Physics · Electromagnetic induction

Motional emf

The magnitude of a straight rod’s motional emf contribution is Bℓv when field, rod and velocity are pairwise perpendicular.

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Formula

Calculate

∣ E_{mot} ∣ = B ℓ v

ℰ in volts, B in tesla, ℓ in metres and v in metres per second.

Variables & units

Variables & units
Symbol	Meaning	Unit
\|ℰmot\|	Magnitude of the motional emf contribution in perpendicular geometry	V
$B$	Uniform magnetic flux density as a nonnegative magnitude	T
$ℓ$	Active straight rod length within the field	m
$v$	Instantaneous translational speed perpendicular to both field and rod	m/s

LaTeX

|\mathcal E_{\mathrm{mot}}|=B\ell v

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For a rotating radial rod, the motional integrand Bωr rises linearly from zero to 1.296 volts per metre at 0.24 metres. Triangle area gives emf 0.15552 volts; the tip value does not apply along the entire rod. — Original radial-rod example: B = 0.300 T, ω = 18.0 rad/s, L = 0.240 m. B and ω point along +z; integration runs from axis to tip. Triangle area: +0.15552 V. Tip value times full length would incorrectly double it. OpenStax §13.3, MIT 8.02 Ch. 10.

Motional emf calculator

Find the induced emf of a rod sliding at right angles through a uniform magnetic field, plus current, braking force and pulling power with a resistor.

Calculation model

Find

Magnetic flux density B

Magnetic flux density at right angles to rod and motion, e.g. 0.5 T between strong magnets.

Active rod length in the field ℓ

Length of rod inside the field, e.g. 10 cm; parts outside the field do not count.

Speed v at right angles to the field

Speed of the rod at right angles to the field, e.g. 2 m/s.

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

Result: Induced emf |ℰ|

|ℰ| = 0.1 V

With B = 0.5 T, ℓ = 10 cm and v = 2 m/s, |ℰ| = 0.1 V.

Working · rearrange for |ℰ|, substitute, evaluate

1. Rearrange for |ℰ|
$|\mathcal E|=B\ell v$
2. Substitute values (in SI units)
- ℓ = 10 cm = 0.1 m
$|\mathcal E|=0.5\cdot 0.1\cdot 2$
3. Result
|ℰ| = 0.1 V

Quick tips

U = B · ℓ · v: twice the speed or twice the rod length in the field gives twice the emf. Unit check: 1 T · 1 m · 1 m/s = 1 V.
Energy check: the pulling power F · v equals the electrical power U · I in the resistor. If the two differ, there is a mistake in the calculation.

Assumptions and limits

Straight translating rod in a uniform stationary field. Rod, velocity and field are pairwise perpendicular; B, ℓ and v are nonnegative magnitudes. This gives instantaneous motional contribution Bℓv, not polarity, general loop emf or an undefined measured voltage. Current and power use the separate load model.

Derivation & background

Generally, $\mathcal E_{a\to b}^{\mathrm{mot}}=\int_a^b(\vec v\times\vec B)\cdot\mathrm d\vec\ell$ . Magnitude and sign require full geometry; one angle between field and velocity is insufficient. For variable motion use instantaneous velocity. A measured voltage requires a defined measurement path.

Exam blueprint

Validity range

A uniform stationary field and straight translating rod in pairwise perpendicular directions give magnitude Bℓv. Constant speed is not required for the instantaneous motional contribution. Specify circuit and measurement models separately.

Derivation steps

An oriented integral of magnetic force per charge gives the motional contribution; only perpendicular special geometry reduces it to the product.

1Specify reference frame, conductor direction and local velocity.
2Calculate $\vec v\times\vec B$ , magnetic force per positive charge.
3Project onto local conductor direction; a contribution transverse to it does not count.
4Integrate $\mathcal E_{a\to b}^{\mathrm{mot}}=\int_a^b(\vec v\times\vec B)\cdot\mathrm d\vec\ell$ .
5For a straight translating rod, field and velocity are constant along it.
6Pairwise perpendicular geometry gives $|\mathcal E|=B\ell v$ ; the cross product still determines direction.
7For rails, check flux change; under load also check mechanical and electrical power.

Rearrangements

Speed from motional emf

v=\frac{|\mathcal E_{\mathrm{mot}}|}{B\ell}

Magnitude form for uniform flux density and pairwise perpendicular rod, field and velocity directions. Use instantaneous speed; a constant waveform is not required.

Task variant

A straight rod of $\ell=0.25\,\mathrm m$ moves at $v=3.0\,\mathrm{m/s}$ in $B=0.40\,\mathrm T$ ; the three directions are pairwise perpendicular. Find the magnitude of motional emf.

$|\mathcal E|=B\ell v=(0.40\,\mathrm T)(0.25\,\mathrm m)(3.0\,\mathrm{m\,s^{-1}})=0.30\,\mathrm V$ For mutually perpendicular rod, field and motion.
Flux check for a stationary rail circuit: in 1 s area increases by 0.75 m² and flux magnitude by 0.30 Wb.

Common mistakes

Ignoring the motion direction or angle.

$|\mathcal E_{\mathrm{mot}}|=B\ell v$ applies in perpendicular special geometry. Generally project $\vec v\times\vec B$ onto local conductor direction and integrate; cross-product magnitude alone is insufficient.

Exam context

Generators, rail conductors and energy conversion through motion in a magnetic field.

Formula cluster

Faraday’s law through motion

Changing magnetic flux explains the same emf as the Lorentz force on moving charges.

