Name……………….
Class……….
Plymstock School Physics
Department
Module G485.1 Electric and
Magnetic Fields
student booklet
Lesson 1 notes – Electric Fields
Objectives
(a) state that electric fields are created by electric charges;
(b) define electric field strength as force per unit positive charge;
(c) describe how electric field lines represent an electric field;
Outcomes
Be able to state that electric fields are created by electric charges;
Be able to define electric field strength as force per unit positive charge;
Be able to describe how electric field lines represent an electric field;
Electric field
A gravitational field is the force per unit mass.
An electric field is the force per unit positive charge.
E=F/Q
Gravitational fields are always attractive.
Electric fields can be attractive or repulsive.
Electrical charges exert forces upon one another. Like charges repel, unlike
charges attract. A field is set up by a charge, and any other charge in that field
will experience a force due to the field and we represent fields with field lines.
Field Lines
Rules about field lines:
Lines are drawn from positive to negative.
They never start or stop in empty space – they stop or start either on a charge
or “at infinity”.
They never cross – if they did, a small positive charge placed there would feel
forces in different directions, which could be resolved into the one true direction
of the field line there.
The density of field lines on a diagram is indicative of the strength of the field.
The second diagram above also shows a point exactly between two like
charges where no field exists (since the forces on a charge placed there would
be exactly equal and opposite in direction). Such a point is called a neutral
point.
Equipotentials
The lines on the hills on this map are called
contour lines. They show areas of a hill that
are the same height above sea level.
The gravitational potential energy along this
line would also be the same.
Gravitational equipotentials
We could say that along these lines there
would be an equal potential wherever we
were on that line; Or an equipotential.
If we drew these lines in the air around the Earth as shown then we would have
an equal potential all along them.
49.0 J/kg
39.2 J/kg
29.4 J/kg
19.6 J/kg
9.8 J/kg
Electric field equipotentials
We can do the same for electric field equipotentials.
(resourcefulphysics.org)
The dotted lines show the equipotentials for the fields caused by the charges
shown.
Lesson 2 notes – Coulomb’ Law
Objectives
(d) select and use Coulomb’s law in the form
F = Qq
4πo r2
(e) select and apply
E =Q
4πo r2
for the electric field strength of a point charge;
(f) select and use
E =V
d
for the magnitude of the uniform electric field strength between charged parallel
plates;
Outcomes
Be able to select and use Coulomb’s law in the form
F = Qq
4πo r2
Be able to select and apply
E =Q
4πo r2
for the electric field strength of a point charge;
Be able to select and use
E =V
d
for the magnitude of the uniform electric field strength between charged parallel
plates;
Coulomb’s law
We know that a field exists around a charge that exerts force on other charges
placed there, but how can we calculate the force? The force will be dependent
upon the sizes of the charges, and their separation. In fact the force follows an
inverse square law, and is very similar in form to Newton’s Law of Universal
Gravitation. It is known as Coulomb’s law, and it is expressed as:
F
where
kQ1Q2
r2
F = force on each charge (N)
Q1 and Q2 are the interacting charges (C)
r = separation of the charges (m)
The k is a constant of proportionality (like G in Newton’s Law of Universal
Gravitation). In a vacuum, and to all intents and purposes, in air, we have
k = 9.0 x 109 N m2 C-2
(units obtained by rearranging the original equation)
More traditionally, Coulomb’s law is written:
F
Q1Q 2
4 o r 2
where 0 is known as the “permittivity of free space”; 0 = 8.85 x 10-12 F m-1
(farads per metre). Permittivity is a property of a material that is indicative of
how well it supports an electric field, but is beyond the scope of these notes.
Thus, we have k = 1/ (4π 0). Different materials have different permittivities,
and so the value of k in Coulomb’s law also changes for different materials.
Points to bring out about Coulomb’s law:
The form is exactly the same as Newton’s Law of Universal Gravitation; in
particular, it is an inverse-square law.
This force can be attractive or repulsive. The magnitude of the force can be
calculated by this equation, and the direction should be obvious from the signs
of the interacting charges. (Actually, if you include the signs of the charges in
the equation, then whenever you get a negative answer for the force, there is
an attraction, whereas a positive answer indicates repulsion).
Although the law is formulated for point charges, it works equally well for
spherically symmetric charge distributions. In the case of a sphere of charge,
calculations are done assuming all the charge is at the centre of the sphere.
