---
title: Magnetic Force on Conductors
module: Magnetic Field
moduleNumber: 6
lessonNumber: 3
order: 603
summary: >
  A magnet pushes on a current-carrying wire even though the wire is electrically
  neutral. The reason is that each moving carrier feels the Lorentz force, and those
  microscopic pushes add up to a force the wire's supports must hold. We sum them
  into $\d\vec F=I\,\d\vec\ell\times\vec B$, collapse it to $\vec F=I\vec L\times\vec B$
  for a straight segment in a uniform field, and see exactly when that shortcut fails
  and the full path integral is needed. The same law runs backward as a measurement:
  a force-versus-current slope weighs a magnetic field against a known length.
topics: [Magnetic Field]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 26 — The Magnetic Field; §§26-1–26-3"
---

## From carrier force to the wire-force law

The magnetic force on a conductor is the sum of Lorentz forces on its mobile charge
carriers. Consider a straight element of conducting wire with vector length
$\d \vec\ell$ directed with conventional current. In a time $\d t$, the charge
crossing a section is $\d q=I\,\d t$. The carriers in that charge packet move through
the element by $\d \vec\ell=\vec v_d\,\d t$ in the conventional-current
direction. Adding the carrier forces gives

$$
\d \vec F=\d q\,\vec v_d\times\vec B
=I\,\d \vec\ell\times\vec B.
$$

This relation concerns the force transmitted to the conducting material. In a metal,
the mobile carriers are electrons, whose drift velocity is opposite
$\d \vec\ell$. Their negative charge reverses the carrier force, leaving the
same conventional-current result for the wire.

$$
% caption: Carrier forces add to a mechanical force on the wire. In a metal the electrons drift opposite the conventional current, and their negative charge flips the Lorentz force so the net push follows $I\,\d\vec\ell\times\vec B$.
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A straight segment in a uniform field has the integrated force

$$
\vec F=I\vec L\times\vec B,
\qquad
F=ILB\sin\theta,
$$

where $\vec L$ points from the segment's start to its end in the conventional
current direction and $\theta$ is the angle from $\vec L$ to $\vec B$.
The force is zero for a wire parallel or antiparallel to the field and is largest
for a perpendicular segment. Reversing either current or field reverses
$\vec F$; reversing both leaves it unchanged.

## Orientation and signed components

Choose coordinate axes before assigning a page direction. If current is along $+x$
and the magnetic field is along $+z$, then

$$
\vec F=IL\,\hat\imath\times B\hat k
=-ILB\,\hat\jmath.
$$

The conductor force begins toward negative $y$. The result follows from the ordered
cross product. Use a right-hand rule after the current and field arrows have been
fixed; component multiplication is less prone to reversal when axes are
unusual or the field is drawn into the page.

$$
% caption: Force on a straight current segment. With current along $+x$ and field out of the page, $\vec F=I\vec L\times\vec B$ points toward $-y$; reversing either the current or the field reverses the force.
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The current arrow is a reference direction. If a circuit calculation later gives a
negative current, the physical current is opposite the arrow and the magnetic force
must be reversed. The magnitude expression $ILB\sin\theta$ cannot supply this
direction by itself. State the vector result or identify the coordinate component
whose sign has been found.

Steady state carries the same total current across every complete section of a series
conductor. The external support or the rest of the circuit provides the mechanical
reaction required to hold the conductor in place. A freely supported segment
accelerates under the combined magnetic, gravitational, contact, and elastic forces.

> **Worked example (Force on a straight wire at an angle).** A straight wire of
> length $L=2.0\ \mathrm{m}$ carries $I=15\ \mathrm{A}$ in a uniform field
> $B=0.50\ \mathrm{T}$, with the wire making a $30^\circ$ angle to the field. The
> force magnitude is
>
> $$
> F=BIL\sin\theta=(0.50)(15)(2.0)\sin30^\circ=(0.50)(15)(2.0)(0.50)=7.5\ \mathrm{N}.
> $$
>
> Its direction is perpendicular to both the wire and the field, given by
> $\vec F=I\vec L\times\vec B$. Aligning the wire with the field ($\theta\to0$) kills
> the force; the maximum $BIL=15\ \mathrm{N}$ occurs at $\theta=90^\circ$, twice this
> value.

