---
title: Magnetic Flux
module: Electromagnetic Induction
moduleNumber: 8
lessonNumber: 1
order: 801
summary: >
  A magnetic field threading a loop collapses to one signed number, the flux, and
  every induced voltage in this module turns out to be a rate of change of that number
  — so defining the flux and its sign comes first. We define it as the surface
  integral of $\vec B$ over an oriented surface, reduce it to $BA\cos\theta$ for a
  uniform field on a flat loop, and carry the flux linkage $N\Phi_B$ of a coil. The
  chosen normal fixes the sign; reversing it flips the sign without touching the field.
  Nonuniform fields and curved surfaces force the integral, so we also build the
  numerical estimate and the checks that separate a reliable value from a nominal
  field-times-area product.
topics: [Electromagnetic Induction]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 27 — Magnetism; §27-3 Gauss's Law for Magnetism"
  - book: Tipler & Mosca
    ref: "Ch. 28 — Magnetic Induction; §28-1 Magnetic Flux"
---

Magnetic flux measures the component of a magnetic field that passes through a chosen
surface. The surface must be specified: the same wire loop can bound either of two
oppositely oriented surfaces, and the selected normal determines the sign. A small
surface element of area $\d A$ with unit normal $\hat n$ has

$$
\d\Phi_B=\vec B\cdot\hat n\,\d A.
$$

The dot product retains both the field component normal to the surface and the signed
orientation. Parallel field and normal vectors give positive flux; opposite vectors
give negative flux; a field tangent to the surface gives zero flux. Flux is a scalar,
although its sign depends on an orientation convention.

$$
% caption: Magnetic flux through one oriented surface element. The patch has unit normal $n$; the field $B$ makes an angle $\theta$ with $n$, and only the projection of $B$ onto $n$ (the dashed drop to the normal axis) contributes to the element flux $\d\Phi_B = B\cos\theta\,\d A$.
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  \draw[->, acc, very thick] (0,0) -- (1.18,1.48) node[above right] {$B$};
  \draw[black, dashed] (1.18,1.48) -- (0.53,1.685);
  \draw[black] (0.428,0.528) arc[start angle=51, end angle=72, radius=0.68];
  \node[black, anchor=north] at (0,-0.72) {area element};
\end{tikzpicture}
$$

Summing the element contributions over $S$ gives

$$
\Phi_B=\int_S\vec B\cdot \d\vec A
=\int_S\vec B\cdot\hat n\,\d A.
$$

Here $\d\vec A=\hat n\,\d A$ is an oriented area element. Changing the
normal from $\hat n$ to $-\hat n$ changes $\Phi_B$ to
$-\Phi_B$. Neither choice is inherently preferred. A circuit convention fixes one
normal, and the right-hand rule then fixes the positive direction around the circuit:
curl the fingers in the chosen traversal direction and the thumb gives the normal.
The chosen loop traversal fixes the induced-emf sign in Faraday's law.

## Surface Orientation and Signed Flux

Every two-sided surface has two possible normals. A flat disk has one normal pointing
out of one face and the other out of the opposite face. The geometrical area is
the same for both choices, but the vector area differs by a sign. A uniform field
passing upward through a horizontal disk has positive flux under an upward-normal
convention and negative flux under a downward-normal convention. The magnetic field
has not changed; the bookkeeping convention has.

$$
% caption: The same horizontal disk under two normal conventions. An upward field gives positive flux for the upward normal (left) and negative flux for the downward normal (right); the physical field is identical in both panels.
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  \draw[->, acc, very thick] (0,-0.10) -- (0,1.45) node[above] {$B$};
  \draw[->, black, thick] (0.34,0.30) -- (0.34,0.96) node[right] {$n$};
  \node[black, anchor=north] at (0,-1.20) {upward normal};
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  \draw[->, black, thick] (5.54,-0.30) -- (5.54,-0.85) node[right] {$n$};
  \node[black, anchor=north] at (5.20,-1.20) {downward normal};
\end{tikzpicture}
$$

The sign convention should be written before a calculation begins. A diagram with a
loop, a normal arrow, and a field arrow provides enough information to determine the
sign without relying on an ambiguous phrase such as "through the loop." In a
multiturn coil, the normal must be chosen consistently for every turn. Reversing the
winding direction reverses the associated loop traversal and therefore reverses the
sign of the flux linkage.

A curved surface can have a normal that varies from point to point. A sphere has
radial outward normal, so field components normal to the
surface differ around the sphere. The integral definition handles that variation by
using the local $\d\vec A$ at every surface element. A single area vector can replace
the full integral only for a flat surface, or under a symmetry condition that makes a
more general calculation simple.

$$
% caption: Local normals on a curved surface. Each patch carries its own outward normal, so a uniform vertical field has positive, zero, and negative local normal components on different parts of the surface.
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  \draw[->, black, thick] (1.30,0) -- (1.98,0) node[right] {$n$};
  \draw[->, black, thick] (0,-1.30) -- (0,-1.98) node[below] {$n$};
  \draw[->, black, thick] (-1.30,0) -- (-1.98,0) node[left] {$n$};
  \node[acc, anchor=west] at (1.05,1.58) {$B$};
  \node[black, anchor=west] at (2.20,-0.80) {curved surface};
\end{tikzpicture}
$$

A closed surface reverses its net flux when every local normal is reversed. In a
uniform magnetic field, the positive flux entering one side of a closed surface is balanced by
negative flux leaving the other side, so the net flux is zero. This result agrees with
the absence of isolated magnetic monopoles in classical electromagnetism. It concerns
the total over the closed surface; individual patches can still have substantial local
flux.

