Magnetic Flux and Flux Linkage

Magnetic flux is a measure of the total magnetic field passing through a given area. While magnetic flux density (B) tells you how strong the field is at a point, magnetic flux (Φ) tells you the total amount of field threading through a surface. This distinction is crucial for understanding electromagnetic induction, which relies on changes in flux.

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Magnetic Flux

Magnetic flux (Φ, the Greek letter phi) through a surface of area A in a uniform magnetic field B is:

Φ = BA cosθ

where:

• Φ is the magnetic flux (Wb — weber)
• B is the magnetic flux density (T)
• A is the area of the surface (m²)
• θ is the angle between the magnetic field direction and the normal (perpendicular) to the surface

Understanding the Angle θ

The angle θ is measured between the field lines and the normal to the surface — not between the field and the surface itself.

• When the field is perpendicular to the surface (i.e. along the normal), θ = 0° and cosθ = 1, giving Φ = BA — this is the maximum flux
• When the field is parallel to the surface, θ = 90° and cosθ = 0, giving Φ = 0 — no flux passes through the surface
• At intermediate angles, the flux is between zero and BA

Think of it like rain falling on a tilted tray. When the tray is horizontal (θ = 0°), it catches the maximum rain. When it is vertical (θ = 90°), no rain is caught.

Common Confusion: The Angle

Many students get confused about which angle to use. Here is a quick check:

Situation	θ (angle to normal)	cosθ	Flux
Field perpendicular to surface (maximum flux)	0°	1	BA
Field at 30° to normal	30°	0.866	0.866BA
Field at 45° to normal	45°	0.707	0.707BA
Field at 60° to normal	60°	0.500	0.500BA
Field parallel to surface (zero flux)	90°	0	0

Units

The unit of magnetic flux is the weber (Wb):

1 Wb = 1 T m² = 1 V s

Worked Example 1

A square coil of side 0.05 m is placed in a uniform magnetic field of 0.80 T. The normal to the coil makes an angle of 60° with the field. Calculate the flux through the coil.

A = 0.05 × 0.05 = 2.5 × 10⁻³ m²

Φ = BA cosθ = 0.80 × 2.5 × 10⁻³ × cos 60° = 0.80 × 2.5 × 10⁻³ × 0.50 = 1.0 × 10⁻³ Wb = 1.0 mWb

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Flux Linkage

For a coil with N turns, the flux linkage is defined as:

Flux linkage = NΦ = NBA cosθ

where N is the number of turns. The unit is also the weber (Wb), though sometimes written as Wb-turns for clarity.

Flux linkage is the total flux through all the turns. If a coil with 500 turns has a flux of 2.0 × 10⁻³ Wb through each turn, the flux linkage is:

NΦ = 500 × 2.0 × 10⁻³ = 1.0 Wb

This is the quantity that appears in Faraday’s law — it is the change in flux linkage that determines the induced EMF.

Worked Example 2

A coil of 150 turns and area 0.030 m² is in a magnetic field of 0.25 T with its face perpendicular to the field. Calculate the flux through one turn and the total flux linkage.

Flux through one turn: Φ = BA cos 0° = 0.25 × 0.030 × 1 = 7.5 × 10⁻³ Wb

Flux linkage: NΦ = 150 × 7.5 × 10⁻³ = 1.125 Wb ≈ 1.1 Wb

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Changing Flux

An EMF is induced in a coil whenever the flux linkage through it changes. The flux linkage NΦ = NBA cosθ can change if:

1. B changes — the magnetic field strength varies (e.g., moving a magnet towards or away from the coil)
2. A changes — the area of the coil within the field changes (e.g., pulling a coil out of a magnetic field)
3. θ changes — the coil rotates relative to the field direction (this is how generators work)
4. N changes — though this is less common in practice

The rate of change of flux linkage determines the magnitude of the induced EMF (Faraday’s law, covered in the next lesson).

Worked Example 3: Changing B

A coil of 100 turns and area 0.050 m² is in a field that decreases uniformly from 0.60 T to 0.20 T in 0.10 s. The coil face is perpendicular to the field. Calculate the change in flux linkage.

