Why rotor core loss is negligible

The core loss (hysteresis plus eddy current) in a magnetic material depends on the frequency of the alternating flux in it:

$$P_h\propto f\,B_m^{1.6}, \qquad P_e\propto f^2B_m^2$$

Reason:

1. In an induction motor the stator field rotates at synchronous speed, and the rotor turns at nearly the same speed. The flux therefore moves relative to the rotor at the slip speed only, so the rotor core frequency is $f_r = sf$.
2. At normal running the slip is only 2–5%. For a 50 Hz supply, $f_r = (0.02\ \text{to}\ 0.05)\times 50 = 1$–$2.5$ Hz.
3. The core loss is proportional to $f$ (hysteresis) and $f^2$ (eddy current), so at 1–2.5 Hz the rotor core loss is a tiny fraction of the stator core loss, which occurs at the full supply frequency of 50 Hz.

Example: at $s = 0.04$, $f_r = 2$ Hz. The rotor eddy-current loss is $(2/50)^2 = 0.0016$ of the loss at 50 Hz, i.e. 0.16%.

Hence in the equivalent circuit the core-loss branch is shown only on the stator side, and the rotor iron loss is neglected. The rotor copper loss $I_2^2R_2$ is the only significant rotor loss. (At starting, $s = 1$ and the rotor frequency is 50 Hz, but the rotor core loss does not matter then because the starting period is short.)