---
title: Motional EMF
module: Electromagnetic Induction
moduleNumber: 8
lessonNumber: 4
order: 804
summary: >
  Push a wire through a magnetic field and its free charges feel a sideways magnetic
  force that piles them up at the ends — a battery made of motion. Motional emf is that
  effect: the work per unit charge a moving conductor supplies is the line integral of
  $\vec v\times\vec B$ along it, which for a rod moving perpendicular to both its length
  and the field collapses to $B\ell v$. We chase where the energy comes from — the hand
  or motor fighting the magnetic drag, never the magnetic force itself — solve the
  sliding-rail circuit from both flux and carrier forces, and carry the idea into
  rotating rods, homopolar disks, generators, and the back emf of a motor.
topics: [Electromagnetic Induction]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 28 — Magnetic Induction; §28-4 Motional EMF"
---

Motional emf arises when a conductor moves through a magnetic field and its mobile
charges experience the magnetic part of the Lorentz force,

$$
\vec F_B=q\,\vec v\times\vec B.
$$

The velocity in this expression is the velocity of a charge through the magnetic field.
For charges embedded in a rigid rod translating through a static field, the rod carries
the charges with it, so their initial velocity is the rod velocity. Positive and negative
carriers are pushed toward opposite ends of the rod. Charge separation builds an
electrostatic field inside the material, and the resulting voltage difference is the
open-circuit motional emf.

The magnetic force changes carrier direction but does no work on an individual charge
because it is perpendicular to the charge velocity. The electrical energy associated
with a closed motional-emf circuit comes from the external force that maintains the
conductor's motion against the magnetic reaction force. The energy account must include
the moving conductor, the circuit current, and the external drive together.

## Translating Rods and Charge Separation

Let a straight rod of length $\ell$ move rightward through a uniform magnetic field
into the page. Its length points upward. A positive carrier moving with the rod has
$\vec v$ to the right and $\vec B$ into the page, so
$\vec v\times\vec B$ points upward. Positive charge accumulates at the upper
end, negative charge accumulates at the lower end, and the electric potential is higher
at the upper end.

$$
% caption: A rod of length $\ell$ moves right through $\vec B$ into the page. The magnetic force $q\vec v\times\vec B$ on positive carriers points up the rod, so positive charge gathers at the top and negative at the bottom, setting up the open-circuit rod voltage.
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$$

The separated charges produce an electrostatic field from the positive end toward the
negative end. Equilibrium inside an open rod occurs when its electric force cancels the
magnetic force on a carrier:

$$
q\vec E_{\rm sep}+q\,\vec v\times\vec B=0,
\qquad
\vec E_{\rm sep}=-\vec v\times\vec B.
$$

With the geometry above, the separation field points downward and has magnitude $vB$.
The endpoint potential difference has magnitude

$$
|\Delta V|=E_{\rm sep}\ell=vB\ell.
$$

The equality applies to an open rod in the uniform perpendicular geometry. No steady
current crosses the gap, but a high-resistance voltmeter connected between the ends can
measure the voltage. Connecting a low-resistance circuit allows the separated charges
to circulate; a continuing external force must then move the rod.

The polarity depends on the carrier charge used in the Lorentz-force calculation. The
internal electrostatic field and the terminal voltage are properties of the conductor,
so the upper end remains at higher potential in the geometry shown even when electrons
are the mobile carriers. Electrons receive a magnetic force downward, leaving an excess
of negative charge at the lower end and an electron deficit at the upper end. Carrier
motion and conventional current point in opposite directions in a metal.

## Sliding Rails and Energy

Two conducting rails and a sliding rod form a closed loop whose area changes as the rod
moves. Take rail separation $\ell$, rod position $x$, and a uniform magnetic field into
the page. With an into-page normal, the flux through the loop is

$$
\Phi_B=B\ell x.
$$

With rod speed $v=\d x/\d t$,

$$
\mathcal E=-\frac{\d\Phi_B}{\d t}=-B\ell v.
$$

The sign refers to the positive traversal associated with the chosen into-page normal.
Its magnitude is $B\ell v$. Lenz's law gives the same direction: rightward rod motion
increases into-page flux, so the induced current produces an out-of-page magnetic
contribution and runs counterclockwise as viewed from the page.

$$
% caption: A rod slides right on rails, enlarging the loop area inside $\vec B$ into the page. Positive carriers are driven up the rod, so the induced conventional current runs counterclockwise around the loop; the resistor $R$ limits it, and the motion supplies the emf.
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$$

The microscopic and flux descriptions agree. In the rod, positive charges are pushed
upward by $q\vec v\times\vec B$. The resulting conventional current goes upward
through the rod, leftward across the upper rail, downward through the resistor, and
rightward along the lower rail. The loop direction is counterclockwise. In the flux
description, the growing into-page area determines the same current through Lenz's law.
The two descriptions emphasize different parts of one electromagnetic process.

