Magnetic Fields

Question 1

What is a magnetic field?

Answer 1

A region in which a force acts on a magnetic material.

Question 2

What are the key features of a magnetic field around a bar magnet?

Answer 2

Field lines go from north to south pole.
Field strongest at poles, gets weaker with distance.
Denser field lines -> stronger field.

Question 3

Describe the key features of the magnetic field around a current carrying straight piece of wire

Answer 3

Concentric circles around the wire that get further apart with distance from wire as field gets weaker.

Question 4

Explain how to predict the direction of the magnetic field around a current carrying straight piece of wire

Answer 4

Use right hand thumb rule:
Thumb = conventional current (+ to -).
Curled fingers = direction of circular field lines.

Question 5

Describe the key features of the magnetic field around a current carrying solenoid

Answer 5

Outside solenoid - same as a bar magnet.
Inside solenoid – straight parallel lines.

Question 6

Explain how to predict the direction of the magnetic field around a current carrying solenoid.

Answer 6

Use right hand grip rule:
Curled fingers = conventional current (+ to -).
Thumb = points towards north pole.

Question 7

Why does a current carrying wire experience a force when placed in an external magnetic field?

Answer 7

The field around the wire and the external magnetic field are added together.
The resultant field produces a force on the wire.

Question 8

What does each finger represent in Fleming’s left-hand rule?

Answer 8

Thumb = F - force on wire
First finger = B - field (N to S)
Second finger = I – conventional current (+ to -)

Question 9

In what position will the force on a current carrying wire be: a. a maximum? b. a minimum (zero)?

Answer 9

a. When the current is perpendicular to the field.
b. When the current is parallel to the field.

Question 10

What happens if an AC is passed through the wire in a magnetic field?

Answer 10

Current constantly changes direction.
So, force constantly changes direction according to Flemings’ left-hand rule.
Causes wire to vibrate.

Question 11

Define magnetic flux density (B)

Answer 11

The force on 1 m of wire carrying a current of 1 A perpendicular to the magnetic field.

Question 12

Give the units for magnetic flux density (B) and define this unit. State whether it is scalar or vector.

Answer 12

Units = tesla
The strength of a magnetic field that produces a force of 1 N on a 1 m wire carrying a current of 1 A flowing perpendicular to the magnetic field. (1 T = 1 N A-1 m-1)
Vector quantity.

Question 13

Give the equation for the force on a current carrying wire in a magnetic field (when I and B are perpendicular).

Answer 13

F = BIL

Question 14

Give the equation for calculating the force on a charged particle moving perpendicular to a magnetic field.

Answer 14

F = BQv

Question 15

Give two conditions that would not result in a force on a charged particle in a magnetic field.

Answer 15

It is not moving (i.e. stationary).
It is moving parallel to the magnetic field.

Question 16

How can Fleming’s left-hand rule be applied to charged particles moving magnetic fields?

Answer 16

Thumb = F - force on charged particle
First finger = B - field (N to S)
Second finger = I – velocity direction of positive charges (swap the direction for negative charges)

Question 17

Why do charged particles move in circular motion in a magnetic field?

Answer 17

The force is always perpendicular to the velocity.
This provides a centripetal force towards the centre of the circle.

Question 18

Derive the equation for the radius of the circle for a charged particle in circular motion in a magnetic field

Answer 18

Centripetal force = force on charged particle in B field.

Question 19

Derive the equation for the time period for a charged particle in circular motion in a magnetic field.

Answer 19

Time period = circumference / velocity

Question 20

What does a cyclotron do?

Answer 20

It accelerates charged particles (e.g. for use in radiotherapy).

Question 21

Describe the structure of a cyclotron.

Answer 21

Two hollow semi-circular electrodes (‘dees’).
Magnetic field applied perpendicular to plane of electrodes.
Alternating electric field (p.d.) applied between electrodes.

Question 22

Explain the motion of the particles when they are inside the semi-circular electrodes.

