Physics · Electromagnetic induction

Induced emf of an ideal generator

Q: What is the peak value?

Peak emf $\mathcal E_0=NBA|\omega|$ is the largest magnitude of the ideal sinusoidal emf. Instantaneous values range between its negative and positive values. RMS emf instead equals $\frac{\mathcal E_0}{\sqrt2}$, giving the same mean heating power in an ideal resistor as a corresponding DC voltage. Phase shifts the waveform, not its peak. Area and turn count refer to identical active turns. Loaded terminal voltage need not equal ideal emf: winding resistance, inductance and back reaction can modify it. In a purely resistive equivalent circuit, instantaneous internal drop is $i(t)r_i$. AC signs and the load model must therefore be specified before claiming that every instantaneous terminal value is simply smaller.

Q: When is the emf zero?

For the selected initial orientation, $\mathcal E(t)=NBA\omega\sin(\omega t)$. Emf vanishes at $\omega t=k\pi$, where k is an integer. Flux is then at a maximum or minimum and its derivative is zero. At a flux zero crossing, change is fastest and emf magnitude is maximal. A different initial phase shifts these times together. This describes generated emf for prescribed rotation. Load-current phase additionally depends on resistance, inductance and capacitance; it cannot be obtained from flux differentiation alone. In a loaded real machine, field back reaction can also matter. Zero instantaneous voltage does not imply zero mean energy transfer over a whole period.

A planar coil rotating uniformly in a uniform magnetic field generates sinusoidal emf when its area normal sweeps the angle to the field uniformly.

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Formula

Calculate

E (t) = NB A ω sin (ω t)

ℰ in volts, B in tesla, A in m², ω in rad/s and t in seconds; N is dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ℰ(t)$	Instantaneous induced emf	V
$N$	Number of turns	dimensionless
$B$	Magnetic flux density	T
$A$	Coil area	m²
$ω$	Signed angular velocity relative to the chosen rotation direction	rad/s
$t$	Time	s

LaTeX

\mathcal{E}(t)=N B A \omega\sin(\omega t)

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Normalised cosine and sine curves over one period. Dashed flux starts at plus one, blue emf starts at zero and rises. Time marks 0, T/4, T/2, 3T/4 and T. — Uniform rotation with θ₀ = 0: normalised flux follows cos(ωt), while normalised emf follows sin(ωt). At maximum flux the emf is zero; a quarter period later it reaches its positive maximum.

Generator induced voltage calculator

Find the peak, rms or instantaneous induced voltage of a coil rotating steadily in a uniform magnetic field, with no load connected.

Calculation model

Find

Turn count N

Number of turns on the coil, e.g. 80; every turn contributes equally.

Magnetic flux density B

Strength of the uniform magnetic field (flux density), e.g. 0.25 T between strong magnets.

Area of one turn A

Area of a single turn, e.g. 120 cm² for a rectangle of 10 cm by 12 cm.

Angular velocity ω

Angular velocity of the rotation, e.g. 100 rad/s; only its magnitude matters for the peak.

Filled with example values. Type your own and the result updates straight away.

Result: Peak induced voltage ℰ₀

ℰ₀ = 24 V

With N = 80, B = 0.25 T, A = 120 cm² and ω = 100 rad/s, ℰ₀ = 24 V.

Working · rearrange for ℰ₀, substitute, evaluate

1. Rearrange for ℰ₀
$\mathcal E_0=NBA|\omega|$
2. Substitute values (in SI units)
- A = 120 cm² = 0.012 m²
$80\cdot 0.25\cdot 0.012\cdot\left|\left(100\right)\right|=24$
3. Result
ℰ₀ = 24 V

Quick tips

RMS value = peak ÷ √2, about 0.71 × the peak. A 24 V peak gives about 17 V rms.
ω from the rotation rate: ω = 2π · n. 3000 revolutions per minute are 50 per second, so ω ≈ 314 rad/s. Twice the speed, twice the voltage.

Assumptions and limits

N equally oriented planar turns of equal area in uniform time-independent B; fixed uniform rotation with axis in the coil plane and perpendicular to B. Ideal open-circuit induced emf, not loaded terminal voltage. Specify area normal and circulation together.

Derivation & background

For equal area per turn, $\Phi(t)=BA\cos(\omega t)$ per turn when the area normal points along the field at t=0. Faraday gives $\mathcal E=-N\frac{\mathrm d\Phi}{\mathrm dt}$ , with peak $\mathcal E_0=NBA|\omega|$ . Sine or cosine depends on the time origin. The sign of $\omega$ is retained in the time function obtained from the flux derivative. For one electrical period per revolution, $f=\frac{|\omega|}{2\pi}$ . Rotation about the area normal does not change flux. Loaded terminal voltage requires a load and internal-impedance model.

Exam blueprint

Validity range

Planar coil with N equal-area turns, constant uniform B and uniform rotation about an axis perpendicular to B in the coil plane. The angle is between B and the area normal. The formula gives induced emf; loaded terminal voltage requires a load model.

Derivation steps

Projected coil area determines flux. Its time derivative gives emf and the quarter-period shift.

1Define the area normal and matching loop direction; θ is the angle from the normal to the magnetic field.
2For uniform rotation in this geometry, $\theta=\omega t+\theta_0$ .
3Flux per turn is $\Phi=BA\cos(\omega t+\theta_0)$ ; equal turns have flux linkage $\Psi=N\Phi$ .
4Apply Faraday: $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ . The negative sign and cosine derivative together give the sine sign.
5 $\mathcal E(t)=NBA\omega\sin(\omega t+\theta_0)$ . Setting $\theta_0=0$ gives the displayed formula.
6Check $\mathcal E_0=NBA\omega$ , $f=\frac{\omega}{2\pi}$ and, for a sinusoid, $\mathcal E_{\mathrm{rms}}=\frac{\mathcal E_0}{\sqrt2}$ . At maximum flux the emf must be zero.

