Revolving magnetic field

Switch on a three-phase induction motor and something remarkable happens before the rotor ever moves: inside the stator, a magnetic field starts sweeping around the air gap like a lighthouse beam, at a fixed speed, without a single moving part making it happen. No commutator, no brushes, no mechanical trick — just three fixed windings and a three-phase supply. This revolving magnetic field is the whole reason an induction motor turns at all; everything else in the chapter (slip, induced rotor currents, torque) is the rotor's response to this field chasing it around the gap. This article extends the Pass-1 treatment: after the core idea and the synchronous-speed formula, it derives the constant $1.5\,\Phi_m$ result properly, works a second full example, and adds exam-style traps and quiz checks so you can use $N_s = 120f/P$ with real confidence.

The core insight

Three windings, $120^\circ$ apart in space, carrying three currents $120^\circ$ apart in time, add to a single flux wave of constant magnitude that rotates at synchronous speed $N_s = 120f/P$ .

The core idea

The setup. A three-phase stator carries three identical windings — call them R, Y, B — physically placed $120^\circ$ apart around the bore. A balanced three-phase supply drives them, so the three phase currents are equal in amplitude but displaced $120^\circ$ from one another in time:

i_R = I_m\sin(\omega t), \qquad i_Y = I_m\sin(\omega t - 120^\circ), \qquad i_B = I_m\sin(\omega t - 240^\circ)

where: $I_m$ is the peak current in each phase and $\omega = 2\pi f$ is the supply angular frequency.

Each winding alone. Current $i_R$ alone produces a flux that pulsates along the fixed R-axis — it grows, falls to zero, reverses, and grows again, but it never leaves that one axis. The same is true for $i_Y$ along the Y-axis and $i_B$ along the B-axis. Individually, none of the three fluxes rotates; each is a stationary axis with a magnitude that oscillates in time.

Adding them together. The trick is that the combination behaves completely differently from any one piece. At every instant, you add the three pulsating flux vectors (each along its own fixed spatial axis, each with the instantaneous magnitude set by its own current) as vectors. Because the spatial displacement ( $120^\circ$ ) matches the time displacement ( $120^\circ$ ) of the driving currents, the vector sum turns out to have constant magnitude — it never changes size as $t$ varies — while its direction sweeps steadily around the air gap. That combined object is the revolving magnetic field. For a simple 2-pole winding, it completes one full mechanical revolution in exactly one cycle of the supply.

Figure

Three winding axes 120 degrees apart with a single resultant flux arrow rotating around the gap over one cycle

Speed of rotation. A 2-pole field turns once per supply cycle. A $P$ -pole field only needs to sweep through $2/P$ of a full mechanical turn to complete one magnetic cycle (matching one electrical cycle), so it turns proportionally slower. This gives the synchronous speed,

N_s = \frac{120f}{P}\ \text{rpm}

where: $f$ is the supply frequency in hertz and $P$ is the total number of poles (always an even number: 2, 4, 6, ...). This is the single most-used formula in the induction-motor chapter — every later calculation of slip and rotor speed starts from this $N_s$ .

Derivation AWhy the resultant flux has constant magnitude 1.5 times the maximum single-phase flux

Show that adding the three time-varying, spatially-displaced pulsating fluxes produced by a balanced three-phase winding gives a single resultant flux vector whose magnitude is fixed at $1.5\,\Phi_m$ (independent of time) while its direction advances uniformly with $\omega t$ , which is exactly the rotating-field behaviour asserted above.

Step 1. Represent each phase's flux as a vector lying along its own fixed spatial axis, with instantaneous magnitude equal to its own sinusoidal current's instantaneous value times a constant $\Phi_m$ (the peak flux one phase alone can produce). Take the R-axis as the reference direction ( $0^\circ$ ), the Y-axis at $120^\circ$ , and the B-axis at $240^\circ$ around the gap, matching the physical layout of the windings.

Step 2. Write the three flux magnitudes using the same time functions as the currents, since flux is proportional to current in a linear magnetic circuit:

\phi_R = \Phi_m\sin(\omega t), \qquad \phi_Y = \Phi_m\sin(\omega t - 120^\circ), \qquad \phi_B = \Phi_m\sin(\omega t - 240^\circ)

Step 3. Resolve all three flux vectors along a fixed horizontal and vertical pair of axes and add the components. Choosing $t = 0$ as a convenient instant to check the claim (any other instant gives the same magnitude, only rotated), the horizontal component works out to $\phi_R + \phi_Y\cos120^\circ + \phi_B\cos240^\circ$ and the vertical component to $\phi_Y\sin120^\circ + \phi_B\sin240^\circ$ . Carrying through the algebra with $\sin120^\circ = \sin240^\circ = \sqrt3/2$ and $\cos120^\circ=\cos240^\circ=-1/2$ for the general instant $t$ , the time-dependent terms in the sum cancel in a way that leaves the magnitude of the resultant independent of $t$ .

Step 4. Evaluate the surviving magnitude: the algebra collapses to a resultant of fixed length $1.5\,\Phi_m$ , i.e. one and a half times the peak flux a single phase alone could produce, while the angle of that resultant vector, measured from the R-axis, increases linearly as $\omega t$ — so the tip of the resultant sweeps around the gap at the uniform angular rate $\omega$ .

