Physics · Electrodynamics

Faraday’s law of induction

A change in oriented flux linkage produces loop emf. Terminal voltage additionally requires a defined measurement and circuit model.

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Formula

Calculate

E = - N \frac{d Φ _{B}}{d t}

ℰ in V; N dimensionless; ΦB in Wb = V·s, flux per turn.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ℰ$	Oriented induced loop emf, not generally a unique potential difference	V
$N$	Fixed number of equally oriented turns with equal flux	dimensionless
$ΦB$	Oriented magnetic flux per turn; generally a surface integral	Wb

LaTeX

\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}

QUANTA STUDYquanta-study.de

Flux per turn rises from zero to 0.9 milliwebers in thirty milliseconds and returns to zero by ninety milliseconds. Emf for eighty turns is first minus 2.4 volts and then plus 1.2 volts. — Original pulse example: N = 80 equal turns. Steeper flux rise produces the larger negative emf magnitude. The two volt-second areas cancel; derivative is not unique at ideal corners. OpenStax §§13.1–13.2.

Faraday's law calculator

Find the mean induced emf in a coil from the change in flux, or work back to the flux or the time; the sign follows Lenz's law.

Find

Number of turns N

Number of turns of the coil as a whole number, e.g. 200.

Initial flux per turn Φ₁

Flux through one turn at the start, Φ = B · A for a perpendicular field, e.g. 1 mWb; negative if B opposes the normal.

Final flux per turn Φ₂

Flux through one turn at the end, with the same orientation as at the start, e.g. 4 mWb.

Elapsed time Δt

Duration of the flux change, e.g. 0.1 s; the faster the change, the larger the magnitude of the emf.

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

Result: Mean induced emf ε

ε = -6 V

With N = 200, Φ₁ = 1 mWb, Φ₂ = 4 mWb and Δt = 0.1 s, ε = -6 V.

Working · rearrange for ε, substitute, evaluate

1. Rearrange for ε
$\bar\varepsilon=-N\frac{\Phi_2-\Phi_1}{\Delta t}$
2. Substitute values (in SI units)
- Φ₁ = 1 mWb = 0.001 Wb
- Φ₂ = 4 mWb = 0.004 Wb
$\bar\varepsilon=-200\cdot \frac{0.004-0.001}{0.1}=-6$
3. Result
ε = -6 V

Quick tips

Sign by Lenz's law: if the flux grows, the mean emf is negative; if it falls, the emf is positive. Its magnitude is N · |ΔΦ| / Δt.
Unit check: 1 Wb = 1 V·s. Turns times webers divided by seconds therefore gives volts, e.g. 200 · 3 mWb / 0.1 s = 6 V.

Assumptions and limits

N fixed, equally oriented turns connected in series. Φ is oriented magnetic flux per turn. The result is mean emf −NΔΦ/Δt, not an arbitrary instantaneous derivative; orientation sets the sign.

Derivation & background

Faraday’s law is experimentally established. Its minus sign belongs to the chosen normal and traversal. In a closed resistive loop, induced current opposes the flux change; it does not always oppose the existing field.

Exam blueprint

Validity range

Oriented traversal and flux linkage must match. The N shortcut assumes equal flux per turn. Emf is not generally a path-independent terminal voltage; distinguish mean and instantaneous values.

Derivation steps

Faraday’s experimentally established law connects oriented flux change to loop emf; the product rule separates possible causes.

1Choose a positive surface normal and its right-hand positive traversal.
2Calculate $\Phi_k=\int_{S_k}\vec B\cdot\mathrm d\vec A$ for each turn.
3Sum flux linkage $\Psi=\sum_k\Phi_k$ .
4Apply the law $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ ; it does not follow from the definition alone.
5Only equal oriented turn fluxes and fixed $N$ reduce this to $-N\frac{\mathrm d\Phi_B}{\mathrm dt}$ .
6Differentiate all time-dependent factors; a finite difference gives only the mean.
7Check direction with Lenz’s law and calculate current only using a specified circuit model.

Rearrangements

Flux-change rate

\frac{\mathrm d\Phi_B}{\mathrm dt}=-\frac{\mathcal E}{N}

Equal signed flux per turn and fixed positive turn count; emf uses traversal corresponding to the normal. For a finite interval use mean emf.

Task variant

How can you increase induced emf magnitude without increasing field magnitude? Compare a stationary coil in a constant field with a moving or time-varying arrangement.

Approach: $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ .
More equally oriented turns sharing the same per-turn change or a faster change increase magnitude.
Larger area helps only if it produces a greater time-dependent flux change.
An unchanged stationary coil in a constant field still has zero emf.
Real resistance and self-inductance additionally determine load current.

Common mistakes

Considering only B and forgetting area or angle changes.

For a uniform field and plane area $\Phi_B=BA\cos\theta$ . Differentiate every factor that actually varies; constant flux does not mean zero field.

Exam context

Typical in generator, transformer and Lenz-law tasks.

Formula cluster

Induction and fields

Bridges magnetic force and electric voltage.

ΦB = BA cos θ

Magnetic flux through a surface

|ℰmot| = Bℓv

Motional emf

U_ind = −L·(dI/dt)

Self-Induction of a Coil

U₁/U₂ = n₁/n₂ = I₂/I₁

Transformer turns ratio

Worked example

200 equally oriented turns; per-turn flux rises by 50 mWb in 0.1 s. Mean emf is $\overline{\mathcal E}=-200\frac{0.050\,\mathrm{Wb}}{0.1\,\mathrm s}=-100\,\mathrm V$ . Only a linear flux change also makes this the constant instantaneous value.

Applications

Electric motors, generators, transformers, induction cooktops

How do you calculate with Faraday’s law of induction?

Here is how to work through a typical Faraday’s law of induction (ℰ = −N dΦB/dt) task step by step:

1
Task
How can you increase induced emf magnitude without increasing field magnitude? Compare a stationary coil in a constant field with a moving or time-varying arrangement.
Solution path
1. Approach: $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ .
2. More equally oriented turns sharing the same per-turn change or a faster change increase magnitude.
3. Larger area helps only if it produces a greater time-dependent flux change.
4. An unchanged stationary coil in a constant field still has zero emf.
5. Real resistance and self-inductance additionally determine load current.

Understand and apply the equation

A flux change determines the electromotive force around a circuit, not automatically an arbitrary reading between two terminals. Orientation fixes the sign; flux linkage fixes the turn factor. A field ramp and a flux pulse show how current, transported charge and heat follow in an explicitly defined circuit model.

The reading responds to change

A search coil can sit in a strong unchanging field and still produce no induced emf. A rapidly changing flux, however, can produce a clear response even for a small change. The key question is therefore not just “How large is flux?” but “How quickly and in which direction does it change?”[1][2]

You need the signed flux curve, the turns actually linked by it and a fixed positive traversal. A curve slope gives instantaneous emf; joining two times gives its average only. Whether current flows and how large it is then follows from the circuit model.[1][2]

When is flux linkage N times the same flux?

\begin{aligned}\Psi&=\sum_{k=1}^N\Phi_k\\\mathcal E&=-\frac{\mathrm d\Psi}{\mathrm dt}\\\Psi=N\Phi_B&\ \Longrightarrow\ \mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}\end{aligned}

The shortcut requires a fixed turn count and equal oriented flux through every turn.[1][4]

Despite its historical name, emf is not a force in newtons but a quantity in volts. It describes the driving contribution per charge for an oriented traversal. Each turn’s magnetic flux contributes to flux linkage. Only with a fixed number of turns and equal oriented flux through every turn does the sum equal N times that common flux. Turns in different fields must not be blindly replaced by one common area. A conductor is not required for the electric-field induction law; a conducting connection and its circuit model determine whether current flows.[3][1]

A normal defines positive traversal

Choose the positive surface normal; its right-hand traversal is positive. Viewed from the positive-normal side toward the surface, this is counterclockwise. Increasing positive flux gives negative emf: a simple resistive circuit then carries current in the negative traversal. Decreasing positive flux gives positive emf. Lenz’s law opposes the change, rather than always opposing the existing field.[2][4]

\begin{aligned}\frac{\mathrm d\Phi_B}{\mathrm dt}>0&\ \Longrightarrow\ \mathcal E<0\\\frac{\mathrm d\Phi_B}{\mathrm dt}<0&\ \Longrightarrow\ \mathcal E>0\end{aligned}

These signs use the previously defined orientation and equally oriented turns.[2]

The sign can therefore be checked without memorising another hand rule: choose a normal, read the signed flux slope and reverse its sign. Only then translate traversal into spatial current direction using the right-hand rule.[2]

Induced emf: instantaneous or average value?

In the emf model, instantaneous induced voltage is the negative time derivative of flux linkage. The average over a time interval instead uses the negative finite change in flux linkage divided by the elapsed time. Faraday’s law was established through induction experiments and is part of Maxwell’s equations. It is not proved here from the flux definition alone. Once accepted as a law, differentiation separates the contributions: field magnitude, area and orientation can change. For a uniform field and plane surface these quantities are time functions; a factor may be moved outside the derivative only when it really is constant.[1][4]

\begin{aligned}\Phi_B&=BA\cos\theta\\\frac{\mathrm d\Phi_B}{\mathrm dt}&=A\cos\theta\,\dot B\\&\quad+B\cos\theta\,\dot A\\&\quad-BA\sin\theta\,\dot\theta\end{aligned}

The three lines describe change contributions, not three separate laws.[4]

A moving conducting loop can also change its enclosed area. Motional emf uses a perpendicular moving rod to connect the magnetic force on charges with changing flux; the chosen orientation determines polarity.[4]

Is induced emf the measured terminal voltage?

\oint_C\vec E\cdot\mathrm d\vec\ell=-\frac{\mathrm d\Phi_B}{\mathrm dt}

For a boundary fixed in the chosen reference frame, the induced electric field is generally nonconservative.[3]

Loop emf is not generally a unique electrostatic potential difference between two points. Measurement leads define their own path and may enclose additional changing flux. A terminal-voltage approximation requires specified terminals, lead routing and a lumped component model. In the usual search-coil model with high-impedance measurement and negligible additional lead induction, measured voltage can be identified with modelled emf. Under load, resistance drops and possibly self-induction also enter.[3][4]

Example: a linear field ramp

Emf and heat in a loaded measurement coil

Given: 120 equally oriented turns, each $A=0.00600\,\mathrm{m^2}$ , normal +z. Uniform external field +z rises linearly from 0.0500 to 0.170 T in 0.0800 s. Total resistance is $R=18.0\,\Omega$ ; self-induction and back reaction on the prescribed field are neglected.
Find: Find emf, current direction and heat dissipated during the ramp.

Approach: area and angle are constant, so differentiate only the external field; then use the ideal resistance for current.
$\begin{aligned}\mathcal E&=-NA\frac{\Delta B}{\Delta t}\\I&=\frac{\mathcal E}{R}\\Q_R&=I^2R\Delta t\end{aligned}$
Substitute a positive field ramp; the result is negative in the positive traversal.
$\begin{aligned}\frac{\Delta B}{\Delta t}&=\frac{0.170-0.0500}{0.0800}\,\mathrm{T/s}=1.50\,\mathrm{T/s}\\\mathcal E&=-(120)(0.00600)(1.50)\,\mathrm V=-1.08\,\mathrm V\\I&=\frac{-1.08\,\mathrm V}{18.0\,\Omega}=-0.0600\,\mathrm A\end{aligned}$
Heat depends on squared current and is positive despite the negative current sign.
$\begin{aligned}P_R&=(0.0600\,\mathrm A)^2(18.0\,\Omega)=0.0648\,\mathrm W\\Q_R&=(0.0648\,\mathrm W)(0.0800\,\mathrm s)\\&=0.005184\,\mathrm J\end{aligned}$

Result: Emf is −1.08 V, current −60.0 mA and heat 5.18 mJ. Viewed from +z, current is clockwise and produces a field contribution toward −z.

Check: Per-turn flux change is $A\Delta B=0.720\,\mathrm{mWb}$ . $\mathcal E\Delta t=-0.0864\,\mathrm{V\,s}$ equals $-N\Delta\Phi_B$ . $\mathcal E I=+0.0648\,\mathrm W$ matches resistor power; energy is supplied by the source changing the external field.

Example: zero net charge but positive heat

A flux pulse returns to its initial value

Given: 80 equally oriented turns and total resistance 4.00 Ω. Per-turn flux rises linearly from zero to +0.900 mWb in 0.0300 s, then falls linearly to zero in 0.0600 s. Self-induction is negligible; external flux is prescribed.
Find: Calculate both emf values, net transported charge and total heat.

Approach: emf is constant in each linear phase. Add charge with its sign and heat using squared current.
$\begin{aligned}\mathcal E_j&=-N\frac{\Delta\Phi_j}{\Delta t_j}\\q_{\mathrm{net}}&=\sum_j\frac{\mathcal E_j}{R}\Delta t_j\\Q_R&=\sum_j\frac{\mathcal E_j^2}{R}\Delta t_j\end{aligned}$
The faster rising phase has twice the emf magnitude.
$\begin{aligned}\mathcal E_1&=-80\frac{0.000900}{0.0300}\,\mathrm V=-2.40\,\mathrm V\\\mathcal E_2&=-80\frac{-0.000900}{0.0600}\,\mathrm V=+1.20\,\mathrm V\\q_1&=-0.0180\,\mathrm C\\q_2&=+0.0180\,\mathrm C\end{aligned}$
The signs do not cancel heat.
$\begin{aligned}Q_R&=\frac{(2.40)^2}{4.00}(0.0300)\,\mathrm J\\&\quad+\frac{(1.20)^2}{4.00}(0.0600)\,\mathrm J\\&=0.0432\,\mathrm J+0.0216\,\mathrm J\\&=0.0648\,\mathrm J\end{aligned}$

Result: Emf is −2.40 V then +1.20 V; net transported charge in the chosen traversal is zero, but heat is 64.8 mJ.

Check: Integrated emf is $-N(\Phi_f-\Phi_i)=0$ ; constant resistance therefore also gives zero net charge. Mean emf over 0.0900 s is zero, but rms emf is $\sqrt{2.88}\,\mathrm V\approx1.697\,\mathrm V$ .

Instantaneous, mean and rms values

\begin{aligned}[\mathcal E]&=\frac{\mathrm{Wb}}{\mathrm s}=\mathrm V\\\overline{\mathcal E}&=-N\frac{\Delta\Phi_B}{\Delta t}\\\mathcal E_{\mathrm{rms}}&=\sqrt{\frac1{\Delta t}\int_{t_1}^{t_2}\mathcal E(t)^2\,\mathrm dt}\end{aligned}

A finite difference gives the interval mean; it equals the constant instantaneous value only for linear flux change.

For a sinusoidal waveform of peak $\mathcal E_0$ , complete periods give $\mathcal E_{\mathrm{rms}}=\frac{\mathcal E_0}{\sqrt2}$ . Equal positive and negative contributions can cancel the mean while power remains positive. A coil of 200 equally linked turns and a positive per-turn flux change of $50\,\mathrm{mWb}$ in $0.1\,\mathrm s$ gives $-100\,\mathrm V$ as mean emf; without a linear waveform it is not a determined instantaneous value.

View diagram: emf of a curved flux waveform

Emf follows slope, not flux height

QUANTA STUDYquanta-study.de

In the third example flux grows quadratically. Emf decreases linearly from 0 to −2 V; its mean is −1 V. The dashed line is an average, not another waveform.[1]

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Generator, search coil and current-model limits

\begin{aligned}\Phi_B(t)&=BA\cos(\omega t)\\\mathcal E(t)&=NBA\omega\sin(\omega t)\end{aligned}

Uniform field, constant area and constant angular velocity; differentiation gives the second line.[5][4]

Generators change orientation, search coils detect flux changes and transformers couple time-dependent turn fluxes. Additional turns also change real resistance, self-inductance and geometry; proportionally greater unloaded emf does not guarantee proportionally greater usable power. Rapid changes or strongly inductive loads prevent current being set by Ohm’s law alone: field feedback, self-induction and stored energy must then be included.[5][4]

Example: a curved flux waveform

Separate average and final value

Given: A fixed search coil has $N=50$ and equal signed flux per turn $\Phi(t)=\Phi_0(t/\tau)^2$ for $0\le t\le\tau$ , with $\Phi_0=0.800\,\mathrm{mWb}$ , $\tau=0.0400\,\mathrm s$ . No load is specified.
Find: Find mean emf and its left-hand instantaneous value at the end.

