Magnetic flux, flux linkage, Faraday's law and electromagnetic induction
AQA A-Level Physics 7
🌀Define magnetic flux: Φ = BA cos θ
🔗Calculate flux linkage: NΦ = NBA cos θ
⚡State and apply Faraday's law: ε = −ΔNΦ/Δt
🔄Apply Lenz's law to determine the direction of induced EMF
📐Calculate induced EMF from a conductor moving in a field
🔬Investigate electromagnetic induction
Magnetic Flux
Magnetic flux (Φ): A measure of the total magnetic field passing through a given area. It depends on the field strength, the area, and the angle between the field and the normal to the area.
Φ = BA cos θ
Φ = magnetic flux (webers, Wb = T m²)
B = magnetic flux density (tesla, T)
A = area of the surface (m²)
θ = angle between B and the normal to the surface
Key cases:
θ = 0° (field perpendicular to surface, parallel to normal): Φ = BA (maximum flux)
θ = 90° (field parallel to surface, perpendicular to normal): Φ = 0 (no flux)
Magnetic flux represents "how much" field is threaded through an area — like counting field lines passing through. Flux is maximum when the field is perpendicular to the plane of the coil.
Flux Linkage
Flux linkage (NΦ): The total flux threaded through all turns of a coil. For a coil of N turns each with flux Φ:
NΦ = NBA cos θ
NΦ = flux linkage (weber-turns, Wb or Wb turns)
N = number of turns in the coil
B = magnetic flux density (T)
A = area of each turn (m²)
θ = angle between B and normal to the coil
Flux linkage is proportional to N because each turn of the coil "links" with the same flux — the effects add. A coil with 100 turns in a given field has 100 times the flux linkage of a single-turn loop.
Flux Φ refers to a single turn. Flux linkage NΦ refers to the whole coil. When applying Faraday's law, use flux linkage (not just flux) for a coil with N turns.
Electromagnetic Induction
Electromagnetic induction: The production of an EMF (and hence a current in a closed circuit) whenever the flux linkage through a circuit changes.
An EMF is induced whenever there is a change in flux linkage. This can happen by:
Moving a conductor in a magnetic field
Moving a magnet relative to a coil
Changing the current (and hence field) in a nearby coil
Rotating a coil in a magnetic field
Induced EMF for a straight conductor moving in a field:
ε = Bvl
B = flux density (T), v = velocity (m s⁻¹), l = length of conductor (m)
(Applies when v is perpendicular to both B and l)
No change in flux linkage = no induced EMF. A conductor stationary in a field, or moving parallel to the field lines, has no EMF induced.
Faraday's Law and Lenz's Law
Faraday's Law: The induced EMF is equal to the rate of change of flux linkage.
ε = −ΔNΦ/Δt (or −dNΦ/dt)
Magnitude: |ε| = ΔNΦ/Δt
Lenz's Law: The direction of the induced current is such that it opposes the change in flux that caused it. (This is the physical meaning of the negative sign in Faraday's law.)
Lenz's law is a consequence of energy conservation: if the induced current helped the change rather than opposing it, we'd get a self-amplifying system violating energy conservation. The induced current always acts to reduce the change — slowing a moving magnet, or creating a field that opposes an increasing external flux.
To find the direction of the induced current: use the right-hand rule. Curl the fingers of the right hand in the direction of the induced current — the thumb points in the direction of the magnetic field produced by that current. Lenz's law says this field must oppose the change in the external flux.
A circular coil of 50 turns, area 0.04 m², is placed in a uniform magnetic field of 0.30 T with the plane of the coil at 30° to the field (i.e. θ = 60° to the normal). Calculate the flux linkage.
1θ = angle between B and normal to coil = 60°
2Φ = BA cos θ = 0.30 × 0.04 × cos 60° = 0.012 × 0.5 = 6.0 × 10⁻³ Wb
A coil of 200 turns and area 15 cm² is placed perpendicular to a magnetic field. The field changes from 0.40 T to 0.10 T in 0.050 s. Calculate the induced EMF.
2Change in flux linkage: ΔNΦ = N × ΔΦ = 200 × (−4.5 × 10⁻⁴) = −0.090 Wb
3|ε| = |ΔNΦ/Δt| = 0.090/0.050 = 1.80 V
Induced EMF = 1.80 V
A straight conductor of length 0.25 m moves at 3.5 m s⁻¹ perpendicular to a magnetic field of 0.80 T. Calculate the induced EMF.
1ε = Bvl = 0.80 × 3.5 × 0.25
2ε = 0.70 V
Induced EMF = 0.70 V
A bar magnet is pushed into a solenoid and the induced current is observed to flow anticlockwise (when viewed from the magnet end). (a) State which pole of the solenoid faces the magnet. (b) Explain how Lenz's law applies.
1Anticlockwise current (viewed from magnet end) → using right-hand rule, the magnetic field of the coil points towards the magnet → this face is a North pole
2Lenz's law: the induced North pole of the solenoid repels the incoming North pole of the magnet, opposing the increase in flux. This opposes the motion — consistent with energy conservation.
The solenoid presents a North pole to the magnet, repelling it and opposing the increase in flux (Lenz's law).
