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(A-Level) Physics > Module 6: C23 - Magnetic Fields > Flashcards

Module 6: C23 - Magnetic Fields Flashcards

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1
Q

What is a Magnetic Field

A

Magnetic field is the region around a permanent magnet or a current carrying conductor in which other magnetic objects will experience a force.

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2
Q

How are Magnetic Field lines drawn

A

Magnetic field lines are drawn in the direction a ‘free’ north pole would move.

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3
Q

How can you show an uniform magnetic field?

A

Equally spaced and parallel magnetic field lines represent a uniform field, that is, the strength of the magnetic field does not vary.

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4
Q

How can you show a magnetic field is stronger at a certain point

A

The magnetic field is stronger when the magnetic field lines are closer. For a bar magnet, the field is strongest at its north (N) and south (S) poles.

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5
Q

How is a magnetic field produced?

A

When a charged particle move, it creates a magnetic field. Therefore, when a current pass through a wire a magnetic field is produced.

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6
Q

How do you explain a magnetic field of a bar magnet?

A

In fact, it is created by the electrons whizzing around the iron nuclei. You can visualise the iron atoms as tiny magnets, all aligned in the same direction.

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7
Q

How can you find the direction the magnetic field when you know the direction of the current?

A

Use the right hand grip rule

  • Thumb points along direction of current
  • Other fingers give the direction of the field
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8
Q

How do you find the direction of a magnetic field around a current carrying wire?

A

The direction of a magnetic field around a current carrying wire can be calculated using the right hand grip rule. Imagine your thumb represents the direction of the current, your fingers will give the direction of the field lines.

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9
Q

State 2 methods of producing a uniform magnetic field

A

• Putting the North and South Pole of 2 bar magnets next to each other
• Inside a solenoid

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10
Q

Suggest how the magnetic field pattern for a solenoid within its core would change when the current is both reversed and increased

A

The field direction would be reversed and there would be more and closer magnetic field lines.

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11
Q

What is Fleming’s Left Hand Rule used for?

A

Fleming’s left hand rule is used to determine the direction of the force on a current carrying conductor in an external magnetic field.

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12
Q

What happens if the magnetic field and current are parallel?

A

If the magnetic field and the current are
parallel to each other then no force will be experienced.

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13
Q

What does your first finger, second finger, and thumb show with Fleming’s left hand rule

A
  • First finger gives the direction of the external magnetic field
  • Second finger gives the direction of the conventional current
  • Thumb gives the direction of motion (force) of the wire
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14
Q

Example Questions:

  1. How will the force change if the field strength is increased?
  2. How will the force change if the current is increased?
  3. How will the force change if the length of the wire is increased?
A

Force will increase for in all of these scenarios.

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15
Q

What does the size of the force on a current carrying wire depend on?

A

● the strength of the magnetic field B (Magnetic flux density)
● the current in the wire I
● the length of the wire in the field L

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16
Q

Equation for Force in a current carrying wire (using Magnetic Flux Density)

A

F = BILsinθ

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17
Q

What is the SI unit for the Magnetic Flux Density

A

The SI unit for the Magnetic flux density (B) is tesla.

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18
Q

Definition of a Tesla

A

One tesla, 1 T, is the magnetic flux density when a wire of length one metre and carrying a current of one ampere at a right angle to the field experiences a force of one newton.

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19
Q

Write the unit of Magnetic flux density, B (T) based on SI base unit

A

B = F / IL

T = N / Am
T = kgms^-2 / Am
T = kgms^-2A^-1m^-1
T = kgs^-2A^-1

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20
Q

Worked Example:

A straight piece of wire of length 3m carrying a current of 2A experiences a force of 12N when placed perpendicular to a uniform magnetic field. Calculate the value of the magnetic flux density.

A

l = 3
I = 2A
F = 12N
θ = 90°

F = BIL
12 = 3 x 2 x B
12 = 6B
2 = B

B = 2T

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21
Q

Is Magnetic Flux Density a Scalar or Vector quantity

A

Magnetic Flux Density is a Vector Quantity

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22
Q

Example Question:

A 2.8XZ length of copper wire carrying a current of 0.80A is placed in a uniform magnetic field. The angle between the wire and the magnetic field is 38°. It experiences a force of 4.0mN. Calculate the magnetic flux density of the field.