In all realistic cases, the electric force between 2 charges objects absolutely
dwarfs the gravitational force between them.
Field strength
For a gravitational field, the Field strength is defined at a point in the
gravitational field “As the force per unit mass placed at that point in the field” –
with units therefore of N kg-1.
What would therefore be the natural way to extend this definition to the electric
field?
As the force per unit charge. Thus it would have units of N C-1.
We thus define the electric field strength at a point in a field as:
E = F/Q
where
E = electric field strength (N C-1)
F = force on charge Q at that point if the field
Important notes:

The field strength is a property of the field and not the particular charge
that is placed there. For example, at a point where the field strength is
2000 N C-1, a 1 C charge would feel a force of 2000 N whereas a 1 mC
charge would feel a force of 2 N; the same field strength, but different
forces due to different charges.

The field strength is a vector quantity. By convention, it points in the
direction that a positive charge placed at that point in the field would feel
a force.
Now, for the non-uniform field
due to a point (or spherical)
charge, we can use Coulomb’s
law to find an expression for the
field strength. Consider the force
felt by a charge q in the field of
another charge Q, where the
charges are separated by a
distance r:
F
kQq
r2
by Coulomb’s Law.
But E = F/q and so
E
kQ
r2
This is our result for the field strength at a distance r from a (point or spherical)
charge Q.
Lesson 2 questions – Coulombs Law
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Lesson 3 questions – Coulomb’s Law
ALL
1
This question is about electric forces.
A very small negatively-charged conducting sphere is suspended by an insulating
thread from support S. It is placed close to a vertical metal plate carrying a positive
charge. The sphere is attracted towards the plate and hangs with the thread at an angle
of 20 degrees to the vertical as shown in fig 1.1.
fig 1.1
a)
Draw at least five field lines on fig 1.1 to show the pattern of the field
between the plate and the sphere.
(3)
MOST
b)
The sphere of weight 1.0 x 10-5N carries a charge of –1.2 x 10-9C.
i)
Show that the magnitude of the attractive force between the
sphere and the plate is about 3.6 x 10-6N.
ii)
(3)
Hence show that the value of the electric field strength at the
sphere, treated as a point charge, is 3.0 x 103 in SI units. State the
unit.
Unit for electric field strength is ……………… (3)
c)
The plate is removed. Fig 1.2 shows an identical sphere carrying a
charge of +1.2 x 10-9C, mounted on an insulating stand. It is placed so
that the hanging sphere remains at 20 degrees to the vertical.
fig 1.2
Treating the spheres as point charges, calculate the distance r between their centres.
r = ………………….m (3)
ALL
d)
On fig 1.2, sketch the electric field pattern between the two charges. By
comparing this sketch to your answer to (a), suggest why the distance between the plate
and the sphere in fig1.1 is half of the distance between the two spheres in fig 1.2.
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2
a)
Define electric field strength at a point in space.
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……………………………………………………………………………………… (2)
b)
Fig 2.1 shows two point charges of equal magnitude, 1.6 x 10-19C, and
opposite sign, held a distance 8.0 x 10-10m apart at points A and B. The charge A is
positive.
8.0 x 10-10m
A
B
Fig 2.1
i)
On Fig 2.1, draw the electric field lines to represent the field in
the region around the two charges.
(3)
ii)
Calculate the magnitude of the electric field strength at the mid
point between the charges. Give a suitable unit for your answer.
Electric field strength = ……………………….. unit ……………….. (5)
MOST
c)
Imagine two equal masses, connected by a light rigid link, carrying equal but
opposite charges. This is a system called a dipole. Fig 2.2 and 2.3 show the dipole
placed in different orientations between two uniformly and oppositely charged plates.
fig 2.2
fig 2.3
Any effects of gravity are negligible.
i)
Describe the electric forces acting on the charges by drawing
suitable arrows on the diagrams.
ii)
Explain the motion, if any, of the dipole when it is released from
rest
in fig 2.2
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in fig 2.3
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……………………………………………………………………………………… (5)
Lesson 4 notes – Electric field strength and potential
Objectives
Select and use E=V/d for the magnitude of the uniform electric field strength
between charged parallel plates
Outcomes
Be able to select and use
E =V
d
for the magnitude of the uniform electric field strength between charged parallel
plates;
Uniform electric fields
What do we mean by a “uniform” field?
One that does not vary from place to place.
