Final answer:
The speed at which the magnetic field of the stator is rotating is 1500 rpm, and the speed of the rotor when the slip is 0.04 is 1440 rpm.
Step-by-step explanation:
The subject of this question is Engineering, specifically relating to electrical engineering and the operation of induction motors. The rotational speed of a magnetic field in a stator for a 4-pole, 3-phase induction motor operating at a supply frequency of 50 Hz can be calculated using the formula for synchronous speed, which is N_s = 120f / P, where N_s is the synchronous speed in revolutions per minute (rpm), f is the frequency in hertz (Hz), and P is the number of poles. Thus, the synchronous speed N_s is 1500 rpm.
The speed of the rotor is given by N = N_s (1 - s), where s is the slip. With a slip of 0.04, the speed of the rotor N would be 1440 rpm.
The speed at which the magnetic field of the stator is rotating can be calculated using the equation:
Speed = (120 * Frequency) / Number of poles
In this case, the frequency is 50 Hz and the number of poles is 4. Plugging in these values into the equation, we get:
Speed = (120 * 50) / 4 = 1500 RPM
When the slip is 0.04, the speed of the rotor can be calculated as:
Speed of rotor = (1 - Slip) * Speed of magnetic field
Plugging in the values, we get:
Speed of rotor = (1 - 0.04) * 1500 = 1440 RPM