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AQA A Level Physics > Section 11 - Magnetic Fields > Flashcards

Section 11 - Magnetic Fields Flashcards

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1
Q

What is a magnetic field?

A

A region where a force is exerted on magnetic materials.

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2
Q

How can magnetic fields be represented?

A

Using field lines.

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3
Q

What is another name for magnetic field lines?

A

Flux lines

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4
Q

Which direction do magnetic field lines go?

A

North to south pole of a magnet (OUTSIDE of it).

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5
Q

How is the strength of a magnetic field represented using field lines?

A

The closer together the lines, the stronger the field.

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6
Q

What is a neutral point in magnetic fields?

A

Where magnetic fields cancel out, so there are no field lines.

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7
Q

What happens when a current passes through a wire?

A

A magnetic field is induced around the wire.

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8
Q

Describe the magnetic field induced around a current-carrying wire.

A

Concentric circles centred around the wire

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9
Q

What rule is used to work out the direction of the magnetic field around a current-carrying wire and how does it work?

A

• Right-hand grip rule
How it works:
• Thumb = Current (conventional)
• Fingers = Magnetic field

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10
Q

What happens to the shape of the magnetic field around a current-carrying wire when it is looped into a single coil?

A

Doughnut-shaped

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11
Q

What happens to the shape of the magnetic field around a current-carrying wire when it is looped into a solenoid?

A

Like a bar magnet.

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12
Q

What is the rule for working out the direction of the magnetic field around a current-carrying solenoid?

A

The right-hand rule, except the thumb and fingers are reversed.

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13
Q

What happens when you place a current-carrying wire in a magnetic field and why?

A
  • The field around the wire and the magnetic field are added together
  • This causes a resultant field so there is a force exerted on the wire

These bunched lines causes a pushing effect on the wire

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14
Q

What direction does the current have to be to the magnetic field lines?

A

Right angles - see in Flemings left hand

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15
Q

What rule is used to work out the direction of the force exerted on a current-carrying wire in a magnetic field and how does it work?

A

• Fleming’s Left-Hand Rule
How it works:
• Thumb = Force (motion) direction
• First finger = Magnetic field direction
• Second finger = Current (conventional) direction

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16
Q

What things is it important to remember when using Fleming’s left-hand rule?

A
  • Magnetic field goes from north to south

* Current is conventional, so it goes from positive to negative terminals

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17
Q
A
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18
Q

When a current-carrying wire is in a magnetic field, what happens if the current is parallel to the field lines?

A

There is no force exerted.

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19
Q

What happens when you pass an AC current through a wire?

A

wire vibrates:

Direction of force in perpendicular to direction of current.

When current is reversed, direction of force is reversed.

Constant alternating current = constantly alternating direction of force

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20
Q

What is magnetic flux density?

A

The force on a current carrying wire at a right angle to a magnetic field is proportional to magnetic flux density, B.

So the definition is:

“The force on one metre of wire carrying a current of one amp at right angles to the magnetic field.”

i.e. It is the magnetic field strength

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21
Q

In magnetic fields, what is B?

A

Magnetic flux density (a.k.a. Magnetic field strength)

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22
Q

What are the units for magnetic field density?

A

Tesla (T)

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23
Q

What is magnetic flux?

A

The number of field lines passing through an area.

It is how much the field is affecting the area.

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24
Q

In magnetic fields, what is φ?

A

Magnetic flux

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25
Q

What are the units for magnetic flux?

A

Wb

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26
Q

Is magnetic flux density scalar or vector?

A

Vector

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27
Q

Describe a experiment to investigate the effect of current on the force exerted on a current-carrying wire in a magnetic field.

A

1) Set up a top pan balance with a square loop of wire fixed to it, so that it is standing up and that the top of the loop passes through a magnetic field, perpendicular to it.
2) Connect the wire in a circuit with a variable resistor, ammeter and dc power supply. Zero the top pan balance when no current is flowing.
If the reading is negative swap the crocodile clips over
3) Vary the current using the variable resistor. At each current value, record the current and mass. Repeat 3 times at each current value and average.
4) Convert into force using F = mg.
5) Plot a graph of force F against current I. Draw a line of best fit.
6) Since F = BIl, the gradient of the rain is equal to B x l.
7) Alternatively, you could vary “l” by varying the length of wire that is perpendicular to the field or vary “B” by changing the strength of the magnets.