Step 5. Because one full sweep of this resultant ( $360^\circ$ electrical) corresponds to $360^\circ/(P/2)$ mechanical degrees for a $P$ -pole winding, and the sweep takes one supply period $T = 1/f$ , the mechanical rotational speed is $N_s = \dfrac{120f}{P}$ rpm, matching the formula quoted above and confirming that constant-magnitude, uniformly-rotating flux is exactly what a balanced three-phase winding produces.

Worked example

Problem. A three-phase induction motor runs on a 50 Hz supply at a full-load speed close to 1450 rpm. Find (a) its number of poles and the exact synchronous speed, and (b) if the same stator were instead wound for 8 poles and re-connected to a 60 Hz supply, find the new synchronous speed and confirm the electrical phase displacement needed between currents to keep the field rotating correctly.

Step 1. Since a real induction-motor rotor always runs slightly below synchronous speed (the small gap is called slip, covered in the next article), the correct pole count is the one whose $N_s$ sits just above 1450 rpm. Try $P = 4$ :

N_s = \frac{120\times 50}{4} = 1500\ \text{rpm}

This is just above 1450 rpm, consistent with a small slip, so $P = 4$ is correct.

Step 2. Rule out the neighbouring standard choice, $P = 6$ :

N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}

That is below the stated running speed, which is impossible since the rotor can never exceed synchronous speed, so $P = 6$ is rejected and $P = 4$ , $N_s = 1500$ rpm stands.

Step 3. For the second part, rewind for $P = 8$ poles on a $f = 60$ Hz supply:

N_s = \frac{120\times 60}{8} = 900\ \text{rpm}

Step 4. Regardless of pole count or supply frequency, a balanced three-phase source always delivers currents $120^\circ$ apart in electrical (time) degrees — this never changes with $P$ . It is the fixed $120^\circ$ time displacement combined with the $120^\circ$ spatial displacement of the windings that keeps producing a constant-magnitude rotating field at any $P$ or $f$ .

Answer: $P = 4$ poles with $N_s = 1500$ rpm for the first machine; for the 8-pole, 60 Hz rewind, $N_s = 900$ rpm, with phase currents always $120^\circ$ apart electrically.

Competitive stretch

A quick sanity check often asked in problems: if a nameplate only gives $N_s$ and $f$ , recover $P$ by rearranging the formula as $P = 120f/N_s$ , and always round to the nearest even integer — pole counts can only be 2, 4, 6, 8, ... never odd, because poles come in north-south pairs.

Common traps

Mixing up electrical degrees and mechanical degrees: for a $P$ -pole machine, one full mechanical revolution of the field equals $(P/2)\times360^\circ$ electrical degrees. The $120^\circ$ phase spacing used throughout is always electrical.
Thinking the resultant flux magnitude "pulses" the way each phase's individual flux does — it does not; the vector sum of three $120^\circ$ -displaced sinusoidal fluxes is constant at $1.5\,\Phi_m$ and only its direction changes.
Forgetting the balanced requirement: a balanced three-phase supply feeding balanced, symmetrically-spaced windings is essential; imbalance produces an elliptical, not circular, rotating field (beyond this course).
Confusing $N_s$ (the field's synchronous speed) with the rotor's actual speed $N_r$ — the two are equal only in the unreachable zero-slip limit; a real induction motor's rotor is always slightly slower.

Ask the tutor: if you want a step-by-step vector-diagram walkthrough of the $1.5\,\Phi_m$ derivation at several instants of time, or more practice converting between pole count, frequency and synchronous speed, ask the AI tutor about this topic.

Why does the field need exactly three phases spaced 120 degrees apart, not two or four

Three phases displaced $120^\circ$ in space and time is the minimum arrangement that produces a perfectly constant-magnitude rotating flux with a simple, symmetric winding. Two-phase windings (displaced $90^\circ$ ) can also produce a rotating field, but three-phase is the industry standard because it is more efficient to generate, transmit and use, and the resulting torque is smoother. The specific $120^\circ$ figure is what makes the algebra in the derivation collapse the time-dependent terms exactly to zero, leaving a constant $1.5\,\Phi_m$ .

Does the revolving field actually carry the rotor around physically, like a rotating magnet dragging a compass needle

Not directly by contact, but the analogy is close in spirit: the revolving field induces currents in the rotor conductors (by Faraday's law, since the rotor "sees" a changing flux as the field sweeps past it), and those induced currents in turn experience a force in the field. That interaction is what actually produces torque. This mechanism — the rotor chasing a field it can never quite catch — is the subject of the next article on induction-motor principle and slip.

If I increase the supply frequency, does the synchronous speed always increase proportionally

Yes, for a fixed pole count $P$ , $N_s = 120f/P$ is directly proportional to $f$ — doubling the frequency doubles $N_s$ . This is exactly why variable-frequency drives (VFDs) can control induction-motor speed: by electronically varying $f$ fed to the stator, the synchronous speed — and hence the motor's operating speed — can be adjusted smoothly without changing the winding or pole count.

Revolving magnetic field

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads