Fields Booklet

7/31/2019 Fields Booklet

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Name.

Class.

Plymstock School PhysicsDepartment

Module G485.1 Electric andMagnetic Fields

student booklet


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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;

OutcomesBe able to state that electric fields are created by electric charges;Be able to define electric field strength as force per unit positivecharge;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, unlikecharges attract. A field is set up by a charge, and any other charge in that fieldwill experience a force due to the field and we represent fields with field lines.

Field Lines


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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 chargeor at infinity.

They never cross if they did, a small positive charge placed there would feelforces 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 likecharges where no field exists (since the forces on a charge placed there wouldbe exactly equal and opposite in direction). Such a point is called a neutralpoint.

Equipotentials

The lines on the hills on this map are called

contour lines. They show areas of a hill thatare the same height above sea level.

The gravitational potential energy along thisline would also be the same.

Gravitational equipotentials

We could say that along these lines therewould be an equal potential wherever wewere on that line; Or an equipotential.

If we drew these lines in the air around the Earth as shown then we would havean equal potential all along them.

Electric field equipotentials

We can do the same for electric field equipotentials.

49.0 J/kg

39.2 J/kg

29.4 J/kg

19.6 J/kg

9.8 J/kg


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The dotted lines show the equipotentials for the fields caused by the chargesshown.

(resourcefulphysics.org)


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Lesson 2 notes Coulomb Law

Objectives

(d) select and use Coulombs law in the formF= Qq

4or2

(e) select and apply

E = Q4or

2

for the electric field strength of a point charge;(f) select and useE = V

dfor the magnitude of the uniform electric field strength between charged parallelplates;

Outcomes

Be able to select and use Coulombs law in the formF= Qq

4or2

Be able to select and applyE = Q

4or2

for the electric field strength of a point charge;Be able to select and useE = V

dfor the magnitude of the uniform electric field strength between charged parallel

plates;

Coulombs law

We know that a field exists around a charge that exerts force on other chargesplaced there, but how can we calculate the force? The force will be dependentupon the sizes of the charges, and their separation. In fact the force follows aninverse square law, and is very similar in form to Newtons Law of UniversalGravitation. It is known as Coulombs law, and it is expressed as:

2

21

r

QkQ

F=

where 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 Newtons Law of UniversalGravitation). 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, Coulombs law is written:


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2

21

4 r

QQF

o

=

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/ (4 0). Different materials have different permittivities,and so the value of k in Coulombs law also changes for different materials.

Points to bring out about Coulombs law:

The form is exactly the same as Newtons Law of Universal Gravitation; inparticular, it is an inverse-square law.

This force can be attractive or repulsive. The magnitude of the force can becalculated 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 inthe equation, then whenever you get a negative answer for the force, there isan attraction, whereas a positive answer indicates repulsion).

Although the law is formulated for point charges, it works equally well forspherically 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 absolutelydwarfs the gravitational force between them.


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Field strength

For a gravitational field, the Field strength is defined at a point in thegravitational 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 electricfield?

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 fieldand not the particular chargethat is placed there. For example, at a point where the field strength is2000 N C-1, a 1 C charge would feel a force of 2000 N whereas a 1 mCcharge 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 thedirection that a positive charge placed at that point in the field would feela force.

Now, for the non-uniform fielddue to a point (or spherical)charge, we can use Coulombslaw to find an expression for thefield strength. Consider the forcefelt by a charge q in the field of

another charge Q, where thecharges are separated by adistance r:

2r

kQqF= by Coulombs Law.

But E= F/q and so

2r

kQE=

This is our result for the field strength at a distance rfrom a (point or spherical)charge Q.


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Lesson 2 questions Coulombs LawName ( /10)..%

Class

ALL

1


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Name ( /29)%

Class.

Lesson 3 questions Coulombs LawALL

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. Thesphere 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.

(3)ii) 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)


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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.

(2)


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2 a) Define electric field strength at a point in space.

..

..

(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.

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

8.0 x 10

-10

m


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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

..

..

..

..

..in fig 2.3

..

..

..

..