Rearrangements

Calculate instantaneous emf or peak value of a generator

\mathcal E_0=NBA|\omega|

Uniformly rotating N-turn coil in a uniform field; load and internal resistance are not included in the open-circuit emf formula. Peak magnitude uses the absolute value of signed angular velocity; the time function retains its sign.

Task variant

An ideal coil has $N=100$ , $B=0.20\,\mathrm T$ , $A=0.010\,\mathrm{m^2}$ , $\omega=50\,\mathrm{rad/s}$ . Its normal points along the field at t=0. Find peak emf and instantaneous emf at $\omega t=\frac\pi6$ .

Approach: $\mathcal E_0=NBA\omega$ and $\mathcal E=\mathcal E_0\sin(\omega t)$ .
Substitution gives $\mathcal E_0=10\,\mathrm V$ and $\mathcal E=5.0\,\mathrm V$ .

Check: at thirty degrees, instantaneous emf is half the peak; maximum magnitude remains 10 V.

Common mistakes

Confusing RMS and peak values or using frequency in hertz for ω.

$\omega$ is angular speed in rad/s. Given frequency, first calculate $\omega=2\pi f$ ; identify peak and RMS values separately.

Exam context

Sinusoidal generator voltage, frequency control and energy conversion.

Formula cluster

Rotating coil

Rotation changes magnetic flux and produces the alternating emf.

ℰ = −N dΦB/dt

Faraday’s law of induction

|ℰmot| = Bℓv

Motional emf

U₁/U₂ = n₁/n₂ = I₂/I₁

Transformer turns ratio

Worked example

For $N=100$ , $B=0.20\,\mathrm T$ , $A=0.010\,\mathrm{m^2}$ and $\omega=50\,\mathrm{rad/s}$ , $\mathcal E_0=NBA\omega=10\,\mathrm V$ . At $\omega t=\frac\pi6$ , $\mathcal E=10\,\mathrm V\cdot\sin\frac\pi6=5.0\,\mathrm V$ . Check: the maximum magnitude is 10 V.

Applications

AC generators, dynamos, power generation and the relation between rotation speed, emf and frequency.

How do you calculate with Induced emf of a rotating generator?

Here is how to work through a typical Induced emf of a rotating generator (ℰ(t) = NBAω sin(ωt)) task step by step:

1
Task
An ideal coil has $N=100$ , $B=0.20\,\mathrm T$ , $A=0.010\,\mathrm{m^2}$ , $\omega=50\,\mathrm{rad/s}$ . Its normal points along the field at t=0. Find peak emf and instantaneous emf at $\omega t=\frac\pi6$ .
Solution path
1. Approach: $\mathcal E_0=NBA\omega$ and $\mathcal E=\mathcal E_0\sin(\omega t)$ .
2. Substitution gives $\mathcal E_0=10\,\mathrm V$ and $\mathcal E=5.0\,\mathrm V$ .
Check: at thirty degrees, instantaneous emf is half the peak; maximum magnitude remains 10 V.

Understand and apply the equation

A rotating coil produces voltage because its magnetic flux changes with time. The decisive quantity is the rate of change, rather than the largest flux value. The ideal generator relation connects speed, turn count and area to the voltage waveform. With a resistive load it also determines electrical power and the required mechanical drive.

Choose angle and reference direction first

A rotating coil does not produce voltage simply because a large magnetic flux passes through it. Voltage appears when that flux changes. Uniform rotation changes the area orientation periodically: flux follows a cosine and generated voltage a sine. The generator converts supplied mechanical energy. Without a connected load, voltage alone does not specify delivered electrical power.[1]

\begin{aligned}\theta(t)&=\omega t+\theta_0\\\Phi(t)&=BA\cos\theta(t)\\\Psi(t)&=N\Phi(t)\end{aligned}

θ is the angle between field and area normal; Φ is flux through one turn and Ψ is total flux linkage.[1][2]

The model uses a planar coil with N equally oriented turns of equal area A. It rotates at constant angular speed about an axis that periodically changes the angle between area normal and uniform field. The rotation axis lies in the coil plane and is perpendicular to the field. Rotation about the area normal, or about an axis parallel to the field, does not automatically produce the same sinusoidal flux change. The following peak-amplitude shorthand uses $\omega$ as a nonnegative angular-speed magnitude. For signed angular velocity, nonnegative peak amplitude is $\mathcal E_0=NBA|\omega|$ ; the time function instead retains the signed factor $\omega$ and angle $\omega t+\theta_0$ .[1][2]

The sign of electromotive force $\mathcal E$ refers to the chosen circulation direction around the coil. In a simple ideal generator model it corresponds to the open-circuit terminal voltage. Under load, winding resistance and additional components can alter measured terminal voltage. Flux is measured in webers; its rate of change has units of volts.[1][2]

How do generator emf, zero crossings and peak amplitude arise?

\begin{aligned}\mathcal E(t)&=-\frac{d\Psi}{dt}\\&=-NBA\frac{d}{dt}\cos(\omega t+\theta_0)\\&=NBA\omega\sin(\omega t+\theta_0)\end{aligned}

Faraday’s minus sign and the derivative of cosine together give the chosen sine sign.[1][2]

For the following sequence we choose $\theta_0=0$ : at time zero flux is at its positive maximum and its rate of change is zero. One quarter turn later flux is zero but voltage is at its positive maximum. After half a turn flux is at its negative maximum and voltage is zero again. Choosing another time origin changes phase, rather than peak value or frequency. Faraday’s law of induction relates the chosen area normal and circulation direction to the flux derivative.[1][2]

