Magnetic Fields of Magnets and Currents

Beginner

Hang a bar magnet from a thread and it settles with one end towards the north. That end is its north pole, the other its south pole. Two like poles push each other away; a north and a south pull together.

A compass needle is just a small magnet on a pivot. Wherever you place it, its north end shows the direction of the magnetic field at that spot. Move it around a magnet and you can trace the field lines: they leave the north pole, curve round and enter the south pole.

Earth is a giant magnet too, with a twist. Since unlike poles attract, the magnetic pole hidden under the Arctic must really be a south pole: that is why it pulls the north end of every compass.

In 1820 Hans Christian Ørsted found that a wire carrying a current also makes a magnetic field. The animation repeats his experiment, seen from above: small compasses sit around an upright wire. With no current they all point north. Switch on, and they swing into a ring around the wire; reverse the current and the ring turns the other way. To remember which way, grip the wire with your right hand, thumb along the current: your fingers curl the way the field goes.

Wind the wire into a coil and the rings add up inside it: you get an electromagnet that you can switch off. Soft iron inside makes it far stronger. Even a fridge magnet works this way, its electrons circling and spinning like tiny currents, all lined up.

Check yourself: a compass sits just north of an upright wire whose current flows upward. Where does its needle point? Answer: west, because seen from above, the field circles anticlockwise.

Intermediate

What a magnetic field is, and how strong

A wire with no magnet in it turns a compass needle once a current flows through it. A magnetic field is what turns a compass needle near a magnet or a current. It has a direction, the way a compass's north end settles, and a size, the magnetic flux density $B$, measured in tesla (T): in 1 T, a wire carrying 1 A across the field is pushed with 1 N on every metre.

A tesla is a lot: a fridge magnet gives a few millitesla and an MRI scanner 1.5–3 T, while Earth's field is only 25–65 µT. A compass lying flat feels only its horizontal part, about 20 µT in mid-latitudes, the green value in the animation.

Field lines run the way a compass points, crowd where the field is strong and never cross; round a bar magnet they leave the north pole and curve into the south pole. Earth's magnetic pole under the Arctic attracts the north end of every compass, so it is a south pole.

The field of a long straight wire

In 1820 Hans Christian Ørsted saw a compass needle swing when he switched on a current in a nearby wire: electric currents make magnetic fields. The field lines of a long straight wire are circles centred on it, so the needle turns sideways, and iron filings on a card round the wire form rings. At a distance $r$ from the wire
$B=\frac{{\mu }_{0}I}{2\pi r}$

where $I$ is the current and ${\mu }_0$ is the magnetic constant,
${\mu }_0=4\pi \times 10^{-7}\mathrm{T}\cdot m/A$
Double $I$ and $B$ doubles; double $r$ and $B$ halves. A handy shortcut is ${\mu }_0/(2\pi )=2\times 10^{-7}\mathrm{T}\cdot m/A$, so that $B=2\times 10^{-7}I/r$ with $I$ in amperes and $r$ in metres.

Which way it goes: reading the diagram

The direction comes from the right-hand grip rule: grip the wire with your right hand, thumb along the current, and your fingers curl the way the field lines go. The rule uses the conventional current $I$; in a metal the electrons drift the other way.

The animation looks down on the card, so the wire is seen end-on: a dot (•) means current out of the screen, like an arrow's tip, and a cross (×) current into it, like its feathers. With $I=20\mathrm{A}$ out of the screen your thumb points at you and the field circles anticlockwise: the blue $B$ arrows on the east side point north. When the current reverses, the arrows flip and every needle swings round; the $B$ readouts, which give only the size of the field, stay the same.

The inner needles, in the stronger field, swing faster than the outer ones, because the twist a field gives a magnet grows with $B$.

Loops and solenoids

Bend the wire into a circular loop of radius $R$ and every part of it sends its field through the middle in the same direction. At the centre
$B=\frac{{\mu }_{0}I}{2R}\text{(centre of a loop)}$
which is $\pi$ times what a straight wire gives at the distance $R$; a flat coil of $N$ turns gives $N$ times as much.

