---
title: "Fleming's right and left hand rules"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:fleming-hand-rules"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/fleming-hand-rules
site: BCT Engineering
---

# Fleming's right and left hand rules

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2074 Bhadra, 3 marks
- Video: [Fleming's Rules | Fleming's Left Hand rule | Fleming's Right Hand rule | Direction of DC machine](https://www.youtube.com/watch?v=xswvejVQAX4) — Learning Concepts (Kanhai Kumar)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/fleming-hand-rules
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

Explain Fleming's right and left hand rules.

## Short answer

Fleming's right-hand rule (generator): thumb=motion, forefinger=field, middle=induced emf. Left-hand rule (motor): thumb=force, forefinger=field, middle=current. Both give orthogonal directions of motion, field, and current/emf.

## Answer

Fleming's rules give the directions of the three quantities involved in generator and motor action. Hold the thumb, forefinger and middle finger of the hand at right angles to each other.

| | Fleming's right-hand rule | Fleming's left-hand rule |
|---|---|---|
| Used for | Generators (induced emf) | Motors (force on a conductor) |
| Thumb | Direction of motion of the conductor | Direction of motion (force) on the conductor |
| Forefinger | Direction of the magnetic field (N to S) | Direction of the magnetic field (N to S) |
| Middle finger | Direction of induced emf / current | Direction of current in the conductor |
| Principle | Electromagnetic induction (e = Blv) | Force on a current-carrying conductor (F = BIl) |

**Memory aid:** the **F**orefinger is the **F**ield, the middle finger is the **C**urrent (or emf), and the thu**M**b is the **M**otion (or thrust).

Example: a conductor moves upward in a field pointing from left to right (N on the left). The right-hand rule gives an induced current flowing into the page. Applying the left-hand rule to this current gives a force on the conductor acting downward, which opposes its motion, as Lenz's law requires.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which rule applies to generators and which to motors?
   A. Right hand for both
   B. Left hand for generators, right hand for motors
   C. Left hand for both
   D. Right hand for generators, left hand for motors
   Answer: D. Right-hand rule gives induced emf direction (generator); left-hand rule gives force on a current conductor (motor). (Key point: Which hand for which)

2. What does the middle finger represent in the left-hand rule?
   A. Direction of motion
   B. Direction of the field
   C. Direction of the current in the conductor
   D. Direction of the induced emf
   Answer: C. In the left-hand rule the forefinger is the field, middle finger the current, thumb the force or motion. (Key point: Middle finger)

3. Which pairing of rule and formula is correct?
   A. Both rules with e = Blv
   B. Right-hand rule with F = BIl
   C. Right-hand rule with e = Blv, left-hand rule with F = BIl
   D. Left-hand rule with e = Blv
   Answer: C. Generator action is e = Blv (right hand); motor action is F = BIl (left hand). (Key point: Underlying formulas)

## Related in Electrical Machine

- [Faraday's laws; Fleming's rules](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Afaraday-fleming-rules) — Electrical Machine, chapter 1

## Related in other subjects

- [Setting out right hand side composite curve](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aright-hand-composite-curve-setting-out) — Surveying II, chapter 6
- [Left and right focal point ratios](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afocal-point-ratios-left-right) — Theory of Structures II, chapter 4
- [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4×
- [Suspension 100 m: UDL on right half, SF and BM at 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-5knm2-10knm2-right-half) — Theory of Structures I, chapter 7, asked 2×
