Physics · Electromagnetism

Magnetic flux

Magnetic flux integrates the oriented normal field component over a selected surface.

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Formula

Φ_{B} = B A cos θ

ΦB in Wb=T·m²; B in T; A in m²; θ is the angle from the surface normal.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ΦB$	Magnetic flux	Wb
$B$	Uniform magnetic flux density	T
$A$	Area	m²
$θ$	Angle between field and surface normal	rad

LaTeX

\Phi_B = B A \cos\theta

QUANTA STUDYquanta-study.de

Oriented flux Φ divided by BA is the cosine of angle to the positive surface normal: plus one at zero degrees, zero at ninety and minus one at one hundred eighty. — Uniform field, plane area: measure angle to the normal. Oriented flux is +0.5 BA at 60° and −0.5 BA at 120°. Original diagram of the flux definition in OpenStax §13.1.

Magnetic flux calculator

Find the magnetic flux, flux density, area or angle for a uniform field through a flat surface, plus the flux linkage of a coil with N turns.

Calculation model

Find

Magnetic flux density B

Strength of the uniform magnetic field (flux density), e.g. 40 mT between two magnet poles.

Area A

Area of the surface or wire loop, e.g. 50 cm².

Angle between B and the surface normal θ

Angle between the field and the surface normal, not the surface itself, from 0° to 180°, e.g. 60°.

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

Result: Magnetic flux Φ

Φ = 100 µWb

With B = 40 mT, A = 50 cm² and θ = 60 °, Φ = 100 µWb.

Working · rearrange for Φ, substitute, evaluate

1. Rearrange for Φ
$\Phi=BA\cos\theta$
2. Substitute values (in SI units)
- B = 40 mT = 0.04 T
- A = 50 cm² = 0.005 m²
- θ = 60 ° = 1.0472 rad
$\Phi=0.04\cdot 0.005\cdot\cos\left(1.0471975512\right)$
3. Result
Φ = 100 µWb

Quick tips

Check the angle: θ is measured from the surface normal, the line at right angles to the surface. If the angle to the surface is given, use 90° minus that value.
Check with limiting cases: a field straight through the surface (θ = 0°) gives the largest flux B · A, a field parallel to the surface (θ = 90°) gives zero.

Assumptions and limits

Uniform magnetic field and a plane surface. B and A are nonnegative magnitudes. θ from 0 to 180° is the angle from the field to the explicitly chosen positive surface normal, not the plane. Φ may be negative. The inverse angle requires BA > 0 and |Φ| ≤ BA; it gives this normal angle, not a unique spatial orientation. No nonuniform surface integral or induced emf.

Derivation & background

Generally, $\Phi_B=\int_S\vec B\cdot\hat{\vec n}\,\mathrm dA$ . A uniform field and plane surface give $BA\cos\theta$ . Normal reversal changes the sign; a field parallel to the surface gives zero flux, not zero field. Multiple turns have flux linkage equal to the sum of their individual fluxes.

Exam blueprint

Validity range

The shortcut applies to an oriented plane area with a constant field component normal to it; a uniform field is sufficient. Measure angle from the chosen normal. Otherwise integrate locally.

Derivation steps

Flux is an oriented surface sum; projection and a constant normal component give the shortcut.

1Choose a positive surface normal and retain its orientation.
2A small element contributes $\mathrm d\Phi_B=\vec B\cdot\hat{\vec n}\,\mathrm dA$ .
3Sum using $\Phi_B=\int_S\vec B\cdot\hat{\vec n}\,\mathrm dA$ .
4On a plane surface the normal is the same everywhere.
5For a uniform field, $B\cos\theta$ is constant and can be factored out.
6The remaining integral is $A$ , giving $\Phi_B=BA\cos\theta$ .
7Check sign, the tangential limit and webers as the unit.

Rearrangements

Flux density from flux

B=\frac{\Phi_B}{A\cos\theta}

This inverse requires A > 0 and cosθ ≠ 0. θ is measured to the positive surface normal. For nonnegative field magnitude, ΦB and cosθ must have the same sign, or ΦB must be zero. With a tangential field, ΦB = 0 does not determine field magnitude.

Task variant

$B=0.30\,\mathrm T$ through $A=0.040\,\mathrm{m^2}$ ; field makes 60° with the normal. Find Φ.

$\Phi_B=BA\cos60^\circ=(0.30\,\mathrm T)(0.040\,\mathrm{m^2})\frac12=0.0060\,\mathrm{Wb}$

Common mistakes

Measuring the angle from the surface instead of its normal.

For a field parallel to the area, Φ must be zero; use this limit to check the angle.

Exam context

Area orientation, induced emf via Faraday’s law and rotating coils.

Formula cluster

Magnetic flux and induction

A changing flux is the basis of electromagnetic induction.

ℰ = −N dΦB/dt

Faraday’s law of induction

|ℰmot| = Bℓv

Motional emf

ℰ(t) = NBAω sin(ωt)

Induced emf of a rotating generator

B = μ₀·(N/l)·I

Magnetic Field of a Long Coil

Worked example

For $B=0.40\,\mathrm T$ through $A=0.020\,\mathrm{m^2}$ with the field normal to the surface, $\theta=0^\circ$ and $\Phi_B=0.0080\,\mathrm{Wb}$ . If the field lies in the plane, $\theta=90^\circ$ and flux is zero. Check: T·m²=Wb.