Flux linkage and voltage: different scales

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Original curves for the first example: flux linkage divided by 0.240 Wb and emf by 24.0 V, hence separate dimensionless normalizations. Shared height does not mean equal physical units.[1][2]

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\begin{aligned}\mathcal E_0&=NBA\omega\\\omega&=2\pi f\\T&=\frac1f\end{aligned}

With one full electrical period per mechanical revolution, f equals the rotation frequency.[1]

The unit check gives teslas times square metres times inverse seconds, hence webers per second. An area stated in square centimetres needs the squared length conversion. Angular measure in radians is dimensionless, yet angular speed is explicitly entered in rad/s rather than revolutions per minute. Machines with several pole pairs have a different relation between electrical frequency and mechanical rotation rate.[1][4]

How do peak, mean and RMS emf differ?

\begin{aligned}\langle\mathcal E\rangle_T&=0\\\mathcal E_{\mathrm{eff}}&=\sqrt{\frac1T\int_0^T\mathcal E^2(t)\,dt}=\frac{\mathcal E_0}{\sqrt2}\end{aligned}

A sinusoidal voltage has zero signed mean but a nonzero RMS value.[3]

RMS is the DC voltage that would deliver the same average heating power to the same ideal resistor. It is neither the mean absolute value nor the largest instantaneous value. For a pure sinusoid its factor follows from the period mean of squared sine; distorted waveforms require evaluation of the integral itself. The RMS value of an alternating voltage is therefore determined from the square of its actual time history.[3]

Track one coil through a revolution

Calculate peak and instantaneous voltage

Given: $N=80$ , $B=0.25\,\mathrm T$ , $A=0.012\,\mathrm{m^2}$ , $\omega=100\,\mathrm{rad/s}$ . Flux is at its positive maximum at t=0.
Find: Find peak voltage, RMS voltage and voltage at $t=\frac{\pi}{600}\,\mathrm s$ .

Approach: First determine peak voltage and phase symbolically.
$\begin{aligned}\mathcal E_0&=NBA\omega\\\theta&=\omega t\end{aligned}$
Substitute the coil data.
$\mathcal E_0=80\cdot0{,}25\,\mathrm T\cdot0{,}012\,\mathrm{m^2}\cdot100\,\mathrm{s^{-1}}=24\,\mathrm V$
Calculate phase and instantaneous voltage.
$\begin{aligned}\theta&=100\cdot\frac{\pi}{600}=\frac{\pi}{6}\\\mathcal E(t)&=24\,\mathrm V\sin\frac{\pi}{6}=12\,\mathrm V\end{aligned}$
Convert peak voltage to RMS voltage.
$\mathcal E_{\mathrm{eff}}=\frac{24}{\sqrt2}\,\mathrm V\approx16{,}97\,\mathrm V$

Result: Peak voltage is 24 V, RMS voltage about 17.0 V and instantaneous voltage 12 V after about 5.24 ms.

Check: The phase is 30°, so instantaneous voltage is half the peak. Frequency is $\frac{100}{2\pi}\approx15.9\,\mathrm{Hz}$ ; the selected time precedes the first maximum at about 15.7 ms.

A load requires mechanical driving power

Voltage, load power and average opposing torque

Given: An ideal generator has $N=100$ , $B=0.20\,\mathrm T$ , $A=0.010\,\mathrm{m^2}$ and $f=25\,\mathrm{Hz}$ . Its sole load is $R=100\,\Omega$ ; winding and mechanical losses are neglected.
Find: Determine mean electrical power and magnitude of the required mean driving torque.

Approach: Connect angular speed to voltage and power.
$\begin{aligned}\omega&=2\pi f\\\mathcal E_0&=NBA\omega\\\overline P&=\frac{\mathcal E_0^2}{2R}\end{aligned}$
Substitute frequency and coil data.
$\begin{aligned}\omega&=50\pi\,\mathrm{rad/s}\\\mathcal E_0&=100\cdot0{,}20\cdot0{,}010\cdot50\pi\,\mathrm V=10\pi\,\mathrm V\end{aligned}$
Calculate mean power and RMS current.
$\begin{aligned}\overline P&=\frac{(10\pi\,\mathrm V)^2}{2\cdot100\,\Omega}=\frac{\pi^2}{2}\,\mathrm W\approx4{,}935\,\mathrm W\\I_{\mathrm{eff}}&=\frac{10\pi}{100\sqrt2}\,\mathrm A\approx0{,}2221\,\mathrm A\end{aligned}$
In the lossless model driving torque provides the same mean power.
$\begin{aligned}\overline\tau&=\frac{\overline P}{\omega}\\&=\frac{\frac{\pi^2}{2}\,\mathrm W}{50\pi\,\mathrm{s^{-1}}}\\&=\frac\pi{100}\,\mathrm{N\,m}\\&\approx0{,}03142\,\mathrm{N\,m}\end{aligned}$

Result: Mean load power is about 4.93 W; the required mean driving torque is about 0.0314 N·m.

Check: $I_{\mathrm{eff}}^2R$ gives the same power. Energy is transferred although signed mean voltage is zero. At constant speed the drive balances electromagnetic opposing torque; its instantaneous value is not constant in this single-coil model.

From the model to a machine

The relation explains why more active turns, larger effective area, stronger field or greater speed increase ideal peak voltage. In a real machine these variables cannot be increased independently without limit: space, copper resistance, field geometry, mechanical strength and cooling constrain the design. A commutator can periodically switch terminal direction; this changes the external waveform rather than Faraday’s induction principle.[1][4]

In a motor the same induction appears as back EMF. Generator and motor therefore represent different operating directions of electromagnetic energy conversion rather than two contradictory force laws. An electrically loaded generator demands additional mechanical work. An EMF source does not imply free energy.[1][4]

Five exercises on magnitude, phase and power

Calculate · Foundation

For $N=60$ , $B=0.40\,\mathrm T$ , $A=0.0050\,\mathrm{m^2}$ and $\omega=120\,\mathrm{rad/s}$ , find peak and RMS voltage.

