--- title: "Rotating magnetic field, slip and standstill torque equation" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:rmf-synchronous-speed-slip-standstill-torque" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/rmf-synchronous-speed-slip-standstill-torque site: BCT Engineering --- # Rotating magnetic field, slip and standstill torque equation Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2071 Magh, 3+5 marks - Video: [Electrical Machines | Lec 6C | Three-Phase Induction Motor | Rotor Quantities: Slip, Torque, etc](https://www.youtube.com/watch?v=s-wCI5jDu3g) — Engineering Devotion - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/rmf-synchronous-speed-slip-standstill-torque - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question What do you mean by rotating magnetic field, synchronous speed and slip of a 3-φ induction motor? Derive the standstill torque equation Ts = K1E2²R2/(R2² + X2²) for 3-φ induction motor. ## Short answer RMF: balanced 3-phase currents in 120° displaced windings produce constant-magnitude rotating field. Ns = 120f/P. Slip s = (Ns − N)/Ns. Standstill torque: T ∝ φ I2 cosφ2; φ ∝ E2, I2 = E2/√(R2²+X2²), cosφ2 = R2/√(R2²+X2²). Hence Ts = K1 E2² R2/(R2²+X2²). ## Answer #### Rotating magnetic field, synchronous speed and slip - **Rotating magnetic field (RMF):** when a balanced 3-phase supply is given to a 3-phase winding whose phases are displaced by 120° in space, the three pulsating fluxes (displaced 120° in time) combine into a resultant flux of **constant magnitude** $1.5\,\phi_m$ that **rotates** around the air gap at constant speed. - **Synchronous speed:** the speed of the RMF, $N_s = \dfrac{120 f}{P}$ rpm. For 50 Hz, 4 poles: 1500 rpm. - **Slip:** the rotor runs at $N < N_s$. The relative speed as a fraction of $N_s$ is the slip: $$ s = \frac{N_s - N}{N_s} $$ At standstill $s = 1$; at synchronous speed $s = 0$; full-load slip is 2–5%. Rotor frequency $f_2 = sf$. #### Derivation of standstill torque Torque is proportional to the flux, the rotor current and the rotor power factor: $$ T \propto \phi\, I_2 \cos\phi_2 $$ At standstill ($s = 1$), per phase: rotor emf $E_2$, resistance $R_2$, reactance $X_2 = 2\pi f L_2$. ```text R2 X2 +--[===]----[~~~]--+ | | (~) E2 I2 --> | | | +------------------+ ``` Rotor impedance and current: $$ Z_2 = \sqrt{R_2^2 + X_2^2}, \qquad I_2 = \frac{E_2}{\sqrt{R_2^2 + X_2^2}} $$ Rotor power factor: $$ \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + X_2^2}} $$ The rotor emf is produced by the stator flux, so $E_2 \propto \phi$, i.e. $\phi = E_2/k_1$. Substituting: $$ \begin{aligned} T_s &= k\,\phi\, I_2\cos\phi_2 \\ &= k\cdot\frac{E_2}{k_1}\cdot\frac{E_2}{\sqrt{R_2^2 + X_2^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + X_2^2}} \\ &= K_1\,\frac{E_2^2 R_2}{R_2^2 + X_2^2} \end{aligned} $$ where $K_1 = k/k_1$. From the power method, $T = P_{ag}/\omega_s$ with $P_{ag} = 3I_2^2R_2$ at $s = 1$, so $K_1 = \dfrac{3}{2\pi N_s/60}$. Notes: 1. $T_s \propto E_2^2 \propto V^2$: starting torque is very sensitive to supply voltage. 2. Starting torque is maximum when $\dfrac{d T_s}{dR_2} = 0$, i.e. $R_2 = X_2$ — achieved by adding rotor resistance in slip-ring motors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the resultant flux of a balanced 3-phase winding with phase flux amplitude phi_m? A. Constant magnitude 1.5 phi_m rotating at synchronous speed B. A pulsating flux of magnitude 3 phi_m in a fixed position C. Constant magnitude phi_m rotating at synchronous speed D. A flux of 1.5 phi_m that stays fixed and alternates in sign Answer: A. The three pulsating fluxes combine into a field of constant magnitude 1.5 phi_m whose axis rotates around the air gap. (Key point: Rotating magnetic field) 2. A rotor runs at N below synchronous speed Ns. What is the slip, and what is its value at standstill? A. s = (Ns - N)/N, equal to 0 at standstill B. s = (Ns - N)/Ns, equal to 1 at standstill C. s = N/Ns, equal to 1 at synchronous speed D. s = (Ns - N)/Ns, equal to 0 at standstill Answer: B. Slip is the relative speed as a fraction of Ns; at standstill N = 0 so s = 1, and at Ns it is 0. (Key point: Slip definition) 3. On what does the induction motor torque depend in the derivation? A. Only the rotor current, T proportional to I2 B. Flux and rotor reactance only C. Flux, rotor current and rotor power factor: T proportional to phi I2 cos(phi2) D. Rotor frequency and slip only Answer: C. Torque is the product of flux, rotor current and the in-phase (power factor) component, T ~ phi I2 cos(phi2). (Key point: Torque proportionality) 4. According to the answer, starting torque Ts = K1 E2^2 R2/(R2^2 + X2^2) is maximum when: A. R2 is very large compared with X2 B. R2 = 0 C. X2 = 0 D. R2 = X2 Answer: D. Setting dTs/dR2 = 0 gives R2 = X2, which is why adding rotor resistance in slip-ring motors improves starting torque. 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