Electromagnetic Induction

Beginner

Connect a coil of wire to a sensitive current meter. There is no battery anywhere, so the needle rests at zero. Now push a bar magnet into the coil: the needle kicks to one side. Hold the magnet still inside, and the needle drops back to zero. Pull it out, and the needle kicks the other way.

This is electromagnetic induction, discovered by Michael Faraday in 1831. The rule is short: a current flows only while the magnetism passing through the coil is changing. A huge magnet lying still does nothing; a small one moved quickly does something. The animation shows the three stages: magnet moving in, magnet resting, magnet moving out.

You have already met the reason. A magnet pushes sideways on moving charges. Move a wire across a magnet's field and the charges inside the wire are carried along with it, so they get pushed along the wire: that push is the induced current. It does not matter what moves, the magnet or the wire.

The induced current always fights the change that made it. As the magnet comes in, the coil turns into a little magnet that pushes it back; as it leaves, the coil tries to hold it. So you must do work to move the magnet, and that work is exactly where the electrical energy comes from.

Moving faster, using a stronger magnet or winding more turns all make a bigger current. Power-station generators, bicycle dynamos and wireless chargers all rely on this.

Check yourself: a magnet rests inside a coil that is joined to a meter. What does the meter show? Answer: zero, because nothing is changing.

Intermediate

Magnetic flux: how much field threads a loop

Most of the world's electricity comes from generators that turn magnets past coils, and every one of them relies on the same rule. Induction is driven not by a magnetic field as such but by a change in how much field passes through a circuit: the magnetic flux. For a flat loop of area $A$ in a uniform field $B$,
$\Phi =BA\cos \theta$
where $\theta$ is the angle between the field and the normal, the line perpendicular to the plane of the loop. A field straight through the loop gives $\theta =0$ and $\Phi =BA$; a field lying in the plane gives $\Phi =0$, because no field line crosses the loop.

Flux is measured in webers, $1\mathrm{Wb}=1\mathrm{T}\cdot {\mathrm{m}}^2$, and it has a sign: reverse the field or turn the loop over and the flux changes sign. A coil of $N$ turns, like the animation's 200, responds to the flux linkage $N\Phi$, since each turn carries the same flux.

Faraday's law: only change counts

Faraday's law states that a changing flux induces an electromotive force (EMF) in a coil of $N$ turns:
$\epsilon =-\frac{N\Delta \Phi }{\Delta t}$
The EMF, measured in volts, is the energy given to each coulomb that goes once round the circuit, and the formula gives its average over the interval $\Delta t$. Through a circuit of resistance $R$ it drives a current $I=\epsilon /R$.

The flux $\Phi =BA\cos \theta$ can change in three ways, and each of them induces an EMF:

• Change the field $B$: move a magnet, or switch an electromagnet on or off.
• Change the area $A$: stretch the loop, or slide a rod along rails.
• Change the angle $\theta$: rotate the coil, which is how a generator works.

The size of the flux itself does not matter. While the magnet rests inside the coil, from 2.6 s to 4.6 s, the $\Phi$ readout holds at 0.20 mWb and the $\epsilon$ readout shows 0 mV.

Lenz's law: the minus sign

The minus sign is Lenz's law: the induced current opposes the change of flux that produced it. As the north pole approaches, the flux grows, so the current makes a field pointing the other way: the end of the coil facing the magnet becomes a north pole and repels it. As the magnet leaves, the flux falls, the current reverses, and that end becomes a south pole that pulls the magnet back. Your right hand gives the direction: thumb along the field the current must make, fingers curling the way it flows.

Lenz's law is energy conservation in disguise. Because the coil pushes back, you must do work to move the magnet, and that work becomes the electrical energy in the circuit. If the coil helped the motion instead, a nudge would send the magnet running away on its own: energy from nothing.

Worked example: the coil in the animation

With the magnet fully inside, the field $B=0.20\mathrm{T}$ runs along the axis of the coil ($\theta =0$) through an area $A=10\mathrm{c}{\mathrm{m}}^2=1.0\times 10^{-3}{\mathrm{m}}^2$:
$\Phi =BA=0.20\cdot 1.0\times 10^{-3}=2.0\times 10^{-4}\mathrm{Wb}=0.20\mathrm{mWb}$
The $\Phi$ readout reaches this value at 2.6 s. It starts at 0.00 because, from 125 mm away, the magnet sends only about 0.0025 mWb through the coil.

The push lasts 2.0 s, from 0.6 s to 2.6 s, so the average EMF over the push is
$|\epsilon |=\frac{N\Delta \Phi }{\Delta t}=\frac{200\cdot 0.20\times 10^{-3}}{2.0}=0.020\mathrm{V}$
or 20 mV. Yet the $\epsilon$ readout dips to about $-89\mathrm{mV}$ at $t\approx 2.2\mathrm{s}$, because the flux does not grow at a steady rate. At 1.7 s it is still only about 0.02 mWb; then, as the north end crosses into the coil, it climbs to about 0.18 mWb by 2.3 s. Over that 0.6 s window the average is $200\cdot 0.16\times 10^{-3}/0.6\approx 53\mathrm{mV}$, and an average can never exceed the peak.

The sign follows Lenz's law: the readout is negative while the flux rises and reaches $+89\mathrm{mV}$ while it falls, as the magnet leaves.

Worked example: a rod sliding on rails

Two parallel rails 25 cm apart are joined at one end through a 0.50 Ω resistor, and a metal rod lies across them in a uniform 0.40 T field perpendicular to their plane. You pull the rod along the rails at a steady 2.0 m/s.

In a time $\Delta t$ the rod, of length $L=0.25\mathrm{m}$ between the rails, adds an area $Lv\Delta t$ to the circuit, so $\Delta \Phi =BLv\Delta t$, and Faraday's law with $N=1$ gives
$\epsilon =BLv$
$\epsilon =0.40\cdot 0.25\cdot 2.0=0.20\mathrm{V}$
and a current $I=\epsilon /R=0.20/0.50=0.40\mathrm{A}$.

As a current in a field, the rod feels a force $F=BIL=0.40\cdot 0.40\cdot 0.25=0.040\mathrm{N}$, which by Lenz's law points backwards, against the motion. Keeping the rod at 2.0 m/s takes a power $Fv=0.080\mathrm{W}$, exactly the $I^{2}R=0.080\mathrm{W}$ that the resistor turns into heat: every joule came from your pull.

Reading the meter, the arrows and the graphs

The green $\Phi$ readout climbs from 0.00 to 0.20 mWb in an S-shaped rise: slow, then fast, then slow again as the magnet settles. The $\epsilon$ readout, the angle of the meter needle and the length of the purple arrow $I$ all follow the EMF in proportion. The N or S beside the left end of the coil is the pole that the induced current makes.

On a graph of $\Phi$ against $t$, the EMF is $-N$ times the slope: negative on a rise, zero on any flat stretch however high it lies, positive on a fall. The steepest point of the rise, near 2.2 s, has a slope of about 0.45 mWb/s, and $200\cdot 0.45\approx 89\mathrm{mV}$ is the peak reading.

Common mistakes

• A strong field is not a large EMF. With the magnet resting inside, 0.20 mWb threads every turn and the EMF is exactly zero.
• The induced current opposes the change, not the field. As the magnet leaves, the induced field points the same way as the magnet's own field, propping up the falling flux.
• The angle $\theta$ is measured from the normal, not from the plane. A coil whose plane makes 30° with the field has $\theta =60^{\circ }$, so with the animation's numbers $\Phi =BA\cos 60^{\circ }=0.10\mathrm{mWb}$, not 0.17 mWb.
• Keep the factor $N$ and convert with $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$. Dropping $N$ turns the animation's 20 mV average into 0.1 mV; taking 10 cm² as $10^{-2}{\mathrm{m}}^2$ turns it into 200 mV.
• An EMF is not a current. A coil with open ends has the full EMF across them but carries no current, so it feels no braking at all.