Show hint

First calculate NBAω, then divide by the square root of two.

Learning goal: Calculate peak and RMS sinusoidal voltage separately.

Show solution and calculation

Approach: $\mathcal E_0=NBA\omega$ .
Peak voltage is 14.4 V; RMS voltage is about 10.2 V.
RMS must be smaller than peak.

\begin{aligned}\mathcal E_0&=60\cdot0{,}40\cdot0{,}0050\cdot120\,\mathrm V=14{,}4\,\mathrm V\\\mathcal E_{\mathrm{eff}}&=\frac{14{,}4}{\sqrt2}\,\mathrm V\approx10{,}18\,\mathrm V\end{aligned}

[1][3]

Rearrange · Application

With $N=150$ , $B=0.20\,\mathrm T$ and $\omega=100\,\mathrm{rad/s}$ , peak voltage should be 18 V. What A is required in the ideal model?

Show hint

Divide target peak voltage by NBω.

Learning goal: Determine required turn area from a peak-voltage target.

Show solution and calculation

Approach: $A=\frac{\mathcal E_0}{NB\omega}$ .
The required area is $0.0060\,\mathrm{m^2}=60\,\mathrm{cm^2}$ .
This is area per turn rather than all turn areas added together.

A=\frac{\mathcal E_0}{NB\omega}=\frac{18}{150\cdot0{,}20\cdot100}\,\mathrm{m^2}=0{,}0060\,\mathrm{m^2}

[1]

Interpret · Application

Flux is at its positive maximum at t=0. For $0,\frac{T}{4},\frac{T}{2},\frac{3T}{4}$ , identify the signs or extremal values of flux and EMF.

Show hint

Flux follows cosine; its negative derivative gives the sine waveform.

Learning goal: Distinguish the times of maximum flux and maximum voltage.

Show solution and calculation

Approach: $\mathcal E=-N\frac{\mathrm d\Phi}{\mathrm dt}$ .
Flux sequence: $+BA,0,-BA,0$ .
EMF sequence: $0,+\mathcal E_0,0,-\mathcal E_0$ .
Largest voltage occurs at a flux zero crossing rather than at maximum flux.

\Phi=BA\cos\frac{2\pi t}{T},\qquad\mathcal E=\mathcal E_0\sin\frac{2\pi t}{T}

[1][2]

Error analysis · Application

A calculation uses $A=10\,\mathrm{cm^2}$ , $N=100$ , $B=0.20\,\mathrm T$ , $\omega=50\,\mathrm{rad/s}$ . It replaces A with 0.10 m² and obtains 100 V. Correct the calculation.

Show hint

Square the centimetre-to-metre conversion before substitution.

Learning goal: Locate an error in squared area-unit conversion.

Show solution and calculation

Approach: $1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2}$ . 10 cm² equals $10^{-3}\,\mathrm{m^2}$ .
The correct peak voltage is 1.0 V.
The erroneous area, and therefore the result, was too large by a factor of 100.

\begin{aligned}A&=10(10^{-2}\,\mathrm m)^2=10^{-3}\,\mathrm{m^2}\\\mathcal E_0&=100\cdot0{,}20\cdot10^{-3}\cdot50\,\mathrm V=1{,}0\,\mathrm V\end{aligned}

[1]

Transfer · Challenge

An ideal generator supplies an unchanged resistor. Field, area and turn count stay fixed while angular speed doubles. How do RMS voltage, mean load power and mean driving torque change?

Show hint

Voltage is proportional to angular speed; power in a fixed resistor depends on voltage squared.

Learning goal: Connect voltage, power and torque scaling when rotation speed changes.

Show solution and calculation

Approach: $\overline P=\frac{\mathcal E_0^2}{2R}$ .
RMS voltage and current double.
Mean power quadruples.
Mean driving torque doubles because four times the power is transmitted at twice the angular speed.
A real machine or another load requires additional checks.

\mathcal E_{\mathrm{eff}}\propto\omega,\qquad\overline P\propto\omega^2,\qquad\overline\tau=\frac{\overline P}{\omega}\propto\omega

[1][3][4]

Example: speed changes during rotation

A generator accelerating from rest

Given: An ideal coil has $N=40$ , $B=0.300\,\mathrm T$ , $A=0.0100\,\mathrm{m^2}$ . It starts from rest at maximum positive flux with constant angular acceleration $\alpha=20.0\,\mathrm{rad/s^2}$ . Field and geometry stay fixed.
Find: Find instantaneous emf after 0.500 s.

Approach: Faraday’s law still holds, but angle now grows quadratically rather than linearly in time.
$\begin{aligned}\theta(t)&=\frac{\alpha t^2}{2}\\\omega(t)&=\alpha t\\\mathcal E(t)&=NBA\alpha t\sin\left(\frac{\alpha t^2}{2}\right)\end{aligned}$
Substitution: phase is 2.50 radians and instantaneous angular speed is 10.0 rad/s.
$\begin{aligned}\mathcal E(0.500\,\mathrm s)&=(40)(0.300)(0.0100)(10.0)\sin(2.50)\,\mathrm V\\&\approx0.718167\,\mathrm V\end{aligned}$

Result: Instantaneous emf is about +0.718 V. Voltage is not a sinusoid with fixed frequency and peak amplitude.

Check: Check: instantaneous envelope magnitude is $NBA\omega=1.20\,\mathrm V$ , larger than the calculated value. At startup both angular speed and emf vanish. The previous sinusoidal RMS shortcut cannot be applied unchanged to this startup interval.

