular to the plate is at a constant acceleration as provided by the constant electrostatic force of the parallel plates.
a = F/m = EQ/m
u = 0
a = 0
u = v
t = L/v
where L = horizontal distance moved parallel to plates
use suvat to solve
Textbook p 438
Questions
Q 24
Textbook p 442
Questions
About V: Q 14 & 6
About capacitors: Q 5
Textbook p 4435
MORE Questions
Q 1, 2 & 5
About capacitors: Q 3 & 4
Questions
6.4 questions 13
Electric potential energy
Charges repel each other so you have to do work to decrease the separation between the charges.
+
+
More force is needed to push the charges together as the separation decreases. All the work done is stored as electric potential energy  this energy is recoverable, just let go!
force
0
0
area = work done
F is inversely proportional to r
2
r
separation
r
+q
+Q
The force between the two positive charges varies with their separation.
The total work done to bring the particles from infinity to a separation r is the total area under
the graph.
It is the same as the electric potential energy.
W =
kQq
r
work done = electric potential energy
If one of the charges is negative, then the value for W will also be negative  this represents the amount of external energy required to completely separate the charged particles to infinity.
This is what happens when atomic or molecular bonds are broken.
Electric potential
The electric potential, V, at a point is defined as the work done per unit charge in bringing a positive charge from infinity to that point. If the test charge is q, the equation for V can be determined by dividing the electric potential energy W by q.
V = =
W
q
kQ
r
The unit for electric potential is unit JC or volts.
1
Electric potential difference (p.d.) is the work done per unit charge between two points around the particle of charge Q.
+Q
A
B
r
A
r
B
The electric p.d. between point A and B is the difference in the potentials at these two points.
Placing a voltmeter between A and B would show the work done per unit charge  if a voltmeter could be placed in empty space around charged spheres.
V =
kQ
r
A
V =
kQ
r
B
Wider CONTEXT:
Parallel Plate Capacitor
A capacitor stores charge, as does an isolated charged sphere of radius, R.
The capacitance of a charged sphere is the ratio of the charge it stores, Q, to the potential, V, at its surface.
C =
Q
V
R
k
V =
kQ
r
far away from other charged objects
=
Imagine the Earth as a huge capacitor floating is space.
If R = 6400km,
calculate
Earth's capacitance.
7.1 x 10 F
4
C =
Equally spaced parallel lines
Magnetic field lines tell us the strength and the direction of the field
 N to S  the direction that
a free North pole would move.
created by the electrons moving within the wire
magnetic field created by electrons moving around iron nuclei
the iron atoms act as tiny magnets, all alligned in the same direction
direction determined by right hand grip rule
Questions
Textbook p 447
Q 3 & 4
Questions
Questions
Textbook p 451
Q 3 to 5
Questions
Questions
Textbook p 456
Q 2 to 5
Questions
7
q V = ½ m v
v = 9.4 x 10 ms
PAG 11.3
• To determine the magnetic field strength of a magnet
• To demonstrate investigative skills
• To research, plan and implement a practical activity
• To make quantitative measurements
• To relate observations to research or theory
AIM:
a) Calculate the electric field strength between the plates.
b) A tiny sphere of weight 3.3 x 10 N has acquired a charge so that it is held in equilibrium midway between the plates by the electric field.
14
(i) State the magnitude and direction of the electric force on the sphere.
(ii) Calculate the charge on the sphere.
(iii) Describe the motion
of the sphere if V doubles.
You need to demonstrate and apply knowledge and understanding of...
magnetic fields are due to moving charges or permanent magnets
magnetic field lines to map magnetic fields
magnetic field patterns for a long straight current
carrying conductor, a flat coil and a long solenoid
You need to demonstrate and apply knowledge and understanding of...
Fleming’s lefthand rule
force on a currentcarrying conductor
magnetic flux density; the unit tesla
techniques and procedures used to determine the uniform magnetic flux density between the poles of a magnet using a currentcarrying wire and digital balance
You need to demonstrate and apply knowledge and understanding of...
force on a charged particle travelling at right angles to a uniform magnetic field; F = BQv
charged particles moving in a uniform magnetic field; circular orbits of charged particles in a uniform magnetic field
You need to demonstrate and apply knowledge and understanding of...
charged particles moving in a region occupied by both electric and magnetic fields; velocity selector
You need to demonstrate and apply knowledge and understanding of...
Faraday’s law of electromagnetic induction
You need to demonstrate and apply knowledge and understanding of...
Lenz’s law
emf =  rate of change of magnetic flux linkage
techniques and procedures used to investigate magnetic flux using search coils
You need to demonstrate and apply knowledge and understanding of...
magnetic flux, the unit weber
magnetic flux linkage
You need to demonstrate and apply knowledge and understanding of...
simple a.c. generator
You need to demonstrate and apply knowledge and understanding of...
simple laminated ironcored transformer
techniques and procedures used to investigate transformers