- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# Rotating Magnetic Field in a 3-Phase Induction Motor

Consider three identical coils located on axes physically at 120° in space and each coil is supplied from one phase of a balanced 3-phase supply. Thus, each coil will produce an alternating flux along its own axis. Let the instantaneous values of the fluxes are given by,

$$\mathrm{\varphi_{1} = \varphi_{m} sin \omega t … (1)}$$

$$\mathrm{\varphi_{2} = \varphi_{m} sin(\omega t − 120°) … (2)}$$

$$\mathrm{\varphi_{3} = \varphi_{m} sin(\omega t + 120°) … (3)}$$

The waveforms of the fluxes produced by the three coils are shown in Figure-1. The resultant flux (ϕ_{r}) at any instant is equal to the phasor sum of the fluxes due to three phases. To determine the values of resultant flux (ϕ_{r}), we consider four point viz. 0, 1, 2 and 3 which are 60° apart.

## Case I – When ωt = 0°

This instant is corresponds to the point 0 in the waveform. Putting ωt = 0° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} = \varphi_{m} sin 0° = 0}$$

$$\mathrm{\varphi_{2} = \varphi_{m} sin(0° − 120°) = −\frac{\sqrt{3}}{2}\varphi_{m}}$$

$$\mathrm{\varphi_{3} = \varphi_{m} sin(0° + 120°) =\frac{\sqrt{3}}{2}\varphi_{m}}$$

Refer Figure-2, the phasor for ϕ_{2} is shown along OY and the phasor for ϕ_{3} is shown along OB. Thus the resultant flux ϕ_{r} is the phasor sum of OY and OB which is shown along OA. The magnitude of the resultant flux is given by,

$$\mathrm{\varphi_{r} = 𝑂𝐴 = 2𝑂𝐸 = 2 \:𝑂𝐵\: cos 30° = 2 \times\frac{\sqrt{3}}{2}\varphi_{m} \times\frac{\sqrt{3}}{2}=\frac{3}{2}\varphi_{m}}$$

## Case II – When ωt = 60°

This instant is corresponds to point 1. Putting ωt = 60° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} = \varphi_{m} sin 60° =frac{\sqrt{3}}{2}\varphi_{m}}$$

$$\mathrm{\varphi_{2} = \varphi_{m} sin(60° − 120°) = −frac{\sqrt{3}}{2}\varphi_{m}}$$

$$\mathrm{\varphi_{3} = \varphi_{m} sin(60° + 120°) = 0}$$

The phasors ϕ_{1}, ϕ_{2} and ϕ_{r} are shown in Figure-3. The value of resultant flux is given by,

$$\mathrm{\varphi_{r} = 𝑂𝐴 = 2\: 𝑂𝑅 \:cos 30° = 2 \times\frac{\sqrt{3}}{2}\varphi_{m} \times\frac{\sqrt{3}}{2}=\frac{3}{2}\varphi_{m}}$$

Hence, it can be seen that the resultant flux is again $(\frac{3}{2} \varphi_m)$ but has rotated through an angle of 60° in the clockwise direction.

## Case III - When ωt = 120°

This instant is corresponds to point 2. Putting ωt = 120° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} = \varphi_{m} sin 120° =\frac{\sqrt{3}}{2}\varphi_{m}}$$

$$\mathrm{\varphi_{2} = \varphi_{m} sin(120° − 120°) = 0}$$

$$\mathrm{\varphi_{3} = \varphi_{m} sin(120° + 120°) = −\frac{\sqrt{3}}{2}\varphi_{m}}$$

Thus, the resultant flux is given by,

$$\mathrm{\varphi_{r} = 𝑂𝐴 = 2\:𝑂𝑅\:cos 30° = 2 \times \frac{\sqrt{3}}{2}\varphi_{m} \times\frac{\sqrt{3}}{2}=\frac{3}{2}\varphi_{m}}$$

Hence, the resultant flux is again $(\frac{3}{2} \varphi_m)$ obtained but has further rotated by an angle of 60° form point 1 in the clockwise direction (see Figure-4).

## Case IV – When ωt = 180°

This instant is corresponds to point 3. Putting ωt = 180° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} = \varphi_{m} sin 180° = 0}$$

$$\mathrm{\varphi_{2} = \varphi_{m} sin(180° − 120°) =\frac{\sqrt{3}}{2}\varphi_{m}}$$

$$\mathrm{\varphi_{3} = \varphi_{m} sin(180° + 120°) = −\frac{\sqrt{3}}{2}\varphi_{m}}$$

Thus, the resultant flux is given by,

$$\mathrm{\varphi_{r} = 𝑂𝐴 = 2\:𝑂𝑌\:cos 30° = 2 \times\frac{\sqrt{3}}{2}\varphi_{m} \times\frac{\sqrt{3}}{2}=\frac{3}{2}\varphi_{m}}$$

Again, the resultant flux is equal to $(\frac{3}{2} \varphi_m)$ but has further rotated by an angle of 60° form the point 2 in the clockwise direction (see Figure-5).

From the above discussion, it is clear that a three phase balanced supply produces a rotating magnetic field.

- Related Articles
- Rotating Magnetic Field in Three-Phase Induction Motor
- Starting Torque of 3-Phase Induction Motor; Torque Equation of 3-Phase Induction Motor
- Rotating Magnetic Field Produced by Two-Phase Supply
- Construction of 3-Phase Induction Motor
- Torque Slip Characteristics of 3-Phase Induction Motor
- Synchronous Speed and Slip of a 3-Phase Induction Motor
- Three Phase Induction Motor Starting Methods
- Three Phase Induction Motor – Working Principle
- 3-Phase Induction Motor – Definition, Working Principle, Advantages and Disadvantages
- Winding EMFs in a 3-Phase Induction Motor; Stator EMF and Rotor EMF
- 3-Phase Induction Motor Rotor Frequency, EMF, Current and Power Factor
- Three-Phase Induction Motor Torque-Speed Characteristics
- Split-Phase Induction Motor – Operation and Characteristics
- Methods of Starting Single-phase Induction Motor
- Running Torque of Three-Phase Induction Motor