Ampere's Law

The integral you would rather not do

In the last lesson you built the field of a current from scratch. Biot-Savart chops the wire into pieces $d\vec{\ell}$, works out the tiny field each piece makes at your point, and adds them up. It always works. It is also brutal: for anything but a straight wire or a loop on its own axis, the integral is a three-line mess of square roots.

Now picture a copper wire two millimetres across carrying ten amps, and ask for the field halfway between the axis and the surface. With Biot-Savart you would have to add up the contributions of every filament of current in that cross-section, at a point that is off-axis for almost all of them. It takes hours. The answer, once you have it, is embarrassingly simple.

That mismatch (vicious integral, simple answer) is a signal. It means the symmetry of the problem is doing most of the work, and there should be a way of using the symmetry directly. That way is Ampere's law. It is to magnetism what Gauss's law is to electrostatics: a statement that is always true, and that becomes a one-line calculation whenever the geometry is symmetric enough.

Circulation, and what the law says

Take any closed path in space: any loop at all, it does not have to be a wire or follow a field line. Walk around it, and at each step multiply the component of $\vec{B}$ along your direction of travel by the length of the step. Add up all those products as you go round once. That sum is written

$\oint \vec{B}\cdot d\vec{\ell}$

and it is called the circulation of $\vec{B}$ around the loop. The circle on the integral sign means the path closes. If the field pushes you forward everywhere, the circulation is large and positive. If the field is always sideways to your path, every term is zero and so is the total.

Ampere's law says the circulation only depends on the current threading the loop:

$\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{enc}$

Here $\mu_0 = 4\pi\times10^{-7}$ T·m/A is the permeability of free space, the same constant that appeared in Biot-Savart, and $I_{enc}$ is the net current passing through any surface bounded by the loop. That is all. The shape of the loop does not matter, the distance to the wires does not matter, and (this is the surprising part) currents outside the loop do not matter.

A current outside your loop certainly makes a field at every point on the loop. But as you walk round, it helps you along one part of the path by exactly as much as it fights you on another, and its net contribution to the circulation is zero. Only threading current survives.

The sign comes from your right hand. Point your fingers the way you walk around the loop; your thumb then points in the positive direction for current. Current threading the other way counts as negative. So if a loop encircles $8.0$ A and $5.0$ A coming out of the page and $3.0$ A going in, traversed counterclockwise, then $I_{enc} = 8.0 + 5.0 - 3.0 = 10.0$ A, and

$\oint \vec{B}\cdot d\vec{\ell} = (4\pi\times10^{-7})(10.0) = 1.26\times10^{-5}\ \text{T·m}$

You never had to know where the wires were inside the loop, or how far apart.

Getting a number out: the straight wire

The law is always true and usually useless. It becomes a tool only when you can find a path on which you already know the shape of $\vec{B}$, so that the integral collapses.

For a long straight wire, argue the shape from symmetry before drawing anything. Rotate the whole setup about the wire's axis and nothing changes, so $B$ can only depend on the distance $r$ from the axis. Slide everything along the axis and nothing changes, so $B$ cannot depend on position along the wire. And Biot-Savart's cross product puts $d\vec{B}$ perpendicular to the current, so the field wraps around the wire in circles.

Now choose your path to match: a circle of radius $r$, centred on the wire, in the plane perpendicular to it. On that circle $\vec{B}$ is everywhere parallel to $d\vec{\ell}$ and everywhere the same size, so the circulation is just $B$ times the length of the path:

$B\,(2\pi r) = \mu_0 I \quad\Longrightarrow\quad B = \frac{\mu_0 I}{2\pi r}$

That took three lines instead of a page. For $I = 5.0$ A at $r = 2.0$ cm:

$B = \frac{(4\pi\times10^{-7})(5.0)}{2\pi(0.020)} = \frac{2\times10^{-7}\times 5.0}{0.020} = 5.0\times10^{-5}\ \text{T}$

The combination $\mu_0/2\pi = 2\times10^{-7}$ is worth memorising; it turns these into mental arithmetic.