Where you meet it

Most of the world's electricity comes from generators that turn magnets past coils to make alternating current at 50 or 60 Hz. Transformers change voltages by induction, wireless phone chargers pass energy across a gap at about 100–200 kHz, induction hobs heat a pan through eddy currents, and guitar pickups and metal detectors sense moving or hidden metal. Some trains brake with eddy currents, and electric cars recover energy by running their motors as generators.

The machines are the subject of Motors, Generators and Transformers. Faraday's law is also half of the mechanism of light: changing magnetic and electric fields make each other and travel together as Electromagnetic Waves.

Advanced

EMF and flux, defined precisely

Faraday's law links two quantities, the EMF round a circuit and the magnetic flux through it, so both need exact definitions before the law can be stated. Walk once round a closed path $C$ in small steps $d\vec{l}$. The EMF $\epsilon$ of the path is the work done per unit charge, over that one trip, by whatever pushes the charge along. If a charge $q$ feels the force $\vec{F}$, then
$\epsilon =\oint (\vec{F}/q)\cdot d\vec{l}$

An EMF is not a potential difference. The electric field of charges at rest does no net work on a charge taken round a closed loop, because the charge ends at the potential it started from: that field has zero circulation. So whenever a closed path has an EMF, some other push is at work, and this lesson is about where that push comes from.

Legend

$C$	The closed path, here the circuit, walked in small steps.
$S$	A surface whose rim is $C$; every such surface carries the same flux.
$d\vec{l}$	One small step along $C$, in the sense shown by the arrowheads.	$\mathrm{m}$
$d\vec{A}$	A small patch of $S$ as a vector: its length is the area of the patch, it is perpendicular to the patch, and the right-hand rule picks its side.	${\mathrm{m}}^2$
$\vec{B}$	Magnetic field at the patch.	$\mathrm{T}$
$\theta$	Angle between $\vec{B}$ and $d\vec{A}$.	$\mathrm{rad}$

Figure 1 sets up the geometry. The arrowheads on $C$ give the sense in which the path is walked, and one step $d\vec{l}$ is drawn. To measure the flux we need a surface $S$ whose rim is $C$; the figure uses the flat one. Cut $S$ into small patches and describe each patch by a vector $d\vec{A}$: its length is the area of the patch, and it stands at right angles to the patch. A patch has two sides, and the right-hand rule picks one: curl the fingers of the right hand along $d\vec{l}$, and the thumb gives the direction of $d\vec{A}$. This rule ties the sign of the flux to the sense in which the EMF is counted.

Now look at the patch itself. The field $\vec{B}$ there makes the angle $\theta$ with $d\vec{A}$, so only its component $B\cos \theta$ crosses the patch, and the patch adds $BdA\cos \theta =\vec{B}\cdot d\vec{A}$ to the flux. Adding up all the patches gives the flux through $C$:
$\Phi =\int \vec{B}\cdot d\vec{A}$
Why does the choice of surface not matter? Because $\nabla \cdot \vec{B}=0$: field lines never start or end, so every line that crosses the flat surface also crosses a dome standing on the same rim. Every surface bounded by $C$ carries the same flux, so $\Phi$ belongs to the path itself.

With both quantities defined, Faraday's law says that the EMF is minus the rate of change of the flux. For a path at rest the only push is the electric field, and the law reads
$\oint \vec{E}\cdot d\vec{l}=-\frac{d}{dt}\int \vec{B}\cdot d\vec{A}$
A coil of $N$ turns is $N$ such paths in series, each threaded by the same flux, so its EMF is
$\epsilon =-N\frac{d\Phi }{dt}$
The minus sign is Lenz's law. A positive EMF drives current along $d\vec{l}$, and by the right-hand rule the field of that current threads $S$ along $d\vec{A}$, adding positive flux. A rising flux therefore gets a negative EMF, whose current removes flux and so opposes the rise.