## General Motional EMF

An arbitrarily shaped conductor moving through a static magnetic field has motional
emf contribution

$$
\mathcal E_{\rm mot}=\oint_C(\vec v\times\vec B)\cdot \d\vec\ell.
$$

The velocity can vary from one part of the conductor to another. A rigid rod in pure
translation has one velocity everywhere; a rotating wire has velocity magnitude and
direction that vary with radius. Only the component of $\vec v\times\vec B$
along the wire element contributes. Motion parallel to $\vec B$ gives no magnetic
force, and motion perpendicular to a wire can produce no endpoint voltage if the force
has no component along the wire.

$$
% caption: When motional emf appears. Left: the rod moves perpendicular to both its length and $\vec B$ into the page, so $\vec v\times\vec B$ lies along the rod and separates charge. Right: the rod moves parallel to $\vec B$ (both up the page), so $\vec v\times\vec B$ vanishes and there is no motional emf.
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$$

For the perpendicular straight rod, $\vec v\times\vec B$ is constant and
parallel to $\d\vec\ell$, so the integral reduces to $B\ell v$. The formula
also prevents an overgeneralization: a moving closed loop wholly inside a uniform static
field can have zero net motional emf when every element shares the same translation.
The local magnetic forces may separate charge in different pieces of the loop, yet the
closed line integral can cancel.

### Current, force, and electrical power on sliding rails

Let the total resistance of the sliding-rail circuit be $R$, including the resistor,
rails, and rod. The induced current magnitude is

$$
I=\frac{B\ell v}{R}.
$$

For the rightward-moving rod in an into-page field, conventional current travels upward
through the rod. The force on that current-carrying rod is

$$
\vec F_{\rm mag}=I\,\vec\ell\times\vec B.
$$

Here $\vec\ell$ points upward and $\vec B$ points into the page, so the
force points left. Its magnitude is

$$
F_{\rm mag}=IB\ell=\frac{B^2\ell^2}{R}v.
$$

The magnetic force resists the rod motion. A person or motor pulling the rod at a
constant rightward speed must exert an equal rightward external force. The word
“resists” is an energy statement: the external force does mechanical work that the
circuit transfers to electrical and thermal energy.

$$
% caption: Force balance in the closed rail circuit. Upward current $I$ in the rod and $\vec B$ into the page give a leftward magnetic drag; an equal rightward pull holds a steady speed, and the resistor $R$ receives the mechanical power.
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$$

Multiplying magnetic-force magnitude by speed gives the mechanical input power

$$
P_{\rm mech}=F_{\rm ext}v
=\frac{B^2\ell^2v^2}{R}.
$$

The electrical power dissipated in the resistance is

$$
P_R=I^2R
=\left(\frac{B\ell v}{R}\right)^2R
=\frac{B^2\ell^2v^2}{R}.
$$

Thus $P_{\rm mech}=P_R$ for steady motion in the simple resistive model. The equality
holds without assigning work to the magnetic force on an individual carrier. The drive
force transfers energy to the moving charges through the rod's lattice and maintains the
charge separation that produces the circuit emf. The resistor converts that electrical
energy into internal energy.

> **Worked example (power in a rail generator).** Take $B=0.600\ \mathrm T$,
> $\ell=0.150\ \mathrm m$, $v=8.00\ \mathrm{m\,s^{-1}}$, and $R=25.0\ \Omega$. The emf
> and current magnitudes are
>
> $$
> |\mathcal E|=B\ell v=0.720\ \mathrm V,
> \qquad
> I=\frac{0.720}{25.0}=2.88\times10^{-2}\ \mathrm A.
> $$
>
> The external-force magnitude is
>
> $$
> F_{\rm ext}=IB\ell
> =(2.88\times10^{-2})(0.600)(0.150)
> =2.59\times10^{-3}\ \mathrm N,
> $$
>
> and both power routes give $F_{\rm ext}v=I^2R=2.07\times10^{-2}\ \mathrm W$. The small
> force still supplies measurable power because the rod travels several metres per second.
> Units check: tesla times metre times metre per second is volt, and ampere times tesla
> times metre is newton.

### Magnetic drag after the drive is removed

Suppose the rod begins with rightward speed $v_0$ and the external pull is removed.
The induced current persists while the rod moves, and the magnetic force is proportional
to speed:

$$
F_x=-\frac{B^2\ell^2}{R}v.
$$

Newton's second law for a rod of mass $m$ becomes

$$
m\frac{\d v}{\d t}=-\frac{B^2\ell^2}{R}v.
$$

Separation and integration give

$$
v(t)=v_0e^{-t/\tau},
\qquad
\tau=\frac{mR}{B^2\ell^2}.
$$

The speed decays exponentially rather than linearly because the current and force both
decrease as the rod slows. At late times the motion becomes very slow, the induced emf
becomes very small, and the magnetic drag weakens by the same factor.

$$
% caption: Speed of a freely sliding rod after the drive is removed. The induced current and magnetic drag are both proportional to $v$, so the speed decays exponentially with time constant $\tau=mR/B^2\ell^2$; larger $R$ or mass lengthens it, larger $B$ or rail spacing shortens it.
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The rod displacement approaches the finite limit

$$
x(t)-x(0)=v_0\tau\left(1-e^{-t/\tau}\right),
\qquad
\Delta x_{\infty}=v_0\tau.
$$

Its initial kinetic energy becomes heat in the resistance. Integrating the resistor
power confirms the energy balance,

$$
\int_0^\infty I^2R\,\d t
=\frac12mv_0^2.
$$

The equality uses the speed solution and the resistance-dominated model. Friction,
air drag, or rail contact losses add further terms to the mechanical energy balance;
they do not reverse the motional-emf current direction.