## Uniform Fields and Coils

When $\vec B$ is uniform over a flat surface of area $A$, the normal is constant
and can be factored out of the integral:

$$
\Phi_B=\vec B\cdot\hat n\,A
=BA\cos\theta,
$$

where $\theta$ is the angle from the selected normal to the field direction. The
angle is not measured from the plane of the loop. If the field makes an angle $\alpha$
with the plane, then $\theta=90^\circ-\alpha$ and

$$
\Phi_B=BA\sin\alpha.
$$

Using the plane angle directly in a cosine formula gives the wrong normal component.
A loop viewed edge-on by the field has $\theta=90^\circ$ and zero
flux even though the magnetic field may be large.

$$
% caption: Angle in the flat-surface flux formula. The angle $\theta$ used in $BA\cos\theta$ is measured between $B$ and the normal $n$, not between $B$ and the plane of the loop. A field parallel to the loop plane has no normal component.
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  \draw[black, dashed] (-1.70,0.13) -- (1.25,0.54) node[right] {loop plane};
\end{tikzpicture}
$$

The projected-area interpretation gives the same formula. The surface presents an
effective area $A_\perp=A\cos\theta$ to the field, so $\Phi_B=BA_\perp$. A broad
loop tilted far from face-on presents a smaller projected silhouette, while a narrow
edge-on silhouette has vanishing projected area. Use projected area for flat-surface
estimates; use $\vec B\cdot \d\vec A$ for signed flux.

The flux magnitude is largest when the field is perpendicular to the surface and
smallest when it is tangent. A plot against $\theta$ has the cosine shape: it starts at
$+BA$, crosses zero at $90^\circ$, and reaches $-BA$ when the field opposes the chosen
normal. The negative half of the graph represents the same field magnitude passing
through the opposite side of the oriented surface.

$$
% caption: Signed flux versus the angle from $B$ to the selected normal. The curve starts at $+BA$, crosses zero when $B$ lies in the surface plane, and reaches $-BA$ when the field opposes the normal.
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  \node[black, anchor=north] at (2.925,-1.28) {right angle};
  \node[black, anchor=west] at (0.25,1.42) {positive};
  \node[black, anchor=west] at (4.45,-1.35) {negative};
\end{tikzpicture}
$$

Magnetic flux has SI unit weber,

$$
1\ \mathrm{Wb}=1\ \mathrm{T\,m^2}.
$$

Tesla measures field strength; multiplying by square metres gives the required flux
dimensions.

> **Worked example (flux through a perpendicular loop).** A field of
> $0.20\ \mathrm T$ perpendicular to a $3.0\times10^{-2}\ \mathrm{m^2}$ loop gives
>
> $$
> \Phi_B=(0.20)(3.0\times10^{-2})
> =6.0\times10^{-3}\ \mathrm{Wb}.
> $$
>
> If the field opposes the chosen normal, the same geometry gives
> $-6.0\times10^{-3}\ \mathrm{Wb}$: the magnitude is unchanged and the sign records
> orientation.

State the selected surface and normal whenever a flux value is tabulated. Reversing
that normal changes the reported sign without changing the physical magnetic field
or the surface geometry.

### Flux linkage of a multiturn coil

A coil containing $N$ closely spaced turns samples the same field through each turn
when its winding is uniform and the field does not vary appreciably across the coil.
The single-turn flux is $\Phi_{B,1}=BA\cos\theta$. Adding the signed contributions of
all turns gives the flux linkage

$$
\Lambda_B=N\Phi_{B,1}=NBA\cos\theta.
$$

The symbol $\Lambda_B$ keeps the total linkage distinct from the flux through one
turn. Introductory treatments sometimes use $\Phi_B$ for the total $N$-turn quantity;
the physical calculation is determined by whether the factor $N$ has been included.
Writing the single-turn area and the number of turns separately avoids a factor-of-$N$
error in induction calculations.

Every turn must share the same orientation. A tightly wound coil has a consistent
normal set by the winding direction, so all turn contributions add. Connecting one
turn in the opposite sense gives a negative contribution relative to the others and
reduces the net linkage. This is the same sign structure used for a series connection
of loops, expressed through the sign of oriented area.

$$
% caption: Flux linkage in a four-turn coil. The same axial field crosses each turn with the same selected normal, so the signed single-turn flux is repeated four times and the total linkage is four times the one-turn value.
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  \node[black, anchor=west] at (1.45,0.55) {four turns};
  \node[black, anchor=west] at (1.45,-0.05) {same area};
  \node[black, anchor=west] at (1.45,-0.65) {same normal};
\end{tikzpicture}
$$

> **Worked example (flux linkage of a 250-turn coil).** A coil of area
> $A=4.0\times10^{-4}\ \mathrm{m^2}$ with $N=250$ turns sits in a uniform
> $0.12\ \mathrm T$ field at orientation $\theta=0$. Its single-turn flux is
>
> $$
> \Phi_{B,1}=BA
> =(0.12)(4.0\times10^{-4})
> =4.8\times10^{-5}\ \mathrm{Wb},
> $$
>
> and the linkage is
>
> $$
> \Lambda_B=N\Phi_{B,1}=250(4.8\times10^{-5})
> =1.2\times10^{-2}\ \mathrm{Wb}.
> $$
>
> The unit stays the weber because a turn count is dimensionless. The larger linkage
> reflects multiple loops encircling the same field region; the field strength has not
> been multiplied by $N$.