Answer 22

• Move in circular motion.
  Force from magnetic field is always perpendicular to velocity -> provides a centripetal force.
  Radius of the circle increases each time due to increased velocity as 𝑟 = 𝑚𝑣 / 𝐵𝑄.

Question 23

Explain the motion of the particles when they are moving across the gap between the semi-circular electrodes.

Answer 23

Accelerated across the gap by the applied p.d..
Applied p.d. is alternating as it has to repeatedly change direction.

Question 24

Explain why the alternating p.d. has a fixed frequency.

Answer 24

The time period of the circular motion is independent of velocity as 𝑇 = 2𝜋𝑚 / 𝐵𝑄.
Particle spends the same amount of time in each electrode.

Question 25

Define magnetic flux φ in words and with an equation. Give the units and state whether it is scalar or vector.

Answer 25

The product of the magnetic flux density (B) and the area (A) (when B is normal to A). φ = BA - Units = webers (Wb) Scalar quantity.

Question 26

Define magnetic flux linkage Nφ in words and with an equation. Give the units and state whether it is scalar or vector.

Answer 26

The product of the magnetic flux (BA) and the number of turns (N) (when B is normal to A). Nφ = BAN Units = weber turns (Wb turns) Scalar quantity.

Question 27

Give the equation for magnetic flux linkage when magnetic flux is not perpendicular to area. Define the angle θ.

Answer 27

Nφ = BAN cosθ Where θ is the angle between the magnetic field and the normal to the plane of the coil.

Question 28

When is magnetic flux linkage a maximum and a minimum?

Answer 28

Maximum when θ = 0, cos θ = 1. Minimum when θ = 90, cos θ = 0

Question 29

What is electromagnetic induction?

Answer 29

Relative motion between a conducting rod and a magnetic field. Conducting rod cuts the magnetic field lines. Causes a change in magnetic flux. Induces an emf in the conducting rod. Which induced a current if there is a complete circuit.

Question 30

State Faraday’s law of electromagnetic induction and give the equation.

Answer 30

The induced emf is directly proportional to the rate of change of flux linkage. Magnitude of induced emf: ε = N Δφ / Δt

Question 31

What can you calculate from a graph of magnetic flux linkage Nφ against time t?

Answer 31

Gradient = magnitude of induced emf.

Question 32

What can you calculate from a graph of emf ε against time t?

Answer 32

Area under graph = change in flux linkage.

Question 33

State Lenz’s law of electromagnetic induction

Answer 33

The induced emf is always in such a direction as to oppose the change of flux that caused it. (This agrees with the principle of conservation of energy.)

Question 34

Use Lenz’s law to explain the acceleration of a magnet entering a vertical coil (North pole first) as: a. enters the coil. b. leaves the coil.

Answer 34

a. Induced emf gives north pole at top of coil -> repels magnet, opposing its motion, accelerates less than g. b. Induced emf gives north pole at bottom of coil -> attracts magnet, opposing its motion, accelerates less than g.

Question 35

Sketch a graph to show the emf induced in the vertical coil (when a magnet enters it North pole first) and explain its shape.

Answer 35

Induced emf in opposite direction as it enters compared to as it leaves (Lenz’s law). Greater emf when leaving due to falling faster and greater rate of change of flux linkage (Faraday’s law). Positive and negative areas must be equal -> represents change in flux linkage.

Question 36

Describe the phase difference between magnetic flux linkage and emf induced for a coil rotating in a magnetic field.

Answer 36

Out of phase by 90 degrees.

Question 37

Sketch the oscilloscope trace for D.C. and A.C. when the time base is switched on.

Answer 37

DC - horizontal line AC - sinosoidal curve

Question 38

Sketch the oscilloscope trace for D.C. and A.C. when the time base is switched off.

Answer 38

DC - single point AC - vertical line

Question 39

Sketch an A.C. trace with the time base switched on and label: a. time period T b. peak voltage V₀ c. peak-to-peak voltage (2V₀)

Answer 39

a. Distance between two peaks/ troughs b. Amplitude of the wave c. 2 x amplitude of the wave

Question 40

What does the y-gain setting of an oscilloscope control?