In terms of the field lines, this means that
they are parallel and evenly spaced. Also
as field strength = – (potential gradient), the
equipotentials should also be evenly
spaced.
Where have we seen such as field?
The field between two parallel charged
plates.
How will a charge move in such a field?
The force is given by
F = EQ
Since E is constant, the force will be constant and therefore, by
F = ma
the acceleration will also be constant and directed along the field lines for a
positive charge and opposite to the field lines for a negative charge.
Now, W=Fd which can be written F=W/d
So E = F/Q becomes E = W/Qd
W/Q is the definition of potential difference V,
So we get for uniform fields that:
E = V/d
Lesson 5 notes – projectile comparisons
Objectives
(g) explain the effect of a uniform electric field on the motion of charged
particles;
(h) describe the similarities and differences between the gravitational fields of
point masses and the electric fields of point charges.
Outcomes
Be able to explain the effect of a uniform electric field on the motion of charged
particles;
Be able to describe the similarities and differences between the gravitational
fields of point masses and the electric fields of point charges.
Comparison of gravitational and electric fields
We finish this topic by discussing the parallels between gravitational and
electric fields.
We have seen the following similarities:
Both fields follow an inverse square law for both force and therefore field
strength.
For both types of field, potential and potential energy are inversely proportional
to distance.
We define the zero of potential to be at infinity
For attractions (and therefore always for gravitational fields) potential energies
are always negative.
The major difference is that repulsions occur in electrostatic fields between
charges of the same sign, whereas as far as we know, gravity is always
attractive.
The similarity of the fields may be brought home further by comparing the forms
of the equations:
Gravitational fields
Electric fields
field strength = -(potential gradient)
g = –V/d (a uniform field)
E = –V/d (a uniform field)
2
F = Gm1m2/r
F = kQ1Q2/r2
2
g = GM/r (a non-uniform field)
E = kQ/r2 (a non-uniform field)
GPE = –Gm1m2/r
EPE = kQ1Q2/r
V = –GM/r
V = kQ/r
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Lesson 5 questions – projectile comparisons
MOST
1
This questions is about changing the motion of electrons using electric fields.
Fig 1.1 shows a horizontal beam of electrons moving in a vacuum. The electrons pass
through a hole in the centre of the metal plate A. At B is a metal grid through which the
electrons can pass. At C is a further metal sheet. The three vertical conductors are
maintained at voltages of +600V at A, 0V at B and +1200V at C. the distance from
plate A to grid B is 40mm.
Fig 1.1
a)
On fig 1.1 draw electric field lines to represent the fields in the regions
between the three plates.
b)
Show that the magnitude of the electric field strength between plate A
and grid B is 1.5 x 104Vm-1.
(2)
c)
hole in A.
Calculate the horizontal force on an electron after passing through the
Force = ………….. N (2)
SOME
d)
Show that the minimum speed that an electron in the beam must have at
the hole in A to reach the grid at B is about 1.5x107 ms-1.
(2)
e)
Calculate the speed of these electrons when they collide with sheet C.
Speed = ……………….. ms-1 (1)
f)
Describe and explain the effect on the current detected at C when the
voltage of the grid B is increased negatively.
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ALL
2
a)
Calculate the (i) gravitational force (ii) the electrical force between two
protons that are 2.0 x 10-10 m apart. Take the mass of a proton to be 1.7 x 10-27kg.
i)
Force = …………….N
ii)
b)
Force = …………….N (4)
What is the ratio of the electrical to gravitational force?
Ratio = …………………….. (2)
Lesson 6 notes - Magnetic Fields
Objectives
(a) describe the magnetic field patterns of a long straight current-carrying
conductor and a long solenoid;
Outcomes
Be able to describe the magnetic field patterns of a long straight currentcarrying conductor and a long solenoid.
A magnetic field is a region in which a particle with magnetic properties
experiences a force, and in which a moving charge experiences a force.
There are two main classes of magnet:
1. Permanent magnets
2. Electromagnets
Field Shapes
Permanent Magnets:
Permanent magnets are common and are made of iron, cobalt or nickel alloys.
To represent the field around a magnet we use a diagram which needs to obey
some rules so that whoever uses it can interpret it correctly.
Here is an example:
The points to note are:
We draw lines to represent magnetic fields. These lines are called lines of flux.
The arrow shows the direction of the force that a free north pole, for instance a
North pole with no South pole (which doesn’t exist!) would feel.