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28
Q

In the experiment to investigate F = BIl, how can you vary each of the variables?
What would need to keep constant when investigating length?

A
  • B -> Use different magnets to vary field strength
  • I -> Use variable resistor to vary the current
  • l -> Use different loop sizes with different lengths perpendicular to the magnetic field (keep I constant with variable resistor, and also B)
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29
Q

What is 1 tesla equal to? Explain this.

A

1 Wb/m².
This is because magnetic flux density is the number of flux lines (Wb) per unit area.

Also 1 N/Am.

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30
Q

What is the equation for the force exerted on a current-carrying wire in a magnetic field?

A

F = B x I x l

Where:
• F = Force (N)
• B = Magnetic flux density (T)
• I = Current (A)
• l = Length of wire in the field (m)

(NOTE: This only applies when the current is at 90° to the magnetic field.)

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31
Q

Will a force still act on a wire even if it isn’t at 90 degrees?

A

Yes - but smaller (as long as it isn’t parallel)

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32
Q

What kind of force will F = BIl give you (For 90 degrees)?

A

Maximum force the wire could feel

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33
Q

Why does a current-carrying wire experience a force in a magnetic field?

A

A force act on the charged particles (electron) moving through it.

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34
Q

Do charged particles experience a force in a magnetic field?

A

Only if they are moving.

NOTE: Check this!

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35
Q

Derive the equation for the force on a charged particle in a magnetic field.

A 
• F = B x I x l
• I = Q / t
• l = v x t
Therefore:
• F = B x (Q / t) x (v x t)
• F = B x Q x v
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36
Q

When do F = BIl and F = BQv apply?

A

When the flow of charge is at 90° to the magnetic field.

And perpendicular to direction of travel (Flemings left hand)

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37
Q

And electron travels at a velocity of 2.00 x 10⁴ m/s perpendicular to a uniform magnetic field with a magnetic flux density of 2.00 T. What is the magnitude of the force acting on the electron?

A

F = BQv = 2.00 x (1.60 x 10⁻¹⁹) x (2.00 x 10⁴) = 6.40 x 10⁻¹⁵ N

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38
Q

What happens to moving charge in a magnetic field and why?

A
  • They are deflected in a circular path
  • Because Fleming’s left hand rule states that the force on a moving charge is always perpendicular to its direction of travel
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39
Q

How can Fleming’s left-hand rule be used for charged particles?

A

Like normal, except the second finger (normally used for current) is the direction of motion for a positive charge.

(i.e. This is intuitive)

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40
Q

When using Fleming’s left hand rule for moving charged particles, what happens if the charge is negative?

A

Point your second finger in the direction opposite to its motion.

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41
Q

How do you carry out an experiment investigating forces on a wire?

A
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42
Q

Is the force experienced by a moving charged particle in a magnetic field affected by mass?
What does depend on mass?

A

No, but the centripetal acceleration caused by it depends on the mass (since F = ma).

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43
Q

What is the equation for the force on a particle in circular orbit?

A

F = mv²/r

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44
Q

What is the equation for the radius of the circular path of a charged particle in a magnetic field?

A

r = mv/BQ

Where:
• r = Radius (m)
• m = Mass (kg)
• v = Velocity (m/s)
• B = Magnetic flux density (T)
• Q = Charge (C)
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45
Q

Derive the equation for the radius of the circular path of a charged particle in a magnetic field.

A 
Force on a charged particle in a magnetic field:
• F = BQv
Force on a particle in circular orbit:
• F = mv²/r
Therefore:
• BQv = mv²/r
• r = mv/BQ
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46
Q

How is a magnetic field going into the paper symbolised?

A

Circles with crosses in them

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47
Q

What happens to the radius of a charged particle moving in a magnetic field if the mass is increased?

