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A-Level Physics · guided topic map

Magnetism for Cambridge International AS & A Level Physics

Magnetism for A-Level Physics, organized into 1 syllabus topic and 8 mapped concept guides.

Syllabus topics
1
Mapped concept guides
8
Educational level
Cambridge International AS & A Level

Syllabus to lesson

Choose the exact concept

Work in order or jump to the concept named in your specification, course outline, or assignment.

20

Magnetic fields

A Level extension

8 guides
  1. 01Magnetic fields and field linesMapped lesson
  2. 02Force on a current-carrying conductorMapped lesson
  3. 03Force on moving chargesMapped lesson
  4. 04Hall effect and velocity selectionMapped lesson
  5. 05Magnetic fields due to currentsMapped lesson
  6. 06Magnetic flux and flux linkageMapped lesson
  7. 07Faraday's lawMapped lesson
  8. 08Lenz's lawMapped lesson

Diagrams

Magnetism as A-Level Physics draws it

The figures from the A-Level Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 5.1Magnetic fieldsA Level
An electron entering a uniform magnetic field at right anglesuniform magnetic field into the page−electronv

Figure comment

Fig. 5.1An electron, drawn as a small circle marked with a negative sign, travels horizontally to the right, and an arrow labelled v gives its velocity. Ahead of it lies a large rectangular region with a dashed boundary, filled with a regular array of crosses and labelled as a uniform magnetic field directed into the page. The electron is shown outside that region, at the instant before it crosses the boundary, with its velocity lying in the plane of the page.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

20 Magnetic fieldsOpen A Level question · A Level extension Q5
Guided questions 5 parts

Reading cue. Crosses mean B points into the page, and the particle is negative, so the conventional current is opposite to the arrow v — apply the left-hand rule to that reversed direction.

  1. aState State the direction of the force on the electron at the instant it crosses the dashed boundary in Fig. 5.1.

    recall2 marks

    Check answer 2 marks
    1. vertically downwards in the plane of the page
    2. perpendicular to both v and B, from the left-hand rule applied with the conventional current opposite to v

  2. bExplain Explain why the electron travels at constant speed inside the field region even though its velocity is changing.

    routine3 marks

    Check answer 3 marks
    1. the magnetic force is always perpendicular to the velocity
    2. so no work is done on the electron and its kinetic energy is unchanged
    3. a constant-magnitude force perpendicular to the motion changes only the direction of the velocity, giving a circular path

  3. cDetermine The electron crosses the boundary at 2.4 × 10⁷ m s⁻¹ into a field of flux density 8.5 mT. Determine the radius of its path, and hence the least distance the field region must extend beyond the boundary if the electron is to leave the region travelling at right angles to its original direction.

    demanding3 marks

    Check answer 3 marks
    1. r = mv/(eB) = (9.11 × 10⁻³¹ × 2.4 × 10⁷) / (1.60 × 10⁻¹⁹ × 8.5 × 10⁻³)
    2. r = 1.6 × 10⁻² m
    3. after turning through 90° the electron has advanced one radius along its original direction, so the region must extend at least 1.6 cm

  4. dDeduce A proton crosses the boundary at the same point and with the same velocity as the electron. Deduce how its path differs from that of the electron.

    top of the paper3 marks

    Check answer 3 marks
    1. the proton is positive, so the magnetic force acts in the opposite direction and the path curves upwards instead of downwards
    2. r = mv/(qB) and the charge magnitudes are equal, so the radius is greater in the ratio of the masses, about 1.8 × 10³
    3. r ≈ 29 m, so within the region drawn the proton's path is almost straight

Transfer challenge

A beam of protons travelling at 3.0 × 10⁵ m s⁻¹ passes undeflected through a region containing a uniform magnetic field of flux density 0.12 T perpendicular to the beam together with a uniform electric field. Determine the electric field strength, and state its direction relative to the magnetic force on the protons.

