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A Level Physics 9702 · A Level · Topic 20 · ~6 min read

Magnetic fields

Magnetic field patterns, the force on a current-carrying conductor F = BIL sin θ, forces on moving charges F = BQv, circular paths, velocity selection and the Hall effect, forces between wires, and electromagnetic induction with Faraday’s and Lenz’s laws.

Field patternsF = BIL sin θTeslaF = BQvCircular orbits r = mv/BQVelocity selectorHall voltageForces between wires
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Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Force on a wire: F = BIL sin θ, with direction from Fleming's left-hand rule. This defines the tesla: 1 T gives 1 N on 1 m carrying 1 A at right angles.
  • Force on a moving charge: F = BQv sin θ, always perpendicular to the velocity, so a charge moving across a uniform field moves in a circle: r = mv/(BQ).
  • Crossed fields: a charge passes undeflected when qE = Bqv, so v = E/B (velocity selector). The same balance produces the Hall voltage V_H = BI/(ntq).
  • Induction: e.m.f. = rate of change of flux linkage, NΔΦ/Δt (Faraday), in a direction that opposes the change (Lenz).

By the end you can

  • Sketch field patterns for wires, flat coils, solenoids and bar magnets.
  • Use F = BIL sin θ and define magnetic flux density.
  • Use F = BQv sin θ and analyse charged particles in magnetic fields, including circular paths, velocity selection and the Hall effect.
  • Explain forces between current-carrying wires.
  • Define magnetic flux and flux linkage, and use Faraday's and Lenz's laws.

1Forces on currents and charges

F = BIL sin θ

Magnetic flux density — the force per unit current per unit length on a straight conductor placed at right angles to the field: B = F/(IL).

NSI outFF = BIL sin θF ⟂ to both B and I(left-hand rule:First finger Field,seCond finger Current,thuMb Motion)
Fleming's left-hand rule: First finger = Field, seCond finger = Current, thuMb = Motion (force).
Worked example

A 0.25 m wire carries 3.0 A at right angles to a 0.040 T field. Find the force.

  1. F = BIL = 0.040 × 3.0 × 0.25 = 0.030 N A1
  • Two parallel wires: each sits in the other's field. Currents in the same direction attract; opposite directions repel.

Moving charges

−vFB into page (×)electron: F = qvBalways ⟂ to v,so circular motionqvB = mv²/rr = mv/(qB)
The magnetic force is always perpendicular to v, so it provides a centripetal force.
F = BQv ; r = mv/(BQ)
the speed doesn't change, because the force does no work
Worked example

A proton moves at 3.0 × 10⁶ m s⁻¹ perpendicular to a 0.20 T field. Find the radius of its path.

  1. BQv = mv²/r ⇒ r = mv/(BQ) C1
  2. r = 1.67 × 10⁻²⁷ × 3.0 × 10⁶/(0.20 × 1.60 × 10⁻¹⁹) = 0.16 m A1

Velocity selection and the Hall effect

+−××××××××××××××××××v+qF_E = qEF_B = Bqvundeflected ifqE = Bqv⇒ v = E/B× = magnetic field into the page; electric field from + plate to − plate.Only particles with speed E/B pass straight through.
Crossed E and B fields: only particles with v = E/B pass straight through.
Worked example

In a velocity selector E = 4.0 × 10⁴ V m⁻¹ and B = 0.050 T. Find the speed of undeflected ions.

  1. v = E/B = 8.0 × 10⁵ m s⁻¹ A1 (it doesn't depend on the charge or mass)
IcurrentB (downwards,through the slab)electrons pushed to the back face+ + + + +− − − −V_H between the front and back facesV_H = BI/(ntq)thin slab andsmall n ⇒ big V_HCharge builds up on one side until the electric force balances the magnetic force.
Charge carriers are pushed sideways until the Hall field balances the magnetic force.
V_H = BI/(ntq) DATA & FORMULAE
n = number density of carriers, t = thickness in the direction of B
Worked example

A semiconductor slab 0.20 mm thick with n = 5.0 × 10²⁴ m⁻³ carries 0.10 A in a 0.50 T field. Find V_H.

  1. V_H = 0.50 × 0.10/(5.0 × 10²⁴ × 2.0 × 10⁻⁴ × 1.6 × 10⁻¹⁹) C1
  2. V_H = 3.1 × 10⁻⁴ V A1 (a copper slab would give only about 10⁻⁸ V, because n is far larger)
Common mistakeUsing the width instead of the thickness t in V_H. t is measured parallel to B.

2Electromagnetic induction

Flux and Faraday's law

Magnetic flux — the product of the flux density and the area perpendicular to the field: Φ = BA.

Flux linkage — NΦ for a coil of N turns.

normalθBΦ = BA cos θθ: between B andthe normal to A
Flux is the field through an area; tilting the coil reduces it.
Φ = BA ; E = −Δ(NΦ)/Δt
Faraday's law, with the minus sign showing Lenz's law
SNmagnet incoil face nearest the magnet becomes N:it repels the approaching N poleinduced current
The induced current makes a field that opposes the change in flux.
Worked example

A coil of 200 turns and area 4.0 × 10⁻⁴ m² is in a field of 0.30 T, which falls to zero in 0.050 s. Find the average induced e.m.f.

  1. Δ(NΦ) = 200 × 0.30 × 4.0 × 10⁻⁴ = 0.024 Wb C1
  2. E = 0.024/0.050 = 0.48 V A1
  • Lenz's law is conservation of energy: if the induced current helped the change, you'd get energy from nothing.
Common mistakeSaying the field "cuts" the coil without describing a change of flux linkage, or omitting "rate of change" in Faraday's law. The 2024 Paper 4 report flagged both.
Linking questionCircular motion is in Motion in a circle; electric fields are in Electric fields; and a.c. from rotating coils is in Alternating currents.

Test yourself — cover the answers first

Quick check

  1. State the unit of magnetic flux density.
  2. What is the force on a charge moving parallel to a magnetic field?
  3. Why doesn't a magnetic field change the speed of a charge?
  4. Two parallel wires carry currents in the same direction. Do they attract or repel?
  5. State Lenz's law.

Exam-style

  1. An electron (m = 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁷ m s⁻¹ in a circle of radius 0.12 m. Find B.
  2. Explain why the Hall voltage is larger in a semiconductor than in a metal.
  3. A 50-turn coil of area 2.0 × 10⁻³ m² is turned from perpendicular to parallel to a 0.12 T field in 0.10 s. Find the average e.m.f.
Show answers
  1. tesla (T)
  2. zero
  3. the force is perpendicular to the velocity, so it does no work
  4. attract
  5. the induced e.m.f. acts to oppose the change producing it
  6. B = mv/(er) = 9.5 × 10⁻⁴ T
  7. n is much smaller, so the carriers move faster for the same current, and V_H = BI/(ntq) is larger
  8. Δ(NΦ) = 50 × 0.12 × 2.0 × 10⁻³ = 0.012 Wb ⇒ 0.12 V
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.