Initial flux linkage: NΦ₁ = 100 × 0.60 × 0.050 = 3.0 Wb

Final flux linkage: NΦ₂ = 100 × 0.20 × 0.050 = 1.0 Wb

Change in flux linkage: Δ(NΦ) = 3.0 − 1.0 = 2.0 Wb

The rate of change = 2.0 / 0.10 = 20 Wb s⁻¹ = 20 V (by Faraday’s law)

Worked Example 4: Changing A

A single rectangular loop of wire (0.10 m × 0.05 m) is being pulled out of a magnetic field of 0.40 T at a steady speed of 0.20 m s⁻¹. The 0.05 m side is perpendicular to the boundary of the field. How long does it take to fully exit the field, and what is the change in flux?

Time to exit: t = 0.10 / 0.20 = 0.50 s

Initial flux: Φ = BA = 0.40 × 0.10 × 0.05 = 2.0 × 10⁻³ Wb

Final flux: 0 Wb (completely outside the field)

Change in flux: 2.0 × 10⁻³ Wb over 0.50 s

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The Search Coil

A search coil is a small coil used to measure magnetic flux density. It works on the principle of electromagnetic induction:

1. The search coil (of known area A and number of turns N) is placed in the magnetic field with its face perpendicular to the field
2. The coil is quickly removed from the field (or the field is switched off), so the flux through it changes from NΦ = NBA to zero
3. The coil is connected to an oscilloscope or a ballistic galvanometer that measures the induced EMF or the total charge flow
4. From the measured EMF and the known values of N, A, and the rate of change, the flux density B can be calculated

Alternatively, the search coil can be rotated at a known frequency in a steady field. The peak EMF is related to B by:

ε₀ = NBAω

where ω is the angular frequency of rotation. This method gives a continuous sinusoidal output.

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Flux Through a Rotating Coil

When a coil rotates in a uniform magnetic field, the angle θ changes continuously. If the coil rotates at angular velocity ω:

θ = ωt

The flux linkage varies sinusoidally:

NΦ = NBA cos(ωt)

At time t = 0, the coil face is perpendicular to the field (θ = 0), and the flux linkage is at its maximum: NBA. As the coil rotates, the flux linkage decreases, reaches zero when the coil is parallel to the field (θ = 90°), then becomes negative, and so on.

This sinusoidal variation in flux linkage is the basis of alternating current (AC) generation.

Worked Example 5: Flux Linkage in a Rotating Coil

A coil of 200 turns and area 0.040 m² rotates at 50 revolutions per second in a uniform field of 0.15 T.

(a) What is the maximum flux linkage?

Maximum flux linkage = NBA = 200 × 0.15 × 0.040 = 1.2 Wb

This occurs when the coil face is perpendicular to the field (cos 0° = 1).

(b) What is the flux linkage at t = 0.005 s?

ω = 2π × 50 = 100π rad s⁻¹

NΦ = 1.2 × cos(100π × 0.005) = 1.2 × cos(π/2) = 1.2 × 0 = 0 Wb

At t = 0.005 s (one quarter of a revolution at 50 Hz), the flux linkage is zero — the coil is parallel to the field.

(c) At what rate is the flux linkage changing at this instant?

d(NΦ)/dt = −NBAω sin(ωt) = −1.2 × 100π × sin(π/2) = −1.2 × 314 × 1 = −377 V

The induced EMF at this instant is 377 V. This is the peak EMF — it occurs when the flux linkage is zero but changing most rapidly.

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Real-World Applications

Wireless charging: A transmitter coil creates an alternating magnetic field. A receiver coil in the phone experiences changing flux linkage, inducing an EMF that charges the battery. The efficiency depends on the alignment of the coils (affecting the flux linkage).

Metal detectors: A coil produces an alternating magnetic field. When a metal object enters the field, it changes the flux pattern and induces eddy currents in the metal, which in turn alter the impedance of the detector coil. The detector senses this change.

Electromagnetic flowmeters: Used in medicine and industry. A magnetic field is applied across a pipe carrying a conducting fluid (such as blood). The moving charges in the fluid experience a force (F = BQv), creating a measurable voltage across the pipe proportional to the flow rate.

