# Magnetic flux

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u24-magnetic-flux/

This sub-topic introduces magnetic flux, the core foundational quantity for all electromagnetic induction calculations. You will learn how to calculate flux and flux linkage for different coil orientations, and master key conventions for exam questions.

**Prerequisites:** [Magnetic flux density (B)](https://www.owlsprep.com/study/cie-9702-u23-magnetic-flux-density/)

## Learning objectives

- Define magnetic flux and magnetic flux linkage correctly
- Calculate magnetic flux for different coil orientations
- Calculate change in flux and flux linkage for moving/rotating coils
- Recognize common exam traps around flux angle conventions

## Definition of Magnetic Flux

**Magnetic Flux** — The product of the component of magnetic flux density perpendicular to an area and the magnitude of the area itself. It measures how many magnetic field lines pass through the area.

*Notation:* \Phi

*Example:* A flat coil aligned parallel to a uniform B field has zero flux, as no field lines pass through the coil.

$$\Phi = BA \cos\theta$$

In this formula, $\theta$ is the angle between the magnetic field vector and the **normal (perpendicular) to the plane of the area**, not the angle between the field and the plane itself. This is the most commonly tested convention in exams.

**Worked example:** A rectangular coil of area $2.0 \times 10^{-4}$ m² is placed in a uniform magnetic field of flux density 0.15 T. The normal to the coil makes an angle of 30° with the magnetic field. Calculate the magnetic flux through the coil.

1. 1. Recall the flux formula:

   $$\Phi = BA \cos\theta$$
2. 2. Substitute the given values:

   $$\Phi = (0.15)(2.0 \times 10^{-4})(\cos 30^\circ)$$
3. 3. Calculate the final value, $\cos 30^\circ \approx 0.866$:

   $$\Phi = 2.6 \times 10^{-5} \text{ Wb}$$

> **Exam tip:** If the question gives the angle between the field and the plane of the coil, subtract this angle from 90° to get $\theta$ for the formula.

## Magnetic Flux Linkage

**Magnetic Flux Linkage** — The total flux linked with a coil of N identical turns, equal to the product of the number of turns and the magnetic flux through one turn of the coil.

*Notation:* N\Phi

*Example:* A 100-turn coil with $1 \times 10^{-5}$ Wb flux per turn has a total flux linkage of $1 \times 10^{-3}$ Wb turns.

Flux linkage is the quantity that directly appears in Faraday's Law of electromagnetic induction. It accounts for the fact that each turn of the coil cuts magnetic field lines independently, so more turns produce a larger induced emf for the same rate of change of flux per turn.

**Worked example:** A 50-turn circular coil of radius 1.5 cm is placed perpendicular to a uniform magnetic field of 0.20 T. Calculate the total flux linkage of the coil.

1. 1. Calculate the area of the coil:

   $$A = \pi r^2 = \pi (0.015)^2 = 7.07 \times 10^{-4} \text{ m}^2$$
2. 2. Calculate flux per turn, $\theta = 0^\circ$ so $\cos 0^\circ = 1$:

   $$\Phi = BA = (0.20)(7.07 \times 10^{-4}) = 1.41 \times 10^{-4} \text{ Wb}$$
3. 3. Multiply by number of turns for flux linkage:

   $$N\Phi = 50 \times 1.41 \times 10^{-4} = 7.1 \times 10^{-3} \text{ Wb turns}$$

> **Exam tip:** Always write the unit of flux linkage as Wb turns, even though some mark schemes accept Wb. This avoids losing unnecessary marks.

## Change in Magnetic Flux

Most exam questions on magnetic flux ask for the change in flux or flux linkage when a coil rotates, moves, or the magnetic field strength changes. Change in flux is calculated as $\Delta\Phi = \Phi_{final} - \Phi_{initial}$, and change in flux linkage is $\Delta(N\Phi) = N\Delta\Phi$.

> **warning**
>
> Flux can be negative depending on orientation, so always account for sign when calculating change. Examiners almost always ask for the magnitude of the change, so take the absolute value at the end if required.

**Worked example:** A flat 100-turn coil is initially placed with its plane parallel to a uniform 0.10 T magnetic field. It is rotated 90° so its plane is now perpendicular to the field. The coil area is $4.0 \times 10^{-3}$ m². Calculate the change in flux linkage.

1. 1. Find initial flux per turn: when plane is parallel, normal is perpendicular to B, so $\theta_{initial} = 90^\circ$:

   $$\Phi_{initial} = BA\cos 90^\circ = 0$$
2. 2. Find final flux per turn: when plane is perpendicular, normal is parallel to B, so $\theta_{final} = 0^\circ$:

   $$\Phi_{final} = BA\cos 0^\circ = (0.10)(4.0 \times 10^{-3}) = 4.0 \times 10^{-4} \text{ Wb}$$
3. 3. Calculate change in flux linkage:

   $$\Delta(N\Phi) = N(\Phi_{final} - \Phi_{initial}) = 100(4.0 \times 10^{-4} - 0) = 4.0 \times 10^{-2} \text{ Wb turns}$$

**Check your understanding**

Test your understanding of angle conventions:

1. The plane of a coil makes an angle of 20° with a uniform magnetic field. What is $\theta$ (for the flux formula)?

   - 20°
   - 70°
   - 90°
   - 0°

   *Answer:* 70°

   *Why:* Correct: $\theta$ is the angle between the normal (perpendicular to the plane) and the field, so $90^\circ - 20^\circ = 70^\circ$.

## Common pitfalls

- **Wrong:** Taking $\theta$ as the angle between the plane of the coil and the magnetic field
  - Why it fails: The flux formula is defined using the angle between the magnetic field and the normal to the plane, not the plane itself
  - Correct: If given the angle to the plane, subtract it from 90° to get $\theta$ before substituting into $\Phi = BA\cos\theta$
- **Wrong:** Forgetting to multiply by the number of turns when calculating flux linkage
  - Why it fails: Flux linkage describes total flux across all turns of a coil, not just flux through one turn
  - Correct: Always multiply flux per turn by $N$ when asked for flux linkage for Faraday's law calculations
- **Wrong:** Getting a change of zero when a coil flips 180°
  - Why it fails: Flips reverse the sign of flux: flux changes from $+BA$ to $-BA$, not from $BA$ to $BA$
  - Correct: Calculate $\Delta\Phi = -BA - BA = -2BA$, so magnitude of change is $2BA$ for a 180° flip
- **Wrong:** Claiming flux is maximum when the plane of the coil is parallel to the magnetic field
  - Why it fails: Maximum flux occurs when the maximum number of field lines pass through the coil area
  - Correct: Flux is maximum when the plane is perpendicular to the magnetic field, and zero when it is parallel

## Cheatsheet

| Quantity | Symbol | Formula/Rule | Unit |
| --- | --- | --- | --- |
| Magnetic flux | $\Phi$ | $BA\cos\theta$, $\theta$ = angle to normal | Weber (Wb) |
| Magnetic flux linkage | $N\Phi$ | $N \times \Phi$ (per turn) | Weber turns (Wb turns) |
| Change in flux | $\Delta\Phi$ | $\Phi_{final} - \Phi_{initial}$ | Weber (Wb) |
| Change in flux linkage | $\Delta(N\Phi)$ | $N \times \Delta\Phi$ | Weber turns (Wb turns) |
| Maximum flux | $\Phi_{max}$ | $BA$ (normal parallel to B) | Weber (Wb) |
| Zero flux | $\Phi = 0$ | Normal perpendicular to B | Weber (Wb) |

## What's next

Magnetic flux is the foundational quantity for the entire topic of electromagnetic induction, which contributes 10-15% of total marks across CIE A-Level Physics papers 2 and 4. Mastery of flux conventions and calculations is essential to avoid losing marks on Faraday's law and induced emf problems, which are frequent high-mark questions. Next, you will build on this knowledge to learn how changing flux produces induced emf, and apply Faraday's and Lenz's laws to solve a wide range of exam problems. This concept also underpins later topics including alternating current, transformers, and electromagnetic generators.

- [Faraday's law and Lenz's law](https://www.owlsprep.com/study/cie-9702-u24-faraday-s-law-and-lenz/)
- [Induced e.m.f.](https://www.owlsprep.com/study/cie-9702-u24-induced-e-m-f/)

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