Applications

Faraday’s induction law, coils, generators and magnetic flux measurements.

How do you calculate with Magnetic flux through a surface?

Here is how to work through a typical Magnetic flux through a surface (ΦB = BA cos θ) task step by step:

1
Task
$B=0.30\,\mathrm T$ through $A=0.040\,\mathrm{m^2}$ ; field makes 60° with the normal. Find Φ.
Solution path
$\Phi_B=BA\cos60^\circ=(0.30\,\mathrm T)(0.040\,\mathrm{m^2})\frac12=0.0060\,\mathrm{Wb}$

Understand and apply the equation

Magnetic flux connects a field to a selected, oriented surface. It can vanish in a strong field or change sign when the normal is reversed. Projection explains the shortcut; spatial integration shows when one field value is insufficient. A change in flux over time, rather than flux alone, gives induction around the corresponding boundary.

Why turning changes the result

Imagine a wire loop in a uniform magnetic field. Turning it leaves its area unchanged yet changes flux. What matters is how much of the surface faces perpendicular to the field. We first understand that projection and its sign; only a spatially varying field requires an integral.[1]

The shortcut needs area, field magnitude and angle to the surface normal. The normal is a perpendicular direction arrow defining the positive sign. “Tilted by 30°” is incomplete: establish whether the angle is measured to the surface or to that arrow.[1]

A field value is not yet flux

\Phi_B=\int_S\vec B\cdot\hat{\vec n}\,\mathrm dA

The normal assigns a positive crossing direction to every surface element.[1][2]

Flux density describes the field at a point; flux sums its normal component over a surface. The surface need not be a physical sheet: it may be an imaginary surface spanning a wire loop. The result is a signed number, not a force, current or field vector. Drawn field lines help show orientation but are not countable physical objects carrying a universal flux per line.[1]

Normal, angle and sign

\begin{aligned}\vec A&=A\hat{\vec n}\\\Phi_B&=\vec B\cdot\vec A=BA\cos\theta\end{aligned}

The second line applies to a plane surface in a uniform field; the angle is between field and normal.[1]

Choose one of the two normal directions and retain it throughout the calculation. Parallel field and normal give positive flux; antiparallel directions give negative flux. A field within the plane gives zero flux although the field itself remains. Reversing the chosen normal changes the description: the same physical state receives the opposite sign. Physically rotating the surface with its attached normal instead changes the geometry.[2]

Turning changes projection and sign

QUANTA STUDYquanta-study.de

For the same plane area in a uniform field: parallel +1, tangential 0, antiparallel −1. Angle is measured to the normal.[1]

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From small surface elements to the shortcut

\begin{aligned}\Delta\Phi_i&\approx\vec B_i\cdot\hat{\vec n}_i\,\Delta A_i\\\Phi_B&=\lim_{\max\Delta A_i\to0}\sum_i\Delta\Phi_i\\&=\int_S\vec B\cdot\hat{\vec n}\,\mathrm dA\end{aligned}

This is a definition as the limit of an oriented sum, not an empirical force equation.[1]

On a plane surface the normal is constant. If the field is also spatially constant, its normal component can be taken outside the integral; the remaining area integral gives the area. Different normals or field values can be handled by suitable surface pieces whose signed contributions are added. Only the normal component needs to be constant to be factored out; a varying tangential component contributes no flux.[1]

Webers, teslas and a safe rearrangement

\begin{aligned}1\,\mathrm{Wb}&=1\,\mathrm{T\,m^2}=1\,\mathrm{V\,s}\\B&=\frac{\Phi_B}{A\cos\theta}\\A&=\frac{\Phi_B}{B\cos\theta}\end{aligned}

The rearrangements require a nonzero denominator and a known orientation.[1]

Squared length units require squared conversion: one square centimetre is one ten-thousandth of a square metre. For a tangential field, zero flux determines neither field magnitude nor area. If field magnitude or area is also unknown alongside the angle, a single flux measurement cannot determine those quantities separately. When solving for a nonnegative field magnitude, flux and cosine signs must agree; a negative inferred magnitude signals inconsistent orientation.

If $B$ and $A$ are independently known and positive, $\Phi_B$ determines the normal angle from 0 to 180° in the uniform plane case. This requires $|\Phi_B|\le BA$ . The sign distinguishes acute and obtuse normal angles, but does not specify a unique spatial orientation of the surface. With $BA=0$ , $\Phi_B=0$ leaves the angle undetermined.[1]

\theta=\arccos\!\left(\frac{\Phi_B}{BA}\right),\qquad BA>0,\quad |\Phi_B|\le BA,\quad 0\le\theta\le\pi

Normal angle from known field magnitude, area and oriented flux in the uniform plane case.[1]

Example: a tilted measurement surface

Surface tilt with a sign

Given: Uniform field $B=0.0750\,\mathrm T$ , plane area $A=0.0180\,\mathrm{m^2}$ , angle to the chosen normal $\theta=30^\circ$ .
Find: Find flux and the result with the reversed normal.