### Terminal voltage, internal resistance, and measurement

The motional emf of a rod is a source term around a complete circuit. It is distinct
from the terminal potential difference measured across the rod ends when current flows.
The moving rod has a finite resistance $r$. A circuit current creates an ohmic potential
drop along the rod, so the terminal voltage magnitude is smaller than the open-circuit
motional emf under a load:

$$
V_{\rm top}-V_{\rm bottom}=B\ell v-Ir
$$

for the geometry in which the upper end has higher open-circuit potential. The first
term comes from charge separation maintained by motion. The second term comes from the
electric field required to carry current through the resistive rod. With an open circuit,
$I=0$ and the voltmeter reads $B\ell v$. With a short external path, the terminal
voltage can become small even though the motional emf remains $B\ell v$ around the
source segment.

$$
% caption: Equivalent circuit of a moving rod. The ideal motional source has emf $B\ell v$ in series with the rod's internal resistance $r$; a load $R$ closes the loop. An open meter reads the source emf, while a load draws current $I$ and drops the terminal voltage by $Ir$.
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The distinction matters most in apparatus with long leads. A potential difference is
defined between terminals along a specified measuring path. In a static electrostatic
circuit, the path can often be ignored because the electric field is conservative. In a
motional-emf apparatus, moving sections and magnetic forces add a non-electrostatic
contribution to the loop emf. A voltmeter must be modeled as part of the physical path;
its leads can themselves move through a magnetic field or enclose changing flux.

Consider a rod moving through a uniform field with no closed rail path. A meter attached
to the rod ends by leads that move with the rod measures the charge-separation voltage
between the ends. A meter connected by fixed leads routed around the magnetic region can
sample a different emf distribution if the combined path encloses time-varying flux.
Careful diagrams state the rod velocity, the field region, and the lead route before
identifying a terminal voltage with an emf.

The sign of a measured terminal voltage follows the electrostatic field from separated
charges. In the standard rightward-rod case, the top terminal is positive. Reversing
the rod velocity reverses the magnetic force on positive carriers and swaps the
polarity. Reversing the external field also swaps polarity. Reversing both leaves the
same terminal polarity, because $\vec v\times\vec B$ retains its direction.

The middle panel represents one reversal and should have opposite terminal signs. In a
written calculation, draw $\vec v$, draw $\vec B$, compute the cross-product
direction, then mark the positive-carrier force. That four-step procedure avoids using
the same clockwise arrow from a different viewing side or from a reversed field.

### Uniform translation and the zero-emf case

A complete rigid loop can translate through a perfectly uniform static magnetic field
without a net motional emf. Every segment moves with the same $\vec v$. The integral

$$
\oint_C(\vec v\times\vec B)\cdot \d\vec\ell
=(\vec v\times\vec B)\cdot\oint_C\d\vec\ell=0
$$

because the directed line elements around a closed loop sum to zero. The loop may have
nonzero static flux, and individual segments can acquire charge separation, but the
algebraic contributions around the complete path cancel. A static uniform magnetic
field cannot deliver a sustained emf to a rigid loop solely because the loop is moving
sideways through it.

$$
% caption: A rigid loop translating entirely within a uniform $\vec B$ region. Every side shares the same velocity, so $\vec v\times\vec B$ points the same way (up) on both vertical sides; their emf contributions cancel around the closed loop and the net current is zero despite the nonzero flux.
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The zero result changes when any part of the geometry changes the flux integral. A loop
crossing into or out of a finite field region has an exposed area that changes. A loop
moving through a nonuniform field samples different field values at successive
positions. A deforming loop changes its area, and a rotating loop changes its
orientation. Each case can have a nonzero motional emf even with a time-independent
magnetic source.

A small rectangular loop of width $w$ moving along $x$ through a slowly varying
perpendicular field $B(x)$ has flux approximately $B(x)A$. Its rate is

$$
\frac{\d\Phi_B}{\d t}=A\,v\frac{\d B}{\d x}.
$$

The induced current direction follows the sign of $v\,\d B/\d x$ together with the chosen
normal. A loop moving rightward into a stronger into-page field has increasing into-page
flux and therefore produces an out-of-page induced contribution. The same loop moving
rightward into a weaker region produces the opposite current.

### Arbitrary rod orientation

The compact result $B\ell v$ applies only when the rod length, rod velocity, and
magnetic field are mutually perpendicular. A general straight rod has an emf obtained
by projecting the magnetic-force direction onto the rod:

$$
\mathcal E_{ab}
=\int_a^b(\vec v\times\vec B)\cdot \d\vec\ell.
$$

Let $\hat\ell$ point from endpoint $a$ to endpoint $b$ for a rigid rod
with uniform $\vec v$ and uniform $\vec B$. Then

$$
\mathcal E_{ab}
=\ell(\vec v\times\vec B)\cdot\hat\ell.
$$

The scalar triple product contains all orientation signs. It vanishes when the
magnetic-force direction is perpendicular to the rod, even if both $v$ and $B$ are
large. It changes sign when the endpoint order is reversed. It also changes sign when
either the velocity or magnetic field reverses.

$$
% caption: A tilted rod samples only the component of $\vec v\times\vec B$ along its own length. The accent arrow is the carrier force; its projection onto the rod (dashed) sets the endpoint emf, while the perpendicular part presses carriers into the rod wall without adding endpoint voltage.
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$$

> **Worked example (tilted-rod emf).** A rod points from $a$ to $b$ along
> $\hat{u}=(\hat x+\hat y)/\sqrt2$, has length $0.40\ \mathrm m$, moves with
> $\vec v=5.0\hat x\ \mathrm{m\,s^{-1}}$, and sits in $\vec B=0.30\hat z\ \mathrm T$. The
> magnetic-force direction on a positive carrier is
>
> $$
> \vec v\times\vec B
> =-1.50\hat y\ \mathrm{V\,m^{-1}},
> $$
>
> so the endpoint emf is
>
> $$
> \mathcal E_{ab}
> =\ell(\vec v\times\vec B)\cdot\hat u=(0.40)(-1.50\hat y)\cdot
> \frac{\hat x+\hat y}{\sqrt2}
> =-0.424\ \mathrm V.
> $$
>
> The negative value says endpoint $b$ is lower in potential than $a$ for the stated
> order. The magnitude is below $B\ell v=0.600\ \mathrm V$ because the rod is tilted
> relative to the carrier-force direction. Labelling the endpoints on a diagram keeps the
> sign from being lost when the rod is redrawn.

## Rotating Conductors

Rigid rotation makes the conductor velocity position dependent.

> **Worked example (emf of a rotating rod).** A rod of length $L$ rotates in its plane
> about one end with angular speed $\omega$, in a uniform field perpendicular to the
> plane. An element at radius $r$ has speed $v=\omega r$, so its motional contribution
> along the radial element is $\d\mathcal E=B\omega r\,\d r$. Integrating from pivot to
> free end,
>
> $$
> |\mathcal E|
> =\int_0^L B\omega r\,\d r
> =\frac12B\omega L^2.
> $$
>
> The factor one-half records the speed variation: the midpoint moves at half the tip
> speed, so the mean speed along the rod is $\omega L/2$. Substituting the tip speed
> $v_{\rm tip}=\omega L$ gives the equivalent form $|\mathcal E|=B L v_{\rm tip}/2$.

$$
% caption: A rod rotates about its lower end in a perpendicular $\vec B$. An element at radius $r$ moves at speed $\omega r$, so its motional contribution grows linearly with $r$; integrating from pivot to tip gives $\tfrac12 B\omega L^2$.
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The polarity follows $\vec v\times\vec B$ locally. With the field into the page
and counterclockwise rotation, the carrier force points radially inward; the pivot end
becomes positive for positive carriers. Reversing the rotation or field reverses the
polarity. A brush contact at the pivot and another at the rim can carry this voltage to
an external circuit while the rod rotates.

A conducting disk rotating about its center follows the same radial integration. Between
the center and a rim of radius $R$,

$$
|\mathcal E_{\rm disk}|=\frac12B\omega R^2.
$$

The homopolar-generator voltage is steady for steady rotation in a steady field.
Unlike a rotating coil with alternating flux linkage, the radial conductor samples a
fixed motional-force direction at every radius. Sliding electrical contacts are required
to connect the rotating disk to a stationary load.

### Moving-circuit form of Faraday's law

A circuit whose boundary moves through electromagnetic fields has general emf

$$
\mathcal E
=\oint_C\left(\vec E+\vec v\times\vec B\right)
\cdot \d\vec\ell.
$$

The electric-field term describes transformer emf from a time-varying magnetic field.
The velocity term describes magnetic forces on charges carried by the moving wire. A
sliding rod in a static uniform field has negligible induced circulating electric field
in the laboratory frame and obtains its emf from the velocity term. A fixed loop in a
changing field has $\vec v=0$ and obtains its emf from the electric-field term.
Many laboratory systems have both terms.

A material loop moving in a magnetic field has a moving-boundary form consistent
with the flux rule,

$$
\oint_C\left(\vec E+\vec v\times\vec B\right)
\cdot \d\vec\ell
=-\frac{\d}{\d t}\int_{S(t)}\vec B\cdot \d\vec A.
$$

The surface $S(t)$ shares the instantaneous wire boundary. The right side changes when
the field itself changes, when the surface moves into a different field region, or when
the surface deforms or rotates. The line-integral expression identifies the physical
force per unit charge around the actual conductor. Both forms give the same emf when
their orientations are chosen consistently.

## Generators and Motors

A rotating coil converts mechanical rotation into a time-varying motional emf. Take a
coil of $N$ turns and area $A$ rotating at angular speed $\omega$ in a uniform magnetic
field. The coil normal makes angle $\theta=\omega t$ with the field when it starts
aligned. The flux linkage and induced emf are

$$
\Lambda_B=NBA\cos(\omega t),
\qquad
\mathcal E=NBA\omega\sin(\omega t).
$$

The flux calculation is compact, while the motional-force picture locates the source in
the two long sides of the coil. Those sides move through the magnetic field in opposite
directions. Charges in both sides are driven around the same circuit sense, so their
individual motional voltages add around the loop. The short sides contribute less or
zero when their motion is parallel to the wire or when the carrier force is perpendicular
to the wire.

$$
% caption: A rectangular coil rotates in a uniform $\vec B$. Its two long sides cut across the field with opposite velocities, and their carrier forces drive conventional current the same way around the loop, so the side emfs add at the brush contacts.
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At a fixed rotation rate, the emf changes sign every half turn. Slip rings connect each
coil end to a stationary brush without reversing the external connections, producing an
alternating voltage. A split-ring commutator reverses the coil connection every half
turn and can make the external current retain one direction. Those connections alter
the terminal waveform; the carrier-force direction in each coil side follows
$\vec v\times\vec B$ at every instant.

The generated current creates a magnetic dipole moment in the coil. The external field
exerts a torque that opposes the imposed rotation while power is delivered to a load.
At steady angular speed, the drive torque must replace the electrical power sent through
the brushes and mechanical losses in bearings and air. The ideal power relation is

$$
\tau_{\rm drive}\omega=\mathcal E I.
$$

The relation is the rotational counterpart of $F_{\rm ext}v=\mathcal EI$ for the
sliding rod. A generator with an open circuit has nearly zero output current and a
small electromagnetic reaction torque. Connecting a lower-resistance load increases
current, increases reaction torque, and requires more mechanical input to preserve the
same rotation rate.

The instantaneous current also depends on circuit impedance, not solely on generated
emf. In a resistive load, current shares the emf sign. An inductive or capacitive load
can shift current in time relative to the emf. The mechanical reaction then depends on
the instantaneous product $\mathcal EI$ and its cycle average. AC circuit phase
relations require separate circuit analysis; the motional source term remains
$NBA\omega\sin(\omega t)$.

### Motor back emf

The same geometry operates in reverse as a motor. An external circuit drives current
through a coil in a magnetic field, and the magnetic torque turns the coil. As the coil gains
angular speed, its moving conductors generate a motional emf. The generated emf has a
polarity that opposes the applied source current, so it is called a back emf. Its
magnitude grows with angular speed:

$$
|\mathcal E_{\rm back}|=k_e\omega
$$

for a motor with fixed geometry, where $k_e$ includes turn count, area, and field
strength. A simple resistive winding equation is

$$
V_{\rm supply}=Ir+k_e\omega.
$$

At startup, $\omega=0$ and the back emf vanishes. The initial current is limited mainly
by winding resistance, so it can be much larger than the running current. As rotation
increases, the back emf reduces the current. A heavy mechanical load slows the motor,
reduces back emf, and allows more current, which raises electromagnetic torque. That
feedback follows from the motional source generated by the rotating conductors.

$$
% caption: Back-emf model of a dc motor. The supply $V$ drives current $I$ through the winding resistance $r$; rotation generates a motional emf of opposite polarity. Slower rotation lowers the back emf and raises the current, faster rotation raises it and lowers the current.
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An unloaded motor can reach a high speed because only friction, air resistance, and
iron losses require torque. Its current becomes small after the back emf nearly matches
the supply voltage. A stalled motor has zero back emf and can draw a destructive current
if the supply and winding resistance permit it. These operating limits come from the
same moving-conductor emf that produces output voltage in generator operation.

Motor torque and back emf can be linked through power. The electrical input to an ideal
motor is $VI$. The winding loss is $I^2r$. The remaining electromagnetic power is
$\mathcal E_{\rm back}I$, which becomes mechanical output $\tau\omega$ in the ideal
model. The result uses the same sign convention as a generator: a back emf opposes the
source-driven current, while its product with current represents energy transferred to
the rotor.

## Frames and Direction Methods

Motional emf depends on charge motion through a magnetic field. In the laboratory frame,
a moving rod has carrier velocity $\vec v$ and the magnetic term
$q\vec v\times\vec B$ separates charge. In a frame moving with the rod, the rod
is stationary, and electromagnetic fields transform so that an electric field accounts
for the same charge separation. The terminal voltage and the current in a closed circuit
agree between frames when the full apparatus is modeled.

