A solenoid bent into a ring
Bend a long solenoid round until its two ends meet, and you get a toroid. It is a ring wound with turns, like a doughnut wrapped in wire. The field of a solenoid escapes at its ends. A toroid has no ends, so its field has nowhere to escape. The lines run round and round inside the ring.
Look down on the ring. Each turn comes up out of the page at the inner edge, runs across the top and goes down at the outer edge. By symmetry, the field lines are circles about the centre of the ring. So Ampere's law needs circles, and the only question is what each circle encloses.
Grow the circle in the figure above. In the hole, the circle encloses no wire at all, so \(B = 0\). Inside the core, it encloses every turn once, at the inner edge. The enclosed current is \(NI\), and \(B\times 2\pi r = \mu_0 NI\). Outside the ring, every turn threads the circle twice, once up and once down. The net current is zero, and so is \(B\).
A toroid of \(500\) turns carries \(2\,\text{A}\). Its core runs from \(r = 0.09\,\text{m}\) to \(0.11\,\text{m}\), with a mean radius of \(0.1\,\text{m}\). Find the enclosed current and the field on a circle of radius \(0.1\,\text{m}\), and the field at the inner and outer edges of the core.
A circle of radius \(r\) inside the core is threaded by all \(500\) turns, each carrying \(2\,\text{A}\) the same way.
\(I_{enc} = NI\), and \(B\times 2\pi r = \mu_0 NI\).
\(I_{enc} = 500\times 2 = 1000\,\text{A}\). At \(r = 0.1\,\text{m}\), \(B = 4\pi\times10^{-7}\times 1000/(2\pi\times 0.1) = 2\times10^{-3}\,\text{T}\). At \(0.09\,\text{m}\), \(2.22\times10^{-3}\,\text{T}\); at \(0.11\,\text{m}\), \(1.82\times10^{-3}\,\text{T}\).
The field is \(2\times10^{-3}\,\text{T}\) on the mean circle. It is about \(10\) % stronger at the inner edge than at the mean circle, and \(10\) % weaker at the outer edge.
Values used: \(\mu_0 = 4\pi\times10^{-7}\,\text{T m/A}\), so \(\mu_0/2\pi = 2\times10^{-7}\,\text{T m/A}\).
The field in a toroid is not quite uniform across the core. It falls as \(1/r\) from the inner edge to the outer edge, because the same \(NI\) is spread over a longer circle. When the core is thin compared with its radius, \(r\) hardly changes across it. The turns per metre along the core are then \(n = N/2\pi r\), and \(B = \mu_0 NI/2\pi r = \mu_0 nI\). A thin toroid is a solenoid bent round, with its field kept inside.
A toroid of \(1000\) turns carries \(1.5\,\text{A}\). What is the field inside its core at a distance of \(0.2\,\text{m}\) from the centre of the ring, in millitesla?
A thin toroid of mean radius \(5\,\text{cm}\) has \(2000\) turns. What current gives a field of \(0.01\,\text{T}\) in its core, in amperes?
What is the magnetic field of an ideal toroid in the hole at its centre, and just outside the ring?
Where the shortcut stops
Ampere's law is true for every steady current and every loop. But it hands you only one number, the integral of \(B\) round the loop. To turn that number into \(B\) at a point, \(B\) must have the same size all round the loop and point along it. Or it must be zero or perpendicular to the path on the parts where it is not.
A single circular loop has no such path. Its field lines are the closed curves of the last lessons, and \(B\) changes along every one of them. The law still holds on each curve, but it cannot tell how the total is shared out along it.
A finite straight wire looks symmetric, but it is only a piece of a circuit. The current must come to it and leave it through other wires, and their fields spoil the symmetry. Ampere's law holds for the whole circuit, not for the piece. For a piece, the Biot-Savart law is the tool, as in the second lesson.
There is one place where the law itself fails as written. Draw a loop round the wire that leads to a charging capacitor. Stretch a surface across the loop so that it passes through the gap between the plates. No current crosses that surface, yet the field on the loop is not zero. James Clerk Maxwell repaired the law with a second term, the displacement current. The chapter Transformers, LC Oscillations and Electromagnetic Waves takes it up.
Ampere's law is true for every loop. Why can it not be used directly to find the field of a single circular loop of current at a point on its axis?
The law gives only \(\oint\vec B\cdot d\vec l\) round a closed path. To get \(B\) at a point, one needs a path on which \(B\) has the same size everywhere and points along the path, so that the integral is \(B\) times a length. The field lines of a single loop are closed curves along which \(B\) changes from point to point. No such path exists, so the field must be found from the Biot-Savart law.
Choosing the tool
Every problem of this chapter is one of two kinds. When the current has enough symmetry, a single Amperian loop gives the field in one line. When it does not, the field is built piece by piece: known straight wires, arcs and loops are added as vectors. A new shape is integrated with the Biot-Savart law.
For which of these can Ampere's law alone give the magnetic field?
A tokamak is a machine that tries to make energy by fusing hydrogen, as the Sun does. The gas is heated to many millions of degrees and must never touch a wall. So it is held in a ring-shaped vacuum chamber inside a set of large toroidal coils. Their field runs round the ring, as in this lesson, and the charged particles spiral along its lines. India's first tokamak, Aditya, was built at the Institute for Plasma Research near Gandhinagar. India is also a partner in ITER, the large international tokamak being built in France.
- Using the outer radius or the mean radius for every point of a toroid's core.\(B = \mu_0 NI/2\pi r\) varies as \(1/r\) across the core; use the \(r\) of the point.
- Expecting a field in the hole of a toroid.A circle in the hole encloses no current, so \(B = 0\) there.
- Using Ampere's law on a finite wire or a single loop.No loop there has constant, tangential \(B\). Use Biot-Savart, or add known pieces.
- Taking the enclosed current of a toroid as \(I\).A circle in the core is threaded by all \(N\) turns, so \(I_{enc} = NI\).