A

F = BILsinθ

4.0x10^-3 = B x 0.80 x 0.028 x sin38

B = 4.0x10^-3 / 0.80 x 0.028 x sin38

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23
Q

How could you find the Magnetic flux density in laboratory

A

Firstly you will need the apparatus for determining the magnetic flux density between two magnets. The magnets are placed on a top-pan balance. The magnetic field between them is almost uniform. A stiff copper wire is held perpendicular to the magnetic field is measure with a ruler. Use crocodile clips, a section of the wire is connected in series with an ammeter and a variable powder supply. The balance is zeroed when there is no current in the wire. With a current I, the wire experiences a vertical upward force (predicting Fleming’s left hand rule).

According to Newton’s 3rd law of motion, the magnets experience an equal downforce, F, which can be calculated from the change in the mass reading, m, using F = mg, where g is the acceleration of free fall. The magnetic flux density B between the magnets can then be determined from the equation B = F/IL

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24
Q

A 2.8cm length of copper wire carrying a current of 0.80A is placed in a uniform magnetic field. The angle between the wire and the magnetic field is 38°. It experiences a force of 4.0mN.

Calculate the magnetic flux density of the field.

A

F = BILsinθ

4x10^-3 = B x 0.80 x 0.028 x sin38

B = 4x10^-3 / 0.80 x 0.028 x sin 38

B = 0.29T

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25
Q

What causes the Northern Lights?

A

This happens when energetic charged particles from the Sun spiral down the Earth’s magnetic field towards a polar region and collide with atoms on the atmosphere, causing them to emit light. In this topic, you will learn how you can model the motion of charged particles in uniform magnetic fields.

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26
Q

Equation for Force on electric charge in a magnetic field (involving velocity)

A

F = Bqv sinØ

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27
Q

How can you derive F = Bqv from F = BIL

A

F = BIL sinØ
F = B(q/t)L sinØ
F = Bq(l/t) sinØ

F = Bqv sinØ

28
Q

Worked Example:

An electron travels perpendicular to a magnetic field of flux density 0.15T. Calculate the acceleration of the electron given its speed is 5x106ms-1.

A

Step 1: Write down the information that you have.
B = 0.15T, v = 5.0x10^6ms^-1, Q = e = 1.6x10^-19 C

Step 2: Select the equations needed
The force acting on the electron is given by F = Bqv and the acceleration a can be calculated from F = ma.
F = Bqv = ma
a = Bqv / m
a = (0.15 x 1.6x10^-19 x 5.0x10^6)/9.11x10^-31
a = 1.3x10^17 ms^-2

29
Q

A beam of electrons, moving at 1.0x10^6ms^-1, is directed through a magnetic field of flux density 0.50T. Calculate the force on each electron when:

a) The beam is at right angles to the magnetic field
b) The beam is at an angle of 45° to the field

A

B = 0.5T
v = 1x10^6 ms^-1
e = 1.6x10^-19

F = Bqv sinØ

a)
F = 0.5 x 1x10^6 x 1.6x10^-19 x sin(90)
F = 8.0x10^-14 N

b)
F = 0.5 x 1x10^6 x 1.6x10^-19 x sin(45)
F = 5.7x10^-14 N

30
Q

How can we find the Radius of Circular Motion?

A

F = mv^2 / r
F = Bqv

Bqv = mv^2 / r
r = mv / Bq

31
Q

Worked Example:

A beam of electrons describes a circular path of radius 15 mm in a uniform magnetic field. The speed of electrons is 8x106ms-1. Calculate the magnetic flux density B of the magnetic field.

A

r = 1.5x10^-2 m
v = 8.0x10^6 ms^-1
Q = e = 1.60x10^-19 C

Bqv = mv^2 / r
B = mv / Qr
B = (9.11x10^-31 x 8.0x10^6) / 1.60x10^-19 x 1.5x10^-2)
B = 3.0 x 10^-3 T

The magnetic flux density is about 3.0mT.

32
Q

An electron beam in a vacuum tube is directed at right angles to a magnetic field, so that it travels along a circular path. Predict the effect on the size and shape of the path that would be produced (separately) by each of the following changes:

a) increasing the magnetic flux density
b) reversing the direction of the magnetic field
c) slowing down the electrons
d) tilting the beam, so that the electrons have a component of velocity along the magnetic field.