## Finite segments and nonuniform fields

The compact form $I\vec L\times\vec B$ requires a straight segment and one
uniform field vector over that segment. A bent wire or a field that changes with
position requires element-by-element addition:

$$
\vec F=I\int_{\rm wire}\d \vec\ell\times\vec B(\vec r).
$$

The element direction follows the actual wire path. In a uniform field, an arbitrary
open path from endpoint $a$ to endpoint $b$ reduces to

$$
\vec F=I(\vec r_b-\vec r_a)\times\vec B.
$$

That reduction fails when $\vec B$ varies along the path. A long lead crossing
the fringe of a magnet can therefore receive a different force from an equal-length
lead placed in the uniform central region. The field map and the wire route both
belong in a force calculation.

A closed loop wholly inside a uniform field has zero vector sum because the net path
displacement is zero. Individual sides can still be strongly loaded. Their
vector contributions cancel in translation. Rotational effects of those separated
forces are a separate topic.

## Force density and a local conductor model

In a bulk conductor, current density gives the local form

$$
\vec f=\vec J\times\vec B,
$$

where $\vec f$ is force per unit volume. The total force on a conducting volume
is $\vec F=\int\vec f\,\d V$. For a uniform wire with cross-sectional area $A$,
$\vec J=I\hat\ell/A$ and $\d V=A\,\d \ell$, recovering
$\d \vec F=I\,\d \vec\ell\times\vec B$.

Force density applies when current is nonuniform. A tapered conductor carrying the
same total current has larger current density in its narrow region. If the magnetic
field is also nonuniform, neither current density nor force density can be inferred
from one average cross section. Integrate the local vector field over the actual
conducting volume.

The local force first acts on the mobile carriers. Collisions and electromagnetic
interactions transfer that momentum to the ion lattice, supports, and attached
structure, so a force sensor mounted to the conductor records the force on the
macroscopic conductor. This transfer keeps the conventional-current form valid even
though electron drift in a metal is oppositely directed. The relation
$\vec f=\vec J\times\vec B$ is a continuum description: it applies when
current density and magnetic field can be defined over volumes large compared with
microscopic carrier spacing.

## Force measurement and analysis

A force measurement must distinguish conductor force from circuit artifacts. Mount a
straight active segment on a force sensor, keep the return lead outside the calibrated
field region when possible, and reverse current while holding the field fixed. The
magnetic contribution reverses with current; gravity and many sensor offsets do not.
Repeating the measurement with the field reversed provides a second sign check.

With a fixed sensor axis, compare readings at $+I$ and $-I$. Half their difference
isolates terms odd in current, including the intended magnetic force, whereas their
average retains many direction-independent offsets. The procedure assumes that
heating and mechanical drift remain small over the reversal interval. A field reversal
gives an independent odd-sign comparison. Neither subtraction removes a force on a
return lead that shares the active field region, so the complete conductor path must
be inspected before assigning a measured force to one segment.

In a perpendicular segment in a mapped uniform field, a plot of measured force
against current should have slope $LB$. Departures can arise from field gradients,
lead forces, heating that changes current, sensor drift, or an inaccurate active
length. Report the current direction, field orientation, active length, sensor axis,
and whether the return path entered the field. Those details determine the vector
quantity that the instrument has measured.

The force law is instantaneous for a specified current and field. A supply that
changes current during a sweep introduces a changing force; a conductor that moves
through a field gradient changes the effective field integral. In either case, record
current and position with the force trace rather than comparing points only by elapsed
time. The simple uniform-segment result is then a calibrated limiting case of the
distributed-force integral.

**Distributed current-density derivation.**

The line-element law follows from the local magnetic force on charge. In a small
conducting volume $\d V$, let $\rho_q$ be the mobile charge density and let $\vec u$
be the local mean carrier velocity. The charge in that volume is
$\d q=\rho_q\,\d V$. Its magnetic force is

$$
\d \vec F=(\rho_q\,\d V)\,\vec u\times\vec B.
$$

Current density is $\vec J=\rho_q\vec u$, including the sign of the mobile
carriers. Electrons in a metal have negative charge and drift opposite conventional
current; the product $\rho_q\vec u$ still points with conventional current.
The electromagnetic body force on the conductor is therefore

$$
\d \vec F=(\vec J\times\vec B)\,\d V,
\qquad
\vec F=\int_{V_c}\vec J(\vec r)\times\vec B(\vec r)\,\d V.
$$

The integration volume contains the conducting material chosen as the system. A
mechanical mount may receive the transmitted force, but it is not part of the
electromagnetic volume integral unless its own current is included. The field is also
local: replacing $\vec B(\vec r)$ by one nominal field value requires the
variation across the active conductor to be negligible at the stated precision.

Take a slender-wire slice of area $A$ perpendicular to the local tangent
$\hat t$. If current density is approximately uniform across that slice,
then $\vec J=(I/A)\hat t$ and $\d V=A\,\d s$. Substitution gives

$$
(\vec J\times\vec B)\d V
=\left(\frac{I}{A}\hat t\times\vec B\right)A\,\d s
=I\,\d \vec\ell\times\vec B,
\qquad
\d \vec\ell=\hat t\,\d s.
$$

The familiar wire expression therefore assumes a tangent current and negligible
cross-sectional variation. A broad strip near a magnet edge can violate the second
assumption. The total current may be measured accurately while force remains
sensitive to how current is distributed across the width. The volume integral is then
the appropriate calculation.

Steady current does not mean uniform $\vec J$. The condition
$\nabla\cdot\vec J=0$ permits streamlines to crowd through a narrow section or
divert around a hole. Where current direction changes, force density changes
direction. The support load is the vector sum of the local forces, not one average
current density multiplied by one average field and the conductor volume. At sharp
contacts or constrictions, a net-force calculation can remain accurate even when
local stress and heating require a finer model.

**Nonuniform-field integration.**

Parameterize the active wire by arc length $s$, with position $\vec r(s)$. Its
force is

$$
\vec F=I\int_{s_a}^{s_b}\frac{\d \vec r}{\d s}\times
\vec B[\vec r(s)]\,\d s.
$$

The limits describe the actual current path included in the model. A return lead in a
fringing field contributes even when it lies outside the pole faces. A section inside
the gap contributes little when it runs nearly parallel to the local field. The path
and the field map must therefore be specified together.

A straight wire along the $x$ axis carrying current toward increasing $x$ in
$\vec B=B_z(x)\hat k$ has force component

$$
F_y=-I\int_a^b B_z(x)\,\d x.
$$

With a linear field map $B_z(x)=B_0+\alpha x$,

$$
F_y=-I\left[B_0(b-a)+\frac{\alpha}{2}(b^2-a^2)\right].
$$

The gradient enters through the integral. The field at the midpoint is sufficient
only in special symmetric cases. With tabulated field data, divide the route into
short intervals and add $I\,\Delta\vec\ell\times\vec B$ as vectors.
Refining that partition should change the force by less than field-map and position
uncertainty before additional digits are reported.

$$
% caption: A straight wire in a field that grows along its length. Each segment feels a different force, so the support load is the vector sum $I\int\d\vec\ell\times\vec B$, not the product of current, length, and one chosen field value.
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Variation across a conductor cross section is a separate issue. If a copper strip
spans a strong gradient, a centreline integral is only an approximation. Evaluate
$\int_V\vec J\times\vec B\,\d V$ using the current distribution and field map
over width and thickness. A narrow high-field fringe at one edge cannot generally be
replaced by the cross-sectional average field without checking the current profile.

The integral is vector-valued. A curved path in a nonuniform field can have local
forces in several directions. Adding their magnitudes gives an upper bound, not the
net support load. Resolve components during integration, particularly near magnet
ends where field direction as well as magnitude changes.

**Shaped finite conductors.**

In a uniform field, an open wire of any shape has

$$
\vec F=I\left(\int_a^b \d \vec\ell\right)\times\vec B
=I(\vec r_b-\vec r_a)\times\vec B.
$$

Its net force depends only on the endpoints of the active path, although local
loading depends on shape. A bowed wire, zigzag trace, and straight lead with the
same endpoints have the same net force in a truly uniform field. They do not have
the same internal force distribution. Each short section is loaded according to its
own direction, and its mounts transmit that distribution to the apparatus.

Apply the endpoint reduction only to the part of a conductor actually within one
uniform region. Treating an entire circuit as if it occupied the magnet gap can hide
a lead force or introduce a cancellation absent from the apparatus. The magnet gap,
active segment, return path, and support locations belong in the same drawing before
a force-balance reading is assigned to one section.

A closed conductor wholly in a uniform field has vanishing endpoint displacement and
zero net translational force. This concerns the vector sum. Individual sides
can still load their mounts strongly. A balance attached to one side responds to the
load through that side and its support, not to the net force on the complete loop.
At a magnet edge, a finite-width bar has part of its current in strong field and part
in weak field. Summing local elements or using the volume integral avoids assigning
the full nominal field to the entire bar.

**Force-balance measurement and uncertainty.**

A force balance measures one component along its sensing axis. Align that axis with
the predicted force or project the calculated vector onto the axis before comparing
values. A small angular misalignment $\delta$ multiplies an otherwise transverse
reading by $\cos\delta$ and can mix gravitational sag or a lead load into the result.

For fixed field orientation, a sensor reading may be represented by

$$
S(I)=S_0+F_B(I)+F_{\rm heat}(I)+F_{\rm drift}(t).
$$

The magnetic term is odd in current. Therefore

$$
F_{\rm odd}=\frac{S(+I)-S(-I)}{2}.
$$

A stable weight and zero offset cancel in this difference. Resistive heating is
usually even in current, but thermal expansion can move a wire through a gradient or
change a contact force. Alternate polarity quickly enough to limit drift while
allowing the balance to settle. Taking all positive-current readings before all
negative-current readings can convert slow zero drift into an apparent magnetic
force.

Field reversal gives a stronger isolation test. With readings labeled by the signs
of applied current and field,

$$
F_{IB}=\frac{S_{++}-S_{-+}-S_{+-}+S_{--}}{4}.
$$

This combination retains the contribution proportional to $IB$ and rejects terms
independent of one reversal. It cannot remove an unaccounted force on a return lead
that enters the same field. Circuit geometry remains part of the measurement model
even with an accurate current measurement.

A uniform perpendicular active length has $F=IG$ with $G=LB$. Independent standard
uncertainties give

$$
u(F)=\sqrt{[G\,u(I)]^2+[I\,u(G)]^2+u_0^2},
\qquad
\frac{u(G)}{G}=\sqrt{\left(\frac{u(L)}{L}\right)^2+
\left(\frac{u(B)}{B}\right)^2}.
$$

Here $u_0$ is the balance uncertainty after zero and repeatability checks. In a
nonuniform field, replace $G=LB$ by the evaluated line or volume integral and
propagate uncertainty of the field map, conductor position, and active-path
definition. Meter resolution alone does not quantify these geometric contributions.

Plot reversal-isolated force against current at fixed field. A linear fit should
intercept zero within combined systematic and statistical uncertainty. Compare its
slope with the calculated field integral per unit current. A nonzero intercept,
curvature, or a slope altered by rerouting a lead identifies a limitation in the
apparatus or model.

Position uncertainty can dominate when the active conductor lies in a steep fringe
field. Record the reference point used to locate the wire, the travel repeatability,
and the spatial resolution of the field map. Repeat readings after a deliberate small
position shift. Agreement with the calculated change in force tests both the mapped
gradient and the definition of active length. A result that changes with clamping
force or lead tension instead indicates a mechanical coupling that must be separated
from the magnetic load.

## Force between parallel current paths

The magnetic field of one current path acts on the current in a second path. For two
long, straight, parallel conductors separated by distance $d$, conductor one produces
at conductor two the field magnitude

$$
B_1(d)=\frac{\mu_0 I_1}{2\pi d}.
$$

If conductor two carries current $I_2$ through a length $L$ that is parallel to the
first path, its magnetic force magnitude is

$$
F_{2\leftarrow1}=I_2LB_1
=\frac{\mu_0 I_1I_2L}{2\pi d}.
$$

The expression applies when the active length and the nearly uniform central part of
each path are large compared with $d$. In that limit, the field from conductor one
changes little along the measured length of conductor two. Near wire ends, bends, or
return connections, the field direction and magnitude require the distributed
integral rather than the long-wire result.

Use the right-hand rule or components to determine the direction before inserting
magnitudes. Parallel currents in the same direction attract: each path is forced
toward the other. Antiparallel currents repel. Reversing either current changes the
sign of the force; reversing both preserves it. The force on path one is equal in
magnitude and opposite in direction to the force on path two when the complete
electromagnetic interaction is included.

> **Worked example (Force between two parallel wires).** Two long parallel wires
> $d=5.0\ \mathrm{cm}$ apart each carry $I=20\ \mathrm{A}$ in the same direction.
> Find the force per unit length. With $\mu_0/2\pi=2.0\times10^{-7}\ \mathrm{T\,m/A}$,
>
> $$
> \frac{F}{L}=\frac{\mu_0 I_1I_2}{2\pi d}
> =(2.0\times10^{-7})\frac{(20)(20)}{0.050}
> =1.6\times10^{-3}\ \mathrm{N/m}.
> $$
>
> The wires attract, since the currents run the same way. A 1 m length feels
> $1.6\ \mathrm{mN}$, small but easily measured on a current balance, and this
> configuration is what historically fixed the definition of the ampere.

$$
% caption: Two parallel wires with current in the same direction. Each wire sits in the other's circular field and is pushed toward it, so the wires attract; reversing either current flips both force arrows.
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$$

The field-plus-force calculation uses the field of one path for the force on the
other. It must not be read as a field acting on a current element of the same ideal
filament. A wire's own field is singular at the filament model's location and does
not supply the net translational force used here. For conductors of finite radius,
internal magnetic stresses exist, but the measured force between two separate paths
is obtained from the field generated by the other path and the current in the selected
path.

