DC and AC motors: HSC Physics Module 6

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.

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.

Without a commutator, the torque on the coil would reverse direction every half-turn (as the coil sweeps through the face-on "dead spot"), so the coil would oscillate rather than spin continuously.

The split-ring commutator reverses the direction of current in the coil at exactly the moment the coil passes through the dead spot, so the torque continues to act in the same rotational sense on each side of the coil every half-turn.

Marks: one for stating that current direction in the coil is reversed each half-turn, one for linking this to keeping the torque in a consistent rotational sense (continuous rotation, not oscillation).

Maximum torque occurs when the plane of the coil is parallel to the field, so $\cos\theta = 1$ and $\tau_{\max} = nBIA$.

$\tau_{\max} = 40 \times 0.25 \times 4.0 \times 0.015 = 0.60\ \text{N m}$.

Marks: one for the correct formula with values substituted, one for the answer $0.60\ \text{N m}$ stated to two significant figures with the unit.

Back EMF is the EMF induced in the rotating armature coils by their own motion through the magnetic field (Faraday's law); by Lenz's law it opposes the supply voltage that drives the current.

The current is $I = \dfrac{V - \varepsilon_{\text{back}}}{R}$.

Marks: one for identifying it as an induced EMF from the coil's rotation, one for stating that it opposes the supply voltage (Lenz's law), one for the correct equation $I = (V - \varepsilon_{\text{back}})/R$.

(a) The graph is a cosine curve: torque is maximum at $\theta = 0^{\circ}$ (plane parallel to the field) and falls smoothly to zero at $\theta = 90^{\circ}$ (plane perpendicular to the field, the "dead spot").

(b) Reading across from $\theta = 60^{\circ}$ on the curve gives $\tau \approx 0.24\ \text{N m}$.

(c) $\tau_{\max} = nBIA = 20 \times 0.40 \times 3.0 \times 0.020 = 0.48\ \text{N m}$, which matches the graph's value at $\theta = 0^{\circ}$.

Marks: one for identifying the cosine shape and $\theta = 0^{\circ}$ for maximum torque, one for reading $\tau \approx 0.24\ \text{N m}$ at $60^{\circ}$ (accept $0.22$-$0.26$), one for the formula $\tau_{\max} = nBIA$ with substitution, one for $\tau_{\max} = 0.48\ \text{N m}$ matching the graph.

(a) At start-up the coil is stationary, so $\varepsilon_{\text{back}} = 0$: $I_{\text{start}} = \dfrac{V}{R} = \dfrac{12}{0.030} = 400\ \text{A}$.

(b) $\tau_{\max} = nBIA = 60 \times 0.45 \times 400 \times 0.025 = 270\ \text{N m}$.

Marks: one for recognising $\varepsilon_{\text{back}} = 0$ at start-up, one for $I_{\text{start}} = 400\ \text{A}$, one for the torque formula with substitution, one for $\tau_{\max} = 270\ \text{N m}$ with the unit.

Synchronous speed: $n_s = \dfrac{120 f}{p} = \dfrac{120 \times 50}{6} = 1000\ \text{rpm}$ (using $p$ as the number of poles).

Slip: $s = \dfrac{n_s - n}{n_s} = \dfrac{1000 - 960}{1000} = 0.040 = 4.0\%$.

Marks: one for the synchronous speed $1000\ \text{rpm}$, one for the slip formula, one for $s = 4.0\%$.

Band-6 plan. (1) Explain the dead-spot problem without a commutator, using $\tau = nBIA\cos\theta$. (2) Explain how the commutator fixes it (reverses current each half-turn). (3) Introduce back EMF from Faraday's/Lenz's law. (4) Explain self-regulation of current/torque with load, using $I = (V - \varepsilon_{\text{back}})/R$. (5) Synthesise: commutator gives continuous one-way torque, back EMF gives automatic speed/current control.

Model answer. A current loop in a magnetic field experiences torque $\tau = nBIA\cos\theta$, where $\theta$ is the angle between the coil's plane and the field. This torque is maximum when the plane is parallel to the field and falls to zero when the plane is perpendicular to the field (the "dead spot"). If the current direction in the coil never changed, the torque would reverse sign as the coil swept past the dead spot, so the coil would oscillate about equilibrium rather than turn continuously.