Approach: secant slope gives the mean; differentiation gives the waveform.
$\begin{aligned}\overline{\mathcal E}&=-N\frac{\Phi_0}{\tau}\\\mathcal E(t)&=-\frac{2N\Phi_0}{\tau^2}t\end{aligned}$
Substitution: instantaneous slope grows linearly and ends at twice the average.
$\begin{aligned}\overline{\mathcal E}&=-50\frac{0.000800}{0.0400}\,\mathrm V=-1.00\,\mathrm V\\\mathcal E(\tau^-)&=-2.00\,\mathrm V\end{aligned}$

Result: Mean emf is −1.00 V; just before the end it is −2.00 V. Current is undetermined without a load model.

Check: Triangular emf-time area is $-\frac12(2.00\,\mathrm V)(0.0400\,\mathrm s)=-0.0400\,\mathrm{V\,s}$ , equal to $-N\Phi_0$ .

Five exercises on flux waveforms and emf

Calculate · Foundation

40 turns of 0.0120 m² each, positive normal toward +z. The uniform external field falls linearly from +0.300 to +0.100 T in 0.0500 s. Find emf and traversal direction for a purely resistive load.

Show hint

The field remains positive while its derivative is negative.

Learning goal: Convert a negative field change into positive oriented emf.

Show solution and calculation

Approach: $\mathcal E=-NA\frac{\Delta B}{\Delta t}$ .
Result is +1.92 V.
Viewed from +z, emf drives counterclockwise current supporting the declining positive flux.

Check: linkage change is −0.0960 Wb, the negative of emf times duration.

\mathcal E=-(40)(0.0120)\frac{0.100-0.300}{0.0500}\,\mathrm V=+1.92\,\mathrm V

[1][2]

Rearrange · Application

Equal oriented flux per turn changes by +0.250 mWb in 0.0400 s. At least 0.200 V magnitude of mean emf is required. How many equally oriented turns are needed?

Show hint

Turn count must be an integer; round upward if necessary.

Learning goal: Find a minimum turn count for a required mean emf.

Show solution and calculation

Approach: $N\ge\frac{|\overline{\mathcal E}|\Delta t}{|\Delta\Phi_B|}$ .
This gives exactly 32, so at least 32 turns.
Mean emf is then −0.200 V.

Check: 31 turns give only 0.19375 V magnitude and miss the requirement.

N_{\min}=\left\lceil\frac{(0.200)(0.0400)}{0.250\cdot10^{-3}}\right\rceil=32

[1]

Interpret · Application

One turn: flux rises linearly from 0 to +2.00 mWb in 0.100 s, remains constant for 0.200 s, then returns to zero in 0.0500 s. Find emf in the three open intervals. Idealize transitions as corners.

Show hint

Use slope, not the height of the flux graph.

Learning goal: Read rising, constant and falling flux as an emf signal.

Show solution and calculation

Approach: take the negative flux slope in each interval.
Emf values are −20.0 mV, zero and +40.0 mV.
Faster decay has twice the magnitude.

Check: the two emf areas cancel because final flux returns to its initial value. Ordinary derivative is undefined at idealized corners.

\begin{aligned}\mathcal E_1&=-\frac{0.00200}{0.100}\,\mathrm V=-0.0200\,\mathrm V\\\mathcal E_2&=0\\\mathcal E_3&=-\frac{-0.00200}{0.0500}\,\mathrm V=+0.0400\,\mathrm V\end{aligned}

[1]

Error analysis · Application

An ideal sinusoidal emf with peak 6.00 V drives a 12.0 Ω resistor. Self-induction is negligible. A report states $\overline P=3.00\,\mathrm W$ over complete periods. Correct it.

Show hint

The period mean of squared sine is one half.

Learning goal: Distinguish peak emf from its heating-equivalent rms value.

Show solution and calculation

Approach: $\overline P=\frac{\mathcal E_{\mathrm{rms}}^2}{R}$ , not peak squared divided by resistance.
Rms emf is about 4.24 V and mean power 1.50 W.
The stated 3.00 W is maximum instantaneous power.

Check: mean absolute current is not needed for this power calculation.

\begin{aligned}\mathcal E_{\mathrm{rms}}&=\frac{6.00}{\sqrt2}\,\mathrm V\\\overline P&=\frac{(6.00\,\mathrm V)^2}{2(12.0\,\Omega)}=1.50\,\mathrm W\end{aligned}

[5]

Transfer · Application

Two equally oriented series turns have $\dot\Phi_1=+0.00300\,\mathrm{Wb/s}$ and $\dot\Phi_2=-0.00100\,\mathrm{Wb/s}$ during an interval. Find total emf and explain why doubling the first contribution is wrong.

Show hint

Sum fluxes before differentiating.

Learning goal: Treat unequal turn fluxes as a linkage sum.

Show solution and calculation

Approach: $\mathcal E=-\frac{\mathrm d}{\mathrm dt}(\Phi_1+\Phi_2)$ .
Total emf is −2.00 mV.
Doubling the first contribution gives −6.00 mV and ignores the second flux’s different change.

Check: individual emfs −3.00 and +1.00 mV sum in oriented series to −2.00 mV.

\mathcal E=-(0.00300-0.00100)\,\mathrm V=-0.00200\,\mathrm V

[1]

In brief

Faraday connects change in total flux linkage with signed emf. Slope gives the instantaneous value; a finite difference gives the average. Lenz checks direction. Current, terminal voltage and heat then require an explicitly stated circuit model.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

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.

Editorial responsibility, sources and methods

Change history

October 1, 2026Developed a DE/EN explanation with an oriented derivation, two original worked examples and five distinct solved exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.
October 1, 2026Distinguished instantaneous and average induced voltage, highlighted the equal-flux condition and terminal-voltage question and linked appropriate follow-up formulas.
October 2, 2026Linked the existing transformer explanation and placed motional emf in context.

