Why does an induction motor draw more current when the load is increased?
The concept of back EMF works well to explain the relationship among speed, load torque and current in a DC motor, but it doesn't work as well for an induction motor. The preferred equivalent circuit is shown below.

R1 and X1 are the resistance of the stator and the part of the inductance that does not provide a useful magnetic field. Bm is the inductance that provides the stator magnetic field. Gc is a resistance that represents the losses in the stator's iron core. X2 represents the inductance of the rotor. R2/s is a variable resistance that represents the resistance of the rotor and also, most importantly, the  the mechanical load.

At no load, the rotor of an induction motor turns at a synchronous speed, the speed of the motor's rotating magnetic field. The speed in revolutions per minute (RPM) is RPM = 120f/P where f is the power frequency (Hz) and P is the number of magnetic poles formed by the stator winding configuration. R can be any even number (poles are always north/south pairs). Usually P is 2 or 4, but 6 pole motors are not uncommon, higher numbers of poles are less common and usually found only in large motors.

When the motor is loaded, the operating speed decreases. The difference between the loaded speed and the synchronous speed is called slip (s). Slip is expressed as a fraction of synchronous speed. At rated load, the slip is generally 2 or 3 percent of the synchronous speed and s = 0.02 or 0.03.

When the slip is zero, R2/s in the diagram is R2/0 or infinite. Therefore, at no load, the current in the motor is determined by R1 and X1 in series with the magnetizing circuit, Gc and Xm. As the motor is loaded, the slip increases and R2/s decreases causing the motor to draw more current.



The diagram represents one phase of a three-phase induction motor. Two similar diagrams are used to represent a single-phase motor.