Validity limits and actual terminal voltage

The sinusoidal formula assumes constant angular speed, an unchanged uniform field and the appropriate rotation geometry. With changing speed, the time-dependent angle must be differentiated. A spatially varying field requires the area integral of the actual field. Several turns can be combined as N times one flux only when their flux histories agree.[2][1]

Under load, internal resistance, inductance and current feedback on the field matter. Friction, eddy currents and additional losses also change required drive power. The ideal relation is therefore a calculable reference model for induced EMF rather than a replacement for a machine characteristic or terminal model.[1][4]

What to take away

Find flux through one turn and its rate of change. Add matching turn contributions and distinguish peak, instantaneous and RMS values. Power and opposing torque additionally require a load model. Nonuniform rotation still obeys induction, but no longer gives a sinusoid of constant frequency.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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October 2, 2026Explicitly distinguished peak magnitude from signed angular velocity; added question headings and paragraph connections to Faraday’s law and RMS values.
October 1, 2026Added an original DE/EN learning article with model limits, two worked examples and five distinct solved exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Induced emf of a rotating generator. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/generator-induktionsspannung

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Frequently asked questions about Induced emf of a rotating generator

What is the peak value?+

Peak emf $\mathcal E_0=NBA|\omega|$ is the largest magnitude of the ideal sinusoidal emf. Instantaneous values range between its negative and positive values. RMS emf instead equals $\frac{\mathcal E_0}{\sqrt2}$ , giving the same mean heating power in an ideal resistor as a corresponding DC voltage. Phase shifts the waveform, not its peak. Area and turn count refer to identical active turns. Loaded terminal voltage need not equal ideal emf: winding resistance, inductance and back reaction can modify it. In a purely resistive equivalent circuit, instantaneous internal drop is $i(t)r_i$ . AC signs and the load model must therefore be specified before claiming that every instantaneous terminal value is simply smaller.

Why is the voltage sinusoidal?+

Uniform rotation in the appropriate geometry makes the angle between area normal and field grow linearly in time. Single-turn flux is $\Phi=BA\cos(\omega t+\theta_0)$ . Faraday’s law takes its negative time derivative, giving $\mathcal E=NBA\omega\sin(\omega t+\theta_0)$ . Sine and cosine describe the same process when their time origins are shifted accordingly. Sign is tied to the selected circulation direction and area normal. The quarter-cycle displacement means that flux and voltage maxima do not occur together. If rotation speed changes, differentiate the actual time-dependent angle. Voltage then generally ceases to be a sinusoid with fixed frequency and constant peak amplitude.

How does rotation rate relate to output frequency?+

The simple generator shown produces one full electrical period per mechanical revolution, so $f=\frac{|\omega|}{2\pi}$ . A speed of 1500 revolutions per minute means 25 revolutions per second and $\omega=50\pi\,\mathrm{rad/s}$ . An idealized machine with p pole pairs instead has $f_{\mathrm{el}}=p f_{\mathrm{mech}}$ : the winding passes through several magnetic cycles per revolution. This is a statement about field geometry, not an additional universal turn count. Check whether a stated frequency is mechanical or electrical. In the simple model, doubling speed also doubles peak emf if field, area and turns remain unchanged. Actual regulation and loading can limit that scaling.

When is the emf zero?+

For the selected initial orientation, $\mathcal E(t)=NBA\omega\sin(\omega t)$ . Emf vanishes at $\omega t=k\pi$ , where k is an integer. Flux is then at a maximum or minimum and its derivative is zero. At a flux zero crossing, change is fastest and emf magnitude is maximal. A different initial phase shifts these times together. This describes generated emf for prescribed rotation. Load-current phase additionally depends on resistance, inductance and capacitance; it cannot be obtained from flux differentiation alone. In a loaded real machine, field back reaction can also matter. Zero instantaneous voltage does not imply zero mean energy transfer over a whole period.

How can generator emf be increased?+

In the ideal model, peak emf grows in proportion to turns, field magnitude, area per turn and angular speed. Doubling one while preserving all other assumptions doubles $\mathcal E_0$ . A larger area means enclosed active turn area, not increased wire cross-section. Higher speed also increases electrical frequency. With an unchanged ideal resistor load, doubling speed would quadruple mean power and demand greater driving torque. In real machines, copper losses, cooling, insulation, mechanical strength and possible core saturation constrain scaling. Changing pole pairs or field geometry requires a revised flux model rather than an extra factor that applies universally to every simple coil formula.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 2, §13.6: Electric Generators and Back Emf

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Induced emf of an ideal generator

A planar coil rotating uniformly in a uniform magnetic field generates sinusoidal emf when its area normal sweeps the angle to the field uniformly.

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Symbol

Meaning

Unit

ℰ(t)

Instantaneous induced emf

N

Number of turns

dimensionless

B

Magnetic flux density

A

Coil area

m²

ω

Signed angular velocity relative to the chosen rotation direction

rad/s

t

Time

Derivation & background

Exam blueprint

Validity range

Derivation steps

Projected coil area determines flux. Its time derivative gives emf and the quarter-period shift.

1Define the area normal and matching loop direction; θ is the angle from the normal to the magnetic field.
2For uniform rotation in this geometry, $\theta=\omega t+\theta_0$ .
3Flux per turn is $\Phi=BA\cos(\omega t+\theta_0)$ ; equal turns have flux linkage $\Psi=N\Phi$ .
4Apply Faraday: $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ . The negative sign and cosine derivative together give the sine sign.
5 $\mathcal E(t)=NBA\omega\sin(\omega t+\theta_0)$ . Setting $\theta_0=0$ gives the displayed formula.
6Check $\mathcal E_0=NBA\omega$ , $f=\frac{\omega}{2\pi}$ and, for a sinusoid, $\mathcal E_{\mathrm{rms}}=\frac{\mathcal E_0}{\sqrt2}$ . At maximum flux the emf must be zero.

How do you calculate with Induced emf of a rotating generator?

