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IB DP Physics · Theme D Fields · D.4 (HL only) · ~6 min read

D.4 Induction

Magnetic flux and flux linkage, Faraday's law, the emf in a moving conductor, Lenz's law and conservation of energy, and the sinusoidal emf from a coil rotating in a magnetic field.

Magnetic fluxFaraday's lawε = BvLLenz's lawAC generatorsEffect of rotation speed
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • D.4 is HL only. Magnetic flux Φ = BA cos θ measures how much field passes through an area.
  • Faraday's law: a changing flux linkage induces an emf, ε = −NΔΦ/Δt. A rod of length L moving at v through a field B has ε = BvL.
  • Lenz's law: the induced current always flows to oppose the change that causes it. That minus sign is energy conservation.
  • A coil rotating in a uniform field gives a sinusoidal emf. Spinning it faster raises both the peak emf and the frequency.

By the end you can

  • Calculate magnetic flux and flux linkage. HL
  • Use Faraday's law to calculate induced emf. HL
  • Use ε = BvL for a conductor moving through a field. HL
  • Apply Lenz's law to find the direction of induced currents, and explain it with energy. HL
  • Describe and explain the emf from a rotating coil, and the effect of changing the rotation frequency. HL

1Magnetic flux HL

normalθBΦ = BA cos θθ: between B andthe normal to A
Only the component of B along the normal passes through the loop.
Φ = BA cos θ DATA BOOKLET
flux in webers (Wb = T m2); θ between B and the normal to the area
  • Flux linkage = NΦ for a coil of N turns.
Worked example

A coil of area 0.050 m2 sits in a 0.40 T field, its normal at 60° to the field. Find the flux.

  1. Φ = BA cos θ = 0.40 × 0.050 × cos 60° M1
  2. = 0.010 Wb A1
Common mistakeUsing the angle between B and the plane of the coil. In Φ = BA cos θ, θ is measured from the normal. Flux is greatest when the field is perpendicular to the plane of the coil.

2Faraday's law HL

ε = −NΔΦΔt DATA BOOKLET
induced emf equals the rate of change of flux linkage
  • Ways to change the flux: move a magnet or coil, change the current in a nearby coil, change the area, or rotate the coil.
Worked example

A 200-turn coil's flux falls from 0.010 Wb to zero in 0.20 s. Find the average emf.

  1. ε = NΔΦ/Δt = 200 × 0.010/0.20 M1
  2. = 10 V A1

A conductor moving through a field

RvLε = BvLI opposes themotion (Lenz)
As the rod moves, the area of the circuit grows, so the flux through it grows.
ε = BvL DATA BOOKLET
emf across a rod moving perpendicular to B
Worked example

A 0.50 m rod slides at 4.0 m s−1 along rails in a 0.20 T field, completing a circuit of resistance 2.0 Ω. Find the emf, the current, the force needed to keep it moving and the power.

  1. ε = BvL = 0.20 × 4.0 × 0.50 = 0.40 V A1
  2. I = ε/R = 0.20 A A1
  3. F = BIL = 0.20 × 0.20 × 0.50 = 0.020 N M1
  4. P = Fv = 0.080 W = I2R ✓: mechanical work becomes electrical energy R1

3Lenz's law HL

Lenz's law — the direction of an induced current is such that its magnetic effect opposes the change producing it.

SNmagnet incoil face nearest the magnet becomes N:it repels the approaching N poleinduced current
Push a north pole in and the coil makes a north pole to push back. Pull it out and the coil makes a south pole to pull it back.
  • Why it must be so: if the induced current helped the change, the magnet would speed up by itself and energy would be created from nothing. Lenz's law is conservation of energy: you must do work to induce a current.
  • Eddy currents in solid metal also oppose motion: used in induction braking and cooking; reduced in transformers by laminating the core.
Exam tipFor direction questions, say what change happens (flux increasing or decreasing), what the induced field must do (oppose it), then use the right-hand grip rule to get the current.

4The AC generator HL

  • A coil of N turns and area A rotating at angular speed ω in a field B has flux linkage NBA cos ωt. The emf (its rate of change) is sinusoidal: maximum when the plane of the coil is parallel to B (flux zero but changing fastest), zero when it is perpendicular.
tεBlue: coil at f. Red dashed: rotated at 2f. Peak emf AND frequency both double.
Doubling the rotation frequency halves the period and doubles the peak emf.
  • Peak emf ε0 = NBAω. Increase it with more turns, a stronger field, a bigger coil or faster rotation.
Worked example

A generator's peak emf is 12 V at 50 Hz. What is its peak emf and frequency at 100 Hz?

  1. ε0 ∝ ω, so peak emf = 24 V A1
  2. frequency = 100 Hz A1
Common mistakeSaying the emf is greatest when the flux is greatest. emf depends on the rate of change of flux, which is greatest when the flux itself passes through zero.
Linking questionInduction is how almost all electricity is generated: in power stations (fission E.4, fossil fuels A.3) a turbine turns a generator. The B.5 ideas of emf and internal resistance apply to its output.
Idea for your IA / explorationDrop a magnet through a coil connected to a data logger and investigate how the peak emf depends on the drop height (speed), or on the number of turns.

Test yourself — cover the answers first

Quick check

  1. What is the unit of magnetic flux?
  2. State Faraday's law.
  3. Which conservation law does Lenz's law follow from?
  4. A magnet is held still inside a coil. Is there an emf?
  5. When is the emf of a rotating coil greatest?
  6. A rod moves twice as fast through the same field. What happens to ε?

Exam-style

  1. A square coil of 50 turns, side 4.0 cm, is pulled out of a 0.30 T field in 0.10 s. Find the average emf.
  2. An aircraft with a 40 m wingspan flies horizontally at 250 m s−1 where the vertical component of the Earth's field is 4.0 × 10−5 T. Find the emf between the wingtips.
Show answers
  1. weber (Wb = T m2)
  2. the induced emf equals the rate of change of flux linkage
  3. conservation of energy
  4. no: the flux is not changing
  5. when the plane of the coil is parallel to the field
  6. it doubles
  7. ΔΦ = BA = 0.30 × 0.0402 = 4.8 × 10−4 Wb
    ε = NΔΦ/Δt = 50 × 4.8 × 10−4/0.10
    = 0.24 V
  8. ε = BvL = 4.0 × 10−5 × 250 × 40
    = 0.40 V
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.