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The permeable iron alloy cores of induction motor rotors are formed from thin laminated sheets, designed to prevent eddy current flow and thus "reduce losses", but isn't the whole point of an induction motor rotor (e.g., squirrel cage) to have eddy currents?

The eddy currents are what result in the Lorentz force exerted on the moving charges in the rotor and thus cause the motor to spin. Larger currents result in a larger torque, or less current for the same torque. If the core wasn't laminated, the same stator magnetic field would induce a larger rotor current, which would result in a larger magnetic force. More current would also be drawn by the stator too.

What makes eddy currents in the rotor so desirable (they make the motor move), but eddy currents in the rotor core so undesirable? Is it because higher resistance reduces the power losses:

$$P = \frac{\varepsilon^2}{R} \propto \frac{1}{R}\cdot\left|\frac{d\Phi}{dt}\right|^2$$

Increasing the resistance of parts of the rotor decreases power loss there, but also reduces the overall torque for the same stator field. This seems like a trade-off. Why does this trade-off fall in favor of high resistance rotor core and low resistance rotor cage?

To summarise: we clearly want the cage to have low resistance. What justification for low resistance applies to the cage but not the magnetic core? If you didn't use silicon steel, the iron material would be about 5 times less conductive than a copper cage, but the larger cross sectional area might make the resistance still less!

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[an] iron material would be about 5 times less conductive than a copper cage, but the larger cross sectional area might make the resistance still less!

You're missing an important fact: skin depth depends unfavorably on permeability. While the skin depth in copper or aluminum might be around ~1cm, it's a fraction of a mm for steel! If the core were solid, current would only flow in the thin surface layer, at very high resistance, yielding little torque and pitiful efficiency. (It will still spin: solid conductors experience a torque in a rotating magnetic field, period. It doesn't matter that a "squirrel cage" geometry is used, nor that it be filled with a permeable material; those just make the design more economical. Indeed people have demonstrated motors at exceptional speed (10s of kHz!), by simply spinning a small, very smooth rod in a pair of coils; this isn't even hard to do, really, it just needs a very free bearing support and as close to zero torque load as possible!)

There are also reasons not to use a solid shell of metal, of given thickness. It would work, I think, if a foil/sheet were used, but this is harder to manufacture than the spokes of a "squirrel cage" design (particularly as it's done today, with molten aluminum cast directly into the core). In particular, you'd have to bond it to the core somehow, lest it tear off or slip or whatever.

The inductance of the rotor is very important. The difference between driven and slip frequencies is the slip rate in the rotating frame. Which is to say: induced current rotates through the rotor at single-digit Hz -- an impressively low time constant for an assembly of bulk metals! Constructing it with slots (so the core faces into the air gap against the stator) is a necessary part of this achievement.

There are also reasons not to make the rotor too good! Very low resistance makes starting or locked-rotor current exceptionally high; it may affect torque pulsations, or harmonic distortion (in the line current, or output torque); and the torque curve in general may have desirable characteristics that are tuned this way, in part. Many of these are a direct compromise with efficiency, which kinda sucks in some respects, but it is what it is, and, all the more reason to choose the right machine for the job!

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  • \$\begingroup\$ Another thing that I missed is that torque is proportional to rotor current, not voltage, so P=V^2/R as suggested in my question is misleading (suggesting V is fixed and P is inverse to R). For a given torque, the current is fixed and thus losses are proportional to R, not inversely proportional. As you say, even ignoring skin effect, you get very poor torque per power dissipation for higher resistivity material and/or current densities closer to the rotation axis. Motors are surprisingly hard! \$\endgroup\$ Commented Sep 26, 2023 at 3:46
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You actually want only the copper of the rotor to conduct the current because it will do so far more effectively than a solid iron core. An iron/steel core will get very warm if not laminated and act as a parallel load that wastes power. Also a solid lump of metal will not produce an effectively "shaped" counter-flux that can produce torque.

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  • \$\begingroup\$ Side note: If you inspect a rotor of an induction motor and find no copper, then there's aluminum. Aluminum conductors are directly cast into grooves of the laminated steel core in most cases. \$\endgroup\$
    – Ariser
    Commented Sep 24, 2023 at 15:04
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    \$\begingroup\$ @PersonWithName Eddy currents in the middle of the core will not be coupled as strongly to the stator field, so they'll waste energy without doing very much. You want the current to be concentrated around the outer edge of the rotor. \$\endgroup\$
    – Hearth
    Commented Sep 24, 2023 at 15:17
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    \$\begingroup\$ @PersonWithName The loops are smaller closer to the center, so they enclose less area. It's a simple geometric argument. \$\endgroup\$
    – Hearth
    Commented Sep 24, 2023 at 15:19
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    \$\begingroup\$ @PersonWithName No, a good conductor doesn't waste much, because it establishes a voltage with opposite direction to the current in the cage reducing the current greatly. In a massive cylinder the currents can take paths which don't yield any torque while still generating heat. Forget about skin effect at 50 Hz (not true but in first approximation). \$\endgroup\$
    – Ariser
    Commented Sep 24, 2023 at 16:02
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    \$\begingroup\$ @Ariser they are called synchronous induction motors but I like Schleifringmotoren because it is more descriptive and, the English term doesn't produce synchronicity when the slip rings are shorted to each other = i.e. you get "slip" (not to be confused with slip ring!!). \$\endgroup\$
    – Andy aka
    Commented Sep 24, 2023 at 17:01

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