A

a) The radius would increase
b) The direction would be reversed
c) The radius would decrease
d) Electrons will spiral around the field lines because they will have a constant component of velocity in the direction of the field lines.

33
Q

What is a Velocity Selector

A

A velocity selector is a device that uses both electric and magnetic fields to select charged particles of specific velocity. It is a vital part of instruments such as mass spectrometers and some particle accelerators.

34
Q

What does the equation r = mv/BQ show?

A
  • Faster-moving particles travel in bigger circles (r∝v)
  • More massive particles move in bigger circles (r∝m)
  • Stronger magnetic fields make the particles move in smaller circles (r∝1/B)
  • Particles with greater charge move in smaller circles (r∝1/Q).
35
Q

What does the velocity depend on in a Velocity Selector?

A

EQ = Bqv

V depends on E and B only

v = E/B

36
Q

What can show you the direction of Induced current in a wire?

A

Fleming’s Right Hand Dynamo Rule

37
Q

What things represent which quantities in Fleming’s Right Hand Rule?

A

You must use your right hand where the thumb, first finger and second finger represent the same quantities as the left hand.

Thumb - Movement
First Finger - Direction of Field
Second Finger - Current

38
Q

How can you induce EMF (+ how can you create an alternating current)

A

To induce an e.m.f, you need a coil and a magnet. When the magnet is pushed towards the coil, an e.m.f is induced across the ends of the coil, and when the magnet is pulled away, a reverse e.m.f is induced. Repeatedly pushing and pulling the magnet will induce an alternating current in the coil. The faster the magnet is moved, the larger the induced e.m.f.

39
Q

What are 2 ways EMF can be induced?

A

EMF can be induced when you move a magnet through a coil of copper wire, as well as moving a copper wire perpendicular to the magnetic field lines of a magnet

40
Q

What is the magnetic flux for a loop of wire of area, A, at right angles to a magnetic flux density, B, is given by?

A

Φ = BA cosØ

(where the loop of wire is at right angles to the magnetic flux density)

41
Q

What is Magnetic Flux Φ defined as

A

The magnetic flux Φ is defined as the product of the component of the magnetic flux density perpendicular to the area and the cross-sectional area.

42
Q

What is Weber (Wb)

A

The unit of magnetic flux is the weber (Wb).

One weber is defined as being the magnetic flux when a field of magnetic flux density of one tesla passes at right angles through a coil of area one metre squared.

43
Q

Show that:

Wb = TM^2

A

Φ = BA

Wb = 1T x 1m^2

Wb = Tm^2

44
Q

Example Question:

What is the magnetic flux through a loop of area 0.4m^2 placed at right angles to a magnetic field of 2T?

A

A = 0.4m^2
B = 2T
Ø = 0

Φ = BA cosØ
Φ = 0.4 x 2 x 1
Φ = 0.8Wb

45
Q

Example Question:

What is the magnetic flux through a loop of area 0.4m^2 in a magnetic field of 2T if the flux density makes an angle of 30° with the loop?

A

A = 0.4m^2
B = 2T
Ø = 60°

Φ = BA cosØ
Φ = 0.4 x 2 x cos(60)
Φ = 0.4Wb

46
Q

What is Magnetic Flux Linkage

A

Magnetic flux linkage is defined as the product of the number of turns in the coil N and the magnetic flux.

Flux Linkage= NΦ

47
Q

Equation for Magnetic Flux Linkage

A

Flux Linkage= NΦ
(Unit =Wb turns)

48
Q

Worked Example:

A coil has 200 turns and a core of cross-sectional area of 1.0 x 10^-4 m^2. The coil is placed at right angles to a magnetic field of flux density 0.30 T. Calculate the magnetic flux and magnetic flux linkage for the coil.

A

Step 1: Calculate the magnetic flux. At right angles, magnetic flux.
Φ = BA
Φ = 0.30 x 1.0x10^-4 = 3.0x10^-5 Wb

Step 2: The magnetic flux linkage is NΦ. Therefore:
Magnetic flux linkage = NΦ
Magnetic flux linkage = 200 x 3.0x10^-5 = 6.0x10^-3 Wb

49
Q

When is e.m.f induced in a circuit

A

An e.m.f is induced in a circuit whenever there is a change in the magnetic flux linking the circuit. Since Φ = BAcosØ, you can induce an e.m.f by changing B, A, or Ø.

