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Physics > Y2: Magnetic fields > Flashcards

Y2: Magnetic fields Flashcards

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1
Q

What is a magnetic field

A

A region in which a force is exerted on magnetic or magnetically susceptible materials

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2
Q

How are flux lines used to represent a magnetic field

A

Lines go from north to south, and are closer together where the field is stronger

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3
Q

How are magnetic fields induced around a wire

A

When a current flows through a wire, a magnetic field will be induced in concentric circles around it.

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4
Q

How can the direction of the induced magnetic field around a wire be determined

A

Right-hand rule:
- Curl the right hand into a fist with the thumb up
- Point the thumb in the direction that the current is flowing
- The curled fingers will show the direction of the field lines

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5
Q

What is a solenoid and how does it induce a magnetic field

A

A coiled wire, inducing a field that acts like a bar magnet.
(Right hand rule can be used if the fingers coil in the direction of the current, the thumb will point to the north pole.)

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6
Q

What is the motor effect

A
  • If a current carrying wire is placed in an external magnetic field, the induced field around the wire interacts with this external field.
  • This forms a resultant field, causing a force to act on the wire
  • The magnitude of the force depends on the perpendicular component of the magnetic field in relation to the current

(Force acts on current carrying wire, as force acts on charged particles in magnetic field)

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7
Q

How can the direction of a force on a wire in a magnetic field be determined

A

Fleming’s left hand rule:
- On the left hand, point the thumb up, the index finger forward and the middle finger to the right
- First finger = field direction (N→S)
- Middle finger = Conventional Current direction (+→-)
- Thumb = Force direction

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8
Q

What is magnetic flux density (B)

A

Represents the strength of the magnetic field
- The number of field lines per unit area
- The force on a 1m wire carrying a current of 1A, perpendicular to the magnetic field
∴ F∝B
- Unit = Tesla
1T = 1NA^-1m^-1

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9
Q

What is the equation for the Force acting on a wire in a magnetic field

A

F = BIL

B: Magnetic flux density (T)
I: component of the current perpendicular to the field (A)
L: Length of the wire in the field (m)

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10
Q

How can you determine the magnetic flux density of a magnetic field

A
  • Set up a circuit with a section of wire (length L) in a perpendicular magnetic field, on a top-pan balance
  • Zero the balance when there is no current, so any mass readings will represent the force acting on the wire
  • Turn on the dc power supply and swap the wire clips so the mass reading is positive
  • Record the mass and current
  • Use a variable resistor to alter the current and record a new mass
  • Repeat to obtain a mean mass reading for a wide range of currents
  • Convert each mass reading into a force (F=mg)
  • Plot an F-I graph
    F = BIL
    ∴ BL = F/I
    ∴ BL = Gradient
    ∴ B = Gradient/L

The same method can be used to alter different factors (eg. change L and use a variable resistor to keep I constant)
to see the effect on the force

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11
Q

What is the equation for the force acting on a charged particle in a magnetic field

A

F = BQV

B: Magnetic flux density
Q: Charge of particle
V: Velocity of particle

For a wire in a magnetic field, F=BIL
For a current, I=Q/t
For a particle, V=s/t
∴ L=Vt
∴ F = B(Q/t)(Vt)
∴ F = BQV

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12
Q

What is the shape of the path of a charge particle in a magnetic field

A

Circular path:
- Force always acts perpendicular to the direction of travel, resulting in circular motion
- Acceleration towards the centre is due to the centripetal acceleration of the circular motion

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13
Q

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

A

r = (mV)/(BQ)

For circular motion, centripetal acceleration = (V^2)/r
F = ma
∴ F = (mV^2)/r
In a magnetic field, the force on a charge particle = BQV
∴ BQV = (mV^2)/r
∴ r = (mV)/(BQ)

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14
Q

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

A

f = (BQ)/(2πm)

For a rotating object, f = V/(2πr)
r = (mV)/(BQ)
∴ f = V/(2π(mV/BQ))
∴ f = (BQ)/(2πm)

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15
Q

What is the relationship between the velocity of a particle in a magnetic field and the frequency of it’s circular path

A

The frequency of the circular path is independent of the velocity.
However, r ∝ V
∴ As the velocity of a charged particle increases, the radius of the circular path increases, but rotations will still occur at the same frequency.