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Common Exam Mistakes

• Using the wrong angle: θ is between the field and the normal to the surface, NOT between the field and the surface
• Confusing flux and flux linkage: Flux Φ = BA cosθ is for one turn. Flux linkage = NΦ = NBA cosθ is for N turns.
• Forgetting the cos term: If the coil is not perpendicular to the field, you need the cosθ factor
• Thinking flux linkage = 0 means no induction: When the flux linkage is zero but changing (e.g., the coil is parallel to the field while rotating), the induced EMF is actually at its maximum

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Key Points

• Magnetic flux: Φ = BA cosθ, measured in webers (Wb)
• θ is the angle between the field and the normal to the surface
• Flux linkage: NΦ = NBA cosθ for a coil of N turns
• A change in flux linkage (from changing B, A, θ, or N) induces an EMF
• A search coil measures B by detecting changes in flux linkage
• In a rotating coil, flux linkage varies as NΦ = NBA cos(ωt)
• Maximum rate of flux change (and hence maximum EMF) occurs when the flux linkage itself is zero

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A-Level Deep Dive: Magnetic Flux and Flux Linkage

Spec mapping

Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields covers the definition of magnetic flux density $B$B, magnetic flux $\Phi = BA\cos\theta$Φ=BAcosθ, and flux linkage $N\Phi$NΦ for a coil of $N$N turns; it also requires use of $\Phi = BA\cos(\omega t)$Φ=BAcos(ωt) for a rotating coil and links flux linkage to Faraday's and Lenz's laws (refer to the official specification document for exact wording). Although flux linkage sits inside Topic 7, it is examined throughout 9PH0 Paper 2 and synoptically in Paper 3, where graphical, algebraic and experimental treatments meet. The Edexcel 9PH0 formula booklet does list $\Phi = BA\cos\theta$Φ=BAcosθ and $\varepsilon = -\dfrac{d(N\Phi)}{dt}$ε=−dtd(NΦ)​, but candidates must still be fluent enough to apply them under time pressure without re-deriving from first principles.

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Worked example with full mark scheme

Question (8 marks):

A flat rectangular coil of $N = 250$N=250 turns and area $A = 8.0 \times 10^{-3}\ \text{m}^2$A=8.0×10−3 m2 is rotated at constant angular speed $\omega = 100\pi\ \text{rad s}^{-1}$ω=100π rad s−1 about an axis perpendicular to a uniform magnetic field of flux density $B = 0.040\ \text{T}$B=0.040 T.

(a) Show that the EMF induced in the coil can be written as $\varepsilon(t) = NBA\omega\sin(\omega t)$ε(t)=NBAωsin(ωt). (3)

(b) Calculate the peak EMF, the frequency of the induced AC, and the RMS EMF. (5)

Solution with mark scheme:

(a) Step 1 — write down the flux linkage as a function of time.

Take $\theta = \omega t$θ=ωt where $\theta$θ is the angle between the field $\mathbf{B}$B and the normal to the coil. At $t = 0$t=0 the coil is taken with its normal parallel to $\mathbf{B}$B, so:

$N\Phi(t) = NBA\cos(\omega t)$NΦ(t)=NBAcos(ωt)

M1 — correct expression for flux linkage with $\cos(\omega t)$cos(ωt) and the factor $N$N. Common error: omitting $N$N (giving flux instead of flux linkage), which loses both M1 and the subsequent A1.

Step 2 — apply Faraday's law.

$\varepsilon = -\dfrac{d(N\Phi)}{dt} = -\dfrac{d}{dt}\left[NBA\cos(\omega t)\right] = NBA\omega\sin(\omega t)$ε=−dtd(NΦ)​=−dtd​[NBAcos(ωt)]=NBAωsin(ωt)

M1 — differentiating $\cos(\omega t)$cos(ωt) correctly to give $-\omega\sin(\omega t)$−ωsin(ωt), with the leading minus from Faraday's law cancelling the minus from the derivative.

A1 — final printed result $\varepsilon(t) = NBA\omega\sin(\omega t)$ε(t)=NBAωsin(ωt). The factor $\omega$ω must be present; candidates who write $\varepsilon = NBA\sin(\omega t)$ε=NBAsin(ωt) miss the chain-rule step and lose this A1.

(b) Step 1 — peak EMF.

