--- title: "Production of rotating magnetic field" subject: "Electrical Machines" chapter: 5 key: "semester_4/electrical_machines:rotating-magnetic-field-production" source: https://www.bctengineer.com/q/semester_4/electrical_machines/rotating-magnetic-field-production site: BCT Engineering --- # Production of rotating magnetic field Electrical Machines · Chapter 5: Three Phase Induction Machines - Asked in: 2071 Bhadra, 8 marks - Video: [Understanding RMF | The driving force behind every AC machine](https://www.youtube.com/watch?v=wqrGHeuxUvI) — Sabin Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/rotating-magnetic-field-production - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5 ## Question Explain with necessary vector diagram how rotating magnetic field is produced in a three phase induction motor. Also explain how this rotating magnetic field helps the motor to rotate. ## Short answer Three identical stator coils displaced 120 degrees in space carry balanced three-phase currents, producing fluxes 120 degrees apart in time. Their vector sum has constant magnitude 1.5 phi_m and rotates at Ns = 120f/P rpm; it induces emf in short-circuited rotor bars, causing currents and torque by Lenz's law, so rotor follows the field below synchronous speed. ## Answer #### Production of rotating magnetic field A three-phase stator winding has three identical coils a, b, c placed $120^\circ$ apart in space (axes at $0^\circ$, $120^\circ$ and $240^\circ$, measured anticlockwise from the axis of coil a). A balanced three-phase supply makes their fluxes alternate with a $120^\circ$ time displacement: $$ \phi_a = \phi_m \sin\omega t, \quad \phi_b = \phi_m \sin(\omega t - 120^\circ), \quad \phi_c = \phi_m \sin(\omega t - 240^\circ) $$ Each flux acts along the axis of its own coil. The resultant flux is the vector sum at each instant. | Instant | $\phi_a$ | $\phi_b$ | $\phi_c$ | Resultant | |---|---|---|---|---| | $\omega t = 0^\circ$ | 0 | $-0.866\phi_m$ | $+0.866\phi_m$ | $1.5\phi_m$ at $-90^\circ$ | | $\omega t = 60^\circ$ | $+0.866\phi_m$ | $-0.866\phi_m$ | 0 | $1.5\phi_m$ at $-30^\circ$ | | $\omega t = 120^\circ$ | $+0.866\phi_m$ | 0 | $-0.866\phi_m$ | $1.5\phi_m$ at $+30^\circ$ | | $\omega t = 180^\circ$ | 0 | $+0.866\phi_m$ | $-0.866\phi_m$ | $1.5\phi_m$ at $+90^\circ$ | For example, at $\omega t = 60^\circ$ the two fluxes of $0.866\phi_m$ (along $+a$ and along $-b$) are $60^\circ$ apart, so the resultant is $2\times 0.866\phi_m\cos 30^\circ = 1.5\phi_m$, midway between them at $-30^\circ$. ```text Coil axes (anticlockwise from a): b (120 deg) \ \ +--------> a (0 deg) / / c (240 deg) Resultant, magnitude 1.5 phi_m: wt = 0 -> -90 deg wt = 60 -> -30 deg wt = 120 -> +30 deg wt = 180 -> +90 deg (turns anticlockwise, from axis a towards b: sequence a-b-c) ``` The resultant has a **constant magnitude of $1.5\phi_m$** and rotates by $60^\circ$ electrical for every $60^\circ$ of time, so it makes one revolution per cycle for a 2-pole winding. For $P$ poles, its speed is $$ N_s = \frac{120 f}{P}\ \text{rpm} $$ The direction of rotation follows the phase sequence a-b-c; swapping any two supply lines reverses it. #### How the RMF makes the rotor rotate 1. The RMF sweeps across the stationary rotor conductors and, by Faraday's law, induces an emf in them. 2. The rotor bars are short-circuited, so a rotor current flows. 3. A current-carrying conductor in a magnetic field experiences a force ($F = BIl$). The forces on all conductors give a torque. 4. By Lenz's law, the rotor tries to reduce the relative speed between itself and the field, so it rotates **in the direction of the RMF**. 5. The rotor speed $N_r$ stays below $N_s$. If $N_r = N_s$, there would be no relative motion, no emf and no torque. The difference gives slip $s = \frac{N_s - N_r}{N_s}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a balanced three-phase winding, the resultant flux has magnitude: A. phi_m, alternating B. 0.866 phi_m C. 3 phi_m D. 1.5 phi_m, constant Answer: D. The vector sum is always 1.5 phi_m, constant in size, rotating. (Key point: Resultant magnitude) 2. The RMF speed in a P-pole machine is: A. N_s = 120f/P rpm B. N_s = 60P/f rpm C. N_s = f/P rpm D. N_s = 120P/f rpm Answer: A. Speed is 120f/P. (Key point: Speed of RMF) 3. How can the direction of rotation be reversed? A. Swapping any two supply lines B. Changing the supply frequency C. Reducing the voltage D. Adding rotor resistance Answer: A. Direction follows phase sequence, so swapping two lines reverses it. (Key point: Direction reversal) 4. Why does the rotor rotate in the direction of the RMF? A. Because of Lenz's law the rotor tries to reduce relative speed with the field B. Because the rotor is magnetised permanently C. Because the field pushes bars sideways away from it D. Because the rotor current is DC Answer: A. Induced current opposes the relative motion, so the rotor follows the field. 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