Stokes's theorem turns the circulation of $\vec{E}$ round $C$ into the flux of $\nabla \times \vec{E}$ through $S$. Since the law holds for every loop, however small, it holds point by point:
$\nabla \times \vec{E}=-\partial \vec{B}/\partial t$
A changing magnetic field is surrounded by a curling electric field, whether or not a wire is there to feel it.

Motional EMF: a circuit moving in a steady field

The flux through a circuit can change for two quite different reasons: the circuit moves through a field that stays put, or the field changes while the circuit stays put. Faraday's law covers both with one formula, but the physics behind them differs. Take the moving circuit first, because there the law follows from a force you already know.

Legend

$\vec{B}$	Magnetic field, pointing into the page (the crosses).	$\mathrm{T}$
$\vec{v}$	Velocity of the rod, to the right.	$m/s$
$L$	Length of the rod between the rails.	$\mathrm{m}$
$\vec{v}dt$	How far the rod moves in the short time $dt$.	$\mathrm{m}$
$d\vec{l}$	A small element of the rod, pointing in the sense of the circuit: clockwise, so down the rod.	$\mathrm{m}$
$d\vec{A}$	The strip swept by $d\vec{l}$ in the time $dt$, as an area vector $\vec{v}dt\times d\vec{l}$; it points into the page.	${\mathrm{m}}^2$
$\vec{F}$	Magnetic force $q\vec{v}\times \vec{B}$ on a positive charge $q$ moving with the rod; it points up the rod.	$\mathrm{N}$

Figure 2 shows the simplest moving circuit. A rod of length $L$ lies across two rails joined at the left end, and it slides to the right with velocity $\vec{v}$. The field $\vec{B}$ points into the page everywhere (the crosses). The charges in the rod move with it, so each feels the magnetic force $q\vec{v}\times \vec{B}$; for a positive charge this is the force $\vec{F}$, which points up the rod. The EMF is the work per unit charge of this force round the circuit:
$\epsilon =\oint (\vec{v}\times \vec{B})\cdot d\vec{l}$
Only the rod moves, so only the rod contributes. Along it $\vec{v}\times \vec{B}$ has the size $vB$ and points along the rod, so the size of the EMF is $BLv$.

Now connect this to the flux. In the short time $dt$ the element $d\vec{l}$ moves through $\vec{v}dt$ and sweeps out the shaded strip, whose area vector is $d\vec{A}=\vec{v}dt\times d\vec{l}$; the whole rod sweeps the area $Lvdt$. Because $\nabla \cdot \vec{B}=0$, the flux through the circuit grows by exactly the flux through these swept strips. Dividing by $dt$:
$\frac{d\Phi }{dt}=\oint \vec{B}\cdot (\vec{v}\times d\vec{l})=-\oint (\vec{v}\times \vec{B})\cdot d\vec{l}$
The second step uses the cyclic property of the triple product, $\vec{B}\cdot (\vec{v}\times d\vec{l})=d\vec{l}\cdot (\vec{B}\times \vec{v})$, together with $\vec{B}\times \vec{v}=-\vec{v}\times \vec{B}$. The right-hand side is minus the EMF, so $\epsilon =-d\Phi /dt$. For a circuit moving through a fixed field, the flux rule is not a new law: it is a consequence of the magnetic force on moving charges.

The signs in Figure 2 bear this out. The element $d\vec{l}$ points down the rod, in the sense of travel round the circuit, which is clockwise. By the right-hand rule $d\vec{A}$ then points into the page, along $\vec{B}$, so the flux is positive and growing. The force $\vec{F}$ points up the rod, against $d\vec{l}$, so the EMF is negative: a rising flux and a negative EMF, as Lenz's law requires.