Stack many turns along a tube and you have a solenoid. Inside a long solenoid the fields of the turns add up to a nearly uniform field along the axis, while outside they almost cancel:
$B={\mu }_{0}nI,n=\frac{N}{L}$
Here $n$ is the number of turns per metre, $N$ turns over a length $L$; the radius does not matter if the coil is long compared with its width. The ends act like a bar magnet's poles: curl your right fingers along the current in the turns and your thumb points to the north end. With a soft-iron core inside, the coil becomes an electromagnet.

Worked example: the compasses in the animation

The wire carries $I=20\mathrm{A}$, and the compasses sit 2 cm and 4 cm from it. With the distances in metres:
$B_1=\frac{2\times 10^{-7}\cdot 20}{0.02}=2.0\times 10^{-4}\mathrm{T}=200\mu \mathrm{T}$
$B_2=\frac{2\times 10^{-7}\cdot 20}{0.04}=1.0\times 10^{-4}\mathrm{T}=100\mu \mathrm{T}$

These are the two blue readouts, ten and five times Earth's horizontal 20 µT. A compass, though, shows the total field, the wire's plus Earth's. For the inner compass due north of the wire, the wire's field points west and Earth's north, so the needle turns from the circle towards north by an angle whose tangent is $20/200$: about 5.7°. At 4 cm the ratio is $20/100$, and the outer needles lean towards north by up to about 11°.

Where would the wire's field match Earth's $B_E=20\mu \mathrm{T}$?
$r=\frac{{\mu }_{0}I}{2\pi B_E}=\frac{2\times 10^{-7}\cdot 20}{20\times 10^{-6}}=0.20\mathrm{m}$
With the current out of the screen, the field 20 cm due west of the wire points south and cancels Earth's: a compass there has no preferred direction. Due north, the two fields add at right angles to about 28 µT, and the needle points north-west. Reversing the current moves the cancelling point to the east.

Worked example: winding an electromagnet

You want 10 mT inside a coil 25 cm long, with wire rated for 2 A. Solve $B={\mu }_{0}nI$ for the turns per metre:
$n=\frac{B}{{\mu }_{0}I}=\frac{0.010}{4\pi \times 10^{-7}\cdot 2}\approx 4.0\times 10^3{\mathrm{m}}^{-1}$
That is about 4000 turns per metre, or $N=nL=4000\cdot 0.25=1000$ turns, and $B=4\pi \times 10^{-7}\cdot 4000\cdot 2\approx 1.0\times 10^{-2}\mathrm{T}$, about 500 times Earth's horizontal field.

On a tube of radius 2 cm the coil uses $1000\cdot 2\pi \cdot 0.02\approx 126\mathrm{m}$ of wire. In copper of cross-section 0.5 mm² its resistance is $\rho \ell /A=1.7\times 10^{-8}\cdot 126/(0.5\times 10^{-6})\approx 4.3\Omega$, so 2 A needs about 8.5 V and turns 17 W into heat.

A soft-iron core lines up its atomic magnets with the coil's field and multiplies it many times: to a few tenths of a tesla in a straight rod, and to nearly 2 T, where iron saturates, in a closed iron path such as a lifting magnet's.

Common mistakes

• Field lines of a wire are circles, not spokes: a compass 2 cm north of a wire carrying 20 A upwards points almost west.
• The field of a long wire falls as $1/r$, not $1/r^2$: 100 µT at 4 cm becomes 50 µT at 8 cm, not 25 µT.
• Distances go in metres: $r=2$ instead of 0.02 gives 2 µT instead of 200 µT. Likewise $n$ counts turns per metre: 1000 turns on 25 cm is $n=4000{\mathrm{m}}^{-1}$, not 1000.
• A compass shows the total field, not the wire's alone: 20 cm from the animation's wire the two are equal, and the needles stop following the circles.
• A mains cable carrying 10 A leaves a compass alone: its two conductors carry opposite currents a few millimetres apart, leaving about 1 µT at 10 cm instead of 20 µT for one alone, and the current reverses 100 times a second.

Where you meet it

Electromagnets lift scrap steel, hold fire doors open and switch relays. A loudspeaker drives its cone with a coil in a magnet's field, and an MRI scanner is a large solenoid: 1.5 T needs $nI\approx 1.2\times 10^{6}A/m$, say 2000 turns per metre carrying 600 A, so its coil is cooled until it superconducts.

Next, Magnetic Force on Moving Charges shows what a field does to a current, Electromagnetic Induction how a changing field drives one, and Motors, Generators and Transformers uses both.

Advanced

Two local laws for steady currents

Every steady magnetic field obeys two rules. Everything else in the lesson, from the straight wire to the coaxial cable, follows from them.

For currents that do not change in time, the magnetic field $\vec{B}$ satisfies
$\nabla \cdot \vec{B}=0,\nabla \times \vec{B}={\mu }_0\vec{J}$
Here $\vec{J}$ is the current density, the current per unit area of cross-section (A/m²), and ${\mu }_0=4\pi \times 10^{-7}\mathrm{T}\cdot m/A$ is the magnetic constant. The divergence $\nabla \cdot \vec{B}$ measures how much field flows out of a tiny volume; the curl $\nabla \times \vec{B}$ measures how much it circulates round a tiny loop.

The first law says that no net flux of $\vec{B}$ leaves any closed surface. Field lines therefore never start or end, because there are no magnetic charges. The lines of $\vec{E}$ in Electric Field and Potential, by contrast, start and end on charges.

The second law says that $\vec{B}$ circulates round currents, and only round currents. Where $\vec{J}=0$, the field has no curl at all.

Ampère's law: circulation counts current

The curl law holds point by point. To use it on a real wire we need a form that adds up the circulation round a whole loop. Integrate the curl law over any surface bounded by a closed loop $C$: Stokes' theorem turns the left side into a line integral round $C$, and the right side becomes ${\mu }_0$ times the current $I_{\mathrm{enc}}$ that crosses the surface. The result is Ampère's law:
$\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\mathrm{enc}}$

Legend

$I$	Current in the wire, upwards. It crosses $S_1$ and $S_2$ alike, so it is the $I_{\mathrm{enc}}$ of Ampère's law.	$\mathrm{A}$
$C$	Closed loop round which $\vec{B}$ is added up.
$S_1$	Flat disc bounded by $C$.
$S_2$	Dome bounded by the same loop $C$; the wire pierces it too.
$d\vec{l}$	Short step along $C$, in the sense the right-hand fingers curl when the thumb points along $I$.	$\mathrm{m}$
$\vec{B}$	Magnetic field on $C$; it circles the wire in the same sense as the steps.	$\mathrm{T}$

In Figure 1 the loop $C$ is cut into small steps $d\vec{l}$. At each step, take the part of $\vec{B}$ along the step and multiply it by the step's length. The left side of the law is the sum of these products all the way round $C$.

The direction of travel fixes the sign. Point your right thumb along a current: if your fingers curl the way the steps go, that current counts as positive. Here the wire carries $I$ upwards, so the steps run anticlockwise as seen from above. The field $\vec{B}$ circles the wire the same way, so every product is positive.

$I_{\mathrm{enc}}$ is the current crossing a surface bounded by $C$. The obvious choice is the flat disc $S_1$. But the dashed dome $S_2$ has the same edge $C$, and the law does not say which surface to use, so both must give the same answer. In Figure 1 they do: the wire pierces $S_1$ and $S_2$ alike, so each is crossed by $I$.

This agreement is not automatic. The divergence of any curl is zero, so taking the divergence of the curl law gives $\nabla \cdot \vec{J}=0$: charge never piles up, current flows round closed circuits, and every surface spanning $C$ carries the same current. A charging capacitor breaks this. A dome that passes between its plates is crossed by no current, while the flat disc is crossed by the wire. That is the first sign that changing fields need an extra term, met at the end of this lesson.

The straight wire, outside and inside

Ampère's law gives $B$ in one line, but only when symmetry tells us in advance that the field is the same all round the loop. The long straight wire is the cleanest case, and it shows how the symmetry argument goes.

A long straight wire looks the same after you slide along it or turn about it. So each part of $\vec{B}$ can depend only on the distance $r$ from the wire. Split $\vec{B}$ into three parts: radial (pointing away from the wire), along the wire, and circling it.