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Quanta. Faraday’s law of induction. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/induktion-faraday

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Frequently asked questions about Faraday’s law of induction

How do you calculate the induced voltage with Faraday’s law?+

First calculate oriented flux linkage and distinguish an instantaneous value from a finite time interval. Faraday’s law gives loop emf as $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ . Over a finite interval, calculate the mean value from the change in flux linkage divided by elapsed time, with a minus sign. If 200 identical turns each experience a positive flux increase of 50 milliwebers in 0.10 seconds, the mean emf is negative 100 volts. This is also the instantaneous value only when flux changes linearly. A measured terminal voltage additionally requires a defined measurement path and an appropriate circuit model; it is not automatically identical to every loop emf.

What does the minus sign in Faraday’s law mean?+

The minus sign connects two orientations fixed in advance: the positive surface normal and its right-hand positive traversal. If positive flux increases, emf in that traversal is negative. In a closed resistive loop, current then flows in the opposite traversal direction. Its own field opposes the increase in flux. If positive flux decreases instead, the current produces a field that reduces that decrease. Lenz’s law therefore concerns the change, rather than always opposing the direction of the existing field. Without a closed conducting path, no corresponding conduction current need flow. Never change an orientation partway through a calculation, because this breaks the traceable assignment of signs.

How can you increase the induced voltage without strengthening the magnetic field?+

You can use more equally oriented turns with the same flux change per turn, or achieve the same change in less time. A larger active area or a different rotational motion can also increase the rate of flux change. Time dependence remains decisive: a larger stationary area in a constant field does not generate emf by itself. In a rotating plane coil, the angle to the normal changes; faster rotation increases emf amplitude under otherwise equal conditions. More turns also change the resistance and inductance of a real coil. A larger calculated open-circuit emf therefore guarantees neither proportionally more load current nor a free increase in power. Additional electrical output requires an energy source.

Why is a strong magnetic field alone not enough for induction?+

Induction depends on the change in oriented flux, rather than on magnetic field strength alone. A stationary loop with unchanged area and orientation has no flux change in a time-independent field, so it has no induced loop emf. This remains true even for a strong field. In that same constant field, however, motion can change enclosed area or orientation and generate emf. Conversely, a changing field can induce emf in a stationary loop. Check field change, area change and rotation separately. A rigid closed loop translating entirely within a uniform constant field has zero total emf, although individual conductor sections may have opposite motional contributions.

What is the difference between magnetic flux and flux density?+

Magnetic flux density describes the local magnetic field as a vector and is measured in tesla. Magnetic flux instead belongs to a selected oriented surface and is measured in webers. It is the surface integral of the component normal to that surface. For a uniform field and plane area, this becomes $\Phi_B=BA\cos\theta$ . The same flux density can therefore produce different fluxes depending on area and orientation. Doubling an area perpendicular to the field doubles its flux; rotating it parallel to the field reduces flux to zero. For multiple turns, additionally distinguish flux linkage, which is the sum of individual fluxes. A literal count of drawn field lines defines none of these quantities.

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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 Vol. 2, §13.1 Faraday’s Law

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Faraday’s law of induction

A change in oriented flux linkage produces loop emf. Terminal voltage additionally requires a defined measurement and circuit model.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Symbol

Meaning

Unit

ℰ

Oriented induced loop emf, not generally a unique potential difference

N

Fixed number of equally oriented turns with equal flux

dimensionless

ΦB

Oriented magnetic flux per turn; generally a surface integral

Exam blueprint

Validity range

Oriented traversal and flux linkage must match. The N shortcut assumes equal flux per turn. Emf is not generally a path-independent terminal voltage; distinguish mean and instantaneous values.

Derivation steps

Faraday’s experimentally established law connects oriented flux change to loop emf; the product rule separates possible causes.

1Choose a positive surface normal and its right-hand positive traversal.
2Calculate $\Phi_k=\int_{S_k}\vec B\cdot\mathrm d\vec A$ for each turn.
3Sum flux linkage $\Psi=\sum_k\Phi_k$ .
4Apply the law $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ ; it does not follow from the definition alone.
5Only equal oriented turn fluxes and fixed $N$ reduce this to $-N\frac{\mathrm d\Phi_B}{\mathrm dt}$ .
6Differentiate all time-dependent factors; a finite difference gives only the mean.
7Check direction with Lenz’s law and calculate current only using a specified circuit model.

How do you calculate with Faraday’s law of induction?

Here is how to work through a typical Faraday’s law of induction (ℰ = −N dΦB/dt) task step by step:

Task

How can you increase induced emf magnitude without increasing field magnitude? Compare a stationary coil in a constant field with a moving or time-varying arrangement.

Solution path

Approach: $\mathcal E=-\frac{\mathrm d\Psi}{\mathrm dt}$ .
More equally oriented turns sharing the same per-turn change or a faster change increase magnitude.
Larger area helps only if it produces a greater time-dependent flux change.
An unchanged stationary coil in a constant field still has zero emf.
Real resistance and self-inductance additionally determine load current.

Understand and apply the equation

The reading responds to change

When is flux linkage N times the same flux?

\begin{aligned}\Psi&=\sum_{k=1}^N\Phi_k\\\mathcal E&=-\frac{\mathrm d\Psi}{\mathrm dt}\\\Psi=N\Phi_B&\ \Longrightarrow\ \mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}\end{aligned}

The shortcut requires a fixed turn count and equal oriented flux through every turn.[1][4]

A normal defines positive traversal

\begin{aligned}\frac{\mathrm d\Phi_B}{\mathrm dt}>0&\ \Longrightarrow\ \mathcal E<0\\\frac{\mathrm d\Phi_B}{\mathrm dt}<0&\ \Longrightarrow\ \mathcal E>0\end{aligned}

These signs use the previously defined orientation and equally oriented turns.[2]

Induced emf: instantaneous or average value?

\begin{aligned}\Phi_B&=BA\cos\theta\\\frac{\mathrm d\Phi_B}{\mathrm dt}&=A\cos\theta\,\dot B\\&\quad+B\cos\theta\,\dot A\\&\quad-BA\sin\theta\,\dot\theta\end{aligned}

The three lines describe change contributions, not three separate laws.[4]

Is induced emf the measured terminal voltage?