A changing field: the induced electric field

Now hold the circuit still and change the field instead, for example by moving the magnet rather than the coil. Charges at rest feel no magnetic force, so the magnetic force cannot be the push. The push comes from a genuinely new electric field, one whose curl is not zero: $\nabla \times \vec{E}=-\partial \vec{B}/\partial t$. Nothing in the magnetic force law predicts it. It is a separate law of nature, found by experiment.

Yet the two mechanisms always give the same EMF, so only the relative motion of magnet and coil matters. Einstein's 1905 relativity paper opens with exactly this example. In the frame of the magnet, the charges in the coil move and feel the force $q\vec{v}\times \vec{B}$; in the frame of the coil, they are at rest and feel $q\vec{E}$. Relativity shows that these are one field seen from two frames: how it splits into electric and magnetic parts depends on the observer.

The EMF pulse: its height and its area

When the animation's magnet is pushed into the coil, the EMF arrives as a single pulse. Two features of the pulse carry the physics: its height is set by how fast the flux changes, and its area by how much the flux changes in total.

Legend

$t$	Time.	$\mathrm{s}$
$\Phi$	Flux through one turn of the coil.	$\mathrm{Wb}$
${\Phi }_{\max }$	The largest flux, reached once the magnet rests inside the coil.	$\mathrm{Wb}$
$\Delta \Phi$	Total rise of the flux during the push.	$\mathrm{Wb}$
$dt$	A short time interval: the horizontal leg of the slope triangle.	$\mathrm{s}$
$d\Phi$	Change of $\Phi$ along the tangent during $dt$: the vertical leg.	$\mathrm{Wb}$
$\epsilon$	EMF induced in the whole coil of $N$ turns.	$\mathrm{V}$
${\epsilon }_{\max }$	Depth of the pulse, below the steepest point of $\Phi$; the EMF itself is negative there.	$\mathrm{V}$
$N\Delta \Phi$	Size of the shaded area under the pulse: $N$ turns times the rise $\Delta \Phi$.	$\mathrm{V}\cdot \mathrm{s}$

Figure 3 is drawn from the same model as the animation. The upper graph is the flux $\Phi (t)$ through one turn while the magnet is pushed in: it rises by $\Delta \Phi$ to ${\Phi }_{\max }$ and then stays flat. The lower graph is the EMF $\epsilon (t)$, which by Faraday's law is $-N$ times the slope of the upper graph. Look at the dot on the flux curve: there the tangent is steepest, and its slope triangle has the legs $dt$ and $d\Phi$. Follow the dashed line straight down, and the EMF is at the bottom of its pulse, whose depth is ${\epsilon }_{\max }$. Where $\Phi$ is flat, before the magnet arrives and after it comes to rest inside, the EMF is zero, however large the flux.

Start with the height. In the model, the flux through a turn depends on the distance $s$ between the centres of magnet and coil as
$\Phi (s)={\Phi }_{\max }[1+(s/a{)}^2{]}^{-3/2}$
with ${\Phi }_{\max }=0.20\mathrm{mWb}$ and $a=30\mathrm{mm}$. The flux depends on time only through $s$, so the chain rule splits the EMF into a factor set by the shape of the field and a factor set by the speed:
$\epsilon =-N\frac{d\Phi }{ds}\frac{ds}{dt}$
The slope $d\Phi /ds$ is proportional to $(s/a)[1+(s/a{)}^2{]}^{-5/2}$. Setting the derivative of this to zero gives $1+(s/a{)}^2=5(s/a{)}^2$, so the slope is steepest at $s=\pm a/2$, where its size is $0.86{\Phi }_{\max }/a$. While entering, the magnet moves at a steady 78 mm/s, and it passes $s=-15\mathrm{mm}$ at $t\approx 2.2\mathrm{s}$. The depth of the pulse is therefore
${\epsilon }_{\max }=200\cdot 0.86\cdot (0.20\times 10^{-3}/0.030)\cdot 0.078\approx 0.089\mathrm{V}$
which is the $-89\mathrm{mV}$ trough of the graph. The pulse is sharp: it is deeper than half this value for only 0.37 s of the 2 s push.