The radial part is zero. Take a closed cylinder of radius $r$ round the wire. Its two flat ends cancel, because the field along the wire is the same at both, entering one as it leaves the other. A radial part would then leave the curved side everywhere and give a net flux, which $\nabla \cdot \vec{B}=0$ forbids.

The part along the wire is zero outside it. Take a rectangle in a plane through the wire, with both long sides outside the wire and on the same side of it. It encloses no current. Its short sides are radial, where the field has no part, so the two long sides must cancel: the field along the wire is the same at every distance. Far away it must vanish, so it vanishes everywhere outside.

What remains circles the wire, with a size $B$ that depends only on $r$. Round a circle of radius $r$ centred on the wire, $\vec{B}$ always points along the step, so the left side of Ampère's law is $B\cdot 2\pi r$. Setting this equal to ${\mu }_{0}I$ gives
$B=\frac{{\mu }_{0}I}{2\pi r}$

Inside a wire of radius $a$ that carries its current uniformly, a circle of radius $r<a$ encloses only part of the current. That part is the ratio of the areas, $\pi r^2/(\pi a^2)=r^2/a^2$. So $B\cdot 2\pi r={\mu }_{0}I{r}^2/a^2$, and
$B=\frac{{\mu }_{0}Ir}{2\pi a^2}$
The field is zero on the axis and rises linearly to ${\mu }_{0}I/(2\pi a)$ at the surface: 4.0 mT for 20 A in a wire of radius 1 mm. Outside, it falls as $1/r$. Figure 4, further down, draws Ampère circles in both regions.

The solenoid: a uniform field inside

A solenoid is a long coil of wire, the standard way to make a uniform field. Ampère's law shows why with a single rectangle that lies half inside the coil and half outside.

Legend

$C$	Rectangular loop for Ampère's law, run anticlockwise.
$\ell$	Length of the two sides of $C$ parallel to the axis.	$\mathrm{m}$
$I$	Current in each turn: out of the page at the top, into it at the bottom.	$\mathrm{A}$
$B$	Field inside the solenoid, along the axis; it is zero outside.	$\mathrm{T}$
$n\ell$	Number of turns passing through $C$; $n$ is the number of turns per metre.

Figure 2 cuts a long solenoid along its axis. In the top row the current comes out of the page (dots); in the bottom row it goes into the page (crosses). Curl the fingers of your right hand along the current in the turns, and your thumb gives the field inside: along the axis, to the right, as the arrows $B$ show.

Outside the coil the field is zero. A rectangle lying wholly outside encloses no current, so its two long sides, parallel to the axis, must cancel: the field along the axis is the same at every distance outside. Far from the coil it must die away, so it is zero everywhere outside. That is the $B=0$ below the coil.

Now follow the rectangle $C$ of Figure 2 round. Its inner side, of length $\ell$, runs along the axis inside the coil, in the direction of $B$, and contributes $B\ell$. Its outer side lies where $B=0$ and contributes nothing. Its two short sides cross the axial field at right angles inside the coil and lie in zero field outside, so they add nothing either. The left side of Ampère's law is just $B\ell$.

On the right side, $C$ is threaded by $n\ell$ turns, where $n$ is the number of turns per metre, and each carries $I$. So $I_{\mathrm{enc}}=n\ell I$, and $B\ell ={\mu }_{0}n\ell I$ gives
$B={\mu }_{0}nI$
Nothing in this depends on how far the inner side is from the axis, so the field is the same everywhere inside: uniform. It depends on the turns per metre and the current, not on the radius of the coil.

The B(r) graph and circles of equal flux

At the Advanced level the animation plots the field of the 20 A wire, $B$ against $r$ from 1 to 6 cm. This section reads that graph and shows why Ampère's law fixes its shape.

The curve is the hyperbola $Br={\mu }_{0}I/(2\pi )=4.0\times 10^{-6}\mathrm{T}\cdot \mathrm{m}$, or 400 µT·cm. It passes through 400 µT at 1 cm, the marked 200 µT at 2 cm, 100 µT at 4 cm and 67 µT at 6 cm. On log–log axes it would be a straight line of slope −1, against −2 for a point charge and −3 for a distant magnet.