ℰ = −N dΦB/dt

Faraday’s law of induction

F = q(v × B)

Lorentz Force

U₁/U₂ = n₁/n₂ = I₂/I₁

Transformer turns ratio

Worked example

A straight rod of $\ell=0.25\,\mathrm m$ translates at $v=3.0\,\mathrm{m/s}$ in uniform $B=0.40\,\mathrm T$ ; the three directions are pairwise perpendicular. Approach and result: $|\mathcal E_{\mathrm{mot}}|=B\ell v=0.30\,\mathrm V$ . In a stationary rail circuit area increases by 0.75 m² in 1 s, confirming the flux-change-rate magnitude. Polarity and traversal sign require the directions.

Applications

Generator principle, moving rods on rails, speed sensors and induction experiments with conducting loops.

How do you calculate with Motional emf?

Here is how to work through a typical Motional emf (|ℰmot| = Bℓv) task step by step:

1
Task
A straight rod of $\ell=0.25\,\mathrm m$ moves at $v=3.0\,\mathrm{m/s}$ in $B=0.40\,\mathrm T$ ; the three directions are pairwise perpendicular. Find the magnitude of motional emf.
Solution path
1. $|\mathcal E|=B\ell v=(0.40\,\mathrm T)(0.25\,\mathrm m)(3.0\,\mathrm{m\,s^{-1}})=0.30\,\mathrm V$ For mutually perpendicular rod, field and motion.
2. Flux check for a stationary rail circuit: in 1 s area increases by 0.75 m² and flux magnitude by 0.30 Wb.

Understand and apply the equation

A moving conductor separates charges when the magnetic Lorentz force has a component along it. The familiar shortcut assumes three mutually perpendicular directions; the general statement is an oriented line integral. A loaded rail rod and a rotating rod show why direction, local velocity and mechanical power must be considered together.

First consider charges, then the complete circuit

A metal rod moves across a magnetic field. Its moving charge carriers are deflected sideways and redistribute toward the ends. An open rod develops a polarity; a loaded rod on rails can also supply current. Yet the same product of field, length and speed is not enough for every moving loop: the return path belongs in the calculation.[2][1]

Distinguish from the start the directed contribution along a rod, emf of the closed circuit and a defined voltage reading. Direction needs the cross product; power additionally needs load resistance and a mechanical model.[2][1]

Charge separation caused by motion

\begin{aligned}\vec F_B&=q\vec v\times\vec B\\\mathcal E_{a\to b}^{\mathrm{mot}}&=\int_a^b(\vec v\times\vec B)\cdot\mathrm d\vec\ell\end{aligned}

Local conductor velocity and the orientation from a to b determine the motional contribution.[1]

The magnetic Lorentz force pushes positive charges along the cross product; electrons respond oppositely. After rapid charge redistribution, an open conductor develops an opposing electric field. In the quasistatic limit the net force component along the conductor vanishes while the conductor continues moving. A conducting return connection can instead permit current. The integral describes the magnetic motional contribution; an additional time-varying field also supplies an electric induction contribution.[1][2]

When the integral reduces to the shortcut

\begin{aligned}\vec v&=v\hat{\vec x}\\\vec B&=-B\hat{\vec z}\\\vec\ell&=\ell\hat{\vec y}\\\mathcal E_{a\to b}^{\mathrm{mot}}&=(\vec v\times\vec B)\cdot\vec\ell=B\ell v\end{aligned}

Here a to b points along +y; B and v are positive magnitudes.[1]

The straight rod, its translational velocity and the uniform magnetic field are pairwise perpendicular in this geometry. Positive charge accumulates at end b; in the open quasistatic rod, b is positive relative to a. Other orientations can reduce magnitude, make it zero or reverse the sign. Knowing only the angle between velocity and field is insufficient: the cross product must also be projected onto the rod direction.[1]

Lorentz-force and flux calculations meet

\begin{aligned}E_y&=-vB\\V_b-V_a&=-\int_a^b\vec E\cdot\mathrm d\vec\ell=B\ell v\\\Phi_B&=-B\ell x\\\mathcal E_C&=-\frac{\mathrm d\Phi_B}{\mathrm dt}=B\ell v\end{aligned}

Rail surface with normal +z and positive counterclockwise traversal; the right rod moves along +x.[2][1]

In the open rod, the first line follows from carrier-force balance. In the closed rail circuit, area grows; because the field points toward −z, flux with a +z normal becomes more negative. Positive traversal emf runs upward along the right rod. Both calculations agree in the same geometry. Merely reversing the chosen normal reverses flux and positive traversal together; the rod’s physical polarity remains unchanged.[2][1]

Electrical energy requires mechanical work

\begin{aligned}I&=\frac{B\ell v}{R}\\F_{\mathrm{drag}}&=IB\ell=\frac{B^2\ell^2v}{R}\\P_{\mathrm{ext}}&=F_{\mathrm{drag}}v=\mathcal E I=I^2R\end{aligned}

Constant velocity, stationary external field and a purely resistive total circuit; friction and self-induction are neglected.[2][1]

A generator-load current produces magnetic drag on the rod. At constant speed an external pull must balance it. Equality of powers checks the model; it does not create extra energy from the magnetic field. Magnetic Lorentz force on an individual particle is perpendicular to its instantaneous total velocity and does no particle work by itself. In a moving conductor, forces mediate energy transfer from the mechanical drive through charges to the resistor.[2][3]

View diagram: magnetic drag

Drag increases with speed

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For the first example at constant B, length and R: drag grows linearly and pulling power quadratically. At the endpoint they are 22.815 mN and 57.0375 mW. The graph shows force magnitude, not its negative x component.[2]

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Mechanical power describes energy supplied by the external drive. In this load model it must equal electrical dissipation; open-circuit emf alone does not determine delivered power.[2]

Example: a loaded rail rod

Polarity, load current and pulling power

Given: Rod length $\ell=0.180\,\mathrm m$ along +y, $\vec v=(2.50\,\mathrm{m/s})\hat{\vec x}$ , $\vec B=-(0.650\,\mathrm T)\hat{\vec z}$ . Ideal stationary rails and total resistance $R=1.50\,\Omega$ . Constant velocity, no friction; self-induction and field back reaction are negligible.
Find: Find emf, current direction, drag and required pulling power.

