A rotating coil produces sinusoidal flux linkage

Magnetic Flux & Flux Linkage — AQA A-Level Physics

When a coil rotates at constant angular speed ω in a uniform magnetic field, the angle θ changes linearly with time: $\theta = \omega t$.
Substituting into the 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). equation:

N\Phi = BAN \cos(\omega t)

$NΦ$: 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). at time t (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 coil (m²)
$N$: number of turns
$ω$: angular speed of rotation (rad s⁻¹)
$t$: time (s)

This is a cosine wave. The 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). oscillates between +BAN and −BAN as the coil rotates.
At t = 0 (θ = 0°): the normal is aligned with the field. Flux linkage is maximum: $N\Phi = BAN$.
At θ = 90° (quarter turn): the normal is perpendicular to the field. Flux linkage is zero.
At θ = 180° (half turn): the normal points opposite to the field. Flux linkage is −BAN (negative maximum).
The coil completes one full oscillation for each complete rotation.

Flux linkage and EMF are 90° out of phase

By 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., the induced EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V). equals the negative rate of change of flux linkage. Mathematically, EMF is the negative derivative of NΦ with respect to time.
The derivative of cos(ωt) is −ω sin(ωt). So if $N\Phi = BAN \cos(\omega t)$, then $EMF = BAN\omega \sin(\omega t)$.
When flux linkage is at a maximum (cos = 1), its rate of change is zero, so the EMF is zero.
When flux linkage is zero (cos = 0), its rate of change is at a maximum, so the EMF is at its peak.
The EMF leads the flux linkage by 90° (or equivalently, the flux linkage lags the EMF by 90°).
This is the single most important relationship for understanding AC generators.

Flux linkage and EMF vs angle for a rotating coil

Two graphs stacked vertically sharing the same θ axis (0° to 360°). Top graph: NΦ = BAN cos θ, a cosine curve starting at maximum. Bottom graph: EMF = BANω sin θ, a sine curve starting at zero. Vertical dashed lines at 90° and 270° show that when NΦ = 0, EMF is at maximum, and vice versa.

Worked Example

A coil rotating in a uniform field has a maximum flux linkage of 0.48 Wb turns. At a certain instant, the flux linkage is 0.24 Wb turns. Calculate the angle between the magnetic field and the normal to the coil.

Show Solution

List known values

Maximum flux linkage: $N\Phi_{\text{max}} = BAN = 0.48 \text{ Wb turns}$
Flux linkage at the instant: $N\Phi = 0.24 \text{ Wb turns}$

Write the flux linkage equation

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

Substituting:

$$0.24 = 0.48 \cos \theta$$

Solve for cos θ

$$\cos \theta = \frac{0.24}{0.48} = 0.50$$

Find θ

$$\theta = \cos^{-1}(0.50) = 60°$$

Answer

$\theta = 60°$

Common Mistake MEDIUM

Students often: Thinking that when the coil face is perpendicular to the field (maximum flux), the EMF is also at maximum.
Instead: It's the opposite. Maximum flux linkage means zero rate of change, so zero EMF. Maximum EMF occurs when the flux linkage is zero (coil face parallel to field, maximum rate of change). They're 90° out of phase.