Approach: project the uniform field onto the chosen normal, then multiply by area.
$\Phi_B=(B\cos\theta)A$
Substitution: the angle is already measured from the normal.
$\begin{aligned}\Phi_B&=(0.0750\,\mathrm T)(0.0180\,\mathrm{m^2})\cos30^\circ\\&\approx1.16913\cdot10^{-3}\,\mathrm{Wb}\end{aligned}$
Reversing the normal gives an angle of 150° and changes the cosine sign.
$\Phi_B^{\mathrm{rev}}=-\Phi_B$

Result: Flux is about +1.17 mWb; with the opposite normal it is −1.17 mWb.

Check: Maximum magnitude is $BA=1.35\,\mathrm{mWb}$ . The result is smaller and retains its magnitude under normal reversal; neither field nor surface was physically changed.

Example: a field gradient across the surface

Do not use the edge value everywhere

Given: Rectangle in the xy plane: $0\le x\le L=0.300\,\mathrm m$ , width $w=0.0800\,\mathrm m$ , normal +z. The prescribed static model field is $\vec B(x)=[B_0+kx]\hat{\vec z}$ , with $B_0=0.120\,\mathrm T$ , $k=0.400\,\mathrm{T/m}$ .
Find: Integrate flux and check the edge-value approximation.

Approach: split the rectangle into strips with a constant field along their width.
$\begin{aligned}\mathrm dA&=w\,\mathrm dx\\\Phi_B&=w\int_0^L(B_0+kx)\,\mathrm dx\end{aligned}$
Integrate before substituting; the linear change gives a factor of one half.
$\begin{aligned}\Phi_B&=w\left(B_0L+\frac{kL^2}{2}\right)\\&=(0.0800)\left[(0.120)(0.300)+\frac{(0.400)(0.300)^2}{2}\right]\,\mathrm{Wb}\\&=0.00432\,\mathrm{Wb}\end{aligned}$
For this linear profile, the area mean is the average of the two edge values.
$\begin{aligned}\langle B_z\rangle&=B_0+\frac{kL}{2}=0.180\,\mathrm T\\\Phi_B&=\langle B_z\rangle wL\end{aligned}$

Result: Flux is 4.32 mWb. Using the right-edge value 0.240 T everywhere gives 5.76 mWb, one third too large.

Check: Area $wL=0.0240\,\mathrm{m^2}$ times mean 0.180 T gives 4.32 mWb. This is an admissible prescribed local model: $\nabla\cdot\vec B=\frac{\partial B_z}{\partial z}=0$ ; the required source currents are not claimed to be absent.

Explore further: open and closed surfaces

\oint_{\partial V}\vec B\cdot\hat{\vec n}_{\mathrm{out}}\,\mathrm dA=0

For a closed surface, total oriented contributions vanish in the Maxwell model.[3]

Zero total flux through a closed surface does not mean zero magnetic field inside. Entering and leaving contributions cancel. This also explains equal oriented flux through two surfaces sharing a boundary: join one surface to the reversed second surface to make a closed surface. Keep the same boundary orientation; arbitrary surfaces without a shared boundary need not have equal flux.[1][3]

Maxwell’s equations in integral form place this closed-flux balance alongside electric-field and circulation balances. This section considers only the magnetic balance; zero total flux does not imply a locally vanishing field.[3]

From a flux value to its time derivative

Faraday’s law of induction relates the time change of oriented flux linkage to an emf. Flux linkage is the sum of the oriented fluxes through all turns. Only for $N$ equally oriented turns with the same flux $\Phi_B$ does $\Psi=N\Phi_B$ hold; it is not the flux through a single surface. The comparisons below therefore need time dependence and orientation, not just one instantaneous flux value.[1]

A simple perpendicular-field case: $B=0.40\,\mathrm T$ , $A=0.020\,\mathrm{m^2}$ and zero normal angle give 8.0 mWb. An unchanged loop in an unchanged field nevertheless has zero induced emf. For equal flux through each turn, $\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}$ . Magnetic flux is not stored energy; energy requires an additional field or component model.[1]

A search coil requires known area and orientation. Curved surfaces, field gradients or oppositely oriented regions require the integral. Material, core geometry and leakage fields affect actual flux density; the geometric definition does not replace a calculation of that field distribution.[1][2]

Example: equal flux, different emf

The instantaneous value does not reveal the slope

Given: Two equal fixed plane loops, each $A=0.0100\,\mathrm{m^2}$ , normal +z, one turn each. Loop 1 has constant $B_{z,1}=0.200\,\mathrm T$ . Around $t=0$ , loop 2 has $B_{z,2}(t)=0.200\,\mathrm T+(0.500\,\mathrm{T/s})t$ .
Find: Compare flux and signed emf at $t=0$ .

