How a Squirrel-Cage Rotor Produces Torque in an Induction Motor

A four-pole induction motor connected to a 50 Hz supply has a stator magnetic field rotating at 1500 rpm. Yet its squirrel-cage rotor might settle at just 1440 rpm.

Why doesn’t the rotor catch up? And why would catching up actually stop the motor from producing torque?

The squirrel-cage rotor produces torque through electromagnetic induction. The rotating magnetic field of the stator moves relative to the conducting bars in the rotor, inducing an electromotive force (EMF) and current in the rotor. These currents experience magnetic forces in the stator’s field. The tangential components of those forces combine to produce torque.

The rotor follows the rotating field, but must remain slightly slower while producing torque as an induction motor. The difference in their speeds is called slip.

01What is actually rotating?

Start with the stator, the stationary outer part of the motor.

Inside the stator are three-phase windings connected to an alternating current supply. The three alternating currents are separated in phase by 120°. The windings are also arranged around the stator so their combined magnetic field changes direction continuously.

Rather than simply growing stronger and weaker in one direction, the resultant magnetic field rotates around the motor.

This is called a rotating magnetic field.

The rotor sits inside the stator, separated from it by a small air gap. It is attached to a shaft, allowing it to rotate.

The rotor contains conducting bars running parallel to the shaft. At each end, these bars are connected by conducting rings called end rings.

Together, the bars and end rings form a structure resembling a cage. This is where the name squirrel-cage rotor comes from.

Importantly, the rotor has no external electrical connection. Its current is induced by the rotating magnetic field.

How does the rotating field form?

Imagine looking directly into the motor from the shaft end.

The stator windings are distributed around the rotor. As the three-phase currents change, the magnetic poles produced by those windings change in strength and position.

The combined magnetic field sweeps around the air gap.

The field rotates at a particular speed called the synchronous speed.

For a four-pole motor connected to a 50 Hz supply, the magnetic field completes 25 mechanical revolutions per second, equivalent to 1500 rpm.

Notice that 50 Hz does not mean 50 mechanical revolutions per second. The number of magnetic poles affects the mechanical rotation speed.

We’ll calculate this properly shortly.

02The experiment: why does the rotor lag behind?

Consider a teaching demonstration designed to test whether the rotor actually rotates at the same speed as the stator’s magnetic field.

Setting up the apparatus

The apparatus consists of:

• A stationary stator containing three-phase windings.
• A squirrel-cage rotor positioned concentrically inside the stator.
• A three-phase AC supply connected to the stator windings.
• A tachometer attached to the rotor shaft to measure rotational speed.
• An adjustable mechanical brake attached to the shaft to vary the load.
• A frequency meter to establish the supply frequency.

The rotor receives no direct electrical supply. Only the stator is electrically connected.

What should happen?

Initially, the rotor is stationary.

When the three-phase supply is switched on, the stator produces a rotating magnetic field. The field immediately begins rotating at synchronous speed.

The rotor starts accelerating in the same direction.

A reasonable first prediction is that the rotor will eventually reach exactly the same speed as the magnetic field.

But that prediction is incorrect.

The tachometer shows that the rotor settles at a speed slightly below synchronous speed when the motor reaches a steady operating condition.

Now increase the mechanical load using the brake.

The rotor slows slightly, even though the stator’s magnetic field continues rotating at the same synchronous speed.

This observation gives us an important clue: the speed difference between the field and rotor changes when the mechanical load changes.

Improving the measurement

A basic demonstration that only observes the motor running without a significant load may not reveal the small difference between the two speeds.

A better experiment measures the rotor speed accurately and applies a controlled mechanical load.

The following values are illustrative, not measurements from a particular motor.

Assume a four-pole motor supplied at 50 Hz.

The pattern is clear.

As the mechanical load increases, the rotor slows down. Its slip increases, and the frequency of the induced rotor current increases.

This supports the idea that relative motion between the rotating field and rotor is essential to induction.