Answer 40

Volts per (vertical) division or volts per cm.

Question 41

What does the time base setting of an oscilloscope control?

Answer 41

Time per (horizontal) division or time per cm. Time is usually in milliseconds (ms).

Question 42

How is an oscilloscope set up?

Answer 42

Connect the source to the y-input. The trace can be vertically centred with the y-shift and horizontally centred with the x-shift.

Question 43

What are root mean square (rms) current and voltage?

Answer 43

The rms voltage/current for A.C. give the equivalent D.C. voltage/current that produce the same power (and same heating effect). (It is the square root of the mean of the squares of all the values of the voltage/current in one cycle.)

Question 44

How can the power of A.C. be calculated?

Answer 44

Power = I_rmsV_rms = (I_rms)²R = (V_rms)² / R

Question 45

What does the 230 V for UK mains electricity represent?

Answer 45

The rms voltage.

Question 46

What is the frequency of UK mains electricity?

Answer 46

50 Hz

Question 47

Describe the structure of a transformer.

Answer 47

Made up of a (electrically insulated) primary and secondary coil. Wrapped around a laminated iron core.

Question 48

Describe the function of a transformer.

Answer 48

Make use of electromagnetic induction to change the p.d. of an alternating current.

Question 49

For a step-up transformer, give the key facts about number of turns on the coils, p.d. and current.

Answer 49

More turns on secondary coil than primary coil. Increases p.d.. Decreases current.

Question 50

For a step-down transformer, give the key facts about number of turns on the coils, p.d. and current.

Answer 50

More turns on primary coil than secondary coil. Decreases p.d.. Increases current.

Question 51

Why are step-up transformers used in the national grid?

Answer 51

Used between power stations and transmission cables. For a given power, as P=IV, increasing the p.d. decreases the current. Less heating effect in cables Less energy wastefully dissipated to surroundings. Increases efficiency.

Question 52

Why are step-down transformers used in the national grid?

Answer 52

Used between transmission cables and domestic use. Lowers p.d. so safer for domestic use.

Question 53

Describe the function of the primary coil.

Answer 53

Alternating p.d. applied across primary coil -> causes alternating current to flow. This produces an alternating magnetic field in the iron core causing it to magnetise and demagnetise quickly.

Question 54

Describe the function of the iron core.

Answer 54

Increases the magnetic flux linkage from the primary to the secondary coil by focusing the direction of the magnetic flux.

Question 55

Describe the function of the secondary coil.

Answer 55

Has an alternating magnetic field surrounding it. Secondary coil therefore cuts the field lines. This induces an emf proportional to the rate of change of flux linkage (Faraday’s law) – will depend on the p.d. in the primary coil and the turns ratio between primary and secondary coils. This induces an alternating current in the coil.

Question 56

Why do transformers only work with A.C.?

Answer 56

Only A.C. will produce an alternating magentic field.

Question 57

How are the coils of a transfomer designed to increase efficiency?

Answer 57

Made from thick copper wire. Copper has a lower resistivity. Larger diameter reduces resistance. Reduces power lost due to heating in the coils.

Question 58

How does the core and positioning of coils increase efficiency?

Answer 58

Iron core used with coils as close as possible. Increases magnetic flux linkage from primary to secondary coil as direction of magnetic flux is focused.

Question 59

Why does a core made from iron increase efficiency?

Answer 59

Iron is magnetically soft. Reduces energy required to magnetise and demagnetise the core. Reduces power lost due to heating in the core.

Question 60

Why do eddy currents occur?

Answer 60

The iron core is being cut by the changing magnetic flux. This induces an emf in the core causing looping currents to flow as the iron conducts.

Question 61

Why do eddy currents decrease efficiency?

Answer 61

Eddy currents create a magnetic field that acts against the field that induced them (Lenz’s law). This reduces the overall field strength. Eddy currents also cause power loss due to heating in the core.

Question 62

Why does laminating the iron core increase efficiency?

Answer 62

Laminating the core involves having laminations of iron separated by laminations of insulator (high resistivity material). Reduces the effect of effect current.