#### When a magnet or magnetic pole ( coil or winding ) is rotated, the magnetic field or flux of constant amplitude produced by the magnet or poles will be a rotating magnetic field. But in this case, an external physical force is needed to rotate magnet or coils. In practice, without any external force, a rotating magnetic field can be produced with the help of a polyphase ac supply. When the three-phase uniformly distributed winding suitably wound on the stator is supplied with a 3-phase ac supply, a rotating magnetic field of constant magnitude rotating at synchronous speed is produced. Let us see how it is possible.

Consider 3-phase, 2 pole stator having three similar windings aa', bb', and cc' displaced in space by 120°, supplied with 3-phase ac supply as shown below.

#### Due to the current flowing in each phase displaced from each other by 120° produces an alternating flux which is assumed to be sinusoidal in nature. The fluxes produced by each phase Ï†_{1}, Ï†_{2}, and Ï†_{3} are also displaced by 120° from each other.

The instantaneous values of fluxes with respect to time is given by,
Ï†_{1} = Ï†_{m} Sin Î¸Ï†_{2} = Ï†_{m} Sin (Î¸ - 120⁰)Ï†_{3} = Ï†_{m} Sin (Î¸ - 240⁰)

Where Ï†_{m} = maximum value of the flux due to any phase. The waveforms of these fluxes and their phase representation are shown below.

_{1}= Ï†

_{m}Sin Î¸

_{2}= Ï†

_{m}Sin (Î¸ - 120⁰)

_{3}= Ï†

_{m}Sin (Î¸ - 240⁰)

#### The three fluxes together give rise to the effect of rotating flux called Rotating Magnetic Field. At any instant, the resultant flux, Ï†_{r} is the phasor sum of the fluxes of the three phases ( Ï†_{1}, Ï†_{2}, and Ï†_{3} ) at that instant. Let us see four different instants 0 (0⁰), 1 (60⁰), 2 (120⁰), and 3(180⁰) as shown in the above waveform.

### i. When Î¸ = 0⁰ (At point 0) :

#### At point 0 on the waveform i.e, at 0⁰, the value of Ï†_{1} is,

#### The value of Ï†_{2} is,

#### The value of Ï†_{3} is,

#### Since Ï†_{1} = 0, the resultant of Ï†_{2} and Ï†_{3} can be found out by reversing Ï†_{2} and adding it with Ï†_{3} as shown below.

#### From the figure Resultant flux Ï†_{r},

#### So at 0⁰, the magnitude of resultant flux is 1.5 times the maximum value of individual flux.

### ii. When Î¸ = 60⁰ (At point 1) :

#### At point 1 on the waveform i.e, at 60⁰, the value of Ï†_{1} is,

#### The value of Ï†_{2} is,

#### The value of Ï†_{3} is,

#### Since Ï†_{3} = 0, the resultant of Ï†_{1} and Ï†_{2} can be found out by reversing Ï†_{2} and adding it with Ï†_{1} as shown below.

#### From the figure Resultant flux Ï†_{r},

#### Therefore at 60⁰, the magnitude of resultant flux Ï†_{r} = 1.5 Ï†_{m}. Here we can notice that the magnitude of Ï†_{r} is the same as that of the previous, also the phasor has rotated clockwise through an angle of 60⁰.

### iii. When Î¸ = 120⁰ (At point 2) :

#### At point 2 on the waveform i.e, at 120⁰, the value of Ï†_{1} is,

#### The value of Ï†_{2} is,

#### The value of Ï†_{3} is,

#### Since Ï†_{2} = 0, the resultant of Ï†_{1} and Ï†_{3} can be found out by reversing Ï†_{3} and adding it with Ï†_{1} as shown below.

#### From the figure Resultant flux Ï†_{r},

#### Therefore at 120⁰, the magnitude of resultant flux Ï†_{r} = 1.5 Ï†_{m}. Here also, the magnitude of Ï†_{r} is the same as that at 0⁰ and 60⁰, but the phasor has further rotated clockwise through an angle of 60⁰ or 120⁰ from the start.

### iv. When Î¸ = 180⁰ (At point 3) :

#### At point 3 on the waveform i.e, at 180⁰, the value of Ï†_{1} is,

#### The value of Ï†_{2} is,

#### The value of Ï†_{3} is,

#### Since Ï†_{1} = 0, the resultant of Ï†_{2} and Ï†_{3} can be found out by reversing Ï†_{3} and adding it with Ï†_{2} as shown below.

#### From the figure Resultant flux Ï†_{r},

#### Hence Ï†_{r} = 1.5 Ï†_{m} but has rotated clockwise through an additional angle of 60⁰ or 180⁰ from the start.

### From the above, it is clear that,

- The resultant flux is of constant magnitude and having a value of 1.5 Ï†
_{m} i.e., Ï†_{r} = 1.5 Ï†_{m}.
- The resultant flux rotates at the synchronous speed, N
_{s} is given by

N_{s} = 120f/P rpm

Where f = supply frequency and P = number of poles of the stator winding.

Therefore, when a 3-phase supply is fed to a 3-phase stator winding, a rotating magnetic field of constant magnitude rotating at synchronous speed is produced.

_{m}i.e., Ï†_{r}= 1.5 Ï†_{m}._{s}is given by_{s}= 120f/P rpm