Field direction always goes from North to South. So pop a magnet at X in the
field (see diagram) and it would align itself with its North pole pointing along the
arrow.
The spacing between the lines of flux tells you about the strength of field - as
the lines get closer together, the field becomes stronger - for example, near the
poles.
Look at this field:
The region in between the poles shows equally spaced, parallel lines. This is
called a uniform field. Field strength remains constant as you move around
this area. Move out from the space between the poles and the field strength
reduces. The lines of flux become further apart.
Temporary Magnets:
Around any conductor that has a current flowing through it there is a
magnetic field. Switch off the current and the magnetic field disappears.
The shape of the field around a straight wire is shown below:
Note: The X means that conventional current is flowing through a wire into the
page. (Think of an arrow - going away you see the flights, coming towards you
see the point!)
Remember: Conventional current is the flow of positive charges. So
conventional current goes in the opposite direction to the electron flow.
In a wire with conventional current flowing out of the page you get:
It’s the same field shape as the one above, but the direction of the field is
different.
Notice that in both cases the lines get further apart as you move away from the
wire, this is because the magnetic field is getting weaker.
How do you remember which way the field goes (clockwise or
anticlockwise)?
Answer: Use the corkscrew rule! The problem here is that as many of you are
under 18 you won’t have a clue what a corkscrew is… obviously! This will
become easier once you’re over 18 as you will be allowed to drink wine and will
therefore have knowledge of a corkscrew.
Imagine you are screwing a corkscrew into or out of the page in the same
direction as conventional current. The turning motion of the corkscrew is in
the same direction as the field arrows need to be.
You can use the same idea to work out the shape of the field when the wire is
coiled. Apply the corkscrew rule to different sections of the coil (below) and you
should see that at all points, the field is to the right on the inside of the coil and
to the left outside. In this example, we've cut a coil in half and are looking at it
from the side - so the conventional current comes out of the page at the top and
into the page at the bottom.
If you look at a long coil of wire (called a solenoid) the field shape becomes:
There is a uniform field inside the centre of the coil; outside the field is the
same as the field around a bar magnet.
Right Hand Grip Rule
A quick way to work out the direction of the magnetic field in a solenoid is the
right hand grip rule…
Make a fist and stick your thumb out (as if hitchhiking). Your fingers are
wrapped in a circle, same as the coils in the solenoid. If you make your fingers
point in the same direction as the conventional current around the coil - your
thumb points towards the end of the solenoid that is the North pole.
Extension
Neutral Points:
When two fields coincide they may cancel each other out and produce points
where the magnetic field strength is zero. These points are called neutral
points.
For Example:
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Lesson 6 questions – Magnetic field patterns
ALL
1)
Figs 1.1 and 1.2 illustrate magnetic field patterns caused by current-carrying
conductors in a plane at right angles to the conductors.
fig 1.1
fig 1.2
What shapes of conductor would produce these field patterns above?
Fig. 1.1
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Fig. 1.2 …………………………………………………………………………………
(2)
Total [2]
2)a) Explain why a compass needle placed very close to a wire may deflect when the
current in the wire is switched on.
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…
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(1)
b)
Fig. 2.1 shows a cross section of a current carrying conductor.
current-carrying
conductor (current out
of plane of page)
fig 2.1
On fig 2.1, draw the magnetic field pattern.
(3)
Total [4]
3)
Fig. 3.1 shows a cell connected to two terminals on a plastic box. When switch
S is closed the magnetic field pattern shown is detected around the box.
fig 3.1
State what is inside the box that produces the field pattern.
…………………………………………………………………………………………
(1)
Total [1]
Learning typical shapes of magnetic fields
Use the examples and guidelines suggested below to learn how to make a rough sketch of the
expected shape of the magnetic fields of magnets and coils.
Flux goes with the flow
Inside a magnet or a piece of magnetised material, the flux just follows the direction of
magnetisation. It emerges from, and enters into, the iron at the poles. So start sketching at the
poles, all flux lines are continuous. A line which emerges (conventionally at a north pole) enters
the material again at the south pole. Flux lines never cross. Think of flux as like a fluid pumped
out of N poles and sucked into S poles (although nothing is actually flowing or physically
connected to the magnet etc)
Here is a sketch of the flux from a short bar magnet:
4.
S
Sketch the flux from a longer magnet, like this:
N
5.
