Ampere's Circuital Law

Going round a closed loop

The Biot-Savart law gives the field of any wire, but often at the cost of an integral. For shapes with strong symmetry there is a shortcut. It does not look at one point. It looks at a whole closed path round the currents.

Draw any closed loop in space and choose a sense to go round it. Cut the loop into small steps \(d\vec l\), each along the path. At each step take \(\vec B\cdot d\vec l = B\,dl\cos\theta\), where \(\theta\) is the angle between the field and the step. Add these round the whole loop. The total is the line integral of \(\vec B\) round the loop, written \(\oint\vec B\cdot d\vec l\).

Try it on a circle of radius \(r\) round a long straight wire, gone round in the sense of the field lines. The field is \(\mu_0 I/2\pi r\) at every point and points along the circle, so \(\cos\theta = 1\). The sum is \(B\times 2\pi r = \mu_0 I\). The radius has dropped out. A bigger circle has a weaker field but a longer path, and the product stays \(\mu_0 I\).

That result is not a lucky accident of circles. It holds for every closed loop round the wire, of any shape. André-Marie Ampère's study of currents led to the general law, which carries his name.

The loop chosen for the law is called an Amperian loop. The current that threads it is the enclosed current, the current through any surface spanning the loop. Currents outside the loop still make field on the loop. But they add as much to the sum on one side of the loop as they take away on the other. So they drop out of the total.

In Fig 4.1 the right hand, curled anticlockwise, has its thumb out of the page. The \(5\,\text{A}\) coming out counts \(+5\,\text{A}\) and the \(3\,\text{A}\) going in counts \(-3\,\text{A}\). So \(\oint\vec B\cdot d\vec l = \mu_0\times 2\,\text{A} = 2.51\times10^{-6}\,\text{T m}\). Go round the other way, and every sign flips: both sides of the law change sign together.

The long wire, in one line

Use the law the other way now. Start from the symmetry of a long straight wire and find its field. Turn the wire about its own axis, or slide it along itself. Nothing changes, so the field cannot change either. Its size can depend only on the distance \(r\) from the wire, and its lines must be circles round it.

Take the Amperian loop to be one of those circles, of radius \(r\). On it \(B\) is constant and along the path. So the integral is \(B\times 2\pi r\), and the law gives \(B\times 2\pi r = \mu_0 I\), that is \(B = \mu_0 I/2\pi r\). Two lessons of Biot-Savart work have become one line.

Inside a thick wire

A real wire has a radius \(R\), and the current flows through its whole cross-section. Outside the wire, a circle of radius \(r > R\) encloses all of \(I\). The field there is \(\mu_0 I/2\pi r\), as if all the current flowed along the axis.

Inside, a circle of radius \(r < R\) encloses only the current through its own area. Let the current be spread evenly, with a uniform current density \(J = I/\pi R^2\). The circle encloses \(J\pi r^2 = Ir^2/R^2\). Then \(B\times 2\pi r = \mu_0 I r^2/R^2\).

The field is zero on the axis and rises in proportion to \(r\). It peaks at the surface, where the two formulas agree at \(\mu_0 I/2\pi R\). Beyond the surface it falls as \(1/r\). For \(R = 0.02\,\text{m}\) and \(20\,\text{A}\): \(10^{-4}\,\text{T}\) at \(0.01\,\text{m}\), \(2\times10^{-4}\,\text{T}\) at the surface, \(10^{-4}\,\text{T}\) at \(0.04\,\text{m}\).

Now let the current crowd towards the surface, with a density \(J = J_0 r/R\) that grows with the distance from the axis. A thin ring of radius \(r'\) and width \(dr'\) carries \(J\times 2\pi r'\,dr'\). Adding the rings from \(0\) to \(r\) gives a current proportional to \(r^3\). So a circle of radius \(r\) encloses \(Ir^3/R^3\), and \(B = \mu_0 Ir^2/2\pi R^3\) inside.

Hollow cylinders and coaxial cables

A long thin-walled tube carries a current along its length. By symmetry its field lines are circles about the axis, as before. A circle inside the hollow encloses no current, so \(B\times 2\pi r = 0\). The field inside the tube is zero at every point. Outside, the circle encloses all of \(I\), and \(B = \mu_0 I/2\pi r\).

A thick tube, with radii \(a\) and \(b\), needs the enclosed part of its wall. For a circle of radius \(r\) between \(a\) and \(b\), the enclosed fraction is the area \(\pi(r^2 - a^2)\) over \(\pi(b^2 - a^2)\). A tube of radii \(0.01\,\text{m}\) and \(0.02\,\text{m}\) carrying \(30\,\text{A}\) encloses \(30\times(2.25 - 1)/(4 - 1)\), or \(12.5\,\text{A}\), at \(0.015\,\text{m}\). There \(B = 2\times10^{-7}\times 12.5/0.015\), which is \(1.67\times10^{-4}\,\text{T}\).

A coaxial cable carries a current \(I\) out along a central core and the same current back along a metal shell round it. The cable of a television set-top box is one. Fig 4.2 cuts it across. A circle in the core encloses \(Ir^2/a^2\). A circle in the gap encloses all of \(I\). A circle outside encloses \(I - I = 0\).

So the field of a coaxial cable lives only inside it: \(\mu_0 Ir/2\pi a^2\) in the core, \(\mu_0 I/2\pi r\) in the gap, and nothing outside. For \(5\,\text{A}\) out and back, the field \(0.01\,\text{m}\) from the axis, in the gap, is \(2\times10^{-7}\times 5/0.01 = 10^{-4}\,\text{T}\). The cable neither disturbs its neighbours nor picks up their fields as easily.