Check answer 3 marks
  1. for no deflection the electric force balances the magnetic force: qE = qvB
  2. E = 3.0 × 10⁵ × 0.12 = 3.6 × 10⁴ V m⁻¹
  3. the electric force must oppose the magnetic force, so E acts in the direction opposite to that magnetic force
02Fig. 10.1Electric fields · CapacitanceA Level
Two parallel metal plates in a vacuum, connected to a 2.0 kV supply2.0 kV+−vacuum5.0 mm

Figure comment

Fig. 10.1Two long horizontal metal plates are drawn one directly above the other and connected by wires at their left-hand ends to a battery labelled 2.0 kV; the upper plate carries a plus sign and the lower plate a minus sign. The space between the plates is empty and labelled vacuum, and a dimension at the right-hand end gives the separation of the plates as 5.0 mm. No field lines are drawn between the plates.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

18 Electric fields19 Capacitance20 Magnetic fieldsOpen A Level question · A Level extension Q10
Guided questions 5 parts

Reading cue. 5.0 mm is the gap, not a plate length, and no field lines are drawn — the plus and minus signs on the plates are the only thing in the figure that fixes the field direction.

  1. aState State the direction of the electric field in the gap in Fig. 10.1, and the direction of the force it exerts on an electron placed there.

    recall2 marks

    Check answer 2 marks
    1. the field points vertically downwards, from the positive upper plate to the negative lower plate
    2. the force on an electron is vertically upwards, towards the positive plate

  2. bCalculate Calculate the work done on an electron that moves from the lower plate to the upper plate, and the speed with which it arrives.

    routine3 marks

    Check answer 3 marks
    1. W = eV = 1.60 × 10⁻¹⁹ × 2.0 × 10³ = 3.2 × 10⁻¹⁶ J
    2. ½mv² = 3.2 × 10⁻¹⁶ J
    3. v = 2.7 × 10⁷ m s⁻¹

  3. cDetermine A charged dust particle of weight 1.28 × 10⁻¹³ N is held at rest midway between the plates. Determine the magnitude and sign of its charge, and the number of excess electrons it carries.

    demanding4 marks

    Check answer 4 marks
    1. E = V/d = 2.0 × 10³ / 5.0 × 10⁻³ = 4.0 × 10⁵ V m⁻¹
    2. for equilibrium qE = weight, so q = 1.28 × 10⁻¹³ / 4.0 × 10⁵ = 3.2 × 10⁻¹⁹ C
    3. the electric force must act upwards while the field points downwards, so the charge is negative
    4. 3.2 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2 excess electrons

  4. dDeduce The battery p.d. is suddenly reduced to 1.0 kV while the particle is still midway between the plates. Deduce the acceleration of the particle, and calculate the time it takes to reach a plate.

    top of the paper4 marks

    Check answer 4 marks
    1. new field = 2.0 × 10⁵ V m⁻¹, so the electric force = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N upwards
    2. resultant = 1.28 × 10⁻¹³ − 6.4 × 10⁻¹⁴ = 6.4 × 10⁻¹⁴ N downwards
    3. mass = 1.28 × 10⁻¹³ / 9.81 = 1.30 × 10⁻¹⁴ kg, so a = 4.9 m s⁻² downwards, that is g/2
    4. falling the 2.5 mm to the lower plate: t = √(2 × 2.5 × 10⁻³ / 4.9) = 3.2 × 10⁻² s

Transfer challenge

An electron travelling at 2.0 × 10⁷ m s⁻¹ enters midway between two parallel plates 5.0 cm long, moving parallel to them, in a uniform field of strength 1.2 × 10⁴ V m⁻¹. Calculate the sideways deflection of the electron as it leaves the plates.

Check answer 3 marks
  1. a = eE/m = (1.60 × 10⁻¹⁹ × 1.2 × 10⁴) / 9.11 × 10⁻³¹ = 2.1 × 10¹⁵ m s⁻²
  2. time between the plates t = 0.050 / 2.0 × 10⁷ = 2.5 × 10⁻⁹ s
  3. deflection = ½at² = 6.6 × 10⁻³ m