The constant product $Br$ is exactly what Ampère's law demands. Take a loop that encloses no current: out along a radius from 2 to 4 cm, round the 4 cm circle through an angle $\Delta \phi$, in along another radius, and back round the 2 cm circle. On the radial parts $\vec{B}$ is at right angles to the path, so they contribute nothing. The outer arc runs with the field and gives $100\mu \mathrm{T}\cdot 4\mathrm{cm}\cdot \Delta \phi$; the inner arc runs against it and gives $-200\mu \mathrm{T}\cdot 2\mathrm{cm}\cdot \Delta \phi$. The two cancel, as they must with no current inside. A field falling faster or slower than $1/r$ would fail this test.

The flux through a strip in a plane through the wire, one metre long and reaching from $r_1$ to $r_2$, is the sum of $Bdr$ across it:
$\Phi =\frac{{\mu }_{0}I}{2\pi }\ln \frac{r_2}{r_1}$
It depends only on the ratio $r_2/r_1$, so every doubling of $r$ adds the same $2.8\times 10^{-6}\mathrm{T}\cdot {\mathrm{m}}^2$. Field lines drawn at equal steps of flux are therefore circles of radius 1, 2, 4 and 8 cm, spreading out as the field weakens; the two dashed circles of the animation are neighbours in such a set.

The Biot–Savart law: a loop seen from its axis

Ampère's law gives $B$ only when symmetry keeps the field constant along some loop. Round a single circular coil no such loop exists, so we need a law that works for a circuit of any shape. The Biot–Savart law builds the field from short pieces of current:
$d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2}$
Here $d\vec{l}$ is a short element of wire pointing along the current, $r$ is its distance from the field point and $\hat{r}$ is the unit vector from the element towards that point. Each element adds a contribution that falls as $1/r^2$ and points at right angles to both the current and the line of sight.

Legend

$I$	Current in the loop, anticlockwise as seen from above.	$\mathrm{A}$
$R$	Radius of the loop.	$\mathrm{m}$
$z$	Height of $P$ above the centre of the loop, along the axis.	$\mathrm{m}$
$P$	Point on the axis where the field is found.
$r$	Distance from the element $d\vec{l}$ to $P$, equal to $\sqrt{R^2+z^2}$.	$\mathrm{m}$
$\hat{r}$	Unit vector from the element towards $P$.
$d\vec{l}$	Short element of the loop, on the right where the current runs into the page; drawn as up in this tilted view.	$\mathrm{m}$
$d\vec{B}$	Field of that element at $P$, at right angles to $d\vec{l}$ and to $\hat{r}$.	$\mathrm{T}$
$d{B}_z$	Axial part of $d\vec{B}$; the axial parts of all elements point up and add.	$\mathrm{T}$

Figure 3 applies the law to a circular loop of radius $R$ carrying $I$, at a point $P$ on its axis a height $z$ above the centre. Look first at the element $d\vec{l}$ on the right-hand side of the ring. There the current runs into the page; in this tilted view, into the page is drawn as up.

Two facts make this element easy. Its distance from $P$ is $r=\sqrt{R^2+z^2}$, by Pythagoras on the triangle with sides $R$ and $z$. And $d\vec{l}$, which runs along the ring, is at right angles to the line $r$, so the cross product has its full size: $dB={\mu }_{0}Idl/(4\pi r^2)$. Its direction is at right angles to both $d\vec{l}$ and $\hat{r}$: up and away from the axis, as the arrow $d\vec{B}$ shows.

Now look at the grey element on the opposite side of the ring. Its current runs the other way, so its field is the mirror image: up and away from the axis on its own side. The sideways parts of the two cancel. Pairing every element with its opposite, all the sideways parts cancel, and only the axial parts $d{B}_z$ survive, all pointing up.

How big is $d{B}_z$? The arrow $d\vec{B}$ is at right angles to the line $r$, so the small triangle formed by $d\vec{B}$ and $d{B}_z$ is similar to the large triangle with sides $R$, $z$ and $r$. Hence $d{B}_z=dBR/r$.