Approach: determine the cross product and traversal first, then current and mechanical balance.
$\begin{aligned}\mathcal E&=B\ell v\\I&=\frac{\mathcal E}{R}\\F_{\mathrm{drag}}&=IB\ell\\P_{\mathrm{ext}}&=F_{\mathrm{drag}}v\end{aligned}$
Substitution: the motional drive points along +y in the rod.
$\begin{aligned}\mathcal E&=(0.650)(0.180)(2.50)\,\mathrm V\\&=0.2925\,\mathrm V\\I&=\frac{0.2925\,\mathrm V}{1.50\,\Omega}=0.1950\,\mathrm A\end{aligned}$
Current along +y and field along −z give drag toward −x.
$\begin{aligned}F_{\mathrm{drag}}&=(0.1950)(0.650)(0.180)\,\mathrm N\\&=0.022815\,\mathrm N\\P_{\mathrm{ext}}&=(0.022815)(2.50)\,\mathrm W\\&=0.0570375\,\mathrm W\end{aligned}$

Result: Emf is 0.2925 V in positive counterclockwise traversal, current 0.1950 A, drag 22.8 mN and pulling power 57.0 mW.

Check: $I^2R=(0.1950\,\mathrm A)^2(1.50\,\Omega)=0.0570375\,\mathrm W$ verifies power. In one second area increases by 0.450 m², giving a flux change of −0.2925 Wb and positive emf 0.2925 V.

Example: local speed rather than tip speed

A radial rod rotates in the field

Given: A straight radial rod extends from the axis to $L=0.240\,\mathrm m$ . It rotates in the xy plane at $\omega=+18.0\,\mathrm{rad/s}$ , in $\vec B=(0.300\,\mathrm T)\hat{\vec z}$ . The rod is open with quasistatic charge redistribution; instrument lead paths are outside the problem.
Find: Calculate the motional contribution from axis to tip and compare with incorrectly using tip speed everywhere.

Approach: speed increases radially; the cross product points outward for this orientation.
$\begin{aligned}v(r)&=\omega r\\\mathcal E_{0\to L}^{\mathrm{mot}}&=\int_0^L B\omega r\,\mathrm dr\\&=\frac12B\omega L^2\end{aligned}$
Substitution gives half the value obtained by using tip speed everywhere.
$\begin{aligned}\mathcal E_{0\to L}^{\mathrm{mot}}&=\frac12(0.300)(18.0)(0.240)^2\,\mathrm V\\&=+0.15552\,\mathrm V\\BLv_{\mathrm{tip}}&=B\omega L^2=0.31104\,\mathrm V\end{aligned}$
Mean local speed of this linear profile is half the tip speed.
$\overline v=\frac{\omega L}{2}=2.16\,\mathrm{m/s}$

Result: The directed motional contribution is +0.15552 V outward; in the quasistatic open rod the tip is positive relative to the axis. The tip-speed calculation overestimates by a factor of two.

Check: $BL\overline v=(0.300)(0.240)(2.16)\,\mathrm V=0.15552\,\mathrm V$ . A suitable stationary return path would give a consistent swept-sector area rate $\frac12L^2\omega$ ; the open rod alone does not define a closed measurement traversal.

Units and instantaneous speed

\begin{aligned}[B\ell v]&=\mathrm{T\,m}\frac{\mathrm m}{\mathrm s}=\mathrm V\\v&=\frac{|\mathcal E|}{B\ell}\end{aligned}

The rearrangement belongs to the perpendicular special geometry with positive field magnitude and active length.[2]

Constant speed is not needed to calculate the instantaneous motional contribution of a straight translating rod. For changing speed, use its instantaneous value; actual electrical response can lag through charge redistribution and self-induction. Constant pulling power, braking time or terminal speed additionally require mechanical and circuit models. Magnetic field strength in amperes per metre is not the same quantity as flux density in teslas.[1]

Motion alone does not guarantee loop emf

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

A complete rigidly translating closed loop in a spatially uniform, time-independent field has zero motional traversal contribution when v is the same everywhere.[1]

Individual sides of a moving loop may show charge separation while traversal contributions cancel. Entry into a bounded field region or rotation gives different local contributions. Rail contacts, field gradients and moving return leads must be included explicitly. The simple resistance model does not describe a superconducting loop or a strongly inductive transient. A straight rod with $\ell=0.25\,\mathrm m$ , $v=3.0\,\mathrm{m/s}$ and $B=0.40\,\mathrm T$ in the perpendicular geometry gives emf magnitude $0.30\,\mathrm V$ .[2][1]

Example: an entire loop leaves the field

Only area inside the field counts

Given: A rectangular loop has height $h=0.120\,\mathrm m$ , width $w=0.200\,\mathrm m$ , normal +z. It leaves a field $\vec B=-(0.300\,\mathrm T)\hat{\vec z}$ at $v=0.500\,\mathrm{m/s}$ . Idealise a sharp field boundary: during exit the right side is already outside the field while the left side remains inside. Total resistance is $R=2.00\,\Omega$ ; neglect self-induction and friction.
Find: Find emf, current direction, exit duration and heat.

