IB — physicsHard2 min read

Magnetic Fields and Forces

F = BIL; F = qvB; magnetic fields around wires and solenoids; motor effect

Tutor AI Team
Tutor AI Team

# Magnetic Fields and Forces — IB Physics

1. Magnetic Fields

Magnetic flux density (BB): measured in tesla (T).

Field patterns:

  • Long straight wire: Concentric circles (right-hand grip rule)
  • Solenoid: Uniform inside, like a bar magnet outside

2. Force on a Current-Carrying Wire

F=BILsin⁡θ\boxed{F = BIL\sin\theta}

Maximum when wire ⊥ field (θ=90°\theta = 90°). Zero when parallel.

Direction: Fleming's Left-Hand Rule (First finger: Field, seCond: Current, thuMb: Force).

3. Force on a Moving Charge

F=qvBsin⁡θ\boxed{F = qvB\sin\theta}

The force is always perpendicular to both vv and BB → circular motion: r=mvqBr = \frac{mv}{qB}

4. Applications

  • DC motor: current in coil + magnetic field → rotation
  • Mass spectrometer: r=mv/(qB)r = mv/(qB) → separates ions by mass
  • Particle accelerators: magnetic fields bend charged particles into circles

Worked Example: Example 1

Problem

A 40 cm wire carries 5A perpendicular to a 0.3T field. F=BIL=0.3×5×0.4=0.6F = BIL = 0.3 \times 5 \times 0.4 = 0.6 N

Solution

Worked Example: Circular Motion

Problem

A proton (m=1.67×10−27m = 1.67 \times 10^{-27} kg) at 2×1072 \times 10^7 m/s enters a 0.5T field. r=mv/(qB)=1.67×10−27×2×107/(1.6×10−19×0.5)=0.418r = mv/(qB) = 1.67 \times 10^{-27} \times 2 \times 10^7/(1.6 \times 10^{-19} \times 0.5) = 0.418 m

Solution

6. Practice Questions

    1. State Fleming's Left-Hand Rule. (2 marks)
    1. An electron moves at 3×1063 \times 10^6 m/s perpendicular to a 0.02T field. Find the force and radius. (4 marks)
    1. Explain how a mass spectrometer separates isotopes. (3 marks)

    Answers

    1. First finger: field direction (N→S). Second finger: current direction (conventional). Thumb: force/motion direction.
    1. F=qvB=1.6×10−19×3×106×0.02=9.6×10−15F = qvB = 1.6 \times 10^{-19} \times 3 \times 10^6 \times 0.02 = 9.6 \times 10^{-15} N. r=mv/(qB)=9.11×10−31×3×106/(1.6×10−19×0.02)=8.54×10−4r = mv/(qB) = 9.11 \times 10^{-31} \times 3 \times 10^6/(1.6 \times 10^{-19} \times 0.02) = 8.54 \times 10^{-4} m.

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Summary

  • F=BILsin⁡θF = BIL\sin\theta (wire); F=qvBsin⁡θF = qvB\sin\theta (charge)
  • Fleming's LHR for direction
  • Circular motion: r=mv/(qB)r = mv/(qB)
Tags:
#ib#physics#electricity#magnetic-field#motor-effect#lorentz-force

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