\oint_C\vec E\cdot\mathrm d\vec\ell=-\frac{\mathrm d\Phi_B}{\mathrm dt}

For a boundary fixed in the chosen reference frame, the induced electric field is generally nonconservative.[3]

Example: a linear field ramp

Emf and heat in a loaded measurement coil

Given: 120 equally oriented turns, each $A=0.00600\,\mathrm{m^2}$ , normal +z. Uniform external field +z rises linearly from 0.0500 to 0.170 T in 0.0800 s. Total resistance is $R=18.0\,\Omega$ ; self-induction and back reaction on the prescribed field are neglected.
Find: Find emf, current direction and heat dissipated during the ramp.

Approach: area and angle are constant, so differentiate only the external field; then use the ideal resistance for current.
$\begin{aligned}\mathcal E&=-NA\frac{\Delta B}{\Delta t}\\I&=\frac{\mathcal E}{R}\\Q_R&=I^2R\Delta t\end{aligned}$
Substitute a positive field ramp; the result is negative in the positive traversal.
$\begin{aligned}\frac{\Delta B}{\Delta t}&=\frac{0.170-0.0500}{0.0800}\,\mathrm{T/s}=1.50\,\mathrm{T/s}\\\mathcal E&=-(120)(0.00600)(1.50)\,\mathrm V=-1.08\,\mathrm V\\I&=\frac{-1.08\,\mathrm V}{18.0\,\Omega}=-0.0600\,\mathrm A\end{aligned}$
Heat depends on squared current and is positive despite the negative current sign.
$\begin{aligned}P_R&=(0.0600\,\mathrm A)^2(18.0\,\Omega)=0.0648\,\mathrm W\\Q_R&=(0.0648\,\mathrm W)(0.0800\,\mathrm s)\\&=0.005184\,\mathrm J\end{aligned}$

Result: Emf is −1.08 V, current −60.0 mA and heat 5.18 mJ. Viewed from +z, current is clockwise and produces a field contribution toward −z.

Example: zero net charge but positive heat

A flux pulse returns to its initial value

Given: 80 equally oriented turns and total resistance 4.00 Ω. Per-turn flux rises linearly from zero to +0.900 mWb in 0.0300 s, then falls linearly to zero in 0.0600 s. Self-induction is negligible; external flux is prescribed.
Find: Calculate both emf values, net transported charge and total heat.

Approach: emf is constant in each linear phase. Add charge with its sign and heat using squared current.
$\begin{aligned}\mathcal E_j&=-N\frac{\Delta\Phi_j}{\Delta t_j}\\q_{\mathrm{net}}&=\sum_j\frac{\mathcal E_j}{R}\Delta t_j\\Q_R&=\sum_j\frac{\mathcal E_j^2}{R}\Delta t_j\end{aligned}$
The faster rising phase has twice the emf magnitude.
$\begin{aligned}\mathcal E_1&=-80\frac{0.000900}{0.0300}\,\mathrm V=-2.40\,\mathrm V\\\mathcal E_2&=-80\frac{-0.000900}{0.0600}\,\mathrm V=+1.20\,\mathrm V\\q_1&=-0.0180\,\mathrm C\\q_2&=+0.0180\,\mathrm C\end{aligned}$
The signs do not cancel heat.
$\begin{aligned}Q_R&=\frac{(2.40)^2}{4.00}(0.0300)\,\mathrm J\\&\quad+\frac{(1.20)^2}{4.00}(0.0600)\,\mathrm J\\&=0.0432\,\mathrm J+0.0216\,\mathrm J\\&=0.0648\,\mathrm J\end{aligned}$

Result: Emf is −2.40 V then +1.20 V; net transported charge in the chosen traversal is zero, but heat is 64.8 mJ.

Instantaneous, mean and rms values

\begin{aligned}[\mathcal E]&=\frac{\mathrm{Wb}}{\mathrm s}=\mathrm V\\\overline{\mathcal E}&=-N\frac{\Delta\Phi_B}{\Delta t}\\\mathcal E_{\mathrm{rms}}&=\sqrt{\frac1{\Delta t}\int_{t_1}^{t_2}\mathcal E(t)^2\,\mathrm dt}\end{aligned}

A finite difference gives the interval mean; it equals the constant instantaneous value only for linear flux change.

View diagram: emf of a curved flux waveform

Emf follows slope, not flux height

QUANTA STUDYquanta-study.de

In the third example flux grows quadratically. Emf decreases linearly from 0 to −2 V; its mean is −1 V. The dashed line is an average, not another waveform.[1]

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Generator, search coil and current-model limits

\begin{aligned}\Phi_B(t)&=BA\cos(\omega t)\\\mathcal E(t)&=NBA\omega\sin(\omega t)\end{aligned}

Uniform field, constant area and constant angular velocity; differentiation gives the second line.[5][4]

Example: a curved flux waveform

Separate average and final value

Given: A fixed search coil has $N=50$ and equal signed flux per turn $\Phi(t)=\Phi_0(t/\tau)^2$ for $0\le t\le\tau$ , with $\Phi_0=0.800\,\mathrm{mWb}$ , $\tau=0.0400\,\mathrm s$ . No load is specified.
Find: Find mean emf and its left-hand instantaneous value at the end.