Here is how to work through a typical Induced emf of a rotating generator (ℰ(t) = NBAω sin(ωt)) task step by step:

Task

Solution path

Approach: $\mathcal E_0=NBA\omega$ and $\mathcal E=\mathcal E_0\sin(\omega t)$ .
Substitution gives $\mathcal E_0=10\,\mathrm V$ and $\mathcal E=5.0\,\mathrm V$ .

Check: at thirty degrees, instantaneous emf is half the peak; maximum magnitude remains 10 V.

Understand and apply the equation

Choose angle and reference direction first

\begin{aligned}\theta(t)&=\omega t+\theta_0\\\Phi(t)&=BA\cos\theta(t)\\\Psi(t)&=N\Phi(t)\end{aligned}

θ is the angle between field and area normal; Φ is flux through one turn and Ψ is total flux linkage.[1][2]

How do generator emf, zero crossings and peak amplitude arise?

\begin{aligned}\mathcal E(t)&=-\frac{d\Psi}{dt}\\&=-NBA\frac{d}{dt}\cos(\omega t+\theta_0)\\&=NBA\omega\sin(\omega t+\theta_0)\end{aligned}

Faraday’s minus sign and the derivative of cosine together give the chosen sine sign.[1][2]

Flux linkage and voltage: different scales

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Original curves for the first example: flux linkage divided by 0.240 Wb and emf by 24.0 V, hence separate dimensionless normalizations. Shared height does not mean equal physical units.[1][2]

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\begin{aligned}\mathcal E_0&=NBA\omega\\\omega&=2\pi f\\T&=\frac1f\end{aligned}

With one full electrical period per mechanical revolution, f equals the rotation frequency.[1]

How do peak, mean and RMS emf differ?

\begin{aligned}\langle\mathcal E\rangle_T&=0\\\mathcal E_{\mathrm{eff}}&=\sqrt{\frac1T\int_0^T\mathcal E^2(t)\,dt}=\frac{\mathcal E_0}{\sqrt2}\end{aligned}

A sinusoidal voltage has zero signed mean but a nonzero RMS value.[3]

Track one coil through a revolution

Calculate peak and instantaneous voltage

Given: $N=80$ , $B=0.25\,\mathrm T$ , $A=0.012\,\mathrm{m^2}$ , $\omega=100\,\mathrm{rad/s}$ . Flux is at its positive maximum at t=0.
Find: Find peak voltage, RMS voltage and voltage at $t=\frac{\pi}{600}\,\mathrm s$ .

Approach: First determine peak voltage and phase symbolically.
$\begin{aligned}\mathcal E_0&=NBA\omega\\\theta&=\omega t\end{aligned}$
Substitute the coil data.
$\mathcal E_0=80\cdot0{,}25\,\mathrm T\cdot0{,}012\,\mathrm{m^2}\cdot100\,\mathrm{s^{-1}}=24\,\mathrm V$
Calculate phase and instantaneous voltage.
$\begin{aligned}\theta&=100\cdot\frac{\pi}{600}=\frac{\pi}{6}\\\mathcal E(t)&=24\,\mathrm V\sin\frac{\pi}{6}=12\,\mathrm V\end{aligned}$
Convert peak voltage to RMS voltage.
$\mathcal E_{\mathrm{eff}}=\frac{24}{\sqrt2}\,\mathrm V\approx16{,}97\,\mathrm V$

Result: Peak voltage is 24 V, RMS voltage about 17.0 V and instantaneous voltage 12 V after about 5.24 ms.

Check: The phase is 30°, so instantaneous voltage is half the peak. Frequency is $\frac{100}{2\pi}\approx15.9\,\mathrm{Hz}$ ; the selected time precedes the first maximum at about 15.7 ms.

A load requires mechanical driving power

Voltage, load power and average opposing torque

Given: An ideal generator has $N=100$ , $B=0.20\,\mathrm T$ , $A=0.010\,\mathrm{m^2}$ and $f=25\,\mathrm{Hz}$ . Its sole load is $R=100\,\Omega$ ; winding and mechanical losses are neglected.
Find: Determine mean electrical power and magnitude of the required mean driving torque.

Approach: Connect angular speed to voltage and power.
$\begin{aligned}\omega&=2\pi f\\\mathcal E_0&=NBA\omega\\\overline P&=\frac{\mathcal E_0^2}{2R}\end{aligned}$
Substitute frequency and coil data.
$\begin{aligned}\omega&=50\pi\,\mathrm{rad/s}\\\mathcal E_0&=100\cdot0{,}20\cdot0{,}010\cdot50\pi\,\mathrm V=10\pi\,\mathrm V\end{aligned}$
Calculate mean power and RMS current.
$\begin{aligned}\overline P&=\frac{(10\pi\,\mathrm V)^2}{2\cdot100\,\Omega}=\frac{\pi^2}{2}\,\mathrm W\approx4{,}935\,\mathrm W\\I_{\mathrm{eff}}&=\frac{10\pi}{100\sqrt2}\,\mathrm A\approx0{,}2221\,\mathrm A\end{aligned}$
In the lossless model driving torque provides the same mean power.
$\begin{aligned}\overline\tau&=\frac{\overline P}{\omega}\\&=\frac{\frac{\pi^2}{2}\,\mathrm W}{50\pi\,\mathrm{s^{-1}}}\\&=\frac\pi{100}\,\mathrm{N\,m}\\&\approx0{,}03142\,\mathrm{N\,m}\end{aligned}$

Result: Mean load power is about 4.93 W; the required mean driving torque is about 0.0314 N·m.

From the model to a machine

Five exercises on magnitude, phase and power

Calculate · Foundation

For $N=60$ , $B=0.40\,\mathrm T$ , $A=0.0050\,\mathrm{m^2}$ and $\omega=120\,\mathrm{rad/s}$ , find peak and RMS voltage.