The distinction matters for a moving magnet and a stationary loop versus a stationary
magnet and a moving loop. The external-flux history through the circuit can be the same,
while the local force description on carriers differs between frames. Faraday's law for
the complete circuit and energy conservation give frame-consistent measurable results.
A direction argument should begin with a specified circuit boundary, its motion, and the
magnetic-field region rather than assigning the emf to one object in isolation.

For a moving rod on fixed rails, the rod carriers move through the static field;
charges in the rails mostly do not.
An apparatus with a fixed loop and changing electromagnet has a natural fixed-loop
analysis: the induced electric field circulates through the laboratory space. The two
setups can produce equal numerical emfs while distributing electric and magnetic force
terms differently around their conductors.

### Practical direction record for moving conductors

Record these quantities to keep the carrier-force and flux arguments consistent.

1. **State the velocity of each moving segment.** A rotating conductor has a different velocity at each radius.
2. **Draw the external magnetic field at that segment.** Include its local direction and any field boundary.
3. **Find $\vec v\times\vec B$ for a positive test carrier.** Its component along the conductor determines endpoint charge separation.
4. **Trace the complete circuit.** The direction through each rail, resistor, brush, and moving segment fixes conventional current.
5. **Check force or torque.** A loaded generator or sliding rod requires external mechanical input; a motor produces mechanical output while its back emf opposes the source current.

An open conductor requires only the endpoint polarity. A closed circuit additionally
requires resistance and other circuit elements to obtain current. The emf sign is
set before the resistor law is used. Separating those stages avoids reversing the
current because a circuit diagram chooses a different positive current arrow.

### Several moving segments in one circuit

A circuit can contain more than one moving conductor. The motional emf is the signed
sum of the contributions from all of them:

$$
\mathcal E_{\rm mot}
=\sum_j\int_{C_j}(\vec v_j\times\vec B_j)
\cdot \d\vec\ell.
$$

Each segment needs its own velocity, field, and chosen line direction. Equal-looking
rods can add or cancel depending on their motion and on how they are connected. Two
parallel rods sliding together on the same rails at equal speed preserve the area between
them and can give zero net loop emf. If one rod moves while the other stays fixed, the
area changes and the rod contributions leave a nonzero emf. If the rods move in opposite
directions, the area can change twice as fast and the emf magnitude can double.

Suppose the rail separation is $\ell$ and two boundary rods have positions $x_1$ and
$x_2$, with the circuit area $A=\ell(x_2-x_1)$. In a uniform perpendicular field,

$$
\mathcal E=-B\ell\left(\frac{\d x_2}{\d t}-\frac{\d x_1}{\d t}\right).
$$

The expression is a coordinate form of the line-integral sum. When the rods have equal
speeds in the same direction, their difference is zero. When the right rod moves right
and the left rod moves left with equal speed $v$, the difference is $2v$, giving an emf
magnitude $2B\ell v$. The circuit direction follows the sign of the changing area and
the selected normal.

Connections can reverse a contribution without reversing a physical motion. A coil wound
in the opposite sense has the opposite traversal convention for its flux linkage. A
series connection of two coils can be aiding or opposing. A circuit diagram should mark
terminal polarity or winding sense before numerical emf values are added. Magnitudes
alone do not preserve the orientation information required for a voltage sum.

## Worked Applications

> **Worked example (driven rail rod and its drag decay).** Rails separated by
> $0.250\ \mathrm m$ sit in a $0.800\ \mathrm T$ field into the page. A rod of mass
> $0.120\ \mathrm{kg}$ moves rightward at $3.00\ \mathrm{m\,s^{-1}}$ through a total
> circuit resistance $5.00\ \Omega$. The motional emf magnitude is
>
> $$
> |\mathcal E|=B\ell v
> =(0.800)(0.250)(3.00)
> =0.600\ \mathrm V.
> $$
>
> With an into-page normal, the growing loop area increases positive flux; Faraday's law
> gives a negative emf against the clockwise traversal, so the conventional current is
> counterclockwise as viewed. In the rod the current runs upward and the magnetic force
> is leftward. The current, external force, and power are
>
> $$
> I=\frac{0.600}{5.00}=0.120\ \mathrm A,
> \qquad
> F_{\rm ext}=IB\ell
> =(0.120)(0.800)(0.250)
> =2.40\times10^{-2}\ \mathrm N,
> $$
>
> $$
> F_{\rm ext}v
> =(2.40\times10^{-2})(3.00)
> =7.20\times10^{-2}\ \mathrm W
> =I^2R=(0.120)^2(5.00)\ \mathrm W.
> $$
>
> The direction check follows the flux rate or carrier cross product; the power check
> compares external force with resistor dissipation. Agreement catches a reversed current
> arrow and a missing factor of rail separation. If the drive is removed at
> $v_0=3.00\ \mathrm{m\,s^{-1}}$, the magnetic-drag time constant is
>
> $$
> \tau=\frac{mR}{B^2\ell^2}
> =\frac{(0.120)(5.00)}{(0.800)^2(0.250)^2}
> =15.0\ \mathrm s,
> $$
>
> so the speed after $10.0\ \mathrm s$ is
> $v=3.00\,e^{-10.0/15.0}=1.54\ \mathrm{m\,s^{-1}}$. The result assumes reliable rail
> contact and nearly constant resistance as the resistor warms, and neglects friction;
> adding friction shortens the stopping motion and changes the energy partition.