Approach: flux uses field value; emf uses its time derivative.
$\begin{aligned}\Phi_j&=AB_{z,j}\\\mathcal E_j&=-A\frac{\mathrm dB_{z,j}}{\mathrm dt}\end{aligned}$
Both field values agree at the chosen instant; their rates do not.
$\begin{aligned}\Phi_1(0)=\Phi_2(0)&=(0.0100)(0.200)\,\mathrm{Wb}\\&=2.00\,\mathrm{mWb}\\\mathcal E_1(0)&=0\\\mathcal E_2(0)&=-(0.0100)(0.500)\,\mathrm V\\&=-5.00\,\mathrm{mV}\end{aligned}$

Result: Both fluxes are +2.00 mWb, but only loop 2 has emf: −5.00 mV. One flux value does not determine emf.

Check: Over $20.0\,\mathrm{ms}$ , flux 2 grows by $0.100\,\mathrm{mWb}$ . Emf-time area $-0.100\,\mathrm{V\,ms}$ equals the negative of that change. Flux 1 stays constant.

Five exercises: projection, model and checking

Calculate · Foundation

A uniform field $B=0.250\,\mathrm T$ crosses a plane area $A=160\,\mathrm{cm^2}$ at $60^\circ$ to the positive normal. Find signed flux.

Show hint

Convert area into square metres.

Learning goal: Use area units and the normal angle together.

Show solution and calculation

Approach: $\Phi_B=BA\cos\theta$ .
Area is 0.0160 m².
Substitution gives +2.00 mWb.

Check: without tilt it would be 4.00 mWb; the cosine halves that value.

\begin{aligned}A&=160\cdot10^{-4}\,\mathrm{m^2}\\\Phi_B&=(0.250)(0.0160)\cos60^\circ\,\mathrm{Wb}\\&=0.00200\,\mathrm{Wb}\end{aligned}

[1]

Rearrange · Foundation

Measured flux is $\Phi_B=+3.00\,\mathrm{mWb}$ in uniform $B=0.200\,\mathrm T$ at $\theta=60^\circ$ . What plane area matches it?

Show hint

Isolate area; the cosine is nonzero here.

Learning goal: Find area from known flux at a nonsingular angle.

Show solution and calculation

Approach: rearrange the flux equation for $A$ .
Result is 0.0300 m².
Substitution back into the original equation gives 3.00 mWb.
A tangential geometry would prevent this determination.

A=\frac{\Phi_B}{B\cos\theta}=\frac{0.00300}{(0.200)\cos60^\circ}\,\mathrm{m^2}=0.0300\,\mathrm{m^2}

[1]

Interpret · Application

A surface with normal +z has two parts of $0.00400\,\mathrm{m^2}$ each. Their fields are $B_z=+0.400\,\mathrm T$ and $-0.100\,\mathrm T$ . Is total flux zero, positive or negative? Find it.

Show hint

Include direction in each contribution, not just the final answer.

Learning goal: Add opposing signed surface contributions rather than field magnitudes.

Show solution and calculation

Approach: add the two signed contributions.
They are +1.60 and −0.400 mWb, totalling +1.20 mWb.
Opposite field directions do not guarantee cancellation; equal flux magnitudes would be required.

\Phi_B=(0.400-0.100)(0.00400)\,\mathrm{Wb}=0.00120\,\mathrm{Wb}

[1]

Error analysis · Application

Field magnitude is 0.600 T and area is 0.0200 m². The field lies 30° above the plane on the positive-normal side. A calculation uses $\Phi_B=BA\cos30^\circ$ . Locate and correct the error.

Show hint

What angle separates field and normal?

Learning goal: Translate a plane angle into a normal angle.

Show solution and calculation

Approach: the normal angle is 60°, not 30°.
Correct flux is +6.00 mWb; the mistaken calculation would give about +10.4 mWb.

Check: as the plane angle approaches zero, flux must approach zero too.

\Phi_B=BA\sin30^\circ=(0.600)(0.0200)\frac12\,\mathrm{Wb}=0.00600\,\mathrm{Wb}

[1]

Transfer · Application

60 equally oriented turns of 0.0150 m² each rotate in a uniform 0.0300 T field from normal angle 0° to 90° in 0.120 s. Find mean emf in the positive traversal defined by the normal’s right-hand rule.

Show hint

Calculate both per-turn fluxes before subtracting.

Learning goal: Convert per-turn flux change into mean emf.

Show solution and calculation

Approach: $\overline{\mathcal E}=-N\frac{\Delta\Phi_B}{\Delta t}$ .
Flux falls from +0.450 mWb to zero; mean emf is +0.225 V.

Check: emf times duration is +0.0270 V s, the negative total linkage change. The instantaneous waveform is not determined without a rotation law.

\overline{\mathcal E}=-60\frac{0-(0.0300)(0.0150)}{0.120}\,\mathrm V=+0.225\,\mathrm V

[1][4]

In brief

Flux is the signed sum of the normal component over an area. Check normal and angle meaning first, then uniformity. The simple product replaces the integral only under its assumptions. Flux value and flux change remain different quantities.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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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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Change history

October 2, 2026Added the inverse normal-angle relation with its limits and linked Maxwell and Faraday context; preserved integral cases, exercises and angle diagram.
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.

