TASC TCE 2023-style practice: A laboratory alternator has a coil of 80 turns and area 1.5 × 10^-2\ m^2 spinning in a 0.25\ T field. It produces a peak EMF of 94\ V. Calculate the angular frequency and rotation rate of the coil, then state the RMS value of the output.

The peak EMF is _0 = NBAω, so rearrange for ω: $ω = (_0)/(NBA) = 94(80)(0.25)(1.510^-2) = (94)/(0.30) = 313\ rad s^-1.$ Rotation rate: f = ω2π = 3132π = 49.8\ Hz, about 50\ Hz. RMS value: _rms = _0sqrt(2) = 941.414 = 66.5\ V. Markers want ω = _0NBA, the conversion f = ω2π, and _rms = _0sqrt(2).

§-Syllabus dot point

TASPhysicsSyllabus dot point

How is alternating current generated from rotating motion?

Explain the operation of an AC generator and the sinusoidal EMF it produces.

How a rotating coil in a magnetic field generates a sinusoidal alternating EMF, the role of slip rings, the difference from a DC motor, and the meaning of peak and RMS values.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

This dot point applies Faraday's law to a rotating coil, the device that produces almost all the world's electricity.

How the EMF is generated

A generator has a coil that is spun inside a magnetic field. As the coil turns, the angle between the field and the coil face changes, so the flux $\Phi = BA\cos\theta$ through the coil rises and falls. Faraday's law says the induced EMF is proportional to the rate of change of flux, so a steadily rotating coil produces a continuously varying EMF.

The EMF is greatest when the coil is moving parallel to the field lines, that is when the coil plane is parallel to the field, because then the flux is changing fastest. The EMF is zero when the coil is perpendicular to the field, where the flux is at a maximum but momentarily not changing. The result is a smoothly varying, sinusoidal output.

The sinusoidal output

For a coil of $N$ turns and area $A$ rotating at angular frequency $\omega$ in a field $B$ , the induced EMF varies as:

\varepsilon = N B A \omega \sin(\omega t)

The peak EMF is $\varepsilon_0 = NBA\omega$ , so a faster rotation, a stronger field, more turns or a larger area all raise the peak voltage. The output reverses direction every half turn, which is what makes it alternating current. The angular frequency relates to the rotation rate by $\omega = 2\pi f$ , so a generator turning at $50\ \text{Hz}$ has $\omega \approx 314\ \text{rad s}^{-1}$ .

Slip rings and the link to motors

A generator connects the coil to the external circuit through slip rings, two continuous rings that each stay connected to one end of the coil. Because the connection never swaps, the output naturally alternates as the coil rotates, giving AC. This contrasts with a DC motor's split-ring commutator, which deliberately swaps the connection every half turn.

A generator is structurally the same as a motor: a coil, a magnetic field and a means of connection. A motor uses an input current to produce rotation; a generator uses input rotation to produce a current. This is why the same machine can often run either way, and why a spinning motor produces a back EMF that opposes the supply.

Peak and RMS values

Because AC voltage varies, we describe it by two numbers. The peak value $\varepsilon_0$ is the maximum, while the root-mean-square (RMS) value is the steady value that would deliver the same average power. For a sinusoidal output they are related by:

\varepsilon_\text{rms} = \frac{\varepsilon_0}{\sqrt{2}}.

Mains voltage quoted as $230\ \text{V}$ is the RMS value, with a higher peak of about $325\ \text{V}$ . The RMS value matters because the average power delivered to a resistor is set by $\varepsilon_\text{rms}$ , not the peak.

Worked example: peak EMF of a generator

A generator coil has $50$ turns, area $0.020\ \text{m}^2$ , and spins at $\omega = 314\ \text{rad s}^{-1}$ in a field of $0.30\ \text{T}$ . Find the peak EMF and the RMS output voltage.