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16
Q

What is a cyclotron

A

A particle accelerator composed of 2, hollow, semi-circular electrodes with a uniform magnetic field applied perpendicular to it and an alternating potential difference between them.

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17
Q

How does a cyclotron work as a particle accelerator

A
  • Charged particles are fired into one of the electrodes at the centre, and follow a (semi)circular path before leaving in the opposite direction
  • A potential difference between the electrodes causes the particle to accelerate towards the other, increasing it’s velocity
  • As the particle is travelling faster, the radius of the (semi)circular path is greater in the second electron, whilst the frequency of rotation (time in the electrode) remains the same.
  • The pd is reversed so that the particle accelerates back to the other electrode, increasing the velocity to increase the radius further.
  • This cycle repeats as the particle accelerates, spiralling outwards until it leaves the cyclotron
  • The frequency of the rotation remains the same for all speeds, so the potential difference alternates at a constant frequency.
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18
Q

What is Magnetic flux (ɸ)

A

The total magnetic flux density through an area A, perpendicular to the field
- Unit = Weber (Wb)

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19
Q

What is the equation for magnetic flux

A

ɸ = BAcosθ

ɸ: Magnetic flux (Wb)
Bcosθ: Component of Magnetic flux density perpendicular to area (T)
A: Area (m^2)

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20
Q

What is electromagnetic induction

A
  • If there is relative motion between a conducting rod and a magnetic field, the electrons in the rod will experience a force
  • This causes the electrons to accumulate at on end of the rod, inducing an emf across the ends (and ∴ a current)
  • emf induced when the conductor cuts the flux lines
  • emf can be induced in a coil or solenoid, by passing a magnetic field through it.
  • emf only induced while movement occurs (if movement stops, emf returns to zero, as there is no longer a force)
  • Direction of induced emf/current can be determined by using Fleming’s left hand rule, with the thumb in the opposite direction to the motion (Lenz’s Law)
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21
Q

What is flux linkage (Nɸ)

A

The magnetic flux for a coil perpendicular to a magnetic field

22
Q

What is the equation for the magnetic flux linkage

A

Nɸ = BANcosθ

N: No. of turns in coil
ɸ: Magnetic flux (Wb)
Bcosθ: Component of Magnetic flux density perpendicular to coil(T)
A: Area of coil (m^2)

23
Q

How would you investigate flux linkage with a search coil

A
  • Set up a circuit with a solenoid (or coiled wire) connected to an AC power supply
  • Connect a search coil to an oscilloscope to record its induced emf
  • Use a clamp stand to hold the search coil in the centre of the solenoid (field is strongest)
  • Orientate the search coil so it is perpendicular to the field inside the solenoid (θ=0°)
  • Record the amplitude on the oscilloscope (= induced emf)
  • Rotate the search coil and record the induced emf every 10°, until the coil is parallel to the field at θ=90°
  • Plot a graph of ε-cosθ (radians), to show a directly proportional relationship
    (as θ approaches 90°, induced ε approaches 0)
  • Using the equation ε=BANωcosθ (⇒ Gradient = BANω), determine B
  • Flux linkage can be calculated for each angle θ, with a graph of Nɸ-θ, showing how the flux linkage decreases as θ increases
24
Q

What is Faraday’s Law

A

Induced emf is directly proportional to the rate of change of flux linkage

25
Q

What is the equation for Faraday’s Law

A

ε = -N(Δɸ/Δt)

  • Negative, as the induced emf occurs in a direction to oppose the change (Lenz’s Law)
  • This shows that ε is the gradient of a Nɸ-t graph, and that NΔɸ is the area under an ε-t graph
26
Q

What is Lenz’s Law

A

An induced emf (and current) is always in such a direction to oppose the change that caused it.
- The induced emf acts in a direction to create a resistive force that opposes the motion that caused it
- ∴ The direction of the induced emf can be determined with Fleming’s LHR, with the thumb in the opposite direction to the motion (direction of resistive force)

27
Q

How can Lenz’s law be used to determine the direction of the induced current in a secondary coil

