Flux linkage multiplies flux by the number of turns

Magnetic Flux & Flux Linkage — AQA A-Level Physics

Key Definition

Flux linkage (NΦ) — The product of the number of turns of a coil and the magnetic flux through each turn. Unit: weber turns (Wb turns), though dimensionally equivalent to Wb.

N\Phi = BAN \cos \theta

$NΦ$: flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area.-turns (Wb turns). (Wb turns)
$B$: magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³.The strength of a magnetic field. The force per unit length per unit currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). on a currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).-carrying conductor perpendicular to the field. Measured in teslaThe SI unit of magnetic flux density. One tesla is the flux density when a force of 1 N acts on a 1 m conductor carrying 1 A perpendicular to the field. (T). (T)
$A$: area of each turn (m²)
$N$: number of turns on the coil
$θ$: angle between field and normal to the coil (°)

Why does N matter? Each turn of the coil is threaded by the same flux Φ. When the flux changes, each turn contributes independently to the induced EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V).. More $turns = more$ total EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V).. Flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area.-turns (Wb turns). captures this.
A coil with 200 turns in a flux of 5 mWb has a flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area.-turns (Wb turns). of 200 × 5 $mWb = 1.0 Wb turns. The flux through each turn$ is still 5 mWb, but the total flux linkage is 200 times larger.
Flux linkage is what actually appears in Faraday's lawThe magnitude of the induced EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V). is proportional to the rate of change of magnetic flux linkage.: EMF = −N(ΔΦ/Δt) = −Δ(NΦ)/Δt. It's the rate of change of flux linkage that determines the induced EMF.
The unit 'Wb turns' is sometimes written as just 'Wb' since 'turns' is dimensionless. Both are acceptable in exams.

Worked Example

A coil with 500 turns and cross-sectional area 1.2 × 10⁻³ m² is placed in a uniform magnetic field of 0.34 T, with the plane of the coil perpendicular to the field. Calculate the flux linkage.

Show Solution

List known values

Number of turns: $N = 500$
Area: $A = 1.2 \times 10^{-3} \text{ m}^2$
Flux densityMass per unit volume of a material. Measured in kg m⁻³.: $B = 0.34 \text{ T}$
Coil plane is perpendicular to field, so the normal is parallel to the field: $\theta = 0°$

Select the equation

$$N\Phi = BAN \cos \theta$$

Since $\theta = 0°$ and $\cos 0° = 1$:

$$N\Phi = BAN$$

Substitute values

$$N\Phi = 0.34 \times 1.2 \times 10^{-3} \times 500$$

Evaluate

$$N\Phi = 0.34 \times 1.2 \times 10^{-3} \times 500$$ $$= 0.204 \text{ Wb turns}$$ $$= 0.20 \text{ Wb turns (2 s.f.)}$$

Answer

$N\Phi = 0.20$ Wb turns

Common Mistake MEDIUM

Students often: Forgetting to include N when calculating flux linkage, and just giving $\Phi = BA \cos \theta$ as the answer.
Instead: The question will usually say 'flux linkage' or 'NΦ' — these always need the number of turns multiplied in. If it just says 'flux', then $\Phi = BA \cos \theta$ without N is correct.