The linkage is linear in turn count when the field through each turn remains fixed.
Doubling $N$ doubles $\Lambda_B$. This statement has a geometrical condition: the
coil must remain compact enough that outer turns do not sample a substantially
different field or a substantially different area. A long coil placed in a rapidly
varying field may require a turn-by-turn sum or an integral over the winding.

Opposing windings test the sign convention. If two equal loops share the same
area and field but are connected with opposite normals, their linkages are
$+BA\cos\theta$ and $-BA\cos\theta$. The sum is zero. The cancellation comes from
orientation, not from a disappearance of field at either loop. This arrangement is
the flux analogue of adding vectors with equal magnitude and opposite direction.

## Nonuniform and Solenoidal Fields

The product $BA\cos\theta$ applies only when the normal component of $\vec B$ is
constant over a flat surface. A field can vary in magnitude from one place to another,
can change direction across a curved surface, or both. The surface integral then adds
small contributions,

$$
\Phi_B=\lim_{\Delta A_i\to0}
\sum_i \bigl(\vec B_i\cdot\hat n_i\bigr)\Delta A_i.
$$

Each partition cell has a local field vector, local normal, and small area. Refining
the cells improves the approximation because the field and normal vary less within
each one. The integral is the limiting value of this geometrical sum.

$$
% caption: Surface-partition calculation for a nonuniform field. The rectangular surface is divided into narrow strips; each strip contributes its local normal component of $B$ times its own area, and the total flux is their signed sum.
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  \node[black, anchor=north] at (2.65,-0.15) {equal-area strips};
  \node[acc, anchor=west] at (5.50,1.62) {larger B};
\end{tikzpicture}
$$

> **Worked example (flux of a linearly varying field).** A rectangle of width $L$ in
> $x$ and height $h$ in $y$ has selected normal along $+z$, and the field is
> perpendicular to it with magnitude
>
> $$
> B(x)=B_0\left(1+\frac{x}{L}\right),
> \qquad 0\leq x\leq L.
> $$
>
> The strip between $x$ and $x+\d x$ has area $\d A=h\,\d x$, so its flux is
> $\d\Phi_B=B(x)h\,\d x$. Integrating across the width,
>
> $$
> \begin{aligned}
> \Phi_B
> &=\int_0^L B_0\left(1+\frac{x}{L}\right)h\,\d x\\
> &=B_0h\left[x+\frac{x^2}{2L}\right]_0^L
> =\frac32B_0Lh.
> \end{aligned}
> $$
>
> The average field over the rectangle is $3B_0/2$; multiplying it by the total area
> $Lh$ recovers the same flux. Using the left-edge field $B_0$ underestimates the
> result and the right-edge field $2B_0$ overestimates it. Integrate the local normal
> field over the full area; an edge value is not representative.

For physical magnetic fields, any two oriented surfaces with the same boundary have
equal flux because $\nabla\cdot\vec B=0$. Direct calculations can nevertheless
look different because the two surfaces require different area elements and may
expose different symmetry. Choose the spanning surface that makes
$\vec B\cdot \d\vec A$ easiest to evaluate, then retain the same boundary
orientation.

When a field has both normal and tangential components, the tangential part contributes
zero to the flux integral. A flat surface in the $xy$ plane has

$$
\vec B=B_x\hat\imath+B_y\hat\jmath+B_z\hat k,
\qquad
\d\vec A=\hat k\,\d A.
$$

The dot product becomes $\vec B\cdot \d\vec A=B_z\,\d A$. Components $B_x$ and
$B_y$ can be large without changing the flux through that horizontal surface. The
normal component must be identified from the surface geometry, not from whichever
field component has the largest numerical magnitude.

### Flux linked by a long solenoid

A long solenoid has an approximately uniform interior field parallel to its axis. Let the
solenoid have $N$ turns, length $\ell$, radius $r$, and current $I$. Away from its
ends, the field magnitude is

$$
B=\mu_0\frac{N}{\ell}I.
$$

The area bounded by one circular turn is $A=\pi r^2$, and the field is parallel to
the selected axial normal. The one-turn flux is $BA$; multiplying by $N$ turns gives

$$
\Lambda_B=NBA
=\mu_0\frac{N^2I\pi r^2}{\ell}.
$$

The $N^2$ dependence has two separate origins. One factor of $N$ appears because the
solenoid's current produces a field proportional to turns per length. The second
factor appears because that field links every turn. Confusing field strength with
linkage omits one of these factors and gives the wrong scaling when the winding is
changed.

$$
% caption: Flux geometry inside a long solenoid. In the central region the axial field is nearly uniform and perpendicular to every circular turn, so one turn contributes $BA$ and the full winding contributes $N$ times that value.
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  \node[black, anchor=north] at (2.60,-1.14) {central length};
  \node[acc, anchor=west] at (5.30,0.25) {$B$};
  \node[black, anchor=west] at (5.30,-0.30) {turn area};
\end{tikzpicture}
$$

> **Worked example (flux linked by a solenoid).** Take $\ell=0.40\ \mathrm m$,
> $r=2.5\ \mathrm{cm}$, $N=600$, and $I=7.5\ \mathrm A$. The turn density is
> $N/\ell=1.50\times10^3\ \mathrm{m^{-1}}$, so the central field is
>
> $$
> B=\mu_0\frac{N}{\ell}I=(4\pi\times10^{-7})(1.50\times10^3)(7.5)
> =1.41\times10^{-2}\ \mathrm T.
> $$
>
> The area of one turn is
>
> $$
> A=\pi(0.025)^2
> =1.96\times10^{-3}\ \mathrm{m^2},
> $$
>
> and the linked flux is
>
> $$
> \Lambda_B=NBA=(600)(1.41\times10^{-2})(1.96\times10^{-3})
> =1.67\times10^{-2}\ \mathrm{Wb}.
> $$
>
> The area uses metres, not centimetres. Leaving the radius as $2.5$ rather than
> $0.025$ in SI units would inflate the area by $10^4$ and the flux by the same factor.
> The dimensional check $\mathrm T\,\mathrm{m^2}=\mathrm{Wb}$ catches neither a missing
> turn count nor a length-conversion error, so each physical factor must be traced
> through separately.

