Production of rotating magnetic field

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.

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 $-30^\circ$.

 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 $s = \frac{N_s - N_r}{N_s}$.