Sketch the flux from a thin flat magnet, such as a Magnadur magnet, like this:
N
S
6.
Sketch the flux from a horseshoe magnet, like this:
Use symmetry
Magnetic fields are usually very symmetrical. Think about which parts must be just like others,
or perhaps their mirror reflections when drawn in two dimensions. For example, the field of the
coil below can be divided into four quarters, each a copy (reflected or inverted) of the others. So
you only need to draw one bit of the field.
7.
Identify the similarly shaped regions of the field between a N and a S pole.
S
N
8.
Identify the similarly shaped regions of the field around a pair of coils with currents
going in the same direction round them. Sketch the field around and in between them.
N and S poles of coils
Looking at a coil face on, if the current goes anticlockwise that face is like a N pole and flux
emerges from it. If the current goes clockwise that face is like a S pole and flux goes into it.
Arrows drawn on the letters N and S help to remember this rule.
9.
Identify N and S poles of this long coil:
–
+
10.
Identify N and S poles of this electric motor winding:
Same environment, same flux
If the pattern of current turns around one place is the same as that around another, the flux
pattern in those places will be the same.
11.
State how this principle tells you that the flux in a long narrow coil will be straight and
uniform, like this:
12.
Sketch the flux inside this doughnut shaped coil:
Put it all together
Use all these ideas together to guess the shape of the flux.
13.
Sketch the flux in the air and in the iron of this electric motor:
pole with winding
stator
rotor
pole with winding
Lesson 7 notes – F=BIL and Fleming’s Left hand Rule
Objectives
(b) state and use Fleming’s left-hand rule to determine the force on current
conductor placed at right angles to a magnetic field;
(c) select and use the equations F = BIL and F = BILsinθ;
(d) define magnetic flux density and the tesla;
Outcomes
State Fleming’s left-hand rule to determine the force on current conductor
placed at right angles to a magnetic field;
Select and use the equations F = BIL
Be able to define magnetic flux density.
Use state Fleming’s left-hand rule to determine the force on current conductor
placed at right angles to a magnetic field;
Select and use the equations F = BILsinθ;
We know that a current carrying wire as a magnetic field around it. When it is
placed into another magnetic field it will feel a force. We have seen
demonstrations of this and know applications of this like the electric motor.
Force on a current-carrying wire
l
I
0.003N
To investigate the forces involved we can use a set up as above. A current
carrying wire is clamped between two permanent magnets in a yoke placed on
a top pan balance. The force exerted on the balance can then be recorded.
The amount of current through the wire and the lent of the wire can both be
changed.
We see a pattern as sown in the graphs below:
This means that Force is proportional to the Current through the wire
And that Force is proportional to the length of the wire.
So we can write that:
F=BIL
Where
F is the force felt by the current carrier
I is the current through the wire
L is the length of the wire
And
B is the Magnetic Field strength (or magnetic flux density)
You need to be able to rearrange the equation to get:
B=F/IL
I=F/BL
L=F/BI
Magnetic Flux density is defined as the force per unit length per unit current
on a current carrying wire placed perpendicular to the field lines. The units are
Tesla.
The force F on a wire can be shown to be proportional to
(a) the current on the wire I,
(b) the length of the conductor in the field L,
(c) the sine of the angle q that the conductor makes with the field , and
(d) the strength of the field - this is measured by a quantity known as the
magnetic flux density B of the field.
The force is given by the equation: F = BILsinθ
The units for B are tesla (T).
The greatest force occurs when the angle that the wire is to the field lines is
90o.
Fleming’s Left hand rule
The force, magnetic field and current all need to be at right angles.
thuMb – Motion of current carrier
First Finger – Field direction
seCond finger – Current direction
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Class…………
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Lesson 7 questions – F=BIL and Flemings left hand
rule
ALL
1)
Fig. 1.1 shows an arrangement for measuring the magnetic flux density B
between poles of a magnet.
fig 1.1
fig 1.2
The coil shown in figs 4.1 and 4.2 has 50 turns. Its lower side XY is horizontal and as a
mean length of 30mm. Before the current is switched on, the balance reads 0.850N and
with the current of 2.0A switched on the balance reads 0.815N.
a)
State the rule that can be used to determine the direction of the force acting on
the magnet.
……………………………………………………………………………………… (1)
b)
Determine the magnitude and direction of the electromagnetic force acting on
the magnet.