The long-solenoid approximation excludes end regions where field lines spread and the
axial magnitude falls. A turn near an end links a smaller and less uniform field than
a central turn. Short solenoids and coils with magnetic cores of nonuniform
permeability require a local field profile before use of $NBA$. The compact formula
assumes the same flux through every turn.

Flux linkage is distinct from magnetic energy and inductance. The flux uses a
specified current and geometry; inductance relates the linkage to current for a
particular circuit; magnetic energy depends on both current and inductance.

## Closed and Composite Surfaces

An open surface has a boundary curve. A circular disk bounded by a wire loop is open
because its edge is the loop. The flux through it depends on the selected normal and
is the quantity used in induction. A closed surface has no boundary: a sphere, a
sealed cylinder, or a box encloses a volume. Its conventional normal is outward at
every point, and its flux is written with a closed-surface integral,

$$
\Phi_{B,\mathrm{closed}}
=\oint_S\vec B\cdot \d\vec A.
$$

For magnetic fields in classical electromagnetism,

$$
\oint_S\vec B\cdot \d\vec A=0.
$$

The zero result expresses a balance of entering and leaving field through a closed
surface. It does not require $\vec B$ to vanish on the surface. A strong uniform
field crossing a box has positive flux through one face and equal negative flux
through the opposite face.

$$
% caption: Flux through a closed cylinder in a uniform axial field. The outward normal gives positive flux on the right end cap and negative flux on the left end cap; the curved wall has zero normal field component, so the total closed-surface flux is zero.
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  \draw[->, black, thick] (4.70,1.14) -- (5.55,1.14) node[right] {$n$};
  \node[black, anchor=north] at (0,-0.20) {negative end};
  \node[black, anchor=north] at (4.70,-0.20) {positive end};
  \node[acc, anchor=west] at (5.30,0.55) {$B$};
\end{tikzpicture}
$$

The cylindrical example can be evaluated directly. Suppose the field is constant and
parallel to the cylinder axis, and each end cap has area $A$. The outward normal on
the right cap is parallel to $\vec B$, giving $+BA$. The outward normal on the
left cap is antiparallel, giving $-BA$. The normal is radial along the curved wall,
perpendicular to the axial field, so that wall contributes zero. Adding the three
parts yields $BA-BA+0=0$.

The cancellation is geometrical. Tilting the cylinder changes the flux contribution
of each cap and introduces a nonzero contribution from the curved wall, but the total
remains zero. The integral tracks the complete oriented surface; splitting it into
pieces changes the intermediate terms, not the total.

Magnetic field lines form continuous loops rather than beginning or ending at isolated
magnetic charges. A bar magnet has field lines emerging from one face and entering the
other outside the magnet, with a return path through its interior. Any closed surface
surrounding the magnet therefore has as much signed flux entering as leaving. The
field near individual patches can be large even though the total is zero.

The closed-surface law is a global constraint. It cannot, by itself, determine the
magnetic field at an arbitrary point, because many nonzero field patterns have zero
net flux through every closed surface. Symmetry and Ampere's law provide additional
information for special current distributions. Here its role is narrower: it verifies
that flux bookkeeping through complementary pieces of a closed surface is internally
consistent.

For induction, the relevant surface is usually open and bounded by the circuit. The
surface can be selected for calculation convenience, but its boundary orientation must
remain tied to the circuit orientation. The closed-surface cancellation result should
not be substituted for the flux through one open loop: an open disk in a uniform field
can have nonzero flux even though adding a complementary surface to close it produces
zero total flux.

## Flux from Geometry and Data

For a flat surface, define the oriented area vector

$$
\vec A=A\hat n.
$$

The flux is

$$
\Phi_B=\vec B\cdot\vec A.
$$

The vector $\vec A$ combines surface area, normal direction, and sign. Cartesian
components then give the flux directly. A rectangular surface lying in the $xy$ plane
with selected normal $+\hat k$ has

$$
\vec A=A\hat k,
\qquad
\Phi_B=B_zA.
$$

The $x$ and $y$ components of the field disappear from the dot product. Their absence
is a geometrical result: they run parallel to the surface rather than through it.

> **Worked example (flux from field components).** A $0.30\ \mathrm m$ by
> $0.40\ \mathrm m$ rectangle lies in the $xy$ plane, so $A=0.120\ \mathrm{m^2}$. At
> the surface,
>
> $$
> \vec B=(0.20\hat\imath-0.30\hat\jmath
> +0.50\hat k)\ \mathrm T.
> $$
>
> For the upward normal, $\vec A=(0.120\hat k)\ \mathrm{m^2}$, so
>
> $$
> \Phi_B
> =\vec B\cdot\vec A
> =(0.50)(0.120)
> =6.0\times10^{-2}\ \mathrm{Wb}.
> $$
>
> The in-plane components, of combined magnitude
> $\sqrt{0.20^2+0.30^2}=0.36\ \mathrm T$, contribute nothing. Reversing the normal
> changes $\vec A$ to $-0.120\hat k\ \mathrm{m^2}$ and the flux to
> $-6.0\times10^{-2}\ \mathrm{Wb}$.

