Motional EMF — Physica

Equivalent forms

\varepsilon = -\frac{d\Phi_B}{dt}

\varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}

BLv is Faraday's law stripped to its kinematic core: the rod sweeps out area at rate Lv, so flux changes at rate BLv — geometry alone fixes the voltage.

Symbol	Name	SI unit	Dimension	Range
$ε$	Motional EMFoutput Voltage induced across the moving conductor	V	M·L²·T⁻³·I⁻¹	0 – 10
$B$	Magnetic field Uniform field perpendicular to the rod's motion	T	M·T⁻²·I⁻¹	0.01 – 2
$L$	Rod length Length of the conductor in the field	m	L	0.05 – 1
$v$	Velocity Speed of the rod perpendicular to its length and to B	m/s	L·T⁻¹	0 – 10

Unit systems

Symbol	SI	CGS	Imperial
$ε$	V	statV	V
$B$	T	G	T
$L$	m	cm	ft
$v$	m/s	cm/s	ft/s

Where it holds

Valid for rigid conductors moving at v ≪ c through a uniform field with v ⊥ B ⊥ L. For non-perpendicular geometry

use \varepsilon = BLv\cdot \sin \theta

; for v approaching c use the full relativistic field transformations.

Dimensional analysis

$[B][L][v] = (kg\cdot s^{-2}\cdot A^{-1})(m)(m\cdot s^{-1}) = kg\cdot m^{2}\cdot s^{-3}\cdot A^{-1} = V \checkmark$

Discovery

Michael Faraday · 1831

Faraday's copper-disc generator (1831) — a disc spun between magnet poles — produced the first continuous motional EMF, proving mechanical motion through a field makes electricity. Every power station since is a refinement of that disc.

Try this

A jet airliner flying through Earth's magnetic field is a 60-meter generator — how many volts appear across its wings?

A conducting rod of length 0.2 m slides at 3 m/s along rails in a 0.5 T field. What EMF does it generate, and which way does the induced current flow?

Research status: stable

Real-world applications

Alternators and dynamos in power plants and cars
Electromagnetic flowmeters measuring blood or liquid-metal flow
Rail guns and electromagnetic launchers (run in reverse)
Space tethers generating power from orbital motion through Earth's field

Common misconceptions

The magnetic force does no net work — the EMF energy comes from whatever pushes the rod, not from the field
EMF appears even with the circuit open; current needs a closed loop but voltage does not
Lenz's law sets the current direction so its force opposes the motion — you must push to keep v constant

Experimental verification

Slide a brass rod along parallel rails between magnet poles with a voltmeter across the rails: the reading tracks BLv linearly in all three parameters. Faraday's 1831 disc generator and every modern alternator verify it continuously.

Derivation

Each charge q in the rod feels magnetic force

F = qv\times B

directed along the rod, equivalent to an effective field

E_eff = v\times B

.

The work per unit charge moving the rod's full length

is \varepsilon = \int (v\times B)\cdot dl = BLv

.

Equivalently: the rod sweeps area

dA = Lv dt

, so

d\Phi /dt = BLv

, matching Faraday's law.

Limiting cases

v \to 0

⟶

\varepsilon \to 0

A stationary rod cuts no flux — no charge separation, no EMF.

v ∥ B⟶

\varepsilon = 0

Motion along the field gives

v\times B = 0

; only the perpendicular velocity component counts.

Circuit open

(R \to \infty )

⟶

\varepsilon = BLv

with

I = 0

Charge separates until equilibrium, but no current flows and no drag force opposes the motion.

What if…

What if the rod moves parallel to the field instead of across it?

No EMF: $v\times B$ vanishes when v ∥ B. Only flux-cutting motion generates voltage.

What if the circuit is closed through a resistance R?

Current $I = BLv/R$ flows, the field exerts a retarding force $F = B^{2}L^{2}v/R$ on the rod, and the mechanical power Fv exactly equals the dissipated power $I^{2}R$ .

What if you double both B and v?

EMF quadruples — it is linear in each factor, so the product scales by 4.

1

Sliding rod on rails

Given ·

B:: 0.5
L:: 0.2
v:: 3

Find ·

\varepsilon

Steps

Check geometry: v ⊥ B ⊥ L, $so \varepsilon = BLv$ applies directly
$\varepsilon = (0.5\,\mathrm{T})(0.2\,\mathrm{m})(3\,\mathrm{m/s}) = 0.3\,\mathrm{V}$
By Lenz's law the induced current opposes the flux increase — it flows so its force brakes the rod

Result ·

\varepsilon = 0.5 \times 0.2 \times 3 = 0.3\,\mathrm{V}

2

Airliner wingtips

Given ·

B:: 0.00005
L:: 60
v:: 250

Find ·

\varepsilon

Steps

Earth's vertical field component $\approx 5\times 10^{-5} T$
$\varepsilon = (5\times 10^{-5})(60)(250) = 0.75\,\mathrm{V}$ across the wingspan
No current flows — the surrounding air is an open circuit, so it's harmless

Result ·

\varepsilon = 5\times 10^{-5} \times 60 \times 250 = 0.75\,\mathrm{V}

Faraday's Law of Induction→Lorentz Force Law→Magnetic Force on a Current-Carrying Wire→