A

Increases

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48
Q

What happens to the radius of a charged particle moving in a magnetic field if the velocity is increased?

A

Increases

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49
Q

What happens to the radius of a charged particle moving in a magnetic field if the magnetic flux density is increased?

A

Decreases

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50
Q

What happens to the radius of a charged particle moving in a magnetic field if the charge is increased?

A

Decreases

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51
Q

What does a cyclotron make use of?

A

A particle accelerator that makes use of the circular deflection of charged particles in a magnetic field.

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52
Q

What are some uses of cyclotrons?

A
  • Producing radioactive tracers

* Producing high-energy beams of radiation for radiotherapy

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53
Q

What is the equation for the force experienced by a charged particle in a magnetic field?

A

F = BQv

Where:
• F = Force (N)
• B = Magnetic flux density (T)
• Q = Charge on particle (C)
• v = Velocity of particle (m/s)
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54
Q

What is the equation for frequency of rotation of charged particles?
What is is independent and dependent of?

A
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55
Q

Describe the structure of a cyclotron.

A
  • Two hollow semi-circular electrodes with alternating potential difference.
  • Slight gap between them.
  • Uniform magnetic field applied perpendicular to the plane of the electrodes.
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56
Q

Describe how a cyclotron works.

A
  • Particle (charged) is fired into one of the electrodes
  • The magnetic field makes it flow a semi-circular path and return to the gap between electrodes
  • The potential difference between them creates an electric field that accelerates the particle across the gap
  • The velocity in now higher, so the particle takes a path with a larger radius before leaving the other electrode
  • As it exits, the potential difference is reversed at this point so that the electric field is reversed and therefore the particle can accelerate across the gap
  • This repeats as the particle spirals outwards, increasing in speed, before exiting the cyclotron
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57
Q

If p.d wasn’t alternating what would happen to the particle in a cyclotron?

A

It would slow down after leaving the second electrode

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58
Q

Since frequency of circular motion doesn’t depend on the radius, how long will the particle stay in each electrode?
What does this mean for the frequency of the alternating p.d?

A

Always spend the same time in the electrode (if B, m and Q are constant).
Alternating p.d has a fixed frequency

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59
Q

Remember to practise writing out the functioning and structure of a cyclotron.
Also research cyclotrons and alternating pd

A
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60
Q

What is the equation for magnetic flux?

A

φ = BA

Where:
• φ = Magnetic flux (Wb)
• B = Magnetic flux density (T)
• A = Area (m²)

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61
Q

What happens when a conductor is moved in a magnetic field?

A

If it cuts through field lines, an emf is induced in the conductor:

The electrons in the rod will experience a force, which causes them to accumulate at one end of the rod - this induces an emf across the ends of the rod = electromagnetic induction.

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62
Q

Why is an emf induced in conductor when it cuts through magnetic field lines?

A
  • The electrons experience a force, so they accumulate at one end of the rod
  • This induces an emf between the positive and negative ends of the rod
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63
Q

What is electromagnetic induction?

A

When an emf is induced in a conductor that cuts through magnetic field lines.

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64
Q

How can you induce an emf in a flat coil or solenoid?

A
  • Moving the coil towards it away from the poles of the magnet
  • Moving a magnet towards or away from the coil
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65
Q

An emf is induced in a conductor when what is changing?

A

The magnetic field

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66
Q

When is a current induced in a conductor that has had an emf induced in it?

A

When it is connected in a circuit.

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67
Q

What happens when the switch closes?

What will the ammeter do?

A
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68
Q

How do a change in magnetic field induce an emf and cause current flow?

A

A magnetic filed interacting with an electric field will generate a force.

When there is a change in magnetic field, force changes.

When force changes, energy changes.

This induces an EMF and current will flow.

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69
Q

What is the difference between flux and flux linkage?

Why is flux linkage used?

A
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70
Q

What is magnetic flux linkage?

A

The product of the magnetic flux passing through the coil and the number of turn in the coil they cut the flux.

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71
Q

What are the units for magnetic flux linkage?

A

Wb turns

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72
Q

What is the equation for magnetic flux linkage when the coil is perpendicular to the field?