## Limits and Measurement

The simple formulas assume a rigid conductor, nonrelativistic speeds, a specified
magnetic field, and a circuit whose charge distribution reaches a quasisteady state.
Charge separation in a metal occurs on a very short electromagnetic relaxation time
compared with ordinary mechanical motion, so the internal electric field usually tracks
the rod velocity closely. Rapidly varying fields, high frequencies, or transmission-line
dimensions require a distributed electromagnetic model rather than a single circuit
emf and resistance.

The magnetic field in a rail experiment is seldom uniform at the edges of a magnet. A
rod far from the central region can sample smaller $B$, so its motional emf is the line
integral of the local field rather than a nominal central value times $\ell v$. A finite
rod can also have a field gradient along its length. Integrating
$(\vec v\times\vec B)\cdot \d\vec\ell$ handles both variations.

Brush and rail contacts introduce contact resistance. Their voltage drops add to the
resistance term used to predict current and heat. Poor contacts can produce intermittent
current even while the rod emf remains smooth. A voltage probe across the rod and a
current probe in the closed loop therefore answer different experimental questions.
The first tracks endpoint potential; the second tracks charge flow around the complete
conducting path.

The magnetic field applies no work directly to a point charge because the instantaneous
magnetic force is transverse to its velocity. Mechanical work enters through the agency
that forces the conductor through the field. In a rail generator, the external pull
exerts work on the rod. In a rotating generator, a turbine or motor exerts torque. In a
motor, the source delivers electrical energy and the magnetic torque transfers energy to
the rotor. Energy accounting must include the conductor and the agency that drives or
powers it; a magnetic force on an isolated carrier does not supply this work.

### Direction and magnitude distinctions

Motional-emf problems often contain three separate quantities.

- **Endpoint polarity** comes from the carrier force $q\vec v\times\vec B$ in an open conductor.
- **Loop emf** comes from integrating electric and motional force per unit charge around the specified circuit path.
- **Current and power** require resistance, inductance, capacitance, and the state of switches or loads after the emf is known.

The rod can have a large endpoint voltage with a high-resistance open circuit and
negligible current. A closed low-resistance circuit can have substantial current and a
small terminal voltage across the rod because its internal resistance creates a drop.
Both cases can share the same motional source $B\ell v$. Recording which quantity is
measured prevents an emf, a terminal voltage, and an ohmic drop from being substituted
for one another.

Every sign statement should include a chosen path or named endpoint pair. Reversing
that convention reverses the reported emf while leaving the physical charge motion,
current, forces, and power transfer unchanged.

### Three direction cases with the same rod

Use a vertical rod and an observer facing the page. Let the rod move rightward. An
into-page magnetic field produces an upward force on positive carriers, so the top end
is positive. An out-of-page field produces a downward force, so the bottom end is
positive. If the field points parallel to the rod, the carrier force lies horizontal and
has no component along the rod; the two rod endpoints acquire no motional voltage from
that field component.

The vector calculation separates these cases. Let the rod direction be
$+\hat y$ and its velocity be $+\hat x$. With
$\vec B=-B\hat z$ into the page,

$$
\vec v\times\vec B
=vB\hat y.
$$

The component along the upward rod is positive. With
$\vec B=+B\hat z$ out of the page, the result is
$-vB\hat y$ and the polarity reverses. With
$\vec B=B\hat y$ along the rod,

$$
\vec v\times\vec B
=vB\hat z,
$$

which has zero dot product with the rod direction. The cross product gives a direct
answer without memorizing a separate verbal rule for each drawing orientation.

A rod can move through a region with an oblique field. Decompose the field into a
component perpendicular to the plane containing the rod and velocity and a component
that does not generate carrier force along the rod. The endpoint emf follows the
triple-product projection, rather than the full field magnitude. This projection explains
why turning a pickup rod by ninety degrees can reduce a voltage to zero without changing
its speed or the magnetic source.

### Flux rate and carrier force in the rail geometry

The sliding-rod circuit allows two independent derivations of one result. The flux route
uses the changing enclosed area:

$$
\Phi_B=B\ell x,
\qquad
\mathcal E=-B\ell\frac{\d x}{\d t}.
$$

The carrier route begins inside the rod. Positive charges move with the rod and feel
upward magnetic force. The upper end becomes positive, driving conventional current
from the rod top through the upper rail, resistor, lower rail, and rod bottom. Both
routes produce the same counterclockwise current for a rod moving right in an into-page
field.

