# Induced e.m.f.

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u24-induced-e-m-f/

This module covers the origin of induced electromotive force (e.m.f.) from changing magnetic flux. You will learn Faraday’s and Lenz’s laws, and calculate magnitude and direction of induced e.m.f. for common CIE exam scenarios.

**Prerequisites:** [Understanding of magnetic flux and flux linkage](https://www.owlsprep.com/study/cie-9702-u24-magnetic-flux-flux-linkage/)

## Learning objectives

- State Faraday's law and Lenz's law of electromagnetic induction
- Calculate the magnitude of induced e.m.f. from rate of change of flux linkage
- Apply Lenz's law to find the direction of induced e.m.f./current
- Derive and use the formula for induced e.m.f. in a moving conductor

## Faraday's Law of Induction

An induced e.m.f. is created whenever there is a change in the magnetic flux linkage through a conductor or coil. This effect, called electromagnetic induction, is the operating principle for generators, transformers and many other electrical devices.

**Faraday's Law of Induction** — The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit.

*Notation:* |\varepsilon| = \frac{\Delta (N\Phi)}{\Delta t}

**Worked example:** A coil with 200 turns has cross-sectional area $2.0 \times 10^{-3} \, \text{m}^2$. A uniform magnetic field perpendicular to the coil changes from $0.10 \, \text{T}$ to $0.45 \, \text{T}$ in $0.20 \, \text{s}$. Calculate the magnitude of the induced e.m.f.

1. Calculate the change in magnetic flux through one turn: $\Delta\Phi = A \Delta B$
2. $$\Delta \Phi = (2.0 \times 10^{-3})(0.45 - 0.10) = 7.0 \times 10^{-4} \, \text{Wb}$$
3. Substitute into Faraday's law, where flux linkage change is $N \Delta\Phi$
4. $$|\varepsilon| = \frac{N \Delta \Phi}{\Delta t} = \frac{(200)(7.0 \times 10^{-4})}{0.20} = 0.70 \, \text{V}$$

## Lenz's Law and Direction of Induced e.m.f.

Faraday's law calculates the magnitude of induced e.m.f., while Lenz's law gives its direction. Lenz's law is a consequence of the principle of conservation of energy.

**Lenz's Law** — The direction of the induced e.m.f. (and induced current, for a closed circuit) is such that it opposes the change in magnetic flux that produced it.

**Worked example:** The north pole of a bar magnet is pushed towards a stationary coil connected to a galvanometer. Find the direction of induced current as viewed from the side of the magnet.

1. The magnetic flux through the coil pointing away from the magnet (into the closest coil face) is increasing.
2. By Lenz's law, the induced current must create a magnetic field that opposes this increase. This means the induced field points out of the coil face towards the magnet.
3. Use the right-hand grip rule for coils: a field pointing out of the front face means the current flows anticlockwise when viewed from the magnet side.

> **tip**
>
> Induced e.m.f. opposes the *change* in flux, not the flux itself. If flux is decreasing, induced current will act to increase it back.

## Induced e.m.f. in a Moving Conductor

When a straight conductor moves through a uniform magnetic field, charge carriers inside the conductor experience a Lorentz force that separates charge, creating an e.m.f. across the ends of the conductor. We can derive this result directly from Faraday's law.

**Derivation:** Derive the induced e.m.f. for a straight conductor moving perpendicular to a uniform magnetic field

*Starting from:* Faraday's law of induction

1. A conductor of length $l$ moves at speed $v$ perpendicular to a uniform field $B$. In time $\Delta t$, it sweeps an area $A = l v \Delta t$.
2. The change in flux cut by the conductor is $\Delta \Phi = B A = B l v \Delta t$.
3. Substitute into Faraday's law for a single conductor ($N=1$): $|\varepsilon| = \frac{\Delta \Phi}{\Delta t}$.

*Conclusion:* The induced e.m.f. for mutually perpendicular $B$, $l$ and $v$ is:

$$\varepsilon = B l v$$

**Worked example:** A 15 cm long metal rod moves at $8.0 \, \text{m/s}$ perpendicular to a uniform magnetic field of flux density $0.50 \, \text{T}$. Calculate the induced e.m.f. across the rod ends.

1. Convert length to SI units: $l = 15 \, \text{cm} = 0.15 \, \text{m}$. All quantities are mutually perpendicular, so $\varepsilon = Blv$ applies directly.
2. $$\varepsilon = (0.50)(0.15)(8.0) = 0.60 \, \text{V}$$

## Common pitfalls

- **Wrong:** Using absolute flux instead of change in flux in Faraday's law
  - Why it fails: Constant flux (even very large flux) produces zero induced e.m.f. Only changing flux generates an e.m.f.
  - Correct: Always calculate the change in flux over the time interval, do not use the absolute flux value.
- **Wrong:** Claiming Lenz's law says induced current opposes the original magnetic field
  - Why it fails: This leads to wrong direction when flux is decreasing. Lenz's law opposes the change, not the field itself.
  - Correct: Always identify whether flux is increasing or decreasing first, then find the direction of induced field that opposes this change.
- **Wrong:** Using $\varepsilon = Blv$ when the conductor moves parallel to the magnetic field
  - Why it fails: The formula $\varepsilon = Blv$ only applies when $B$, $l$ and $v$ are all mutually perpendicular. No flux is cut when moving parallel to $B$, so no e.m.f. is induced.
  - Correct: Check the orientation of all three quantities before using the formula. Induced e.m.f. is zero for motion parallel to the magnetic field.
- **Wrong:** Forgetting to multiply flux by number of turns $N$ for a coil
  - Why it fails: Faraday's law uses flux linkage ($N\Phi$), not flux through a single turn. Missing $N$ gives an incorrect magnitude by a factor of $N$.
  - Correct: Always calculate flux linkage as $N \times \Phi$ for multi-turn coils when finding induced e.m.f.

## Cheatsheet

| Concept | Formula/Rule | Key Point |
| --- | --- | --- |
| Faraday's Law | $\|\varepsilon\| = \frac{\Delta(N\Phi)}{\Delta t}$ | Magnitude of induced e.m.f. = rate of change of flux linkage |
| Lenz's Law | N/A | Direction of induced e.m.f. opposes the change in flux that created it |
| Moving conductor (perpendicular) | $\varepsilon = Blv$ | E.m.f. across straight conductor cutting flux |
| Open circuit | N/A | Induced e.m.f. exists, but no induced current flows |

## What's next

Induced e.m.f. is the core concept for all applications of electromagnetic induction, which makes up a significant portion of CIE A-Level Physics exam questions. The principles you learned here are the foundation for understanding alternating current generation in rotating coils, and the operation of transformers, which are common extended response topics. Mastery of Faraday's and Lenz's laws is also required for many other topics including electromagnetic braking and particle acceleration.

- [Alternating currents](https://www.owlsprep.com/study/cie-9702-u25-overview/)
- [AC characteristics](https://www.owlsprep.com/study/cie-9702-u25-ac-characteristics/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-9702-u24-induced-e-m-f/