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……………………………………………………………………………………… (2)
a)
Calculate the magnetic flux density B between the poles of the magnet.
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Total [6]
2)
Fig 2.1 shows a wire placed at righ angles to a magnetic field. The wire rests on
two metal supports.
fig 2.1
The length of the wire between the supports is 4.5 cm, and this length has weight 6.0 x
10-2N. The current in the wire is slowly increased from zero until the wire starts to lift
off the metal supports.
a)
Calculate the current I in the wire. The magnetic flux density is 0.36T.
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b)
Suggest why the overhead cables for the National Grid cannot be freely
supported by the Earths magnetic field.
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……………………………………………………………………………………… (2)
Total [5]
3)
Fig. 3.1 shows a current-carrying metal rod that can roll freely on two parallel
metal rails. The rod is at right angles to the magnetic field lines.
fig 3.1
a)
Determine the direction of the force experienced by the rod. Explain how you
determined this direction.
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
b)
The current in the metal rod is 2.0A and it has a length 5.0cm between the two
metal rails. Calculate the force experienced by the metal rod given the magnetic flux
density is 1.8 x 10-3T.
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……………………………………………………………………………………… (3)
Total [5]
Lesson 9 notes – Charged particles in a magnetic field
Objectives
(e) select and use the equation F = BQv for the force on a charged particle
travelling at right angles to a uniform magnetic field;
(f) analyse the circular orbits of charged particles moving in a plane
perpendicular to a uniform magnetic field by relating the magnetic force to the
centripetal acceleration it causes;
Outcomes
Be able to select and use the equation F = BQv for the force on a charged
particle travelling at right angles to a uniform magnetic field.
Be able to analyse the circular orbits of charged particles moving in a plane
perpendicular to a uniform magnetic field by relating the magnetic force to the
centripetal acceleration it causes.
Be able to derive the equation F = BQv from the definition of current and
magnetic flux density.
When a wire carrying a current through a field feels a force it is because the
magnetic field pushes the electrons inside the wire to one edge of the wire.
These electrons actually then apply force to the wire.
The same effect occurs if the electrons are not inside a piece of wire - for
example, if they are in a beam crossing a vacuum.
F=BIl
v=l/t
So l=vt
So, F=BIvt
But, I=q/t
So q=It
Therefore, F=Bqv
For an electron with charge e,
F=Bev
So,
F = B q v sin θ
Where: F = force (N)
B = magnetic field strength (T)
q = charge on the particle (C)
v = velocity of the particle (m/s)
Note: Angle θ is between the direction of the beam and the magnetic field
direction.
Use Fleming’s left hand rule to work out the direction of the force. Align your
second finger with the beam of particles remembering that it points the way
positive particles flow, the opposite way to electron flow.
The diagrams show a charged particle in a magnetic field. The force is always
at right angles to the motion of the charged particle so if the speed of the
particle is right then a circular path can be created.
Finding the Charge-to-Mass Ratio of a Particle
When a charged particle enters a magnetic field we now know it will be forced
to change direction. If it stays in the field it will continue to change direction and
will move in a circle. The force produced will provide the centripetal force on the
moving particle.
½ mv2=Vq (since V=W/q)
And mv2/r=Bqv
So mv2=Bqvr
½ Bqv r=Vq
½ Bvr =V
v = 2V/Br
So q/m = 2V/r2B2
This idea is used in velocity-selectors, where particles of different mass-tocharge ratio will rotate in circles with a different radius. These are used in mass
spectrometers.
Lesson 9 questions – Moving charge in a magnetic field
Name…………………. Class……………..
MOST
1
(
/36)…….%…….
2
3
Lesson 12 notes – Electromagnetic induction
Objectives
(a) define magnetic flux;
(b) define the weber.
(c) select and use the equation for magnetic flux φ=BAcosθ;
Outcomes
Be able to define magnetic flux;
Be able to define the weber.
Be able to select and use the equation for magnetic flux φ=BAcosθ;
Be able to rearrange and use the equation for magnetic flux φ=BAcosθ.
Magnetic Flux
Around a magnet there is a magnetic field with an assiciated flow of magnetic
energy around it.
The flow of energy is called magnetic flux (φ).
If B is at 90° to the area, A, then the total magnetic flux, f "passing through" A is
defined to be:
φ = AB
Magnetic flux is given the symbol φ and is measured in units called Webers
(Wb).
