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Next exam on November 14 – 15, 2026:
All concepts

Magnetic Field, Ampère Force and Lorentz Force

磁场、安培力与洛伦兹力

The defining equation and vector superposition of magnetic flux density, the magnitude and direction of the Ampère force, the Lorentz force and its direction, and charged-particle motion in a magnetic field.

13 real-exam questions · 4 sittings22 in mock papers26 practice

Formulas

  • Definition of magnetic flux density

    Definition
    B=FILB=\frac{F}{IL}

    Applies when：A ratio definition requiring the wire to be perpendicular to the field; B is set by the field itself

  • Ampère force

    Formula
    F=BILsin⁡θF=BIL\sin\theta

    Applies when：A straight current-carrying wire in a uniform field; θ is the angle between wire and field, zero when parallel, direction by the left-hand rule

  • Lorentz force

    Formula
    F=qvBsin⁡θF=qvB\sin\theta

    Applies when：θ is the angle between velocity and field, zero when v∥B; the force is always perpendicular to the velocity, so it does no work on the particle

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Physics December 2025 Past Paper
    A straight conductor 0.5 m long is placed perpendicular to a uniform magnetic field. When a current of 10 A flows through it, the force exerted on the conductor is 1.5 N . The magnitude of the magnetic field's magnetic induction is ( )
    • A．0.3 T
    • B．0.2 T
    • C．0.1 T
    • D．0.4 T
    Show answer & solution

    Answer: A

    1 Magnetic force on a straight conductor: F=BIL(sin⁡90∘=1)\mathrm { F } = \mathrm { BIL } \left( \sin 90 ^ { \circ } = 1 \right). 2  B=F/(IL)=1.5/(10×0.5)=0.3 T\mathrm {~B} = \mathrm { F } / ( \mathrm { IL } ) = 1.5 / \left( 10 \times 0.5 \right) = 0.3 \mathrm {~T}.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    In a uniform magnetic field, a 0.2 m wire perpendicular to the field carries 4 A．The Ampère force is 0.8 N . The magnetic flux density is
    • A．0.1 T
    • B．0.5 T
    • C．1.0 T
    • D．2.0 T
    Show answer & solution

    Answer: C

    Ampère force (perpendicular): F=BILF = B I L. So 0.8=B×4×0.2,B=1.0 T0.8 = B \times 4 \times 0.2 , B = 1.0 \mathrm {~T}. If not perpendicular, sin⁡θ\sin \theta is needed. Answer: C.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    A 0.5 m wire is placed in a magnetic field B=0.4 TB = 0.4 \mathrm {~T}. The maximum magnetic force on the wire is 0.2 N . The current in the wire is
    • A．2 A
    • B．0.5 A
    • C．1 A
    • D．4 A
    Show answer & solution

    Answer: C

    The maximum magnetic force occurs when the wire is perpendicular to BB : F=BILF = B I L . I=FBL=0.20.4×0.5=1 AI = \frac { F } { B L } = \frac { 0.2 } { 0.4 \times 0.5 } = 1 \mathrm {~A}.
    Take the full CSCA Physics April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

13 questions across 4 sittings1 of 4 sittings available to verify free online

22 questions across 6 mock papers — our own papers, so a coverage signal rather than real-exam evidence

26 related practice questions availablePractise free: Magnetic Fields

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Glossary

electric current

Related concepts

Electromagnetic Induction