A

Nφ = BAN

Where:
• Nφ = Magnetic flux linkage (Wb turns)
• B = Magnetic flux density (T)
• A = Area of coil (m²)
• N = Number of turns
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73
Q

What is the equation for magnetic flux when the coil is not perpendicular to the field?

A

φ = BAcosθ

Where:
• φ = Magnetic flux (Wb)
• B = Magnetic flux density (T)
• A = Area of coil (m²)
• θ = Angle between the field and the normal of the plane of the loop

(NOTE: Not given in exam!)

74
Q

What is the equation for magnetic flux linkage when the coil is not perpendicular to the field?

A

Nφ = BANcosθ

Where:
• Nφ = Magnetic flux linkage (Wb turns)
• B = Magnetic flux density (T)
• A = Area of coil (m²)
• θ = Angle between the field and the normal of the plane of the loop
75
Q

Why is BANcos used instead of BAN?

What rule do we have to take into account?

A

When flux linkage isn’t perpendicular to coil of wire, we need to resolve for horizontal component because that is where you get to max force (perpendicular):

76
Q

What does the rate of change of flux tell you?

A

How strong the emf will be in volts:

77
Q

What can’t you do with Nφ = BAN

A
78
Q

In Nφ = BANcosθ, what is θ? Explain this.

A
  • It is the angle between the magnetic field lines and the perpendicular to the coil.
  • cosθ is used because Bcosθ is essentially resolving the magnetic field vector into the component that is perpendicular to the area.
79
Q

Describe what the difference is between:
• Magnetic flux density
• Magnetic flux
• Magnetic flux linkage

A
  • Magnetic flux density - The number of magnetic field lines per unit area in a magnetic field
  • Magnetic flux - The total number of field lines passing through a given area.
  • Magnetic flux linkage - The magnetic flux multiplied by the number of coils.
80
Q
A
81
Q

Describe an experiment to investigate how the angle of a coil in a magnetic field affects the induced emf.

A

1) Set up solenoid connected to an alternating power supply. This will act to provide an alternating magnetic field.
2) Place a search coil on a podium in the centre. It should have a known area and number of loops. Place the podium on a protractor and connect the search coil to an oscilloscope.
3) Turn the time base off on the oscilloscope so that only the amplitude is shown as a vertical line.
4) Orientate the search coil so that it is perpendicular to the podium and record the emf.
5) Keep rotating the search coil by 10° and record the emf each time. Do this until the coil has been rotated by 90°.
6) The emf should decrease gradually until the search coil is parallel to the field lines.
7) Plot a graph of induced emf against θ, where the emf is at a maximum at 0° and zero at 90°.

82
Q

Remember to practise drawing out the setup for the investigation of how the angle of a coil in a magnetic field affects the induced emf.

A

See diagram pg 145 of revision guide or 341 in textbook

83
Q

What is the symbol for magnetic flux linkage?

A

Nφ

84
Q

State Faraday’s law.

A

The magnitude of the emf induced in a conductor is directly proportional to the rate of change of flux linkage.

85
Q

Give the equation for Faraday’s Law.

A

ε = NΔφ/Δt

Where:
• ε = Magnitude of induced emf (V)
• Nφ = Magnetic flux linkage (Wb turns)
• t = Time (s)

86
Q

How is Faraday’s derived?

A
87
Q

What will induce a larger emf?

A

A larger change in flux and the faster you do it (change in t is lower if you do it quicker)

88
Q

When does Faraday’s law apply?

A

Only for an object that has a changing flux - 1 flux linkage to another.

89
Q

On a graph of flux linkage against time, what does the gradient give?

A

The induced emf.

90
Q

On a graph of emf against time, what does the area under the line give?

A

Flux linkage change

91
Q

A conducting rod of length l moves through a perpendicular uniform magnetic field, B, at a constant velocity, v. Show that the magnitude of the emf induced in the rod is equal to Blv.