Now the area. Integrating Faraday's law over the whole pulse gives
$\int \epsilon dt=-N\int d\Phi =-N\Delta \Phi$
The area depends only on the total change of flux, not on how quickly it happened. Here the flux rises from 0.0025 mWb, with the magnet 125 mm away, to 0.20 mWb, so the area is $-200\cdot (0.20-0.0025)\mathrm{mWb}\approx -39.5\mathrm{mV}\cdot \mathrm{s}$ (a volt-second is a weber). Pulling the magnet out gives a pulse of the same area and the opposite sign. A push twice as fast gives a pulse twice as tall and half as wide, with the same area.

The area has a practical use. Through a circuit of resistance $R$ the charge that flows is $Q=\int \epsilon dt/R$, so $Q=N|\Delta \Phi |/R$: about 2.0 mC per pulse with 20 Ω. This stays exact even when the coil's own inductance matters, because the current is zero before and after the pulse, so the inductive EMF integrates to zero. A search coil connected to a charge meter can therefore measure a field whatever the speed of the hand that moves it.

The rotating coil: a generator

A generator changes the flux by rotation: a coil turns steadily in a fixed field. Let it turn at angular speed $\omega$ in a uniform field $B$, so that the angle $\theta$ of Figure 1, between $\vec{B}$ and the coil's area vector, grows as $\theta =\omega t$. Each turn, of area $A$, then carries $\Phi =BA\cos \omega t$, and Faraday's law gives
$\epsilon =-N\frac{d\Phi }{dt}=NBA\omega \sin \omega t$
The EMF is zero when the coil faces the field square on and the flux is largest, because the cosine is flat at its peak. The EMF is largest a quarter turn later, when the coil lies edge-on to the field and the flux passes through zero, changing fastest. Flux and EMF are a quarter cycle apart, like position and velocity in simple harmonic motion.

Spin the animation's coil, with $N=200$ and $A=1.0\times 10^{-3}{\mathrm{m}}^2$, in its 0.20 T field 50 times a second, so that $\omega =2\pi \cdot 50\approx 314rad/s$. The peak EMF is
$NBA\omega =200\cdot 0.20\cdot 1.0\times 10^{-3}\cdot 314\approx 12.6\mathrm{V}$
The average form of the law checks this. In each half turn the flux goes from $+0.20$ to $-0.20\mathrm{mWb}$ in 0.010 s, so the average EMF is $200\cdot 0.40\times 10^{-3}/0.010=8.0\mathrm{V}$. That is $2/\pi$ of the peak, the mean value of a half sine wave, as it should be. The peak is 140 times the 89 mV of the hand-pushed magnet, for two reasons: the flux now reverses completely instead of merely appearing, and it does so a hundred times a second.

Induced currents push back

Once an induced EMF drives a current, that current sits in a magnetic field and feels a force. By Lenz's law the force always opposes the motion that caused it, so induction acts as a brake. The rod on rails makes this quantitative.

Push the rod and let go. At speed $v$ the EMF $BLv$ drives the current $I=BLv/R$ through the circuit's resistance $R$, and the field pushes back on the rod with the force $BIL=B^2{L}^{2}v/R$. Newton's second law for a rod of mass $m$ is then
$m\frac{dv}{dt}=-\frac{B^2{L}^2}{R}v$
The drag is proportional to the speed, so the speed decays exponentially, $v=v_0{e}^{-t/\tau }$, with the time constant $\tau =mR/(B^2{L}^2)$.