Every element is at the same distance $r$ and gives the same ratio $R/r$, and the element lengths add up to the circumference $2\pi R$. The sum is therefore just a product:
$B=\frac{{\mu }_{0}I}{4\pi r^2}\cdot \frac{R}{r}\cdot 2\pi R=\frac{{\mu }_{0}I{R}^2}{2(R^2+z^2{)}^{3/2}}$

Two limits check the result. At the centre, $z=0$, so $r=R$ and every element is side-on, and $B={\mu }_{0}I\cdot 2\pi R/(4\pi R^2)={\mu }_{0}I/(2R)$. For 20 A in a loop of radius 2 cm that is 628 µT, $\pi$ times the 200 µT of a straight wire at the same 2 cm.

Far along the axis, for $z\gg R$, the bracket becomes $z^3$ and $B\approx {\mu }_{0}I{R}^2/(2z^3)$: the field falls as $1/z^3$. At 20 cm the same loop gives only 0.62 µT, where a straight wire still gives 20 µT. A small loop is a magnetic dipole, with the same far field as a bar magnet. Every real circuit, return wires included, is a closed loop, so far enough away its field also falls as the cube of the distance.

Finite wires and short solenoids

The same sum handles straight pieces of wire and coils of finite length, and shows how good the ideal formulas are.

A straight segment of length $\ell$, seen from a point at perpendicular distance $r$ opposite its midpoint, gives
$B=\frac{{\mu }_{0}I}{2\pi r}\cdot \frac{\ell /2}{\sqrt{(\ell /2{)}^2+r^2}}$
The first factor is the infinite-wire value; the second, always less than 1, is the correction for the missing ends. For a 1 m wire seen from 4 cm it is 0.9968, so the field falls only 0.3 % short.

For a solenoid six radii long, like the animation's, adding up the fields of its turns gives 95 % of ${\mu }_{0}nI$ at the centre and 49 % at the ends, where the computed field lines fan out. The ideal $B={\mu }_{0}nI$ is the limit of a very long coil.

Worked problem: a coaxial return path

A coaxial cable sends a current out along a central rod and brings it back along a tube round it. It is a good test of Ampère's law, because the enclosed current changes from one circle to the next.

Let the animation's 20 A flow out of the screen along a copper rod of radius $a=1\mathrm{cm}$ and back into it through a thin tube of radius $b=3\mathrm{cm}$ round the rod. What do the compasses at 2 cm and 4 cm feel, and what is the field 0.5 cm from the axis?

Legend

$I$	Current in the rod, out of the page; the tube carries the same current back.	$\mathrm{A}$
$a$	Radius of the rod.	$\mathrm{m}$
$b$	Radius of the tube.	$\mathrm{m}$
$B$	Size of the field on a circle of radius $r$ round the axis.	$\mathrm{T}$
$r$	Distance from the axis.	$\mathrm{m}$

The left half of Figure 4 is the cross-section. The dot at the centre is the current $I$ coming out of the page along the rod of radius $a$; the crosses are its return into the page along the tube of radius $b$. The dashed circles 1, 2 and 3 are Ampère loops at 0.5 cm, 2 cm and 4 cm. The drawing squeezes the radii to make room for the labels, so it is not to scale.

The cable looks the same after turning about its axis or sliding along it, just like the straight wire, so the field again circles the axis with a size that depends only on $r$. On each circle, Ampère's law gives $B\cdot 2\pi r={\mu }_0{I}_{\mathrm{enc}}$, or $B=({\mu }_0/2\pi )I_{\mathrm{enc}}/r$ with ${\mu }_0/(2\pi )=2\times 10^{-7}\mathrm{T}\cdot m/A$. The only work is finding $I_{\mathrm{enc}}$.

Circle 1, inside the rod, encloses $(0.5/1{)}^2$, a quarter, of the rod's area and so a quarter of its current. Circle 2 encloses all of the rod's current and none of the tube's. Circle 3 encloses both, and they cancel:
$I_{\mathrm{enc}}(0.5\mathrm{cm})=20\cdot (0.5/1{)}^2=5\mathrm{A},B=\frac{2\times 10^{-7}\cdot 5}{0.005}=2.0\times 10^{-4}\mathrm{T}$
$I_{\mathrm{enc}}(2\mathrm{cm})=20\mathrm{A},B=200\mu \mathrm{T}$
$I_{\mathrm{enc}}(4\mathrm{cm})=20-20=0,B=0$

The right half of Figure 4 plots $B$ against $r$ for the whole cable. The field climbs in a straight line to a peak of 400 µT at $r=a$, follows the animation's $1/r$ hyperbola out to $r=b$, then drops to zero. Circles 1 and 2 sit at the same height: 200 µT inside the rod at 0.5 cm, and 200 µT outside it at 2 cm.