Approach: negative flux becomes less negative during exit. Positive traversal runs downward on the left side.
$\begin{aligned}\Phi&=-Bh(w-vt)\\\mathcal E&=-\dot\Phi=-Bhv\\I&=\frac{\mathcal E}{R}\\\Delta t&=\frac wv\end{aligned}$
Substitution: the minus sign indicates clockwise traversal viewed from +z.
$\begin{aligned}\mathcal E&=-(0.300)(0.120)(0.500)\,\mathrm V\\&=-0.0180\,\mathrm V\\I&=-9.00\,\mathrm{mA}\\\Delta t&=0.400\,\mathrm s\\Q_R&=I^2R\Delta t=64.8\,\mathrm{\mu J}\end{aligned}$

Result: During exit: −18.0 mV, 9.00 mA clockwise and 64.8 µJ heat over 0.400 s. Entirely inside or outside the constant field, loop emf is zero.

Check: Flux changes from $-7.20\,\mathrm{mWb}$ to zero, so $\mathcal E\Delta t=-7.20\,\mathrm{mWb}$ . Drag magnitude is $0.324\,\mathrm{mN}$ ; its work over 0.200 m is 64.8 µJ.

Five exercises on direction, load and motion

Calculate · Application

Straight rod from a to b along +y with $\ell=0.200\,\mathrm m$ . $\vec v=(3.00\hat{\vec x}+4.00\hat{\vec y})\,\mathrm{m/s}$ , $\vec B=-0.500\hat{\vec z}\,\mathrm T$ . Find the motional contribution toward b.

Show hint

Only the cross product’s y component contributes to the rod integral.

Learning goal: Determine the complete geometric projection from vectors.

Show solution and calculation

Approach: $\mathcal E=(\vec v\times\vec B)\cdot\vec\ell$ .
The cross product is $(-2.00,1.50,0)\,\mathrm{V/m}$ .
Projecting onto 0.200 m along +y gives +0.300 V.

Check: speed along the rod does not contribute; using total speed 5.00 m/s incorrectly gives 0.500 V.

\mathcal E_{a\to b}^{\mathrm{mot}}=(1.50\,\mathrm{V/m})(0.200\,\mathrm m)=+0.300\,\mathrm V

[1]

Rearrange · Foundation

A perpendicular rod with active length 0.150 m must generate emf magnitude 0.180 V in a uniform 0.400 T field. What speed is needed?

Show hint

Isolate the positive speed magnitude.

Learning goal: Calculate required speed in perpendicular geometry.

Show solution and calculation

Approach: $v=\frac{|\mathcal E|}{B\ell}$ .
Result is 3.00 m/s.
Substitution into the original product gives 0.180 V.
Required speed is determined only for the prescribed perpendicular geometry; polarity additionally requires directional information.

v=\frac{0.180\,\mathrm V}{(0.400\,\mathrm T)(0.150\,\mathrm m)}=3.00\,\mathrm{m/s}

[2]

Interpret · Application

A rigid rectangular loop translates entirely at 2.00 m/s along +x through an everywhere uniform, time-independent field toward −z. Both vertical sides are 0.100 m high; field magnitude is 0.500 T. Every side moves at the same speed. Find motional traversal contribution for normal +z.

Show hint

The right and left sides are traversed in opposite directions.

Learning goal: Distinguish local charge separation from total loop emf.

Show solution and calculation

Approach: integrate the constant cross product around the complete closed path.
The right side contributes +0.100 V and the left −0.100 V; horizontal sides give zero.
Total emf is zero.

Check: area and orientation stay fixed, so flux is constant. One side’s contribution is not the whole circuit’s generator emf.

\mathcal E_C=+B\ell v-B\ell v=0

[1]

Error analysis · Application

An ideal loaded rail rod has constant field, length and total resistance. Initially $\mathcal E=0.200\,\mathrm V$ , $R=2.00\,\Omega$ . Speed doubles. A report says pulling power rises from 20.0 to 40.0 mW. Correct it, neglecting friction and self-induction.

Show hint

Both emf and current change; power is their product.

Learning goal: Scale the power balance quadratically when speed doubles.

Show solution and calculation

Approach: $P=\frac{\mathcal E^2}{R}$ with emf proportional to speed.
Doubling gives emf 0.400 V, current 0.200 A and power 80.0 mW.
Initial power was correct; its scaling factor was wrong.

Check: drag and speed both double, so their product quadruples.

\begin{aligned}P_1&=\frac{(0.200\,\mathrm V)^2}{2.00\,\Omega}=0.0200\,\mathrm W\\P_2&=\frac{(0.400\,\mathrm V)^2}{2.00\,\Omega}=0.0800\,\mathrm W\end{aligned}

[2]

Transfer · Challenge

A rail rod of mass 0.0500 kg, length 0.250 m and initial speed 2.00 m/s moves in a perpendicular 0.800 T field. Total resistance is 2.00 Ω. No external pull after $t=0$ ; neglect friction, self-induction and field back reaction. Find the time constant, speed and emf magnitude after 2.50 s.

Show hint

Insert speed-dependent drag into Newton’s equation.

Learning goal: Combine magnetic drag with mechanical exponential decay.

Show solution and calculation

Approach: $m\dot v=-\frac{B^2\ell^2}{R}v$ .
This gives exponential decay with time constant 2.50 s.
After one time constant, speed is about 0.73576 m/s and emf magnitude 0.14715 V.