(5)


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Lesson 4 notes Electric field strength and potential

Objectives

Select and use E=V/d for the magnitude of the uniform electric field strengthbetween charged parallel plates

Outcomes

Be able to select and useE = V

d

for the magnitude of the uniform electric field strength between charged parallelplates;

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 thatthey are parallel and evenly spaced. Alsoas field strength = (potential gradient), theequipotentials should also be evenlyspaced.

Where have we seen such as field?

The field between two parallel chargedplates.

How will a charge move in such a field?

The force is given by

F= EQ

Since Eis constant, the force will be constant and therefore, by

F= ma

the acceleration will also be constant and directed along the field lines for apositive 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


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Lesson 5 notes projectile comparisons

Objectives

(g) explain the effect of a uniform electric field on the motion ofcharged particles;(h) describe the similarities and differences between the gravitationalfields of point masses and the electric fields of point charges.

OutcomesBe able to explain the effect of a uniform electric field on the motion of chargedparticles;Be able to describe the similarities and differences between the gravitationalfields 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 andelectric fields.

We have seen the following similarities:Both fields follow an inverse square law for both force and therefore fieldstrength.

For both types of field, potential and potential energy are inversely proportionalto distance.

We define the zero of potential to be at infinity

For attractions (and therefore always for gravitational fields) potential energiesare always negative.

The majordifference is that repulsions occur in electrostatic fields betweencharges of the same sign, whereas as far as we know, gravity is alwaysattractive.

The similarity of the fields may be brought home further by comparing the formsof the equations:Gravitational fields Electric fields

field strength = -(potential gradient)g= V/d (a uniform field) E= V/d (a uniform field)F= Gm1m2/r2 F = kQ1Q2/r2

g= GM/r2 (a non-uniform field) E= kQ/r2 (a non-uniform field)GPE= Gm1m2/r EPE = kQ1Q2/rV= GM/r V= kQ/r


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Name. ( /15).%

Class.

Lesson 5 questions projectile comparisonsMOST

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 aremaintained 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 10

4

Vm

-1

.

(2)

c) Calculate the horizontal force on an electron after passing through the

hole in A.

Force = .. N (2)


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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.

(2)

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)

Force = .N (4)

b) What is the ratio of the electrical to gravitational force?

Ratio = .. (2)

Lesson 6 notes - Magnetic Fields


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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 straightcurrent-carrying conductor and a long solenoid.

A magnetic field is a region in which a particle with magnetic propertiesexperiences a force, and in which a moving charge experiences a force.

There are two main classes of magnet:

1. Permanent magnets

2. Electromagnets

Field ShapesPermanent 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 obeysome 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 aNorth pole with no South pole (which doesnt exist!) would feel.Field direction always goes from North to South. So pop a magnet at X in thefield (see diagram) and it would align itself with its North pole pointing along thearrow.The spacing between the lines of flux tells you about the strength of field - asthe lines get closer together, the field becomes stronger - for example, near the

poles.


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Look at this field:

The region in between the poles shows equally spaced, parallel lines. This iscalled a uniform field. Field strength remains constant as you move around

this area. Move out from the space between the poles and the field strengthreduces. The lines of flux become further apart.Temporary Magnets:Around any conductor that has a current flowing through it there is amagnetic 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 thepage. (Think of an arrow - going away you see the flights, coming towards yousee the point!)

Remember: Conventional current is the flow of positive charges. Soconventional current goes in the opposite direction to the electron flow.

In a wire with conventional current flowing out of the page you get:

Its the same field shape as the one above, but the direction of the field isdifferent.Notice that in both cases the lines get further apart as you move away from thewire, this is because the magnetic field is getting weaker.How do you remember which way the field goes (clockwise oranticlockwise)?

Answer: Use the corkscrew rule! The problem here is that as many of you areunder 18 you wont have a clue what a corkscrew is obviously! This willbecome easier once youre over 18 as you will be allowed to drink wine and willtherefore have knowledge of a corkscrew.


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Imagine you are screwing a corkscrew into or out of the page in the samedirection as conventional current. The turning motion of the corkscrew is inthe 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 iscoiled. Apply the corkscrew rule to different sections of the coil (below) and youshould 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 itfrom the side - so the conventional current comes out of the page at the top andinto 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 thesame as the field around a bar magnet.