So the inner compasses, at 2 cm, notice nothing new: the tube's current lies outside their circle, and a uniform tube makes no field inside itself. The outer compasses, at 4 cm, feel only Earth's horizontal field of about 20 µT and point north as if the current were off.

If 5 A leaked to earth elsewhere, so that only 15 A came back through the tube, circle 3 would enclose 5 A. The outer compasses would then feel $2\times 10^{-7}\cdot 5/0.04=25\mu \mathrm{T}$ from the cable, more than Earth's field. A residual-current device uses exactly this: a ring round both conductors senses only their net current, the $I_{\mathrm{enc}}$ of Ampère's law, and cuts off the circuit when it is not zero.

One field seen from two frames

Magnetism is not a separate force. This section shows that the magnetic force of a wire is an electric force seen from another frame.

Let the wire be copper with a cross-section of 2.5 mm². Its mobile electrons carry about $3.4\times 10^4\mathrm{C}$ per metre of wire, balanced exactly by the positive ions, so the wire is neutral. To carry 20 A the electrons drift at only $v\approx 0.59mm/s$.

Now let a charge move alongside the wire, parallel to the electrons and at their speed $v$. In the lab it feels a magnetic force, because it moves through the field $B$. In its own frame it is at rest, and a magnetic field exerts no force on a charge at rest. The force cannot vanish by a change of viewpoint, so in that frame it must be electric.

It is. In the charge's frame the ions move, so their spacing is length-contracted and they are more crowded than in the lab. The electrons are now at rest, at their own natural spacing, which is wider than the contracted spacing seen in the lab, so they are less crowded. The wire is no longer neutral.

The two densities differ by a fraction $v^2/c^2\approx 4\times 10^{-24}$: a net charge of less than one electron per metre of wire. Yet at 2 cm this tiny charge makes an electric field of $1.2\times 10^{-7}V/m$, the same as the $vB$ of a charge moving at $v$ through 200 µT. Electric and magnetic fields are two aspects of one field, mixed by a change of frame, with ${\mu }_0{\epsilon }_0=1/c^2$ as the rate of exchange.

The right-hand rule is only a convention. $\vec{B}$ appears both in the field law and in the force $q\vec{v}\times \vec{B}$, so a left-hand rule used in both would reverse $\vec{B}$ and change no prediction.

Permanent magnets obey the same equations. Their currents are inside atoms, above all the spins of electrons, lined up across whole domains in iron.

Limits of validity

These are the laws of steady currents. When currents change, Maxwell's term ${\mu }_0{\epsilon }_0\partial \vec{E}/\partial t$ joins ${\mu }_0\vec{J}$ in the curl law. It mends the charging capacitor of Figure 1, and it makes changes of the field spread at $c=1/\sqrt{{\mu }_0{\epsilon }_0}\approx 3.00\times 10^{8}m/s$.

The static formulas still hold at distances much smaller than the wavelength. At 50 Hz the wavelength is 6000 km, so household wiring is magnetostatic; at 100 MHz it is 3 m, and a wire radiates.

The laws also assume empty space or air, whose permeability differs from ${\mu }_0$ by less than one part per million. The strongest steady laboratory fields are about 45 T, and magnetars, a kind of neutron star, reach about $10^{10}\mathrm{T}$.

The magnetism of matter is quantum at heart. Classical physics predicts no magnetisation at all for charges in thermal equilibrium, and the alignment of spins in iron is a quantum effect.

Where this leads

Magnetic Force on Moving Charges adds the force $q\vec{v}\times \vec{B}$, the attraction between parallel currents and the circular orbits of charged particles. Electromagnetic Induction shows that a changing $\vec{B}$ drives an electric field round a loop, the counterpart of Ampère's law, and Motors, Generators and Transformers uses coils and iron cores. With Maxwell's term, the two curl laws give the waves of Electromagnetic Waves, and the change of frame above is developed in Special Relativity.