define magnetic flux as the product of the magnetic flux density and the cross-sectional area perpendicular to the direction of the magnetic flux density

Electromagnetic Induction – Magnetic Flux

Learning Objective

To define magnetic flux, relate it to magnetic flux density and the area it penetrates, and apply the concept in quantitative descriptions of electromagnetic induction (Cambridge IGCSE/A‑Level 9702 – Topic 20.5).

Key Symbols and Units

SymbolQuantitySI Unit
\(\Phi\)Magnetic fluxweber (Wb)
\(\mathbf{B}\)Magnetic flux density (field strength)tesla (T)
\(\mathbf{A}=A\mathbf{n}\)Area vector (magnitude \(A\), direction \(\mathbf{n}\) normal to the surface)square metre (m²)
\(\theta\)Angle between \(\mathbf{B}\) and the normal \(\mathbf{n}\)degrees (°) or radians
\(N\)Number of turns in a coildimensionless
\(\varepsilon\)Induced emfvolt (V)

1. Definition of Magnetic Flux

  • Vector form: \(\displaystyle \Phi = \int_{S}\mathbf{B}\cdot d\mathbf{A}\)
  • Here \(\mathbf{B}\) is the magnetic flux density (T) and \(d\mathbf{A}=dA\,\mathbf{n}\) is an infinitesimal area element whose direction \(\mathbf{n}\) is perpendicular (normal) to the surface.
  • Scalar (uniform‑field) form: When \(\mathbf{B}\) is uniform and the surface is flat, the integral reduces to the simple product

    \[

    \Phi = B\,A\cos\theta

    \]

    where \(\theta\) is the angle between \(\mathbf{B}\) and the normal \(\mathbf{n}\).

  • Special cases

    • \(\theta =0^{\circ}\) (field perpendicular to the surface) → \(\Phi = BA\).
    • \(\theta =90^{\circ}\) (field parallel to the surface) → \(\Phi =0\) (no flux passes through).

2. Magnetic Flux Linkage

If a coil contains \(N\) identical turns, each turn experiences the same flux \(\Phi\). The total flux linkage is

\[

\Lambda = N\Phi

\]

Flux linkage \(\Lambda\) has the unit weber‑turn (Wb·turn) and appears directly in Faraday’s law for a coil.

3. Faraday’s Law of Electromagnetic Induction

The induced emf \(\varepsilon\) in a coil is proportional to the rate of change of its flux linkage:

\[

\varepsilon = -\frac{d\Lambda}{dt}= -N\frac{d\Phi}{dt}

\]

For a uniform change over a finite interval \(\Delta t\):

\[

\varepsilon = -\,N\,\frac{\Delta\Phi}{\Delta t}

\]

The negative sign expresses Lenz’s law: the induced emf always opposes the change that produces it.

4. Lenz’s Law – Direction of the Induced emf

  • The induced current creates a magnetic field that opposes the original change in flux (increase or decrease).
  • Use the right‑hand rule for the induced current and the “thumb‑fingers” rule for the resulting magnetic field to determine the direction.

Diagram placeholder: a rectangular loop entering a uniform magnetic field, showing the induced current direction that opposes the increase of flux.

5. Factors that Influence the Magnitude of the Induced emf

From \(\varepsilon = -N\Delta\Phi/\Delta t\) we see that the emf increases when any of the following is increased:

  1. Rate of change of flux** (\(\Delta\Phi/\Delta t\)) – faster motion of the conductor or faster variation of \(B\).
  2. Number of turns** (\(N\)) – more conductors give a larger total linkage.
  3. Area** (\(A\)) – a larger loop cuts through more field lines.
  4. Magnetic flux density** (\(B\)) – a stronger field provides more flux per unit area.

6. Worked Examples

Example 1 – Static coil in a uniform field

Calculate the magnetic flux through a circular coil of radius \(r = 0.10\ \text{m}\) placed in a uniform magnetic field \(B = 0.5\ \text{T}\) with the field perpendicular to the plane of the coil.

  1. Area of the coil: \(\displaystyle A = \pi r^{2}= \pi (0.10)^{2}=3.14\times10^{-2}\ \text{m}^{2}\).
  2. \(\theta =0^{\circ}\) → \(\cos\theta =1\).
  3. Flux: \(\displaystyle \Phi = B A = 0.5\ \text{T}\times3.14\times10^{-2}\ \text{m}^{2}=1.57\times10^{-2}\ \text{Wb}\).

If the coil has \(N = 20\) turns, the flux linkage is \(\displaystyle \Lambda = N\Phi = 20\times1.57\times10^{-2}=3.14\times10^{-1}\ \text{Wb·turn}\).

Example 2 – Rotating coil (generator)

A single‑turn rectangular coil of area \(A = 0.02\ \text{m}^{2}\) rotates at a constant angular speed \(\omega = 300\ \text{rad s}^{-1}\) in a uniform magnetic field \(B = 0.8\ \text{T}\). Find the peak induced emf.

  1. Flux as a function of time: \(\displaystyle \Phi(t)=B A\cos(\omega t)\).
  2. Differentiate: \(\displaystyle \varepsilon(t)= -\frac{d\Phi}{dt}= B A\omega\sin(\omega t)\).
  3. Peak emf (when \(\sin(\omega t)=1\)): \(\displaystyle \varepsilon_{\max}= B A\omega = 0.8\times0.02\times300 = 4.8\ \text{V}\).

Increasing any of the four factors (\(B, A, \omega, N\)) would raise the generated voltage, illustrating the points in Section 5.

7. Summary Checklist for Exams (AO1–AO3)

  • State the definition \(\Phi = BA\cos\theta\) (or \(\Phi = \int\mathbf{B}\cdot d\mathbf{A}\)) and give the unit (Wb). – AO1
  • Explain flux linkage \(N\Phi\) and its role in Faraday’s law. – AO2
  • Write Faraday’s law with the correct sign and apply it to calculate induced emf. – AO2
  • Use Lenz’s law and right‑hand rules to determine the direction of the induced current. – AO3
  • Identify the four factors that affect the magnitude of the emf and show how they appear in the formula. – AO2
  • Solve numerical problems involving static and rotating coils, and moving conductors. – AO3

8. Diagrammatic Illustrations (placeholders)

Diagram (a): magnetic flux through a tilted rectangular area showing \(\mathbf{B}\), the normal \(\mathbf{n}\), and the angle \(\theta\).

Diagram (b): a rotating coil in a uniform field illustrating the sinusoidal variation of emf with time.