The inverse-distance dependence is conditional. A pair of thin paths with separation
much greater than their radii follows $F/L\propto1/d$ over a central region. At
separations comparable with the conductor size, nonuniform current distribution,
insulation thickness, and finite-width geometry affect the force. At large
separation, the return portions of the circuit can no longer be ignored because their
fields may be comparable with the field from the nominally active path.

**Suspended-conductor balance.**

A suspended conductor converts a magnetic force into a change of support load. Let a
horizontal active segment of length $L$ lie perpendicular to a horizontal magnetic
field, so that its magnetic force is vertical. If upward is positive and the support
reading is $N$, static balance gives

$$
N+F_{B,y}-mg=0.
$$

With a downward magnetic force, the support load exceeds the weight by its magnitude;
with an upward force, it is reduced. Reversing current reverses $F_{B,y}$ while the
weight remains unchanged. A pair of readings at equal current magnitudes therefore
gives

$$
F_{B,y}=\frac{N(+I)-N(-I)}{2},
$$

after the sign convention is fixed. The formula removes a stable weight offset but
does not remove drift, wire heating, or a force on a return conductor that is
mechanically coupled to the same balance.

> **Worked example (Current to levitate a wire).** A horizontal wire of linear mass
> density $\lambda=5.0\ \mathrm{g/m}$ lies perpendicular to a horizontal field
> $B=0.40\ \mathrm{T}$. What current makes the magnetic force support the wire's own
> weight? Balancing force per unit length against weight per unit length,
>
> $$
> BI=\lambda g
> \quad\Longrightarrow\quad
> I=\frac{\lambda g}{B}=\frac{(5.0\times10^{-3})(9.81)}{0.40}=0.12\ \mathrm{A}.
> $$
>
> The current direction must make $I\vec L\times\vec B$ point up; reversing it doubles
> the downward load instead. Only $\lambda$ and $B$ enter, not the wire's length, since
> both weight and magnetic force scale with $L$.

$$
% caption: A suspended segment in a horizontal field. The vertical magnetic force adds to or subtracts from the weight at the support, so reversing the current isolates the magnetic part as an odd-in-current difference.
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A balance experiment must state which object is weighed. If the magnet is mounted on
the balance and the conductor is fixed externally, the force on the magnet is the
opposite of the force on the conductor. The numerical magnitude agrees, but the
recorded sign differs. If both magnet and conductor are mounted to the same rigid
frame, their mutual force is internal to that assembly and may not appear as a net
load change. The system boundary determines which reaction force reaches the sensor.

Static balance also requires negligible acceleration and elastic settling. A flexible
wire can change position as current changes, thereby sampling a different field.
Measure or constrain the position of the active segment. A force trace acquired while
the supply ramps is a dynamic record; compare readings only after current, position,
and sensor output have reached the stated steady values.

## Circuit closure and return-path effects

Current cannot exist only in the active segment. Every force measurement contains a
complete closed path, including supply leads and return conductors. The force assigned
to a named segment is valid only after the other path contributions have been shown
small, calculated, or excluded from the sensing assembly.

In a uniform field, a closed path has zero net translational magnetic force. This
does not make the force on one exposed side vanish. A rectangular circuit can have a
strong force on the segment inside the magnet gap and an equal opposite force on a
return segment elsewhere in the same uniform field. If both segments share the same
mount, their forces cancel in the mount load. If the return is routed outside the
field or supported separately, the sensor can register the force of the active
segment.

The same caution applies to parallel-path measurements. A two-wire transmission line
contains equal and opposite currents separated by a small distance. Far from the
pair, their fields partially cancel; near either wire, the other wire and its return
path both contribute. Replacing the full geometry with one isolated long wire can
give the wrong force sign or magnitude when conductor spacing is not large compared
with the active dimensions.

Define the active region geometrically. Record the endpoints at which the field map
is applied, the physical route of each lead, and the mechanical connection of each
lead to the balance. A lead that enters a fringe field can contribute little to the
circuit's net force while still transferring a measurable load to one mount. The
force calculation and the mechanical load path must agree.

**Quantitative model checks.**

For parallel paths, test the long-wire model by holding $I_2$, $L$, and $d$ fixed
while varying $I_1$. The predicted force is linear in $I_1$. Reversing one current
tests the sign, and reversing both checks that the measured force returns to its
original sign. At fixed currents, a separation sweep tests the approximate
$1/d$ dependence. Use only the range in which the length-to-separation ratio supports
the long-wire approximation.