Check: kinetic energy falls to $e^{-2}$ of its initial value; lost energy becomes resistor heat entirely in this ideal model.

\begin{aligned}\tau&=\frac{mR}{B^2\ell^2}=2.50\,\mathrm s\\v(\tau)&=v_0e^{-1}\approx0.73576\,\mathrm{m/s}\\|\mathcal E(\tau)|&=B\ell v(\tau)\approx0.14715\,\mathrm V\end{aligned}

[1][2]

In brief

The motional contribution is an oriented line integral of the cross product. The shortcut assumes perpendicular special geometry. Check the entire circuit and all moving segments; one side’s contribution is insufficient. Under load the mechanical budget must match electrical power.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 2, 2026Linked Lorentz force and mechanical power in the explanatory context; preserved oriented calculations, load cases, exercises and diagram.
October 1, 2026Developed a DE/EN explanation with an oriented derivation, two original worked examples and five distinct solved exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

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Frequently asked questions about Motional emf

Why does a moving conductor develop voltage?+

Charges in the conductor move with it through the magnetic field. The magnetic Lorentz force can have a component along the conductor and shift charges towards different ends. In an open rod, this creates an electric field that balances the longitudinal magnetic force in quasistatic equilibrium. Determine polarity using the cross product of conductor velocity and magnetic field, rather than a positive magnitude alone. A moving conductor therefore need not develop an end voltage in every geometry. A force transverse to the considered conductor path contributes nothing to its emf. Under load, electrical energy comes from mechanical work by the drive; the magnetic force itself does no work on an individual charge.

When may I use ℰ=Bℓv?+

The shortcut gives magnitude for a straight rod in a uniform field when rod, velocity and magnetic field are pairwise perpendicular. Field and translational velocity must be constant along the active rod segment. Use instantaneous velocity: constant speed throughout the whole motion is not required for this motional contribution. In a more general geometry, integrate $\mathcal E_{a\to b}^{\mathrm{mot}}=\int_a^b(\vec v\times\vec B)\cdot\mathrm d\vec\ell$ . A single angle between velocity and field does not replace projection onto conductor direction. During rotation, local velocity varies along a radial rod; applying tip speed to the full length overestimates the result. Measured voltage additionally requires a specified measurement path.

How does the formula fit Faraday’s law?+

Consider a rod moving on fixed rails and changing the area of a closed loop. In a uniform stationary field, the rate of flux change equals field magnitude times active rod length times speed, with the sign set by the chosen normal. Faraday’s law gives the same loop emf as integrating magnetic force per charge along the moving rod. The fixed rails have no motional contribution. An entirely rigid translating loop in a uniform field is different: its area and orientation remain unchanged. Contributions from opposite sides cancel, so total emf is zero. Check the complete loop before interpreting one rod’s contribution as a generator emf.

What changes when a resistor is connected?+

A connected resistor allows current when a closed conducting path exists. In the simple quasistatic model without significant self-inductance, $I=\frac{\mathcal E}{R}$ . Current in the moving rod produces a magnetic opposing force; its magnitude in perpendicular rail geometry is $BI\ell$ . A drive must work against that force to maintain constant speed. Mechanical input power then equals Joule dissipation. Without a drive, the rod can slow down, and emf and current decrease with speed. Real wire resistance, inductance and additional friction modify the simplified model. Resistor power follows from actual current, rather than automatically from an arbitrary measured open-circuit voltage.

How do I check the unit?+

First check the input quantities: tesla for magnetic flux density, metres for active rod length and metres per second for the appropriate velocity component. Using $1\,\mathrm T=1\,\frac{\mathrm{V\,s}}{\mathrm{m^2}}$ , their product has units of volts. This confirms dimension, but not geometry or polarity. For example, 0.50 tesla, 0.20 metres and a perpendicular velocity component of three metres per second give an emf magnitude of 0.30 volts. Additional motion along the rod does not belong in that product. For a closed circuit, also check power: volts times amperes and force times speed must give both the same watt unit and the same numerical value in the ideal loaded case.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 2, §13.3: Motional Emf

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Motional emf

The magnitude of a straight rod’s motional emf contribution is Bℓv when field, rod and velocity are pairwise perpendicular.

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Symbol

Meaning

Unit

|ℰmot|

Magnitude of the motional emf contribution in perpendicular geometry

B

Uniform magnetic flux density as a nonnegative magnitude

ℓ

Active straight rod length within the field

v

Instantaneous translational speed perpendicular to both field and rod

m/s

Derivation & background

Exam blueprint

Validity range

Derivation steps

An oriented integral of magnetic force per charge gives the motional contribution; only perpendicular special geometry reduces it to the product.

1Specify reference frame, conductor direction and local velocity.
2Calculate $\vec v\times\vec B$ , magnetic force per positive charge.
3Project onto local conductor direction; a contribution transverse to it does not count.
4Integrate $\mathcal E_{a\to b}^{\mathrm{mot}}=\int_a^b(\vec v\times\vec B)\cdot\mathrm d\vec\ell$ .
5For a straight translating rod, field and velocity are constant along it.
6Pairwise perpendicular geometry gives $|\mathcal E|=B\ell v$ ; the cross product still determines direction.
7For rails, check flux change; under load also check mechanical and electrical power.

How do you calculate with Motional emf?

Here is how to work through a typical Motional emf (|ℰmot| = Bℓv) task step by step:

Task

A straight rod of $\ell=0.25\,\mathrm m$ moves at $v=3.0\,\mathrm{m/s}$ in $B=0.40\,\mathrm T$ ; the three directions are pairwise perpendicular. Find the magnitude of motional emf.

Solution path

$|\mathcal E|=B\ell v=(0.40\,\mathrm T)(0.25\,\mathrm m)(3.0\,\mathrm{m\,s^{-1}})=0.30\,\mathrm V$ For mutually perpendicular rod, field and motion.
Flux check for a stationary rail circuit: in 1 s area increases by 0.75 m² and flux magnitude by 0.30 Wb.