1. A coil of area 0.010 m² is placed perpendicular to a 0.50 T field. What is the magnetic flux through the coil?
Φ = BA cos θ = 0.50 × 0.010 × cos 0° = 0.50 × 0.010 = 5.0 × 10⁻³ Wb. (θ = 0° because field is perpendicular to the coil plane, parallel to normal.)
2. A coil in a magnetic field has zero flux linkage. What is the angle between the magnetic field and the normal to the coil?
Φ = BA cos θ = 0 when cos θ = 0 → θ = 90°. The field is parallel to the plane of the coil (perpendicular to the normal). No field lines pass through the coil.
3. According to Lenz's law, the direction of the induced current is such that:
Lenz's law (the physical meaning of the minus sign in Faraday's law): the induced current opposes the change that caused it — conservation of energy.
4. A conductor of length 0.5 m moves at 2 m s⁻¹ perpendicular to a 0.4 T field. What EMF is induced?
ε = Bvl = 0.4 × 2 × 0.5 = 0.4 V.
5. A 400-turn coil of area 20 cm² has its flux linkage change from 0.16 Wb to 0.04 Wb in 0.040 s. Calculate the magnitude of the induced EMF.
1. A rectangular loop of dimensions 0.15 m × 0.08 m rotates at 50 rev s⁻¹ in a uniform magnetic field of 0.35 T. (a) Calculate the maximum flux linkage if the coil has 80 turns. (b) Calculate the peak induced EMF.
(a) Area A = 0.15 × 0.08 = 0.012 m². Maximum flux linkage = NBA (when θ = 0°) = 80 × 0.35 × 0.012 = 0.336 Wb. (b) Angular frequency ω = 2πf = 2π × 50 = 314.2 rad s⁻¹. The flux linkage varies as NΦ = NBAcos(ωt). The induced EMF = −d(NΦ)/dt = NBAω sin(ωt). Peak EMF = NBAω = 0.336 × 314.2 = 105.6 V ≈ 106 V.
2. A metal rod of length 1.2 m slides along horizontal rails in a uniform vertical magnetic field of 0.25 T at 4.0 m s⁻¹. The rails are connected at one end by a resistor of 6.0 Ω. The rod has resistance 2.0 Ω. Calculate: (a) the induced EMF; (b) the current in the circuit; (c) the force needed to maintain constant velocity.
(a) ε = Bvl = 0.25 × 4.0 × 1.2 = 1.2 V. (b) Total resistance = R_rod + R_load = 2.0 + 6.0 = 8.0 Ω. I = ε/R_total = 1.2/8.0 = 0.15 A. (c) The current-carrying rod in the magnetic field experiences a force F = BIl opposing the motion (Lenz's law). F = BIl = 0.25 × 0.15 × 1.2 = 0.045 N. To maintain constant velocity, an external force of 0.045 N must be applied in the direction of motion. This equals the power check: P = Fv = 0.045 × 4.0 = 0.18 W = εI = 1.2 × 0.15 = 0.18 W ✓ (all mechanical power converted to electrical).
3. Explain why a falling magnet takes longer to pass through a copper tube than through a cardboard tube of the same dimensions. Include Lenz's law, induced currents and forces in your answer.
As the magnet falls through the copper tube, its changing flux induces eddy currents in the copper (which is a good electrical conductor). By Lenz's law, these induced currents create magnetic fields that oppose the change in flux — i.e. they oppose the motion of the magnet. The eddy current below the magnet creates a north pole (repelling the falling magnet's south end, slowing descent). The eddy current above the magnet creates a south pole (attracting the north end of the magnet, also slowing descent). These forces combined slow the magnet significantly — it reaches terminal velocity much earlier than free fall. In a cardboard tube, no conductive material is present, so no eddy currents form, and the magnet falls nearly freely. The energy of the slower descent is converted to heat in the copper via the eddy currents (I²R heating).
Objective: Investigate how the magnitude and direction of an induced EMF depend on the rate of change of flux linkage.
Equipment
Search coil (known N and A) and large solenoid with AC supply
Bar magnet
Galvanometer or millivoltmeter (centre-zero)
Datalogger with voltage probe (optional)
Method A — Bar Magnet and Coil
Connect the coil to the galvanometer. Push the north pole of the magnet into the coil and observe the deflection (direction and magnitude).
Hold the magnet stationary inside the coil. Observe that there is no deflection (no change in flux).
Withdraw the magnet. Observe that the deflection reverses.
Push the magnet in more rapidly — observe a larger deflection (greater rate of change of flux → greater EMF).
Repeat with the south pole — deflections are reversed in direction (Lenz's law).
Method B — Varying Current in Solenoid
Place the search coil inside the solenoid. Connect both to datalogger channels.
Apply a sinusoidal AC to the solenoid. Record the flux through the search coil and the induced EMF simultaneously.
Verify that the induced EMF is 90° out of phase with the flux (EMF = −dΦ/dt).
Safety
Keep magnets away from computer screens, credit cards, and electronic equipment. Use low-voltage AC supplies only. Ensure no trailing wires near rotating equipment.