The tachometer does not directly measure rotor current. Rather, the rotor current frequency can be calculated from the measured slip.

03How does induced current actually produce torque?

We now have two important pieces of information:

• The stator produces a rotating magnetic field.
• The rotor contains conducting bars connected into closed electrical circuits.

The missing step is understanding how these produce a turning force.

Step 1: The rotating field induces an EMF

As the stator’s magnetic field sweeps past the rotor bars, the magnetic flux through the conducting loops changes.

According to Faraday’s law of electromagnetic induction, a changing magnetic flux induces an EMF.

Because the rotor bars are connected by end rings, there are closed conducting paths through which current can flow.

For example, current can travel along one rotor bar, around an end ring, and back through another bar.

This produces induced currents throughout the rotor cage.

Step 2: The induced current experiences a magnetic force

A current-carrying conductor in a magnetic field experiences a force.

For a straight conductor perpendicular to a uniform magnetic field:

\[ F=BIL \]

Where:

• \(F\) is the magnetic force in newtons (N).
• \(B\) is the magnetic flux density in tesla (T).
• \(I\) is the current in the conductor in amperes (A).
• \(L\) is the length of the conductor inside the magnetic field in metres (m).

The rotor bars carry current while sitting inside the stator’s magnetic field.

The interaction between the current and magnetic field produces forces on the bars.

Because the bars run parallel to the shaft and the magnetic field has a radial component, the resulting forces have tangential components around the rotor.

Step 3: The forces produce a turning effect

The rotor bars are positioned at a distance from the central axis.

A tangential force acting at this radius produces torque.

For a force acting perpendicular to the radius:

\[ \tau=rF \]

Where:

• \(\tau\) is torque in newton metres (Nm).
• \(r\) is the distance from the rotation axis in metres (m).
• \(F\) is the tangential force in newtons (N).

The forces acting on the different rotor bars combine to produce a net electromagnetic torque.

This torque accelerates the rotor in the direction of the rotating magnetic field.

Why does the rotor move in the same direction as the field?

This is where Lenz’s law becomes important.

Lenz’s law states that an induced current flows in a direction such that its magnetic effect opposes the change in magnetic flux that produced it.

In this situation, the induced rotor currents produce forces that tend to reduce the relative motion between the rotor and the rotating magnetic field.

The rotor is therefore pulled in the direction of the rotating field.

A common misconception is that Lenz’s law means the rotor must oppose the magnetic field’s rotation.

That is not quite right.

The rotor opposes the relative motion responsible for the induction. When the rotor is moving slower than the field, the resulting torque accelerates it forwards.

This distinction becomes particularly useful when we consider what happens if the rotor moves faster than the field.

04Understanding slip

Slip measures the difference between the synchronous speed of the stator’s magnetic field and the actual rotational speed of the rotor.

It is essential to understanding how an induction motor produces torque.

1. Calculate synchronous speed

The synchronous speed is given by:

\[ n_s=\frac{120f_s}{P} \]

Where:

• \(n_s\) is synchronous speed in revolutions per minute (rpm).
• \(f_s\) is the stator supply frequency in hertz (Hz).
• \(P\) is the number of magnetic poles.

For example, consider a four-pole motor connected to a 50 Hz supply.

\[ n_s=\frac{120\times50}{4} \]

\[ \boxed{n_s=1500\text{ rpm}} \]

The magnetic field rotates at 1500 rpm.

This does not mean the rotor must rotate at 1500 rpm.

2. Calculate slip

Slip is calculated using:

\[ s=\frac{n_s-n_r}{n_s} \]

Where:

• \(s\) is slip, expressed as a decimal or percentage.
• \(n_s\) is synchronous speed in rpm.
• \(n_r\) is actual rotor speed in rpm.

For a motor operating normally, the rotor speed is below synchronous speed, so slip is positive.

If the rotor is stationary:

\[ s=\frac{1500-0}{1500}=1 \]

The slip is 100%.