A second check compares the force per active length with the field measured at the
second path. The prediction $F/L=I_2B_1$ isolates the field calculation from the
current-force calibration. Agreement in both this comparison and the
$\mu_0I_1I_2/(2\pi d)$ comparison supports the geometry and field model. A difference
between them can indicate finite-length effects, an inaccurate separation, or return
path fields rather than an error in the conductor-force law.

For the product $q=I_1I_2L/d$, independent standard uncertainties give

$$
\left(\frac{u(q)}{q}\right)^2=
\left(\frac{u(I_1)}{I_1}\right)^2+
\left(\frac{u(I_2)}{I_2}\right)^2+
\left(\frac{u(L)}{L}\right)^2+
\left(\frac{u(d)}{d}\right)^2.
$$

This estimate omits correlated errors such as a common current meter scale or a
separation measured from insulated surfaces rather than current centrelines. State
such choices explicitly. Residuals that reverse with current are evidence for a
missing magnetic force contribution; residuals that remain at zero current point
instead to sensor offset, weight drift, or mechanical preload.

Repeat selected measurements after exchanging the physical roles of the two paths.
The force magnitude should be unchanged when current magnitudes, active length, and
centreline separation are unchanged, while the load recorded by a given support can
change sign because its mechanical boundary has changed. This comparison detects an
asymmetric lead route or a support that carries part of the return-path load. It also
separates a geometric bias from a current-meter calibration error, since the latter
remains when the paths are exchanged.

Record ambient temperature when thermal drift correlates with current polarity.

## Force distribution in coils and multi-turn paths

A coil is a continuous current path with many differently oriented segments. Its
magnetic loading follows the same local law as a straight conductor,

$$
\d \vec F=I\,\d \vec\ell\times\vec B,
\qquad
\vec F_{\rm coil}=I\oint \d \vec\ell\times\vec B(\vec r).
$$

The closed-path symbol records that the current returns through every turn and every
lead. In a uniform magnetic field, the vector integral over a complete closed turn
vanishes. Individual sections can still carry substantial forces. Opposite sides may
be loaded in opposite directions, and the wire insulation, winding form, and clamps
must transmit those local loads even when the net translational force of the complete
turn is zero.

A straight active path concentrates most of the net measurable force into one
identified segment. A coil distributes force along its turns. If only one side of each turn lies
inside a strong field region while the return side is outside or in weak field, the
forces on the active sides can add to a nonzero translational load. If both sides
occupy equal and opposite field regions, cancellation can be nearly complete. The
field map and the full winding route determine which situation applies.

$$
% caption: A rectangular turn crossing a localized field region. The side inside the region (shaded) carries a large local force while the return sides sample weak field; the net coil load is the vector sum around the whole path.
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$$

For $N$ closely packed turns in the same field distribution, each turn follows the
same path integral to a good approximation. The total force is then

$$
\vec F_{\rm total}=N I\oint_{\rm one\ turn}
\d \vec\ell\times\vec B(\vec r).
$$

This proportionality requires equal current in all series turns and similar field
sampling. A thick winding pack can span a gradient, so outer turns may see a
different field from inner turns. Parallel branches also require care: the total
current divides among branches, and multiplying one branch force by the total circuit
current overestimates the load.

The wire does not need to be circular for the integral to apply. A printed spiral,
rectangular winding, or bent busbar can be treated as connected line elements if its
cross section is small relative to the field-variation scale. A broad coil conductor
or a densely packed winding near an iron pole edge may instead require
$\int_V\vec J\times\vec B\,\d V$. The volume form also identifies local force
concentrations that a centreline model hides.

Mechanical supports respond to the distribution as well as the net force. A coil can
show little net displacement while its former, adhesive, or terminal tabs carry
large internal loads. Calculate the translational support load from the vector sum,
then assess individual segments separately where wire clearance or structural stress
matters. The two calculations answer different design questions.

**Current-balance measurements.**

A current balance compares the electromagnetic force between fixed and movable
current paths with a known mechanical force. The movable path is mounted to a balance
or flexure; the fixed path is attached to a separate support. Let the instrument axis
be the direction in which the movable path is constrained. The reading is then the
component of the magnetic force along that axis, after weight and other constant
loads have been removed.

For two circuit paths driven independently, many geometries can be summarized over a
limited operating range by

$$
F_{\rm axis}=K I_1 I_2,
$$

where $K$ contains the path geometry, the field distribution generated by path one,
and the projection onto the sensor axis. The sign of $K I_1I_2$ is fixed by the
current directions and the selected positive axis. Reversing one current changes the
force sign. Reversing both currents restores the original sign.