Understand and apply the equation

First consider charges, then the complete circuit

Charge separation caused by motion

\begin{aligned}\vec F_B&=q\vec v\times\vec B\\\mathcal E_{a\to b}^{\mathrm{mot}}&=\int_a^b(\vec v\times\vec B)\cdot\mathrm d\vec\ell\end{aligned}

Local conductor velocity and the orientation from a to b determine the motional contribution.[1]

When the integral reduces to the shortcut

\begin{aligned}\vec v&=v\hat{\vec x}\\\vec B&=-B\hat{\vec z}\\\vec\ell&=\ell\hat{\vec y}\\\mathcal E_{a\to b}^{\mathrm{mot}}&=(\vec v\times\vec B)\cdot\vec\ell=B\ell v\end{aligned}

Here a to b points along +y; B and v are positive magnitudes.[1]

Lorentz-force and flux calculations meet

\begin{aligned}E_y&=-vB\\V_b-V_a&=-\int_a^b\vec E\cdot\mathrm d\vec\ell=B\ell v\\\Phi_B&=-B\ell x\\\mathcal E_C&=-\frac{\mathrm d\Phi_B}{\mathrm dt}=B\ell v\end{aligned}

Rail surface with normal +z and positive counterclockwise traversal; the right rod moves along +x.[2][1]

Electrical energy requires mechanical work

\begin{aligned}I&=\frac{B\ell v}{R}\\F_{\mathrm{drag}}&=IB\ell=\frac{B^2\ell^2v}{R}\\P_{\mathrm{ext}}&=F_{\mathrm{drag}}v=\mathcal E I=I^2R\end{aligned}

Constant velocity, stationary external field and a purely resistive total circuit; friction and self-induction are neglected.[2][1]

View diagram: magnetic drag

Drag increases with speed

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Mechanical power describes energy supplied by the external drive. In this load model it must equal electrical dissipation; open-circuit emf alone does not determine delivered power.[2]

Example: a loaded rail rod

Polarity, load current and pulling power

Given: Rod length $\ell=0.180\,\mathrm m$ along +y, $\vec v=(2.50\,\mathrm{m/s})\hat{\vec x}$ , $\vec B=-(0.650\,\mathrm T)\hat{\vec z}$ . Ideal stationary rails and total resistance $R=1.50\,\Omega$ . Constant velocity, no friction; self-induction and field back reaction are negligible.
Find: Find emf, current direction, drag and required pulling power.

Approach: determine the cross product and traversal first, then current and mechanical balance.
$\begin{aligned}\mathcal E&=B\ell v\\I&=\frac{\mathcal E}{R}\\F_{\mathrm{drag}}&=IB\ell\\P_{\mathrm{ext}}&=F_{\mathrm{drag}}v\end{aligned}$
Substitution: the motional drive points along +y in the rod.
$\begin{aligned}\mathcal E&=(0.650)(0.180)(2.50)\,\mathrm V\\&=0.2925\,\mathrm V\\I&=\frac{0.2925\,\mathrm V}{1.50\,\Omega}=0.1950\,\mathrm A\end{aligned}$
Current along +y and field along −z give drag toward −x.
$\begin{aligned}F_{\mathrm{drag}}&=(0.1950)(0.650)(0.180)\,\mathrm N\\&=0.022815\,\mathrm N\\P_{\mathrm{ext}}&=(0.022815)(2.50)\,\mathrm W\\&=0.0570375\,\mathrm W\end{aligned}$

Result: Emf is 0.2925 V in positive counterclockwise traversal, current 0.1950 A, drag 22.8 mN and pulling power 57.0 mW.

Check: $I^2R=(0.1950\,\mathrm A)^2(1.50\,\Omega)=0.0570375\,\mathrm W$ verifies power. In one second area increases by 0.450 m², giving a flux change of −0.2925 Wb and positive emf 0.2925 V.

Example: local speed rather than tip speed

A radial rod rotates in the field

Given: A straight radial rod extends from the axis to $L=0.240\,\mathrm m$ . It rotates in the xy plane at $\omega=+18.0\,\mathrm{rad/s}$ , in $\vec B=(0.300\,\mathrm T)\hat{\vec z}$ . The rod is open with quasistatic charge redistribution; instrument lead paths are outside the problem.
Find: Calculate the motional contribution from axis to tip and compare with incorrectly using tip speed everywhere.

Approach: speed increases radially; the cross product points outward for this orientation.
$\begin{aligned}v(r)&=\omega r\\\mathcal E_{0\to L}^{\mathrm{mot}}&=\int_0^L B\omega r\,\mathrm dr\\&=\frac12B\omega L^2\end{aligned}$
Substitution gives half the value obtained by using tip speed everywhere.
$\begin{aligned}\mathcal E_{0\to L}^{\mathrm{mot}}&=\frac12(0.300)(18.0)(0.240)^2\,\mathrm V\\&=+0.15552\,\mathrm V\\BLv_{\mathrm{tip}}&=B\omega L^2=0.31104\,\mathrm V\end{aligned}$
Mean local speed of this linear profile is half the tip speed.
$\overline v=\frac{\omega L}{2}=2.16\,\mathrm{m/s}$

Result: The directed motional contribution is +0.15552 V outward; in the quasistatic open rod the tip is positive relative to the axis. The tip-speed calculation overestimates by a factor of two.

Units and instantaneous speed

\begin{aligned}[B\ell v]&=\mathrm{T\,m}\frac{\mathrm m}{\mathrm s}=\mathrm V\\v&=\frac{|\mathcal E|}{B\ell}\end{aligned}

The rearrangement belongs to the perpendicular special geometry with positive field magnitude and active length.[2]

Motion alone does not guarantee loop emf

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

A complete rigidly translating closed loop in a spatially uniform, time-independent field has zero motional traversal contribution when v is the same everywhere.[1]

Example: an entire loop leaves the field

Only area inside the field counts

Given: A rectangular loop has height $h=0.120\,\mathrm m$ , width $w=0.200\,\mathrm m$ , normal +z. It leaves a field $\vec B=-(0.300\,\mathrm T)\hat{\vec z}$ at $v=0.500\,\mathrm{m/s}$ . Idealise a sharp field boundary: during exit the right side is already outside the field while the left side remains inside. Total resistance is $R=2.00\,\Omega$ ; neglect self-induction and friction.
Find: Find emf, current direction, exit duration and heat.