The split-ring commutator solves this: as the coil rotates through the dead spot, the brushes cross the gap in the split ring and the current in the coil reverses direction. Because the field direction has not changed, reversing the current keeps the force on each side of the coil acting in the same rotational sense, so the coil is driven around continuously in one direction.

As the coil turns, the changing magnetic flux through it induces an EMF by Faraday's law; by Lenz's law this induced EMF opposes the supply voltage that is driving the rotation, so it is called a back EMF $\varepsilon_{\text{back}}$. The current actually drawn is $I = (V - \varepsilon_{\text{back}})/R$, where $R$ is the armature resistance. At start-up $\varepsilon_{\text{back}} = 0$ so the current (and torque) is large; as the motor speeds up, $\varepsilon_{\text{back}}$ rises toward $V$ and the current falls to a small running value.

This gives automatic self-regulation: if the mechanical load increases, the motor slows, $\varepsilon_{\text{back}}$ drops, and $I$ (and hence torque, since $\tau \propto I$) rises to meet the extra load; as the load eases, the motor speeds up, $\varepsilon_{\text{back}}$ rises, and the current falls back down. Together, the commutator makes continuous one-directional rotation possible, and back EMF makes the resulting speed and current respond automatically to the load, without any external control circuit.

Marker's note: the top band connects the dead-spot torque problem explicitly to $\tau = nBIA\cos\theta$, explains the commutator's timing (reversal exactly at the dead spot), gives the back-EMF equation and reasons through both the start-up and loaded cases, and ends with the synthesising point that these are two separate mechanisms working together (mechanical switching for direction, electromagnetic feedback for speed/current). A response describing only one of the two mechanisms caps in the middle band.

Band-6 plan. (1) State how each motor produces continuous torque (DC: commutator + brushes; AC induction: rotating field + slip, no rotor contacts). (2) Compare reliability/maintenance (brush wear vs brushless). (3) Compare starting/control behaviour (DC: simple speed-torque control via back EMF; induction: robust, self-starting, needs a drive for variable speed). (4) Weigh the claim and reach a judgement, not just a list.

Model answer. A DC motor produces continuous one-directional torque using a split-ring commutator: as the coil sweeps through the dead spot where $\tau = nBIA\cos\theta$ would otherwise fall to zero and reverse sign, the commutator reverses the current in the coil so the torque stays in the same rotational sense. This mechanism relies on brushes maintaining sliding electrical contact with the rotating commutator.

An AC induction motor instead sets up a rotating magnetic field in the stator (from multi-phase AC) and induces currents in short-circuited conducting bars on the rotor (Faraday's law); the force on these induced currents, sitting in the rotating field, drags the rotor around at a speed slightly below the field's synchronous speed (Lenz's law requires this slip, since zero relative motion would induce no current and no torque). Crucially, there is no electrical contact to the rotor at all.

For most industrial applications, the induction motor is superior: with no brushes or commutator there is no wear-prone sliding contact, so it needs less maintenance, generates no brush sparking (important in dusty or explosive environments), and is mechanically simpler and more robust, which is why it is the workhorse in pumps, fans, compressors and conveyor drives. Its main limitation is that its speed is set by the supply frequency and pole count, so variable-speed operation needs an added variable-frequency drive.

The DC motor's advantage is that back EMF gives simple, direct, wide-range speed and torque control from the supply voltage alone, and it can develop very high starting torque, useful in traction and precision-control applications (starter motors, some robotics). So the claim is broadly correct for typical continuous industrial duty, where brushless robustness and low maintenance dominate, but it is not universal: applications needing simple, wide-range, precise speed/torque control without a separate drive still favour a DC motor.

Marker's note: the top band explains the actual continuous-torque mechanism for both motor types (not just "one has brushes"), uses the slip concept correctly for the induction motor, and reaches a genuine assessment (superior for most industrial duty, with a stated exception) rather than a flat list of pros and cons.