Show hint

First calculate NBAω, then divide by the square root of two.

Learning goal: Calculate peak and RMS sinusoidal voltage separately.

Show solution and calculation

Approach: $\mathcal E_0=NBA\omega$ .
Peak voltage is 14.4 V; RMS voltage is about 10.2 V.
RMS must be smaller than peak.

\begin{aligned}\mathcal E_0&=60\cdot0{,}40\cdot0{,}0050\cdot120\,\mathrm V=14{,}4\,\mathrm V\\\mathcal E_{\mathrm{eff}}&=\frac{14{,}4}{\sqrt2}\,\mathrm V\approx10{,}18\,\mathrm V\end{aligned}

[1][3]

Rearrange · Application

With $N=150$ , $B=0.20\,\mathrm T$ and $\omega=100\,\mathrm{rad/s}$ , peak voltage should be 18 V. What A is required in the ideal model?

Show hint

Divide target peak voltage by NBω.

Learning goal: Determine required turn area from a peak-voltage target.

Show solution and calculation

Approach: $A=\frac{\mathcal E_0}{NB\omega}$ .
The required area is $0.0060\,\mathrm{m^2}=60\,\mathrm{cm^2}$ .
This is area per turn rather than all turn areas added together.

A=\frac{\mathcal E_0}{NB\omega}=\frac{18}{150\cdot0{,}20\cdot100}\,\mathrm{m^2}=0{,}0060\,\mathrm{m^2}

[1]

Interpret · Application

Flux is at its positive maximum at t=0. For $0,\frac{T}{4},\frac{T}{2},\frac{3T}{4}$ , identify the signs or extremal values of flux and EMF.

Show hint

Flux follows cosine; its negative derivative gives the sine waveform.

Learning goal: Distinguish the times of maximum flux and maximum voltage.

Show solution and calculation

Approach: $\mathcal E=-N\frac{\mathrm d\Phi}{\mathrm dt}$ .
Flux sequence: $+BA,0,-BA,0$ .
EMF sequence: $0,+\mathcal E_0,0,-\mathcal E_0$ .
Largest voltage occurs at a flux zero crossing rather than at maximum flux.

\Phi=BA\cos\frac{2\pi t}{T},\qquad\mathcal E=\mathcal E_0\sin\frac{2\pi t}{T}

[1][2]

Error analysis · Application

A calculation uses $A=10\,\mathrm{cm^2}$ , $N=100$ , $B=0.20\,\mathrm T$ , $\omega=50\,\mathrm{rad/s}$ . It replaces A with 0.10 m² and obtains 100 V. Correct the calculation.

Show hint

Square the centimetre-to-metre conversion before substitution.

Learning goal: Locate an error in squared area-unit conversion.

Show solution and calculation

Approach: $1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2}$ . 10 cm² equals $10^{-3}\,\mathrm{m^2}$ .
The correct peak voltage is 1.0 V.
The erroneous area, and therefore the result, was too large by a factor of 100.

\begin{aligned}A&=10(10^{-2}\,\mathrm m)^2=10^{-3}\,\mathrm{m^2}\\\mathcal E_0&=100\cdot0{,}20\cdot10^{-3}\cdot50\,\mathrm V=1{,}0\,\mathrm V\end{aligned}

[1]

Transfer · Challenge

An ideal generator supplies an unchanged resistor. Field, area and turn count stay fixed while angular speed doubles. How do RMS voltage, mean load power and mean driving torque change?

Show hint

Voltage is proportional to angular speed; power in a fixed resistor depends on voltage squared.

Learning goal: Connect voltage, power and torque scaling when rotation speed changes.

Show solution and calculation

Approach: $\overline P=\frac{\mathcal E_0^2}{2R}$ .
RMS voltage and current double.
Mean power quadruples.
Mean driving torque doubles because four times the power is transmitted at twice the angular speed.
A real machine or another load requires additional checks.

\mathcal E_{\mathrm{eff}}\propto\omega,\qquad\overline P\propto\omega^2,\qquad\overline\tau=\frac{\overline P}{\omega}\propto\omega

[1][3][4]

Example: speed changes during rotation

A generator accelerating from rest

Given: An ideal coil has $N=40$ , $B=0.300\,\mathrm T$ , $A=0.0100\,\mathrm{m^2}$ . It starts from rest at maximum positive flux with constant angular acceleration $\alpha=20.0\,\mathrm{rad/s^2}$ . Field and geometry stay fixed.
Find: Find instantaneous emf after 0.500 s.

Approach: Faraday’s law still holds, but angle now grows quadratically rather than linearly in time.
$\begin{aligned}\theta(t)&=\frac{\alpha t^2}{2}\\\omega(t)&=\alpha t\\\mathcal E(t)&=NBA\alpha t\sin\left(\frac{\alpha t^2}{2}\right)\end{aligned}$
Substitution: phase is 2.50 radians and instantaneous angular speed is 10.0 rad/s.
$\begin{aligned}\mathcal E(0.500\,\mathrm s)&=(40)(0.300)(0.0100)(10.0)\sin(2.50)\,\mathrm V\\&\approx0.718167\,\mathrm V\end{aligned}$

Result: Instantaneous emf is about +0.718 V. Voltage is not a sinusoid with fixed frequency and peak amplitude.

Validity limits and actual terminal voltage

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Explicitly distinguished peak magnitude from signed angular velocity; added question headings and paragraph connections to Faraday’s law and RMS values.
October 1, 2026Added an original DE/EN learning article with model limits, two worked examples and five distinct solved exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Induced emf of a rotating generator. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/generator-induktionsspannung

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Create a curated FSRS exam set for ℰ(t) = NBAω sin(ωt): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Induced emf of an ideal generator

Variables & units

Generator induced voltage calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Rotating coil

Worked example

Applications

How do you calculate with Induced emf of a rotating generator?