The agreement follows because the moving rod changes the circuit boundary and its
enclosed area. A wire moving within a field while preserving the same total flux requires
the complete motional line integral. Local carrier forces can create charge separation,
while contributions to the closed motional line integral around a completed circuit
cancel. The flux derivative tests the full circuit. The carrier calculation tests each
moving segment. A complete solution keeps both levels of description distinct.

The force balance also provides a sign test. Current upward in the moving rod and field
into the page produce force left. A rightward external drive must oppose that force.
If a proposed current gives a rightward magnetic force on a rod already moving rightward
while the resistor receives positive power, the current direction has been reversed.
The apparent energy gain would have no mechanical input.

### Scaling laws for apparatus design

The straight-rod source magnitude scales as

$$
|\mathcal E|\propto B\ell v.
$$

Doubling the magnetic field, active rod length, or speed doubles the open-circuit emf.
At fixed total resistance, current follows the same proportionality. Magnetic drag
at fixed resistance has a stronger dependence,

$$
F_{\rm mag}=\frac{B^2\ell^2}{R}v.
$$

Doubling $B$ or $\ell$ quadruples the drag force at fixed speed. Lowering resistance
increases current and drag, while it decreases the internal voltage loss fraction in a
source with fixed rod resistance. These changes also affect heating: the resistor power
scales as $B^2\ell^2v^2/R$ in the ideal rail model.

A generator designer cannot raise speed without mechanical consequences. Higher speed
raises emf and potential output power, but it also increases reaction torque for a
given load. Faster sliding contacts can wear and heat; higher current raises winding
losses. The electromagnetic formulas identify the first scaling step, while thermal,
mechanical, and insulation limits set practical operating ranges.

A rotating rod has source magnitude scaling as $B\omega L^2/2$. Length has a
quadratic effect because longer rods add conductor length and reach greater speed near
the tip. A rotating coil has source magnitude proportional to $NBA\omega$, so turn
count and area offer additional ways to raise emf. The associated reaction torque must
be supplied by the prime mover whenever the circuit draws power.

### Conditions for zero output

Several different physical conditions can lead to zero current, and they should be kept
separate.

- A **zero motional emf** occurs when $(\vec v\times\vec B)\cdot \d\vec\ell$ integrates to zero around the circuit. Examples include motion parallel to $\vec B$ and rigid translation of a closed loop through a uniform field.
- A **nonzero emf with an open circuit** produces endpoint charge separation but no steady current around the missing path.
- A **nonzero emf with large resistance** produces a small current. The emf remains set by the geometry and motion.
- A **zero net emf from cancellation** can occur when multiple moving segments contribute equal and opposite line integrals.

Each condition produces a different experimental signature. A voltmeter can detect
open-circuit endpoint polarity. An ammeter requires a closed path and reports current.
A force sensor on a driven rod detects magnetic drag when current flows. A stationary
rod in a static field may show no signal in all three instruments despite the presence
of substantial magnetic flux through a nearby loop.

The distinction also prevents an incorrect statement that a magnetic field alone acts
as a battery. The source term requires relative charge motion through the field or a
time-varying field that generates circulating electric force. Geometry, circuit closure,
and resistance determine how that source appears at terminals and in a load.

### Measurement boundaries and source characterization

Characterize a motional source with separate open-circuit and loaded measurements.
With the external circuit open, measure the endpoint voltage using a voltmeter whose
input resistance is large compared with the conductor and contact resistances. This
measurement estimates the emf under the stated speed, field, and geometry. Then
connect known load resistances and measure terminal voltage and current. A linear
terminal-voltage-versus-current plot can estimate effective internal resistance,
provided speed and magnetic field remain stable during each point.

Mechanical measurements complete the source description. Record driving force and
speed for a sliding rod, or torque and angular speed for a rotating generator.
Mechanical input power should match electrical load power plus resistive, contact,
and mechanical losses within the measurement uncertainty. A mismatch can locate a
boundary error. Counting only the external load misses winding loss; counting a
motor's electrical input and a generator's electrical output without the shaft
coupling double-counts the conversion path.

Field calibration matters because emf and drag have different field dependence.
Open-circuit emf scales with the field magnitude, while the loaded rail drag scales
approximately with field squared at fixed resistance and speed. Repeat a test at
several calibrated field settings. A linear voltage trend with a quadratic force
trend supports the ideal model. Saturation of a magnetic core, a changing air gap,
or field nonuniformity produces systematic departures that a single operating point
cannot identify.

Contacts require their own diagnostics. A brush or sliding rail contact can add
resistance, intermittent current, and a voltage drop that changes with speed or
pressure. Measure the rod voltage and the load voltage separately, and inspect the
current trace for spikes or missing intervals. A smooth emf with a noisy load current
identifies a circuit-contact limitation, whereas a fluctuating emf can reflect speed
or field variation. These distinctions preserve the link between the motion that
generates emf and the circuit that delivers usable electrical power.