Analyse the operation of DC and AC motors, including the torque on a current loop tau = n B I A cos theta, the role of the commutator, back EMF, and the AC induction motor principle

What this dot point is asking

NESA wants you to derive the torque on a current loop in a uniform field, identify what changes in a DC motor (the commutator) versus an AC motor (the rotating field or slip rings), discuss back EMF as the natural consequence of a rotating coil obeying Faraday's law, and outline how an AC induction motor uses a rotating magnetic field to drag the rotor along.

The answer

Torque on a current loop

Consider a rectangular coil of $n$ turns, side lengths $a$ and $b$ (so area $A = ab$), carrying current $I$ in a uniform field $\vec{B}$. The two sides of length $a$ that lie perpendicular to $\vec{B}$ experience forces $F = nBIa$ in opposite directions, forming a couple. The lever arm is $(b/2) \cos \theta_{\text{plane-to-field}}$ on each side, giving a total torque:

where $\theta$ is the angle between the plane of the coil and the field. (Equivalently, with $\phi$ as the angle between the area normal and the field, $\tau = n B I A \sin \phi$.)

Special cases:

• $\theta = 0$ (plane parallel to field, normal perpendicular to field): $\tau_{\max} = nBIA$.
• $\theta = 90°$ (plane perpendicular to field, normal aligned with field): $\tau = 0$.

So the maximum torque occurs when the coil is edge-on to the field, and zero torque when the coil is face-on to the field. The plane-of-coil-parallel-to-field position is the driving position; the face-on position is the dead spot.

DC motors and the commutator

If you simply attach a DC supply to a coil in a magnetic field, the torque drives the coil toward the face-on position, decelerating as it approaches and then reversing direction past it. The coil would oscillate about the equilibrium, not rotate.

A split-ring commutator solves this. The commutator is a metal ring split into two halves, each connected to one end of the coil, with carbon brushes contacting it from outside. As the coil rotates through the face-on dead spot, the brushes cross the split in the ring and the current direction in the coil reverses. The torque now drives the coil away from the dead spot in the same rotational sense as before.

Equivalent statement: the commutator ensures that the side of the coil moving up always carries current in the direction that produces an upward force from the field, so torque is always in the same rotational sense.

Real DC motors use many coils at different angles, each with its own commutator segment, so that some coil is always near the maximum-torque position. This smooths out the torque ripple and avoids dead spots entirely.

Back EMF

When the coil rotates in the field, the changing flux through it induces an EMF (Faraday's law). By Lenz's law this induced EMF opposes the supply voltage that is causing the rotation: it is a back EMF $\varepsilon_{\text{back}}$.

The net voltage driving current through the armature resistance $R$ is:

Consequences:

• Start-up. $\varepsilon_{\text{back}} = 0$, so $I_{\text{start}} = V/R$ is very large. Large motors use starter resistors that are progressively switched out as the motor accelerates.
• Running. $\varepsilon_{\text{back}}$ is close to $V$, current is small, and the motor draws just enough to overcome friction and the mechanical load.
• Loaded. If you push down on the shaft, the motor slows, $\varepsilon_{\text{back}}$ drops, and $I$ rises to supply more torque. The motor self-regulates.
• Stalled. If the rotor cannot turn, $\varepsilon_{\text{back}} = 0$ and the full $V/R$ current flows, potentially burning out the motor.

Power balance: $V I = I^2 R + \varepsilon_{\text{back}} I$. The first term is heat dissipated in the windings; the second term is the mechanical power delivered to the shaft.

AC motors: synchronous vs induction

AC motors split into two broad families. Both rely on the same idea: produce a rotating magnetic field in the stator (the stationary part) by feeding multi-phase AC into a set of coils arranged around the rotor.

Synchronous motor. The rotor is a magnet (a permanent magnet or an electromagnet fed by slip rings). It locks onto the rotating stator field and spins at exactly the same frequency (the synchronous speed, $f_{\text{rotor}} = f_{\text{supply}}$). Used in clocks, turntables and precision applications.