Right Hand Grip RuleA quick way to work out the direction of the magnetic field in a solenoid is theright hand grip ruleMake a fist and stick your thumb out (as if hitchhiking). Your fingers arewrapped in a circle, same as the coils in the solenoid. If you make your fingerspoint in the same direction as the conventional current around the coil - yourthumb 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 pointswhere the magnetic field strength is zero. These points are called neutralpoints.For Example:


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Name.. ( /27)..%

Class

Lesson 6 questions Magnetic field patternsALL

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

..

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.

(1)

b) Fig. 2.1 shows a cross section of a current carrying conductor.

current-carrying

conductor (current out

of plane of page)


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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 theexpected 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 thepoles, all flux lines are continuous. A line which emerges (conventionally at a north pole) entersthe material again at the south pole. Flux lines never cross. Think of flux as like a fluid pumpedout of N poles and sucked into S poles (although nothing is actually flowing or physicallyconnected to the magnet etc)

Here is a sketch of the flux from a short bar magnet:

4. Sketch the flux from a longer magnet, like this:

S N


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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, orperhaps their mirror reflections when drawn in two dimensions. For example, the field of the coilbelow can be divided into four quarters, each a copy (reflected or inverted) of the others. So youonly need to draw one bit of the field.

7. Identify the similarly shaped regions of the field between a N and a S pole.


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SN

8. Identify the similarly shaped regions of the field around a pair of coils with currents goingin 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 fluxemerges 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:

+


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10. Identify N and S poles of this electric motor winding:


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Same environment, same flux

If the pattern of current turns around one place is the same as that around another, the fluxpattern 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 anduniform, like this:

12. Sketch the flux inside this doughnut shaped coil:


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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:

rotor

pole with winding

pole with winding

stator


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Lesson 7 notes F=BIL and Flemings Left hand Rule

Objectives

(b) state and use Flemings left-hand rule to determine the force oncurrent conductor placed at right angles to a magnetic field;(c) select and use the equationsF=BIL andF=BILsin ;(d) define magnetic flux densityand the tesla;

Outcomes

State Flemings left-hand rule to determine the force on currentconductor placed at right angles to a magnetic field;

Select and use the equationsF=BIL

Be able to define magnetic flux density.

Use state Flemings left-hand rule to determine the force on currentconductor placed at right angles to a magnetic field;Select and use the equationsF=BILsin ;

We know that a current carrying wire as a magnetic field around it. When it isplaced into another magnetic field it will feel a force. We have seendemonstrations 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 currentcarrying wire is clamped between two permanent magnets in a yoke placed ona 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 bechanged.

We see a pattern as sown in the graphs below:


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This means that Force is proportional to the Current through the wireAnd that Force is proportional to the length of the wire.


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So we can write that:

F=BIL

Where F is the force felt by the current carrier

I is the current through the wireL is the length of the wire

And B is the Magnetic Field strength (ormagnetic flux density)You need to be able to rearrange the equation to get:

B=F/ILI=F/BLL=F/BI

Magnetic Flux density is defined as the force per unit length per unit currenton 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= BILsinThe units for B are tesla (T).The greatest force occurs when the angle that the wire is to the field lines is90o.

Flemings Left hand rule

The force, magnetic field and current all need to be at right angles.thuMb Motion of current carrier

First Finger Field directionseCond finger Current direction


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Name.. ( /16)..%

Class

Lesson 7 questions F=BIL and Flemings left hand ruleALL

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 andwith 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.

(2)

a) Calculate the magnetic flux density B between the poles of the magnet.

(3)

Total [6]

2) Fig 2.1 shows a wire placed at righ angles to a magnetic field. The wire rests on

two metal supports.


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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.

(3)

b) Suggest why the overhead cables for the National Grid cannot be freely supported

by the Earths magnetic field.

(2)

Total [5]


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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.

(3)

Total [5]


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Lesson 9 notes Charged particles in a magnetic field

Objectives

(e) select and use the equation F= BQvfor the force on a charged particletravelling at right angles to a uniform magnetic field;(f) analyse the circular orbits of charged particles moving in a planeperpendicular 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= BQvfor the force on a chargedparticle travelling at right angles to a uniform magnetic field.Be able to analyse the circular orbits of charged particles moving in a planeperpendicular to a uniform magnetic field by relating the magnetic force to thecentripetal acceleration it causes.Be able to derive the equation F= BQvfrom the definition of current andmagnetic flux density.