A
  • A coil flows in coil 1, resulting in a magnetic field through it
  • If coil 2 is placed next to coil 2, the change in flux through coil 2 induces a current in the wire
  • According to Lenz’s law, the current in coil 2 with flow in the opposite direction to coil 1, inducing an opposing magnetic field.
  • If AC power is used, an inverse alternating current will be induced in coil 2
  • This is similar to when a bar magnet is place in a solenoid, as the field induces a current that flows in a direction to cause an opposing magnetic field to resist the change
28
Q

What is the equation for the emf induced in a coil rotating in a magnetic field

A

ε = BANωsin(ωt)

When a coil rotates uniformly in a magnetic field, the coil ‘cuts’ the flux lines, so an alternating emf is induced
Nɸ = BANcosθ
For a rotating coil, θ = ωt
∴ Nɸ = BANcos(ωt)
∴ Differentiate: N(Δɸ/Δt) = -BANωsin(ωt)
∴ -N(Δɸ/Δt) = BANωsin(ωt)
ε = -N(Δɸ/Δt)
∴ ε = BANωsin(ωt)
∴ The amplitude of the induced emf can be altered by changing the speed of rotation, or the magnetic flux density of the field

29
Q

What is a generator and how does it induce a current

A
  • A wire coil is manually turned in a magnetic field
  • This induces an emf in the wire that alternates direction each time the coil is parallel to the field during it’s rotation (every half turn)
  • A Slip ring commutator allows this induced AC to be connected to and power an external circuit
30
Q

What is a motor

A
  • A wire coil with a dc current through it inside a magnetic field.
  • The current and the field result in a force that rotates the coil until it is parallel to the magnetic field
  • A Split ring commutator alternates the direction of the coil in the wire, so that the force always continues to in the same direction and he rotation in continuous
  • The momentum of the rotating coil allows it move through the points where the perpendicular component of the field to the motion is 0
31
Q

What is the difference between a slip ring and a split ring commutator

A

Slip ring commutator:
- 2 solid rings, one attached to each end of the coiled wire, with brushes connected, allowing an AC current to be obtained from a generator

Split ring commutator:
- 2 half circles, each attached to one end of the coiled wire, with brushes attached, so a DC power supply can alternate direction in the rotating coil of a motor

32
Q

What is an alternating current

A

Current that changes direction with time, as the voltage goes up and down (+/-) across a resistance.

33
Q

What is an oscilloscope

A

A computer that can display AC and DC voltage, as a trace is made by a beam of electrons moving across the screen
- Y-axis = Voltage (set to specific V per division)
- X-axis = Time base (can be switched off)

34
Q

How is AC displayed on an oscilloscope

A

Sine wave oscillating between positive and negative peak voltage
(if time base is switched off, will appear as vertical line on the y axis)

35
Q

How is DC displayed on an oscilloscope

A

Horizontal line at the value of V
(if time base is switched, will appear as a dot on the y-axis)

36
Q

What are the root mean square voltage and current

A

The average magnitude of the voltage and current, as accounts for the negative values with AC.
(Power output for 2V AC will be less than 2V DC, as the alternating current oscillates below this peak value)

37
Q

What is the equation for the root mean square voltage for AC

A

For a sign wave of V-t:
V(rms) = Vo/√2

V(rms): Root mean square voltage
Vo: Peak voltage

38
Q

What is the equation for the root mean square current for AC

A

For a sign wave of V-t:
I(rms) = Io/√2

I(rms): Root mean square voltage
Io: Peak voltage

39
Q

What is the equation for the average power output for AC

A

Average power = I(rms) x V(rms)
∴ = (IoVo)/2

40
Q

What is the root mean square voltage for UK mains

A

V(rms) = 230V

41
Q

What is the peak voltage for UK mains

A

Vo = 330V

42
Q

What is a transformer, and how do they work

A

A laminated iron core used to change the voltage size for alternating current
- Primary coil wrapped around one side, and secondary coil wrapped around the other
- AC in the primary coil induces a magnetic field that alternates direction
- Iron core is magnetised in a constantly alternating direction
- Changing magnetic flux across the core induces an alternating emf in the secondary coil
- This causes an AC in the secondary coil
- The AC alternates at the same frequency, but the magnitude of the voltage in the secondary coil depends on the ratio of the number of turns in each coil.