A parallelogram generated by two edge vectors $\vec u$ and $\vec v$ has area
vector

$$
\vec A=\vec u\times\vec v.
$$

Its magnitude is the parallelogram area and its direction follows the right-hand rule
from $\vec u$ to $\vec v$. Exchanging the order reverses the area vector:

$$
\vec v\times\vec u=-\vec u\times\vec v.
$$

Reversing the vector order selects the opposite surface normal algebraically. A triangle
spanned by the same vectors has half the area vector,

$$
\vec A_{\rm triangle}=\frac12\vec u\times\vec v.
$$

Coordinate vectors supply a sign audit. If a result is positive, the
calculated normal component of $\vec B$ points along the selected area vector. If a
result is negative, it points opposite. The magnitude cannot identify the sign by
itself; it must be paired with the normal convention shown in the geometry.

The area-vector method applies only to a planar patch represented by one normal. A
curved surface can be approximated by small planar patches, each with its own area
vector. Summing those vectors without their local field values is insufficient when
$\vec B$ varies across the surface. The full surface integral keeps each local
dot product associated with the appropriate patch.

### Flux from a measured field map

Many field configurations do not have a simple symbolic expression. A measured map
of the normal component can still determine flux. Divide the surface into cells of
area $\Delta A_i$, measure or calculate the normal field at a representative point in
each cell, and form

$$
\Phi_B\approx\sum_i B_{n,i}\Delta A_i,
\qquad
B_{n,i}=\vec B_i\cdot\hat n_i.
$$

The summation is a numerical surface integral. The sign remains in $B_{n,i}$, so cells where
the field points against the selected normal subtract from the total. A map of field
magnitudes alone is insufficient when the field direction varies.

Suppose a rectangular surface is divided into six equal cells, each with area
$1.0\times10^{-2}\ \mathrm{m^2}$. The measured normal components in tesla are

| | left | middle | right |
|---|---:|---:|---:|
| upper row | $0.30$ | $0.38$ | $0.46$ |
| lower row | $0.34$ | $0.42$ | $0.50$ |

> **Worked example (numerical flux from a cell map).** The six measured normal
> components above sum to $2.40\ \mathrm T$, so the flux estimate is
>
> $$
> \Phi_B
> \approx(2.40\ \mathrm T)(1.0\times10^{-2}\ \mathrm{m^2})
> =2.4\times10^{-2}\ \mathrm{Wb}.
> $$
>
> The mean normal field is $0.40\ \mathrm T$ over a total area
> $6.0\times10^{-2}\ \mathrm{m^2}$, giving the same result by $\Phi_B=\bar B_nA$. The
> average-field shortcut is valid because the mean was formed from the same
> area-weighted cell data.

Cell resolution controls the numerical error. A coarse grid may miss a narrow region
of rapid field variation, especially near a wire, a magnet edge, or the end of a
solenoid. Subdividing each cell reduces the error if the representative values are
sampled consistently. Comparing a coarse-grid estimate with a refined-grid estimate
gives a practical convergence check. The two estimates should approach a stable value
as the largest cell dimension becomes small relative to the field-variation length.

For nonidentical cells, each area must remain attached to its own field value. An
unweighted average of $B_n$ values is incorrect if some cells cover more surface than
others. The proper average is

$$
\bar B_n
=\frac{\sum_iB_{n,i}\Delta A_i}{\sum_i\Delta A_i}.
$$

The numerator is the flux estimate itself. The estimator also handles curved surfaces when
each cell is small enough to be treated as planar and its local normal is known.

The numerical method is also a model check. If the cell values appear inconsistent
with a stated symmetry, revisit the normal convention, coordinate registration, and
field calibration before averaging. A sign reversal in one region may be a physical
feature or a misplaced normal arrow. The signed cell sum displays the distinction;
an absolute-value average loses it.

### Composite, annular, and piecewise surfaces

Flux is additive over a surface partition. If a surface is composed of disjoint pieces
$S_1,S_2,\ldots$, all carrying normals chosen consistently, then

$$
\Phi_B(S)=\Phi_B(S_1)+\Phi_B(S_2)+\cdots.
$$

The piecewise-sum rule applies whether the pieces are selected by material boundaries, by field
regions, or by a convenient mathematical partition. A surface that contains a hole is
handled by subtracting the missing area with the same normal convention. The boundary
of the hole has the opposite traversal direction from the outer boundary, consistent
with the oriented-area subtraction.

An annular loop with outer radius $b$ and inner radius $a$, placed perpendicular to
a uniform magnetic field, has

$$
A_{\rm annulus}=\pi(b^2-a^2),
$$

so

$$
\Phi_B=B\pi(b^2-a^2).
$$

The missing central disk contributes no flux because it is not part of the chosen
surface. Replacing the annulus by the outer disk alone would overcount the physical
area bounded by the loop arrangement.