50
Q

What is Faraday’s Law of Electromagnetic Induction

A

The magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.

ϵ ∝ Δ(NΦ) / Δt

ϵ = e.m.f
Δ(NΦ) = change in Magnetic Flux
Δt = change in time

51
Q

What equation for e.m.f is found after combining Faraday’s Law and Lenz’s Law

A

ϵ = - Δ(NΦ) / Δt

52
Q

What is Lenz’s Law of Electromagnetic Induction

A

The direction of any induced e.m.f or current is always in a direction that opposes the change that producing it.

(Lenz’s law is an expression of conservation of energy.)

53
Q

Equation for Flux Linkage

A

Flux Linkage = NΦ = N(BAcosØ)

Flux Linkage = BAN cosØ

54
Q

According to Faraday’s Law, the induced e.m.f. ϵ = - Δ(BAN cosØ) / Δt

  • For a graph of Magnetic Flux Linkage against time, what is the magnitude of the gradient equal to?
  • What is is the e.m.f, ϵ, proportional to in the equation
A
  • The magnitude of the gradient from the magnetic flux linkage against time graph is equal to the induced e.m.f. ϵ.
  • For a given generator, B, A, and N, are all constant, therefore, ϵ ∝ -Δ(cosØ) / Δt
55
Q

For a graph that shows the variation of e.m.f, ϵ, with time, t, what is the maximum induced e.m.f directly proportional to

A
  • The magnetic flux density (B)
  • The cross-sectional area (A) of the coil
  • The number of turns (N)
  • The frequency (f) of the rotating coil
56
Q

How are the input and output voltages related in a transformer

A

The input voltage Vp and output voltage Vs are related to the number np of turns on the primary coil and number ns of turns on the secondary coil by the turn ratio equation:

Ns/Np = Vs/Vp
(for an ideal transformer)

57
Q

Where does a Step-Up Transformer have more turns

A

A step-up transformer has more turns on the secondary than on the primary coil, and Vs>Vp.

58
Q

Where does a Step-Down Transformer have more turns

A

A step-down transformer has fewer turns on the secondary than on the primary coil, and Vs>Vp.

59
Q

Worked Example:

A step-down transformer changes 230V mains voltage to 5V. The transformer has 920 turns on its primary coil. Calculate the number of turns on its secondary coil.

A

Step 1: Rearrange the turn-ratio equation

Ns/Np = Vs/Vp
Ns = VsNp/Vp

Step 2: Calculate the number of turns on the secondary coil

Ns = VsNp / Vp
Ns = 5.0 x 920 / 230 = 20 turns

60
Q

What is the Power Output for Efficient Transformers

A

For a 100% efficient transformer, the output power from the secondary coil is equal to the output power from the primary coil

61
Q

How does the equation show that Power in the primary coil is equal to power in the secondary coil:

Vs/Vp = Is/Ip

A

Vs/Vp = Is/Ip
VsIs = VpIp

P = IV

Ps = Pp

62
Q

What happens to Voltage and Current in a Step-Up Transformer

A

In a step up transformer, the voltage is stepped up, but the current is stepped down. Increasing the voltage by a factor of 100 will decrease the output current by a factor of 100.

63
Q

What happens to Voltage and Current in a Step-Down Transformer

A

In a step down transformer, the voltage is stepped down, but the current is stepped up. Decreasing the voltage by a factor of 100 will increase the output current by a factor of 100.

64
Q

How can Transformers be made efficient

A

Transformers can be made efficient by using low-resistance windings to reduce power losses due to the heating effect of the current. Making a laminated core with layers of iron separated by an insulator helps to minimise currents induced in the core itself (eddy currents), so this too minimises losses due to heating. The core is mass of soft iron, which is very easy to magnetise and demagnetise, and this also helps to improve overall efficiency of the transformer.

65
Q

What 2 things result in energy loss in transformers

A
  1. Due to heating of primary and secondly coil
  2. Due to eddy currents (current induced in coil) set up in the core of a transformer.

(A-Level) Physics (27 decks)

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