Take the rails of the Intermediate example, with $B=0.40\mathrm{T}$, $L=0.25\mathrm{m}$ and $R=0.50\Omega$, and a 20 g rod: $\tau =0.020\cdot 0.50/(0.40^2\cdot 0.25^2)=1.0\mathrm{s}$. Launched at 2.0 m/s, the rod coasts a distance $v_0\tau =2.0\mathrm{m}$ (the integral of $v$ over all time), and all of its 0.040 J of kinetic energy ends up as heat in the resistor.

In solid metal there is no wire to guide the current: the induced electric field drives eddy currents round closed loops inside the metal itself. They heat the pan on an induction hob, brake trains without wear and make a magnet sink slowly down a copper tube. Transformer cores are built from thin insulated sheets, called laminations, which break these loops and cut the losses. A coil also opposes changes in its own current, because the flux of that current threads the coil itself. This self-induction is why breaking the current in a large coil can draw a spark.

Worked problem: a magnet dropped through the coil

Drop the animation's magnet from rest, axis vertical, with its centre 20 cm above the centre of the coil, so that it falls straight through. First leave the coil open: no current flows, so nothing brakes the fall. How do the entry and exit pulses compare? Then close the coil through 20 Ω: how much braking does that add?

Peak heights. Each pulse is deepest where the slope $d\Phi /ds$ is steepest, at $s=\pm 15\mathrm{mm}$, so
$|{\epsilon }_{\max }|\approx N\cdot 0.86({\Phi }_{\max }/a)v=200\cdot 0.86\cdot (0.20\times 10^{-3}/0.030)v\approx 1.15v$
in volts for $v$ in m/s. The centre of the magnet reaches $s=-15\mathrm{mm}$ after falling 0.185 m, at $v=\sqrt{2\cdot 9.81\cdot 0.185}\approx 1.91m/s$, and $s=+15\mathrm{mm}$ after 0.215 m, at about 2.05 m/s. So the entry pulse peaks near $-2.2\mathrm{V}$ and the exit pulse near $+2.35\mathrm{V}$, only 15 ms later.

Widths. Each pulse has the area $N{\Phi }_{\max }\approx 40\mathrm{mV}\cdot \mathrm{s}$, because the flux rises from almost nothing to ${\Phi }_{\max }$ and then falls back. With equal areas, the taller exit pulse must be narrower: area divided by height gives about 18 ms for the entry pulse and 17 ms for the exit pulse. Both pulses are about 25 times taller and 25 times narrower than in the animation, the ratio of the speeds.

Braking. Now close the coil through 20 Ω. The power turned into heat is ${\epsilon }^2/R$, and its integral $\int {\epsilon }^{2}dt/R$, worked out numerically over both pulses, is about 5.8 mJ. A 50 g magnet arrives with $mgh=0.050\cdot 9.81\cdot 0.20\approx 0.098\mathrm{J}$ of kinetic energy, so it loses about 6% of its energy. Its speed goes as the square root of the energy, so it drops by only about 3%: ignoring the braking in the first part was fair. In a copper tube the resistance is tiny, the currents are large, and the braking is strong enough to hold the magnet to a slow terminal speed.

Limits of validity: when the flux rule fails

The flux rule works when the path of the current moves with the material that carries it, as the rod on rails does. A spinning disc shows what goes wrong when it does not, and what to use instead.

Legend

$\vec{B}$	Magnetic field along the axle, pointing out of the page (the dots).	$\mathrm{T}$
$\omega$	Angular speed of the disc.	$rad/s$
$r$	Radius of the disc: the distance from the axle to the rim.	$\mathrm{m}$
$d\rho$	A short piece of the radius, at distance $\rho$ from the axle.	$\mathrm{m}$
$\vec{v}$	Velocity of the metal at a point of the radius, of size $\omega \rho$: it grows in proportion to the distance from the axle.	$m/s$
$\vec{F}$	Magnetic force $q\vec{v}\times \vec{B}$ on a positive charge in the metal: along the radius, away from the axle.	$\mathrm{N}$
$\epsilon$	EMF between the brushes, read by the meter.	$\mathrm{V}$