$$
% caption: Flux through an annular surface. A uniform normal field crosses the material between radii $a$ and $b$; the central hole is excluded, so the effective area is the outer disk minus the inner disk.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \fill[acc!10] (0,0) circle (1.60);
  \fill[white] (0,0) circle (0.68);
  \draw[black, thick] (0,0) circle (1.60);
  \draw[black, thick] (0,0) circle (0.68);
  \foreach \x in {-1.15,-0.575,0,0.575,1.15} {
    \draw[->, acc, thick] (\x,-2.05) -- (\x,2.05);
  }
  \draw[black, dashed] (0,0) -- (1.60,0);
  \draw[black, dashed] (0,0) -- (-0.68,0);
  \node[black, anchor=south] at (0.90,0.14) {$b$};
  \node[black, anchor=south] at (-0.34,0.12) {$a$};
  \node[black, anchor=west] at (2.10,0.62) {material area};
  \node[black, anchor=west] at (2.10,-0.05) {central hole};
  \node[acc, anchor=west] at (2.10,-0.72) {$B$};
\end{tikzpicture}
$$

> **Worked example (flux through an annulus).** Take $a=1.0\ \mathrm{cm}$,
> $b=3.0\ \mathrm{cm}$, and $B=0.80\ \mathrm T$ normal to the annulus. Converting radii
> to metres,
>
> $$
> \begin{aligned}
> \Phi_B
> &=B\pi(b^2-a^2)=(0.80)\pi\left[(0.030)^2-(0.010)^2\right]\\
> &=2.0\times10^{-3}\ \mathrm{Wb}.
> \end{aligned}
> $$
>
> The outer-disk calculation alone gives $2.3\times10^{-3}\ \mathrm{Wb}$, too large by
> the central-disk contribution. The difference is appreciable because the hole occupies
> one ninth of the outer-disk area.

Piecewise fields use the same additivity. Suppose one half of a flat loop lies in a
uniform field $+B\hat n$ and the other half lies in
$-B\hat n$, with equal areas. The two fluxes cancel, even though the field
magnitude is nonzero throughout both halves. If one region has twice the area of the
other, the total has the sign of the larger-area contribution. Signed area and signed
normal field must both remain explicit until the pieces have been added.

An interface where the field changes abruptly requires no special flux rule. Divide
the surface at the interface, use the appropriate expression on each side, and add the
results. The field value on a line interface has zero area measure and does
not affect the surface integral. This is the same logic used for discontinuous density
or pressure functions in an ordinary area integral.

### Flux as a function of field and geometry

Flux through a flat, rigid loop in a uniform field depends on three independently
specified quantities:

$$
\Phi_B=BA\cos\theta.
$$

The field magnitude $B$, the surface area $A$, and the angle $\theta$ can each vary
with time while the others remain fixed. A changing field magnitude may result from a
moving magnet or a changing current in a nearby coil. A changing area may result from
a sliding conductor or a deforming loop. A changing angle may result from rotation of
the loop or rotation of the field source. These are distinct geometrical routes to a
changed flux.

$$
% caption: Three independent routes to a changed flat-loop flux $BA\cos\theta$. Increasing the normal field magnitude (left), increasing the loop area (centre), or rotating the normal toward $B$ (right) each raises the signed flux when the original orientation is positive.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, dashed] (2.55,-1.35) -- (2.55,1.75);
  \draw[black, dashed] (5.35,-1.35) -- (5.35,1.75);
  \draw[black, fill=acc!8] (0.85,-0.55) ellipse (0.70 and 0.22);
  \draw[->, acc, thin] (0.55,-0.55) -- (0.55,0.20);
  \draw[->, acc, very thick] (1.20,-0.55) -- (1.20,1.35);
  \node[black, anchor=north] at (0.85,-0.85) {stronger B};
  \draw[black, fill=acc!8] (3.95,-0.55) ellipse (1.02 and 0.32);
  \draw[->, acc, very thick] (3.95,-0.55) -- (3.95,1.35) node[above] {$B$};
  \node[black, anchor=north] at (3.95,-0.95) {larger area};
  \draw[black, fill=acc!8] (6.15,-0.62) -- (7.75,-0.20) -- (7.55,0.34) -- (5.95,-0.08) -- cycle;
  \draw[->, black, thick] (6.85,-0.13) -- (6.85,1.10) node[above] {$n$};
  \draw[->, acc, very thick] (6.85,-0.13) -- (7.35,1.10) node[right] {$B$};
  \node[black, anchor=north] at (6.90,-0.85) {smaller angle};
\end{tikzpicture}
$$

When all three are time dependent, ordinary differentiation gives

$$
\frac{\d\Phi_B}{\d t}
=A\cos\theta\,\frac{\d B}{\d t}
+B\cos\theta\,\frac{\d A}{\d t}
-BA\sin\theta\,\frac{\d\theta}{\d t}.
$$

Each term has a direct geometrical meaning. The first records a changing normal field
through a fixed surface. The second records a changing projected area in a fixed field.
The third records rotation; its sign follows from whether the angle from $\vec B$ to
the selected normal is increasing or decreasing. Faraday's law connects this flux
rate to induced emf. Here, the derivative identifies which feature of the
field-and-surface configuration is changing.

> **Worked example (flux rate of a rotating loop).** A loop of fixed area
> $A=2.0\times10^{-2}\ \mathrm{m^2}$ rotates in a constant $0.50\ \mathrm T$ field. At
> an instant when $\theta=60^\circ$ and the angle increases at $8.0\ \mathrm{rad/s}$,
> only the rotation term survives:
>
> $$
> \frac{\d\Phi_B}{\d t}
> =-BA\sin\theta\,\frac{\d\theta}{\d t}
> =-(0.50)(2.0\times10^{-2})(0.866)(8.0)
> =-6.9\times10^{-2}\ \mathrm{Wb/s}.
> $$
>
> The negative sign means the positive flux is decreasing under the selected normal, not
> that the field magnitude is negative. A reverse rotation with $\d\theta/\d t<0$ at the
> same instant would give a positive flux rate.

