Operation of Induction Motors and their Back EMF
I must have some fundamental misunderstanding (probably about back-emf)


Yes, the concept of back emf doesn't work very well for induction motors.

The diagram below is used to explain the operation of a three-phase induction motor. It represents one of three phases of a wye-connected motor.


Image from Fitzgerald, Kingsley, Umans, Electric Machinery, 4th ed

R1, X1, R2 and X2 represent the Stator and rotor resistance and reactance.  Bm and Gc represent magnetizing reactance and the mechanism of power loss in the iron core. R2(1-s)/s is a variable resistance that represents the mechanism of converting electrical power to mechanical power. The slip, s is the slip / synchronous RPM. Note that the rotor circuit, is shown as being directly connected to the stator. An induction motor is like a transformer. Here it is assumed to be a 1:1 transformer and the secondary winding has been omitted.

When the motor rotor is not turning, the instant after power is applied to start the motor s = 1 and R2(1-s)/s = 0. The current is determined by R1 plus X1 plus the parallel combination of Gc, Bm and X2 + R2. The current is quite high, typically on the order of six times the motor’s full-load running current. Since the current in the magnetizing branch Im is considerably less than the motor’s full-load current most of the stator current also flows in the rotor. In actuality the effective stator to rotor turns ratio causes the rotor current to be perhaps 10 to 100 times the stator current.

The current in the rotor causes torque even though R2(1-s)/s = 0 and the electrical power into the motor is all converted to heat when the rotor is not turning. Remember mechanical power require both force and motion. No mechanical power is produced until the rotor begins to turn.

When the rotor begins to turn, slip decreases and R2(1-s)/s increases. The current decreases, but since R2(1-s)/s is no longer zero electrical power is converted to mechanical power. When the motor approaches the normal operating speed, s is about 0.03 and R2(1-s)/s is about 32 x R2. At that point most of the electrical power going to the rotor is converted to mechanical power and only about 3% is lost in R2.

See also my answers to the following:

Why does an induction motor draw more current when the load is increased?

When load increases in rotor of induction motor how does stator draws more current?

Torque-Speed Equation for Induction Motor