A
  • Distance travelled, s = vΔt
  • Area of flux it cuts, A = lvΔt
  • Total magnetic flux cut through, Δφ = BA = BlvΔt
  • Faraday’s Law gives, ε = Δ(
92
Q

What is the equation for the emf induced in a rod passing through a magnetic field?

A

ε = Blv

Where:
• ε = Induced emf (V)
• B = Magnetic flux density (T)
• l = Length of rod (m)
• v = Velocity (m/s)

(NOTE: Not given in exam, but can be derived)

93
Q

Derive the equation for the emf induced in a rod moving at constant speed through a magnetic field.

A 
When v is the velocity, A is the area covered by the loop, l is the length of the rod and d is the distance travelled by the rod:
• ε = NΔφ / Δt = ΔBAN / t
• ε = B x ΔA / t
• ε = B x l x (Δd / t)
• ε = B x l x v = Blv
94
Q

What can you think of magnetic flux as being equal to?

A

The flux cut per second.

95
Q

State Lenz’s law.

A

An induced emf is always in such a direction as to oppose the change that caused it.

96
Q

What equation do Lenz’s law and Faraday’s Law combine to give?

A

ε = -NΔφ/Δt

Where:
• ε = Magnitude of induced emf (V)
• Nφ = Magnetic flux linkage (Wb turns)
• t = Time (s)

(NOTE: Not given in exam. The equation for the magnitude of ε is given, but there is no minus sign.)

97
Q

What principle does Lenz’s law agree with?

How?

A
  • Conservation of energy
  • The energy used to pull a conductor through magnetic field against the resistance caused by the magnetic attraction is what produced the induced current
98
Q

How can Fleming’s left hand rule be used to calculate the direction of the current induced in a rod moving through a magnetic field?

A
  • Think about the direction of movement of the rod
  • The resistance force caused by the current must act in the opposite direction. Use this as the “force”.
  • Use Fleming’s left hand rule as usual.
99
Q

How is Lenz’s law and Faraday’s law apparent in a magnet falling through a copper tube?

A

…
The current opposes and slows down the movement of the magnet caused by GPE. The GPE is converted to Electric potential energy.

For there to be Electric potential, energy has to be taken from creating a magnetic field and opposing the filed that it created. (Lenz’s)

100
Q

What happens when a coil rotates in a magnetic field?

A

The coil cuts the flux and an alternating emf is induced?

101
Q

What is the amount of flux cut by the coil know as?

What is it’s equation?

A

Flux linkage:

Nφ = BANcosθ

102
Q

As the coil rotates through a magnetic field, θ changes.

As a result, what varies?

A

Flux Linkage:

Nφ = BANcosθ

103
Q

What does flux linkage vary between?

A

As changes, flux linkage varies sinusoidally (follows same pattern as sin (or cos) curve) between +BAN and -BAN.

104
Q

What is a generator?

A

A machine that uses a rotating coil in a magnetic field to convert kinetic energy into electrical energy.
They induce an electric current by rotating a coil in a magnet.

105
Q

What is another name for a generator?

A

Dynamo

106
Q

What is an alternator?

A

A generator that generates alternating current (a.c.)

107
Q

How does an a.c. generator work?

A

• A spinning rectangular coil is placed in a magnetic field
• Slip rings and brushes (connected to an external circuit) are connected to the coil
• This means the direction of the induced current is
reversed every half turn = produces an alternating current.
The output voltage also changes with every half rotation

108
Q

As a coil in a generator rotates, what does the flux linkage vary between?

A

+BAN and -BAN

109
Q

What does the speed at which θ changes depend on?

A

Angular speed, ω

We use this in the flux linkage equation instead of θ (since θ = ωt)

110
Q

What is the equation for the the flux linkage of a coil in a generator?

A

Nφ = BANcos(ωt)

Where:
• Nφ = Magnetic flux linkage (Wb turns)
• B = Magnetic flux density (T)
• A = Area of coil (m²)
• ω = Angular velocity (rad/s)
• t = Time from coil being perpendicular to magnetic field (s)

(NOTE: Not given in exam, but Nφ = BANcosθ is given and ωt = θ)

111
Q

What does induced emf depend on?