Approach: negative flux becomes less negative during exit. Positive traversal runs downward on the left side.
$\begin{aligned}\Phi&=-Bh(w-vt)\\\mathcal E&=-\dot\Phi=-Bhv\\I&=\frac{\mathcal E}{R}\\\Delta t&=\frac wv\end{aligned}$
Substitution: the minus sign indicates clockwise traversal viewed from +z.
$\begin{aligned}\mathcal E&=-(0.300)(0.120)(0.500)\,\mathrm V\\&=-0.0180\,\mathrm V\\I&=-9.00\,\mathrm{mA}\\\Delta t&=0.400\,\mathrm s\\Q_R&=I^2R\Delta t=64.8\,\mathrm{\mu J}\end{aligned}$

Result: During exit: −18.0 mV, 9.00 mA clockwise and 64.8 µJ heat over 0.400 s. Entirely inside or outside the constant field, loop emf is zero.

Check: Flux changes from $-7.20\,\mathrm{mWb}$ to zero, so $\mathcal E\Delta t=-7.20\,\mathrm{mWb}$ . Drag magnitude is $0.324\,\mathrm{mN}$ ; its work over 0.200 m is 64.8 µJ.

Five exercises on direction, load and motion

Calculate · Application

Straight rod from a to b along +y with $\ell=0.200\,\mathrm m$ . $\vec v=(3.00\hat{\vec x}+4.00\hat{\vec y})\,\mathrm{m/s}$ , $\vec B=-0.500\hat{\vec z}\,\mathrm T$ . Find the motional contribution toward b.

Show hint

Only the cross product’s y component contributes to the rod integral.

Learning goal: Determine the complete geometric projection from vectors.

Show solution and calculation

Approach: $\mathcal E=(\vec v\times\vec B)\cdot\vec\ell$ .
The cross product is $(-2.00,1.50,0)\,\mathrm{V/m}$ .
Projecting onto 0.200 m along +y gives +0.300 V.

Check: speed along the rod does not contribute; using total speed 5.00 m/s incorrectly gives 0.500 V.

\mathcal E_{a\to b}^{\mathrm{mot}}=(1.50\,\mathrm{V/m})(0.200\,\mathrm m)=+0.300\,\mathrm V

[1]

Rearrange · Foundation

A perpendicular rod with active length 0.150 m must generate emf magnitude 0.180 V in a uniform 0.400 T field. What speed is needed?

Show hint

Isolate the positive speed magnitude.

Learning goal: Calculate required speed in perpendicular geometry.

Show solution and calculation

Approach: $v=\frac{|\mathcal E|}{B\ell}$ .
Result is 3.00 m/s.
Substitution into the original product gives 0.180 V.
Required speed is determined only for the prescribed perpendicular geometry; polarity additionally requires directional information.

v=\frac{0.180\,\mathrm V}{(0.400\,\mathrm T)(0.150\,\mathrm m)}=3.00\,\mathrm{m/s}

[2]

Interpret · Application

A rigid rectangular loop translates entirely at 2.00 m/s along +x through an everywhere uniform, time-independent field toward −z. Both vertical sides are 0.100 m high; field magnitude is 0.500 T. Every side moves at the same speed. Find motional traversal contribution for normal +z.

Show hint

The right and left sides are traversed in opposite directions.

Learning goal: Distinguish local charge separation from total loop emf.

Show solution and calculation

Approach: integrate the constant cross product around the complete closed path.
The right side contributes +0.100 V and the left −0.100 V; horizontal sides give zero.
Total emf is zero.

Check: area and orientation stay fixed, so flux is constant. One side’s contribution is not the whole circuit’s generator emf.

\mathcal E_C=+B\ell v-B\ell v=0

[1]

Error analysis · Application

An ideal loaded rail rod has constant field, length and total resistance. Initially $\mathcal E=0.200\,\mathrm V$ , $R=2.00\,\Omega$ . Speed doubles. A report says pulling power rises from 20.0 to 40.0 mW. Correct it, neglecting friction and self-induction.

Show hint

Both emf and current change; power is their product.

Learning goal: Scale the power balance quadratically when speed doubles.

Show solution and calculation

Approach: $P=\frac{\mathcal E^2}{R}$ with emf proportional to speed.
Doubling gives emf 0.400 V, current 0.200 A and power 80.0 mW.
Initial power was correct; its scaling factor was wrong.

Check: drag and speed both double, so their product quadruples.

\begin{aligned}P_1&=\frac{(0.200\,\mathrm V)^2}{2.00\,\Omega}=0.0200\,\mathrm W\\P_2&=\frac{(0.400\,\mathrm V)^2}{2.00\,\Omega}=0.0800\,\mathrm W\end{aligned}

[2]

Transfer · Challenge

A rail rod of mass 0.0500 kg, length 0.250 m and initial speed 2.00 m/s moves in a perpendicular 0.800 T field. Total resistance is 2.00 Ω. No external pull after $t=0$ ; neglect friction, self-induction and field back reaction. Find the time constant, speed and emf magnitude after 2.50 s.

Show hint

Insert speed-dependent drag into Newton’s equation.

Learning goal: Combine magnetic drag with mechanical exponential decay.