Approach: secant slope gives the mean; differentiation gives the waveform.
$\begin{aligned}\overline{\mathcal E}&=-N\frac{\Phi_0}{\tau}\\\mathcal E(t)&=-\frac{2N\Phi_0}{\tau^2}t\end{aligned}$
Substitution: instantaneous slope grows linearly and ends at twice the average.
$\begin{aligned}\overline{\mathcal E}&=-50\frac{0.000800}{0.0400}\,\mathrm V=-1.00\,\mathrm V\\\mathcal E(\tau^-)&=-2.00\,\mathrm V\end{aligned}$

Result: Mean emf is −1.00 V; just before the end it is −2.00 V. Current is undetermined without a load model.

Check: Triangular emf-time area is $-\frac12(2.00\,\mathrm V)(0.0400\,\mathrm s)=-0.0400\,\mathrm{V\,s}$ , equal to $-N\Phi_0$ .

Five exercises on flux waveforms and emf

Calculate · Foundation

40 turns of 0.0120 m² each, positive normal toward +z. The uniform external field falls linearly from +0.300 to +0.100 T in 0.0500 s. Find emf and traversal direction for a purely resistive load.

Show hint

The field remains positive while its derivative is negative.

Learning goal: Convert a negative field change into positive oriented emf.

Show solution and calculation

Approach: $\mathcal E=-NA\frac{\Delta B}{\Delta t}$ .
Result is +1.92 V.
Viewed from +z, emf drives counterclockwise current supporting the declining positive flux.

Check: linkage change is −0.0960 Wb, the negative of emf times duration.

\mathcal E=-(40)(0.0120)\frac{0.100-0.300}{0.0500}\,\mathrm V=+1.92\,\mathrm V

[1][2]

Rearrange · Application

Equal oriented flux per turn changes by +0.250 mWb in 0.0400 s. At least 0.200 V magnitude of mean emf is required. How many equally oriented turns are needed?

Show hint

Turn count must be an integer; round upward if necessary.

Learning goal: Find a minimum turn count for a required mean emf.

Show solution and calculation

Approach: $N\ge\frac{|\overline{\mathcal E}|\Delta t}{|\Delta\Phi_B|}$ .
This gives exactly 32, so at least 32 turns.
Mean emf is then −0.200 V.

Check: 31 turns give only 0.19375 V magnitude and miss the requirement.

N_{\min}=\left\lceil\frac{(0.200)(0.0400)}{0.250\cdot10^{-3}}\right\rceil=32

[1]

Interpret · Application

One turn: flux rises linearly from 0 to +2.00 mWb in 0.100 s, remains constant for 0.200 s, then returns to zero in 0.0500 s. Find emf in the three open intervals. Idealize transitions as corners.

Show hint

Use slope, not the height of the flux graph.

Learning goal: Read rising, constant and falling flux as an emf signal.

Show solution and calculation

Approach: take the negative flux slope in each interval.
Emf values are −20.0 mV, zero and +40.0 mV.
Faster decay has twice the magnitude.

Check: the two emf areas cancel because final flux returns to its initial value. Ordinary derivative is undefined at idealized corners.

\begin{aligned}\mathcal E_1&=-\frac{0.00200}{0.100}\,\mathrm V=-0.0200\,\mathrm V\\\mathcal E_2&=0\\\mathcal E_3&=-\frac{-0.00200}{0.0500}\,\mathrm V=+0.0400\,\mathrm V\end{aligned}

[1]

Error analysis · Application

An ideal sinusoidal emf with peak 6.00 V drives a 12.0 Ω resistor. Self-induction is negligible. A report states $\overline P=3.00\,\mathrm W$ over complete periods. Correct it.

Show hint

The period mean of squared sine is one half.

Learning goal: Distinguish peak emf from its heating-equivalent rms value.

Show solution and calculation

Approach: $\overline P=\frac{\mathcal E_{\mathrm{rms}}^2}{R}$ , not peak squared divided by resistance.
Rms emf is about 4.24 V and mean power 1.50 W.
The stated 3.00 W is maximum instantaneous power.

Check: mean absolute current is not needed for this power calculation.

\begin{aligned}\mathcal E_{\mathrm{rms}}&=\frac{6.00}{\sqrt2}\,\mathrm V\\\overline P&=\frac{(6.00\,\mathrm V)^2}{2(12.0\,\Omega)}=1.50\,\mathrm W\end{aligned}

[5]

Transfer · Application

Two equally oriented series turns have $\dot\Phi_1=+0.00300\,\mathrm{Wb/s}$ and $\dot\Phi_2=-0.00100\,\mathrm{Wb/s}$ during an interval. Find total emf and explain why doubling the first contribution is wrong.

Show hint

Sum fluxes before differentiating.

Learning goal: Treat unequal turn fluxes as a linkage sum.

Show solution and calculation

Approach: $\mathcal E=-\frac{\mathrm d}{\mathrm dt}(\Phi_1+\Phi_2)$ .
Total emf is −2.00 mV.
Doubling the first contribution gives −6.00 mV and ignores the second flux’s different change.

Check: individual emfs −3.00 and +1.00 mV sum in oriented series to −2.00 mV.

\mathcal E=-(0.00300-0.00100)\,\mathrm V=-0.00200\,\mathrm V

[1]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

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 1, 2026Developed a DE/EN explanation with an oriented derivation, two original worked examples and five distinct solved exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.
October 1, 2026Distinguished instantaneous and average induced voltage, highlighted the equal-flux condition and terminal-voltage question and linked appropriate follow-up formulas.
October 2, 2026Linked the existing transformer explanation and placed motional emf in context.