Figure 4 shows a copper disc of radius $r$ turning at angular speed $\omega$ in a field $\vec{B}$ along its axle, out of the page (the dots). One brush touches the axle and one the rim, and wires lead from both to a meter that reads the EMF $\epsilon$. The current runs through the wires, the meter and one radius of the disc between the brushes. That path stays where it is as the disc turns, so the flux through it never changes, and the flux rule predicts no EMF at all. Yet the meter reads one.

The force law explains the reading. The metal of the disc moves, so the charges in it feel $q\vec{v}\times \vec{B}$. At distance $\rho$ from the axle the metal moves at the speed $\omega \rho$, so in Figure 4 the arrows $\vec{v}$ along the radius grow in proportion to $\rho$, and their tips lie on a straight dashed line through the centre. The force $\vec{F}$ on a positive charge points along the radius, away from the axle (it is drawn on the left-hand radius). A short piece $d\rho$ of the radius therefore adds $B\omega \rho d\rho$ to the EMF, and the sum from the axle to the rim is the area under that straight line, a triangle:
$\epsilon ={\int }_0^{r}B\omega \rho d\rho =\frac{1}{2}B\omega r^2$
For a disc of radius 10 cm turning 50 times a second in 0.50 T this is 0.79 V. The rule that never fails is the force law itself, with the force per unit charge integrated along the actual path of the charges:
$\epsilon =\oint (\vec{E}+\vec{v}\times \vec{B})\cdot d\vec{l}$

The circuit picture has limits of its own. It assumes the EMF appears all round the loop at once, which needs a circuit much smaller than the wavelength $c/f$ of the changing field: 6000 km at 50 Hz, but only 30 cm at 1 GHz, where wires start to radiate as aerials. We have also ignored the coil's own field. For the animation's coil, of length $\ell =60\mathrm{mm}$, the long-solenoid estimate of its inductance, ${\mu }_0{N}^{2}A/\ell$, gives under 1 mH, and with 20 Ω a time constant near 40 µs: negligible against the 0.4 s pulse, but dominant at high frequencies.

How the push splits between $\vec{E}$ and $\vec{v}\times \vec{B}$ depends on the frame of reference. At low speeds, a frame moving at velocity $\vec{v}$ sees the electric field ${\vec{E}}^{\prime }=\vec{E}+\vec{v}\times \vec{B}$; at the 78 mm/s of the animation this rule is off by only about $(v/c{)}^2\approx 7\times 10^{-20}$.

At the other extreme is a superconducting ring. Round a loop with zero resistance the total EMF must be zero, or the current would grow without limit, so the flux through the ring, including the flux of its own current, cannot change at all. That flux also comes only in whole multiples of the flux quantum $h/2e\approx 2.07\times 10^{-15}\mathrm{Wb}$. The coil's 0.20 mWb is about $10^{11}$ quanta; SQUIDs, superconducting quantum interference devices, resolve a small fraction of one quantum, enough to record the magnetic fields of the brain.

Where this leads

Magnetic Force on Moving Charges supplied the $\vec{v}\times \vec{B}$ half of this lesson. Motors, Generators and Transformers applies the law to machines: a transformer is two coils sharing one changing flux, and a motor is a generator run backwards, with the back EMF that Lenz's law predicts. Electric Power and Energy at Home follows that energy to the socket, and the eddy-current brake reappears as a linear damper in Damping and Resonance.

Add Maxwell's correction, that a changing electric field also makes a magnetic field, and the two fields regenerate each other in empty space, travelling at $c=1/\sqrt{{\mu }_0{\epsilon }_0}$: the subject of Electromagnetic Waves. And because only the relative motion of magnet and coil matters, the road leads on to Special Relativity.