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T-12: Rotating Magnetic Field (RMF)

Section: B | Priority: 🔴 MUST | Exam Frequency: 5/7 years Sources: Theraja Ch-34 (Art. 34.6-34.8), VK Mehta Ch-8 (Art. 8.3-8.4), Slides L-01, L-02

Why This Topic Matters

The 3-phase RMF proof appeared in 5 out of 7 papers (2017, 2018, 2020, 2021, 2024). It is the foundation of every induction motor. Without a rotating field, there is no torque. This proof carries 5-8 marks and is one of the highest-scoring derivations in Section B. The 2-phase RMF proof appeared in 2021.

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📝 Key Definitions

Induction Motor: "In a.c. motors, the rotor does not receive electric power by conduction but by induction in exactly the same way as the secondary of a 2-winding transformer receives its power from the primary. That is why such motors are known as induction motors. In fact, an induction motor can be treated as a rotating transformer i.e. one in which primary winding is stationary but the secondary is free to rotate." — Theraja, Art. 34.2

Rotating Magnetic Field: "When stationary coils, wound for two or three phases, are supplied by two or three-phase supply respectively, a uniformly-rotating (or revolving) magnetic flux of constant value is produced." — Theraja, Art. 34.6

Synchronous Speed: "The speed at which the rotating magnetic field revolves is called the synchronous speed ($N_s$). For a machine with $P$ poles: $N_s=120f/P$ r.p.m." — VK Mehta, Art. 8.3

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How a 3-Phase Supply Creates a Rotating Field

Three stator windings are placed 120° apart in space. Each carries current that is 120° apart in time. Each winding creates a pulsating flux along its own axis. The vector sum of these three pulsating fluxes is a rotating flux of constant magnitude.

The key insight: no individual flux rotates. Each flux pulsates along a fixed axis. But the vector sum sweeps around the stator at synchronous speed.

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3-Phase RMF: Mathematical Proof

This is the most important proof in Section B. Know every step.

Setup: Three windings 120° apart in space. Balanced 3-phase supply:

$${\Phi }_R={\Phi }_m\sin \omega t$$

$${\Phi }_Y={\Phi }_m\sin (\omega t-120°)$$

$${\Phi }_B={\Phi }_m\sin (\omega t+120°)$$

Step 1: Resolve into X and Y components.

Take R-phase axis as the +X reference. Y-phase axis is at 120° from X. B-phase axis is at 240° from X.

X-component:

$${\Phi }_x={\Phi }_R\cos 0°+{\Phi }_Y\cos 120°+{\Phi }_B\cos 240°$$

$$={\Phi }_m\sin \omega t+{\Phi }_m\sin (\omega t-120°)\left(-\frac{1}{2}\right)+{\Phi }_m\sin (\omega t+120°)\left(-\frac{1}{2}\right)$$

Step 2: Simplify using trig identity $\sin (A-B)+\sin (A+B)=2\sin A\cos B$:

$${\Phi }_x={\Phi }_m\sin \omega t-\frac{1}{2}{\Phi }_m\cdot 2\sin \omega t\cos 120°$$

$$={\Phi }_m\sin \omega t-\frac{1}{2}{\Phi }_m\cdot 2\sin \omega t\cdot \left(-\frac{1}{2}\right)$$

$$={\Phi }_m\sin \omega t+\frac{1}{2}{\Phi }_m\sin \omega t=\frac{3}{2}{\Phi }_m\sin \omega t$$

Y-component:

$${\Phi }_y={\Phi }_Y\sin 120°+{\Phi }_B\sin 240°$$

$$={\Phi }_m\sin (\omega t-120°)\cdot \frac{\sqrt{3}}{2}+{\Phi }_m\sin (\omega t+120°)\cdot \left(-\frac{\sqrt{3}}{2}\right)$$

Using $\sin (A-B)-\sin (A+B)=-2\cos A\sin B$:

$$=\frac{\sqrt{3}}{2}{\Phi }_m\cdot (-2\cos \omega t\sin 120°)=\frac{\sqrt{3}}{2}{\Phi }_m\cdot (-2\cos \omega t)\cdot \frac{\sqrt{3}}{2}$$

$$=-\frac{3}{2}{\Phi }_m\cos \omega t$$

Step 3: Find magnitude.

$${\Phi }_r=\sqrt{{\Phi }_x^2+{\Phi }_y^2}=\sqrt{{\left(\frac{3}{2}{\Phi }_m\right)}^2({\sin }^2\omega t+{\cos }^2\omega t)}$$

$$\boxed{{\Phi }_r=\frac{3}{2}{\Phi }_m=1.5{\Phi }_m=\text{constant}}$$

Step 4: Find rotation speed.

$$\theta ={\tan }^{-1}\left(\frac{{\Phi }_y}{{\Phi }_x}\right)={\tan }^{-1}\left(\frac{-\cos \omega t}{\sin \omega t}\right)=\omega t-90°$$

$$\frac{d\theta }{dt}=\omega =2\pi f⟹N_s=\frac{120f}{P}\text{ rpm}$$

Conclusion: The 3-phase supply produces a rotating magnetic field of constant magnitude $1.5{\Phi }_m$, rotating at synchronous speed $N_s=120f/P$ rpm. (Proved)

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2-Phase RMF: Mathematical Proof

Setup: Two windings placed 90° apart in space. Balanced 2-phase supply:

$${\Phi }_a={\Phi }_m\sin \omega t\text{(along X-axis)}$$

$${\Phi }_b={\Phi }_m\sin (\omega t-90°)=-{\Phi }_m\cos \omega t\text{(along Y-axis)}$$

Magnitude:

$${\Phi }_r=\sqrt{{\Phi }_a^2+{\Phi }_b^2}=\sqrt{{\Phi }_m^2{\sin }^2\omega t+{\Phi }_m^2{\cos }^2\omega t}={\Phi }_m=\text{constant}$$

Space angle:

$$\theta ={\tan }^{-1}\left(\frac{-\cos \omega t}{\sin \omega t}\right)=\omega t-90°$$

$$\frac{d\theta }{dt}=\omega ⟹N_s=\frac{120f}{P}\text{ rpm}$$

Result: 2-phase produces RMF of magnitude ${\Phi }_m$ at synchronous speed. Compare: 3-phase produces $1.5{\Phi }_m$ (50% stronger).