Cite this reference page

Quanta. Magnetic flux through a surface. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetischer-fluss

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Frequently asked questions about Magnetic flux through a surface

What does magnetic flux describe physically?+

Magnetic flux describes the normal component of a magnetic field summed over a selected surface. Field strength, area and orientation therefore matter together. First choose a positive surface normal: a field pointing along it contributes positive flux, while an opposite field contributes negative flux. For a uniform field and plane area, $\Phi_B=BA\cos\theta$ , with the angle measured from the normal. A tangential field gives zero flux even though the field itself remains present. Flux is neither stored magnetic energy nor a literal count of field lines. For a spatially varying field, add the local contributions through a surface integral.

How can I check a calculation involving magnetic flux?+

First fix the positive normal and sketch whether the field mostly points with it or against it. Then determine the angle to the normal, rather than to the plane itself. In a uniform field, the magnitude of the result cannot exceed $BA$ ; a larger value indicates an arithmetic or input error. For example, $B=0.0750\,\mathrm T$ , $A=0.0180\,\mathrm{m^2}$ and thirty degrees to the normal give about positive $1.17\,\mathrm{mWb}$ . Reverse only the chosen normal as an independent check: the magnitude remains unchanged and the sign reverses. With a nonuniform field, check the local normal component and its surface integral instead.

What assumptions and limits apply to magnetic flux?+

The short product formula requires a plane area and a field component normal to that area that is constant across it. A uniform field satisfies this condition, but it is not the only possible case. On a curved surface, the local normal changes; in a nonuniform field, strength or direction may also change. Use the surface integral instead of multiplying an arbitrarily selected field value by total area. Reversing the normal changes only the orientation convention. Physically rotating the surface can change the actual flux. The total magnetic flux through a closed surface is zero. This does not mean that flux through every individual patch is also zero.

How do I check units and scale for magnetic flux?+

Convert area to square metres before substitution. One square centimetre equals $10^{-4}\,\mathrm{m^2}$ , because both length directions must be converted. Tesla times square metre gives weber. For example, 160 square centimetres is $0.0160\,\mathrm{m^2}$ . With 0.25 tesla and sixty degrees to the positive normal, the result is $\Phi_B=0.00200\,\mathrm{Wb}$ , or two milliwebers. Without the angle factor, the maximum possible magnitude would be four milliwebers. This comparison independently checks the scale. The sign has no unit and comes from orientation. A unit check alone therefore cannot detect an incorrectly chosen angle or a reversed normal convention.

When should I combine magnetic flux with another relation?+

For induction problems, combine flux with Faraday’s law. First calculate the time-dependent oriented flux; then determine its change and the corresponding loop emf. For multiple turns, add their fluxes to obtain flux linkage. Multiplication by the number of turns is valid only when every turn has the same flux change. For example, rotating an area from a field-normal position into a field-parallel position removes its initial flux even if field strength stays constant. A current requires an additional circuit model, such as a closed resistive loop with negligible self-inductance. Flux alone determines neither current nor dissipated power nor stored energy.

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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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Symbol

Meaning

Unit

ΦB

Magnetic flux

B

Uniform magnetic flux density

A

Area

m²

θ

Angle between field and surface normal

rad

Derivation & background

Exam blueprint

Validity range

The shortcut applies to an oriented plane area with a constant field component normal to it; a uniform field is sufficient. Measure angle from the chosen normal. Otherwise integrate locally.

Derivation steps

Flux is an oriented surface sum; projection and a constant normal component give the shortcut.

1Choose a positive surface normal and retain its orientation.
2A small element contributes $\mathrm d\Phi_B=\vec B\cdot\hat{\vec n}\,\mathrm dA$ .
3Sum using $\Phi_B=\int_S\vec B\cdot\hat{\vec n}\,\mathrm dA$ .
4On a plane surface the normal is the same everywhere.
5For a uniform field, $B\cos\theta$ is constant and can be factored out.
6The remaining integral is $A$ , giving $\Phi_B=BA\cos\theta$ .
7Check sign, the tangential limit and webers as the unit.

How do you calculate with Magnetic flux through a surface?

Here is how to work through a typical Magnetic flux through a surface (ΦB = BA cos θ) task step by step:

Task

$B=0.30\,\mathrm T$ through $A=0.040\,\mathrm{m^2}$ ; field makes 60° with the normal. Find Φ.

Solution path

$\Phi_B=BA\cos60^\circ=(0.30\,\mathrm T)(0.040\,\mathrm{m^2})\frac12=0.0060\,\mathrm{Wb}$

Understand and apply the equation

Why turning changes the result

A field value is not yet flux

\Phi_B=\int_S\vec B\cdot\hat{\vec n}\,\mathrm dA

The normal assigns a positive crossing direction to every surface element.[1][2]

Normal, angle and sign

\begin{aligned}\vec A&=A\hat{\vec n}\\\Phi_B&=\vec B\cdot\vec A=BA\cos\theta\end{aligned}

The second line applies to a plane surface in a uniform field; the angle is between field and normal.[1]

Turning changes projection and sign

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For the same plane area in a uniform field: parallel +1, tangential 0, antiparallel −1. Angle is measured to the normal.[1]