The units of flux are: Tm² and 1 Tm² = 1 Weber
The flux associated with a magnetic field is therefore a measure of the number
of magnetic field lines penetrating some surface.
The above picture shows the special case of a plane area A and a uniform flux
density B. The normal to the field is at an angle with the field. In this case, the
flux is given by:
φ=BAcosθ
If B is the value of the flux density.
And the angle is between the plane of the surface and the magnetic field lines.
If a conductor is passed through a magnetic field then a voltage will be induced
across it.
Flux Linkage
For N number of coils we use this equation:
φN=BANcosθ
Lesson 14 notes – Faraday and Lenz
Objectives
(e) state and use Faraday’s law of electromagnetic induction;
(f) state and use Lenz’s law;
(g) select and use the equation: induced e.m.f. = –rate of change of magnetic
flux linkage;
Outcomes
Be able to describe the dynamo effect.
Be able to state Faraday’s law of electromagnetic induction.
Be able to state Lenz’s law.
Be able to use Lenz’s law.
Be able to select and use the equation: induced e.m.f. = –rate of change of
magnetic flux linkage;
Be able to derive the equation: induced e.m.f. = –rate of change of magnetic
flux linkage;
The Dynamo Effect
When a wire is moved relative to a
magnetic field, an electric current is
induced (brought about) in the wire. This
is the dynamo effect.
If the wire is stationary then no electric
current is produced.
A dynamo transforms kinetic
(movement) energy into electrical energy
because of this dynamo effect.
Electrical Generators
To easily generate electricity a coil of wire is normally placed inside a fixed
magnet. This can then be rotated in order to induce a current within it.
In this example, the friction driver may be placed against the tyre of a bike.
The same idea is used in most power stations but Instead of a bike wheel
turning the axle, moving turbines spin it round.
Increasing the Dynamo Effect
The current that the dynamo effect induces can be increased in three simple
ways:
 Make the fixed magnetic field larger by using stronger magnets;
 Use more turns of coiled wire increasing the length of the conductor
inside the magnetic field;
 Make the relative movement between the wire and the magnets faster.
In other words, spin the coil quicker.
AC Current
Alternating Current is generated by power stations. If we think about the
dynamo effect we can understand why.
When a wire is moved as shown on
the left the current flows around the
circuit as shown by the red arrows.
When a wire is moved in the opposite
direction, as shown on the right, the
current flows around the circuit in the
opposite direction as shown by the red
arrows.
So if the wire is moved one way and then the other, current will flow around the
circuit in one direction and then the other. This is called alternating current.
Faraday and Lenz
Faradays Law says that the rate of change of flux (or flux linkage) is
proportional to the induced EMF.
Lenz's Law says that the emf induced acts in such a way to oppose the change
producing it. ie E=-d(BAN)/dt
The origin of electromagnetic induction
Electromagnetic induction occurs because of the motor effect.
Draw a diagram of a wire with electrons in it moving down through magnetic
field into the paper.
Use FLR to show the electrons moving to the left
(which direction is the force on it because of the moving electrons? Upwards –
against the force down (This is what Lenz said))
Lesson 14 questions – Faraday and Lenz
Name ……………………
Class…………………
MOST
1
(
/36)……..%……..
2
SOME
3
Lesson 16 notes – AC Generators
Objectives
(h) describe the function of a simple ac generator;
Outcomes
Be able to describe how an a.c. generator works.
Direct Current (DC)
Current is the flow of charge in a specific time. Direct
current flows around a circuit in one direction. Charge
cycling around the loop.
Conventional current flows from the positive terminal
to the negative terminal as shown in the diagram.
Examples of power sources that produce direct current
would be a battery or a photocell.
The Dynamo Effect
When a wire is moved relative to a
magnetic field, an electric current is
induced (brought about) in the wire. This
is the dynamo effect.
If the wire is stationary then no electric
current is produced.
A dynamo transforms kinetic
(movement) energy into electrical
energy because of this dynamo effect.
Electrical Generators
To easily generate electricity a coil of
wire is normally placed inside a fixed
magnet. This can then be rotated in
order to induce a current within it.
In this example, the friction driver may
be placed against the tyre of a bike.
The same idea is used in most power
stations but Instead of a bike wheel
turning the axle, moving turbines spin
it round.
Increasing the Dynamo Effect
The current that the dynamo effect induces can be increased in three simple
ways:
 Make the fixed magnetic field larger by using stronger magnets;
 Use more turns of coiled wire increasing the length of the conductor
inside the magnetic field;
 Make the relative movement between the wire and the magnets faster.
