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AP Physics 2 · guided topic map

Magnetism for AP Physics 2

Magnetism for AP Physics 2, organized into 1 syllabus topic and 7 mapped concept guides.

Syllabus topics
1
Mapped concept guides
7
Educational level
AP Physics 2: Algebra-Based

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

12

Magnetism and Electromagnetism

AP Physics 2: Algebra-Based

7 guides
  1. 01Magnets and magnetic fields12–15%
  2. 02Force on moving charges12–15%
  3. 03Force on currents12–15%
  4. 04Magnetic fields made by currents12–15%
  5. 05Magnetic flux12–15%
  6. 06Faraday's and Lenz's laws12–15%
  7. 07Generators and transformers12–15%

Diagrams

Magnetism as AP Physics 2 draws it

The figures from the AP Physics 2 practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Figure 1Magnetic Fields and ElectromagnetismAP
Straight current-carrying wire in a uniform magnetic field directed into the pageuniform magnetic field B = 0.40 T, into the pageI = 3.0 AL = 0.25 m

Figure comment

Figure 1A rectangular region of uniform magnetic field is shown by a grid of small crosses, marking a field of magnitude 0.40 T directed into the page. A straight wire lies horizontally across the region, in the plane of the page and therefore at right angles to the field, and an arrow drawn along the wire shows the conventional current of 3.0 A flowing to the right. A dimension line below the region, with projection lines dropped from each end of the wire, marks the length of wire lying in the field as L = 0.25 m. No force is shown.

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

Guided questions 5 parts

Reading cue. The crosses mean B points into the page, away from you, and the 0.25 m dimension line spans only the part of the wire inside the field region — that is the length that counts.

  1. aDetermine Determine the direction of the magnetic force acting on the wire.

    recall2 marks

    Check answer 2 marks
    1. applies a right-hand rule with the current directed to the right and the field directed into the page
    2. the force is directed up the page, at right angles to both the current and the field
  2. bDetermine The current in the wire is reversed and at the same instant the field is increased to 0.80 T. Determine the new magnitude and direction of the force on the wire.

    routine3 marks

    Check answer 3 marks
    1. F = BIL with L unchanged, so F = 0.80 × 3.0 × 0.25
    2. F = 0.60 N, double the original magnitude
    3. reversing the current reverses the force, so it now points down the page
  3. cExplain The wire is turned to a new direction while still lying in the plane of the page, with the same 0.25 m of it inside the field region. Explain why the magnitude of the force is unchanged, and state the one change of orientation that would reduce that force to zero.

    demanding3 marks

    Check answer 3 marks
    1. the crosses show that B is perpendicular to the page, so any wire lying in the page makes an angle of 90° with the field and sin θ = 1
    2. B, I and the length inside the field are all unchanged, so the magnitude BIL is identical for every direction that can be drawn in the page; only the direction of the force turns as the wire turns
    3. the force falls to zero only if the wire is turned out of the page to lie along the field itself, pointing straight into the page, where sin θ = 0
  4. dExplain A student says that because the force on the wire is 0.30 N, the magnetic field must do 0.30 J of work on the wire for every metre the wire is pushed sideways. Explain why the magnetic field in fact does no work on the charges in the wire, and indicate what does supply the energy.

    top of the paper4 marks

    Check answer 4 marks
    1. the magnetic force on any charge is qv × B, always perpendicular to that charge's own velocity, so it does no work on the charge
    2. once the wire moves sideways across the field there is a motional emf in it opposing the original current, by Faraday's law and Lenz's law
    3. the source driving the 3.0 A must therefore do extra work against this back-emf to hold the current steady
    4. the mechanical energy gained by the wire comes from that source, with the field acting only as the intermediary that redirects it

Transfer challenge

Two long straight parallel wires 4.0 cm apart each carry a current of 3.0 A in the same direction, with no external field present. Determine the force per unit length that each exerts on the other, and state whether they attract or repel.

Check answer 4 marks
  1. each wire sits in the field of the other: B = μ₀I/(2πd) = (2 × 10⁻⁷ × 3.0)/0.040 = 1.5 × 10⁻⁵ T
  2. force per unit length = BI = 1.5 × 10⁻⁵ × 3.0
  3. F/L = 4.5 × 10⁻⁵ N per metre
  4. currents in the same direction attract, so each wire is pulled towards the other