Choose angle and reference direction first

How do generator emf, zero crossings and peak amplitude arise?

Flux linkage and voltage: different scales

How do peak, mean and RMS emf differ?

Track one coil through a revolution

Calculate peak and instantaneous voltage

A load requires mechanical driving power

Voltage, load power and average opposing torque

From the model to a machine

Five exercises on magnitude, phase and power

For N=60N=60N=60, B=0.40 TB=0.40\,\mathrm TB=0.40T, A=0.0050 m2A=0.0050\,\mathrm{m^2}A=0.0050m2 and ω=120 rad/s\omega=120\,\mathrm{rad/s}ω=120rad/s, find peak and RMS voltage.

With N=150N=150N=150, B=0.20 TB=0.20\,\mathrm TB=0.20T and ω=100 rad/s\omega=100\,\mathrm{rad/s}ω=100rad/s, peak voltage should be 18 V. What A is required in the ideal model?

Flux is at its positive maximum at t=0. For 0,T4,T2,3T40,\frac{T}{4},\frac{T}{2},\frac{3T}{4}0,4T​,2T​,43T​, identify the signs or extremal values of flux and EMF.

A calculation uses A=10 cm2A=10\,\mathrm{cm^2}A=10cm2, N=100N=100N=100, B=0.20 TB=0.20\,\mathrm TB=0.20T, ω=50 rad/s\omega=50\,\mathrm{rad/s}ω=50rad/s. It replaces A with 0.10 m2 and obtains 100 V. Correct the calculation.

An ideal generator supplies an unchanged resistor. Field, area and turn count stay fixed while angular speed doubles. How do RMS voltage, mean load power and mean driving torque change?

Example: speed changes during rotation

A generator accelerating from rest

Validity limits and actual terminal voltage

What to take away

Sources for this explanation

Frequently asked questions about Induced emf of a rotating generator

Subject references and further reading

More formulas: Electricity and magnetism

Retain Induced emf of a rotating generator for exams

Induced emf of an ideal generator

Variables & units

Generator induced voltage calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Rotating coil

Worked example

Applications

How do you calculate with Induced emf of a rotating generator?

Choose angle and reference direction first

How do generator emf, zero crossings and peak amplitude arise?

Flux linkage and voltage: different scales

How do peak, mean and RMS emf differ?

Track one coil through a revolution

Calculate peak and instantaneous voltage

A load requires mechanical driving power

Voltage, load power and average opposing torque

From the model to a machine

Five exercises on magnitude, phase and power

For N=60N=60N=60, B=0.40 TB=0.40\,\mathrm TB=0.40T, A=0.0050 m2A=0.0050\,\mathrm{m^2}A=0.0050m2 and ω=120 rad/s\omega=120\,\mathrm{rad/s}ω=120rad/s, find peak and RMS voltage.

With N=150N=150N=150, B=0.20 TB=0.20\,\mathrm TB=0.20T and ω=100 rad/s\omega=100\,\mathrm{rad/s}ω=100rad/s, peak voltage should be 18 V. What A is required in the ideal model?

Flux is at its positive maximum at t=0. For 0,T4,T2,3T40,\frac{T}{4},\frac{T}{2},\frac{3T}{4}0,4T​,2T​,43T​, identify the signs or extremal values of flux and EMF.

A calculation uses A=10 cm2A=10\,\mathrm{cm^2}A=10cm2, N=100N=100N=100, B=0.20 TB=0.20\,\mathrm TB=0.20T, ω=50 rad/s\omega=50\,\mathrm{rad/s}ω=50rad/s. It replaces A with 0.10 m2 and obtains 100 V. Correct the calculation.

An ideal generator supplies an unchanged resistor. Field, area and turn count stay fixed while angular speed doubles. How do RMS voltage, mean load power and mean driving torque change?

Example: speed changes during rotation

A generator accelerating from rest

Validity limits and actual terminal voltage

What to take away

Sources for this explanation

Frequently asked questions about Induced emf of a rotating generator

Subject references and further reading

More formulas: Electricity and magnetism

Retain Induced emf of a rotating generator for exams

For $N=60$ , $B=0.40\,\mathrm T$ , $A=0.0050\,\mathrm{m^2}$ and $\omega=120\,\mathrm{rad/s}$ , find peak and RMS voltage.

With $N=150$ , $B=0.20\,\mathrm T$ and $\omega=100\,\mathrm{rad/s}$ , peak voltage should be 18 V. What A is required in the ideal model?

Flux is at its positive maximum at t=0. For $0,\frac{T}{4},\frac{T}{2},\frac{3T}{4}$ , identify the signs or extremal values of flux and EMF.

A calculation uses $A=10\,\mathrm{cm^2}$ , $N=100$ , $B=0.20\,\mathrm T$ , $\omega=50\,\mathrm{rad/s}$ . It replaces A with 0.10 m² and obtains 100 V. Correct the calculation.

For $N=60$ , $B=0.40\,\mathrm T$ , $A=0.0050\,\mathrm{m^2}$ and $\omega=120\,\mathrm{rad/s}$ , find peak and RMS voltage.

With $N=150$ , $B=0.20\,\mathrm T$ and $\omega=100\,\mathrm{rad/s}$ , peak voltage should be 18 V. What A is required in the ideal model?

Flux is at its positive maximum at t=0. For $0,\frac{T}{4},\frac{T}{2},\frac{3T}{4}$ , identify the signs or extremal values of flux and EMF.

A calculation uses $A=10\,\mathrm{cm^2}$ , $N=100$ , $B=0.20\,\mathrm T$ , $\omega=50\,\mathrm{rad/s}$ . It replaces A with 0.10 m² and obtains 100 V. Correct the calculation.