If the rotor reaches synchronous speed:

\[ s=\frac{1500-1500}{1500}=0 \]

The slip is zero.

3. Calculate rotor current frequency

The frequency of the induced current in the rotor depends on slip.

\[ f_r=sf_s \]

Where:

• \(f_r\) is rotor current frequency in Hz.
• \(s\) is slip.
• \(f_s\) is stator supply frequency in Hz.

For example, if slip is 4% and the supply frequency is 50 Hz:

\[ f_r=0.04\times50 \]

\[ \boxed{f_r=2\text{ Hz}} \]

This means the induced rotor current has a frequency of 2 Hz, not 50 Hz.

At standstill, the rotor current frequency equals the supply frequency because slip is 100%.

At synchronous speed, the rotor current frequency falls to zero.

Why can’t an induction motor run at synchronous speed?

Imagine the rotor reaches exactly 1500 rpm while the magnetic field also rotates at 1500 rpm.

There is now no relative motion between the rotor and the rotating magnetic field.

Consequently:

1. The magnetic flux through the rotor loops is no longer changing due to relative motion.
2. No EMF is induced in the rotor bars.
3. The induced rotor current falls to zero.
4. The electromagnetic torque produced by the induction mechanism falls to zero.

The rotor can no longer receive the torque needed to maintain its motion against a mechanical load.

In a real motor, the rotor therefore operates slightly below synchronous speed. Even when the external mechanical load is very small, some slip is needed to produce torque that overcomes friction and air resistance.

A useful analogy is a conveyor belt carrying a box. The belt needs relative motion against the box to transfer force through friction. If both move at exactly the same speed, there is no relative sliding.

The analogy has limits: an induction motor transfers torque through electromagnetic forces, not friction. However, it captures why relative motion matters.

What happens when the mechanical load increases?

Suppose the motor is initially operating at 1470 rpm, with a synchronous speed of 1500 rpm.

Now the mechanical load increases.

The rotor briefly slows down. This increases the relative speed between the rotor and magnetic field.

The greater relative motion increases the induced EMF and rotor current, allowing the motor to produce more electromagnetic torque.

The motor settles at a new operating speed where the electromagnetic torque balances the mechanical load and rotational losses.

The important relationship is:

Increasing mechanical load → increasing slip → greater induced rotor current → greater electromagnetic torque.

This describes the normal operating region of an induction motor. Torque does not increase indefinitely with slip, and the exact relationship depends on the motor’s electrical characteristics.

05Worked examples

06How does a squirrel-cage motor differ from other motors?

Not every motor produces rotor torque through induced current.

The distinction is how the electrical current or magnetic field is established in the rotor.

Comparison with a brushed DC motor

In a brushed DC motor, current is supplied directly to the rotor windings.

The interaction between this current and the external magnetic field produces a torque, similar to the current-carrying coil mechanism.

The commutator reverses the current in the armature windings at appropriate points in the rotation, allowing the torque to continue acting in the same rotational direction.

This differs from the squirrel-cage motor, where the rotor current is induced rather than supplied through brushes.

For a closer look at the underlying magnetic turning effect, see Torque on a Current-Carrying Coil for HSC Physics.

Comparison with a synchronous motor

A synchronous motor uses a rotor with its own magnetic field.

This field locks into alignment with the rotating magnetic field produced by the stator.

The rotor then rotates at synchronous speed, even while producing torque.

This is different from the squirrel-cage induction motor, which needs relative motion to induce rotor current.

The distinction is useful in exam questions:

• If rotor current is induced by a rotating magnetic field, think induction motor and slip.
• If the rotor has its own magnetic field that locks to the stator field, think synchronous motor.
• If current is supplied directly to the rotor through brushes and a commutator, think brushed DC motor.

For the general relationship between a turning force, its perpendicular distance from an axis, and torque, see Torque in HSC Physics: Moment Arms and Rotational Effect.

07Questions and solutions

Question 1

A four-pole squirrel-cage induction motor operates on a 50 Hz AC supply. Its rotor rotates at 1425 rpm.