The configuration formula also distinguishes a nonzero flux from a changing flux. A
stationary loop can have a large constant $BA\cos\theta$; a loop edge-on to the field
can have zero flux while a small rotation produces a nonzero rate of change. The
initial value and the rate describe different aspects of the same configuration.

For $N$ identical turns, the flux linkage is

$$
\Lambda_B(t)=N\Phi_B(t).
$$

A coil with fixed $N$ has every linkage change equal to $N$ times the corresponding
single-turn flux change. If the number of active turns changes through a switching
connection, the physical circuit has changed and the linkage expression must be
recomputed for the new winding arrangement. The switched circuit has a different
linkage model from one continuously rotating loop.

The three-factor form $BA\cos\theta$ has a narrow range of validity. It applies to one
flat surface in a uniform field. A nonuniform field requires the surface integral at
each time, and a curved or deforming surface requires its local area vectors. The
geometrical sources of change remain the same, but the compact product is replaced by

$$
\Phi_B(t)=\int_{S(t)}\vec B(\vec r,t)\cdot \d\vec A.
$$

The notation $S(t)$ records that the chosen surface itself may move or change shape.
It keeps the field distribution and surface geometry visible in the calculation.

### Consistency conditions for a flux calculation

A flux calculation requires an oriented surface and the full field distribution.
It should identify four pieces of information in the same
coordinate system:

- **Surface and boundary:** the actual patch, disk, coil turn, or closed surface
  included in the integral.
- **Normal convention:** the selected $\hat n$, or the right-hand-rule loop
  traversal that fixes it.
- **Normal field:** $B_n=\vec B\cdot\hat n$ at every relevant location.
- **Area measure:** $A$ for a uniform planar case, or $\d A$ and integration limits for
  a varying geometry.

The dot product places the sign in the normal field component. A scalar calculation
with $B$, $A$, and a positive cosine yields only a magnitude unless the angle is
explicitly measured from the selected normal. A diagram that shows an arrow into the
surface but reports a positive result can still be correct if the selected normal
points into the page; the sign follows the stated convention, not page orientation.

> **Worked example (signed linkage of a tilted coil).** A $40$-turn flat coil of area
> $5.0\times10^{-2}\ \mathrm{m^2}$ sits in a uniform $0.36\ \mathrm T$ field with its
> selected normal at $120^\circ$ to the field. The one-turn flux is
>
> $$
> \Phi_{B,1}
> =BA\cos\theta=(0.36)(5.0\times10^{-2})\cos120^\circ
> =-9.0\times10^{-3}\ \mathrm{Wb},
> $$
>
> negative because the field has a component opposite the selected normal. The coil
> linkage is
>
> $$
> \Lambda_B=N\Phi_{B,1}=40(-9.0\times10^{-3})
> =-0.36\ \mathrm{Wb}.
> $$
>
> Reversing the normal flips both signs while preserving the magnitudes. Rotating the
> coil by $120^\circ$ in the direction that decreases $\theta$ brings the normal parallel
> to the field and raises the one-turn flux magnitude to
> $1.8\times10^{-2}\ \mathrm{Wb}$.

The dimensional check is

$$
[\Phi_B]=[B][A]
=\mathrm T\,\mathrm{m^2}
=\mathrm{Wb}.
$$

The unit of a time rate of change is

$$
\frac{\mathrm{Wb}}{\mathrm s}=\frac{\mathrm{T\,m^2}}{\mathrm s}=\mathrm V.
$$

The final equality follows from the SI definitions and prepares the dimensional form
of Faraday's law. It does not turn a static flux value into a voltage. A nonzero rate
of flux change is the quantity with unit volts.

Field-line sketches can support a qualitative estimate when their density represents
field magnitude consistently. Doubling the drawn line density through a fixed
perpendicular area represents twice the flux in that visual convention. The drawing
is not a literal count of physical objects. Different diagrams use different line
counts and scales, so numerical flux must be calculated from field values and areas.

The limits of $BA\cos\theta$ check orientation and scale.

$$
\begin{array}{c|c}
\text{condition} & \Phi_B\\ \hline
\theta=0 & +BA\\
\theta=90^\circ & 0\\
\theta=180^\circ & -BA\\
B=0\ \text{or}\ A=0 & 0
\end{array}
$$

These limits apply to a uniform field through a flat surface. A calculation that gives
a nonzero result for a tangent uniform field has used the plane angle in the
wrong trigonometric function or has selected an inconsistent normal. A result whose
magnitude exceeds $BA$ for a single flat surface likewise violates the bound

$$
|\Phi_B|\leq BA.
$$

For nonuniform fields, replace the right side by an area-weighted bound such as
$\int_S|\vec B|\,\d A$. The signed flux can be much smaller than that bound because
oppositely directed normal components cancel. A small net flux therefore does not
imply a weak field everywhere on the surface.

Flux calculations often feed directly into later induction problems, where a sign
error changes the predicted emf direction. Preserving the normal convention from the
first surface sketch through the final numerical result prevents that error from being
introduced at the transition between geometry and circuit analysis.