The balance must not support both interacting paths as one rigid assembly. Their
equal and opposite magnetic forces are then internal to the assembly and do not
produce the intended differential load. A fixed path on the laboratory frame and a
movable path on the sensor provide a clear boundary. Leads can defeat this separation
if they exert force on the sensor frame or if a return path shares the mapped field.

A mechanical calibration can be made with a known mass increment or a calibrated
test force along the sensing axis. Record the zero before and after each current
sequence. A slowly changing zero requires repeated references during the run.
Interleave positive and negative current readings, and repeat the sequence after
reversing the field-producing path. The odd-in-one-current part of the data isolates
the intended interaction more reliably than a single reading.

If one source drives both field and active path in a fixed geometry, the force can
scale as $I^2$ rather than as two independently adjusted currents. The sign is then
not determined by changing the sign of that one source, because both current roles
reverse together. Separate current paths or a separately reversible field are needed
when a sign-reversal check is required.

**Uncertainty propagation and model checks.**

For $F=K I_1I_2$, independent standard uncertainties give

$$
u(F)=\sqrt{(I_1I_2u(K))^2+(K I_2u(I_1))^2+
(K I_1u(I_2))^2+u_0^2},
$$

where $u_0$ represents balance repeatability and residual zero uncertainty. The
relative contribution from $K$ often dominates when path separation, active length,
or coil position is uncertain. Treating $K$ as exact because the current meters have
many digits understates the experimental uncertainty.

A geometric model can express $K$ as an evaluated line integral per current product.
For example, a long parallel-path approximation gives

$$
K=\frac{\mu_0L}{2\pi d}.
$$

Its fractional uncertainty includes length and centreline separation,

$$
\left(\frac{u(K)}{K}\right)^2=
\left(\frac{u(L)}{L}\right)^2+
\left(\frac{u(d)}{d}\right)^2,
$$

before adding uncertainty from finite-length corrections or a nonuniform field map.
Measure $d$ between current centrelines rather than between outer insulation
surfaces. If the windings have appreciable radius, report the geometric convention
used to define the effective separation.

Plot the reversal-isolated force against $I_1I_2$ while holding geometry fixed. A
linear fit tests the proportionality and gives an experimental value of $K$. Its
intercept should agree with zero within the balance uncertainty. Curvature can result
from path motion, heating, magnetic-material response in nearby supports, or an
unaccounted current-dependent lead force. Residuals should be plotted against each
individual current as well as their product; that separates a one-source artifact
from a failure of the two-current model.

**Limit checks for distributed paths.**

Several limits test a coil-force calculation without requiring a separate apparatus.
At $I=0$, every magnetic force term must vanish. Reversing one current in a
two-path interaction reverses the computed force. In a uniform field, the total
translational force on a complete closed path must be zero. A calculation that
predicts a nonzero value in that limit has omitted a return segment or used an
inconsistent field direction.

The straight-path result is recovered when one coil side lies in the active region
and all remaining path segments lie in negligible field. The multi-turn expression
reduces to $N$ times the single-turn result when every turn samples the same field
and carries the same current. It must not be extrapolated to a winding whose thickness
is comparable with the gradient scale.

Dimensional checks are immediate. The coefficient $K$ has units of
$\mathrm{N\,A^{-2}}$, so $K I_1I_2$ has units of force. The parallel-wire form
contains $\mu_0L/d$, which has the same units. A force model that retains a current
product but loses the length-to-separation factor cannot describe a force magnitude.

Finally, compare the calculated direction with the balance convention. A positive
force in the conductor calculation can produce a negative change in the sensor
reading when the balance measures the reaction on the fixed support. Writing the
system boundary and sensing axis beside each data table prevents this sign reversal
from being mistaken for disagreement with the magnetic model.

Compare repeated runs after restoring the same documented path geometry and clamp
positions. Repeatability then tests the complete force model, not current alone.

## Complete current-force experiment

A reproducible current-force experiment begins with a mechanical boundary as well as
an electrical circuit. Mount the active conductor on a balance or flexure, hold the
magnet and return lead on a separate support, and define one sensing axis. Mark the
active endpoints at the boundaries of the mapped field region. The reported length is
the path length between those marks, not the distance between electrical terminals.

Measure the zero reading with current off, then measure at several current magnitudes
in both directions. Record current, sensor output, active-conductor position, and
ambient temperature for every point. Position matters when the conductor lies near a
field gradient; a force change caused by sag is not evidence for a different current
law. Use a rigid guide or an independent position measurement when the expected force
is comparable with the stiffness-limited displacement of the mount.