Calculate:

a) The synchronous speed.

b) The percentage slip.

c) The frequency of the induced rotor current.

Solution 1

The synchronous speed is 1500 rpm, the slip is 5%, and the rotor current frequency is 2.5 Hz.

Part a: Synchronous speed

Use:

\[ n_s=\frac{120f_s}{P} \]

Substitute:

\[ n_s=\frac{120\times50}{4} \]

\[ \boxed{n_s=1500\text{ rpm}} \]

Part b: Percentage slip

Use:

\[ s=\frac{n_s-n_r}{n_s} \]

Substitute:

\[ s=\frac{1500-1425}{1500} \]

\[ s=0.05 \]

Therefore:

\[ \boxed{s=5\%} \]

Part c: Rotor current frequency

Use:

\[ f_r=sf_s \]

\[ f_r=0.05\times50 \]

\[ \boxed{f_r=2.5\text{ Hz}} \]

The rotor current frequency is lower than the stator supply frequency because the rotor is moving in the same direction as the rotating field, reducing their relative speed.

Question 2

A four-pole induction motor is connected to a 50 Hz supply.

A student records the following rotor speeds:

• Light mechanical load: 1485 rpm.
• Increased mechanical load: 1410 rpm.

a) Calculate the slip at each operating condition.

b) Calculate the rotor current frequency at each operating condition.

c) Explain why the rotor slows down when the mechanical load increases.

d) A student claims that because slip increases from 1% to 6%, the rotor current and torque must both increase by exactly a factor of six. Explain why this conclusion is not justified.

Solution 2

The slip increases from 1% to 6%, and the rotor current frequency increases from 0.5 Hz to 3 Hz. The increased slip allows the motor to develop greater torque, but the exact increase in current and torque cannot be determined from slip alone.

Part a: Calculate the slip

First, calculate synchronous speed:

\[ n_s=\frac{120\times50}{4} \]

\[ n_s=1500\text{ rpm} \]

For light load:

\[ s=\frac{1500-1485}{1500} \]

\[ \boxed{s=0.01=1\%} \]

For increased load:

\[ s=\frac{1500-1410}{1500} \]

\[ \boxed{s=0.06=6\%} \]

Part b: Calculate rotor current frequency

For light load:

\[ f_r=0.01\times50 \]

\[ \boxed{f_r=0.5\text{ Hz}} \]

For increased load:

\[ f_r=0.06\times50 \]

\[ \boxed{f_r=3.0\text{ Hz}} \]

Part c: Explain the change in rotor speed

The increased mechanical load initially slows the rotor.

This increases the relative speed between the rotor and rotating magnetic field, producing a greater rate of change of magnetic flux through the rotor loops.

The induced EMF and rotor current increase, allowing the motor to produce greater electromagnetic torque.

The motor reaches a new equilibrium at a lower rotational speed.

Part d: Evaluate the student’s claim

The student’s conclusion is not justified.

Although slip increases by a factor of six, it does not follow that rotor current and torque must also increase by exactly six times.

The relationship between slip, induced current, and torque depends on the rotor’s electrical resistance, reactance, and the motor’s operating conditions.

In the normal operating region, torque often increases approximately with slip. However, this is not a universal linear relationship.

The information provided establishes the direction of change, not an exact torque ratio.

Question 3

A squirrel-cage rotor is modified so that both conducting end rings are completely removed. The individual rotor bars remain in place, but none of the bars is electrically connected to another.

The stator continues producing the same rotating magnetic field.

Assume the rotor core is laminated and any eddy-current effects in the core are negligible.

a) Will an EMF still be induced in the rotor bars?

b) Will significant current flow through the rotor?

c) Will the motor produce significant electromagnetic torque through the normal squirrel-cage induction mechanism? Explain.

Solution 3

An EMF can still be induced in the individual rotor bars, but significant current and electromagnetic torque will not be produced through the normal squirrel-cage mechanism.