When a wire carrying a current through a field feels a force it is because themagnetic 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 - forexample, if they are in a beam crossing a vacuum.

F=BIl

v=l/t

So l=vt

So, F=BIvt

But, I=q/tSo q=It

Therefore, F=Bqv

For an electron with charge e,

F=Bev

So,F = B q v sin

Where: F = force (N)

B = magnetic field strength (T)

q = charge on the particle (C)

v = velocity of the particle (m/s)

Note: Angle is between the direction of the beam and the magnetic field

direction.


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Use Flemings left hand rule to work out the direction of the force. Align yoursecond finger with the beam of particles remembering that it points the waypositive particles flow, the opposite way to electron flow.

The diagrams show a charged particle in a magnetic field. The force is alwaysat right angles to the motion of the charged particle so if the speed of theparticle is right then a circular path can be created.


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Finding the Charge-to-Mass Ratio of a Particle

When a charged particle enters a magnetic field we now know it will be forcedto change direction. If it stays in the field it will continue to change direction andwill move in a circle. The force produced will provide the centripetal force on themoving particle.

mv2=Vq (since V=W/q)

And mv2/r=Bqv

So mv2=Bqvr

Bqv r=Vq

Bvr =V

v = 2V/Br

So q/m = 2V/r2B2

This idea is used in velocity-selectors, where particles of different mass-to-charge ratio will rotate in circles with a different radius. These are used in massspectrometers.


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Lesson 9 questions Moving charge in a magnetic fieldName. Class..( /36).%.

MOST

1


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2


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3


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Lesson 12 notes Electromagnetic induction

Objectives

(a) define magnetic flux;(b) define the weber.(c) select and use the equation for magnetic flux = BAcos;

Outcomes

Be able to define magnetic flux;Be able to define the weber.Be able toselect and use the equation for magnetic flux = BAcos;Be able to rearrange and use the equation for magnetic flux

= BAcos.

Magnetic Flux

Around a magnet there is a magnetic field with an assiciated flow of magneticenergy around it.

The flow of energy is called magnetic flux ().

If B is at 90 to the area, A, then the total magnetic flux, f "passing through" A is

defined to be:

= AB

Magnetic flux is given the symbol and is measured in units called Webers(Wb).

The units of flux are: Tm and 1 Tm = 1 Weber

The flux associated with a magnetic field is therefore a measure ofthe numberof magnetic field lines penetrating some surface.


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The above picture shows the special case of a plane area A and a uniform fluxdensity B. The normal to the field is at an angle with the field. In this case, theflux is given by:

=BAcos

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 inducedacross it.

Flux Linkage

For N number of coils we use this equation:

N=BANcos


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Lesson 14 notes Faraday and LenzObjectives

(e) state and use Faradays law of electromagnetic induction;

(f) state and use Lenzs law;(g) select and use the equation: induced e.m.f. = rate of change of magneticflux linkage;

Outcomes

Be able to describe the dynamo effect.

Be able to state Faradays law of electromagnetic induction.

Be able to state Lenzs law.

Be able to use Lenzs law.

Be able to select and use the equation: induced e.m.f. = rate of change ofmagnetic flux linkage;

Be able to derive the equation: induced e.m.f. = rate of change of magneticflux linkage;

The Dynamo Effect

When a wire is moved relative to amagnetic field, an electric current isinduced (brought about) in the wire. Thisis the dynamo effect.If the wire is stationary then no electriccurrent 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 fixedmagnet. 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 wheelturning the axle, moving turbines spin it round.

Increasing the Dynamo Effect

The current that the dynamo effect induces can be increased in three simpleways:

Make the fixed magnetic field larger by using stronger magnets;

Use more turns of coiled wire increasing the length of the conductorinside the magnetic field;

Make the relative movement between the wire and the magnets faster.In other words, spin the coil quicker.


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AC Current

Alternating Current is generated by power stations. If we think about thedynamo effect we can understand why.

When a wire is moved as shown onthe left the current flows around thecircuit as shown by the red arrows.

When a wire is moved in the oppositedirection, as shown on the right, thecurrent flows around the circuit in theopposite direction as shown by the red

arrows.