43
Q

What is the relationship between the voltage in the primary and secondary coil of a transformer

A

Ns/Np = Vs/Vp

Ns: No. of turns in the secondary coil
Np: No. of turns in the primary coil
Vs: Voltage in secondary coil
Vp: Voltage in primary coil

According to Faraday’s law:
Vp = Np(Δɸ/Δt) ⇒ Vp/Np = (Δɸ/Δt)
Vs = Ns(Δɸ/Δt) ⇒ Vs/Ns = (Δɸ/Δt)
∴ Vp/Np = Vs/Ns
∴ Ns/Np = Vs/Vp

44
Q

What is a step up transformer

A

A transformer that increases the voltage by having more turns in the secondary coil

Ns/Np = Vs/Vp
For step up: Ns > Np
∴ Vs > Vp

45
Q

What is a step down transformer

A

A transformer that decreases the voltage by having more turns in the primary coil

Ns/Np = Vs/Vp
For step down: Ns < Np
∴ Vs < Vp

46
Q

What is true about 100% efficient transformers

A

Power in = Power out
However, in practice, some power is always lost

47
Q

What factors limit the efficiency of transformers, and how is their impact reduced

A

Eddy currents:
- Small circular currents in the core induced by the changing flux, with opposing magnetic fields to that of the core (Lenz’s law)
- Reduce the overall magnetic field strength and generate heat, which dissipates energy
- Effect can be reduced by using laminated core (contains layers of insulation, which stops currents flowing)

Resistance in coils:
- Heat from wires dissipates energy
- To limit this, thick copper wires are used
(Large diameter = lower resistance, Copper = low resistivity)

Energy needed to magnetise core
- Energy dissipated as heat
- To reduce this, magnetically softy material is used (Iron - easily magnetised)

Loss of magnetic flux between coils
- Flux decreases between the coils if they are far apart
- To reduce this, the transformer can be designed so the coils are closer together, or even on the same side of the core.

48
Q

What is the equation for the efficiency of a transformer

A

Efficiency = IsVs / IpVp
(= ratio of input power to output power)

49
Q

How can the energy lost to the surroundings for a transformer be calculated

A

E = Pt
∴ E(loss) = P(loss)t
P(loss) = IpVp - IsVs
∴ E(loss) = (IpVp - IsVs)t

50
Q

How are transformers used in the national grid

A

Step Up:
- Electricity from power stations travels in the national grid as the lowest possible current.
- This is because the power loss can be calculated from P=I^2(R)
- ∴ P∝I^2
- ∴ The lower, the current, the less power lost.
- For the power (IV) to be maintained whilst the current is decreased, the voltage must be increased
- ∴ Step up transformers will increase the voltage to 400,000V for transmission through the national grid

Step down:
- Increasing the voltage creates more safety and insulation risks
- ∴ Step down transformers will reduce the voltage to 230V for domestic use

51
Q

How do you investigate the relationship between the number or turns and the voltage in each coil of a transformer

A
  • Set up a transformer with 5 turns in the primary coil and 10 turns in the secondary coil (∴ Np:Ns = 1:2)
  • Turn on the AC power supply in the primary coil, and record I and V for each coil
  • Keeping Vp constant, repeat with different Np:Ns ratios, recording I and V in each coil
  • Results should show that Ns/Np = Vs/Vp
    (However, this is for 100% efficient transformers, so won’t be fully accurate in reality).
52
Q

How do you investigate the relationship between the current and the voltage in each coil of a transformer

A
  • Set up a transformer with a constant Np:Ns ratio, an a variable resistor in the primary circuit
  • Turn on the AC power supply in the primary coil, and record I and V for each coil
  • With the same Np:Ns ratio (don’t change the coils), adjust the variable resistor to change the input current
  • Record I and V for each coil, repeating for a range of different input currents
  • Results should show that Ns/Np = Vs/Vp = Is/Ip, as for a constant power, I∝1/v
    (However, this is for 100% efficient transformers, so won’t be fully accurate in reality).

Physics (14 decks)

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