The peak occurs when $\sin(\omega t) = 1$sin(ωt)=1, so $\varepsilon_0 = NBA\omega$ε0​=NBAω.

$\varepsilon_0 = 250 \times 0.040 \times 8.0 \times 10^{-3} \times 100\pi = 25.13\ \text{V} \approx 25\ \text{V}$ε0​=250×0.040×8.0×10−3×100π=25.13 V≈25 V

M1 — substitution into $NBA\omega$NBAω. A1 — $\varepsilon_0 \approx 25\ \text{V}$ε0​≈25 V to two significant figures.

Step 2 — frequency.

$\omega = 2\pi f$ω=2πf, so $f = \dfrac{\omega}{2\pi} = \dfrac{100\pi}{2\pi} = 50\ \text{Hz}$f=2πω​=2π100π​=50 Hz.

B1 — $f = 50\ \text{Hz}$f=50 Hz (the UK mains frequency, useful sanity check).

Step 3 — RMS EMF.

For a sinusoidal AC, $\varepsilon_{\text{RMS}} = \dfrac{\varepsilon_0}{\sqrt{2}}$εRMS​=2​ε0​​.

$\varepsilon_{\text{RMS}} = \dfrac{25.13}{\sqrt{2}} \approx 17.8\ \text{V}$εRMS​=2​25.13​≈17.8 V

M1 — correct use of $\varepsilon_0/\sqrt{2}$ε0​/2​. A1 — $\varepsilon_{\text{RMS}} \approx 18\ \text{V}$εRMS​≈18 V.

Total: 8 marks (M4 A3 B1, split as shown).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A circular search coil of 400 turns and radius $r = 1.5\ \text{cm}$r=1.5 cm is placed with its plane perpendicular to a uniform magnetic field. The field is switched off, and the flux linkage falls linearly to zero over $\Delta t = 0.020\ \text{s}$Δt=0.020 s. The mean induced EMF is measured as $0.71\ \text{V}$0.71 V.

(a) Calculate the magnetic flux density before the field was switched off. (4)

(b) Explain why the EMF in (a) is the mean EMF and not the peak EMF. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — using $\varepsilon = \dfrac{\Delta(N\Phi)}{\Delta t}$ε=ΔtΔ(NΦ)​ to find the change in flux linkage: $\Delta(N\Phi) = \varepsilon \Delta t = 0.71 \times 0.020 = 0.0142\ \text{Wb}$Δ(NΦ)=εΔt=0.71×0.020=0.0142 Wb.
• M1 (AO1.1b) — recognising area $A = \pi r^2 = \pi (0.015)^2 = 7.07 \times 10^{-4}\ \text{m}^2$A=πr2=π(0.015)2=7.07×10−4 m2.
• M1 (AO2.1) — equating \Delta(N\Phi) = NBA\ since the field collapses to zero from $B$B.
• A1 (AO1.1b) — $B = \dfrac{0.0142}{400 \times 7.07 \times 10^{-4}} \approx 0.050\ \text{T}$B=400×7.07×10−40.0142​≈0.050 T.

(b)

• B1 (AO2.4) — Faraday's law gives the instantaneous EMF as the rate of change of flux linkage; the measurement here is the average over a finite interval.
• B1 (AO2.4) — only when the flux linkage falls linearly with time does the mean EMF equal the constant instantaneous EMF.

Total: 6 marks split AO1 = 3, AO2 = 3. Edexcel uses search-coil questions to mix procedural calculation (AO1) with conceptual interpretation of mean vs instantaneous and linear vs sinusoidal change (AO2). Candidates who only quote the formula without addressing what "mean" means lose the AO2 marks.

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Synoptic links

Connects to:

1. Faraday's law (Topic 7): the operational definition of induced EMF is $\varepsilon = -\dfrac{d(N\Phi)}{dt}$ε=−dtd(NΦ)​. Flux linkage is the quantity whose rate of change is the EMF; without flux linkage there is no induction.

2. Lenz's law (Topic 7): the minus sign in Faraday's law is Lenz's law in disguise — the induced EMF opposes the change in flux linkage that produced it. This is a direct consequence of conservation of energy: an induced current that aided the change would create a self-amplifying source of free energy.