Part a: Induced EMF

Yes.

The rotating magnetic field still moves relative to the conducting bars.

Therefore, the magnetic flux associated with the rotor’s conducting paths changes, and an EMF can be induced along the individual bars.

Removing the end rings does not stop the magnetic field from inducing an EMF.

Part b: Rotor current

No significant sustained current can flow through the bars.

The original end rings connected the bars into closed conducting paths. Without those connections, the bars are electrically isolated.

Although an EMF exists, there is no complete conducting path for a sustained induced current.

Part c: Electromagnetic torque

The motor will not produce significant electromagnetic torque through the normal squirrel-cage induction mechanism.

A current-carrying conductor in a magnetic field experiences a magnetic force.

However, if there is no significant rotor current, the magnetic forces responsible for the induction motor’s torque are absent.

This demonstrates why the end rings are essential. They do not simply hold the rotor bars together mechanically; they complete the electrical circuits that allow induced current to flow.

Question 4

Two squirrel-cage induction motors operate under different conditions.

Their specifications and operating speeds are:

a) Calculate the synchronous speed of each motor.

b) Calculate the slip of each motor as a percentage.

c) Calculate the rotor current frequency of each motor.

d) Both rotors operate 60 rpm below their respective synchronous speeds. A student concludes that both motors must therefore have the same slip and produce the same torque. Evaluate this reasoning.

Solution 4

Motor A has a slip of 4% and a rotor current frequency of 2 Hz. Motor B has a slip of 5% and a rotor current frequency of 3 Hz. Their equal absolute speed differences do not mean they have equal slip or torque.

Part a: Synchronous speeds

For Motor A:

\[ n_s=\frac{120\times50}{4} \]

\[ \boxed{n_{s,A}=1500\text{ rpm}} \]

For Motor B:

\[ n_s=\frac{120\times60}{6} \]

\[ \boxed{n_{s,B}=1200\text{ rpm}} \]

Part b: Percentage slip

For Motor A:

\[ s_A=\frac{1500-1440}{1500} \]

\[ \boxed{s_A=0.04=4\%} \]

For Motor B:

\[ s_B=\frac{1200-1140}{1200} \]

\[ \boxed{s_B=0.05=5\%} \]

Part c: Rotor current frequency

For Motor A:

\[ f_{r,A}=0.04\times50 \]

\[ \boxed{f_{r,A}=2\text{ Hz}} \]

For Motor B:

\[ f_{r,B}=0.05\times60 \]

\[ \boxed{f_{r,B}=3\text{ Hz}} \]

Part d: Evaluate the student’s reasoning

The student’s reasoning is incorrect.

Both rotors are 60 rpm below their respective synchronous speeds, but slip is calculated as a fraction of synchronous speed, not simply as the difference in rpm.

Motor A has a synchronous speed of 1500 rpm, while Motor B has a synchronous speed of 1200 rpm.

Consequently, the same 60 rpm difference represents different percentages of their respective synchronous speeds.

Their rotor current frequencies also differ because rotor frequency depends on both slip and supply frequency.

Finally, we cannot conclude that either motor produces greater torque.

The exact torque depends on additional factors, including rotor resistance, reactance, magnetic flux, and motor design.

The key lesson is that an absolute speed difference is not enough to compare the operating conditions of different induction motors.

Question 5

A four-pole squirrel-cage induction machine is connected to a 50 Hz three-phase electrical grid.

The stator’s rotating magnetic field moves clockwise at 1500 rpm.

An external turbine drives the rotor clockwise at 1530 rpm.

Assume that:

• The grid supplies the magnetising current required to establish the stator magnetic field.
• The rotor and magnetic field continue rotating clockwise.
• A small constant mechanical load opposes the rotor’s rotation.
• The machine is capable of operating in both induction motor and induction generator modes.

a) Calculate the signed slip.

b) Calculate the rotor current frequency.

c) Determine the direction of the electromagnetic torque on the rotor. Is the machine operating as a motor or generator?

d) The turbine is disconnected while the machine remains connected to the grid. Explain what happens as the rotor slows down through 1500 rpm.

e) Can the machine remain at exactly 1500 rpm while supplying the mechanical load? Explain why or why not.