So if the wire is moved one way and then the other, current will flow around thecircuit in one direction and then the other. This is called alternating current.

Faraday and LenzFaradays Law says that the rate of change of flux (or flux linkage) isproportional to the induced EMF.

Lenz's Law says that the emf induced acts in such a way to oppose the changeproducing 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))


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Lesson 14 questions Faraday and LenzName ( /36)..%..

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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 onedirection. Charge cycling around the loop.

Conventional current flows from the positiveterminal to the negative terminal as shown in thediagram.

Examples of power sources that produce directcurrent would be a battery or a photocell.

The Dynamo Effect

When a wire is moved relative to amagnetic field, an electric current isinduced (brought about) in the wire. Thisis the dynamo effect.If the wire is stationary then no electriccurrent is produced.

A dynamo transforms kinetic(movement) energy into electricalenergy because of this dynamo effect.

Electrical Generators

To easily generate electricity a coil ofwire is normally placed inside a fixed

magnet. This can then be rotated inorder to induce a current within it.

In this example, the friction driver maybe placed against the tyre of a bike.The same idea is used in most powerstations but Instead of a bike wheelturning the axle, moving turbines spinit round.


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Increasing the Dynamo Effect

The current that the dynamo effect induces can be increased in three simpleways:

Make the fixed magnetic field larger by using stronger magnets;

Use more turns of coiled wire increasing the length of the conductorinside 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 thedynamo effect we can understand why.

When a wire is moved as shown onthe left the current flows around thecircuit as shown by the red arrows.

When a wire is moved in the oppositedirection, as shown on the right, thecurrent flows around the circuit in theopposite direction as shown by the redarrows.

So if the wire is moved one way and then the other, current will flow around thecircuit in one direction and then the other. This is called alternating current.


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Alternating Current Generation

The white blob shows that half of the coil spinning round. When it is moving upat A it is cutting through most field lines and so induces a maximum negativecurrent. When the coil does not cut through any field lines at B there is nocurrent. At C the coil is cutting through the maximum amount of field lines againbut in the opposite direction and so there is a positive current. The currentcontinues 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 flowof current and magnitude of voltage. This can be viewed in using a voltage-timegraph. A Cathode Ray Oscilloscope (CRO) can be used to see this graph. They-axis shows the variation of voltage while the x-axis shows how this varies withtime. The graph shows a smooth repeating sinusoidal curve with a set numberof waves in one second.

The frequency of a wave is the number of cycles per second and has the unitHertz (Hz) (or /s). The frequency is controlled by the speed of rotation of thegenerator and is set at 50 rotations per second (50Hz).

Voltage

Time


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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 potentialdifference across them. (an electron gun)The green screen fluoresces when an electron hits it and so if a continuousbeam is shot out at the screen, patterns can be made to form on the screenThe 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 thex-axis) or the voltage per division (the y gain on the y-axis)Using these controls accurate measurements can be taken of frequency andvoltage.


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Lesson 16 questions AC GeneratorsName ( /11)..%..

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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-downtransformers.

Be able to describe how transformers are used in the national Grid andlink to electrical power formula.

Transformers

Transformers are used to transform voltage. Theyuse the dynamo effect to induce current in coils ofwire. By changing the amount of coils, the amount ofvoltage can be varied and it is directly linked.Transformers are of two types: step up or stepdown. Step up transformers increase the output(secondary) voltage and step down decrease theoutput 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 beshown to equal:

Power (Watts) = Current (Amps) x Voltage (Volts)

Whilst a transformer can change the voltage, the power in must equal thepower out and so as the voltage increases, the current must decrease and vice

versa.


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National Grid

The diagram shows how the voltage changes throughout power lines in theNational Grid. Power stations produce electricity of about 20 000 Volts which isthen 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. Ifthey are factories, they may need higher voltages but these would beunnecessary and dangerous for home use and so the voltage is stepped downto 230V for us.

Why cant voltage just be transferred at lower voltages though?

As electricity passes through wires it heats the wire. This heating depends onthe current flowing, if it is large, the wire heats up a lot of energy wasted inheating the air! As we have seen, Power = Current x Voltage, so we candecrease the current and still keep the power the same by increasing thevoltage.


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Lesson 17 questions TransformersName ( /12)..%..

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