A

The rate of change of flux linkage (Faraday’s).

Therefore it is also sinusoidal.

112
Q

What is the considered the “starting point” for a coil in a generator?

A

Perpendicular to magnetic field, so that the normal is parallel to the field

113
Q

What is the equation for the induced emf on a coil in a generator?

A

ε = BANωsin(ωt)

Where:
• ε = Induced emf (V)
• B = Magnetic flux density (T)
• A = Area of coil (m²)
• ω = Angular velocity (rad/s)
• t = Time from coil being perpendicular to magnetic field (s).

(observation = calculus is probably used to form this equation as seen on the graphs of emf and flux linkage)

114
Q

Describe the graph for flux linkage in a coil in a generator, starting with the coil perpendicular to the magnetic field.

A

Cosine wave from +BAN to -BAN

115
Q

Describe the graph for induced voltage in a coil in a generator, starting with the coil perpendicular to the magnetic field.

A

Sine wave

116
Q

How are the graphs for flux linkage and induced voltage in a coil in a generator related?

A

They are 90° out of phase.

Flux linkage is the derivative of emf
cos is the derivative sin

117
Q

Remember to practise drawing out the graphs for flux linkage and emf against time for a generator.

A

Pg 147 of revision guide or pg 345 of textbook.

118
Q

A rectangular coil with 30.0 turns, each with an area of 0.200m² is rotated as shown at 20rad/s in a uniform 1.50mT magnetic field. Calculate the maximum emf induced in the coil.

A
  • ε = BANωsin(ωt)
  • Therefore ε is greatest when sin(ωt) is +/-1
  • This gives ε = 1.50 x 10⁻³ x 0.200 x 30.0 x 20.0 x +/-1 = +/-0.180V
119
Q

What does the shape of the graph of induced emf depend on?

Which factor effects frequency and which doesn’t?

A

Increasing speed of rotation = increased frequency and increases maximum emf.

Increasing magnetic flux density, B will increase the maximum emf but will have no effect on frequency.

120
Q

What is back emf?

A

The emf that opposes the change in current which induced it.

121
Q

Give equations to show how back emf relates to the supply voltage.

A

• V - ε = IR
(Supply voltage - Back emf = IR)
• IV - Iε = I²R
(Input power - Useful mechanical power = Wasted power due to heating)

122
Q

Explain whether an electric motor working with a high or low load is more efficient.

A
  • With low load, the motor spins fast, so the back emf is large. This means that the current is low, so the electrical power wasted is very low.
  • With high load, the motor spins slowly, so the back emf is small. This means that the current is high, so the electrical power wasted is very high.
123
Q

What is an alternating current?

A

One that is constantly changing direction with time.

124
Q

What does ac mean for a voltage across a resistance?

A

It goes up and down in a regular pattern - positive and negative.

125
Q

What device can be used to show how voltage changes over time?

A

Oscilloscope

126
Q

What is on the x-axis and y-axis of an oscilloscope?

A
  • x-axis -> Time

* y-axis -> Voltage

127
Q

Is the y-axis on an oscilloscope fixed?

A

No, there is a control (Y-gain control dial) to change how many volts per division you want the y-axis scale to represent.

E.g. 5V per division.

(On some oscilloscopes, the height of each square on the grid is 1cm, so the scale might be in terms of V per cm)

128
Q

Describe the oscilloscope trace for an ac current.

A

Sine wave

129
Q

Describe the oscilloscope trace for a dc current.

A

Horizontal line at a given voltage

130
Q

Describe the oscilloscope trace for an ac current with timebase turned off.

A

Vertical line

131
Q

Describe the oscilloscope trace for a dc current with timebase turned off.

A

Dot

132
Q

What are the 3 basic pieces of information you can get from an ac oscilloscope trace?

A
  • Time period, T = can give you frequency
  • Peak voltage, V₀
  • Peak-to-peak voltage
133
Q

How can you find the time period from an oscilloscope?

A

It is from one peak to another.

134
Q

How can you find the peak voltage from an oscilloscope?

A

It is from the normal to the top of the sine wave.