Now the wire with thickness. Let the wire have radius $a$ and carry $I$ spread uniformly over its cross-section, and put your circle inside the metal at radius $r < a$. The symmetry argument is unchanged, so the left side is still $B(2\pi r)$. What changes is the right side: your loop no longer catches all the current, only the fraction of the cross-section it encloses,

$I_{enc} = I\,\frac{\pi r^2}{\pi a^2} = I\,\frac{r^2}{a^2}$

so that

$B = \frac{\mu_0 I r}{2\pi a^2} \qquad (r \le a)$

Inside, the field grows straight from zero at the axis. Outside it falls as $1/r$. At $r = a$ the two expressions agree, both giving $\mu_0 I/2\pi a$, which is the largest field anywhere: right at the surface.

That halfway-into-the-copper question from the opening is now one substitution. With $a = 2.0$ mm, $I = 10$ A and $r = 1.0$ mm:

$B = \frac{(4\pi\times10^{-7})(10)(0.0010)}{2\pi(0.0020)^2} = \frac{2\times10^{-7}\times 10 \times 0.0010}{4.0\times10^{-6}} = 5.0\times10^{-4}\ \text{T}$

When the current is not spread evenly

Nothing above required uniform current. The only step that used it was $I_{enc} = I r^2/a^2$. If the current density $J$ varies with radius, replace that step with an integral, and it is a one-dimensional integral over thin rings, not the three-dimensional monster Biot-Savart would have handed you.

A ring of radius $s$ and thickness $ds$ has area $2\pi s\,ds$, so

$I_{enc}(r) = \int_0^r J(s)\, 2\pi s\, ds$

Worked example. A wire of radius $R = 6.0$ mm carries a total of $9.0$ A with density $J(s) = J_0(1 - s/R)$, strongest on the axis and falling to zero at the surface. Find $B$ at $s = R/2 = 3.0$ mm.

First find $J_0$ from the total current, because the problem gave you amps, not a density:

$I = \int_0^R J_0\left(1 - \frac{s}{R}\right)2\pi s\, ds = 2\pi J_0\left(\frac{R^2}{2} - \frac{R^2}{3}\right) = \frac{\pi J_0 R^2}{3}$

So $J_0 = 3I/\pi R^2 = 3(9.0)/[\pi(0.0060)^2] = 27/(1.131\times10^{-4}) = 2.39\times10^{5}$ A/m².

Now the same integral, stopped at $r = R/2$:

$I_{enc} = 2\pi J_0\left(\frac{r^2}{2} - \frac{r^3}{3R}\right) = 2\pi J_0\left(\frac{R^2}{8} - \frac{R^2}{24}\right) = 2\pi J_0 \frac{R^2}{12}$

Compare that with the total, $2\pi J_0 R^2/6$: the inner half of the wire carries exactly half the current, $4.5$ A. Then

$B = \frac{\mu_0 I_{enc}}{2\pi r} = \frac{2\times10^{-7} \times 4.5}{0.0030} = 3.0\times10^{-4}\ \text{T}$

The pattern to take away: find the density, integrate it out to your radius, and only then divide by $2\pi r$. Two-thirds of the errors on problems like this come from skipping the first step and using the total current.

Straight paths: solenoid, sheet, toroid

Circles are not the only useful loops. When the field is straight and uniform, a rectangle is better.

A solenoid is a wire wound in a tight helix, $n$ turns per metre, carrying $I$. Far from the ends, the field inside points along the axis and is uniform; outside a long solenoid it is very nearly zero, because the return field spreads over the whole of space. Draw a rectangle with one side of length $L$ inside along the axis, the opposite side outside, and the two short sides crossing the winding perpendicular to the field.

Go round it. The inside side gives $BL$, since $\vec{B}$ is parallel to the path and constant. The outside side gives zero, since $B \approx 0$ out there. The two short sides give zero, since $\vec{B}$ is perpendicular to $d\vec{\ell}$ along them. So the circulation is $BL$. The rectangle links $nL$ turns, each carrying $I$, so $I_{enc} = nLI$, and

$BL = \mu_0 n L I \quad\Longrightarrow\quad B = \mu_0 n I$

The $L$ cancels, which it had to: the answer cannot depend on how long you chose to draw your rectangle. Note what $B$ does not depend on: the solenoid's radius, and your position inside it. Only turns per metre. A solenoid $0.50$ m long with $800$ turns has $n = 1600$ m⁻¹, so at $1.5$ A, $B = (4\pi\times10^{-7})(1600)(1.5) = 3.0\times10^{-3}$ T.