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From small surface elements to the shortcut

\begin{aligned}\Delta\Phi_i&\approx\vec B_i\cdot\hat{\vec n}_i\,\Delta A_i\\\Phi_B&=\lim_{\max\Delta A_i\to0}\sum_i\Delta\Phi_i\\&=\int_S\vec B\cdot\hat{\vec n}\,\mathrm dA\end{aligned}

This is a definition as the limit of an oriented sum, not an empirical force equation.[1]

Webers, teslas and a safe rearrangement

\begin{aligned}1\,\mathrm{Wb}&=1\,\mathrm{T\,m^2}=1\,\mathrm{V\,s}\\B&=\frac{\Phi_B}{A\cos\theta}\\A&=\frac{\Phi_B}{B\cos\theta}\end{aligned}

The rearrangements require a nonzero denominator and a known orientation.[1]

\theta=\arccos\!\left(\frac{\Phi_B}{BA}\right),\qquad BA>0,\quad |\Phi_B|\le BA,\quad 0\le\theta\le\pi

Normal angle from known field magnitude, area and oriented flux in the uniform plane case.[1]

Example: a tilted measurement surface

Surface tilt with a sign

Given: Uniform field $B=0.0750\,\mathrm T$ , plane area $A=0.0180\,\mathrm{m^2}$ , angle to the chosen normal $\theta=30^\circ$ .
Find: Find flux and the result with the reversed normal.

Approach: project the uniform field onto the chosen normal, then multiply by area.
$\Phi_B=(B\cos\theta)A$
Substitution: the angle is already measured from the normal.
$\begin{aligned}\Phi_B&=(0.0750\,\mathrm T)(0.0180\,\mathrm{m^2})\cos30^\circ\\&\approx1.16913\cdot10^{-3}\,\mathrm{Wb}\end{aligned}$
Reversing the normal gives an angle of 150° and changes the cosine sign.
$\Phi_B^{\mathrm{rev}}=-\Phi_B$

Result: Flux is about +1.17 mWb; with the opposite normal it is −1.17 mWb.

Check: Maximum magnitude is $BA=1.35\,\mathrm{mWb}$ . The result is smaller and retains its magnitude under normal reversal; neither field nor surface was physically changed.

Example: a field gradient across the surface

Do not use the edge value everywhere

Given: Rectangle in the xy plane: $0\le x\le L=0.300\,\mathrm m$ , width $w=0.0800\,\mathrm m$ , normal +z. The prescribed static model field is $\vec B(x)=[B_0+kx]\hat{\vec z}$ , with $B_0=0.120\,\mathrm T$ , $k=0.400\,\mathrm{T/m}$ .
Find: Integrate flux and check the edge-value approximation.

Approach: split the rectangle into strips with a constant field along their width.
$\begin{aligned}\mathrm dA&=w\,\mathrm dx\\\Phi_B&=w\int_0^L(B_0+kx)\,\mathrm dx\end{aligned}$
Integrate before substituting; the linear change gives a factor of one half.
$\begin{aligned}\Phi_B&=w\left(B_0L+\frac{kL^2}{2}\right)\\&=(0.0800)\left[(0.120)(0.300)+\frac{(0.400)(0.300)^2}{2}\right]\,\mathrm{Wb}\\&=0.00432\,\mathrm{Wb}\end{aligned}$
For this linear profile, the area mean is the average of the two edge values.
$\begin{aligned}\langle B_z\rangle&=B_0+\frac{kL}{2}=0.180\,\mathrm T\\\Phi_B&=\langle B_z\rangle wL\end{aligned}$

Result: Flux is 4.32 mWb. Using the right-edge value 0.240 T everywhere gives 5.76 mWb, one third too large.

Explore further: open and closed surfaces

\oint_{\partial V}\vec B\cdot\hat{\vec n}_{\mathrm{out}}\,\mathrm dA=0

For a closed surface, total oriented contributions vanish in the Maxwell model.[3]

From a flux value to its time derivative

Example: equal flux, different emf

The instantaneous value does not reveal the slope

Given: Two equal fixed plane loops, each $A=0.0100\,\mathrm{m^2}$ , normal +z, one turn each. Loop 1 has constant $B_{z,1}=0.200\,\mathrm T$ . Around $t=0$ , loop 2 has $B_{z,2}(t)=0.200\,\mathrm T+(0.500\,\mathrm{T/s})t$ .
Find: Compare flux and signed emf at $t=0$ .

Approach: flux uses field value; emf uses its time derivative.
$\begin{aligned}\Phi_j&=AB_{z,j}\\\mathcal E_j&=-A\frac{\mathrm dB_{z,j}}{\mathrm dt}\end{aligned}$
Both field values agree at the chosen instant; their rates do not.
$\begin{aligned}\Phi_1(0)=\Phi_2(0)&=(0.0100)(0.200)\,\mathrm{Wb}\\&=2.00\,\mathrm{mWb}\\\mathcal E_1(0)&=0\\\mathcal E_2(0)&=-(0.0100)(0.500)\,\mathrm V\\&=-5.00\,\mathrm{mV}\end{aligned}$

Result: Both fluxes are +2.00 mWb, but only loop 2 has emf: −5.00 mV. One flux value does not determine emf.