AC induction motor (squirrel cage). The rotor is a set of conducting bars short-circuited at each end (no slip rings or commutator at all). The rotating stator field sweeps past the rotor, inducing currents in the bars (Faraday's law). These currents, sitting in the rotating field, experience a magnetic force that drags the rotor in the direction of rotation (Lenz's law: the induced current opposes the change, that is, the relative motion of field past rotor). The rotor accelerates but always runs slightly slower than the field; the difference is called the slip. Without slip there would be no induced current and hence no torque.

The AC induction motor is brushless, robust, and self-starting under load. It is the workhorse of industry, used in pumps, fans, compressors, washing machines and electric vehicles (often paired with a variable-frequency drive that adjusts the supply frequency to control speed).

Worked example: car starter motor

A car starter motor has an armature with $50$-turn coils of area $0.030$ m$^2$. The field strength is $0.50$ T and the supply voltage is $12$ V. The armature resistance is $0.040$ ohms.

Starting current (back EMF zero, coil in driving position):

$I_{\text{start}} = V / R = 12 / 0.040 = 300$ A.

Maximum starting torque per coil:

$\tau_{\max} = n B I A = 50 \times 0.50 \times 300 \times 0.030 = 225$ N m.

This huge torque (and current) is why a car battery sags when you turn the key and why the starter motor is engaged only briefly. Once the engine fires and turns the motor faster than required, the back EMF rises and the current drops.

Worked example: AC induction motor slip

A four-pole induction motor running on $50$ Hz mains has a synchronous (stator-field) speed of $1500$ rpm. Under load the rotor turns at $1440$ rpm. Find the slip.

Slip:

$s = (1500 - 1440) / 1500 = 0.040 = 4.0\%$.

Typical industrial induction motors run with a few percent slip at rated load. At no load, slip is near zero; under heavy load, slip increases, the induced currents rise, and the torque rises with it.

Examples in context

Example 1. DC motor in a Snowy Hydro 2.0 sluice-gate actuator. A heritage brushed DC motor drives a sluice-gate valve. The single-turn loop has area $A = 0.040 \text{ m}^2$ in field $B = 0.30 \text{ T}$, carrying $I = 8.0 \text{ A}$. Maximum torque (loop plane parallel to $B$, normal at $90^{\circ}$) is $\tau_{\max} = n B I A = 1 \times 0.30 \times 8.0 \times 0.040 = 0.096 \text{ N m}$. A $50$-turn motor scales this to $\tau = 4.8 \text{ N m}$. Splitring commutators reverse the current each half-turn so the torque stays one-way; without the commutator the loop would oscillate. Back-EMF rises with speed, limiting the steady-state current.

Example 2. AC induction motor in a Sydney Light Rail tram on George Street. A $300 \text{ kW}$ three-phase AC induction motor runs at $1450 \text{ rpm}$ off a 50 Hz supply. Synchronous speed is $f \times 60 / p = 50 \times 60 / 2 = 1500 \text{ rpm}$ for a 4-pole motor. Slip is $(1500 - 1450)/1500 = 3.3\%$. Rotor frequency is $0.033 \times 50 = 1.67 \text{ Hz}$. The rotating stator field induces currents in the squirrel-cage rotor; these currents in turn experience Lorentz forces that drag the rotor along. Trams use this style because they need no brushes, are robust, and slip increases naturally on hills to deliver more torque.

Try this

Q1. Explain the function of the split-ring commutator in a DC motor. [2 marks]

• Cue. Reverses current direction in the coil every half-turn so torque keeps acting in the same rotational sense.

Q2. A rectangular coil of $50$ turns, area $0.020 \text{ m}^2$, carries $3.0 \text{ A}$ in a field $B = 0.40 \text{ T}$. Calculate the maximum torque on the coil. [2 marks]

• Cue. $\tau_{\max} = n B I A = 50 \times 0.40 \times 3.0 \times 0.020 = 1.20 \text{ N m}$.

Q3. A DC motor's coil rotates at $1200 \text{ rpm}$ in a $0.50 \text{ T}$ field. (a) Explain what back EMF is and the direction it acts. (b) Sketch how the supply current changes as the motor accelerates from rest. (c) Justify why DC motors stall (draw very high current) at startup. [2+2+2 marks]

• Cue. (a) EMF induced in coil opposing supply (Lenz). (b) Current high at rest, drops as back EMF rises. (c) At zero speed, back EMF = 0, so $I = V/R$ is large.