The same rectangle trick handles an infinite sheet carrying surface current density $K$, measured in amps per metre of width. By symmetry the field is parallel to the sheet, perpendicular to the current, and reverses sign as you cross. Take a rectangle straddling the sheet with two sides of length $L$ parallel to it. Both those sides contribute $BL$, the short sides contribute nothing, and the loop catches $KL$ of current:

$2BL = \mu_0 K L \quad\Longrightarrow\quad B = \frac{\mu_0 K}{2}$

Independent of distance, exactly like the field of an infinite charged plane. For $K = 50$ A/m, $B = (4\pi\times10^{-7})(50)/2 = 3.14\times10^{-5}$ T. A thick slab of thickness $2d$ carrying density $J$ is the same calculation with $K$ replaced by whatever lies inside your loop: at height $y$ within the slab the loop catches $J(2y)L$ per length $L$, giving $B = \mu_0 J y$, and outside it catches everything, giving $B = \mu_0 J d$.

A toroid is a solenoid bent into a doughnut, $N$ turns in total. Its field lines close on themselves inside the windings, so the natural path is a circle of radius $r$ round the doughnut's hole. That circle threads all $N$ turns once each:

$B(2\pi r) = \mu_0 N I \quad\Longrightarrow\quad B = \frac{\mu_0 N I}{2\pi r}$

Unlike the solenoid, this field is not uniform: it is strongest at the inner edge of the cross-section and weakest at the outer edge. Outside the windings, in the hole or beyond the rim, a similar circle encloses zero net current (each turn crosses it twice, in opposite directions, or not at all), and there the field really is zero.

Worked example. A toroid of square cross-section runs from inner radius $a = 0.10$ m to outer radius $b = 0.14$ m, stands $h = 0.040$ m tall, has $N = 800$ turns and carries $I = 5.0$ A. Find the flux through one cross-section.

You cannot write flux as $B \times \text{area}$, because $B$ changes across the face. Strip the square into vertical slivers of height $h$ and width $dr$, over each of which $B$ is effectively constant:

$\Phi = \int_a^b \frac{\mu_0 N I}{2\pi r} h\, dr = \frac{\mu_0 N I h}{2\pi}\ln\!\frac{b}{a}$

Put the numbers in. $\mu_0 N I h/2\pi = (2\times10^{-7})(800)(5.0)(0.040) = 3.2\times10^{-5}$, and $\ln(0.14/0.10) = \ln 1.4 = 0.3365$, so

$\Phi = (3.2\times10^{-5})(0.3365) = 1.08\times10^{-5}\ \text{Wb}$

That logarithm is the fingerprint of a $1/r$ field integrated across a gap, and it will come back when you compute inductance.

Superposing fields: coax and the drilled hole

Ampere's law needs symmetry. Real problems often break it slightly: a hole bored off-centre, two nested conductors with mismatched currents. The fix is almost always the same: write the awkward object as a sum of symmetric objects, solve each with Ampere's law, and add the fields as vectors.

The gentle case is a coaxial cable: a core carrying $I_1$ out and a shield carrying $I_2$ back. A circle at radius $r$ still has the full rotational symmetry, so $B = \mu_0 I_{enc}/2\pi r$ everywhere; all you have to do is count correctly. Between the conductors, $I_{enc} = I_1$. Outside both, $I_{enc} = I_1 - I_2$, which is zero for a properly matched cable and is why coax does not radiate into the room. If only $5.0$ A returns for $7.0$ A sent, then at $r = 8.0$ mm you enclose $2.0$ A and $B = (2\times10^{-7})(2.0)/0.0080 = 5.0\times10^{-5}$ T. If a miswired cable sends $10$ A and $4.0$ A the same way, you enclose $14$ A, and at $r = 5.0$ mm, $B = (2\times10^{-7})(14)/0.0050 = 5.6\times10^{-4}$ T.