Show solution and calculation

Approach: $m\dot v=-\frac{B^2\ell^2}{R}v$ .
This gives exponential decay with time constant 2.50 s.
After one time constant, speed is about 0.73576 m/s and emf magnitude 0.14715 V.

Check: kinetic energy falls to $e^{-2}$ of its initial value; lost energy becomes resistor heat entirely in this ideal model.

\begin{aligned}\tau&=\frac{mR}{B^2\ell^2}=2.50\,\mathrm s\\v(\tau)&=v_0e^{-1}\approx0.73576\,\mathrm{m/s}\\|\mathcal E(\tau)|&=B\ell v(\tau)\approx0.14715\,\mathrm V\end{aligned}

[1][2]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Linked Lorentz force and mechanical power in the explanatory context; preserved oriented calculations, load cases, exercises and diagram.
October 1, 2026Developed a DE/EN explanation with an oriented derivation, two original worked examples and five distinct solved exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

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Motional emf

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Motional emf calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

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Task variant

Common mistakes

Exam context

Faraday’s law through motion

Worked example

Applications

How do you calculate with Motional emf?

First consider charges, then the complete circuit

Charge separation caused by motion

When the integral reduces to the shortcut

Lorentz-force and flux calculations meet

Electrical energy requires mechanical work

Drag increases with speed

Example: a loaded rail rod

Polarity, load current and pulling power

Example: local speed rather than tip speed

A radial rod rotates in the field

Units and instantaneous speed

Motion alone does not guarantee loop emf

Example: an entire loop leaves the field

Only area inside the field counts

Five exercises on direction, load and motion

A perpendicular rod with active length 0.150 m must generate emf magnitude 0.180 V in a uniform 0.400 T field. What speed is needed?

A rigid rectangular loop translates entirely at 2.00 m/s along +x through an everywhere uniform, time-independent field toward −z. Both vertical sides are 0.100 m high; field magnitude is 0.500 T. Every side moves at the same speed. Find motional traversal contribution for normal +z.

An ideal loaded rail rod has constant field, length and total resistance. Initially E=0.200 V\mathcal E=0.200\,\mathrm VE=0.200V, R=2.00 ΩR=2.00\,\OmegaR=2.00Ω. Speed doubles. A report says pulling power rises from 20.0 to 40.0 mW. Correct it, neglecting friction and self-induction.

A rail rod of mass 0.0500 kg, length 0.250 m and initial speed 2.00 m/s moves in a perpendicular 0.800 T field. Total resistance is 2.00 Ω. No external pull after t=0t=0t=0; neglect friction, self-induction and field back reaction. Find the time constant, speed and emf magnitude after 2.50 s.

In brief

Sources for this explanation

Frequently asked questions about Motional emf

Subject references and further reading

More formulas: Electricity and magnetism

Retain Motional emf for exams

Motional emf

Variables & units

Motional emf calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Faraday’s law through motion

Worked example

Applications

How do you calculate with Motional emf?

First consider charges, then the complete circuit

Charge separation caused by motion

When the integral reduces to the shortcut

Lorentz-force and flux calculations meet

Electrical energy requires mechanical work

Drag increases with speed

Example: a loaded rail rod

Polarity, load current and pulling power

Example: local speed rather than tip speed

A radial rod rotates in the field

Units and instantaneous speed

Motion alone does not guarantee loop emf

Example: an entire loop leaves the field

Only area inside the field counts

Five exercises on direction, load and motion

A perpendicular rod with active length 0.150 m must generate emf magnitude 0.180 V in a uniform 0.400 T field. What speed is needed?

A rigid rectangular loop translates entirely at 2.00 m/s along +x through an everywhere uniform, time-independent field toward −z. Both vertical sides are 0.100 m high; field magnitude is 0.500 T. Every side moves at the same speed. Find motional traversal contribution for normal +z.

An ideal loaded rail rod has constant field, length and total resistance. Initially E=0.200 V\mathcal E=0.200\,\mathrm VE=0.200V, R=2.00 ΩR=2.00\,\OmegaR=2.00Ω. Speed doubles. A report says pulling power rises from 20.0 to 40.0 mW. Correct it, neglecting friction and self-induction.

A rail rod of mass 0.0500 kg, length 0.250 m and initial speed 2.00 m/s moves in a perpendicular 0.800 T field. Total resistance is 2.00 Ω. No external pull after t=0t=0t=0; neglect friction, self-induction and field back reaction. Find the time constant, speed and emf magnitude after 2.50 s.

In brief

Sources for this explanation

Frequently asked questions about Motional emf

Subject references and further reading

An ideal loaded rail rod has constant field, length and total resistance. Initially $\mathcal E=0.200\,\mathrm V$ , $R=2.00\,\Omega$ . Speed doubles. A report says pulling power rises from 20.0 to 40.0 mW. Correct it, neglecting friction and self-induction.

A rail rod of mass 0.0500 kg, length 0.250 m and initial speed 2.00 m/s moves in a perpendicular 0.800 T field. Total resistance is 2.00 Ω. No external pull after $t=0$ ; neglect friction, self-induction and field back reaction. Find the time constant, speed and emf magnitude after 2.50 s.

An ideal loaded rail rod has constant field, length and total resistance. Initially $\mathcal E=0.200\,\mathrm V$ , $R=2.00\,\Omega$ . Speed doubles. A report says pulling power rises from 20.0 to 40.0 mW. Correct it, neglecting friction and self-induction.

A rail rod of mass 0.0500 kg, length 0.250 m and initial speed 2.00 m/s moves in a perpendicular 0.800 T field. Total resistance is 2.00 Ω. No external pull after $t=0$ ; neglect friction, self-induction and field back reaction. Find the time constant, speed and emf magnitude after 2.50 s.