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Quanta. Faraday’s law of induction. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/induktion-faraday

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Frequently asked questions about Faraday’s law of induction

How do you calculate the induced voltage with Faraday’s law?+

What does the minus sign in Faraday’s law mean?+

How can you increase the induced voltage without strengthening the magnetic field?+

Why is a strong magnetic field alone not enough for induction?+

What is the difference between magnetic flux and flux density?+

Retain Faraday’s law of induction for exams

Create a curated FSRS exam set for ℰ = −N dΦB/dt: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Faraday’s law of induction

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Faraday's law calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Induction and fields

Worked example

Applications

How do you calculate with Faraday’s law of induction?

The reading responds to change

When is flux linkage N times the same flux?

A normal defines positive traversal

Induced emf: instantaneous or average value?

Is induced emf the measured terminal voltage?

Example: a linear field ramp

Emf and heat in a loaded measurement coil

Example: zero net charge but positive heat

A flux pulse returns to its initial value

Instantaneous, mean and rms values

Emf follows slope, not flux height

Generator, search coil and current-model limits

Example: a curved flux waveform

Separate average and final value

Five exercises on flux waveforms and emf

40 turns of 0.0120 m2 each, positive normal toward +z. The uniform external field falls linearly from +0.300 to +0.100 T in 0.0500 s. Find emf and traversal direction for a purely resistive load.

Equal oriented flux per turn changes by +0.250 mWb in 0.0400 s. At least 0.200 V magnitude of mean emf is required. How many equally oriented turns are needed?

One turn: flux rises linearly from 0 to +2.00 mWb in 0.100 s, remains constant for 0.200 s, then returns to zero in 0.0500 s. Find emf in the three open intervals. Idealize transitions as corners.

An ideal sinusoidal emf with peak 6.00 V drives a 12.0 Ω resistor. Self-induction is negligible. A report states P‾=3.00 W\overline P=3.00\,\mathrm WP=3.00W over complete periods. Correct it.

Two equally oriented series turns have Φ˙1=+0.00300 Wb/s\dot\Phi_1=+0.00300\,\mathrm{Wb/s}Φ˙1​=+0.00300Wb/s and Φ˙2=−0.00100 Wb/s\dot\Phi_2=-0.00100\,\mathrm{Wb/s}Φ˙2​=−0.00100Wb/s during an interval. Find total emf and explain why doubling the first contribution is wrong.

In brief

Sources for this explanation

Frequently asked questions about Faraday’s law of induction

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Faraday’s law of induction for exams

Faraday’s law of induction

Variables & units

Faraday's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Induction and fields

Worked example

Applications

How do you calculate with Faraday’s law of induction?

The reading responds to change

When is flux linkage N times the same flux?

A normal defines positive traversal

Induced emf: instantaneous or average value?

Is induced emf the measured terminal voltage?

Example: a linear field ramp

Emf and heat in a loaded measurement coil

Example: zero net charge but positive heat

A flux pulse returns to its initial value

Instantaneous, mean and rms values

Emf follows slope, not flux height

Generator, search coil and current-model limits

Example: a curved flux waveform

Separate average and final value

Five exercises on flux waveforms and emf

40 turns of 0.0120 m2 each, positive normal toward +z. The uniform external field falls linearly from +0.300 to +0.100 T in 0.0500 s. Find emf and traversal direction for a purely resistive load.

Equal oriented flux per turn changes by +0.250 mWb in 0.0400 s. At least 0.200 V magnitude of mean emf is required. How many equally oriented turns are needed?

One turn: flux rises linearly from 0 to +2.00 mWb in 0.100 s, remains constant for 0.200 s, then returns to zero in 0.0500 s. Find emf in the three open intervals. Idealize transitions as corners.

An ideal sinusoidal emf with peak 6.00 V drives a 12.0 Ω resistor. Self-induction is negligible. A report states P‾=3.00 W\overline P=3.00\,\mathrm WP=3.00W over complete periods. Correct it.

Two equally oriented series turns have Φ˙1=+0.00300 Wb/s\dot\Phi_1=+0.00300\,\mathrm{Wb/s}Φ˙1​=+0.00300Wb/s and Φ˙2=−0.00100 Wb/s\dot\Phi_2=-0.00100\,\mathrm{Wb/s}Φ˙2​=−0.00100Wb/s during an interval. Find total emf and explain why doubling the first contribution is wrong.

In brief

40 turns of 0.0120 m² each, positive normal toward +z. The uniform external field falls linearly from +0.300 to +0.100 T in 0.0500 s. Find emf and traversal direction for a purely resistive load.

An ideal sinusoidal emf with peak 6.00 V drives a 12.0 Ω resistor. Self-induction is negligible. A report states $\overline P=3.00\,\mathrm W$ over complete periods. Correct it.

Two equally oriented series turns have $\dot\Phi_1=+0.00300\,\mathrm{Wb/s}$ and $\dot\Phi_2=-0.00100\,\mathrm{Wb/s}$ during an interval. Find total emf and explain why doubling the first contribution is wrong.

40 turns of 0.0120 m² each, positive normal toward +z. The uniform external field falls linearly from +0.300 to +0.100 T in 0.0500 s. Find emf and traversal direction for a purely resistive load.

An ideal sinusoidal emf with peak 6.00 V drives a 12.0 Ω resistor. Self-induction is negligible. A report states $\overline P=3.00\,\mathrm W$ over complete periods. Correct it.

Two equally oriented series turns have $\dot\Phi_1=+0.00300\,\mathrm{Wb/s}$ and $\dot\Phi_2=-0.00100\,\mathrm{Wb/s}$ during an interval. Find total emf and explain why doubling the first contribution is wrong.