The predicted force is evaluated from the same geometry used in the experiment. For
a straight perpendicular segment in a uniform region, $F_{\rm axis}=\pm ILB$ after
projection onto the sensor axis. For a mapped region, use the measured path in
$I\int \d \vec\ell\times\vec B$. The sign in either form refers to the
force on the conductor. A balance mounted to the magnet reports the reaction and
therefore has the opposite sign convention.

**Reference subtraction and current reversal.**

A reference configuration estimates forces that remain when the active magnetic
interaction is removed. It may use zero current, a displaced conductor outside the
mapped field, or an otherwise identical path with the active segment absent. The
reference must preserve the same mechanical loading and lead tension. Moving a wire
to a reference position can change its sag or contact force, so the subtraction is
valid only when those changes are negligible or independently measured.

For fixed field direction, form the current-odd difference

$$
F_{\rm odd}(I)=\frac{S(+I)-S(-I)}{2}.
$$

Constant weight, a stable sensor offset, and many even-in-current heating terms
cancel. The average

$$
F_{\rm even}(I)=\frac{S(+I)+S(-I)}{2}-S(0)
$$

is a diagnostic. It should remain small compared with the intended force
after reference subtraction. A growing even term can reveal thermal expansion,
current-dependent cable tension, or a shift of the conductor in a field gradient.

Interleave rather than block the polarities. A sequence containing all positive
currents followed by all negative currents confounds polarity with elapsed time.
Repeat the reference point during the sequence. Its change estimates drift over the
same interval as the force readings. Field reversal provides an additional check:
the magnetic term reverses when either current or field reverses, whereas a gravity
offset does not.

**Calibration residuals and force scaling.**

Calibrate the balance with known loads along the sensing axis before and after a
current run. A linear calibration has $S=a+bF$, but retain the residuals rather than
only the fitted slope. Residual structure can indicate friction, flexure hysteresis,
sensor saturation, or a zero that depends on load history. The calibration range
should bracket the largest expected magnetic force; extrapolation beyond that range
adds an unmeasured model assumption.

After converting readings to force, fit the reversal-isolated values to the predicted
scaling law. At fixed field and geometry, a straight active path gives

$$
F_{\rm axis}=cI+F_0,
\qquad c=\pm LB
$$

for a uniform perpendicular field. The fitted intercept $F_0$ should agree with zero
within the combined calibration and repeatability uncertainty. The residual at each
current is $r_i=F_i-(cI_i+F_0)$. Plotting $r_i$ against current, position, and run
order distinguishes a nonlinear force response from drift or a geometry change.

Scaling conclusions require more than a straight line through a few points. Repeat
the current sweep at a second field magnitude or active length. The slope should
scale with the independently measured $B$ or $L$. Reversing the current should
reverse the force without changing its magnitude at the same absolute current.
Reducing current toward zero should make the force approach the reference-subtracted
zero without a discontinuity.

A valid agreement is limited by the uncertainty budget. For $F=ILB$, independent
fractional uncertainties combine as

$$
\left(\frac{u(F)}{F}\right)^2=
\left(\frac{u(I)}{I}\right)^2+
\left(\frac{u(L)}{L}\right)^2+
\left(\frac{u(B)}{B}\right)^2+
\left(\frac{u_{\rm bal}}{F}\right)^2.
$$

The balance term becomes large at low force, while length and field-position errors
can dominate at high force. State the current range, reference method, calibration
residuals, and active-path geometry with the final slope. Those records make a
force-scaling conclusion testable rather than dependent on one plotted line.

Use a symmetric acquisition schedule when the apparatus is susceptible to thermal
drift. One sequence can be $0,+I,-I,0,-I,+I,0$, with the magnitude held fixed over
the sequence. The surrounding zero readings estimate drift without assuming that it
is linear in time. Repeat the sequence at each current magnitude, then combine the
paired odd differences rather than averaging raw readings of unlike polarity. If the
reference changes by more than the stated repeatability, lengthen the settling time,
reduce self-heating, or include a time-dependent zero correction with its uncertainty.

Before reporting a scaling result, test three limits against the recorded data. The
reference-subtracted force should approach zero as current approaches zero. Equal
and opposite currents should give equal and opposite force readings within
uncertainty. Repeating the same current after repositioning the return lead outside
the mapped region should leave the active-path result unchanged. Failure of the last
test indicates an unmodelled circuit contribution, even if the current-force plot
appears linear.