NSWPhysicsSyllabus dot point

Inquiry Question 4: How are electric and magnetic fields applied in electrical generation, transmission and use?

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

A focused answer to the HSC Physics Module 6 dot point on motors. Torque on a current loop tau = nBIA cos theta, the split-ring commutator in DC motors, back EMF and its role in steady-state current, the rotating-field principle of the AC induction motor, and where each is used.

Generated by Claude OpusReviewed by Better Tuition Academy10 min answerUpdated 2026-05-18

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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:

\boxed{\tau = n B I A \cos \theta}

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:

IMATH_15 (plane parallel to field, normal perpendicular to field): $\tau_{\max} = nBIA$ .
IMATH_17 (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:

V_{\text{net}} = V_{\text{supply}} - \varepsilon_{\text{back}}, \qquad I = \frac{V_{\text{supply}} - \varepsilon_{\text{back}}}{R}

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.

Common traps

Using $\cos \theta$ vs $\sin \theta$ inconsistently. The HSC formula sheet typically gives $\tau = nBIA \cos \theta$ with $\theta$ as the angle between the plane of the coil and the field. If you measure $\theta$ from the area normal instead, the formula becomes $\tau = nBIA \sin \theta$ . Either is fine if you are consistent; check carefully which one the question expects.

Saying back EMF "reduces" the motor's power. It does not reduce the useful mechanical power. The mechanical power delivered to the shaft equals $\varepsilon_{\text{back}} I$ . Back EMF is the channel through which electrical energy becomes mechanical energy.

Confusing the commutator with slip rings. A split-ring commutator (DC motor) reverses the current direction in the coil every half-turn. Slip rings (AC generator, synchronous motor) deliver an unbroken AC signal to or from the coil.

Forgetting that the induction motor needs slip. Without slip there is no relative motion between field and rotor, no induced EMF, and no torque. An induction motor cannot run exactly at synchronous speed.

Treating AC and DC motors as equally interchangeable. A DC motor uses a commutator and runs on DC. A standard AC motor runs on AC and either uses slip rings (synchronous) or no electrical contact to the rotor at all (induction).

Ignoring the $n$ (number of turns) in the torque formula. Like Faraday's law, the turns multiply the effect of a single loop.

In one sentence

A current loop in a uniform field experiences a torque $\tau = nBIA \cos \theta$ that DC motors keep in the same rotational sense using a split-ring commutator, with back EMF self-regulating the current; AC induction motors use a rotating stator field to induce currents in a short-circuited rotor that the field then drags along at slightly less than synchronous speed.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2021 HSC5 marksA rectangular coil of 50 turns and area 0.020 m^2 carries a current of 2.5 A in a uniform magnetic field of 0.30 T. Calculate the maximum torque on the coil and the torque when the plane of the coil is parallel to the field. Explain why a DC motor needs a commutator.

Show worked answer →

Maximum torque occurs when the plane of the coil is parallel to the field (so the angle between the field and the normal is 90 degrees, and $\cos \theta$ in $\tau = nBIA \cos \theta$ is the cosine of the angle the plane makes with the field):

$\tau_{\max} = n B I A = 50 \times 0.30 \times 2.5 \times 0.020 = 0.75$ N m.

When the plane is parallel to the field, $\theta = 0°$ in the standard "plane-to-field" convention, so $\cos \theta = 1$ , and the torque is the maximum value: $0.75$ N m.

When the plane is perpendicular to the field (the normal aligns with $\vec{B}$ ), $\cos \theta = 0$ and the torque is zero.

Commutator: in a DC motor, without a commutator, the torque on the coil would reverse direction every half-turn (when the coil passes through the plane perpendicular to the field), so the coil would oscillate but not rotate continuously. The split-ring commutator reverses the direction of current in the coil at exactly the point where the torque would otherwise change sign, so the torque on the coil stays in the same rotational sense and the motor spins continuously in one direction.

Markers reward correct identification of when torque is maximum (plane parallel to field), the calculation with $n$ included, and a clear commutator explanation tied to torque reversal.

2019 HSC4 marksDefine back EMF in a DC motor and explain how it affects the current drawn by the motor at start-up versus at full operating speed.

Show worked answer →

Back EMF is the EMF induced in the armature coils of a motor as they rotate in the magnetic field. By Faraday's law, the rotation produces $d\Phi / dt$ through the coils, generating an EMF. By Lenz's law, this induced EMF opposes the supply voltage that drives the motor; hence "back EMF."

The net voltage driving current through the armature resistance $R$ is the supply voltage minus the back EMF:

$I = (V - \varepsilon_{\text{back}}) / R$ .

At start-up the coil is stationary, so $\varepsilon_{\text{back}} = 0$ and the starting current is $I_{\text{start}} = V / R$ , which can be very large because $R$ is small. This is why large DC motors include starter resistors that are progressively shorted out as the motor accelerates.

At full operating speed the back EMF is close to the supply voltage, so $(V - \varepsilon_{\text{back}})$ is small and the running current is much lower than the starting current. The motor draws just enough current to overcome friction and supply the mechanical load.

If the mechanical load increases, the motor slows, back EMF drops, and the current rises automatically to provide more torque. This self-regulation is essential to motor operation.

Markers reward the Faraday/Lenz definition, the equation form $I = (V - \varepsilon_{\text{back}})/R$ , and a comparison of start-up versus running current.