135
Q

How can you find the peak-to-peak voltage from an oscilloscope?

A

It is from the very bottom to the very top of the sine wave.

136
Q

Explain what has a higher power supply, a 2V ac supply or 2V dc supply.

A
  • 2V dc supply

* Because in the ac supply, it will be below 2V most of the time.

137
Q

How do you compare ac and dc power output?

What can’t you do?

A

Find the average.

Normal average doesn’t work because the positive and negative bits cancel out.

Instead use root mean square voltage.

138
Q

What is the symbol for the rms voltage?

A

V(rms)

139
Q

What is the symbol for the rms current?

A

I(rms)

140
Q

What is the rms current?

A

The value of the direct current that would give the same heating effect as the alternating current in the same resistor.

141
Q

What is the equation for the rms voltage?

A

V(rms) = V₀ / √2

Where:
• V(rms) = rms voltage (V
• V₀ = Peak voltage (V)
For a sine wave)

142
Q

What is the equation for the rms current?

A

I(rms) = I₀ / √2

Where:
• I(rms) = rms current (A
• I₀ = Peak current (A)
For a sine wave)

143
Q

How can you work out the average power for an ac supply?

A

Work out the V(rms) and I(rms) values, then multiply them together.

144
Q

How can you work out the frequency from an oscilloscope?

A
  • Work out time period by looking at the time from peak to peak
  • f = 1 / T
145
Q

A light is powered by a sinusodial ac power supply with a peak voltage of 2.12V and a root mean square current of 0.40A.

a) Calculate the root mean square voltage of the power supply.
b) Calculate the power of the power supply.

A

a) V(rms) = V₀ / √2 = 2.12 / √2 = 1.499 = 1.50V

b) Power = 1.499 x 0.40 = 0.5996 = 0.60W (to 2 s.f.)

146
Q

What is the rms of UK mains supply?

A

230V

147
Q

Calculate the peak-to-peak voltage of UK mains electricity supply.

A
  • V₀ = V(rms) x √2
  • V₀ = 230 x √2 = 325.26V
  • Peak-to-peak V = 2 x 325.26 = 650V (to 2 s.f.)
148
Q

What are transformers?

A

Devices that make use of electromagnetic induction to change the size of the voltage for an alternating current.

149
Q

Describe the structure of a transformer.

A
  • Primary coil around one side of laminated iron core

* Secondary coil around other side of laminated iron core

150
Q

Describe how a transformer works.

A

An alternating current in the primary coil causes the core to magnetise, demagnetise and remagnetise continuously in opposite directions.

This produces a rapidly changing magnetic flux.

The changing magnetic flux is passed through the iron core to the secondary coil.

This induces an alternating voltage of the same frequency as the input voltage, but different voltage (assuming the number of turns is different).

151
Q

What is the transformer equation relating the number of turns on each coil and the voltage?

A

Ns / Np = Vs / Vp

Where:
• Ns = No. of turns on secondary coil
• Np = No. of turns on primary coil
• Vs = Voltage across secondary coil
• Vp = Voltage across primary coil
152
Q

Derive the transformer equation relating the number of turns on each coil and the voltage.

A 
Faraday’s Law:
• Vp = Np x Δφ / Δt
• Vs = Ns x Δφ / Δt
Therefore:
• Ns / Np = Vs / Vp
153
Q

What do step-up transformers do and how do they work?

A
  • Increase the voltage

* By having more turns on the secondary coil

154
Q

What do step-down transformers do and how do they work?

A
  • Decrease the voltage

* By having fewer turns on the secondary coil

155
Q

What is the output voltage for a transformer with a primary coil of 120 turns, a secondary coil of 350 turns and an input voltage of 230V?

A
  • Ns / Np = Vs / Vp
  • Vs = Vp x Ns / Np
  • Vs = 230 x 350 / 120 = 670.83 = 670V (to 2 s.f.)
156
Q

What are most power losses in a transformer due to? How does this work?

A
  • Eddy currents in the transformer’s iron core are looping currents induced by the changing magnetic flux i the core.
  • These create a magnetic field that acts against the field that produced them, reducing the field strength
  • They also generate heat
157
Q

How can the effects of eddy currents in a transformer be reduced?