## Surface Choice and Scale

An open loop can bound many different surfaces. A circular wire loop, for example,
can be spanned by its flat disk or by a smoothly bulged cap. The surfaces have the
same boundary but different local normals and different area elements. In a magnetic
field, their total fluxes agree when both are oriented consistently with the same loop
traversal. Joining one surface to the oppositely oriented version of the other creates
a closed surface. Gauss's law for magnetism then gives

$$
\oint \vec B\cdot \d\vec A
=\int_{S_1}\vec B\cdot \d\vec A
-\int_{S_2}\vec B\cdot \d\vec A
=0,
$$

and therefore

$$
\int_{S_1}\vec B\cdot \d\vec A
=\int_{S_2}\vec B\cdot \d\vec A.
$$

The minus sign in the intermediate equation comes from the opposite orientation needed
to close the two surfaces. It is an orientation statement, not a claim that either
open-surface flux is negative.

Choose the spanning surface that simplifies the integrand
without changing the loop boundary. If a field is uniform and a flat disk is available,
the disk gives $BA\cos\theta$ directly. If the field has cylindrical symmetry and a
curved surface makes the field normal component constant, that curved surface may be
better. The selected surface may pass through empty space; it need not be a material
membrane or a physical object in the apparatus.

Candidate surfaces must share the same oriented boundary. Replacing a loop by a larger loop changes the boundary and therefore
changes the flux problem. Reversing the boundary traversal reverses the selected normal
and changes the sign of the reported flux. A surface choice can simplify a calculation;
it cannot change the circuit or erase the orientation convention.

A moving or deforming loop can update its chosen spanning surface at each instant.
Its boundary must remain the instantaneous loop, and its normal must remain
connected to the same chosen traversal. This bookkeeping prevents a sign jump when a
loop rotates through an edge-on orientation. The flux passes smoothly through zero as
$\cos\theta$ changes sign; the normal convention remains fixed.

Surface freedom also clarifies why field lines are only a visualization. A different
spanning surface can cut through a different visual pattern of drawn lines, yet its
calculated flux has the same value when the boundary is unchanged. The invariant
quantity is the surface integral of the physical field, not the apparent number of
lines on one particular sketch.

The same argument supports a direct check on a numerical computation. Evaluate the
flux through two convenient meshes sharing a boundary. If both meshes represent the
same magnetic field and orientations, refined numerical results should converge to the
same value. A discrepancy that remains under refinement points to an inconsistent
normal, missing portion of the surface, or an incorrectly mapped field component.

In the induction setting, the circuit boundary carries the physical emf measurement,
while the spanning surface is an auxiliary geometrical construction. The boundary fixes
the orientation; the surface is selected to evaluate the associated flux. Keeping those
roles separate makes later applications of Faraday's law unambiguous.

### Flux density and spatial scale

Magnetic flux is an integrated quantity. The same total flux can arise from a strong
field through a small area or a weak field through a large area. Field strength and
flux therefore answer different questions. The local normal flux density is

$$
B_n=\vec B\cdot\hat n
=\frac{\d\Phi_B}{\d A}
$$

when the surface orientation is fixed and the differential limit is taken locally.
In a uniform perpendicular field, $B_n=B$ everywhere and total flux is $BA$.
In a nonuniform field, $B_n$ varies across the surface and the flux is its
area-weighted accumulation.

Consider two circular loops in the same perpendicular uniform field. If the second
loop has twice the radius of the first, its area is four times as large:

$$
A_2=\pi(2r)^2=4\pi r^2=4A_1.
$$

The second loop therefore has four times the flux, even though both loops sample the
same field strength. If a multiturn coil has twice as many turns as another coil with
the same turn area, it has twice the linkage. Radius scaling and turn-count scaling are
separate effects, so a coil with twice the radius and twice the turns has eight times
the linkage under the same field and orientation.

Spatial scale also determines whether a uniform-field approximation is credible. Let
the normal field change by a characteristic amount $\Delta B_n$ across a loop of
diameter $d$. When $|\Delta B_n|\ll|B_n|$, replacing the field by its value at the
loop center gives a small relative flux error. When the variation is comparable with
the central field, use a surface integral or a measured grid. A compact coil can be
treated as one sampling area in a slowly varying field; a large coil cannot.

The sign and scale can be checked by subdividing. If a surface is split into two equal
pieces in a uniform normal field, each has half the total flux and the same sign. If
the pieces have opposite normals by construction, their signed values cancel. These
tests follow directly from additivity and are more reliable than visual estimates from
field-line density.

Flux density appears in different coordinate descriptions without changing its
meaning. On a horizontal plane, it is the $z$ component of $\vec B$. On a vertical
cylindrical wall, it is the radial component. On a tilted coil, it is the projection
onto the coil normal. The vector dot product selects the correct component in every
case and prevents a coordinate label from being mistaken for a physical direction.

Surface geometry and the local normal field determine a flux result. Coil comparisons
therefore require the size, orientation, and placement near a nonuniform source. Equal
areas can have different flux when their normals differ; equal normals can have different
flux when the source varies across their areas. A stated field value must identify the
location and normal component that it represents.

> **Worked example (field-area trade-off).** A $0.10\ \mathrm T$ field perpendicular
> to a $0.020\ \mathrm{m^2}$ loop gives
>
> $$
> \Phi_B=(0.10)(0.020)=2.0\times10^{-3}\ \mathrm{Wb}.
> $$
>
> Halving the field and doubling the area preserves that flux. Halving the radius
> instead reduces the area by a factor of four and the flux to one quarter, unless the
> field or turn count compensates. The result separates the local field from the
> geometrical collection area of the loop.

A coil whose turns share the same area and orientation has flux linkage equal to the
one-turn flux multiplied by turn count. Windings with different radii or orientations
require a turn-by-turn sum or an equivalent distributed integral. The distinction keeps
geometrical flux and coil linkage separate in later induction calculations.