In other words, spin the coil quicker.
AC Current
Alternating Current is generated by power stations. If we think about the
dynamo effect we can understand why.
When a wire is moved as shown on
the left the current flows around the
circuit as shown by the red arrows.
When a wire is moved in the opposite
direction, as shown on the right, the
current flows around the circuit in the
opposite direction as shown by the red
arrows.
So if the wire is moved one way and then the other, current will flow around the
circuit in one direction and then the other. This is called alternating current.
Alternating Current Generation
The white blob shows that half of the coil spinning round. When it is moving up
at A it is cutting through most field lines and so induces a maximum negative
current. When the coil does not cut through any field lines at B there is no
current. At C the coil is cutting through the maximum amount of field lines again
but in the opposite direction and so there is a positive current. The current
continues to vary as expected until it reaches A again and the cycle continues.
A voltage-time graph has similar characteristics.
Voltage-time Graphs
Alternating current varies the direction of flow of charge and the size of the flow
of current and magnitude of voltage. This can be viewed in using a voltage-time
graph. A Cathode Ray Oscilloscope (CRO) can be used to see this graph. The
y-axis shows the variation of voltage while the x-axis shows how this varies with
time. The graph shows a smooth repeating sinusoidal curve with a set number
of waves in one second.
Voltage
Time
The frequency of a wave is the number of cycles per second and has the unit
Hertz (Hz) (or /s). The frequency is controlled by the speed of rotation of the
generator and is set at 50 rotations per second (50Hz).
The Cathode Ray Oscilloscope (CRO)
A cathode is a negative electrode.
A cathode ray is made up of negative charges (electrons) which are shot from a
cathode towards an anode (a positive electrode) which have a large potential
difference across them. (an electron gun)
The green screen fluoresces when an electron hits it and so if a continuous
beam is shot out at the screen, patterns can be made to form on the screen
The patterns will depend upon the signal that is plugged into one of the inputs.
The settings on the CRO can change the time per division (the timebase on the
x-axis) or the voltage per division (the y gain on the y-axis)
Using these controls accurate measurements can be taken of frequency and
voltage.
Lesson 16 questions – AC Generators
Name ……………………
Class…………….
ALL
1
(
/11)……..%……..
MOST/SOME
2
Lesson 17 notes – Transformers
Objectives
(i) describe the function of a simple transformer;
(j) select and use the turns-ratio equation for a transformer;
(k) describe the function of step-up and step-down transformers.
Outcomes
Be able to make a transformer.
Be able to describe how a transformer works.
Be able to use the turns-ratio equation for a transformer;
Be able to describe the function of step-up and step-down transformers.
Be able to describe how transformers are used in the national Grid and link to
electrical power formula.
Transformers
Transformers are used to transform voltage. They
use the dynamo effect to induce current in coils of
wire. By changing the amount of coils, the amount of
voltage can be varied and it is directly linked.
Transformers are of two types: step up or step
down. Step up transformers increase the output
(secondary) voltage and step down decrease the
output voltage.
Power
Power is the rate of transfer of energy. In other words it is the amount of energy
transferred in a unit time. The equation is:
Power (Watts) = Energy (Joules) / time (seconds)
Electrical power is the rate at which current transfers energy and it can be
shown to equal:
Power (Watts) = Current (Amps) x Voltage (Volts)
Whilst a transformer can change the voltage, the power in must equal the
power out and so as the voltage increases, the current must decrease and vice
versa.
National Grid
The diagram shows how the voltage changes throughout power lines in the
National Grid. Power stations produce electricity of about 20 000 Volts which is
then stepped up to 400 000 Volts to travel large distances in the National Grid.
The voltage is then stepped down depending on the needs of the consumers. If
they are factories, they may need higher voltages but these would be
unnecessary and dangerous for home use and so the voltage is stepped down
to 230V for us.
Why can’t voltage just be transferred at lower voltages though?
As electricity passes through wires it heats the wire. This heating depends on
the current flowing, if it is large, the wire heats up – a lot of energy wasted in
heating the air! As we have seen, Power = Current x Voltage, so we can
decrease the current and still keep the power the same by increasing the
voltage.
Lesson 17 questions – Transformers
Name ……………………
Class…………………
ALL
1
(
/12)……..%……..
MOST