A

Laminating the core with layers of insulation so current can’t flow.

158
Q

How can heat losses due to resistance in the wires in a transformer be reduced?

A

Using thick copper wire (which has low resistance = less heat).

Copper has low resistivity and larger diameter = smaller resistance.

159
Q

Energy is wasted as it heats the core when it magnetises and demagnetises. How can this be reduced?

A

Use magnetically soft material = magnetism disappears after current is removed.

One that magnetises and demagnetises easily.

160
Q

How is the loss of magnetic flux from primary to secondary coil reduced?

A

Make sure the coils are really close together - this can include winding the coils on top of each other around the same part of the core instead of around different parts of the core.

161
Q

What is the equation for an ideal transformer?

A

Power in = Power out:

162
Q

Give the equation for the efficiency of a transformer.

A

Efficiency = IsVs / IpVp

Where:
• Is = Current in the secondary coil (A)
• Vs = Potential difference across secondary coil (V)
• Ip = Current in primary coil (A)
• Vp = Potential difference across primary coil (V)

163
Q

What equation links voltage, number of turns and current?

A
164
Q

A transformer has an input current of 173A and doubles the input voltage to give an output voltage of 36,500V. Calculate the maximum possible current output by the transformer.

A
  • Vp = 35,600 / 2 = 17,800V
  • IpVp = IsVs
  • Is = IpVp / Vs = 173 x 17800 / 35600 = 86.5A
165
Q

The efficiency of a transformer is 0.871. It decreases an initial voltage of 11560V to 7851V and Ip us 195A. Calculate the current output by the transformer.

A
  • Efficiency = IsVs / IpVp

* Is = Efficiency x Ip x Vp / Vs = 0.871 x 195 x 11560 / 7851 = 250.083 = 250A (to 3 s.f.)

166
Q

What is the equation for energy lost?

A

E = Pt

E = energy in, J
P = power, W
t = time, s
167
Q
A
168
Q

Where are transformers used?

A

In the National Grid.

169
Q

What is the national grid?

A

Electricity from power stations is sent round the country in the national grid.

“The National Grid is Great Britain’s electricity transmission network, distributing the electrical power generated in England, Scotland, and Wales”

170
Q

Do you send higher or low current around the country

A

Low = causes less energy loss due to heating in cables.

171
Q

What is the equation for power losses due to resistance?

A

P = (I^2)R

172
Q

What happens if you double to transmitted current?

A

P = (I^2)R.

You quadruple to power lost.

173
Q

At what voltage do pylons send electricity around the country? (High or low?)

A

High, because this helps keep the current low, which reduces power losses.

High voltage means current has to be low in order to conserve power and energy (P = IV).

174
Q

How else do you reduce energy loss?

A

Using cables with lowest possible resistance.

175
Q

Describe the voltage at which electricity moves at various points between the power station and the home.

A 
• Power station to step-up transformer = 25,000V.
Then to pylons.
• Pylons = 400,000V.
Then to step-down transformer.
• Step-down transformer to homes = 230V.
176
Q

How and why is voltage stepped down to 230V before it can be used in homes?

A

Safety and insulation issues.

This is done in stages: Power transferred from overhead lines to underground wires.

177
Q

A current of 1330 A is used to transmit 1340 MW of power through 147 km of cables. The resistance of the transmission is 0.130 Ω per km. Calculate the power wasted.

A
  • Total resistance = 0.130 x 147 = 19.11 Ω

* Power lost = I²R = 1330² x 19.11 = 3.3803 x 10⁷ = 3.38 x 10⁷ W (to 3 s.f.)

178
Q

How do you investigate the relationship between the number of turns and the voltage across the coils of a transformer?

A
179
Q

How do you investigate the relationship between current and voltage of the transformer coil for a given number of turns in the coil?

A
180
Q

What should you do before you set up an experiment with transformers?

A

Risk assessment - Keep voltages and currents at a safe level.

AQA A Level Physics (14 decks)

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