Check: Over $20.0\,\mathrm{ms}$ , flux 2 grows by $0.100\,\mathrm{mWb}$ . Emf-time area $-0.100\,\mathrm{V\,ms}$ equals the negative of that change. Flux 1 stays constant.

Five exercises: projection, model and checking

Calculate · Foundation

A uniform field $B=0.250\,\mathrm T$ crosses a plane area $A=160\,\mathrm{cm^2}$ at $60^\circ$ to the positive normal. Find signed flux.

Show hint

Convert area into square metres.

Learning goal: Use area units and the normal angle together.

Show solution and calculation

Approach: $\Phi_B=BA\cos\theta$ .
Area is 0.0160 m².
Substitution gives +2.00 mWb.

Check: without tilt it would be 4.00 mWb; the cosine halves that value.

\begin{aligned}A&=160\cdot10^{-4}\,\mathrm{m^2}\\\Phi_B&=(0.250)(0.0160)\cos60^\circ\,\mathrm{Wb}\\&=0.00200\,\mathrm{Wb}\end{aligned}

[1]

Rearrange · Foundation

Measured flux is $\Phi_B=+3.00\,\mathrm{mWb}$ in uniform $B=0.200\,\mathrm T$ at $\theta=60^\circ$ . What plane area matches it?

Show hint

Isolate area; the cosine is nonzero here.

Learning goal: Find area from known flux at a nonsingular angle.

Show solution and calculation

Approach: rearrange the flux equation for $A$ .
Result is 0.0300 m².
Substitution back into the original equation gives 3.00 mWb.
A tangential geometry would prevent this determination.

A=\frac{\Phi_B}{B\cos\theta}=\frac{0.00300}{(0.200)\cos60^\circ}\,\mathrm{m^2}=0.0300\,\mathrm{m^2}

[1]

Interpret · Application

A surface with normal +z has two parts of $0.00400\,\mathrm{m^2}$ each. Their fields are $B_z=+0.400\,\mathrm T$ and $-0.100\,\mathrm T$ . Is total flux zero, positive or negative? Find it.

Show hint

Include direction in each contribution, not just the final answer.

Learning goal: Add opposing signed surface contributions rather than field magnitudes.

Show solution and calculation

Approach: add the two signed contributions.
They are +1.60 and −0.400 mWb, totalling +1.20 mWb.
Opposite field directions do not guarantee cancellation; equal flux magnitudes would be required.

\Phi_B=(0.400-0.100)(0.00400)\,\mathrm{Wb}=0.00120\,\mathrm{Wb}

[1]

Error analysis · Application

Field magnitude is 0.600 T and area is 0.0200 m². The field lies 30° above the plane on the positive-normal side. A calculation uses $\Phi_B=BA\cos30^\circ$ . Locate and correct the error.

Show hint

What angle separates field and normal?

Learning goal: Translate a plane angle into a normal angle.

Show solution and calculation

Approach: the normal angle is 60°, not 30°.
Correct flux is +6.00 mWb; the mistaken calculation would give about +10.4 mWb.

Check: as the plane angle approaches zero, flux must approach zero too.

\Phi_B=BA\sin30^\circ=(0.600)(0.0200)\frac12\,\mathrm{Wb}=0.00600\,\mathrm{Wb}

[1]

Transfer · Application

60 equally oriented turns of 0.0150 m² each rotate in a uniform 0.0300 T field from normal angle 0° to 90° in 0.120 s. Find mean emf in the positive traversal defined by the normal’s right-hand rule.

Show hint

Calculate both per-turn fluxes before subtracting.

Learning goal: Convert per-turn flux change into mean emf.

Show solution and calculation

Approach: $\overline{\mathcal E}=-N\frac{\Delta\Phi_B}{\Delta t}$ .
Flux falls from +0.450 mWb to zero; mean emf is +0.225 V.

Check: emf times duration is +0.0270 V s, the negative total linkage change. The instantaneous waveform is not determined without a rotation law.

\overline{\mathcal E}=-60\frac{0-(0.0300)(0.0150)}{0.120}\,\mathrm V=+0.225\,\mathrm V

[1][4]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Added the inverse normal-angle relation with its limits and linked Maxwell and Faraday context; preserved integral cases, exercises and angle diagram.
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.

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Quanta. Magnetic flux through a surface. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetischer-fluss

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Magnetic flux

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How do you calculate with Magnetic flux through a surface?

Why turning changes the result

A field value is not yet flux

Normal, angle and sign

Turning changes projection and sign

From small surface elements to the shortcut

Webers, teslas and a safe rearrangement

Example: a tilted measurement surface

Surface tilt with a sign

Example: a field gradient across the surface

Do not use the edge value everywhere

Explore further: open and closed surfaces

From a flux value to its time derivative

Example: equal flux, different emf

The instantaneous value does not reveal the slope

Five exercises: projection, model and checking

A uniform field B=0.250 TB=0.250\,\mathrm TB=0.250T crosses a plane area A=160 cm2A=160\,\mathrm{cm^2}A=160cm2 at 60∘60^\circ60∘ to the positive normal. Find signed flux.

Measured flux is ΦB=+3.00 mWb\Phi_B=+3.00\,\mathrm{mWb}ΦB​=+3.00mWb in uniform B=0.200 TB=0.200\,\mathrm TB=0.200T at θ=60∘\theta=60^\circθ=60∘. What plane area matches it?