Inside the thickness of the shield, from $b$ to $c$, only part of the return current is caught, in proportion to the area:

$I_{enc}(r) = I_1 - I_2\,\frac{r^2 - b^2}{c^2 - b^2}$

Worked example, the hard kind. A long conductor of radius $R$ carries uniform current density $J$ out of the page. A cylindrical hole of any radius is bored parallel to the axis, its centre offset by $\vec{d}$ from the conductor's axis. Find the field inside the hole.

A drilled cylinder has no symmetry, so Ampere's law cannot touch it directly. But a solid cylinder of density $J$ plus a smaller cylinder of density $-J$ filling the hole is exactly the same current distribution: inside the hole the two densities cancel to zero, and everywhere else the full $J$ survives. Each piece is a perfectly symmetric solid cylinder, and each you already know.

For a solid cylinder of density $J$, the field at an interior point a distance $s$ from its own axis is $B = \mu_0 J s/2$, pointing circumferentially. Written as a vector it is $\vec{B} = \tfrac{1}{2}\mu_0 \vec{J}\times\vec{s}$, where $\vec{s}$ runs from that cylinder's axis to the point.

Take a point $P$ inside the hole. Let $\vec{s}_1$ go from the big axis to $P$ and $\vec{s}_2$ from the hole's axis to $P$. Adding:

$\vec{B} = \tfrac{1}{2}\mu_0\vec{J}\times\vec{s}_1 - \tfrac{1}{2}\mu_0\vec{J}\times\vec{s}_2 = \tfrac{1}{2}\mu_0\vec{J}\times(\vec{s}_1 - \vec{s}_2)$

and $\vec{s}_1 - \vec{s}_2$ is just $\vec{d}$, the offset between the two axes. It does not contain $P$ at all. The field everywhere inside the hole is the same vector:

$B = \frac{\mu_0 J d}{2}$

perpendicular to $\vec{d}$. A uniform field inside a bored cavity: a genuinely surprising result, and one that falls out of two lines of superposition. With $J = 2.0\times10^{6}$ A/m² and $d = 1.0$ cm:

$B = \frac{(4\pi\times10^{-7})(2.0\times10^{6})(0.010)}{2} = 1.26\times10^{-2}\ \text{T}$

The same picture answers a harder question. Suppose $R = 1.0$ cm, the hole has radius $R/2$ with its axis $R/2$ from the conductor's axis, and the remaining metal carries $15$ A. What is the field at the original axis? First get $J$, remembering the metal is what is left after drilling:

$J = \frac{I}{\pi R^2 - \pi (R/2)^2} = \frac{4I}{3\pi R^2} = \frac{4(15)}{3\pi(0.010)^2} = 6.37\times10^{4}\ \text{A/m}^2$

At the original axis, $\vec{s}_1 = 0$, so the full cylinder contributes nothing. The negative cylinder contributes $\mu_0 J s_2/2$ with $s_2 = R/2 = 0.0050$ m:

$B = \frac{(4\pi\times10^{-7})(6.37\times10^{4})(0.0050)}{2} = 2.0\times10^{-4}\ \text{T}$

Knowing when to reach for it

Ampere's law is the right tool when the problem hands you one of a very short list of shapes, in a form that hints the answer should be clean.

Look for these words in the statement: long or infinite straight wire, cylindrical, coaxial, solenoid, toroid, wide sheet or slab. Each of those is a claim about symmetry, and it is there to tell you that one well-chosen loop will do the job.

Look for a question asked at a radius or a depth rather than at a named point: "at $r = 3.0$ mm", "1.0 cm from the central plane". That phrasing means the author expects $B$ to depend on one coordinate only, which is exactly the condition that makes the circulation collapse to $B$ times a length.

Look for a current density given as a function, $J(r)$, or a total current on a thick conductor. That is an invitation to integrate $J$ over the enclosed area and nothing more.

And when the geometry is nearly but not quite symmetric (a hole off the axis, an unbalanced shield) do not abandon the law. Split the object into symmetric pieces and add.

If instead you are given a finite segment of wire, a single circular loop, a point off the axis of a coil, or any shape with no repeating symmetry, Ampere's law is still true but it will not give you $B$. Go back to Biot-Savart and integrate.

Everything above quietly assumed the currents were steady. Next that assumption goes, and you will find that a magnetic field which changes in time drives a current all by itself, with no battery anywhere: that is Faraday's law and Lenz's law.