Solution 5

The signed slip is -2%, the rotor current frequency is 1 Hz, and the machine operates as an induction generator while the turbine drives it above synchronous speed.

After the turbine is disconnected, the rotor slows through synchronous speed and eventually settles below 1500 rpm, operating as a motor.

Part a: Calculate signed slip

Take clockwise rotation as the positive direction.

The synchronous speed is 1500 rpm, and the rotor speed is 1530 rpm.

Using:

\[ s=\frac{n_s-n_r}{n_s} \]

Substitute:

\[ s=\frac{1500-1530}{1500} \]

\[ s=\frac{-30}{1500} \]

\[ \boxed{s=-0.02=-2\%} \]

The negative sign indicates that the rotor is rotating faster than the stator’s magnetic field in the same direction.

This is different from normal induction motor operation, where the rotor rotates more slowly than the field.

Part b: Calculate rotor current frequency

Rotor current frequency depends on the magnitude of slip.

\[ f_r=|s|f_s \]

Substitute:

\[ f_r=0.02\times50 \]

\[ \boxed{f_r=1\text{ Hz}} \]

Even though the rotor is rotating faster than the stator field, the relative speed corresponds to a rotor current frequency of 1 Hz.

Part c: Determine torque direction and machine operation

The electromagnetic torque acts anticlockwise, opposing the clockwise rotor motion.

The rotor is moving faster than the rotating magnetic field, so the relative motion has reversed compared with normal motoring operation.

The induced rotor currents therefore produce a torque that opposes the rotor’s motion.

The external turbine must supply mechanical energy to overcome this opposing torque.

That mechanical energy is converted into electrical energy and delivered to the grid.

The machine is operating as an induction generator.

The grid still supplies the magnetising current required to establish the magnetic field, but the machine delivers real electrical power back to the grid.

Part d: What happens when the turbine is disconnected?

The rotor initially continues rotating clockwise at 1530 rpm due to its rotational inertia.

However, the turbine is no longer supplying mechanical energy.

The electromagnetic generator torque and the mechanical load both oppose the rotor’s motion, so the rotor begins slowing down.

As its speed approaches 1500 rpm, the slip approaches zero.

At exactly synchronous speed:

\[ s=0 \]

Therefore:

\[ f_r=0 \]

There is no induced rotor current through the induction mechanism, so the electromagnetic torque falls to zero.

The rotor then continues slowing because the mechanical load is still opposing its motion.

Once its speed falls below 1500 rpm, the slip becomes positive.

The induced rotor current is re-established, and the electromagnetic torque reverses direction to act clockwise.

The machine now operates as an induction motor.

It eventually reaches a steady operating speed slightly below synchronous speed, where its electromagnetic torque balances the opposing mechanical load and rotational losses.

Part e: Can it remain at exactly synchronous speed under load?

No, not while producing the required induction torque.

At exactly 1500 rpm, the rotor and stator magnetic field have no relative motion.

Without relative motion, no rotor EMF is induced, so the induction mechanism cannot produce the torque required to overcome the mechanical load.

The rotor must slow slightly below synchronous speed to establish positive slip and produce motor torque.

This is the important distinction: an induction machine can operate above or below synchronous speed, but the sign of its slip determines whether it acts as a generator or a motor.

The same rotating magnetic field and squirrel-cage construction can therefore support two different energy conversions.

08Where to go next

The key relationship to remember is that an induction motor needs relative motion between its rotating magnetic field and rotor to induce current and produce torque.

Once you understand this, the next useful step is connecting electromagnetic torque to mechanical power, angular velocity, and the efficiency of an electric motor.

Those relationships let you move beyond explaining how the rotor turns and start analysing how effectively an induction motor transfers electrical energy into mechanical work.