A surface with normal +z has two parts of 0.00400 m20.00400\,\mathrm{m^2}0.00400m2 each. Their fields are Bz=+0.400 TB_z=+0.400\,\mathrm TBz​=+0.400T and −0.100 T-0.100\,\mathrm T−0.100T. Is total flux zero, positive or negative? Find it.

Field magnitude is 0.600 T and area is 0.0200 m2. The field lies 30° above the plane on the positive-normal side. A calculation uses ΦB=BAcos⁡30∘\Phi_B=BA\cos30^\circΦB​=BAcos30∘. Locate and correct the error.

60 equally oriented turns of 0.0150 m2 each rotate in a uniform 0.0300 T field from normal angle 0° to 90° in 0.120 s. Find mean emf in the positive traversal defined by the normal’s right-hand rule.

In brief

Sources for this explanation

Frequently asked questions about Magnetic flux through a surface

Subject references and further reading

More formulas: Electricity and magnetism

Retain Magnetic flux through a surface for exams

Magnetic flux

Variables & units

Magnetic flux calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetic flux and induction

Worked example

Applications

How do you calculate with Magnetic flux through a surface?

Why turning changes the result

A field value is not yet flux

Normal, angle and sign

Turning changes projection and sign

From small surface elements to the shortcut

Webers, teslas and a safe rearrangement

Example: a tilted measurement surface

Surface tilt with a sign

Example: a field gradient across the surface

Do not use the edge value everywhere

Explore further: open and closed surfaces

From a flux value to its time derivative

Example: equal flux, different emf

The instantaneous value does not reveal the slope

Five exercises: projection, model and checking

A uniform field B=0.250 TB=0.250\,\mathrm TB=0.250T crosses a plane area A=160 cm2A=160\,\mathrm{cm^2}A=160cm2 at 60∘60^\circ60∘ to the positive normal. Find signed flux.

Measured flux is ΦB=+3.00 mWb\Phi_B=+3.00\,\mathrm{mWb}ΦB​=+3.00mWb in uniform B=0.200 TB=0.200\,\mathrm TB=0.200T at θ=60∘\theta=60^\circθ=60∘. What plane area matches it?

A surface with normal +z has two parts of 0.00400 m20.00400\,\mathrm{m^2}0.00400m2 each. Their fields are Bz=+0.400 TB_z=+0.400\,\mathrm TBz​=+0.400T and −0.100 T-0.100\,\mathrm T−0.100T. Is total flux zero, positive or negative? Find it.

Field magnitude is 0.600 T and area is 0.0200 m2. The field lies 30° above the plane on the positive-normal side. A calculation uses ΦB=BAcos⁡30∘\Phi_B=BA\cos30^\circΦB​=BAcos30∘. Locate and correct the error.

60 equally oriented turns of 0.0150 m2 each rotate in a uniform 0.0300 T field from normal angle 0° to 90° in 0.120 s. Find mean emf in the positive traversal defined by the normal’s right-hand rule.

In brief

Sources for this explanation

A uniform field $B=0.250\,\mathrm T$ crosses a plane area $A=160\,\mathrm{cm^2}$ at $60^\circ$ to the positive normal. Find signed flux.

Measured flux is $\Phi_B=+3.00\,\mathrm{mWb}$ in uniform $B=0.200\,\mathrm T$ at $\theta=60^\circ$ . What plane area matches it?

A surface with normal +z has two parts of $0.00400\,\mathrm{m^2}$ each. Their fields are $B_z=+0.400\,\mathrm T$ and $-0.100\,\mathrm T$ . Is total flux zero, positive or negative? Find it.

Field magnitude is 0.600 T and area is 0.0200 m². The field lies 30° above the plane on the positive-normal side. A calculation uses $\Phi_B=BA\cos30^\circ$ . Locate and correct the error.

60 equally oriented turns of 0.0150 m² each rotate in a uniform 0.0300 T field from normal angle 0° to 90° in 0.120 s. Find mean emf in the positive traversal defined by the normal’s right-hand rule.

A uniform field $B=0.250\,\mathrm T$ crosses a plane area $A=160\,\mathrm{cm^2}$ at $60^\circ$ to the positive normal. Find signed flux.

Measured flux is $\Phi_B=+3.00\,\mathrm{mWb}$ in uniform $B=0.200\,\mathrm T$ at $\theta=60^\circ$ . What plane area matches it?

A surface with normal +z has two parts of $0.00400\,\mathrm{m^2}$ each. Their fields are $B_z=+0.400\,\mathrm T$ and $-0.100\,\mathrm T$ . Is total flux zero, positive or negative? Find it.

Field magnitude is 0.600 T and area is 0.0200 m². The field lies 30° above the plane on the positive-normal side. A calculation uses $\Phi_B=BA\cos30^\circ$ . Locate and correct the error.

60 equally oriented turns of 0.0150 m² each rotate in a uniform 0.0300 T field from normal angle 0° to 90° in 0.120 s. Find mean emf in the positive traversal defined by the normal’s right-hand rule.