Section 11 - Magnetic Fields Flashcards

You may prefer our related Brainscape-certified flashcards:
1
Q

What is a magnetic field?

A

A region where a force is exerted on magnetic materials.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

How can magnetic fields be represented?

A

Using field lines.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is another name for magnetic field lines?

A

Flux lines

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Which direction do magnetic field lines go?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

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

A

The closer together the lines, the stronger the field.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What is a neutral point in magnetic fields?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What happens when a current passes through a wire?

A

A magnetic field is induced around the wire.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

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

A

Concentric circles centred around the wire

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

See diagram pg 140 of revision guide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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.

See diagram pg 140 of revision guide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
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
• First finger = Magnetic field
• Second finger = Current (conventional)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

What is magnetic flux density?

A
  • The force on 1m of site carrying a current of 1A right angles to the magnetic field.
  • i.e. It is the magnetic field strength
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

In magnetic fields, what is B?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

What are the units for magnetic field density?

A

Tesla (T)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

What is magnetic flux?

A

The number of field lines passing through an area.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

In magnetic fields, what is φ?

A

Magnetic flux

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

What are the units for magnetic flux?

A

Wb

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Is magnetic flux density scalar or vector?

A

Vector

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
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.
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.

(See diagram pg 141 of revision guide)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Remember to practise drawing out the setup for the experiment to investigate F = BIl.

A

Pg 141 of revision guide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

In the experiment to investigate F = BIl, how can you vary each of the variables?

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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
27
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
28
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.)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
30
Q

Do charged particles experience a force in a magnetic field?

A

Only if they are moving.

NOTE: Check this!

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
31
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
32
Q

When do F = BIl and F = BQv apply?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
33
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
34
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
35
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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
36
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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
37
Q

Is the force experienced by a moving charged particle in a magnetic field affected by mass?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
38
Q

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

A

F = mv²/r

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
39
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)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
40
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
41
Q

How is a magnetic field going into the paper symbolised?

A

Circles with crosses in them

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
42
Q

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

A

Increases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
43
Q

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

A

Increases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
44
Q

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

A

Decreases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
45
Q

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

A

Decreases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
46
Q

What is a cyclotron?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
47
Q

What are some uses of cyclotrons?

A
  • Producing radioactive tracers

* Producing high-energy beams of radiation for radiotherapy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
48
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)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
49
Q

Describe the structure of a cyclotron.

A
  • Two hollow semicircular electrodes with alternating p.d.
  • Slight gap between them
  • Uniform magnetic field applied perpendicular to the plane of the electrodes
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
50
Q

Describe how a cyclotron works.

A
  • Particle is fired into one of the electrodes
  • The magnetic field makes it flow a semicircular path and return to the gap between electrodes
  • The potential difference between them creates an electric field that accelerates the particle
  • 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 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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
51
Q

Remember to practise writing out the functioning and structure of a cyclotron.

A

Pg 143 of revision guide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
52
Q

What is the equation for magnetic flux?

A

φ = BA

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
53
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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
54
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
55
Q

What is electromagnetic induction?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
56
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
57
Q

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

A

The magnetic field

58
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.

59
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.

60
Q

What are the units for magnetic flux linkage?

A

Wb turns

61
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
62
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!)

63
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
64
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.

(See diagram pg 144 of revision guide)

65
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.
66
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°.

(See diagram pg 145 of revision guide)

67
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

68
Q

What is the symbol for magnetic flux linkage?

A

69
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.

70
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)

71
Q

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

A

The induced emf.

72
Q

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

A

Flux linkage

73
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(
74
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)

75
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
76
Q

Remember to ask Sir about the ε = Blv confusion. The magnetic flux is not changing.

A

Do it.

77
Q

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

A

The flux cut per second.

78
Q

State Lenz’s law.

A

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

79
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.)

80
Q

What principle does Lenz’s law agree with?

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
81
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.

(See bottom of pg 146 of revision guide)

82
Q

What is a generator?

A

A machine that uses a rotating coil in a magnetic field to convert kinetic energy into electrical energy.

83
Q

What is another name for a generator?

A

Dynamo

84
Q

What is an alternator?

A

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

85
Q

How does an a.c. generator work?

A
  • A spinning rectangular coil is placed in a magnetic field
  • Slip rings and brushes are connected to the coil
  • This means the direction of the induced current is reversed every half turn
86
Q

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

A

+BAN and -BAN

87
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)

88
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

89
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)
90
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

See diagram pg 147 of revision guide

91
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

See diagram of 147 of revision guide

92
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.

See diagrams pg 147 of revision guide

93
Q

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

A

Pg 147 of revision guide

94
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
95
Q

What is back emf?

A

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

96
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)

97
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.
98
Q

What is an alternating current?

A

One that is constantly changing direction

99
Q

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

A

Oscilloscope

100
Q

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

A
  • x-axis -> Time

* y-axis -> Voltage

101
Q

Is the y-axis on an oscilloscope fixed?

A

No, there is a control to change how many volts per division there are.

102
Q

Describe the oscilloscope trace for an ac current.

A

Sine wave

103
Q

Describe the oscilloscope trace for a dc current.

A

Horizontal line at a given voltage

104
Q

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

A

Vertical line

105
Q

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

A

Dot

106
Q

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

A
  • Time period, T
  • Peak voltage, V₀
  • Peak-to-peak voltage
107
Q

How can you find the time period from an oscilloscope?

A

It is from one peak to another.

See diagram of 148 of revision guide

108
Q

How can you find the peak voltage from an oscilloscope?

A

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

109
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.

110
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,

111
Q

What is the symbol for the rms voltage?

A

V(rms)

112
Q

What is the symbol for the rms current?

A

I(rms)

113
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.

114
Q

What is the equation for the rms voltage?

A

V(rms) = V₀ / √2

Where:
• V(rms) = rms voltage (V)
• V₀ = Peak voltage (V)

115
Q

What is the equation for the rms current?

A

I(rms) = I₀ / √2

Where:
• I(rms) = rms current (A)
• I₀ = Peak current (A)

116
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.

117
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
118
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.)

119
Q

What is the rms of UK mains supply?

A

230V

120
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.)
121
Q

What are transformers?

A

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

122
Q

Describe the structure of a transformer.

A
  • Primary coil around one side of iron core

* Secondary coil around other side of iron core

123
Q

Describe how a transformer works.

A
  • An alternating current in the primary coil produces magnetic flux
  • The changing magnetic field is passed through the iron core to the secondary coil
  • This induces an alternating voltage of the same frequency as the input voltage
124
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
125
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
126
Q

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

A
  • Increase the voltage

* By having more turns on the secondary coil

127
Q

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

A
  • Decrease the voltage

* By having fewer turns on the secondary coil

128
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.)
129
Q

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

A
  • Eddy currents in the transformer’s iron core
  • These create a magnetic field that acts against the field that produced them, reducing the field strength
  • They also generate heat
130
Q

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

A

Laminating the core with layers of insulation

131
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)

132
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)

133
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
134
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.)

135
Q

Where are transformers used?

A

In the National Grid.

136
Q

High or low, at what voltage do pylons send electricity around the country?

A

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

137
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
  • Pylons = 400,000V
  • Step-down transformer to homes = 230V
138
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.)