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Reversing Direction of Rotation

To reverse the direction of rotation of a 3-phase IM, interchange any two of the three supply lines. This changes the phase sequence from R-Y-B to R-B-Y. The rotating field now rotates in the opposite direction.

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🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Prove that a 3-phase supply produces a rotating magnetic field of constant magnitude $1.5{\Phi }_m$ at synchronous speed.

Appeared: 2017 Q1(b), 2018 Q6(b), 2024 Q5(a) — (5-8 marks)

Full Answer:

See 3-Phase RMF: Mathematical Proof above for the complete derivation. The proof shows:

1. Three windings 120° apart in space carry currents 120° apart in time
2. Resolve all three pulsating fluxes into X and Y components
3. X-component = $\frac{3}{2}{\Phi }_m\sin \omega t$
4. Y-component = $-\frac{3}{2}{\Phi }_m\cos \omega t$
5. Magnitude = $\sqrt{(\frac{3}{2}{\Phi }_m{)}^2({\sin }^2\omega t+{\cos }^2\omega t)}=1.5{\Phi }_m$ (constant)
6. Space angle $\theta =\omega t-90°$, so field rotates at $\omega =2\pi f$, giving $N_s=120f/P$ rpm

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🎯 Q2: Show that a 2-phase supply produces a rotating magnetic field at synchronous speed.

Appeared: 2021 Q5(c) — (5 marks)

Full Answer:

Two windings placed 90° apart in space. Balanced 2-phase supply creates fluxes:

$${\Phi }_a={\Phi }_m\sin \omega t\text{(X-axis)},{\Phi }_b=-{\Phi }_m\cos \omega t\text{(Y-axis)}$$

Magnitude: ${\Phi }_r=\sqrt{{\Phi }_m^2{\sin }^2\omega t+{\Phi }_m^2{\cos }^2\omega t}={\Phi }_m$ (constant)

Space angle: $\theta ={\tan }^{-1}(-\cos \omega t/\sin \omega t)=\omega t-90°$

Speed: $d\theta /dt=\omega =2\pi f⟹N_s=120f/P$ rpm

The 2-phase supply produces a rotating field of constant magnitude ${\Phi }_m$ (not $1.5{\Phi }_m$ as in 3-phase) at synchronous speed. (Proved)

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🎯 Q3: What is an electrical machine? Describe the principle of operation of a 3-phase induction motor.

Appeared: 2019 Q5(a) — (4 marks)

Full Answer:

Electrical machine: A device that converts electrical energy to mechanical energy (motor) or mechanical energy to electrical energy (generator) using electromagnetic induction.

3-phase IM operating principle:

1. Three-phase balanced AC supply to the stator creates a rotating magnetic field of magnitude $1.5{\Phi }_m$ at synchronous speed $N_s=120f/P$ rpm.
2. The RMF sweeps across the stationary rotor conductors. The relative motion induces EMF in the rotor bars (Faraday's law).
3. The induced EMF drives current through the short-circuited rotor bars (squirrel cage) or through external resistance (slip-ring motor).
4. Current-carrying rotor conductors in the stator magnetic field experience a force (Lorentz force). This force produces torque.
5. The rotor spins in the direction of the RMF (Lenz's law: rotor tries to reduce relative motion).
6. The motor always runs at $N<N_s$ (slip $s>0$). If $N=N_s$, relative motion is zero, EMF is zero, current is zero, torque is zero. So the motor can never reach synchronous speed.

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Exam Variants

Year	Question	Key Points
2017 Q1(b)	Prove 3-phase RMF	${\Phi }_r=1.5{\Phi }_m$, rotates at $N_s$
2018 Q6(b)	Prove 3-phase RMF	Same proof
2019 Q5(a)	IM operating principle	RMF + Faraday + Lenz
2021 Q5(c)	Prove 2-phase RMF	${\Phi }_r={\Phi }_m$, rotates at $N_s$
2024 Q5(a)	Prove 3-phase RMF (8 marks)	Full vector proof

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⚡ Exam Tips & Common Mistakes

1. State assumptions at the start. Write "three identical windings, 120° apart in space, carrying balanced 3-phase currents 120° apart in time." This earns setup marks.
2. Don't confuse $1.5{\Phi }_m$ with $3{\Phi }_m$. All three phases never peak at the same time. The maximum vector sum is $1.5{\Phi }_m$.
3. Show the trig identity you use. The examiner wants to see $\sin (A-B)+\sin (A+B)=2\sin A\cos B$. Don't skip this step.
4. 2-phase produces ${\Phi }_m$, not $1.5{\Phi }_m$. If asked to compare, state this clearly.
5. Know how to reverse rotation. Swap any two supply lines. One sentence, but it carries 1-2 marks.

🔗 Related Topics

• T-13: Slip & Basics — What happens once the RMF starts spinning
• T-22: DFRT & 1-Phase IM — What happens with only one phase (no RMF, just pulsating field)
• T-15c: Torque-Speed Curves — How torque depends on slip

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