Step 1: Identify the formula for peak EMF

When a coil rotates at constant angular frequency in a uniform field, Faraday's law gives a sinusoidal EMF whose peak value occurs when the coil plane is parallel to the field and flux is changing most rapidly. The peak EMF depends on how many turns the coil has, how strong the field is, the area swept by the coil, and how fast it spins:

\varepsilon_0 = NBA\omega

Each factor multiplies the effect: more turns, a larger area, a stronger field, or a faster rotation all increase the maximum output voltage.

Step 2: Substitute the given values and evaluate

Reading off $N = 50$ , $B = 0.30\ \text{T}$ , $A = 0.020\ \text{m}^2$ , and $\omega = 314\ \text{rad s}^{-1}$ , and substituting:

\varepsilon_0 = (50)(0.30)(0.020)(314) = 94\ \text{V}

Step 3: Convert peak to RMS

The RMS value is the equivalent steady voltage that would deliver the same average power to a resistive load. For a sinusoidal waveform the RMS value is always smaller than the peak by a factor of $\sqrt{2}$ :

\varepsilon_\text{rms} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{94}{\sqrt{2}} = 66\ \text{V}

Final answer: Peak EMF $= 94\ \text{V}$ ; RMS output $= 66\ \text{V}$ .

In the exam, link the size of the EMF to how fast the flux is changing, and identify slip rings as the feature that gives alternating output. If asked, state that the EMF is zero when the coil is perpendicular to the field and maximum when the coil plane is parallel to it, and convert between peak and RMS values with the $\sqrt{2}$ factor.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20242 marksA

30

-turn coil is disconnected from a power source and rotated at

50\ \text{Hz}

in a uniform

0.2\ \text{T}

magnetic field. Sketch the potential difference at the outputs against time, starting from the position where the coil plane is parallel to the field, and explain its shape.

Show worked answer →

As the coil rotates the flux varies as $\Phi = BA\cos(\omega t)$ . By Faraday's law the induced EMF is the rate of change of flux, $\varepsilon = -N\dfrac{d\Phi}{dt} = NBA\omega\sin(\omega t)$ , which is sinusoidal.

The graph is a sine curve. Starting with the coil plane parallel to the field, the flux is momentarily minimum but its rate of change is maximum, so the EMF starts at its peak value and traces a smooth sinusoid, crossing zero a quarter-turn later (when the coil plane is perpendicular to the field and flux is maximum but unchanging).

Period: $T = \dfrac{1}{f} = \dfrac{1}{50} = 0.02\ \text{s}$ , so one full cycle occupies $0.02\ \text{s}$ , with equal positive and negative peaks each half cycle (the current reverses every half turn). Markers want a labelled sinusoid of period $0.02\ \text{s}$ , EMF maximum when the coil plane is parallel to the field, and the reasoning that EMF is proportional to the rate of change of flux.

TCE 20233 marksA laboratory alternator has a coil of

80

turns and area

1.5 \times 10^{-2}\ \text{m}^2

spinning in a

0.25\ \text{T}

field. It produces a peak EMF of

94\ \text{V}

. Calculate the angular frequency and rotation rate of the coil, then state the RMS value of the output.

Show worked answer →

The peak EMF is $\varepsilon_0 = NBA\omega$ , so rearrange for $\omega$ :

\omega = \frac{\varepsilon_0}{NBA} = \frac{94}{(80)(0.25)(1.5\times10^{-2})} = \frac{94}{0.30} = 313\ \text{rad s}^{-1}.

Rotation rate: $f = \dfrac{\omega}{2\pi} = \dfrac{313}{2\pi} = 49.8\ \text{Hz}$ , about $50\ \text{Hz}$ .

RMS value: $\varepsilon_\text{rms} = \dfrac{\varepsilon_0}{\sqrt{2}} = \dfrac{94}{1.414} = 66.5\ \text{V}$ .

Markers want $\omega = \dfrac{\varepsilon_0}{NBA}$ , the conversion $f = \dfrac{\omega}{2\pi}$ , and $\varepsilon_\text{rms} = \dfrac{\varepsilon_0}{\sqrt{2}}$ .